Rule of operations of infinite series -- Euler's formula is an approximate

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
a(n) is called the general term, a(0),a(1),... the addend, summand or just
term. n is referred to as the index. Series S is the sum from the first term
a(0) to the last term a(k). The sum of those first terms (n<k) is called the
partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
called an infinite series. Note that there are infinite(NEVER terminate)
addends. I.e. basically, the addition of addends cannot be completed in
finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
(the index is ∞) must be shown to indicate the general term.
Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
<=> S= 1+a*S-a^(∞+1)
<=> S(1-a)=1-a^(∞+1)
<=> S= (1-a^(∞+1))/(1-a)
Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
S= 1+2+3+...+n // (1)
S= n+...+3+2+1 // (2)
2S= n*(n+1) // (1)+(2)
<=> S= n*(n+1)/2
If the last addend is missing, the expanded form is prone to magic tricks,
because the rearrangement of the expanded form may likely change the
S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
= Σ(n=1,∞) n+1 // S is modified
(or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
S=1+2+4+8+... // The last addend is omitted (ill-formed)
<=> S=1+2(1+2+4+8+...)
<=> S=1+2S
<=> S=-1
S=1+2+4+8+...+2^∞
<=> S=1+2(1+2+4+...+2^(∞-1))
<=> S=1+2S-2^(∞+1)
<=> S=2^(∞+1)-1 // Lots of similar "magic calculation" deriving the result
// S=-1 can be found in youtube (from the omission of the
// term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
= a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
Proof: Omitted (Can be derived from the expanded form. Trivial rules are also
omitted)
Basically, formula for finite series are also applicable to infinite series(
but mathematical inducion cannot prove such formula because by definition,
∞ means 'the procedure never terminate' and the Peano axiom is only valid in
finite steps).
Note: Many 'equations' of infinite series (esp. about π,e) can be proved
false by the theorems above. They are actually approximates (limits).
Ex: Σ(n=1,∞) 1/n² ≒ π²/6
Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Chris M. Thomasson

2024-04-24 20:27:09 UTC

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend, summand
or just
   term. n is referred to as the index. Series S is the sum from the
first term
   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to magic
tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation" deriving
the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is only
valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can be
proved
         false by the theorems above. They are actually approximates
(limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

Moebius

2024-04-24 20:58:53 UTC

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend, summand
or just
   term. n is referred to as the index. Series S is the sum from the
first term
   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to magic
tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation"
deriving the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can be
proved
         false by the theorems above. They are actually approximates
(limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

See: https://en.wikipedia.org/wiki/Law_of_large_numbers

(Though I doesn't make any sense in the present context.)

Ross Finlayson

2024-04-24 21:02:44 UTC

Observe the law(s) of large numbers.

Define a large number? What is large to you?

Well, there's potential, practical, effective, and actual infinity.

Then, once upon a time, Zdislav Kovaric used to write to sci.math,
he details some "kinds" of infinity.

Then, there are various sorts law(s) of large numbers,
it's what the each of the laws are, for any or all large
numbers, it's the each of the law(s) of large numbers, for
the usual law of small numbers.

Usually, practically or effectively infinite are large enough.

Not always, ....

It sort of helps to have at least three definitions of continuous
domains, which together form a more replete and complete than the usual
definition of "complete".

Here these infinite expressions and their completions and their closure
of the forms, there was a pretty good discussion about series that
converge or diverge, and series that slowly converge or slowly diverge,
about a bunch of various kinds of criteria of convergence and
divergence, due various operations and their alternations,
in infinite expressions.

Four is a pretty good-sized number, ....

Chris M. Thomasson

2024-04-24 21:22:46 UTC

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend, summand
or just
   term. n is referred to as the index. Series S is the sum from the
first term
   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to magic
tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation"
deriving the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can be
proved
         false by the theorems above. They are actually approximates
(limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

Well, there's potential, practical, effective, and actual infinity.
Then, once upon a time, Zdislav Kovaric used to write to sci.math,
he details some "kinds" of infinity.
Then, there are various sorts law(s) of large numbers,
it's what the each of the laws are, for any or all large
numbers, it's the each of the law(s) of large numbers, for
the usual law of small numbers.
Usually, practically or effectively infinite are large enough.
Not always, ....
It sort of helps to have at least three definitions of continuous
domains, which together form a more replete and complete than the usual
definition of "complete".
Here these infinite expressions and their completions and their closure
of the forms, there was a pretty good discussion about series that
converge or diverge, and series that slowly converge or slowly diverge,
about a bunch of various kinds of criteria of convergence and
divergence, due various operations and their alternations,
in infinite expressions.
Four is a pretty good-sized number, ....

