Discussion:
The Gabriel Polynomial - an advanced concept not possible in mainstream calculus.
(too old to reply)
Eram semper recta
2021-08-20 23:13:19 UTC
Permalink
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.

https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW

I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Uètovaný Wattù
2021-08-20 23:15:01 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the
Gabriel Polynomial in a reasonable period of time. I suggest you move
on. https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not
arrogance, but FACT.
so you admit your IQ range below the 150.
Quantum Bubbles
2021-08-21 18:49:12 UTC
Permalink
Post by Uètovaný Wattù
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the
Gabriel Polynomial in a reasonable period of time. I suggest you move
on. https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not
arrogance, but FACT.
so you admit your IQ range below the 150.
He probably forgot to mention the standard deviation was 40. :-)
Quantum Bubbles
2021-08-21 18:51:51 UTC
Permalink
Post by Uètovaný Wattù
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the
Gabriel Polynomial in a reasonable period of time. I suggest you move
on. https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not
arrogance, but FACT.
so you admit your IQ range below the 150.
He probably forgot to mention the standard deviation was 40. :-)
But I will give JG 115 some credit, he is not an anti-vaxxer. Everyone has a redeeming feature somewhere.
Brain Deitke
2021-08-21 20:04:56 UTC
Permalink
Post by Quantum Bubbles
Post by Quantum Bubbles
Post by Uètovaný Wattù
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot
understand the Gabriel Polynomial in a reasonable period of time. I
suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW I
am a genius and I do know better than YOU or anyone else. This is
not arrogance, but FACT.
so you admit your IQ range below the 150.
He probably forgot to mention the standard deviation was 40.
But I will give JG 115 some credit, he is not an anti-vaxxer. Everyone
has a redeeming feature somewhere.
how the fuck you guys can be this stupid. If you don't want piss in your
shoes this won't make you anti-piss in the shoes.

Also, if even food and water can make you ill, how on earth the crap, you
don't know what it is, a stranger paid to inject that into your body,
kept in a freezer bellow -30⁰ C, can make you other than seriously ill,
damaged for life, and even dead?

You uneducated, severely confused LGBTs.
Jim Burns
2021-08-21 20:58:38 UTC
Permalink
Post by Quantum Bubbles
Post by Uètovaný Wattù
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the
Gabriel Polynomial in a reasonable period of time. I suggest you move
on. https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not
arrogance, but FACT.
so you admit your IQ range below the 150.
He probably forgot to mention the standard deviation was 40. :-)
But I will give JG 115 some credit, he is not an anti-vaxxer.
Everyone has a redeeming feature somewhere.
I have heard it said (of someone elsewhere) that,
even if he had no other use, he could still serve as a bad example.

It seems to me that JG's true calling in life is
to serve as a bad example.
Eram semper recta
2021-08-21 22:09:58 UTC
Permalink
Post by Jim Burns
Post by Quantum Bubbles
Post by Uètovaný Wattù
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the
Gabriel Polynomial in a reasonable period of time. I suggest you move
on. https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not
arrogance, but FACT.
so you admit your IQ range below the 150.
He probably forgot to mention the standard deviation was 40. :-)
But I will give JG 115 some credit, he is not an anti-vaxxer.
Everyone has a redeeming feature somewhere.
I have heard it said (of someone elsewhere) that,
even if he had no other use, he could still serve as a bad example.
It seems to me that JG's true calling in life is
to serve as a bad example.
I have a purpose: I am going to get rid of you and your fellow idiots once and for all. You will in the near future be ashamed to even say that you had anything to do with mathematics.

You lying, dumb cunt.
Ward Ehlers
2021-08-22 13:41:20 UTC
Permalink
Post by Eram semper recta
It seems to me that JG's true calling in life is to serve as a bad
example.
I have a purpose: I am going to get rid of you and your fellow idiots
once and for all. You will in the near future be ashamed to even say
that you had anything to do with mathematics.
1. Following intramuscular injection, it must be expected that the gene-
based vaccines will reach the bloodstream and disseminate throughout the
body [1]. We request evidence that this possibility was excluded in pre-
clinical animal models with all three vaccines prior to their approval
for use in humans by the EMA.

2. If such evidence is not available, it must be expected that the
vaccines will remain entrapped in the circulation and be taken up by
endothelial cells. There is reason to assume that this will happen
particularly at sites of slow blood flow, i.e. in small vessels and
capillaries [2]. We request evidence that this probability was excluded
in pre-clinical animal models with all three vaccines prior to their
approval for use in humans by the EMA.

3. If such evidence is not available, it must be expected that during
expression of the vaccines’ nucleic acids, peptides derived from the spike
protein will be presented via the MHC I — pathway at the luminal surface
of the cells. Many healthy individuals have CD8-lymphocytes that recognize
such peptides, which may be due to prior COVID infection, but also to
cross- reactions with other types of Coronavirus [3; 4] [5]. We must
assume that these lymphocytes will mount an attack on the respective
cells. We request evidence that this probability was excluded in pre-
clinical animal models with all three vaccines prior to their approval
for use in humans by the EMA.

4. 1f such evidence is not available, it must be expected that endothelial
damage with subsequent triggering of blood coagulation via platelet
activation will ensue at countless sites throughout the body. We request
evidence that this probability was excluded in pre-clinical animal models
with all three vaccines prior to their approval for use in humans by the
EMA.

5. If such evidence is not available, it must be expected that this will
lead to a drop in platelet counts, appearance of D-dimers in the blood,
and to myriad ischaemic lesions throughout the body including in the
brain, spinal cord and heart. Bleeding disorders might occur in the wake
of this novel type of DIC-syndrome including, amongst other
possibilities, profuse bleedings and haemorrhagic stroke. We request
evidence that all these possibilities were excluded in pre-clinical
animal models with all three vaccines prior to their approval for use in
humans by the EMA.

6. The SARS-CoV-2 spike protein binds to the ACE2 receptor on platelets,
which results in their activation [6]. Thrombocytopenia has been reported
in severe cases of SARS-CoV-2 infection [7]. Thrombocytopenia has also
been reported in vaccinated individuals [8]. We request evidence that the
potential danger of platelet activation that would also lead to
disseminated intravascular coagulation (DIC) was excluded with all three
vaccines prior to their approval for use in humans by the EMA.

7. The sweeping across the globe of SARS-CoV-2 created a pandemic of
illness associated with many deaths. However, by the time of consideration
for approval of the vaccines, the health systems of most countries were no
longer under imminent threat of being overwhelmed because a growing
proportion of the world had already been infected and the worst of the
pandemic had already abated. Consequently, we demand conclusive
evidence that an actual emergency existed at the time of the EMA granting
Conditional Marketing Authorisation to the manufacturers of all three
vaccines, to justify their approval for use in humans by the EMA,
purportedly because of such an emergency.


Should all such evidence not be available, we demand that approval for
use of the gene-based vaccines be withdrawn until all the above issues
have been properly addressed by the exercise of due diligence by the
EMA.
Dan Christensen
2021-08-21 19:03:50 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
JG here claims to have a discovered as shortcut to mastering calculus without using limits. Unfortunately for him, this means he has no workable a definition of the derivative of a function. It blows up for functions as simple f(x)=|x|. Or even f(x)=0. As a result, he has had to ban 0, negative numbers and instantaneous rates of change rendering his goofy little system quite useless. What a moron!

Forget calculus. JG has also banned all axioms because he cannot even derive the most elementary results of basic arithmetic, e.g. 2+2=4. Such results require the use of axioms, so he must figure he's now off the hook. Again, what a moron!

Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:

"There are no points on a line."
--April 12, 2021

"Pi is NOT a number of ANY kind!"
--July 10, 2020

"1/2 not equal to 2/4"
--October 22, 2017

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

"3 =< 4 is nonsense.”
--October 28, 2017

"Zero is not a number."
-- Dec. 2, 2019

"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017

“There is no such thing as an empty set.”
--Oct. 4, 2019

“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019

No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.

Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog a http://www.dcproof.wordpress.com
Eram semper recta
2021-08-21 20:18:02 UTC
Permalink
Post by Dan Christensen
"There are no points on a line."
Lie. I never said that. What I did say is that a line does not consists of points. When we talk about points on a line, we really mean distances that are indicated much like road signs do for distances travelled along a road.

