Discussion:
Jack Huizenga and his "critique" of my New Calculus. One learns more about Huizenga than my new calculus...
(too old to reply)
John Gabriel
2017-03-29 01:32:36 UTC
Permalink
9/21/2016 A Harvard alumnus comments! | The New Calculus

Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.

The question was:

“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.

Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.

One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.

In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!

There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga

The great mathematics historian Carl Boyer had this to say:
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.

That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer

I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.

Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga

One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.

Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga

The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.

Needless to say, if calculus were already extremely well understood, we would not have other mathematics PhDs making statements as follows:

Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)

Knisley goes on to say:

However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)

As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.

There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga

It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga

Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.

In the following comment I debunk the concept of Dedekind cuts and Cauchy sequences:

https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU

Our buffoon continues:

The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga

It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.

In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.

His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga

Of course we know the slope of the tangent line, provided we know the slope of a parallel secant line. This is grade 8 mathematics! What he states in parenthesis is even more amusing. It demonstrates clearly that he, like many of his colleagues never understood calculus:

If a given function is not continuous and smooth, then any of the methods of calculus are null and void.

One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.

He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga

I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.

It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.

That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article

https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E

The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!

Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga

Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.

The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.

The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.

(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.

And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.

This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.


Comments are unwelcome and will be ignored.

Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

***@gmail.com (MIT)
***@psu.edu (HARVARD)
***@mit.edu (MIT)
***@math.okstate.edu (David Ullrich)
***@clarku.edu
***@gmail.com
b***@gmail.com
2017-03-29 09:07:56 UTC
Permalink
John Gabriel thinks Cauchy did e-d Calculus:

Epsilonic Blues - Topologabriely of Mythnifolds
https://drive.google.com/file/d/0B-mOEooW03iLX1Z5eUpkdHZKaUE/view

Others think this is not historically adequate:

On the history of epsilontics, Sinkevich Galina
https://arxiv.org/abs/1502.06942
Post by John Gabriel
9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
b***@gmail.com
2017-03-29 09:14:09 UTC
Permalink
And relating to mean value theorem, as JG does:

f'(x) = (f(x+h)-f(x))/h

doesn't make sense. We would need to write:

f'(c) = (f(x+h)-f(x))/h

And say there is c in (x,x+h). The below arvix also
says something about mean value theorem and cauchy.

But how long will it take that JG gets anything right?
Post by b***@gmail.com
Epsilonic Blues - Topologabriely of Mythnifolds
https://drive.google.com/file/d/0B-mOEooW03iLX1Z5eUpkdHZKaUE/view
On the history of epsilontics, Sinkevich Galina
https://arxiv.org/abs/1502.06942
Post by John Gabriel
9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
7777777
2017-04-14 06:57:23 UTC
Permalink
Post by b***@gmail.com
f'(x) = (f(x+h)-f(x))/h
doesn't make sense.
it does not make sense as long as h is finite.
Post by b***@gmail.com
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h)
yes, f'(x) and f'(c) are not the same as long as h is finite.
j4n bur53
2017-04-14 11:49:20 UTC
Permalink
h is from R, its not a set, its some urelement.
It does not have a set size property finite, infinite
countable or infinite uncountable.

I guess you mean "not infinitessimal small", i.e.
not outside of R, since R is archimedian, it doesn't
have anyway some infinitessimals in it.

Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
Post by 7777777
Post by b***@gmail.com
f'(x) = (f(x+h)-f(x))/h
doesn't make sense.
it does not make sense as long as h is finite.
Post by b***@gmail.com
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h)
yes, f'(x) and f'(c) are not the same as long as h is finite.
7777777
2017-04-14 13:20:29 UTC
Permalink
Post by j4n bur53
h is from R, its not a set, its some urelement.
It does not have a set size property finite, infinite
countable or infinite uncountable.
I guess you mean "not infinitessimal small", i.e.
not outside of R, since R is archimedian, it doesn't
have anyway some infinitessimals in it.
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
I am not exactly sure what you want me to demonstrate, but I will try anyway.
Consider first
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
if m and n are finite then f'(c) is not equal to f'(x)
So lets use the infinitesimal dx = m = n

We end up with
f'(c) ={ f(c+dx) - f(c-dx) } / 2dx

And we need to demonstrate
f'(x) = (f(x+h)-f(x))/h = (f(x+dx)-f(x))/dx

so that f'(c) = f'(x) also
{ f(c+dx) - f(c-dx) } / 2dx = { f(x+dx) - f(x-dx) } / 2dx

in other words, we need to demonstrate

(f(x+dx)-f(x))/dx = { f(x+dx) - f(x-dx) } / 2dx

also the same as

f(x + dx) - f(x) = 1/2 f(x + dx) - 1/2 f(x - dx)

f(x) = 1/2 [ f(x + dx) + f(x - dx)]

if that's a valid formula remains to be seen...
John Gabriel
2017-04-14 13:40:19 UTC
Permalink
Post by 7777777
Post by j4n bur53
h is from R, its not a set, its some urelement.
It does not have a set size property finite, infinite
countable or infinite uncountable.
I guess you mean "not infinitessimal small", i.e.
not outside of R, since R is archimedian, it doesn't
have anyway some infinitessimals in it.
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
I am not exactly sure what you want me to demonstrate, but I will try anyway.
Consider first
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
if m and n are finite then f'(c) is not equal to f'(x)
Yes idiot. { f(c+n) - f(c-m) } / (m+n) is equal to f'(c). What is x? If you are talking about the x in f'(x) = (f(x+h)-f(x))/h, then these are two different things.

f'(x) = (f(x+h)-f(x))/h is WRONG. You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.

<More Moron shit ignored>
7777777
2017-04-14 14:06:19 UTC
Permalink
Post by John Gabriel
f'(x) = (f(x+h)-f(x))/h is WRONG.
what happens if h = dx <-------- the infinitesimal ?
Post by John Gabriel
You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
but do you need dx, the infinitesimal ?
Post by John Gabriel
<More Moron shit ignored>
for how long time the moron keeps ignoring the infinitesimal dx ?
John Gabriel
2017-04-14 16:00:46 UTC
Permalink
Post by 7777777
Post by John Gabriel
f'(x) = (f(x+h)-f(x))/h is WRONG.
what happens if h = dx <-------- the infinitesimal ?
There are NO infinitesimals dummy. For the thousandth time:

THERE ARE NO INFINITESIMALS!!!!

