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Anonymous Wed 27 Oct 15:46:51 2021 No.13795141 View Reply Original Report

Quoted By: >>13795150 >>13795159 >>13795163 >>13795186 >>13795230 >>13795300 >>13795390 >>13795603 >>13795646

Can someone explain why this is true?

Anonymous

Anonymous Wed 27 Oct 2021 15:48:20 No.13795145 Report

Quoted By: >>13795230

It's not

Anonymous

Anonymous Wed 27 Oct 2021 15:50:07 No.13795150 Report

Quoted By:

>>13795141
it's not, just some mental juggling

Anonymous

Anonymous Wed 27 Oct 2021 15:53:51 No.13795159 Report

Quoted By:

>>13795141
It's more-less a quark of applying certain logic to a situation, where it by definition, isn't applicable.

Anonymous

Anonymous Wed 27 Oct 2021 15:55:30 No.13795163 Report

Quoted By:

>>13795141
Where is the starting point for your sigma.

Anonymous

Anonymous Wed 27 Oct 2021 16:05:36 No.13795186 Report

Quoted By:

>>13795141
Ramanujan was "dilating" when he wrote that letter to Hardy. He says so right in the text as a preface to his description of that summation. "I dilate on this"

Anonymous

Anonymous Wed 27 Oct 2021 16:27:43 No.13795230 Report

Quoted By:

>>13795141
>>13795145
You are confusing analytic continuation and the original operator. It's like confusing the Gamma function and the original factorial definition. They aren't the same and it's meaningless to say it is in a formula like that (although there is a lot of good reasons, much better than this -1/12 crap to accept the Gamma function as the canonical extension of the factorial)

Anonymous

Anonymous Wed 27 Oct 2021 17:01:47 No.13795300 Report

Quoted By:

>>13795141
Presenting this without context is like asking for a raisin and someone giving you a grape because le raisin means grape in French.

Anonymous

Anonymous Wed 27 Oct 2021 17:39:25 No.13795390 Report

Quoted By:

>>13795141
watch the 3blue1brown video on the Riemann zeta function and he explains this visually, the TLDW is basically that at -1 the function becomes the sum of all natural numbers but the graph isn't defined there, through some fuckery you can extend the parts of the graph that are defined into the undefined domain and at -1 this extension is equal to -1/12

susssy imposter baka

susssy imposter baka Wed 27 Oct 2021 18:15:35 No.13795476 Report

Quoted By:

fuck math

Anonymous

Anonymous Wed 27 Oct 2021 19:06:37 No.13795603 Report

Quoted By:

>>13795141
https://www.youtube.com/watch?v=YuIIjLr6vUA

Anonymous

Anonymous Wed 27 Oct 2021 19:22:43 No.13795646 Report

Quoted By:

>>13795141
The closest you get in real analysis is

$I(f):=\int_0^\infty f\, {\rm d}n$

$\sum_{n=0}^\infty q^n - I(q^n) = \dfrac{1}{1-q}+\dfrac{1}{\log(q)} = \dfrac{1}{2} - \dfrac{1}{12} (q-1) + {\mathcal O}((q-1)^2)$ for $q<1$

$\implies$

$\lim_{q\to 1}\big(\sum_{n=0}^\infty n \, q^n - I(n\,q^n)\big) = - \dfrac{1}{12}$

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