/sci/ - Science & Math » Thread #10839504

1MiB, 1440x808, 1404236448233.png

View Same Google iqdb SauceNAO

Anonymous Fri 26 Jul 22:42:38 2019 No.10839504 View Reply Original Report

Quoted By: >>10839520 >>10839555 >>10840422 >>10842485 >>10842487 >>10842505 >>10842508 >>10842511 >>10842521 >>10842526 >>10842640

Does the sum of all numbers really equals -1/12? How is it possible?

Anonymous

Anonymous Fri 26 Jul 2019 22:44:39 No.10839512 Report

Quoted By: >>10842505

Anonymous

Anonymous Fri 26 Jul 2019 22:46:36 No.10839520 Report

Quoted By: >>10842253 >>10842505

>>10839504
No, where did this misunderstanding come from?

Anonymous

Anonymous Fri 26 Jul 2019 22:57:19 No.10839552 Report

Quoted By:

If $\sum_{n=0}^\infty$ is the notation for the sum in analysis, then $\sum_{n=0}^\infty n$ is divergent.

It can be pointer out that Peano arithmetic doesn't enable you to compute infinite sums, so e.g. to prove things like $\sum_{1=0}^\infty \frac{1}{n} = 2$ , you need uncountable sets such as ${\mathbb R}$ and a metric on it - i.e. I huge ballast of mathematical gadgetry.

There are other theories of infinite sums where the result are true, e.g. the Ramanujan sum. In the theory of (different) infinity sums, you can define properties your notion of sum should fulfill, and the sum over n is one that it's particularly hard to give finite meaning to. In comparison, for example, the sum over (-1)^n can be given a value in many theories of infinity sums.
Not all theories giving a value to n are algebraic in that sense, e.g. analytic continuation suggests a value for it too.
But again, the theory with most applications - analysis - doesn't assign a value to the sum over n.

-1/12 pops up in varous related forms also in analysis, e.g. Lie theory and regulatizations.

Anonymous

Anonymous Fri 26 Jul 2019 22:57:49 No.10839553 Report

Quoted By: >>10842659

If this is cause of numberphile don't believe them. They're known for bs info and clickbait. There's a german dude on youtube who shows it's bs.

Anonymous

Anonymous Fri 26 Jul 2019 22:58:53 No.10839555 Report

Quoted By: >>10840475 >>10842505

>>10839504
NO. It's just a bunch of algebraic hocus pocus

Anonymous

View Same Google iqdb SauceNAO d.png, 73KiB, 893x352

Anonymous Fri 26 Jul 2019 23:00:05 No.10839557 Report

Quoted By:

If $\sum_{n=0}^\infty$ is the notation for the sum in analysis, then $\sum_{n=0}^\infty n$ is divergent.

It can be pointer out that Peano arithmetic doesn't enable you to compute infinite sums, so e.g. to prove things like $\sum_{1=0}^\infty \frac{1}{2^n} = 2$ or $\sum_{1=0}^\infty \frac{1}{n^2} = \frac{1}{6} \pi^2$ , you need uncountable sets such as ${\mathbb R}$ and a metric on it - i.e. I huge ballast of mathematical gadgetry.

There are other theories of infinite sums where the result are true, e.g. the Ramanujan sum. In the theory of (different) infinity sums, you can define properties your notion of sum should fulfill, and the sum over n is one that it's particularly hard to give finite meaning to. In comparison, for example, the sum over (-1)^n can be given a value in many theories of infinity sums.
Not all theories giving a value to n are algebraic in that sense, e.g. analytic continuation suggests a value for it too.
But again, the theory with most applications - analysis - doesn't assign a value to the sum over n.

-1/12 pops up in various related forms also in analysis, e.g. Lie theory and regularizations. The source is most often the second term in pic related.

Anonymous

Anonymous Sat 27 Jul 2019 06:56:15 No.10840422 Report

Quoted By: >>10841759

>>10839504
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 Sat 27 Jul 2019 07:32:14 No.10840475 Report

Quoted By:

>>10839555
Is it truth?

Anonymous

Anonymous Sat 27 Jul 2019 19:48:46 No.10841759 Report

Quoted By: >>10842534 >>10842612

>>10840422
You have kept repeating this for years now but you never pointed to that smoothed function that -1/12 is supposedly the limit to 0 of.