Grow a brain and stop being a moron. You keep holding onto this shit even though I have shown you over and over again on STATU that there is no such thing as an infinitesimal.

Look you moron: It is either ZERO or it is NOT. What part of this do you not understand? Do you subscribe to the stupidity of the BIG STUPID?

If you claim infinitesimal, then what you are saying is that there is a number that follows ZERO. There isn't. Get it?!! No, don't give me your shit about ineffable or ourobouros.
Post by 7777777
Post by John Gabriel
You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
but do you need dx, the infinitesimal ?
No stupid. In mainstream mythmatics, that is the LIMIT of the finite difference ratios as dx gets close to 0.


h or dx is NEVER zero in the mainstream formulation except when they determine the derivative and then use it in the self-referential limit verifinition.
Post by 7777777
Post by John Gabriel
<More Moron shit ignored>
for how long time the moron keeps ignoring the infinitesimal dx ?
I don't think that even a hundred slaps on the back of your head will help you to put some brains.
7777777
2017-04-14 16:48:37 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
f'(x) = (f(x+h)-f(x))/h is WRONG.
what happens if h = dx <-------- the infinitesimal ?
THERE ARE NO INFINITESIMALS!!!!
Grow a brain and stop being a moron. You keep holding onto this shit even though I have shown you over and over again on STATU that there is no such thing as an infinitesimal.
you have not shown anything. you have only shown that you are a moron.
It is here: dx <---------- the infinitesimal
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple someone would have already solved the problem a long time ago. But no, unfortunately things are not simple. What I have shown (and not just me, btw), over and over again, it maybe be both 0 and non-zero at the same time.
Post by John Gabriel
What part of this do you not understand? Do you subscribe to the stupidity of the BIG STUPID?
There is no such thing. There is no such thing as "BIG STUPID". What there is
is a problem, a problem common to all, but people like you just pretend
that the problem simply does not exist. While at the same time, the rest of us
have to struggle solving the very hard problems. You just keep ridiculing us,
without understanding that it is you who is in error.
Post by John Gabriel
If you claim infinitesimal, then what you are saying is that there is a number that follows ZERO. There isn't. Get it?!! No, don't give me your shit about ineffable or ourobouros.
No-one takes you seriously, you just keep repeating the same nonsense over and over again, basically telling us what can exist and what can't, but you can't
convince anyone, because what exists and what doesn't are just your beliefs.
Post by John Gabriel
Post by 7777777
Post by John Gabriel
You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
but do you need dx, the infinitesimal ?
No stupid. In mainstream mythmatics, that is the LIMIT of the finite difference ratios as dx gets close to 0.
It is h that gets close to 0.

I am talking about dx. Do you ever pay attention to detail?
dx does not necessarily get close to anything, and if it does, then the situation is different than when h gets close to 0.
Post by John Gabriel
h or dx is NEVER zero in the mainstream formulation except when they determine the derivative and then use it in the self-referential limit verifinition.
in the mainstream formulation dx is equal to 0 because according to them
dx = 0.000...001 = 0
Post by John Gabriel
Post by 7777777
Post by John Gabriel
<More Moron shit ignored>
for how long time the moron keeps ignoring the infinitesimal dx ?
I don't think that even a hundred slaps on the back of your head will help you to put some brains.
are you talking to me? It seems to me that it is you who needs to grow brains.
You can't have calculus based only on finite difference.
b***@gmail.com
2017-04-14 16:52:26 UTC
Permalink
Who is them? The deamons that hunt you in the night?
There is no them, you are responsible for yourself to
do what you want to do in math.
Post by 7777777
in the mainstream formulation dx is equal to 0 because according to them
dx = 0.000...001 = 0
b***@gmail.com
2017-04-14 16:55:16 UTC
Permalink
Post by j4n bur53
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
Who is them? The deamons that hunt you in the night?
There is no them, you are responsible for yourself to
do what you want to do in math.
Post by 7777777
in the mainstream formulation dx is equal to 0 because according to them
dx = 0.000...001 = 0
7777777
2017-04-14 17:16:45 UTC
Permalink
Post by j4n bur53
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
I am still wondering what exactly are you talking about. I already tried to answer it. And here I go again:

dx = 0.000...001

f'(x) = (f(x+dx)-f(x))/dx = (f(x+0.000...001)-f(x))/0.000...001
b***@gmail.com
2017-04-14 18:06:26 UTC
Permalink
How does this exactly work?

Take f(x) = x^2.
What is (x + 0.000...001) ?

Take f(x) = ln(x).
What is ln(x + 0.000...001) ?

Take f(x) = g(h(x)).
What is g(h(x + 0.000...001)) ?

I don't think it works so easily... You neeed to
define the arithmetic for infinitessimals and show us.

Basically the arithmetic of mixture your "finite" and
"non-finite" numbers. And transcendental functions as well.

Can you do that, do you have such a sound system?
Where is it written, or is in your mind only?
Post by 7777777
dx = 0.000...001
f'(x) = (f(x+dx)-f(x))/dx = (f(x+0.000...001)-f(x))/0.000...001
7777777
2017-04-14 18:33:06 UTC
Permalink
Post by b***@gmail.com
How does this exactly work?
Take f(x) = x^2.
What is (x + 0.000...001) ?
f(x) = x^2
f(x + dx) = (x + dx)^2 = x^2 + 2xdx + (dx)^2
now insert dx = 0.000...001
b***@gmail.com
2017-04-14 18:49:02 UTC
Permalink
And? What is the value of (0.000...001)^2, you
know for to obtain f'(x) it must vanish.

But you said 0.000...001 != 0.

This is a contradiction. Its similar to the contradiction
that BKK had, when he said 1.4142... != sqrt(2).

I am pretty sure we end up with:

Garbage in, Garbage out
Post by b***@gmail.com
f(x) = x^2
f(x + dx) = (x + dx)^2 = x^2 + 2xdx + (dx)^2
now insert dx = 0.000...001
b***@gmail.com
2017-04-14 19:01:52 UTC
Permalink
More precisely it must exactly happen 0.000...001 == 0
to obtain f'(x). Please observe for f(x)=x^2, we have

f'(x) = 2*x

Now lets try your method of plugging in. First of all
you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.

