/sci/ - Science & Math

[Anonymous No.7509864]

Hello /sci/!

I frequently find debate about the proof of how the sum of all natural integers, equals to -1/12.
Since discussions about said topic are usually completely and utterly retarded and based on emotions, I came to you.

So? Is this something that's actually valid or just a meme formula?

[Anonymous No.7509873]

When an equation results in a contradition, it means at least one of the assumptions was invalid.

The assumption that is invalid in this case probably has to do with the way infinity/infinite series work.

[Anonymous No.7509874]

>>7509864
You could just google sum of all naturals and read the wikipedia page.

[Anonymous No.7509875]

>>7509864
It's valid, but its not a sum in the sense that you'd normally understand it, its been regulated, which is a rigorous way of assigning a value to a sum that would otherwise diverge. Google regularisation and analytic continuation.

[Anonymous No.7509881]

>>7509875

not OP, but what you are describing is what I'd to to kind of sidestep the actual problem and see if it holds any kind of merit

[Anonymous No.7509894]

>>7509881
I don't get what your saying, that is what the "sum" is. Regulators are used all the time in QFT.

[Anonymous No.7509899]

>>7509894

I'll have to look into it more. I'm a math pleb, not much of a physicist. I think it's pretty intuitive personally.

[Anonymous No.7509900]

>>7509881
There is no sidestepping, the problem has been solved, the series is divergent. Now can be assign any useful value to it? Yes, -1/12

[Anonymous No.7509906]

>>7509864
When we write:

<span class="math"> \sum_{n=1}^{\infty} a_n = L [/spoiler]

This is really shorthand for

<span class="math">\lim_{h\to\infty} \sum_{n=1}^{h} a_n = L
[/spoiler]

That is, an infinite sum is really the limit of the partial sums. What the partial sums approach when you use an arbitrarily large number of terms. In this sense, the limit is not equal to -1/12, and clearly diverges.

However, there is another way to define what
<span class="math"> \sum_{n=1}^{\infty} a_n = L [/spoiler]
should mean. This is done via zeta function regularization. In order to really understand what is going on here, you'll need to study math higher mathematics than just Calculus. But suffice it to say, using such a process on the natural numbers results in a "sum", that is, a regularized sum, of -1/12.

[Anonymous No.7509920]

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

All of the ramanujan summation stuff, cutoff and zeta regularization do is look at the smoothed curve at x = 0. What sums usually do is to 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.

[Anonymous No.7509927]

>>7509906
Here's a more ad hoc way of smoothing than analytical continuation, that leads to the value.

http://www.wolframalpha.com/input/?i=Limit[z%2F%281-z%29^2-1%2FLog[z]^2%2Cz-%3E1]

>>7509864
Purely classically,

<span class="math"> \sum_{k=0}^\infty k [/spoiler]

is the same as the limit of z to 1 of

<span class="math"> \sum_{k=0}^\infty k z^k [/spoiler]

namely undefined (or infinity).

On the other hand, for any |z|<1, you have the expansion

<span class="math"> \sum_{k=0}^\infty k z^k = \frac{z}{(z-1)^2} [/spoiler]

http://www.wolframalpha.com/input/?i=Sum[k+z^k%2C{k%2C0%2CInfinity}]

Now if you define a forward average <span class="math"> \langle f(k) \rangle [/spoiler] of a function via

<span class="math"> \langle f(k) \rangle := \int_{k}^{k+1} f(k') \, dk' [/spoiler]

you might consider

<span class="math"> \sum_{k=0}^\infty ( k z^k - \langle k z^k \rangle ) = \frac {z} {(z-1)^2} - \frac {1} {\log(z)^2} [/spoiler]

Both functions <span class="math"> \frac {z} {(z-1)^2} [/spoiler] and <span class="math"> \frac {1} {\log(z)^2} [/spoiler] are singular at z=1 themselves, but the later is actually growing faster by a slight amount:

Purely classically, -1/12 is only the limit as z approaches 1 value of this,
i.e. the sum, but only "regularized" by subtracting the smooth surroundings.
The limit is in the first link, but you can actually draw the sharply rising graphs of the two and see how the difference close to z=1 remains -1/12.

[Anonymous No.7509929]

By the way zeta regularization has been used on the sum
<div class="math">\sum_{n=1}^\infty n^3 = \frac{1}{120}</div>
in the derivation of the casimir effect, which has been experimentally verified.

[Anonymous No.7509938]

Welp, my concept of magnitude is thoroughly fucked for the day

[Anonymous No.7509951]

>>7509929
You can (classically) compute those numbers (or any other stemming from k^n) using the average I've stated in >>7509927.

