Explain this meme to me

Your pic is false in analysis, but some well studied theories of infinite sums get this result.
Look up zeta-function regularization and the then Riemann zeta function reflection formula to get that value.

I give you a less outrageous ideas. In analysis, an infinite sum is defined as a limit of finite sums
[math] \sum_{k=0}^\infty a_k = \lim_{n\to \infty} \sum_{k=0}^n a_k [/math]
The limit is defined using a more or less complicated formula involving the norm on the real numbers, which are defined on some very complicated way.
Using this theory, you can prove statements like the following

[math] \sum_{k=0}^\infty \dfrac{1}{2^k} = 2 [/math]

and you can also prove the following three results:

for all z with |z|

What should I study to be able to understand those equations?
Anything I could do somewhere like khan academy/udacity/whatever?

>the guy who does periodic videos also does numberphile
I never noticed this.

Which ones? For the bulk I posted, you just need real analysis. I'm sure there is something on khan academy, but I'm more a fan of books. If you want it all down to straight definitions, you'd want to learn some predicate logic too.

But those I posted in particular happen to be very straight forward:

Consider what you get if you multiply out
[math] (1+z+z^2+z^3+z^4) \cdot (1-z) [/math]
according to the scheme
[math](a+b)\cdot (u+v) = (a+b)\cdot u + (a+b)\cdot v [/math]

You get
[math] (1+z+z^2+z^3+z^4) - (z+z^2+z^3+z^4+z^5) [/math]
in which many terms cancel, so that you get
[math] 1-z^5 [/math]
So
[math] (1+z+z^2+z^3+z^4) \cdot (1-z) = 1-z^5 [/math]
or, rewritten when z isn't one,
[math] 1+z+z^2+z^3+z^4 = \dfrac {1-z^5} {1-z} [/math]

And indeed, this doesn't just work for the number n=4, but for any n. Not surprisingly, by induction, you get the formula
[math] (1+z+z^2+z^3+...+z^n) \cdot (1-z) = 1-z^{n+1} [/math]
This is even a theorem in a theory integers, and if you consider rational numbers, you get, when z isn't one,
[math] 1+z+z^2+z^3+...+z^n= \dfrac {1-z^{n+1}}{1-z}[/math]

Now take z=1/2.
You know [math] (1/2)^3 = 1/8 < 1/2[/math]
Indeed, for every exponenet m that's bigger than 1, you have
[math] (1/2)^n < 1/2 [/math]
and the bigger the n, the smaller the number.
For n arbitrary large, [math] (1/2)^n [/math] becomes arbitrary small.
We write
[math] \lim_{n \to \infty} (1/2)^n = 0[/math].
And thise holds not just for z=1/2 but any z with absolute value smaller than 1.

Now as for those z,
[math] \lim_{n \to \infty} z^n = 0[/math].
the above formula becomes
[math] \lim_{n \to \infty} (1+z+z^2+z^3+...+z^n) = \dfrac {1}{1-z} [/math]
For example
1+1/2+1/4+1/8+1/16+... = 2

The left hand side is also written
[math] \sum_{k=0}^\infty = z^k [/math]

Quite similarly, for a particular well behaved infiite sum R(z), you can derive results like

[math] \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + (z-1) \cdot R(z) [/math]

or let's write the last line as

[math] \sum_{n=0}^\infty n \, (1-\epsilon)^n - \dfrac {1} { \log(1-\epsilon)^2} = - \dfrac{1}{12} + \epsilon \cdot S(\epsilon) [/math]

Very nice explanation, thanks

np

You can now use the formula for [math] (1-z)^{-1} [/math] derived above to show

[math] \sum_{n=0}^\infty n \, z^n - \left( \sum_{k=1}^\infty \dfrac{1}{k} (1-z)^k \right)^{-2} = - \dfrac{1}{12} + (z-1) R(z) [/math]

or

[math] \lim_{z\to 1} \sum_{n=0}^\infty n \, z^n - (\sum_{k=1}^{\infty}\dfrac{1}{k}(1-z)^k)^{-2} = - \dfrac{1}{12} [/math]

Or consider the maybe surprising formula

[math] \frac{1}{3}(4^3-2^3) = \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) [/math]

This can be derived with the rather complicated formula by Euler and MacLaurin

[math] \int_a^b f(n)\,{\mathrm d}n = \sum_{n=a}^{b-1} f(n)+\left(\lim_{x\to b}-\lim_{x\to a}\right)\left(\dfrac{1}{2}-\dfrac{1}{12}\dfrac{d}{dx}+\dots\right)f(x) [/math]

(where ... represents a sum of differential operators determined by Bernoullli numbers)

Ramanujan summation is a theory of infinite sum that mirrors the result of the above formulas, except some terms are systematically dropped.

You probably know that the sum of the first n non-negative numbers is n(n+1)/2. Now verify with epsilon-delta arguments that the sequence n(n+1)/2 approaches -1/12.

it is a value that is merely assigned (this is the key word here) to the series, it is not actually equal to it (it is divergent after all)

I actually tried to brige the two results, but I didn't get anything nice.