But the question is for (f(x+dx)-f(x))/dx. Which is
now (for you "finite" quantities, and you didn't tell
me how to compute it for infinitessimals):

x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx

Now you said your idea is just plugging in infinitessimal
in pace of a finite quantity, so I do plugging in, i.e.
setting dx = 0.000...001:

g(x) = 2*x + 0.000...001

But since you said 0.000...001 != 0, we don't have g(x)
= f'(x). But sometimes you say is even both, 0.000...001 != 0
and 0.000...001 == 0.

This is very confusing, you need a clear framework how
everything works. The limes framework is much clearer,
we all know that "f(x)" and "lim x->a f(x)" are two different
expressions.

Please also show:

f(x) = ln(x)
f(x) = g(h(x))
Post by b***@gmail.com
And? What is the value of (0.000...001)^2, you
know for to obtain f'(x) it must vanish.
But you said 0.000...001 != 0.
This is a contradiction. Its similar to the contradiction
that BKK had, when he said 1.4142... != sqrt(2).
Garbage in, Garbage out
Post by b***@gmail.com
f(x) = x^2
f(x + dx) = (x + dx)^2 = x^2 + 2xdx + (dx)^2
now insert dx = 0.000...001
7777777
2017-04-14 19:14:03 UTC
Permalink
Post by b***@gmail.com
More precisely it must exactly happen 0.000...001 == 0
to obtain f'(x). Please observe for f(x)=x^2, we have
f'(x) = 2*x
Now lets try your method of plugging in. First of all
you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
But the question is for (f(x+dx)-f(x))/dx. Which is
now (for you "finite" quantities, and you didn't tell
x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx
Now you said your idea is just plugging in infinitessimal
in pace of a finite quantity, so I do plugging in, i.e.
g(x) = 2*x + 0.000...001
But since you said 0.000...001 != 0, we don't have g(x)
= f'(x). But sometimes you say is even both, 0.000...001 != 0
and 0.000...001 == 0.
This is very confusing
It is not confusing.
You need to consider (dx)^2 = (0.000...001)^2 can be equal to 0 even though
0.000...001 is non-zero

And then you end up with f'(x) = 2x
b***@gmail.com
2017-04-14 20:06:16 UTC
Permalink
Thats actually not true. Can you please demonstrate
step by step? I didn't use (0.000...001)^2 = 0.

If you use (0.000...001)^2 = 0, you don't compute
(f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x).

And for this computation you get:

f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2

Now if you alow 0.000...001^2 = 0, you only get:

g(x) = 2*x*0.000...001

Which is still not f'(x). You are completely confused
what f'(x) is and what it is not. Maybe you are inspired

by this paper:

An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
Post by 7777777
It is not confusing.
You need to consider (dx)^2 = (0.000...001)^2 can be equal to 0 even though
0.000...001 is non-zero
And then you end up with f'(x) = 2x
b***@gmail.com
2017-04-14 20:11:54 UTC
Permalink
For my taste setting e^2 = 0 is a cludge, which
leads to nowhere. It just covers up the division
by e, and the limes e = 0.

The results are similar to Edward Studys duals,
but one could probably do them without the
e^2 = 0 cludige.
Post by b***@gmail.com
Thats actually not true. Can you please demonstrate
step by step? I didn't use (0.000...001)^2 = 0.
If you use (0.000...001)^2 = 0, you don't compute
(f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x).
f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2
g(x) = 2*x*0.000...001
Which is still not f'(x). You are completely confused
what f'(x) is and what it is not. Maybe you are inspired
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
Post by 7777777
It is not confusing.
You need to consider (dx)^2 = (0.000...001)^2 can be equal to 0 even though
0.000...001 is non-zero
And then you end up with f'(x) = 2x
7777777
2017-04-16 06:15:56 UTC
Permalink
Post by b***@gmail.com
Thats actually not true. Can you please demonstrate
step by step? I didn't use (0.000...001)^2 = 0
you did not use (0.000...001)^2 = 0 and therefore ended up with
Post by b***@gmail.com
x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx
If you use (0.000...001)^2 = 0, you don't compute
(f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x)
no. you must be very confused to say so. If I use (0.000...001)^2 = 0 then
I do compute (f(x+h)-f(x))/h as expected. I tried to understand your confusion,
to understand what you are doing. First of all I want to use dx instead of
h. So you are telling that I am computing only f(x+dx)-f(x). Also
f(x+dx)-f(x) = x^2 + 2xdx + (dx)^2 - x^2
Post by b***@gmail.com
f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2
g(x) = 2*x*0.000...001
Which is still not f'(x).
Of course it is not f'(x) but that's not what I am doing.

It is very strange why do you keep telling this thing, and it is wrong, because that's not what I am doing. I am computing (f(x+dx)-f(x))/dx as expected.
Why did you omit division by dx?
Post by b***@gmail.com
You are completely confused what f'(x) is and what it is not.
nope. I am not confused. It is you who made an error. You omitted division
by dx claiming that it is me who is doing so when I use (0.000...001)^2 = 0,
but that's not something that I do. It is something that you do, and I don't
know your reasons for doing so except that you just made an error.


The correct way is, and this is what I do:

f'(x) = (f(x+dx)-f(x))/dx = (x^2 + 2xdx + (dx)^2 - x^2)/dx =

(2xdx + (dx)^2)/dx

now I use (dx)^2 = (0.000...001)^2 = 0

so that

(2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x

which is f'(x)

there is no division by 0 because 0.000...001 is nonzero
John Gabriel
2017-04-16 07:18:43 UTC
Permalink
Post by 7777777
Of course it is not f'(x) but that's not what I am doing.
99.99% of the time, you have no clue what you are saying. Pure idiot is all you are.
b***@gmail.com
2017-04-16 08:52:35 UTC
Permalink
Post by 7777777
(2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x
But it is:
(2*x*dx+dx^2)/dx=2*x+dx

And you say:
(2*x*dx+dx^2)/dx=2*x

Therefore dx=0
Post by 7777777
there is no division by 0 because 0.000...001 is nonzero
But the above gives 0.000...001=0.
7777777
2017-04-16 19:53:55 UTC
Permalink
Post by b***@gmail.com
Post by 7777777
(2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x
(2*x*dx+dx^2)/dx=2*x+dx
(2*x*dx+dx^2)/dx=2*x
Therefore dx=0
Post by 7777777
there is no division by 0 because 0.000...001 is nonzero
But the above gives 0.000...001=0
I don't know where you want to get at, what's your agenda.
Do you want to deny the infinitesimal difference ratio f'(x) = dy/dx ?
And if so, what's your suggestion what should be used instead of it?
I don't necessarily have answers to all of your questions, and you
should also yourself struggle finding them yourself, just like everyone else
here. Do you expect that I am some kind of answering machine?
I have only shown that the finite difference quotient is not enough to
define the derivative, and I have told why, and I have also told that
the infinitesimal difference ratio is needed. Now, if you want to deny
also this, what do you have got left?
John Gabriel
2017-04-16 20:00:06 UTC
Permalink
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.

No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
Post by 7777777
and I have told why, and I have also told that
the infinitesimal difference ratio is needed.
Yep. You've written a lot of nonsense and continue to do so.
7777777
2017-04-16 20:28:54 UTC
Permalink
Post by John Gabriel
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
you can't seem to understand that the problem is, what you have is

f'(c) = { f(c+n)-f(c-m) } / (m+n)

and it is not f'(x)

You are misleading by telling that x = c, two different points are not the one
and the same point. Furthermore, you are unable to deal with the infinitesimal
range, I seems that just because you deny them so you can't work with them.
John Gabriel
2017-04-17 12:41:13 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
you can't seem to understand that the problem is, what you have is
f'(c) = { f(c+n)-f(c-m) } / (m+n)
and it is not f'(x)
No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.

f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n).
Post by 7777777
You are misleading by telling that x = c, two different points are not the one
and the same point. Furthermore, you are unable to deal with the infinitesimal
range, I seems that just because you deny them so you can't work with them.
b***@gmail.com
2017-04-17 12:56:07 UTC
Permalink
Thats not the mean value theorem, and its not true
for arbitrary m,n. Trivially if you only consider

m,n so that f'(x) = { f(x+n)-f(x-m) } / (m+n) is
true, then also f'(c) = { f(c+n)-f(c-m) } / (m+n)

with x = c is true.
Post by John Gabriel
f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n).
7777777
2017-04-17 16:44:02 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
you can't seem to understand that the problem is, what you have is
f'(c) = { f(c+n)-f(c-m) } / (m+n)
and it is not f'(x)
No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n)
What is f'(x) for f(x) = sin(x)?
John Gabriel
2017-04-17 21:07:48 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
you can't seem to understand that the problem is, what you have is
f'(c) = { f(c+n)-f(c-m) } / (m+n)
and it is not f'(x)
No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n)
What is f'(x) for f(x) = sin(x)?
You've been shown many times you moron.
7777777
2017-04-18 05:25:42 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
I have only shown that the finite difference quotient is not enough to
define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
you can't seem to understand that the problem is, what you have is
f'(c) = { f(c+n)-f(c-m) } / (m+n)
and it is not f'(x)
No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n)
What is f'(x) for f(x) = sin(x)?
You've been shown many times you moron.
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
dear genius, what is c?
dear genius, am I right to assume that f'(c) = {Delta f}/ {Delta x} ?
dear genius, what is m, what is n?
dear genius, why do you want to go looking for c?
dear genius, why do you want to go looking for m and n?
dear genius, am I right to assume that the point x is not the same point as the point c if they are two different points?

Dear genius, why do you keep insisting that two different points x and c
are one and the same point contrary to the facts?

Dear genius, would you nevertheless like to write that x=c?
Dear genius, would this not lead to
dy/dx = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h ?
Dear genius, what do you tell to people who say that it is correct to tell that dy/dx = Lim_Delta x ->0 {Delta f}/ {Delta x}= Lim_h->0(f(x+h)-f(x))/h and not
dy/dx = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h?
Dear genius, do you think that
f'(x) = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h
makes sense?
Dear genius, don't you think that we would need to write:

f'(c) = (f(x+h)-f(x))/h

And say there is c in (x,x+h)?
John Gabriel
2017-04-18 14:39:31 UTC
Permalink
Post by 7777777
Post by John Gabriel
You've been shown many times you moron.
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
Yes.
Post by 7777777
dear genius, what is c?
dear genius, am I right to assume that f'(c) = {Delta f}/ {Delta x} ?
dear genius, what is m, what is n?
dear genius, why do you want to go looking for c?
dear genius, why do you want to go looking for m and n?
dear genius, am I right to assume that the point x is not the same point as the point c if they are two different points?
Moron. f'(c) = { f(c+n)-f(c-m) } / (m+n) =/= {Delta f}/ {Delta x}
Post by 7777777
Dear genius, why do you keep insisting that two different points x and c
are one and the same point contrary to the facts?
Because they are you infinitely stupid idiot!
Post by 7777777
f'(x) = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h makes sense?
No. This is shit.
Post by 7777777
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h)?
No moron. You would need to write:

f'(c) = { f(c+n)-f(c-m) } / (m+n)
b***@gmail.com
2017-04-18 15:28:46 UTC
Permalink
But the below is not correct, m,n are not independent.
Take f(x)=sin(x) at x=pi/6 and n=pi/30 and m=pi/30, does

the equality hold? No.

sin(pi/6+pi/30) = 0.58778525229247312916870595463907
sin(pi/6-pi/30) = 0.41887902047863909846168578443727
pi/30+pi/30 = 0.20943951023931954923084289221863
cos(pi/6) = 0.86602540378443864676372317075294

f(x+m)-f(x-m)
------------- = 0.80646785136591709963875719845893 <> f'(x)
m+n

But choosing m,n so that the equation holds doesn't
make any sense, how do you have f'(x) from the beginning?

By Geogebra auto diff CAS? Not very helpful...
Post by 7777777
f'(c) = { f(c+n)-f(c-m) } / (m+n)
7777777
2017-04-19 05:36:47 UTC
Permalink
Post by 7777777
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
Yes.
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
But you used to tell that f'(0) does not exist, because according to you inflection points have no derivative, so did you change your mind?
Or would you again like to invent an exception to the rules, in order to deny the undeniable?
John Gabriel
2017-04-19 12:27:09 UTC
Permalink
Post by 7777777
Post by 7777777
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
Yes.
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
Post by 7777777
But you used to tell that f'(0) does not exist, because according to you inflection points have no derivative, so did you change your mind?
Idiot. The one who must change his mind is YOU.
Post by 7777777
Or would you again like to invent an exception to the rules, in order to deny the undeniable?
The New Calculus does not change you moron. It is the first and only rigorous formulation of calculus in history.
b***@gmail.com
2017-04-19 14:34:04 UTC
Permalink
But teachel, you said all men have too many holes
in head, and that this cancelation and equality is
korrekt:

P = P * (x-k) / (x-k) /* JGs rule */

Evem when x - k = 0. Now I take x=m and k=-n, and
compute your secant limes for an inflection point.

Take the case f(x)=x^3. We have:

f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x-n^3
--------------- = -----------------------------------------------
m+n m+n

3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2+m*n+n^2)
= -----------------------------------------------
m+n

Now we apply JGs cancellation rule:

= 3*x^2 - 3*(m-n)*x + (m^2+m*n+n^2)

Now if we evaluate this for m=0,n=0 which is allowed
by JGs cancellation rule, we get:

= 3*x^2 - 3*0*x + 0

= 3*x^2

JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.

Therefore we get f'(x)=3*x^2 even for inflection points.

P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
Post by John Gabriel
Post by 7777777
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
b***@gmail.com
2017-04-19 14:42:42 UTC
Permalink
Corr.:
Sorry, I made two little sign errors:

f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x+n^3
--------------- = -----------------------------------------------
m+n m+n

3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2-m*n+n^2)
= -----------------------------------------------
m+n

Now we apply JGs cancellation rule:

= 3*x^2 - 3*(m-n)*x + (m^2-m*n+n^2)
Post by b***@gmail.com
But teachel, you said all men have too many holes
in head, and that this cancelation and equality is
P = P * (x-k) / (x-k) /* JGs rule */
Evem when x - k = 0. Now I take x=m and k=-n, and
compute your secant limes for an inflection point.
f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x-n^3
--------------- = -----------------------------------------------
m+n m+n
3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2+m*n+n^2)
= -----------------------------------------------
m+n
= 3*x^2 - 3*(m-n)*x + (m^2+m*n+n^2)
Now if we evaluate this for m=0,n=0 which is allowed
= 3*x^2 - 3*0*x + 0
= 3*x^2
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
Post by John Gabriel
Post by 7777777
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
7777777
2017-04-19 19:09:10 UTC
Permalink
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."

Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
Post by b***@gmail.com
P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
Post by John Gabriel
Post by 7777777
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
Yes.
John Gabriel
2017-04-20 15:49:37 UTC
Permalink
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.

Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.


Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?

It is this: f'(x)= -x/sqrt(1-x^2)

Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?

Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?

Of course not, you are a nincompoop!!
Post by 7777777
Post by b***@gmail.com
P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
Post by John Gabriel
Post by 7777777
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
Yes.
7777777
2017-04-20 16:02:13 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
hey apeman, did you think that no-one would notice your lie:
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
John Gabriel
2017-04-20 19:11:28 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
I copy and paste the previous comment for you. Reread it because you are too stupid and you missed the answer.

Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?

It is this: f'(x)= -x/sqrt(1-x^2)

Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?

Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
7777777
2017-04-20 19:46:36 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
I copy and paste the previous comment for you. Reread it because you are too stupid and you missed the answer.
Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?
It is this: f'(x)= -x/sqrt(1-x^2)
Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?
Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
What has this example got to do with finding the derivative of sin(x) at
x = 0 ?

Do you think that the derivative of sin(x) at
x = 0 is undefined, just like the derivative of f(x)=sqrt(1-x^2) at x = 1
and x = -1 ?
John Gabriel
2017-04-20 23:43:54 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
I copy and paste the previous comment for you. Reread it because you are too stupid and you missed the answer.
Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?
It is this: f'(x)= -x/sqrt(1-x^2)
Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?
Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
What has this example got to do with finding the derivative of sin(x) at
x = 0 ?
Do you think that the derivative of sin(x) at
x = 0 is undefined, just like the derivative of f(x)=sqrt(1-x^2) at x = 1
and x = -1 ?
YES. A derivative is by definition the expression of the tangent line slope. A tangent line NEVER crosses another curve otherwise it cannot be a tangent line.

If Newton thought the way idiots of today think, there wouldn't even be the finite difference quotient f(x+h)-f(x) / h.

That "limit" of that difference quotient which is a secant line that most resembles a tangent line, is what is called the "derivative".
7777777
2017-04-21 04:12:54 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
I copy and paste the previous comment for you. Reread it because you are too stupid and you missed the answer.
Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?
It is this: f'(x)= -x/sqrt(1-x^2)
Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?
Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
What has this example got to do with finding the derivative of sin(x) at
x = 0 ?
Do you think that the derivative of sin(x) at
x = 0 is undefined, just like the derivative of f(x)=sqrt(1-x^2) at x = 1
and x = -1 ?
YES. A derivative is by definition the expression of the tangent line slope. A tangent line NEVER crosses another curve otherwise it cannot be a tangent line.
Your mistake is you have tried to re-define what is the tangent line.
You think that there can't be a tangent line at an inflection point.
The truth is only that there is no secant at an inflection point, therefore your
"method" does not work there. It does not mean that there is no tangent line at an inflection point. You have incorrectly assumed that because
your "method" does not work at an inflection point, you need to re-define what is the tangent line, without considering that it is your "method" which is wrong. You don't have a correct definition for the derivative, as I have said
over and over again, you only have the mean value theorem, that is the finite
difference quotient, but you need the infinitesimal difference ratio dy/dx.
Your mistake derives from your rejection of infinitesimals.
You have completely ignored the possibility that you can be wrong.
Instead, you have chosen to attack and ridicule people who tell the truth.
Post by John Gabriel
If Newton thought the way idiots of today think, there wouldn't even be the finite difference quotient f(x+h)-f(x) / h.
That "limit" of that difference quotient which is a secant line that most resembles a tangent line, is what is called the "derivative".
The limit of the difference quotient is called the derivative f'(x):

