Please, /sci/. Help me understand the meme.
Is this statement true? Is the sum of all whole numbers really equal to -1/12? Is the equation in the picture even equal to the sum of all whole numbers? What am I missing? The meme has been coated with so many layers of irony that I don't know where it ends.
>>8116448
>What is analytic continuation
It works if you accept that [math]0.5 = 1 - 1 + 1 - 1 + 1 ...[/math]
>>8116459
You don't have to accept that. It's the Cesaro sum.
>>8116470
>accepting Cesaro sums
do you even constructivism, bro?
i kinda like the sum of all powers of 2 being -1 and the two's complement thing.
dont know about this -1/12 tho
>>8116448
No it's not you dumb fuck.
Stop baiting and look up Ramanujan summation.
No one else need reply.
Remember to sage.
Your first mistake is believing that math is logical. If we had different rules (called axioms) then 1 + 1 could very well be 0. With our current rules 1 + 2 + 3 + 4... = -1/12. That's just the way it is. If you had a calculator and started going 1 + 2 + 3 to eternity eventually you would have -1/12.
>>8116569
No you definitely would not get -1/12 that way
>>8116569
Wrong
>>8116569
>eventually
what is infinity
analytic continuation
>>8116448
this """"proof"""" is the best irl shitpost in maths history
>>8116448
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/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#/media/File:Sum1234Summary.svg
https://en.wikipedia.org/wiki/Ramanujan_summation
>Ramanujan summation essentially is a property of the partial sums, rather than a property of the entire sum, as that doesn't exist.
Regarding the analytical continuation:
Euler introduces the function
[math] \sum_{n=1}^\infty\, n^{-s} [/math]
for real s>1 and proved that
[math] \zeta(s) = \sum_{n=1}^\infty\, n^{-2} =\dfrac{\pi^2}{6} [/math]
That's the Basel problem, 1735, see here:
https://en.wikipedia.org/wiki/Basel_problem
The expression [math] n^{-s} = e^{ \log{ n^{ -s } } } = e^{ -s\,\log{n}} [/math] is pretty unhandy when you take s to be complex and so it's hard to pass the sum to the complex plane. Harder than with a power series anyway.
Riemann does this in 1859 by asserting that any smooth function over the complex numbers, which equals Eulers sum on the positive real line, will fulfill a mirroring relation
[math] \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s) [/math]
With this, all values on the left of s=1/2 are determined by the value on the right.
That's on page 2 and 3 of his famous paper:
(The Riemann hypothesis uttered on page 4 btw.)
https://upload.wikimedia.org/wikipedia/commons/c/cb/Ueber_die_Anzahl_der_Primzahlen_unter_einer_gegebenen_Gr%C3%B6sse.pdf
For s=1-2m, you get
[math] \zeta(1-2m) = \dfrac{2(2m-1)!}{(4\pi^2)^m} \cos(m\pi) \zeta(2m) [/math].
Using the Basel problem, you find [math] \zeta(-1) = -\dfrac{1}{12} [/math].
>>8116790
this this this
>>8117042
To continue on how he derives the reflection formula for the analytic continuation on the left of s=1/2... it's worth writing down because it's pretty cute:
Start with the Gamma funciton defined as
[math] \Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/math]
Substitution x to n·x let's you write
[math] \Gamma(s) = n^s \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]
or
[math] {\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]
Recognizing the geometric series
[math] \sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} [/math]
(which btw. expands as [math] \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2) [/math])
gives you
[math] \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x [/math]
Which is the famous integral representation.
The next step is a bit more elaborate, but you can see where the sinus pops in.
He takes the integral into the complex plane, where the [math] \frac{1} { {\mathrm e}^x-1}[/math] diverges periodically in steps of [math]2\pi\,i [/math].
He pics up the sum again on the other side and that's how the ingrediences to
[math] \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s) [/math]
come together.
The integer values are all determined by Bernoulli numbers, essentially stemming from the integrand above.
If you look at the more general polylog [math] \sum_{n=1}^\infty n^{-s}\,z^n [/math], there is a more general integral representation just as well.
No, it is not true.