Only a little zoom on a Newton Set:

Moebius

2024-04-24 21:04:12 UTC

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend, summand
or just
   term. n is referred to as the index. Series S is the sum from the
first term
   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to magic
tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation"
deriving the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can be
proved
         false by the theorems above. They are actually approximates
(limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

See: https://en.wikipedia.org/wiki/Law_of_large_numbers

(Though it doesn't make any sense in the present context.)

Chris M. Thomasson

2024-04-24 21:22:12 UTC

Post by Moebius

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend,
summand or just
   term. n is referred to as the index. Series S is the sum from the
first term
   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to
magic tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation"
deriving the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can be
proved
         false by the theorems above. They are actually approximates
(limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

See: https://en.wikipedia.org/wiki/Law_of_large_numbers
(Though it doesn't make any sense in the present context.)

Can I say that the inverse of this very small number seems interesting
wrt a larger number? Keep in mind that it's unbounded:

In the abstract, beyond our finite self's, this does go on forever....
Fair enough?

Moebius

2024-04-24 22:54:58 UTC

Post by Moebius

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
   a(n) is called the general term, a(0),a(1),... the addend,
summand or just
   term. n is referred to as the index. Series S is the sum from
the first term
   a(0) to the last term a(k). The sum of those first terms (n<k)
is called the
   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S is
   called an infinite series. Note that there are infinite(NEVER
terminate)
   addends. I.e. basically, the addition of addends cannot be
completed in
   finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
   (the index is ∞) must be shown to indicate the general term.
   Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
     S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
     <=> S= 1+a*S-a^(∞+1)
     <=> S(1-a)=1-a^(∞+1)
     <=> S= (1-a^(∞+1))/(1-a)
   Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
     S= 1+2+3+...+n // (1)
     S= n+...+3+2+1 // (2)
     2S= n*(n+1)     // (1)+(2)
     <=> S= n*(n+1)/2
   If the last addend is missing, the expanded form is prone to
magic tricks,
   because the rearrangement of the expanded form may likely change the
        S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
        = Σ(n=1,∞) n+1 // S is modified
          (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
     S=1+2+4+8+...     // The last addend is omitted (ill-formed)
     <=> S=1+2(1+2+4+8+...)
     <=> S=1+2S
     <=> S=-1
     S=1+2+4+8+...+2^∞
     <=> S=1+2(1+2+4+...+2^(∞-1))
     <=> S=1+2S-2^(∞+1)
     <=> S=2^(∞+1)-1   // Lots of similar "magic calculation"
deriving the result
                       // S=-1 can be found in youtube (from the
omission of the
                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                     = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
   Proof: Omitted (Can be derived from the expanded form. Trivial
rules are also
          omitted)
   Basically, formula for finite series are also applicable to
infinite series(
   but mathematical inducion cannot prove such formula because by
definition,
   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in
   finite steps).
   Note: Many 'equations' of infinite series (esp. about π,e) can
be proved
         false by the theorems above. They are actually
approximates (limits).
         Ex: Σ(n=1,∞) 1/n² ≒ π²/6
             Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
             Σ(n=0,∞) k^n/n! ≒ e^k
----------

Observe the law(s) of large numbers.

Define a large number? What is large to you?

See: https://en.wikipedia.org/wiki/Law_of_large_numbers
(Though it doesn't make any sense in the present context.)

Can I say that the inverse of this very small number seems interesting
http://youtu.be/0jGaio87u3A

Holy shit, I swear, I only took a tiny piece!!!

Post by Chris M. Thomasson
In the abstract, beyond our finite self's, this does go on forever....
Fair enough?

Sure - at least in the "mathematical reality". :-P

Chris M. Thomasson

2024-04-24 21:55:23 UTC

On 4/24/2024 12:00 PM, wij wrote:
[...]

