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Anonymous Wed 04 Mar 18:44:29 2020 No.11439850 View Reply Original Report

Quoted By: >>11439883 >>11439885 >>11441849 >>11442777

Somebody please explain why negative 2 in the riemann zeta function is 0? That means that 1+4+9+16+25...=0.
On what planet is that true?

Explain how 1/2 +14something times i is 0 as well, how did he prove it? I dont get it right now it makes no sense. Did he find equivalent expressions to the zeta function and use those instead? Talk...

Anonymous

Anonymous Wed 04 Mar 2020 18:59:52 No.11439883 Report

Quoted By: >>11442810

>>11439850
Of course a real math thread gets no replies but a greta or race thread is top page

Anonymous

Anonymous Wed 04 Mar 2020 19:00:17 No.11439885 Report

Quoted By: >>11439904 >>11440036

>>11439850
t. hasn't taken introductory complex analysis
what the fuck is analytic continuation anyway?

Anonymous

Anonymous Wed 04 Mar 2020 19:07:15 No.11439904 Report

Quoted By: >>11439909

>>11439885
I think that riemann found an equation that has all the same results as the zeta function but whenever the zeta function is infinity the other equation has a finite value

Whenever equation one converges equation two converges to the same number

But if equation one diverges then equation two converges to some number

Thats it i think

Anonymous

Anonymous Wed 04 Mar 2020 19:08:31 No.11439909 Report

Quoted By:

>>11439904
And when equation two converges to Zero that counts as a Zero for equation one even though equation one doesnt converge there

Anonymous

Anonymous Wed 04 Mar 2020 19:13:45 No.11439925 Report

Quoted By:

Does a single person on sci know why the squares are 0?

Anonymous

Anonymous Wed 04 Mar 2020 19:19:34 No.11439933 Report

Quoted By:

Bump

Anonymous

Anonymous Wed 04 Mar 2020 19:38:20 No.11439976 Report

Quoted By:

Bump

Anonymous

Anonymous Wed 04 Mar 2020 19:42:07 No.11439988 Report

Quoted By: >>11440005 >>11440031

Sounds like a chinese pasted into translate.

Anonymous

Anonymous Wed 04 Mar 2020 19:46:08 No.11440005 Report

Quoted By:

>>11439988
Im Just autistic or something it doesnt affect intelligence Just clearness and coherence of speach

Anonymous

Anonymous Wed 04 Mar 2020 19:55:55 No.11440031 Report

Quoted By:

>>11439988
Any explaination for powers equal to zero

Anonymous

Anonymous Wed 04 Mar 2020 19:58:24 No.11440036 Report

Quoted By: >>11440060 >>11440067

>>11439885
isn't the whole point of 1+2+3...=-1/12 either that
>the Riemann zeta is broken
or
>analytical continuation if bullshit
?

Anonymous

Anonymous Wed 04 Mar 2020 20:07:13 No.11440060 Report

Quoted By: >>11440098

>>11440036
Well, if an analític function that equals the riemann zeta in every way except for values where the riemann zeta is infinity, and the other function is -1/12 for -1 which the riemann zeta equals infinity for, but -1/12 has other examples where its 1+2+3+4 proving its not bull

Anonymous

Anonymous Wed 04 Mar 2020 20:09:44 No.11440067 Report

Quoted By: >>11440070

>>11440036
0*x is equivalent to x^0 -1

Anonymous

Anonymous Wed 04 Mar 2020 20:10:49 No.11440070 Report

Quoted By:

>>11440067
Imagine an equivalent equation to the riemann zeta function that equals -1/12 for -1 though

Anonymous

Anonymous Wed 04 Mar 2020 20:17:21 No.11440091 Report

Quoted By:

Somebody answer if ur smart

Anonymous

Anonymous Wed 04 Mar 2020 20:22:34 No.11440098 Report

Quoted By:

>>11440060
>but -1/12 has other examples where its 1+2+3+4 proving its not bull
source? Either way, it's objectively false, which means there's gotta be some function that can transform infinity to a real number, while at all other values remain as identities

Anonymous

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Anonymous Wed 04 Mar 2020 20:29:41 No.11440123 Report

Quoted By:

$(1-z)\sum_{k=0}^{n-1} z^k = (1-z)\cdot (1+z+z^2+z^3+\cdot z^{n-1}) = 1-z^n$

So for all z not 1,

$\sum_{k=0}^{n-1} z^k = \dfrac{1-z^n}{1-z}$

For for all z with |z|<1,

$\sum_{k=0}^\infty z^k = \dfrac{1}{1-z}$

This is just the geometric series.

For for d in the interval (0, 2) and letting z=-1+d

$\sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2-d}$

So:

$\lim_{d\to 0} \sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2}$

Let me make a point that this isn't to claim:

$1-1+1-1+1-1+ ... = \dfrac{1}{2}$

But the algebraic properties of the sum relate "1-1+1-1+1-1+" to "1/2" in many context and that's also the case with 1+4+9+16+25 and 1+2+3+4+... It's not random, and it's not real or complex analysis either.