You cannot arbitrarily rearrange an non absolutely convergent infinite sum otherwise you run into problems like
>>8116459
>>8117071
>problems like
He said "if you accept that", and it's exactly like that. There is no problem. It's derivavle in a chosen context, just like any standard infinite sum has a value in the standard context. And any context is men made
$\zeta (-1)=-1/12$ because the Zeta function is a well defined function that is analytically continued from its definition on x>1
However $\Sigma n$ does not converge as a series, so it has no limit.
>>8116448
The statement in the picture is true.
>>8116569
lol
I still don't see how the integral form goes to the mirror relationship, can you explain further?
>>8117061
and i think you meant 1/(e^(-x)-1) in the last integrand
>>8117061
woop i made a mistake
I still don't see how the integral form goes to the mirror relationship, can you explain further?
>>8117071
what this guy said.
>>8116448
https://www.youtube.com/watch?v=w-I6XTVZXww
>>8117859
In that video why is S_2 = 1/4 and not 1/2 since S_2 = 1+-1+1+-1... like S_1? What am I missing?
>>8117865
I think I see, he added 2 copies of S2 to itself to get 1/2 and then divided by 2 to get what S2 should be which is 1/4. I'm curious why he shifted the addition of the second copy by S2 by 1 position. I guess it was to get the summation of S1, but seems weird.
>>8117865
dont even bother its stupid anyway
>>8117071
Rearranging sums has nothing to do with the analytic continuation of the Riemann zeta function.
>>8117901
k thanks
>>8117788
Not better than Riemann. It's only two pages of a 10 page paper and an englisch transcription is to be found on Wikipedia
http://www.claymath.org/sites/default/files/ezeta.pdf
>>8116448
no, it isn't
the proof for it treats infinity like a variable
>I swear I tried, but now I leave it as an exercise to the reader
>>8118069
>the proof for it treats infinity like a variable
kek
wtf is up with fucking ramanujan creds to the guy for being an insane autodidact but the only stuff anyone knows from him is what he did wrong jesus fucking imbiciles
>>8116459
>He actually, unironically and willingly believes this is how [math]\zeta \left(-1 \right)[/math] is calculated
>>8119758
>he actually believes that he knows what the words he use means
>>8118154
>>8116448
>>8116448
Idk if relevant, but during my metric spaces course I saw a theorem saying that if you change the position of elements of a sequence that you'll get a different adherence point. Unless the series is absolute convergent. So you could probably play around and get answers like -5/7 or 1/12 ..., by switching the terms.
(That's how I understand it)
Like I said idk if the same applies here, it was one of those funny moments where you find a connection between random things.
>>8121296
a series that converges but doesn't converge absolutely can be arrange to converge to anything you want it to.
This has nothing to do with metric spaces (where the concept of summation and absolute values don't exist) and certainly nothing with the (simply non-convergent) sum talked of in this thread.
>>8117071
>You cannot arbitrarily rearrange an non absolutely convergent infinite sum otherwise you run into problems like
I can arrange anything I like. it is the outcome that displeases you
>>8121327
Metric spaces and differential calculus was the full name of the course. We started off with metric spaces, went to rows and sequences etc and into multivariate calculus.
How does the concept of summation and absolute values not exist in metric spaces? We defined like 3 main different metrics for R^n, and all had summation or abs in them... was everything a lie?
>>8121362
(not the same anon)
This is what you're on about (I think): https://en.wikipedia.org/wiki/Riemann_series_theorem
>>8121362
a metric is simpy a distance function (symmetric in its arguments, non-negative and respecting the triangle inequality for any three points).
Obviously you have a distance on Euclidean n-space. But not every set with a distance function is Euclidean n-space.
Do you see how that works?
So just because because there are spaces in which you can do calculus that are metric spaces, doesn't mean you can do calculus in every metric space.
Do you understand, or should I give you an example?
>>8121394
Yes I think that's what I'm talking about, we didn't give it a name in the course notes.
>pic related
The first point says that if a series is convergent but not absolutely, then you can find a permutation to blablabla..the same as in the wiki.
>>8121398
Yes I understand that metric is simply a distance function which has to satisfy 3 conditions. I also know that a metric doesn't imply a norm, but a norm does imply a metric. (If this has any relevance here)
Btw, non-negative, wasn't that a condition for the norm? Metric as far as I remember had symmetry, triangle inequality and separability (idk what the correct name is in English, but d(x,y) = 0 <=> x = y, or is this the same?)
As an example, if I remember correctly, is the cross product. It's only defined on R^3 (and R^7 if I remember this random remark from my prof correctly.)
But please explain though how abs and summation might not be defined in a certain metric space, you've poked my curiosity. The courses I had were only limited to Euclidean spaces.
>>8121432
Not the guy you're talking to but I'll try to explain.
As you said the metric doesn't imply a norm which, in R^n, is the absolute value. This means you can define a metric space that doesn't have a norm or abs or whatever you want to call it.
The same thing is true for summation, an example I remember is the trivial metric: define d(x,y)=0 if x=y and d(x,y)=1 otherwise. This satisfies all the conditions and you can define it on any set even if summation isn't possible.
As for non-negativity, I think the condition is usually stated as d(x,y)>=0 with equality when x=y as you said but I'm not sure of this.
The sum of all numbers is 0
>>8121362
That's because a generic metric space might not have an obvious operation resembling addition, that is compatible with the topology (ie. such that addition is continuous wrt the topology).
A more general setting for studying series is that of normed vector spaces, that is vector spaces (such as R^n) endowed with a norm (look up the definition if you don't know it). These are metric spaces but, in a sense, are much more than just metric spaces, and the extra structure makes it possible to talk about series in that context.
>>8121495
>The sum of all numbers is 0
All real numbers are just the integers?
>>8121492
Yes I've seen that example and used it to show that a metric doesn't imply a norm.
>...a norm which, in R^n, is the absolute value.
Never really thought about it like this, i.e. the norm = absolute value. Always thought abs was some tool used.
Now abs has like 2 meanings for me. The positive value of a number and the norm. In Euclidean space I assume they're both the same thing, but couldn't you have the abs in its first meaning in some space where there's no norm?
Yeah we defined norms to be positive definite so d(x,y) >= 0, but we didn't demand this from metrics. (Unless it's implied by the other conditions)
>>8121544
Yeah I remember some exercises during linear algebra where the prof would define some form of addition and some form of scalar multiplication and we would have to check if it was a vector field or not. You had to go through all the axioms etc. defined for a field, or specific shortcuts related to vector fields.
I remember a situation where he defined addition in such a way that the neutral element wasn't 0 but something else. Pretty weird but amazing nonetheless.
>>8121546
That's what I said no?
Anyway, thanks for helping guys!
>>8121559
>In Euclidean space I assume they're both the same thing.
Well.. not really, in R yes, in R^n the norm is the length of the vector which is what I thought you meant when you said absolute value, there's no "positive value" of a vector.
>but couldn't you have the abs in its first meaning in some space where there's no norm?
I think that the properties of the absolute value satisfy all the conditions of a norm, so no
>>8121597
Yes I believe you're correct then. It makes sense to just have a norm || || defined, where | | is simply a special case of the norm in R.
Is my understanding correct now?
>>8121616
Yeah, that was the point I was trying to make
>>8121624
Yeah sorry, sometimes it takes a detour to get to me :).
Tyvm
>>8116448
how the fuck can all integers add up to a fraction that doesnt make any sense at all
>>8121637
Why?
No, it's not true that the sum (limit of partial sums) is -1/12. But I'd be astonished if you got a convincing reason for it not being a fraction, if the limit existed.
We got
[math] \sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6} [/math]
so a sum of rationals can take you to something that isn't a rational.
The arithmetical theory of natural numbers, say via the Peano axioms, concerns finite sums (and only those), and there you can prove that this operation is closer - a finite sum of integers is an integer.
But there is no converging infinite sum non-zero integers, so it's not like we have a lot of examples that suggest the infinite sum of numbers should be an integer or something.
That was tl;dr for "it doesn't make sense" is just your feeling.