[OHP No.7510033]

>>7509899
You think <span class="math"> \Sum_{n=1}^{\infty} n = -\frac{1}{12} [/spoiler] is intuitive somehow, without considering analytic continuation of <span class="math"> \zeta [/spoiler]? Please, do tell...

[Anonymous No.7510047]

>>7509864
Clearly meme bullshit. Mathematics has lost it's way. Nothing new to discover so it just makes up shit now.

[Anonymous No.7510050]

>I frequently find debate about the proof of how the sum of all natural integers, equals to -1/12.
>So? Is this something that's actually valid or just a meme formula?
How can you even debate about something if you don't know it to be true?

[Anonymous No.7510053]

>>7510033

I think it's fairly intuitive to ascribe values to divergent sums in order to talk about them in different ways and compare them.

Clearly, I am not talking about the specific mechanics of analytic continuation, but it seems like a pretty neat concept.

>inb4 deconstructive post on how much mathematics I don't know yet

[Anonymous No.7510060]

>>7509929
Hmm ok you've got me listening now

[Anonymous No.7510064]

>>7509951
>apple
klelk

[Anonymous No.7510087]

>>7510064
>actually caring what type of computer someone uses
kek

[Anonymous No.7510095]

It's pretty much a meme formula. It comes from misception of Riemann zeta function.

[OHP No.7510096]

>>7510053
No no, that's cool. I just wanted to make sure you weren't being haughty by saying, "yeah, 1+2+3+... is obviously -1/12... you guys are pleb for thinking analytic continuation is necessary to see it!"

[Anonymous No.7510099]

>>7510087
Shut up applefag, Windows is the best. Apple is just a paid Linux, where with Windows you can do everything and more.

[Anonymous No.7510133]

>>7510096
We can try to collect ingredients.

I think <span class="math"> \frac {1} {12} [/spoiler] should always be read as <span class="math"> \frac {1} {1!} \frac {1} {2!} \frac {1} {3!} [/spoiler] here and then we want to motivate the factors in the Euler MacLaurin series.

For

<span class="math"> f(z) = \sum_{k=0}^\infty a_k z^k [/spoiler]

the approximation over real deal expansion

<span class="math"> \frac {f(0)} {f(f(0)\,z)} = \frac{1} { 1 + \sum_{k=0}^\infty \ frac{a_k} {a_0} (a_0\,z)^k} = 1-a_1\,z + (a_1\, a_1-a_0\, a_2)\, z^2- (a_1\, a_1\, a_1 - 2\, a_0\,a_1\,a_2+a_0\,a_0\,a_3)\,z^3 + O(z^3). [/spoiler]

carries a combinatorical pattern.
If the "real deal" is something smooth, you get exponentials (like with exponential flows vs. finite steps, or Laplace transformations vs. discrete sequences generating functions). In particular, <span class="math"> a_{k-1}=\frac {1} {k!} [/spoiler] gives

<span class="math"> a_1=\frac {1} {2!} [/spoiler]

<span class="math"> a_1\, a_1 - a_0 \, a_2 = \frac {1} {2!} \frac {1} {2!} - \frac {1} {1!} \frac {1} {3!} = \frac {1} {1!} \frac {1} {2!} \frac {1} {3!} (3-2) = \frac {1} {12} [/spoiler]

Consider the finite difference <span class="math"> \Delta_h f(x) [/spoiler] of a function f about x, when shifted by h. With the Taylor expansion

<span class="math"> \Delta_h f(x) = f(x+h) - f(x) = f'(x) \, h+ \sum_{k=2}^\infty \frac{1}{k!} f^{(k)}(x)\,h^k [/spoiler]

we can compute the correction factor, by which the first order approximation fails to capture the finite difference:

<span class="math"> \frac {f'(x)\,h} {\Delta_h f(x)} = 1 - \frac {f''(x)} {2!} (\frac {h} {f'(x)}) + (\frac {f''(x)\,f''(x)} {2!\,2!} - \frac {f'(x)\,f'''(x)} {1!\,3!}) (\frac{h}{f'(x)})^2 + O(h^3). [/spoiler]

so for example

http://www.wolframalpha.com/input/?i=Series[1%2F%28Exp[h]-1%29%2C{h%2C0%2C2}]

The Ramanujan terms then come up in

<span class="math"> \int_a^b f(n)\, dn = \sum_{n=a}^{b-1} f(n)+ (lim_{x\to b} - lim_{x\to a}) ( \frac {1} {2} - \frac {1} {12} \frac{d}{dx} + … ) f(x) [/spoiler]

and for f(x)=x, so that the sum is over 1, 2, 3, 4, 5, the second term sees exactly the first derivative with the -1/12

[Anonymous No.7510150]

>>7510133
E.g. the formula gives us identities such as

<span class="math"> \int_2^4 n^2 dn=(2^2+3^2)+ \frac {1} {2} (4^2-2^2) - \frac{1} {12} 2 (4^1-2^1) [/spoiler]

or for finite sums of 1,2,3,4,5 you find

<span class="math"> \sum_{n=0}^{b-1} n = (\frac {b^2} {2} - \frac{b} {2} + \frac{1} {12} ) - ( 0 - 0 + \frac{1} {12} )=\dfrac{b(b-1)}{2} [/spoiler]

Ramanujan summation up to infinity is literally droppin one of the two brackets, leaving -1/12 in that case.

and if one does this integration vs. sum game often enough, the factor would appear natural.