FromTheRafters

2024-04-24 23:40:28 UTC

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+

Post by wij
Infinite Series |

+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
a(n) is called the general term, a(0),a(1),... the addend, summand or just
term. n is referred to as the index. Series S is the sum from the first
term a(0) to the last term a(k). The sum of those first terms (n<k) is
called the partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S
is called an infinite series. Note that there are infinite(NEVER terminate)
addends. I.e. basically, the addition of addends cannot be completed in
finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
(the index is ∞) must be shown to indicate the general term.
Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
<=> S= 1+a*S-a^(∞+1)
<=> S(1-a)=1-a^(∞+1)
<=> S= (1-a^(∞+1))/(1-a)
Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
S= 1+2+3+...+n // (1)
S= n+...+3+2+1 // (2)
2S= n*(n+1) // (1)+(2)
<=> S= n*(n+1)/2
If the last addend is missing, the expanded form is prone to magic tricks,
because the rearrangement of the expanded form may likely change the
S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
= Σ(n=1,∞) n+1 // S is modified
(or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
S=1+2+4+8+... // The last addend is omitted (ill-formed)
<=> S=1+2(1+2+4+8+...)
<=> S=1+2S
<=> S=-1
S=1+2+4+8+...+2^∞
<=> S=1+2(1+2+4+...+2^(∞-1))
<=> S=1+2S-2^(∞+1)
<=> S=2^(∞+1)-1 // Lots of similar "magic calculation" deriving the
result // S=-1 can be found in youtube (from the
omission of the // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
= a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
Proof: Omitted (Can be derived from the expanded form. Trivial rules are
also omitted)
Basically, formula for finite series are also applicable to infinite
series( but mathematical inducion cannot prove such formula because by
definition, ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in finite steps).
Note: Many 'equations' of infinite series (esp. about π,e) can be proved
false by the theorems above. They are actually approximates (limits).
Ex: Σ(n=1,∞) 1/n² ≒ π²/6
Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
Σ(n=0,∞) k^n/n! ≒ e^k
----------

For a finite sequence, add the finitely many terms by simple addition.
If the sequence is infinite, you need a series to obtain a sum. The
word series already implies an attempt at summing infinitely many
terms.

wij

2024-04-25 14:13:08 UTC

Post by FromTheRafters

Post by wij
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
....
+-----------------+

Post by wij
Infinite Series |

+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
a(n) is called the general term, a(0),a(1),... the addend, summand or just
term. n is referred to as the index. Series S is the sum from the first
term   a(0) to the last term a(k). The sum of those first terms (n<k) is
called the   partial sum. "a(0)+...+a(k)" is called expanded form.
Infinite Series::= If the series S refers to infinite terms/addend (n=∞), S
is   called an infinite series. Note that there are infinite(NEVER terminate)
addends. I.e. basically, the addition of addends cannot be completed in
finite steps by definition.
Operation Principle of Infinite Series: The last addend of the expanded form
(the index is ∞) must be shown to indicate the general term.
Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
    S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
    <=> S= 1+a*S-a^(∞+1)
    <=> S(1-a)=1-a^(∞+1)
    <=> S= (1-a^(∞+1))/(1-a)
Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
    S= 1+2+3+...+n // (1)
    S= n+...+3+2+1 // (2)
    2S= n*(n+1)     // (1)+(2)
    <=> S= n*(n+1)/2
If the last addend is missing, the expanded form is prone to magic tricks,
because the rearrangement of the expanded form may likely change the
       S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
       = Σ(n=1,∞) n+1 // S is modified
         (or S=(1+2)+(3+4)+... = Σ(n=1,∞) 4*n-1)
    S=1+2+4+8+...     // The last addend is omitted (ill-formed)
    <=> S=1+2(1+2+4+8+...)
    <=> S=1+2S
    <=> S=-1
    S=1+2+4+8+...+2^∞
    <=> S=1+2(1+2+4+...+2^(∞-1))
    <=> S=1+2S-2^(∞+1)
    <=> S=2^(∞+1)-1   // Lots of similar "magic calculation" deriving the
result                       // S=-1 can be found in youtube (from the
omission of the                       // term containing ∞).
Theorem1: s1=s2 <=> s1-s2=0
Theorem2: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
                    = a(∞)+ Σ(n=0,∞-1) a(n)
Theorem3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem4: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
Proof: Omitted (Can be derived from the expanded form. Trivial rules are
also          omitted)
Basically, formula for finite series are also applicable to infinite
series(   but mathematical inducion cannot prove such formula because by
definition,   ∞ means 'the procedure never terminate' and the Peano axiom is
only valid in   finite steps).
Note: Many 'equations' of infinite series (esp. about π,e) can be proved
        false by the theorems above. They are actually approximates (limits).
        Ex: Σ(n=1,∞) 1/n² ≒ π²/6
            Σ(n=0,∞) (-1^n)*(1/(2n+1)) ≒ π/4
            Σ(n=0,∞) k^n/n! ≒ e^k
----------

Not sure what you were talking about.

FromTheRafters

2024-04-25 16:19:24 UTC