Going on..
The analytic continuation F of a function f is, roughly, a complex differentiable function and which, whether f is defined, has the functions agreeing F(z)=f(z). Complex differentiation is a more strict condition than multivariate differentiability (complex differentiable means the Jacobian has a certain form).
Long story short, $F(z)=\dfrac{1}{1-z}$ is a continuation of $f(z)\sum_{k=0}^\infty z^k$ and the continuation of the function
$\sum_{n=1}^\infty\, n^{-z}$
happens to fulfill
$F_\zeta(z) = 2\,(2\pi)^{z-1}\sin{\left(\pi\,z/2\right)}\,\Gamma(1-z)\,F_\zeta(1-z)$ .
From the latter formula (and knowledge of the positive vaues), you can read off the "spooky" equations.
But again, it's not an equality of analysis in the same way that proving

$\lim_{d\to 0} \sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2}$

isn't proving

$1-1+1-1+1-1+ ... = \dfrac{1}{2}$

Anonymous

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Anonymous Wed 04 Mar 2020 20:33:30 No.11440131 Report

Quoted By: >>11441844

$(1-z)\sum_{k=0}^{n-1} z^k = (1-z)\cdot (1+z+z^2+z^3+\cdot z^{n-1}) = 1-z^n$

So for all z not 1,

$\sum_{k=0}^{n-1} z^k = \dfrac{1-z^n}{1-z}$

For for all z with |z|<1,

$\sum_{k=0}^\infty z^k = \dfrac{1}{1-z}$

This is just the geometric series.

For for d in the interval (0, 2) and letting z=-1+d

$\sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2-d}$

So:

$\lim_{d\to 0} \sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2}$

Let me make a point that this isn't to claim:

$1-1+1-1+1-1+ ... = \dfrac{1}{2}$

But the algebraic properties of the sum relate "1-1+1-1+1-1+" to "1/2" in many context and that's also the case with 1+4+9+16+25 and 1+2+3+4+... It's not random, and it's not real or complex analysis either.

Going on form that example..
The analytic continuation F of a function f is, roughly, a complex differentiable function and which, where f is defined, has the functions agreeing F(z)=f(z). Complex differentiation is a more strict condition than multivariate differentiability (complex differentiable means the Jacobian has a certain form).

Long story short, $F(z)=\dfrac{1}{1-z}$ is a continuation of $f(z) = \sum_{k=0}^\infty z^k$ and the continuation of the function

$\sum_{n=1}^\infty\, n^{-z}$

happens to fulfill

$F(z) = 2\,(2\pi)^{z-1}\sin{\left(\pi\,z/2\right)}\,\Gamma(1-z)\, F(1-z)$ .
(F is zeta, here)

From the latter formula (and knowledge of the positive vaues), you can read off the "spooky" equations.
But again, it's not an equality of analysis in the same way that proving

$\lim_{d\to 0} \sum_{k=0}^\infty (-1+d)^k = \dfrac{1}{2}$

isn't proving

$1-1+1-1+1-1+ ... = \dfrac{1}{2}$

Anonymous

Anonymous Wed 04 Mar 2020 20:37:11 No.11440135 Report

Quoted By: >>11441844

Although, for what it's worth, we see that

$1-1+1-1+1-1+1-1+ \lim_{d\to 0} \sum_{k=2m+1}^\infty (-1+d)^k = \dfrac{1}{2}$

for any $m$ . For what it's worth. Of course, the first 2m terms always are just zero.

Anonymous

Anonymous Wed 04 Mar 2020 20:43:24 No.11440149 Report

Quoted By: >>11441844

Also, the d limit is a one-sided limit, but I'm not sure how to make the LaTeX

Anonymous

Anonymous Thu 05 Mar 2020 12:58:28 No.11441844 Report

Quoted By: >>11442590 >>11442605

>>11440131
>>11440135
>>11440149
Translate into English? I dont speak symbol math

Anonymous

Anonymous Thu 05 Mar 2020 13:01:59 No.11441849 Report

Quoted By: >>11441852

>>11439850
>On what planet is that true?
Not in real analysis.

The equality shouldn't be taken literally (or equivalent to the one used elsewhere in analysis).

Anonymous

Anonymous Thu 05 Mar 2020 13:03:13 No.11441852 Report

Quoted By: >>11442208 >>11442616

Then why is it important? >>11441849

Anonymous

Anonymous Thu 05 Mar 2020 16:31:14 No.11442208 Report

Quoted By:

>>11441852
If it's important, then I guess that implies the continuation is correct, and the zeta function (as OP wrote it) is broken.

Anonymous

Anonymous Thu 05 Mar 2020 18:52:00 No.11442590 Report

Quoted By:

>>11441844
What do you not understand?

Consider this:
If z is a real number between -1 and 1, then the sum over all numbers of the form z^n equals 1/(1-z). For example, 1+1/2+1/4+1/9+1/16+.. equals 1/(1-1/2)=2.

The function defined by 1/(1-z) agrees everywhere with that sum over all z^n, under the condition that z is between -1 and 1.
Moreover, the function 1/(1-z) also has a return value for z=9001, namely -1/9000.
On the other hand, the sum over all terms of the form 9000^n does not converge in real analysis.

Suary so far: Where the sum over z^n is defined, the function defined by 1/(1-z) agrees with it. But the function 1/(1-z) also has values where that sum doesn't.

The function 1/(1-z) continues the sum over z^n's in the above sense.

Anonymous

Anonymous Thu 05 Mar 2020 18:54:13 No.11442605 Report

Quoted By:

>>11441844
1-1+1.... = 1/2

Anonymous

Anonymous Thu 05 Mar 2020 18:56:00 No.11442616 Report

Quoted By:

>>11441852
"Importance" isn't well defined. I can't answer that question.

Anonymous

Anonymous Thu 05 Mar 2020 19:29:56 No.11442777 Report

Quoted By:

>>11439850
>the absolute state of /sci/

s Thu 05 Mar 2020 19:38:09 No.11442810 Report

Quoted By:

>>11439883
>real math thread
threads about the zeta function or ramanujan summation are always the worst. they solely get posted by schizos or idiots with a highschool level understanding of math.

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