dy/dx = Lim h-> (f(x+h)-f(x))/h means that the finite difference quotient
f(x+h)-f(x)/h is never equal to dy/dx.
In other words, dy/dx is never finite, it is never equal to a finite difference quotient.
Because you have refused to deal with dy/dx, and instead work only with
the finite difference quotient f(x+h)-f(x)/h, you have introduced an error
into your calculation. That error manifests in your attempt to calculate
the derivative of sin(x) at x=0. You can never get the correct answer,
instead you start looking for the point c. The error disappears only
when the range becomes infinitesimal, also with the introduction of the infinitesimal dx.
John Gabriel
2017-04-21 14:23:09 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
Post by b***@gmail.com
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
I copy and paste the previous comment for you. Reread it because you are too stupid and you missed the answer.
Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?
It is this: f'(x)= -x/sqrt(1-x^2)
Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?
Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
What has this example got to do with finding the derivative of sin(x) at
x = 0 ?
Do you think that the derivative of sin(x) at
x = 0 is undefined, just like the derivative of f(x)=sqrt(1-x^2) at x = 1
and x = -1 ?
YES. A derivative is by definition the expression of the tangent line slope. A tangent line NEVER crosses another curve otherwise it cannot be a tangent line.
Your mistake is you have tried to re-define what is the tangent line.
Fail stupid. It was not I who tried to re-define the tangent line but Judith Grabiner and the BIG STUPID. It actually goes back to a fucked up school principal in the UK (someone with the same shit brains as Theresa May):

Grabiner writes:

"How then was the tangent to be defined at the point (0,O) for a curve like y =x^3 (FIGUREl), or to a point on a curve with many turning points (FIGURE 2)? "

Page 2 of
https://www.maa.org/sites/default/files/0025570x04690.di021131.02p02223.pdf

Grabiner is part of the BIG STUPID and in her September 1983 publication, she admitted that it was not possible to have a tangent line at x=0 when f(x)=x^3. If it were, then there wouldn't be a need to "define" a tangent.

See, there's the first problem with morons of the BIG STUPID: if the facts don't fit the theory, change the facts.

By definition, a tangent line intersects a curve in one point, extends finitely to both sides and crosses the curve nowhere.

Ironically, in their redefinition, the BIG STUPID used the difference quotient to define the tangent line slope, not noticing the circularity of their actions. Till that point, a derivative was derived from a tangent line, but now the tangent line was derived from the derivative. Circular? Of course.

Grabiner claims the Greeks didn't study many curves. Evidently she never read the Works of Archimedes - just like most of you morons haven't.

Pull up your pants and give your mouth a chance you moron. You do not have even a tiny bit of my intelligence.


The rest of your comment is shit.
John Gabriel
2017-04-14 20:14:43 UTC
Permalink
Post by 7777777
Post by b***@gmail.com
More precisely it must exactly happen 0.000...001 == 0
to obtain f'(x). Please observe for f(x)=x^2, we have
f'(x) = 2*x
Now lets try your method of plugging in. First of all
you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
But the question is for (f(x+dx)-f(x))/dx. Which is
now (for you "finite" quantities, and you didn't tell
x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx
Now you said your idea is just plugging in infinitessimal
in pace of a finite quantity, so I do plugging in, i.e.
g(x) = 2*x + 0.000...001
But since you said 0.000...001 != 0, we don't have g(x)
= f'(x). But sometimes you say is even both, 0.000...001 != 0
and 0.000...001 == 0.
This is very confusing
It is not confusing.
You need to consider (dx)^2 = (0.000...001)^2 can be equal to 0 even though
0.000...001 is non-zero
And then you end up with f'(x) = 2x
See, even the local idiot Jan Burse disagrees with you? Chuckle.
b***@gmail.com
2017-04-14 20:19:42 UTC
Permalink
bird brain John Gabriel birdbrain, I don't disagree with 777777,
since he didn't yet fully explain his method. He says this and that.

As a bird brain John Gabriel birdbrain, you cannot distingzish
inquiry from opinion. Anyway this is a funny read, much more

funnier than the new fake calculoose:

An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
Post by John Gabriel
See, even the local idiot Jan Burse disagrees with you? Chuckle.
b***@gmail.com
2017-04-14 20:23:12 UTC
Permalink
The big advantage of limes is that it doesn't break
trichotomy, since you still evaluate in the finite Quantities,

you can also compute limes of piecewise functions easily, etc..
Post by b***@gmail.com
bird brain John Gabriel birdbrain, I don't disagree with 777777,
since he didn't yet fully explain his method. He says this and that.
As a bird brain John Gabriel birdbrain, you cannot distingzish
inquiry from opinion. Anyway this is a funny read, much more
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
Post by John Gabriel
See, even the local idiot Jan Burse disagrees with you? Chuckle.
John Gabriel
2017-04-14 20:13:58 UTC
Permalink
Post by 7777777
you have not shown anything. you have only shown that you are a moron.
It is here: dx <---------- the infinitesimal
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple someone would have already solved the problem a long time ago.
It's not simple and that's why they couldn't solve it. It took genius - ME.
Post by 7777777
What I have shown (and not just me, btw), over and over again, it maybe be both 0 and non-zero at the same time.
Idiot. Nothing can be 0 and non-zero. Either one or the other. Stupid boy.
Post by 7777777
There is no such thing. There is no such thing as "BIG STUPID".
There is and YOU are part of it.
Post by 7777777
You just keep ridiculing us, without understanding that it is you who is in error.
I ridicule you because you are a thick headed idiot.
Post by 7777777
No-one takes you seriously, you just keep repeating the same nonsense over and over again, basically telling us what can exist and what can't, but you can't convince anyone, because what exists and what doesn't are just your beliefs.
Liar. Many have been convinced. Even mainstream professors who hold PhDs in mathematics.
Post by 7777777
in the mainstream formulation dx is equal to 0 because according to them
dx = 0.000...001 = 0
FAIL. Even the local morons on this site will not agree with you. Please just show me one moron who agrees with you. Chuckle.
Post by 7777777
are you talking to me? It seems to me that it is you who needs to grow brains. You can't have calculus based only on finite difference.
Yes idiot. You can have calculus based only on finite difference - it's called The New Calculus. Good bye stupid.
7777777
2017-04-16 04:51:41 UTC
Permalink
Post by John Gabriel
Post by 7777777
are you talking to me? It seems to me that it is you who needs to grow brains. You can't have calculus based only on finite difference.
Yes idiot. You can have calculus based only on finite difference - it's called The New Calculus. Good bye stupid.
you don't have such a thing. you have only the mean value theorem:

your
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
is the same as
f'(c) = {f(b) - f(a)}/(b-a)
because
c+n = b
c-m = a
meaning
b-a = c+n-(c-m) = c+n-c+m = m+n
John Gabriel
2017-04-16 07:17:06 UTC
Permalink
Post by 7777777
Post by John Gabriel
Post by 7777777
are you talking to me? It seems to me that it is you who needs to grow brains. You can't have calculus based only on finite difference.
Yes idiot. You can have calculus based only on finite difference - it's called The New Calculus. Good bye stupid.
your
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
is the same as
f'(c) = {f(b) - f(a)}/(b-a)
because
c+n = b
c-m = a
meaning
b-a = c+n-(c-m) = c+n-c+m = m+n
Moron. It's called the secant theorem (a geometrical result and first proved constructively by me only) in the new calculus. But yes, it is directly related to the mean value theorem and quite funny that no one ever understood it.