Anonymous

Anonymous Sat 27 Jul 2019 22:12:42 No.10842060 Report

Quoted By: >>10842520

theres a funky sum for complex numbers ? and you can find the formula, actually its an infinite series so its a sum to infinity
anyway you put complex numbers and viola you get some thing
anyway if you put some numbers you get like normal numbers for example if you put -1 you basipalli get reciptrotic value of number

funky sum(x) = 1/1^x + 1/2^x ..
so put x = -1 and you reverse up and down numbers to get
1 + 2 + ...
BUT
the problem is the funky sum doesnt defined for negative numbers, infact not for 0 either
but you can extend the graph of the sum on a plane but you can do it in many ways but the only aprooved one is the one that retains some stuff and that way is to flip it
and it turns out that when you map the flipped extended graph of funky sum, on x=-1 you get y = -1/12

Anonymous

Anonymous Sat 27 Jul 2019 23:55:04 No.10842253 Report

Quoted By:

>>10839520
probably something like this
https://www.youtube.com/watch?v=kIq5CZlg8Rg

Anonymous

Anonymous Sun 28 Jul 2019 01:43:02 No.10842485 Report

Quoted By:

>>10839504
Yes, but you have to shit in the street between each digit.

Anonymous

Anonymous Sun 28 Jul 2019 01:43:58 No.10842487 Report

Quoted By:

>>10839504
what?

Anonymous

Anonymous Sun 28 Jul 2019 01:50:09 No.10842505 Report

Quoted By: >>10842508 >>10845223 >>10845280

>>10839504
Yes it does, see https://math.stackexchange.com/questions/1327812/limit-approach-to-finding-1234-ldots for one example
>>10839512
>>10839520
>>10839555
>t. undergrads
Yes it does, retards

Anonymous

Anonymous Sun 28 Jul 2019 01:51:11 No.10842508 Report

Quoted By:

>>10839504
>>10842505
sum of all positive integers*

Anonymous

Anonymous Sun 28 Jul 2019 01:51:49 No.10842511 Report

Quoted By:

>>10839504
these are really smart guys that came up with this anon. don't try to question them, when you are a simple everyday citizen. the magnets repel each other. don't ask why

Anonymous

Anonymous Sun 28 Jul 2019 01:57:32 No.10842520 Report

Quoted By:

>>10842060
The so-called “funky sum” is defined, you’re just extending the definition of summation in a natural way, much like we come up with natural extensions of addition, multiplication, exponentiation, etc. of naturals, integers, rationals, reals, complex numbers, vector spaces, functions, etc.

Anonymous

Anonymous Sun 28 Jul 2019 01:57:45 No.10842521 Report

Quoted By:

>>10839504
Fuzzy maths tricks, which I here from this day dub "Methematics"

Anonymous

View Same Google iqdb SauceNAO 8F871E99-AA54-483E-B9D0-6B01D7C0 (...).png, 4KiB, 295x210

Anonymous Sun 28 Jul 2019 02:01:21 No.10842526 Report

Quoted By: >>10842535

>>10839504
What is true of the partial sums is not always true of the total sum. See pic related for example, where the partial sums are all rational but the total sum is irrational.

Anonymous

Anonymous Sun 28 Jul 2019 02:03:52 No.10842534 Report

Quoted By: >>10843419

>>10841759
not him but here you go
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Cutoff_regularization

Anonymous

Anonymous Sun 28 Jul 2019 02:05:18 No.10842535 Report

Quoted By: >>10842578 >>10842583

>>10842526
true, but this has no bearing on the question

Anonymous

Anonymous Sun 28 Jul 2019 02:35:32 No.10842578 Report

Quoted By: >>10842583 >>10842608

>>10842535
It has bearing because people frequently use the erroneous argument that such a sum cannot be negative because the partial sums are positive.

Anonymous

Anonymous Sun 28 Jul 2019 02:39:27 No.10842583 Report

Quoted By:

>>10842535
>>10842578
The correct statement would be: IF the ordinary limit of partial sums exists AND the partial sums are positive, then the total sum must be positive.

Anonymous

Anonymous Sun 28 Jul 2019 02:49:45 No.10842608 Report

Quoted By:

>>10842578
except that statement is true, the set of nonnegative reals is closed in the metric topology, therefore the limit of any such sequence in that set is contained in that set. in particular, if the partial sums of a series are positive, then the infinite sum is nonnegative or is not defined.
your rational sum counterexample works because the rationals are not a closed subset of the reals.

