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"Dirchlet regularization"

Anonymous Mon 11 May 06:06:27 2020 No.11658939 View Reply Original Report

Quoted By: >>11658948 >>11658950 >>11658951 >>11659038 >>11659087

NEGATIVE 1/12. NEGATIVE 1/12. NEGATIVE FUCKING 1/12. WHAT THE FUCK IS THIS SHIT?

Anonymous

Anonymous Mon 11 May 2020 06:07:48 No.11658940 Report

Quoted By:

it’s infinity

Anonymous

Anonymous Mon 11 May 2020 06:11:20 No.11658948 Report

Quoted By: >>11660145

>>11658939
It's easy to confuse yourself with this shit but it's quite simple.

All that the ramanujan summation stuff, cutoff and zeta regularization does, is look at the smoothed curve at x = 0.
What sums usually do is look at the value as x->inf.

It's just a unique value you can assign to a sum, really they have many such values.

https://en.wikipedia.org/wiki/File:Sum1234Summary.svg

$\displaystyle<br />\zeta \neq \Sigma<br />$

https://youtu.be/sD0NjbwqlYw?t=10m

Anonymous

Anonymous Mon 11 May 2020 06:13:11 No.11658950 Report

Quoted By:

>>11658939

this is your brain on hinduism

Anonymous

Anonymous Mon 11 May 2020 06:14:10 No.11658951 Report

Quoted By:

>>11658939
See
https://math.stackexchange.com/questions/3664350/ramanujan-summations-1-12-is-not-an-element-of-the-group-of-all-positive-integ

https://math.stackexchange.com/questions/3065870/ramanujan-summation-series

Anonymous

Anonymous Mon 11 May 2020 07:08:19 No.11659038 Report

Quoted By: >>11659098

>>11658939
it's not the sum of the series
once again, it's NOT the fucking sum of the series
it's a number assigned to the series based on a more general pattern, which would agree with the sum IF the series was summable
but this series is divergent, and so it's NOT the sum. just a number assigned to it.

Anonymous

Anonymous Mon 11 May 2020 07:30:44 No.11659087 Report

Quoted By:

>>11658939
That sum "equals" -1/12 in the sense that sums that truly do converge (eg 1/n^2) equal the Zeta function, and there exists only one possible "analytic continuation" of that function for sums that don't truly converge

Anonymous

Anonymous Mon 11 May 2020 07:35:59 No.11659092 Report

Quoted By:

where does this meme come from, this is retarded

Anonymous

Anonymous Mon 11 May 2020 07:37:47 No.11659098 Report

Quoted By: >>11659102

>>11659038
It IS the sum of the series you fucking dunce.

Anonymous

Anonymous Mon 11 May 2020 07:38:48 No.11659102 Report

Quoted By: >>11660129

>>11659098
do you know how the sum of a series is defined ?

Anonymous

Anonymous Mon 11 May 2020 17:12:03 No.11660129 Report

Quoted By: >>11660216

>>11659102
There are many different definitions. Who are you to say which one is "the" definition?

Anonymous

Anonymous Mon 11 May 2020 17:15:24 No.11660145 Report

Quoted By:

>>11658948
wtf is that sign?
A snake?

Anonymous

Anonymous Mon 11 May 2020 17:33:59 No.11660216 Report

Quoted By: >>11660435

>>11660129
that would be true if we were talking about "a sum". "the sum" means one thing and one thing only.

Anonymous

Anonymous Mon 11 May 2020 18:43:11 No.11660435 Report

Quoted By: >>11660562

>>11660216
>"the sum" means one thing and one thing only.
And what would that be? You mean the infinite sum is the limit of partial sums if an only if it converges w.r.t. the metric over ${\mathbb R}$ ? That's the most useful sum, but it has caveats like relying on ${\mathbb R}$ .
It's noteworthy that Peano arithmetic is not a language that's capable of even expressing infinite sums.

Anonymous

Anonymous Mon 11 May 2020 19:27:21 No.11660562 Report

Quoted By: >>11661085

>>11660435
>You mean the infinite sum is the limit of partial sums
yes

Anonymous

Anonymous Mon 11 May 2020 21:49:59 No.11661085 Report

Quoted By:

>>11660562
1. The Cauchy limit of partial sums is just ONE definition of summation that is SUBSUMED by more general definitions/strict extensions.
2. The P-ADIC limit of the partial sums of 1+2+4+8+16+... is -1.
Take that you fucking idiot. You haven't even specified which limit you're working with.

Anonymous

Anonymous Mon 11 May 2020 22:04:16 No.11661121 Report

Quoted By:

It's misleading notation but based on something real: the analytic continuation of the Riemann zeta function $\zeta(s)$

Anonymous

Anonymous Mon 11 May 2020 22:12:38 No.11661150 Report

Quoted By:

>there are actual people in this thread who think OP doesn't understand that the value is obtained by analytically continuing the Dirichlet series when he specifically makes sure to call it Dirichlet regularization
Y'all niggas genuinely need to reconsider your lives.

Anonymous

Anonymous Mon 11 May 2020 22:49:58 No.11661253 Report

Quoted By:

people never cry about the geometric series and start going 1 + 2 + 4 + 8.... = -1, why does this much more obscure sum generate so much controversy?

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