Yes, the mean value theorem is about a special kind of finite difference - one called a "natural arithmetic mean". You are too stupid to waste my time.
7777777
2017-04-16 07:37:11 UTC
Permalink
Post by John Gabriel
Post by 7777777
Post by John Gabriel
Post by 7777777
are you talking to me? It seems to me that it is you who needs to grow brains. You can't have calculus based only on finite difference.
Yes idiot. You can have calculus based only on finite difference - it's called The New Calculus. Good bye stupid.
your
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
is the same as
f'(c) = {f(b) - f(a)}/(b-a)
because
c+n = b
c-m = a
meaning
b-a = c+n-(c-m) = c+n-c+m = m+n
Moron. It's called the secant theorem (a geometrical result and first proved constructively by me only) in the new calculus. But yes, it is directly related to the mean value theorem and quite funny that no one ever understood it.
Yes, the mean value theorem is about a special kind of finite difference - one called a "natural arithmetic mean". You are too stupid to waste my time.
troll boy, for how long time are you continuing claiming

f'(c) = {Delta f}/{Delta x} <---------- finite difference quotient

is the definition of the derivative?

What is m ? What is n? Why do you want to look for them? Why do you want to look
for c? Why you never tell what is the derivative at point x, the point at which
we want to know the value of the derivative?
Me
2017-04-14 20:32:17 UTC
Permalink
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple ...
John is right. If x e IR then x is either ZERO or it is NOT ZERO, at least in the context of classical mathematics.

Moreover, if x e IR and |x| < 1/n for all n IN, then x = 0 (since the reals are archimedian).
it maybe be both 0 and non-zero at the same time.
Not in the context of classical mathematics / classical logic.

Btw. In one may call 0 e IR the only infitesimal in IR.

See:
https://math.stackexchange.com/questions/1086713/is-0-an-infinitesimal
b***@gmail.com
2017-04-14 20:53:39 UTC
Permalink
Post by Me
Post by 7777777
it maybe be both 0 and non-zero at the same time.
Not in the context of classical mathematics / classical logic.
Yes without further context, where it changes its
value, like an operator applied to it,

or two different equality signs, we usually
get a contradiction

e = 0 & ~ e = 0 /* is a contradiction */

e =_1 0 & ~ e =_2 0 /* is not per se a contradiction */

And of course (for the first sample, we need that
we are not in R, some other domain D, for the second
sample we assume n > 0):

e^2 = 0 & ~ e = 0 /* is not per se a contradiction */

lim 1/n = 0 & ~ 1/n = 0 /* is not per se a contradiction */

The later cases are case where an operator is involved.
The last case is very problematic for JG, since he
believes ~ 0.999... = 1.

But ~ 0.999... = 1 is possible when we are outside of R,
some other domain D. Its value could be 1 - e, with e^2 = 0
and ~ e = 0. Right?

Don't ask me how ~ 0.999... = 1 exactly works, just flunking
it. I tried duals once, but problem is to make it really
injective, i.e. from f : R -> R, to get an injective

f : D -> D, this was the obstactle. Basically when f'(x)=0,
it didn't work the extension. Bell mentions this when he
talks about stationary points I guess.
Me
2017-04-14 21:00:01 UTC
Permalink
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple ...
John is right. For every x: x is either ZERO or it is NOT ZERO, at least in the context of classical mathematics: x = 0 v ~(x = 0).

Moreover, if x e IR and |x| < 1/n for all n IN, then x = 0 (since the reals are archimedian).
it maybe be both 0 and non-zero at the same time.
Not in the context of classical mathematics / classical logic: here we have for each and every x: ~(x = 0 & ~(x = 0))

Btw.: One may call 0 e IR the only infitesimal in IR.

See:
https://math.stackexchange.com/questions/1086713/is-0-an-infinitesimal
7777777
2017-04-15 03:57:03 UTC
Permalink
Post by Me
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple ...
John is right. For every x: x is either ZERO or it is NOT ZERO
Let's get this straight:

is 0.000...001 either ZERO or is it NOT ZERO ?
Python
2017-04-15 08:50:40 UTC
Permalink
Post by 7777777
Post by Me
Post by John Gabriel
Look you moron: It is either ZERO or it is NOT.
If it would be that simple ...
John is right. For every x: x is either ZERO or it is NOT ZERO
is 0.000...001 either ZERO or is it NOT ZERO ?
It is nothing, it is meaningless. It is an abuse in notation.

0.1374 is a shortcut for a mapping from {1,2,3,4,...}
to {0,1,2,3,4,5,6,7,8,9} :
1 -> 1
2 -> 3
3 -> 7
4 -> 4
any n > 4 -> 0

similarly, 0.99999... is a shortcut for:
1 -> 9
2 -> 9
any n -> 9

being well defined, from this mapping the value it represents
is the sum for all n of the terms m(n)/10^n

On the contrary, when you write :
0.0000....001

there is NO WAY to consider it as a shortcut for such a mapping.

Ask yourself, Mr Seven, take 0.000...001, let n be ANY integer,
what is the nth digit of 0.000...001?
b***@gmail.com
2017-04-14 15:40:18 UTC
Permalink
Post by 7777777
I am not exactly sure what you want me to demonstrate, but I will try anyway.
Consider first
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
if m and n are finite then f'(c) is not equal to f'(x)
Yes and no.

Yes:
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) does hold in general.

No:
But I think calling m and n finite doesn't make any
sense, they are not sets, they are reals.