You have

[math] 1+2+3+4+... m = \lim_{z\to 1} \sum_{k=0}^m k \, z^k [/math]
but the limit m to infinity doesn't exist.

Now introduce a function smoothening

[math] \langle f \rangle (k) := \int_k^{k+1} f(t)\, dt [/math]

If you consider

[math] \sum_{k=0}^m \left( k \, z^k - (1-q) \langle k \, z^k \rangle \right) [/math]

you get an expression which for z to 1 and then q to 1 goes to
[math] m(m+1)/2[/math]
as it should
and if you take the same expression and go with m to infinity and with q to 0 and THEN with z to 1, you get
[math] -1/12[/math]

I found the regularized sum has a closed expression, but it it's not worth looking at much, as it looks as follows:

[math] \dfrac{(1-q) (z-1)^2 \left(z^{m+1}-1\right)+z \log (z) \left(z^m \left((m+1) (q-1)(z-1)^2+(m (z-1)-1) \log(z)\right)+\log (z)\right)}{(z-1)^2 \log ^2(z)} [/math]

...

whos this girl

cara "bushiest eyebrows since brezhnev" delevingne

watch this and ignore everyone in this thread OP:
youtu.be/sD0NjbwqlYw

Thats SO retarded. Watch this one instead, much better:

youtube.com/watch?v=jcKRGpMiVTw

I recommend you his channel if you are interested in mathematics. I really like his approach to problems.

Glad to see 3blue1brown getting some cred on Veeky Forums

The dude's production quality is top-tier

what's really amazing is that this somehow works for physics, why? why does it work? last time I checked it was something about tests on virtual particle energy density. It's just incredible.

How long until there is a program that does what he does?

An AI that teaches? Or what do you mean by "program"?
He has his Python code that he uses to generate his videos on github

Why not? It's just math.

Here you go buddy:

[math]
\newline
S = 1 + 2 + 3 + 4 + 5 + 6 + ... \newline
A = 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2}
\newline
B = 1 - 2 + 3 - 4 + 5 - 6 + ...
\newline
B = 0 + 1 - 2 + 3 - 4 + 5 - ...
\newline
2B = 1 - 1 + 1 - 1 + 1 - 1 + ... \newline
B = \frac{1}{4} \newline
S - B = 0 + 4 + 0 + 8 + 0 + 12 + ... \newline
S - B = 4 + 8 + 12 + 16 + 20 + ... \newline
S - B = 4S \newline
- B = 3S \newline
\frac{-1}{12} = S \newline
S = 1 + 2 + 3 + 4 + 5 + 6 + ... = \frac{-1}{12}

[/math]

terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

Can someone explain to me why this shit fails to render half the time for me

I fear the day the Jude masters analytic continuation.

Use \dots.
\newline maybe? I don't know.

[math]\text{because you're using}\\\texttt{\\newline}~\text{rather than}~\texttt{\\\\}\,.[/math]

/thread

[math] \\ S = 1 + 2 + 3 + 4 + 5 + 6 + ... \\ A = 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2} \\ B = 1 - 2 + 3 - 4 + 5 - 6 + ... \\ B = 0 + 1 - 2 + 3 - 4 + 5 - ... \\ 2B = 1 - 1 + 1 - 1 + 1 - 1 + ... \\ B = \frac{1}{4} \\ S - B = 0 + 4 + 0 + 8 + 0 + 12 + ... \\ S - B = 4 + 8 + 12 + 16 + 20 + ... \\ S - B = 4S \\ - B = 3S \\ \frac{-1}{12} = S \\ S = 1 + 2 + 3 + 4 + 5 + 6 + ... = \frac{-1}{12}\\ S = 1 + 2 + 3 + 4 + 5 + 6 + ... \\ A = 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2} \\ B = 1 - 2 + 3 - 4 + 5 - 6 + ... \\ B = 0 + 1 - 2 + 3 - 4 + 5 - ... \\ 2B = 1 - 1 + 1 - 1 + 1 - 1 + ... \\ B = \frac{1}{4} \\ S - B = 0 + 4 + 0 + 8 + 0 + 12 + ... \\ S - B = 4 + 8 + 12 + 16 + 20 + ... \\ S - B = 4S \\ - B = 3S \\ \frac{-1}{12} = S \\ S = 1 + 2 + 3 + 4 + 5 + 6 + ... = \frac{-1}{12} [/math]

make sure every "\" has a space before it

>No, 1+1+1+1+... is equal to -1/2.

Yes, if you average it.

But this wouldn't work with OP's scenario.

same desu

p=2*2*2*...
p=2p
p=0
QED

Does:

[math]-12 -24 -36 -48-... = 1[/math]

yes

Presupposes infinity is symmetric about the origin

What does 1 + 1 + 1 + 1+... equal according to this meme?