A line is one of innumerable distances between any two points.
A straight line is the shortest distance between two points.
Post by Dan Christensen
"Pi is NOT a number of ANY kind!"
True. Pi is merely a symbol for an incommensurable magnitude - apparently a concept too advanced for an imbecile like Dan Christensen.
Post by Dan Christensen
"1/2 not equal to 2/4"
Lie. I have NEVER said this. What I have talked about is the difference in the process of measure.
What does this mean? Well, 1/2 is the name given to a measure done by enumerating 1 of two equal parts of the unit.
2/4 is the name given to a measure done by enumerating 2 of four equal parts of the unit.

There is the case in geometry where 1/2 is not necessarily equal to 2/4. For example:

_ / _ _
_ _ / _ _ _ _

The length _ is not equal to the length _ _ .
Post by Dan Christensen
“1/3 does NOT mean 1 divided by 3 and never has meant that”
True. My brilliant article on how a genius mind discovers number and indeed how my brilliant ancestors (Ancient Greeks) realised number explains in detail:

https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w

Also, my article on pi not being a number of any kind:

https://drive.google.com/file/d/1FFg_9XCkIwTZ9N1jbU4oMYfHHHuFHYf3

The true story of how we got numbers:

https://drive.google.com/file/d/0B-mOEooW03iLYTg1TGY4RTIwakU

No such thing as a "real number" or a "real number line":

https://drive.google.com/file/d/0B-mOEooW03iLMHVYcE8xcmRZRnc

There is no valid construction of "real number" - it's a myth:

https://drive.google.com/file/d/0B-mOEooW03iLSTROakNyVXlQUEU
Post by Dan Christensen
"3 =< 4 is nonsense.”
True. In mathematics, it is called an invalid disjunction.

3 <= 4 means EITHER 3 < 4 OR 3 = 4

Actually, there is no "OR" part, so the logical disjunction is invalid.
Post by Dan Christensen
"Zero is not a number."
True. While not a number of any kind, it is very useful in mathematics.

https://drive.google.com/file/d/1w2tt7IgoIu-ychDCoYi-4jOAzToy0ViM
Post by Dan Christensen
"0 is not required at all in mathematics, just like negative numbers."
Half-truth. While negative numbers are not required in mathematics, they are extremely useful.
Post by Dan Christensen
“There is no such thing as an empty set.”
True. Even the father of all mainstream mathematical cranks rejected the idea of empty set. But let's not go too far ... there isn't even a definition of "set" in set theory!




Post by Dan Christensen
“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
True. These are propositions that are implied by the given equations. For example, my historic geometric identity states:

[f(x+h)-f(x)]/h = dy/dx + Q(x,h)

And so, f(x+h)-f(x)]/h <=> dy/dx + Q(x,h)

The theorem:

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

How it provides a rigorous definition of integral for the flawed mainstream calculus:

https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

The day will come when this vicious anonymous troll Dan Christensen is convicted in a court of law.

Download for free the most important mathematics book ever written:

https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
Dan Christensen
2021-08-22 04:16:07 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math

On Saturday, August 21, 2021 at 4:18:08 PM UTC-4, I am Super Rectum (aka John Gabriel, Troll Boy) wrote...
Post by Eram semper recta
Post by Dan Christensen
"1/2 not equal to 2/4"
Lie. I have NEVER said this.
A direct quote from October 22, 2017 here at sci.math
Post by Eram semper recta
What I have talked about is the difference in the process of measure.
What does this mean? Well, 1/2 is the name given to a measure done by enumerating 1 of two equal parts of the unit.
2/4 is the name given to a measure done by enumerating 2 of four equal parts of the unit.
When will you learn, Troll Boy? 1/2 is ALWAYS EQUAL to 2/4.

[snip]
Post by Eram semper recta
Post by Dan Christensen
“1/3 does NOT mean 1 divided by 3 and never has meant that”
True. My brilliant article on how a genius mind discovers number and indeed how my brilliant ancestors (Ancient Greeks) realised number explains in detail...
If you can't dazzle them brilliance, baffle them with bullshit, right, Troll Boy?
Post by Eram semper recta
Post by Dan Christensen
"3 =< 4 is nonsense.”
True. In mathematics, it is called an invalid disjunction.
Nothing "invalid" about it, Troll Boy.
Post by Eram semper recta
3 <= 4 means EITHER 3 < 4 OR 3 = 4
It means 3 < 4 OR 3 = 4, which is always truly.

See: https://en.wikipedia.org/wiki/Truth_table#Logical_disjunction_(OR)

[snip]
Post by Eram semper recta
Post by Dan Christensen
"Zero is not a number."
True. While not a number of any kind, it is very useful in mathematics.
It really is a number, Troll Boy. Deal with it.
Post by Eram semper recta
Post by Dan Christensen
"0 is not required at all in mathematics, just like negative numbers."
Half-truth.
Nope. Completely false.
Post by Eram semper recta
While negative numbers are not required in mathematics, they are extremely useful.
<yawn!>
Post by Eram semper recta
Post by Dan Christensen
“There is no such thing as an empty set.”
True.
Umm... What about the set of all your brilliant mathematical discoveries? Empty.

[snip]
Post by Eram semper recta
Post by Dan Christensen
“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions”
True.
Nope. The biconditional is logical connective. See https://en.wikipedia.org/wiki/Logical_biconditional

3 is not a logical proposition or a statement that is true or false. 3 is a number. So 3 <=> 2+1 would be an error in syntax. Deal with it, Troll Boy.


[snip]
Post by Eram semper recta
Post by Dan Christensen
Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at
https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

Also, all direct quotes from you, Troll Boy. To the extent that you will be remembered at all, history will not be kind to you. Time to cut your losses and move on Troll Boy.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Eram semper recta
2021-08-22 11:56:13 UTC
Permalink
Post by Dan Christensen
"There are no points on a line."
Lie. I never said that. What I did say is that a line does not consists of points. When we talk about points on a line, we really mean distances that are indicated much like road signs do for distances travelled along a road.

A line is one of innumerable distances between any two points.
A straight line is the shortest distance between two points.
Post by Dan Christensen
"Pi is NOT a number of ANY kind!"
True. Pi is merely a symbol for an incommensurable magnitude - apparently a concept too advanced for an imbecile like Dan Christensen.
Post by Dan Christensen
"1/2 not equal to 2/4"
Lie. I have NEVER said this. What I have talked about is the difference in the process of measure.
What does this mean? Well, 1/2 is the name given to a measure done by enumerating 1 of two equal parts of the unit.
2/4 is the name given to a measure done by enumerating 2 of four equal parts of the unit.

There is the case in geometry where 1/2 is not necessarily equal to 2/4. For example:

_ / _ _
_ _ / _ _ _ _

The length _ is not equal to the length _ _ .
Post by Dan Christensen
“1/3 does NOT mean 1 divided by 3 and never has meant that”
True. My brilliant article on how a genius mind discovers number and indeed how my brilliant ancestors (Ancient Greeks) realised number explains in detail:

https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w

Also, my article on pi not being a number of any kind:

https://drive.google.com/file/d/1FFg_9XCkIwTZ9N1jbU4oMYfHHHuFHYf3

The true story of how we got numbers:

https://drive.google.com/file/d/0B-mOEooW03iLYTg1TGY4RTIwakU

No such thing as a "real number" or a "real number line":

https://drive.google.com/file/d/0B-mOEooW03iLMHVYcE8xcmRZRnc

There is no valid construction of "real number" - it's a myth:

https://drive.google.com/file/d/0B-mOEooW03iLSTROakNyVXlQUEU
Post by Dan Christensen
"3 =< 4 is nonsense.”
True. In mathematics, it is called an invalid disjunction.

3 <= 4 means EITHER 3 < 4 OR 3 = 4

Actually, there is no "OR" part, so the logical disjunction is invalid.
Post by Dan Christensen
"Zero is not a number."
True. While not a number of any kind, it is very useful in mathematics.

https://drive.google.com/file/d/1w2tt7IgoIu-ychDCoYi-4jOAzToy0ViM
Post by Dan Christensen
"0 is not required at all in mathematics, just like negative numbers."
Half-truth. While negative numbers are not required in mathematics, they are extremely useful.
Post by Dan Christensen
“There is no such thing as an empty set.”
True. Even the father of all mainstream mathematical cranks rejected the idea of empty set. But let's not go too far ... there isn't even a definition of "set" in set theory!

http://youtu.be/KvxjOMW6Q9w

http://youtu.be/1CcSsOG0okg
Post by Dan Christensen
“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
True. These are propositions that are implied by the given equations. For example, my historic geometric identity states:

[f(x+h)-f(x)]/h = dy/dx + Q(x,h)

And so, f(x+h)-f(x)]/h <=> dy/dx + Q(x,h)

The theorem:

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

How it provides a rigorous definition of integral for the flawed mainstream calculus:

https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

The day will come when this vicious anonymous troll Dan Christensen is convicted in a court of law.