Anonymous

Anonymous Sun 28 Jul 2019 02:50:47 No.10842612 Report

Quoted By: >>10843419

>>10841759
lrn2read

Anonymous

Anonymous Sun 28 Jul 2019 03:12:17 No.10842640 Report

Quoted By:

>>10839504
It is infinity

Anonymous

Anonymous Sun 28 Jul 2019 03:20:17 No.10842651 Report

Quoted By:

no, the sum is divergent.
there is a sense in which one can say this, in particular, the riemann zeta function is defined for all complex inputs whose real part is greater than 1. analytic continuation lets us "finish off" the rest of the function in other areas of the complex plane in a single, unique way. it happens that the analytic continuation of the riemann zeta function sends -1 to -1/12. if you pop -1 into the typical riemann zeta function as it is defined to the right of Re z = 1, then it looks like you're saying 1 + 2 + 3 + 4 + ... = -1/12. but you're not actually doing that anymore, instead you're popping -1 into a function which works like an infinite sum on some parts of the complex plane, and like any other function on other parts of the plane.
>then how can you rearrange the sum to get this value!!!
by assuming something nonsensical about the sum 1 - 1 + 1 - ... and letting that coincidentally take you to the value of the riemann zeta function at -1.

Anonymous

Anonymous Sun 28 Jul 2019 03:25:49 No.10842659 Report

Quoted By:

>>10839553
Mathologer, that guy is awesome

Anonymous

Anonymous Sun 28 Jul 2019 12:27:06 No.10843419 Report

Quoted By: >>10843528 >>10845252

>>10842534
>>10842612
What's the function?

Yes, it must be of the form $-1/12 + O(n)$ but what is it?

Anonymous

Anonymous Sun 28 Jul 2019 13:45:58 No.10843528 Report

Quoted By:

>>10843419
google it

Anonymous

View Same Google iqdb SauceNAO shocking.jpg, 4KiB, 225x160

Anonymous Mon 29 Jul 2019 01:10:47 No.10845223 Report

Quoted By:

>>10842505
> negative result with only positive addends

Anonymous

View Same Google iqdb SauceNAO 1513373994755.png, 34KiB, 408x450

Anonymous Mon 29 Jul 2019 01:20:47 No.10845252 Report

Quoted By: >>10845334

>>10843419
did you actually read that link
>"The constant term of the asymptotic expansion does not depend on f: it is necessarily the same value given by analytic continuation, ? 1/12.[1]"
>click on the cited article [1]
>"However, these issues can be resolved by replacing the abruptly truncated partial sums $\sum_{n=1}^N n^s$ with smoothed sums $\sum_{n=1}^\infty \eta(n/N) n^s$ , where $\eta: {\bf R}^+ \rightarrow {\bf R}$ is a cutoff function, or more precisely a compactly supported bounded function that equals 1 at 0"
>look at equation 12: $\sum_{n=1}^\infty n \eta(n/N) = -\frac{1}{12} + C_{\eta,1} N^2 + O(\frac{1}{N})$ with eta as above

wow that was hard

Anonymous

View Same Google iqdb SauceNAO b560496b85e4f6ae8b87437295964d09.jpg, 120KiB, 599x800

Anonymous Mon 29 Jul 2019 01:31:17 No.10845280 Report

Quoted By: >>10845368

>>10842505
>t. pretending to not be an undergrad when he clearly is
?n is a divergent sum no matter how you look at it. It does not equal anything. The -1/12 result comes from analytic continuation, which is something entirely different from the sum "equalling" -1/12.

Anonymous

Anonymous Mon 29 Jul 2019 01:58:28 No.10845334 Report

Quoted By: >>10845367

>>10845252
What is the big E doing?

Anonymous

Anonymous Mon 29 Jul 2019 02:09:13 No.10845367 Report

Quoted By:

>>10845334
just read the blog post, he spells it all out in excruciating detail

Anonymous

Anonymous Mon 29 Jul 2019 02:09:32 No.10845368 Report

Quoted By:

>>10845280
this. i mean even as an undergraduate you acquire awareness of this in calculus II

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches