Explain this meme to me So if someone gave me a dollar everyday,

https://www.youtube.com/watch?v=w-I6XTVZXww

u lose so little so who cares haha

>>8799941
You're not allowed to re-arrange the terms of an infinite series within the standard theory. If you can get these meme results and also the riemann zeta function is = to that at -1 or whatever I don't fucking know

>>8799956
ayyytake the money

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

Infinite is not a number, so you can't assign a value to a series like this in a traditional way. The same method that we use to assign a value to an infinite series that adds up to something, can also be used to assign a value to an infinite series that can't add up to anything. The value of -1/12 is the ONLY value that can ever be associated to the series of 1+2+3+4+... . It's not a sum of that series, but the method used to assign that value is the same method that is used to assign sums to series that DO add up to something.

>>8799941
It's a kind of jokes mathematicians tell each other, mostly for fun. And when you understand it it's a bit funny the first time (on the scale of mathematical jokes, keep that in mind).

Some ppl who understand maths badly will come with circonvoluted concepts of why in some sense it might, with enough rewriting and redefinition, be true, but it's actually really, really not.

Really don't bother with that OP, there's a shittons of lots of things of interest in the realm of mathematics, and meme series are not the greatest of them.

>>8799975
> i saw a nuberphile video and now i know math

>No, 1+1+1+1+... is equal to -1/2.
Nope.
>Infinite is not a number, so you can't assign a value to a series like this in a traditional way.
Except u can, actually.

>The same method that we use to assign a value to an infinite series that adds up to something, can also be used to assign a value to an infinite series that can't add up to anything.
There exists a loads of methods to add up non-absolutely convergent series, but none of these are "the same" as the one we use for ACS.

>The value of -1/12 is the ONLY value that can ever be associated to the series of 1+2+3+4+... .
This degree of falsehood should only be allowed on /b/
>It's not a sum of that series, but the method used to assign that value is the same method that is used to assign sums to series that DO add up to something.
Thanks for making quite clear you have no idea of what you're talking about. Like, really, between taking the smallest upper bound of a set, and defining a function, analitycally expands it, wrongly assume said function keeps all it's property over the extension, and apply it to a point, then write an = sign for the lulz, there's quite a margin.

>>8800365
are you just massively autistic or something? although the part about it being unique isnt true in general his explanation was perfectly fine for the OP who obviously hast studies math at this level.

>>8799955

fpbp

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>>8799941
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|<1 you have,

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

so fall epsilon with 0<epsilon<2 you have,

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

Thus

[math] \lim_{\epsilon \to 0} \sum_{k=0}^\infty (-1)^k \cdot (1 - \epsilon)^k = \dfrac{1}{2} [/math]

On the other hand, the expression [math] \lim_{\epsilon \to 0} \sum_{k=0}^\infty (-1)^k [/math] can't be computed within the theory of analysis.

Now there are other theories of infinite sums in which it can be computed. Some of those turn out to make the following result provable:

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

>>8800483
What should I study to be able to understand those equations?
Anything I could do somewhere like khan academy/udacity/whatever?

>>8799955
>the guy who does periodic videos also does numberphile
I never noticed this.

>>8800489
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]

>>8800518
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]

>>8800525
>>8800518
Very nice explanation, thanks

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>>8800541
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)

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>>8800573
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]

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>>8799941

>>8800570
whos this girl

>>8800802
cara "bushiest eyebrows since brezhnev" delevingne

>>8799941
watch this and ignore everyone in this thread OP:
https://youtu.be/sD0NjbwqlYw

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>>8799941
>>8799955
Thats SO retarded. Watch this one instead, much better:

https://www.youtube.com/watch?v=jcKRGpMiVTw

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

>>8800850
Glad to see 3blue1brown getting some cred on /sci/

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.

>>8801742
How long until there is a program that does what he does?

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>>8801783
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

>>8801759
Why not? It's just math.

>>8799941
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]

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

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>>8803154
Can someone explain to me why this shit fails to render half the time for me

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>>8799941
I fear the day the Jude masters analytic continuation.

>>8803154
Use \dots.
>>8803164
\newline maybe? I don't know.

>>8803164
[math]\text{because you're using}\\\texttt{\\newline}~\text{rather than}~\texttt{\\\\}\,.[/math]

>>8800850
/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]

>>8803164
make sure every "\" has a space before it

>>8799975
>No, 1+1+1+1+... is equal to -1/2.

Yes, if you average it.

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

>>8801660

same desu

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

Does:

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

>>8804626
yes

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>>8799941
Presupposes infinity is symmetric about the origin

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