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1/12

Anonymous Tue 22 Sep 07:13:00 2020 No.12148108 View Reply Original Report

Quoted By: >>12148174 >>12149799

is it real, or meme?

seems real to me:
...4-3-2-1-0 = 1/12

Anonymous

Anonymous Tue 22 Sep 2020 07:45:00 No.12148174 Report

Quoted By: >>12148184

>>12148108
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 Tue 22 Sep 2020 07:53:27 No.12148184 Report

Quoted By: >>12148190

>>12148174
did you reply to the wrong thread?

what's a "ramanujan summation" stuff? I'm confused. also, what's a "zeta regularization", "smoothed curve" ? you're confusing me.

In the pic, I'm just using a summation series, an analytic function of that series, and then an integral of that function from 0 to 1. and I get 1/12. Are you saying there are other ways to get 1/12?

I don't see where x is infinity in the pic, either.

Anonymous

Anonymous Tue 22 Sep 2020 07:55:40 No.12148190 Report

Quoted By: >>12148199

>>12148184
>you're confusing me
shocker

Anonymous

Anonymous Tue 22 Sep 2020 08:00:09 No.12148199 Report

Quoted By:

>>12148190
Sorry.
I'm not an expert in mathematics. I only know basic stuff like integrals so that's all I used to get 1/12.
In my pic, I only went from x=-4 to x=1.5
I didn't use x = infinity.
You sound very smart though because you said all that stuff I didn't use.

Anonymous

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Anonymous Tue 22 Sep 2020 19:42:40 No.12149610 Report

Quoted By: >>12149738

Sorry for phone posting. What you are doing generalizes. See pic related. Zeta(-3)=-1/120

Anonymous

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Anonymous Tue 22 Sep 2020 20:28:58 No.12149738 Report

Quoted By:

>>12149610
very interesting.
I honestly find this fascinating. It's like we live in a tiny fragment of multiple infinities that somehow doesn't cancel out.

Anonymous

Anonymous Tue 22 Sep 2020 20:52:29 No.12149799 Report

Quoted By:

>>12148108
1.Take the gauss formula for sum of the first n integers: (n(n+1))/2
2. find for which values the formula equals 0. (hint: 0 and -1).
3. Compute the definite integral for the formula between -1 and 0.
Repeat for all negative odd values of the zeta function.

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