The finite "sum n^m" formulas more concisely (where the 2nd Bernoulli number 1/3! brings about the 1/12) and the relation to the zeta function is touched upon here
https://en.wikipedia.org/wiki/Faulhaber's_formula#Relationship_to_Riemann_Zeta_Function

[Anonymous No.7510153]

>>7510133
E.g. the formula gives us identities such as

<span class="math"> \int_2^4 n^2 dn=(2^2+3^2)+ \frac {1} {2} (4^2-2^2) - \frac{1} {12} 2 (4^1-2^1) [/spoiler]

or for finite sums of 1,2,3,4,5 you find

<span class="math"> \sum_{n=0}^{b-1} n = (\frac {b^2} {2} - \frac{b} {2} + \frac{1} {12} ) - ( 0 - 0 + \frac{1} {12} )=\frac {b \, (b-1)} {2} [/spoiler]

Ramanujan summation up to infinity is literally droppin one of the two brackets, leaving -1/12 in that case.
And if one does this integration vs. sum game often enough, the factor would appear natural.

The finite "sum n^m" formulas more concisely (where the 2nd Bernoulli number 1/3! brings about the 1/12) and the relation to the zeta function is touched upon here
https://en.wikipedia.org/wiki/Faulhaber's_formula#Relationship_to_Riemann_Zeta_Function

[Anonymous No.7510491]

>>7509920
/thread

[Anonymous No.7510504]

>>7509864
It doesn't intuitively make sense because it's nothing more than a theory, and doesn't apply to reality. It only equals -1/12 if you go to infinity, and infinity is not applicable to the real world.
Though it applies to areas of theoretical physics, theoretical physics is still simply a theory, hence the name.
Infinity isn't real.

[Anonymous No.7510521]

>>7510504
>infinity is not applicable to the real world

sure, since achilles *never* catches up with the tortoise

[Anonymous No.7510620]

>>7509864
>Since discussions about said topic are usually completely and utterly retarded and based on emotions, I came to you.
you came to the wrong place motherfucker

[Uhhhhhhh no stop it with the s(...) No.7510661]

IF that series actually did converge, then depending on the CONTEXT of which the series is in its answer would be -1/12. If i remember correctly, there are certain conexts (methods of assigning a value to divergent series) which the sum can not be assigned -1/12. By itself, the answer to that series is obviously infinity. If you want to distinguish between the people that know what theyre talking about and those that are copying and pasting pictures from wikipedia and saying "All of the ramanujan summation stuff, cutoff and zeta regularization do is look at the smoothed curve at x = 0", i cant help you there.

To answer your question, a proof doesn't exist for your question, but if you give us a little more context than yeah there are possibly several proofs of equality IF that series actually converged to a finite number, and the best proofs you'll find here are probably about as rigorous as Euler was under the generality of algebra (including myself if i wanted to give you one) when he first wrote about this and related sums.

P.S. If anyone tells you "*snort* Riemann series convergence theorem, you cant rearrange the terms" then just ignore them and do it anyway, you only get problems when dealing with Dir. L-functions that dont have a pole at L(1).

[Anonymous No.7510839]

>>7509920
>>7510491
>/thread
It would be more convincing if you could say what that curve is. I've seen these pictures on those Wikipedia pages for years, but it's never mentioned how to obtain them, and I also don't get it from the Tau article.
You just see the pic and the accommodating explanation and decide to call it a day.

[Anonymous No.7510854]

>>7510839
It's a curve that looks like it might reach -1/12 at one point, isn't that proof enough?

[Anonymous No.7510868]

>>7510854
Of course not, only a rigorous proof with cat ear titty girls is good enough for a 4chan professor. Harrumph.

[Anonymous No.7510945]

>>7510854
No, how do I know how the interpolation comes about, so I use the procedures for other series.
In fact the forward average in >>7509927 is a result of my attempt to come up with that interpolation.

[Anonymous No.7510946]

>>7510854
I wonder how the interpolation comes about, so I use the procedures for other series.
In fact the forward average in >>7509927 is a result of my attempt to come up with that interpolation.