Download for free the most important mathematics book ever written:

https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view

The New Calculus is proof that you CAN DO calculus without the use of LIMIT THEORY.

Don't believe me? Study it. I am a genius and the greatest mathematician alive today.
Dan Christensen
2021-08-22 14:59:46 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math

On Sunday, August 22, 2021 at 7:56:19 AM UTC-4, I am Super Rectum (aka John Gabriel, Troll Boy) wrote...

Troll Boy here seems to be at a loss for words and can only mindlessly repeat his already failed arguments. What a moron!

Dan
Eram semper recta
2021-08-22 17:14:10 UTC
Permalink
Post by Dan Christensen
"There are no points on a line."
Lie. I never said that. What I did say is that a line does not consists of points. When we talk about points on a line, we really mean distances that are indicated much like road signs do for distances travelled along a road.

A line is one of innumerable distances between any two points.
A straight line is the shortest distance between two points.
Post by Dan Christensen
"Pi is NOT a number of ANY kind!"
True. Pi is merely a symbol for an incommensurable magnitude - apparently a concept too advanced for an imbecile like Dan Christensen.
Post by Dan Christensen
"1/2 not equal to 2/4"
Lie. I have NEVER said this. What I have talked about is the difference in the process of measure.
What does this mean? Well, 1/2 is the name given to a measure done by enumerating 1 of two equal parts of the unit.
2/4 is the name given to a measure done by enumerating 2 of four equal parts of the unit.

There is the case in geometry where 1/2 is not necessarily equal to 2/4. For example:

_ / _ _
_ _ / _ _ _ _

The length _ is not equal to the length _ _ .
Post by Dan Christensen
“1/3 does NOT mean 1 divided by 3 and never has meant that”
True. My brilliant article on how a genius mind discovers number and indeed how my brilliant ancestors (Ancient Greeks) realised number explains in detail:

https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w

Also, my article on pi not being a number of any kind:

https://drive.google.com/file/d/1FFg_9XCkIwTZ9N1jbU4oMYfHHHuFHYf3

The true story of how we got numbers:

https://drive.google.com/file/d/0B-mOEooW03iLYTg1TGY4RTIwakU

No such thing as a "real number" or a "real number line":

https://drive.google.com/file/d/0B-mOEooW03iLMHVYcE8xcmRZRnc

There is no valid construction of "real number" - it's a myth:

https://drive.google.com/file/d/0B-mOEooW03iLSTROakNyVXlQUEU
Post by Dan Christensen
"3 =< 4 is nonsense.”
True. In mathematics, it is called an invalid disjunction.

3 <= 4 means EITHER 3 < 4 OR 3 = 4

Actually, there is no "OR" part, so the logical disjunction is invalid.
Post by Dan Christensen
"Zero is not a number."
True. While not a number of any kind, it is very useful in mathematics.

https://drive.google.com/file/d/1w2tt7IgoIu-ychDCoYi-4jOAzToy0ViM
Post by Dan Christensen
"0 is not required at all in mathematics, just like negative numbers."
Half-truth. While negative numbers are not required in mathematics, they are extremely useful.
Post by Dan Christensen
“There is no such thing as an empty set.”
True. Even the father of all mainstream mathematical cranks rejected the idea of empty set. But let's not go too far ... there isn't even a definition of "set" in set theory!

http://youtu.be/KvxjOMW6Q9w

http://youtu.be/1CcSsOG0okg
Post by Dan Christensen
“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
True. These are propositions that are implied by the given equations. For example, my historic geometric identity states:

[f(x+h)-f(x)]/h = dy/dx + Q(x,h)

And so, f(x+h)-f(x)]/h <=> dy/dx + Q(x,h)

The theorem:

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

How it provides a rigorous definition of integral for the flawed mainstream calculus:

https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

The day will come when this vicious anonymous troll Dan Christensen is convicted in a court of law.

Download for free the most important mathematics book ever written:

https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view

The New Calculus is proof that you CAN DO calculus without the use of LIMIT THEORY.

Don't believe me? Study it. You will be pleasantly surprised.

I am a genius and the greatest mathematician alive today.
Eram semper recta
2021-08-21 20:22:43 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Refreshed due to troll activity.
Quantum Bubbles
2021-08-21 20:31:33 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Refreshed due to troll activity.
You consider your own posts to be troll, activity?

On another note, have you booked that IQ test yet or are you going to cringe in a corner and wuss out?

Regards
Eram semper recta
2021-08-21 22:11:24 UTC
Permalink
Post by Quantum Bubbles
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Refreshed due to troll activity.
You consider your own posts to be troll, activity?
On another note, have you booked that IQ test yet or are you going to cringe in a corner and wuss out?
Oh dear. Anonymous coward and prolific troll Quantum Babbling is baaaacck!

You've nothing to say about the Gabriel Polynomial, do you? Oh wait, you are simply too fucking stupid to understand it. LMAO.
Post by Quantum Bubbles
Regards
Quantum Bubbles
2021-08-22 07:16:11 UTC
Permalink
Post by Quantum Bubbles
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Refreshed due to troll activity.
You consider your own posts to be troll, activity?
On another note, have you booked that IQ test yet or are you going to cringe in a corner and wuss out?
Oh dear. Anonymous coward and prolific troll Quantum Babbling is baaaacck!
You've nothing to say about the Gabriel Polynomial, do you? Oh wait, you are simply too fucking stupid to understand it. LMAO.
Post by Quantum Bubbles
Regards
I'm back for a little while, then I will be gone for a while again. Work stuff, medical stuff ... having fun taking a Mensa Test which you probably don't have the stones to take yourself, chuckle, these things are all better uses of my time than wasting further neurotransmitters reading about your wingless bird trying to fly again.

Let's see... Quantum Theory ... Thermodynamics ... Electromagnetism ... Non-Linear Dynamics ... General Relativity (has yet another observation vindicated yet this week? there were so many already...), modern coding and cryptography. Modern 'mainstream' (i.e. correct) mathematics has given us all of this and they are all very successful, a fact which is hard to rationally explain if your negative views on mathematics are true. By contrast early 17th century style methods have given us ...?????... oh well.

What is it you would like people to say? Why should other readers care about your unlettered vanity project with no serious prospects? What is it you imagine is going to happen? We already have a brilliant branch of mathematics called Real Analysis and some lovely associated subjects like the Calculus of Finite Differences and Approximation Theory.

But anyway:

1) have you got a rigorous derivation of the Euler-Lagrange Equation from the Calculus of Variations yet, or is your system still a waste of internet data storage for anyone interested in applied mathematics?

2) After laughably claiming an IQ 4 standard deviations above average, when are you going to provide a shred of evidence for this in the form of passing the Mensa Exam?

3) Have you read a biology textbook about evolution yet, or is your theory of concepts still partly based on an ignorance of this area?

4) Have you followed my advice and tried gardening yet?

Embarrassing yourself on here day after day, cannot be good for you and is not remotely necessary. Educate yourself about the damaging effects of NPD, make adjustments, find new hobbies and try to enjoy your 60s, rather than urinating them away against the brick wall of internet forums used by your betters in mathematics. Do you want Archimedes Plutonium to be indicative of your future? Because that is where you are currently headed. But there is hope. You can put a stop to it.

Go for it champ.

QB
Chris M. Thomasson
2021-08-22 08:47:49 UTC
Permalink
Post by Quantum Bubbles
Post by Quantum Bubbles
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Refreshed due to troll activity.
You consider your own posts to be troll, activity?
On another note, have you booked that IQ test yet or are you going to cringe in a corner and wuss out?
Oh dear. Anonymous coward and prolific troll Quantum Babbling is baaaacck!
You've nothing to say about the Gabriel Polynomial, do you? Oh wait, you are simply too fucking stupid to understand it. LMAO.
Post by Quantum Bubbles
Regards
I'm back for a little while, then I will be gone for a while again. Work stuff, medical stuff ... having fun taking a Mensa Test which you probably don't have the stones to take yourself, chuckle, these things are all better uses of my time than wasting further neurotransmitters reading about your wingless bird trying to fly again.
Let's see... Quantum Theory ... Thermodynamics ... Electromagnetism ... Non-Linear Dynamics ... General Relativity (has yet another observation vindicated yet this week? there were so many already...), modern coding and cryptography. Modern 'mainstream' (i.e. correct) mathematics has given us all of this and they are all very successful, a fact which is hard to rationally explain if your negative views on mathematics are true. By contrast early 17th century style methods have given us ...?????... oh well.
What is it you would like people to say? Why should other readers care about your unlettered vanity project with no serious prospects? What is it you imagine is going to happen? We already have a brilliant branch of mathematics called Real Analysis and some lovely associated subjects like the Calculus of Finite Differences and Approximation Theory.
1) have you got a rigorous derivation of the Euler-Lagrange Equation from the Calculus of Variations yet, or is your system still a waste of internet data storage for anyone interested in applied mathematics?
2) After laughably claiming an IQ 4 standard deviations above average, when are you going to provide a shred of evidence for this in the form of passing the Mensa Exam?
3) Have you read a biology textbook about evolution yet, or is your theory of concepts still partly based on an ignorance of this area?
4) Have you followed my advice and tried gardening yet?
gardening can teach one about fractals. A plant is a fractal, and they
need to be cared for...
Post by Quantum Bubbles
Embarrassing yourself on here day after day, cannot be good for you and is not remotely necessary. Educate yourself about the damaging effects of NPD, make adjustments, find new hobbies and try to enjoy your 60s, rather than urinating them away against the brick wall of internet forums used by your betters in mathematics. Do you want Archimedes Plutonium to be indicative of your future? Because that is where you are currently headed. But there is hope. You can put a stop to it.
Go for it champ.
QB
Eram semper recta
2021-08-21 22:18:29 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Quiz on Gabriel Polynomial:

Choose either True or False for all of the following statements.

[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public

If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
Eram semper recta
2021-08-21 22:21:15 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h

You up to it?! Chuckle.
Dan Christensen
2021-08-22 04:22:31 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, JG here claims to have a discovered as shortcut to mastering calculus without using limits. Unfortunately for him, this means he has no workable a definition of the derivative of a function. It blows up for functions as simple f(x)=|x|. Or even f(x)=0. As a result, he has had to ban 0, negative numbers and instantaneous rates of change rendering his goofy little system quite useless. What a moron!

Forget calculus. JG has also banned all axioms because he cannot even derive the most elementary results of basic arithmetic, e.g. 2+2=4. Such results require the use of axioms, so he must figure he's now off the hook. Again, what a moron!

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog a http://www.dcproof.wordpress.com
Eram semper recta
2021-08-22 11:58:18 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h

I see. No attempts at the Quiz. Of course I did not expect any, but I included it to prove to you that Anonymous trolls like Quantum Bubbles and Dan Christensen can't grasp this level of mathematics because they can't even grasp the basics of mathematics.
Eram semper recta
2021-08-22 13:56:39 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
I see. No attempts at the Quiz. Of course I did not expect any, but I included it to prove to you that Anonymous trolls like Quantum Bubbles and Dan Christensen can't grasp this level of mathematics because they can't even grasp the basics of mathematics.
Refreshed due to Czech troll activity.
Eram semper recta
2021-08-22 17:15:11 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Students:

Study my article. As you can see, the morons on this newsgroup can't refute anything therein.

Don't miss out by listening to fools and trolls.
Dan Christensen
2021-08-22 18:21:41 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math

On Sunday, August 22, 2021 at 1:15:16 PM UTC-4, I am Super Rectum (aka John Gabriel, Troll Boy) wrote...
Post by Eram semper recta
Study my article.
A complete waste of time.
Post by Eram semper recta
As you can see, the morons on this newsgroup can't refute anything therein.
The moronic and delusional JG here cannot refute the fact that his goofy little system simply doesn't work. See my previous postings in this thread for proof.

Dan
Eram semper recta
2021-08-22 22:16:04 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
So the Gabriel Polyomial is far more advanced than anything you've ever seen in your bogus formulation of mainstream calculus.

Quiz on Gabriel Polynomial:

Choose either True or False for all of the following statements.

[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public

If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!

For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
zelos...@gmail.com
2021-08-23 04:59:15 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
Eram semper recta
2021-08-23 12:14:09 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.

Quiz on Gabriel Polynomial:

Choose either True or False for all of the following statements.

[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public

If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!

For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Eram semper recta
2021-08-23 19:56:52 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Awww. No takers? Tsk, tsk.
zelos...@gmail.com
2021-08-24 05:09:53 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it? It is entirely pointless.

You think highschool mathematics is "advanced"
Eram semper recta
2021-08-24 12:58:28 UTC
Permalink
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
zelos...@gmail.com
2021-08-25 05:33:04 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.

A polynomial is not advanced.
Eram semper recta
2021-08-25 15:29:35 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Lola Caine
2021-08-25 18:52:09 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
*climate_change* stands for _depopulation_, friend. Don't be fooled.
zelos...@gmail.com
2021-08-26 05:16:38 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.

They aren't, polynomials never are.
Eram semper recta
2021-08-26 10:04:58 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!

"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM

The above is a gem, Malum. An absolute gem! LMAO.

As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.

Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
zelos...@gmail.com
2021-08-27 05:08:05 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)

https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ

Read it, I am saying that is what YOU are saying, not what I think.
Eram semper recta
2021-08-27 10:31:22 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).

That is what YOU said! You are a crank! Here it is again with link following:

***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************

https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
zelos...@gmail.com
2021-08-30 09:25:29 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).
***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
I pointed out that is how YOU think. Not that I think it you dishonest lying sack of shit :)
Eram semper recta
2021-08-30 11:27:57 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).
***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
I pointed out that is how YOU think. Not that I think it you dishonest lying sack of shit :)
The lying sack of SHIT is YOU, you vile bastard.

I said the exact opposite of what you and Python were saying.

IF [f(x+h)-f(x)]/h = f'(x) + Q(x,h) THEN f'(x)= [f(x+h)-f(x)]/h - Q(x,h)

and then the two fucking morons (YOU and Python) tried to insinuate that it was circular. LMAO.

You're the world's greatest mainstream crank!
zelos...@gmail.com
2021-08-31 05:03:06 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).
***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
I pointed out that is how YOU think. Not that I think it you dishonest lying sack of shit :)
The lying sack of SHIT is YOU, you vile bastard.
I said the exact opposite of what you and Python were saying.
IF [f(x+h)-f(x)]/h = f'(x) + Q(x,h) THEN f'(x)= [f(x+h)-f(x)]/h - Q(x,h)
and then the two fucking morons (YOU and Python) tried to insinuate that it was circular. LMAO.
You're the world's greatest mainstream crank!
You did when we discussed D and T, derivative and tangent, functions.

D was defined in terms of limit
T was defined in terms of D

yet you claim it was circular because D uses T after some re-arrangement. Just like the illustration I gave.
Eram semper recta
2021-08-31 11:48:58 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).
***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
I pointed out that is how YOU think. Not that I think it you dishonest lying sack of shit :)
The lying sack of SHIT is YOU, you vile bastard.
I said the exact opposite of what you and Python were saying.
IF [f(x+h)-f(x)]/h = f'(x) + Q(x,h) THEN f'(x)= [f(x+h)-f(x)]/h - Q(x,h)
and then the two fucking morons (YOU and Python) tried to insinuate that it was circular. LMAO.
You're the world's greatest mainstream crank!
You did when we discussed D and T, derivative and tangent, functions.
Lying again, crank? Nowhere did I say this. Provide the link, you vile bastard!
Post by ***@gmail.com
D was defined in terms of limit
I have NEVER defined the derivative in terms of a limit. Newsflash: I do not use the bullshit of limit theory in my historic geometric theorem. It is 100% geometry:

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
T was defined in terms of D
That's EXACTLY what YOU do in your bullshit MAINSTREAM calculus and it is CIRCULAR.
Post by ***@gmail.com
yet you claim it was circular because D uses T after some re-arrangement. Just like the illustration I gave.
Your drivel was nothing of the sort. You imagined that you were agreeing with your fellow asswipe Jean Pierre Messager (aka Python) and I made a fool of you yet again! LMAO.
zelos...@gmail.com
2021-09-01 05:08:35 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Your polynomial is anything but advance. It is high school level shite.
So take the quiz and let's see if you understood anything Malum! Everyone here knows you are a crank.
Choose either True or False for all of the following statements.
[A] It was inspired by the ideas of Newton
[B] It is far superior to Taylor's polynomial which is never exact
[C] It is always a closed form and can contain a variable number of terms
[D] It allows for systematic integration regardless of a given function
[E] Its core concept is the mean value theorem
[F] Although its purpose is not function approximation, it will outperform Taylor series for ANY given function
[G] There are many things about the polynomial which John Gabriel has not shared with the public
If you score 7/7, then you are a genius!
If you score 4/7, then you are smart!
If you score less than 4/7, you have no hope!
For extra credit, write a short paragraph on why you need the New Calculus derivative for the Gabriel Polynomial and why you cannot use the flawed mainstream definition, ie. f'(x)= lim_{h->0} [f(x+h)-f(x)]/h
Again, your shit is ego stroking so why would we do it?
You do not even attempt it because it is too advanced for you! :)
Post by ***@gmail.com
It is entirely pointless.
Untrue. Opinions and assertions are like shit - they are everywhere.
Post by ***@gmail.com
You think highschool mathematics is "advanced"
ROFLMAO. It is so advanced that it makes your pea brain spin!
I feel no need to stroke your over inflated narcissistic ego.
Nor were you asked to do any such thing ever! :)
Post by ***@gmail.com
A polynomial is not advanced.
The Gabriel Polynomial is advanced. Chuckle.
Yes I was, your "Quizes" is nothing more htan stroking your ego.
They aren't, polynomials never are.
Go easy on the booze. Your responses are getting more incoherent by the day!
"c=f+g <=> g=c-f is CIRCULAR" - ZELOS MALUM
The above is a gem, Malum. An absolute gem! LMAO.
As I've always told you, you do not understand what it means to be a well-formed concept, because if you did, then you would understand what it means to be "circular". Tsk, tsk.
Oops-Allah (Uppsala), the university for silly Swedes!!!! LMAO
You are a liar :)
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
Read it, I am saying that is what YOU are saying, not what I think.
You went along with that delusional psychopath Jean Pierre Messager (aka Python).
***************************************************************
Post by ***@gmail.com
Post by Eram semper recta
He think that stuff you point out is circular
if c=f+g
then g=c-f
so circular!
***************************************************************
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/vAEmniRtBwAJ
I pointed out that is how YOU think. Not that I think it you dishonest lying sack of shit :)
The lying sack of SHIT is YOU, you vile bastard.
I said the exact opposite of what you and Python were saying.
IF [f(x+h)-f(x)]/h = f'(x) + Q(x,h) THEN f'(x)= [f(x+h)-f(x)]/h - Q(x,h)
and then the two fucking morons (YOU and Python) tried to insinuate that it was circular. LMAO.
You're the world's greatest mainstream crank!
You did when we discussed D and T, derivative and tangent, functions.
Lying again, crank? Nowhere did I say this. Provide the link, you vile bastard!
Post by ***@gmail.com
D was defined in terms of limit
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
T was defined in terms of D
That's EXACTLY what YOU do in your bullshit MAINSTREAM calculus and it is CIRCULAR.
Post by ***@gmail.com
yet you claim it was circular because D uses T after some re-arrangement. Just like the illustration I gave.
Your drivel was nothing of the sort. You imagined that you were agreeing with your fellow asswipe Jean Pierre Messager (aka Python) and I made a fool of you yet again! LMAO.
Lying again, crank? Nowhere did I say this. Provide the link, you vile bastard!
https://groups.google.com/g/sci.math/c/5NjpX0It60I/m/nLvlglEMCQAJ
https://groups.google.com/g/sci.math/c/bxHjnK8_00o/m/L5t27UkMCQAJ
https://groups.google.com/g/sci.math/c/QMhqLZH1434/m/a_ShbscMCQAJ
Post by Eram semper recta
I have NEVER defined the derivative in terms of a limit.
I didn't say you did, I said I did. D was defiend in terms of limit, T in terms of D, and you cry circular which is not it.

To be circular T is defined in terms of D, and D in terms of T, but it isn't. D is defined independently of T
Post by Eram semper recta
That's EXACTLY what YOU do in your bullshit MAINSTREAM calculus and it is CIRCULAR.
See, right here you do claim it is circular just like I said you do.

D defined in terms of limit
T defined in terms of D
Is not circular. It is a linear progression of definitions.
Post by Eram semper recta
Your drivel was nothing of the sort. You imagined that you were agreeing with your fellow asswipe Jean Pierre Messager (aka Python) and I made a fool of you yet again! LMAO.
You didn't, you did however manage to confirm in this post exactly what I said you did while denying it. That is quite foolish.
Alan Mackenzie
2021-08-23 13:57:05 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand
the Gabriel Polynomial in a reasonable period of time. I suggest you
move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is
not arrogance, but FACT.
Your polynomial is anything but advanced. It is high school level
shite.
That wouldn't surprise me in the least. But does it have any meaning at
all? What does it do? What does it mean? (I'm not prepared to let
anything from google on my machine, and JG hasn't said anything
substantive about "his polynomial" here.)
--
Alan Mackenzie (Nuremberg, Germany).
markus...@gmail.com
2021-08-27 20:49:00 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
Eram semper recta
2021-08-27 20:52:57 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
LMAO. Reddit is a trash heap and everyone knows this!
markus...@gmail.com
2021-08-27 20:57:57 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
LMAO. Reddit is a trash heap and everyone knows this!
But it has good memes doe. :)
Eram semper recta
2021-08-27 21:09:09 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
LMAO. Reddit is a trash heap and everyone knows this!
But it has good memes doe. :)
Man, there is nothing good about Reddit. It's just one big shit pile.
Python
2021-08-28 00:35:45 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
Good!!! Very good one!
New Age Prophet
2021-08-28 00:47:10 UTC
Permalink
Post by Python
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
Good!!! Very good one!
Says one crank (Python) to another (Markus Klyver).
Serg io
2021-08-28 04:45:12 UTC
Permalink
Post by New Age Prophet
Post by Python
Post by ***@gmail.com
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Perfect time to post this: https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
Good!!! Very good one!
Says one crank (Python) to another (Markus Klyver).
Mr Rectum's math is crank math, it is simply troll food, it has QuAck Function for crispyness, and boy DOES IT SMELL !

all his real math is here;
https://old.reddit.com/r/JohnGabrielMemes/comments/7tmyeg/john_gabriel_starter_pack/
Eram semper recta
2021-09-01 11:28:43 UTC
Permalink
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
Essentially what the Baboons of mainstream academia will have you believe is that:

t(c+h)-t(c)]/h = Lim (h->0)[f(c+h)-f(c)]/h = f'(c) + Q(c,h) and is possible ONLY if Q(c,h)=0, but this happens only in the case of the straight line t(x). It NEVER happens with the finite difference quotients. The mainstream want you to believe that there is some finite difference ratio hovering at infinity which produces f'(c):

t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo

t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!

Can you imagine how embarrassed Newton and Leibniz would be at these idiots today?! They knew that they could not solve the tangent line problem - this had to wait for the great John Gabriel. It is I who revealed to the entire world that:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
Quantum Bubbles
2021-09-01 16:16:01 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
Well I got my Mensa test results today Mr. Gabriel, I suspect my IQ might be higher than yours as I managed to qualify for both Mensa and Intertel (against my own expectations, but then again I tend to be self-critical). I'm going to put my evidence on my YouTube channel at the weekend. When are we going to see official confirmation that your IQ was good enough to pass a Mensa test?

QB
Eram semper recta
2021-09-02 10:13:20 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.

Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
markus...@gmail.com
2021-09-02 15:04:01 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
markus...@gmail.com
2021-09-03 19:37:58 UTC
Permalink
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
Eram semper recta
2021-09-04 06:00:52 UTC
Permalink
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.

Final example and then you will be ignored:

f(x)=x^2 x=1 h=1

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h


[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3

If you vary h, then the above is no longer true for x=1:

h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3

Do you understand now, maximus stupidicus?

You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then

h=0: 2 + 0 = 2 =/= 3

In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
markus...@gmail.com
2021-09-04 12:26:46 UTC
Permalink
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true? That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.

If f=x^2, then the choice Q(h)=h will satisfy this condition since (x+h)^2-x^2=h(2x+h) and lim_{h \to 0} h=0. So we have the identity 2x+h=2x+Q(h).

For proof that Q(h)=0 is continuous at h=0, just consider h small. Then Q(h)=h is small.
Eram semper recta
2021-09-05 10:34:02 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.

[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).

You CANNOT take "limits" because the identity is expressed in terms of CONSTANTS:

[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.

What you do in your bogus mainstream calculus is analogous to the following:

3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.

You are saying:

Lim 3 = 1 + Lim 2

which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".

Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.

<Drivel>
Eram semper recta
2021-09-05 10:39:41 UTC
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Post by ***@gmail.com
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.

It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).

Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
Eram semper recta
2021-09-05 10:41:04 UTC
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Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.
It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).
Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
Algebra itself actually began in Book 7 of calculus. It was not a later invention by the Arabs or anyone else.

https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w
Eram semper recta
2021-09-05 10:46:14 UTC
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Post by Eram semper recta
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Post by Eram semper recta
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Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.
It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).
Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
Algebra itself actually began in Book 7 of calculus. It was not a later invention by the Arabs or anyone else.
https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w
One has to be a "lawyer" to learn mainstream mythmatics because it is based on regulations, not facts.

Imagine a lawyer during prohibition trying to defend a drunken client. Do you think he would win the case? LMAO.

Now imagine a lawyer before or after prohibition trying to defend a drunken client. His chances of winning increase significantly.

Mainstream math academics are similar to vile lawyers. I would have mentioned "Jewish lawyers" but declined - fact is no one beats the Jews at anything involving trickery. I should know! I have Jewish genes. Ha, ha.

See, unlike many vile people in the world, I place a high importance on honesty and truth. Unfortunately, look where it got me? Arguing with stubborn imbeciles like you.
Eram semper recta
2021-09-05 10:47:26 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.
It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).
Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
Algebra itself actually began in Book 7 of calculus.
Oopsie, I meant "Euclid's Elements", not "calculus". Tiredness and lack of oxygen affects my ability to think straight, but even so I am better at it than anyone else.
Post by Eram semper recta
It was not a later invention by the Arabs or anyone else.
https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w
Chris M. Thomasson
2021-09-05 19:50:45 UTC
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Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.
It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).
Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
Algebra itself actually began in Book 7 of calculus.
Oopsie, I meant "Euclid's Elements", not "calculus". Tiredness and lack of oxygen affects my ability to think straight, but even so I am better at it than anyone else.
Fwiw, one of my friends suffers from a somewhat chronic fatigue
syndrome. Iirc, she compared it to a hell, or prison; I cannot exactly
remember right now. Tired all the damn time, low energy. Iirc, she did
not have low oxygen levels. Humm... There is a nasty fire really close
to South Lake Tahoe, where I have a lot of family. I lived there for
around thirteen years. Its the god damn Caldor Fire:

https://www.fire.ca.gov/incidents/2021/8/14/caldor-fire/

I live around Carson City (east of the lake) and the smoke from that
damn fire would damn near block out the sun. It would be a red glow in
the sky. Ash was falling like snow. It almost burned down South Lake Tahoe!

Its hard to breathe here.

If the winds pick up south of the lake, that 43% containment can go down
to 33% really quick. It was crowning in the tops of the trees around 4-5
days ago.
Post by Eram semper recta
Post by Eram semper recta
It was not a later invention by the Arabs or anyone else.
https://drive.google.com/file/d/1hasWyQCZyRN3RkdvIB6bnGIVV2Rabz8w
markus...@gmail.com
2021-09-05 15:33:43 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
[f(x+h)-f(x)]/h ≈ f'(x) for small h (limits) is NOT a solution to the tangent line problem.
It's called hand waving gibberish meant to cover up your ignorance of geometry (which is 100% rigorous and DOES NOT rely on axioms).
Cauchy was a conceited fool to say that the solution to the tangent line problem needed more than algebra (a weak form of geometry made possible by the ABSTRACT UNIT) because the TRUTH is that geometry as realised by my brilliant ancestors was sufficient to solve the problem. Sadly the morons who came after them were not up to the task. It took a descendant (John Gabriel) to prove this!
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Eram semper recta
2021-09-05 23:23:00 UTC
Permalink
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).

Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.

<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.

formal: done in accordance with convention or etiquette

"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.

As I've said, my identity proves mainstream math academics are cranks.

A crank is one who cannot be convinced in the face of overwhelming evidence.

YOU and MALUM, are prime examples.
markus...@gmail.com
2021-09-06 00:09:24 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Eram semper recta
2021-09-06 10:13:02 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
markus...@gmail.com
2021-09-06 15:20:39 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Eram semper recta
2021-09-07 06:22:24 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.

Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO

Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.

Proof: New Calculus

https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view

You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
zelos...@gmail.com
2021-09-07 11:24:15 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
With you bullshit = things you do not like/understand
markus...@gmail.com
2021-09-07 13:08:09 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed you how the derivative is a limit and explained how it works several times. I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Eram semper recta
2021-09-07 18:00:29 UTC
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Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)

LMAO.
markus...@gmail.com
2021-09-07 19:22:58 UTC
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Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
markus...@gmail.com
2021-09-07 21:05:13 UTC
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Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits. Why is my previous comment a such mystery to you?

"We never "drop" the term Q. We take the limit.

Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have

[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.

Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.

Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.

So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.

This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Eram semper recta
2021-09-08 07:37:35 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!

Again moron:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

If f(x) = x^2 and x=1 and h=1 then

[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h

So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3

Thus, 3 = 2 + 1

You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!

The limit of 1 is itself, that is, 1.

If you got away with your monkey business, you would end up with 3 = 2.

What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
zelos...@gmail.com
2021-09-08 09:15:03 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.

Try f(x)=x^3 and say your Q is constant :)
Eram semper recta
2021-09-08 12:43:52 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.

It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.

Grow a brain, baboon Swede boy!
zelos...@gmail.com
2021-09-08 13:17:38 UTC
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Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :) But lets test your claim!

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2

now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?

then Q is anything BUT constant given it has both x and h as variables.
Eram semper recta
2021-09-08 15:52:53 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!

Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).

Because you are such a moron, I suggest you choose a value for x and h and follow the steps:

1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.

The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
markus...@gmail.com
2021-09-08 18:38:17 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
Eram semper recta
2021-09-09 05:58:09 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.

https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>

It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
Eram semper recta
2021-09-09 06:14:52 UTC
Permalink
Post by ***@gmail.com
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x.
In fact, Q(x,h) can be any of the following:

i. Q(x,h) = 0 in the special case of f representing a straight line
ii. Q(x,h) can be an expression in h only
iii. Q(x,h) can be an expression in both x and h

Q(x,h) can NEVER be an expression only in x - as the theorem shows.
However, an open minded person thinks about these things and considers all cases.
Dan Christensen
2021-09-09 06:56:23 UTC
Permalink
STUDENTS BEWARE: Don't be a victim of JG's fake math
Post by Eram semper recta
i. Q(x,h) = 0 in the special case of f representing a straight line
ii. Q(x,h) can be an expression in h only
iii. Q(x,h) can be an expression in both x and h
What is Q(x,h) when f(x)=|x|? Not today??? (Hmmm... What will JG ban next? Absolute values? Hee, hee!)

JG here claims to have a discovered a shortcut to mastering calculus without using limits. Unfortunately for him, this means he has no workable a definition of the derivative of a function. It blows up for functions as simple f(x)=|x|. Or even f(x)=0. As a result, he has had to ban 0, negative numbers and instantaneous rates of change rendering his goofy little system quite useless. What a moron!

Forget calculus. JG has also banned all axioms because he cannot even derive the most elementary results of basic arithmetic, e.g. 2+2=4. Such results require the use of axioms, so he must figure he's now off the hook. Again, what a moron!

Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:

"There are no points on a line."
--April 12, 2021

"Pi is NOT a number of ANY kind!"
--July 10, 2020

"1/2 not equal to 2/4"
--October 22, 2017

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

"3 =< 4 is nonsense.”
--October 28, 2017

"Zero is not a number."
-- Dec. 2, 2019

"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017

“There is no such thing as an empty set.”
--Oct. 4, 2019

“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019

No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.

Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog a http://www.dcproof.wordpress.com
Dan Christensen
2021-09-10 13:04:53 UTC
Permalink
Post by Dan Christensen
STUDENTS BEWARE: Don't be a victim of JG's fake math
Post by Eram semper recta
i. Q(x,h) = 0 in the special case of f representing a straight line
ii. Q(x,h) can be an expression in h only
iii. Q(x,h) can be an expression in both x and h
What is Q(x,h) when f(x)=|x|? Not today??? (Hmmm... What will JG ban next? Absolute values? Hee, hee!)
Still no reply, Troll Boy?
Post by Dan Christensen
JG here claims to have a discovered a shortcut to mastering calculus without using limits. Unfortunately for him, this means he has no workable a definition of the derivative of a function. It blows up for functions as simple f(x)=|x|. Or even f(x)=0. As a result, he has had to ban 0, negative numbers and instantaneous rates of change rendering his goofy little system quite useless. What a moron!
Forget calculus. JG has also banned all axioms because he cannot even derive the most elementary results of basic arithmetic, e.g. 2+2=4. Such results require the use of axioms, so he must figure he's now off the hook. Again, what a moron!
"There are no points on a line."
--April 12, 2021
"Pi is NOT a number of ANY kind!"
--July 10, 2020
"1/2 not equal to 2/4"
--October 22, 2017
“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015
"3 =< 4 is nonsense.”
--October 28, 2017
"Zero is not a number."
-- Dec. 2, 2019
"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017
“There is no such thing as an empty set.”
--Oct. 4, 2019
“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019
No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.
Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog a http://www.dcproof.wordpress.com
markus...@gmail.com
2021-09-09 14:44:11 UTC
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Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
What does a "factor of h" mean? If it means using Taylor's theorem and treat everything as polynomials, then the common definition is better since it doesn't require the function to be analytic.
Eram semper recta
2021-09-10 04:52:13 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
What does a "factor of h" mean?
You should go back to primary school.

<irrelevant drivel as usual>
zelos...@gmail.com
2021-09-10 05:34:17 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
Post by ***@gmail.com
then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
What does a "factor of h" mean?
You should go back to primary school.
<irrelevant drivel as usual>
your typical response when you cannot respond
markus...@gmail.com
2021-09-10 10:06:51 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
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then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
What does a "factor of h" mean?
You should go back to primary school.
<irrelevant drivel as usual>
Can't you just define what you mean instead?
Eram semper recta
2021-09-10 10:16:51 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
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you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
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Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Once again showing you do not understand limits even in its rudimentary form.
Try f(x)=x^3 and say your Q is constant :)
Yes, Q(x,h) is indeed CONSTANT.
It's works exactly the same way no matter what is f. The only requirement is that f is smooth which is a requirement in any case for any of the methods of calculus to work.
Grow a brain, baboon Swede boy!
I got a better brain than yours :)
What "brain" do you have? LMAO. You mean the two brain cells in your rock head skull? Chuckle. You stupid Swede APE!
Post by ***@gmail.com
But lets test your claim!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is yours, now f(x)=x^3
f(x+h)=x^3+3x^2 h+3xh^2+h^3
f(x+h)-f(x)=3x^2 h+3xh^2+h^3
(f(x+h)-f(x))/h=3x^2+3xh+h^2
All good so far.
Post by ***@gmail.com
now what do you call f'?
and Q?
f'(x)=3x^2, Q(x,h)=3xh+h^2 fits your desires as we arbitrarily decide it?
f'(x)=3x^2, Q(x,h)=3xh+h^2 is correct but it is NOT arbitrarily decided. My historic theorem proves that the expression containing h in its terms is Q(x,h) and the expression containing x is f'(x). That's why I called my theorem HISTORIC because it is!
Now if you actually bothered to study my theorem, then you would know these things, but instead you argue like the idiot that you are.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
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then Q is anything BUT constant given it has both x and h as variables.
h is NOT a variable, you idiot!!! I told you several times that for every non-parallel secant line there is exactly one (x,h) pair that satisfies the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
1. Verify that the x and h you chose satisfies [f(x+h)-f(x)]/h = f'(x) + Q(x,h).
2. Now try choosing a different h but keeping f'(x) the same.
The conclusion would be obvious to anyone with a modicum of intelligence. But not you! You're a prize idiot.
You provide no definition to uniquely decide Q.
You're lying. Page 7 of the following document states clearly that Q(x,h) can only be an expression who terms must contain at least one factor in h if f is not a straight line.
https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
<<Therefore, f1/h will be an expression that contains both x and h, whereas f2/h will contain only x.>>
It's hard to write the above to include all cases because Q(x,h) can also contain no factors of x. However, an open minded person thinks about these things and considers all cases.
What does a "factor of h" mean?
You should go back to primary school.
<irrelevant drivel as usual>
Can't you just define what you mean instead?
As I've said, you're an idiot and it would make no difference telling you that a <factor> is a very well known concept since over 2300 years ago AND it's in the DICTIONARY!!!