For the No please consider the definition:

infinitesimal (plural infinitesimals)

(mathematics) A non-zero quantity whose magnitude is smaller than any positive number (by definition it is not a real number).

https://en.wiktionary.org/wiki/infinitesimal

So it reads smaller than any positive number and not smaller
than any finite number. You cannot use the adjective finite
with numbers, its not a set.

You can say:
{0, 1, 2} is finite
or
{2, 4, 6, ...} is not finite

But applying the adjective finite to real numbers doesn't
make any sense to me.
b***@gmail.com
2017-04-14 15:41:58 UTC
Permalink
Corr.:
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) doesn't hold in general.
Post by b***@gmail.com
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) does hold in general.
John Gabriel
2017-04-14 16:01:37 UTC
Permalink
Post by b***@gmail.com
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) doesn't hold in general.
Post by b***@gmail.com
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) does hold in general.
It holds in general you dipshit.
b***@gmail.com
2017-04-14 16:08:55 UTC
Permalink
Compare with the following mismatch:

a) A menstruating woman is sitting in the kitchen.

b) A menstruating man is standing at the door.

Which one has an error?

Now:

a) the set of primes isn't finite

b) if h is finite then ...

On the other hand in the past it was indeed common to
use the adjective finite on quantities. But still it
sounds strange to me. But the 777777 usage, is for example
found in the introduction here:

The Method of Fluxions and Infinite Series: With
Its Application to the ... von Isaac Newton

I easily find the following passage:

.. and introduces non but infinitely little Quantities
that are relatively to; which he arrives at by beginning
with finite Quantities, and proceeding by a gradual
and necessary progress of diminution. His computations
always commence by finite and intelligible quantities, ...

So maybe thats why 77777 uses the adjective finite
for h, m and n, etc.. but its still sounds wrong to
me since in our modern times we have sets and set theory,
and it looks like a confusion error.

But without any clear apparatus given by 77777, where the
notion of finite for a number is exactly defined, and
also what it means for a number to be non-finite, what
Post by j4n bur53
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
So 7777777, what D (?surreals etc..?) are you using?
Post by j4n bur53
So it reads smaller than any positive number and not smaller
than any finite number. You cannot use the adjective finite
with numbers, its not a set.
{0, 1, 2} is finite
or
{2, 4, 6, ...} is not finite
But applying the adjective finite to real numbers doesn't
make any sense to me.
b***@gmail.com
2017-04-14 16:38:23 UTC
Permalink
Although newtons booklet looks a little
shitty compared to the booklet of euler,

On page 51 he left handedly computes the
tangent to a spiral of archimedes, so that

I guess we nevertheless have a genius before us...

BTW: And funny thing, he writes "affirmative
integer" instead of "positive integer".
Post by b***@gmail.com
The Method of Fluxions and Infinite Series: With
Its Application to the ... von Isaac Newton
.. and introduces non but infinitely little Quantities
that are relatively to; which he arrives at by beginning
with finite Quantities, and proceeding by a gradual
and necessary progress of diminution. His computations
always commence by finite and intelligible quantities, ...
b***@gmail.com
2017-04-14 16:47:39 UTC
Permalink
Its Application to the Geometry of Curve-lines. By ... Sir Isaac Newton, ... Translated from the Author's Latin Original Not Yet Made Publick. To which is Subjoin'd, a Perpetual Comment Upon the Whole Work, ... By John Colson, ...

Isaac Newton
Henry Woodfall; and sold by John Nourse, 1736 - 339 pages
Post by b***@gmail.com
Although newtons booklet looks a little
shitty compared to the booklet of euler,
On page 51 he left handedly computes the
tangent to a spiral of archimedes, so that
I guess we nevertheless have a genius before us...
BTW: And funny thing, he writes "affirmative
integer" instead of "positive integer".
Post by b***@gmail.com
The Method of Fluxions and Infinite Series: With
Its Application to the ... von Isaac Newton
.. and introduces non but infinitely little Quantities
that are relatively to; which he arrives at by beginning
with finite Quantities, and proceeding by a gradual
and necessary progress of diminution. His computations
always commence by finite and intelligible quantities, ...
Dan Christensen
2017-04-14 14:41:04 UTC
Permalink
Post by John Gabriel
9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago...
When it comes to calculus, JG is utterly clueless. Using his Wacky New "Calclueless," he cannot even determine the derivative of functions as simple as y=x. He actually claims they are "undefined!" What a moron.

Neither can JG derive even the most elementary results of basic arithmetic. A proof of 2+2=4 is well beyond his meager capabilities.

JG's Wacky New Calclueless is a dead end and a complete waste of time. Interested readers should see my recent posting, "About the spamming troll John Gabriel in his own words (April 2017)."


Dan
William Elliot
2017-04-15 03:57:50 UTC
Permalink
Post by Dan Christensen
When it comes to calculus, JG is utterly clueless.
DC, who didn't take his promissed haitus, is clueles about sci.math
thinking it's a gossip forum. Good bye, you are blacklisted.
Dan Christensen
2017-04-16 22:03:01 UTC
Permalink
Post by William Elliot
Post by Dan Christensen
When it comes to calculus, JG is utterly clueless.
DC, who didn't take his promissed haitus,
Bags are packed. One delay after another. Hurry up and wait. Sigh...
Post by William Elliot
thinking it's a gossip forum. Good bye, you are blacklisted.
OH, NOOOOOO!!!! Please, please, oh please don't "blacklist" me!!! (HA, HA, HA!)

Toodles.


Dan
Eram semper recta
2023-03-19 14:36:40 UTC
Permalink
Post by John Gabriel
9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
I wonder what these fucking morons might have to say about my historic geometric theorem now? Chuckle.

https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020

Still think that calculus requires "limits", you dumb fucks?!
Mostowski Collapse
2023-03-19 14:51:40 UTC
Permalink
I though you invented New Calculus 40-50 years ago,
when you were a tiny wanker?
Post by Eram semper recta
I wonder what these fucking morons might have to say about my historic geometric theorem now? Chuckle.
https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
Still think that calculus requires "limits", you dumb fucks?!
Eram semper recta
2023-03-19 14:37:55 UTC
Permalink
Post by John Gabriel
9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
It's great to lambast the lot of you vile, dishonest, incorrigibly stupid bastards!!!!!

https://www.academia.edu/98758410/Was_Newtons_and_Leibnizs_method_of_setting_h_0_valid_The_answer_is_YES
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