factor: an expression or magnitude that when multiplied with another produces a given expression or magnitude.
Bud Kanguroo
2021-09-10 12:54:24 UTC
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Can't you just define what you mean instead?
As I've said, you're an idiot and it would make no difference telling
you that a <factor> is a very well known concept since over 2300 years
ago AND it's in the DICTIONARY!!!
stop spamming re-quoting crap, idiot. Last chance, then you go on ignore.
Bud Kanguroo
2021-09-10 12:45:57 UTC
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What does a "factor of h" mean?
You should go back to primary school.
quoting 419 lines of crap, just to say this. You must feel like idiot.
zelos...@gmail.com
2021-09-09 05:15:24 UTC
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but it is NOT arbitrarily decided.
It is as you provide no criteria for determining them.
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That's why I called my theorem HISTORIC because it is!
it isn't because we can just as well say f'(x)=0 and Q(x,h)=3x^2+3xh+h^2 and call it a day and have every functions derivative be 0
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h is NOT a variable, you idiot!!!
It is a variable because it has no specific value.
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2. Now try choosing a different h but keeping f'(x) the same.
This makes it a variable cause it can change.
Eram semper recta
2021-09-09 06:00:12 UTC
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Post by Eram semper recta
but it is NOT arbitrarily decided.
It is as you provide no criteria for determining them.
LIE.
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Post by Eram semper recta
That's why I called my theorem HISTORIC because it is!
it isn't because we can just as well say f'(x)=0 and Q(x,h)=3x^2+3xh+h^2 and call it a day and have every functions derivative be 0
No. You can't just be your normal idiot self MALUM. In mathematics we don't define things on a wing and a prayer.
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Post by Eram semper recta
h is NOT a variable, you idiot!!!
It is a variable because it has no specific value.
h has a UNIQUE value for any given non-parallel secant line.
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Post by Eram semper recta
2. Now try choosing a different h but keeping f'(x) the same.
This makes it a variable cause it can change.
No. This makes it ignorance.
zelos...@gmail.com
2021-09-09 09:24:26 UTC
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Post by Eram semper recta
No. You can't just be your normal idiot self MALUM. In mathematics we don't define things on a wing and a prayer.
You're not in mathematics so you cannot use "we" with it.
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h has a UNIQUE value for any given non-parallel secant line.
Post by Eram semper recta
2. Now try choosing a different h but keeping f'(x) the same.
Are contradictory, if you can chose a different h, then it isn't fixed.
markus...@gmail.com
2021-09-08 13:10:31 UTC
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Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Suppose you understood limits.
No one understands the theory of limits as well as I do. Get it moron? It's not that I don't understand. I don't agree. There is a difference - can you tell, idiot?
Post by ***@gmail.com
Why is my previous comment a such mystery to you?
All your comments are gibberish. To attribute "mystery" to anything you say might imply some intelligence. You have none. You are a prize idiot.
Post by ***@gmail.com
"We never "drop" the term Q. We take the limit.
What don't you understand about the fact: "You CANNOT take the limit of Q(x,h) because it is a CONSTANT" ??!!!!!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
If f(x) = x^2 and x=1 and h=1 then
[f(x+h)-f(x)]/h = 2x + h meaning that f'(x)=2x and Q(x,h) = h
So, [f(1+1)-f(1)]/1 = 2(1)+1 = 3
Thus, 3 = 2 + 1
You cannot take the limit of 1, you fucking idiot!!! For the thousandth time!!!!!
The limit of 1 is itself, that is, 1.
If you got away with your monkey business, you would end up with 3 = 2.
What part of this do you not understand, you goddam MORON?!
Post by ***@gmail.com
Fix/fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Conversely, if lim [f(x+h) -f(x)]/h=A=f'(x), then [f(x+h) -f(x)]/h=A+Q(h) holds for some Q as above.
So the definitions are equivalent and neither of them are circular. And we don't "drop" any term either, we take the limit, and Q(h) goes to zero when h does.
This exactly captures the idea that [f(x+h)-f(x)]/h ≈ f'(x) for small h. Because, what does it mean? It means that [f(x+h)-f(x)]/h=f'(x)+(something small) and that small quantity can in general depend on h, so it is a function in h and we can call this function Q. I.e. (something small) is exactly what we call Q(h)."
Why are you evaluating the difference quotient in a point h instead of taking the limit in h?
FromTheRafters
2021-09-08 08:35:56 UTC
Permalink
Hello math people, I want to learn calculus, is this the right place?
Best,
Uwus
No, but any questions you might have can be answered here if you ignore
the cranks.
zelos...@gmail.com
2021-09-08 05:24:18 UTC
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Post by Eram semper recta
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Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
The derivative definition is purely analytic, not geometric.
Absolute Rubbish! The derivative definition is purely geometric. There is NOTHING that comes from algebra or anywhere else.
You can have a geometric picture of it, but the limit definition comes from analysis/topology and not geometry.
Either you are sincerely deceived or just plain stupid or both because the above statement is obviously FALSE.
Analysis and topology are bullshit that came along hundreds of years after Newton and Leibniz. LMAO
Calculus is based PURELY on geometry. THIS AND NOTHING ELSE.
Proof: New Calculus
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
You had better start studying the New Calculus because it won't be long before you will be unemployable if you don't know it!
I mean, I have already pointed
You can't even point to anything but your arse.
Post by ***@gmail.com
you how the derivative is a limit and explained how it works several times.
And I have proved you wrong every time, but you are as thick as a brick.
Post by ***@gmail.com
I've also pointed out how you try to disprove math with dictionaries and making up your own definitions. You're infuriating and extremely tiring to deal with.
Yes, it must be infuriating when you get whipped so badly, isn't it? :)
LMAO.
The only thing you have proven is that you don't understand these concepts.
Chuckle. Any one can read the previous comments and see who it is that has no understanding. It's also easy to spot a crank like you - cranks tend to avoid the OP because they do not understand it and their agenda is to disrupt and cause chaos through libel and ad hominem. This describes YOU and MALUM exactly.
Yes, they all can see it is YOU who do not understand things because again we can cite sources that isn't 2000 years out of date!
zelos...@gmail.com
2021-09-06 05:16:24 UTC
Permalink
Post by Eram semper recta
Post by Eram semper recta
Post by ***@gmail.com
Post by ***@gmail.com
Post by Eram semper recta
Post by Eram semper recta
Post by Eram semper recta
Unless your IQ is at least 140 or higher, then you cannot understand the Gabriel Polynomial in a reasonable period of time. I suggest you move on.
https://drive.google.com/file/d/1MnnnczqCA6vAEEoL1_OQAabDyHHzNcwW
I am a genius and I do know better than YOU or anyone else. This is not arrogance, but FACT.
t(c+h)-t(c)]/h = [f(c+h_1)-f(c)]/h_1 ; [f(c+h_2)-f(c)]/h_2 ; [f(c+h_3)-f(c)]/h_3 ; ... ; [f(c+h_3n-f(c)]/h_n ...; [f(c+h_oo)-f(c)]/h_oo
t(c+h)-t(c)]/h = [f(c+h_oo)-f(c)]/h_oo ?! What?!!! There is no such finite difference. It does not exist! But Baboon mathematics of the mainstream insists there is a limit!
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
It usually takes a genius to realise the most simple concepts that escape the syphilitic brains of highly educated idiots in the mainstream.
The mainstream art of hand waving is an interesting psychological study in its own right.
Just take [f(x+h)-f(x)]/h = f'(x) + Q(x,h) and rewrite is as [f(x+h)-f(x)]/h = f'(x) + Lim_{h->0}Q(x,h) and then pretend that it no longer matters if the left hand side equals the right hand side. The baboons think they are actually doing something by taking the limit of a constant, but it is simply an excuse to cover up their grand ignorance, stupidity and inccompetence.
The LHS equals the RHS because lim_{h->0} Q(x, h)=0 for every x.
Nope. Because lim_{h->0} Q(x, h)= Q(x,h).
Q(x,h) is UNIQUE for any given non-parallel secant line.
https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
If you remove the above link, I stop responding.
We have lim_{h->0} Q(x, h)=0 for every x.
No, you do not have lim_{h->0} Q(x, h)=0 for every x.
f(x)=x^2 x=1 h=1
[f(x+h)-f(x)]/h = f'(x) + Q(x,h)
=> [f(x+h)-f(x)]/h = 2x + h
So, f'(x)=2x and Q(x,h)=h
[f(1+1)-f(1)]/1 = f'(1) + 1 = 2 + 1 = 3
h=2: 2 + 2 = 4 =/= 3
h=1/2: 2 + 1/2 = 5/2 =/= 3
h=1/4 2 + 1/4 = 9/4 =/= 3
Do you understand now, maximus stupidicus?
You cannot have h=0 which is equivalent to lim_{h->0} Q(x, h)=0 because then
h=0: 2 + 0 = 2 =/= 3
In the above example, the non-parallel secant line slope MUST be 3 for h=1. If you change h, then you are no longer talking about the same non-parallel secant line.
Why would varying h not make [f(x+h)-f(x)]/h = f'(x) + Q(x,h) true?
I've explained this to you a thousand times already. Man, you are dumber than rocks.
[f(x+h)-f(x)]/h = f'(x) + Q(x,h) is a FUCKING IDENTITY, you retard!!!! You cannot just drop the CONSTANT term Q(x,h) on the RHS because then you are saying [f(x+h)-f(x)]/h = f'(x) which is true only in one case (straight line).
[f(x+h)-f(x)]/h is CONSTANT for the slope of any given non-parallel line.
f'(x) is CONSTANT for the slope of any given tangent line.
Q(x,h) which is the difference between [f(x+h)-f(x)]/h and f'(x) is a CONSTANT.
3 = 1 + 2 where 3 is the non-parallel secant line slope, 1 is the derivative and 2 is the difference.
Lim 3 = 1 + Lim 2
which changes NOTHING. You are imagining that x is constant and that h varies, but this is DELUSION.
That's the whole point: [f(x+h)-f(x)]/h ≈ f'(x) for small h, which is formalised by adding an error term Q(h) with lim_{h \to 0} Q(h)=0.
Grrr. There is NO "error term", you imbecile!! Your "formalism" is speak for "I never understood Euclid's Elements".
Greek mathematics is 100% sound and FORMAL. No further bullshit is needed in the form of "foundations", "set theory" and all your residual wankery.
<Drivel>
We never "drop" the term Q. We take the limit.
You do exactly this - "drop the term Q".
Fixate x. If for some Q(h) defined on some neighbourhood (-s, s)\{0} which can be extended continuously Q(0)=0, we have
"Fixate x" ? LMAO.
[f(x+h) -f(x)]/h=A+Q(h), then we say that f'(x):=A.
You don't say anything. It was I who revealed [f(x+h) -f(x)]/h=f'(x)+Q(x,h) to you. I, the world's greatest genius.
Taking limits h->0 gives lim [f(x+h) -f(x)]/h=lim A+lim Q(h)=A+0=A.
Nope. Because there is no finite difference f(x+h) -f(x)]/h that will result in lim f'(x) +lim Q(x,h)=f'(x) + 0=f'(x).
Again, f(x+h) -f(x)]/h is CONSTANT for ANY non-parallel secant line so you can't vary any given x or h because then my identity [f(x+h) -f(x)]/h=f'(x)+Q(x,h) would be false, but as we know, my identity results from my brilliant geometric theorem.
<drivel>
I agree that stating [f(x+h)-f(x)]/h ≈ f'(x) for small h is hand-waving, but it is made formal by introducing limits.
Nothing is made "formal" in any sense whatsoever. In fact, this very lingo raises a red flag, because mathematics is not about formality or authoritative orders or decrees. In mathematics, there is only rational thinking and hard cold facts and logic.
formal: done in accordance with convention or etiquette
"In accordance with convention or etiquette" does not prove anything. The Queen of England has rules of etiquette which prohibit any foreign principality from hugging her. ROFLMAO. You see, Klyver, you're nothing but a stupid prick who has been brainwashed and created in the image of your moronic professors and educators.
As I've said, my identity proves mainstream math academics are cranks.
A crank is one who cannot be convinced in the face of overwhelming evidence.
YOU and MALUM, are prime examples.
This comes from the one that cannot be convinced by overwhelming evidence and rejects mainstream that has overwhelming evidence to support it.
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