I challenge you plebs to prove this infinite sum

>>8483819
trivial

>>8483819
Left as an exercise to the newfriends

File: bachelor thesis.png (16KB, 512x693px) Image search: [Google]

16KB, 512x693px

>>8483819
weak bait

>>8483819
I'll start the proof.

Towards a contradiction, suppose that the statement is untrue.

>>8483851
That's not how it works friendo

>>8483861
You can't start a proof by saying it's not true dumbass you have to show it first

>>8483867
let me guess, all you know is "HURR DURR PEDMAS" from high school and you think it's the golden rule for everything

>>8483819
Idk how you are supposed to prove this, but you can do it via the Fourier series of x, x^2, and x^3.

File: photo.jpg (116KB, 900x900px) Image search: [Google]

116KB, 900x900px

>>8483819
just sum all up and see if they equal each other

>>8483868
Surely this is bait. Have you never heard of proof by contradiction?

>>8483819

Pfff it is not that hard, introduce Bernstein polynoms and use the Parseval formula.

>>8483926
>introduce Bernstein polynoms
Surely you mean BerenSTAIN?

>>8483819
limit /sci/ -> infinity = /b/

File: elementary proof.png (25KB, 552x1138px) Image search: [Google]

25KB, 552x1138px

i remember doing this one in high school

>>8483935

No I'm just incredibly retarded : I meaned Bernouilli's numbers :

[math]
B_0(X) = 1
[/math]

[math]
B'_{n+1}(X) = (n+1)B_n(x) [/math]
[math]
\int_{[0,1]} B_n(t)dt


[/math]

defines a sequence of polynoms, and denote [math] b_n = B_n(0) [/math]

Then you have the Bernouilli number sequence. Using the FOurier developpment of [math]B_n(X) [/math] on [math] [0,1] [/math] and the Parseval formula, you can compute :

[eqn]
\sum_{ n \geq 1} \frac{1}{n^{2k} }
[/eqn]

How the fuck would you solve this shit using Fourier series, I can't seem to find a function that will let me solve this fucking sum

>>8484041
I remember proving that the sum over 1/n^2 is pi^2/6 using the Fourier series of |x|
So i guess that would be a starting point

>>8484011
Wait, so it doesn't involve that retarded sine formula nobody bothers proving?

File: k.jpg (33KB, 255x407px) Image search: [Google]

33KB, 255x407px

>>8484011
>>8484048

>>8483819
We know that [math] \sum_{n=1}^{\infty} 1/n^2=\pi^2/6[/math]. So
[math] \sum_{n=1}^{\infty} 1/n^6=(\sum_{n=1}^{\infty} 1/n^2)^3=\pi^6/6^3=\pi^6/945[/math].

>>8484002
Is math a meme?

>>8484011
>>8484165
how do you do formulas in 4chan?

>>8483819
I am a Calc 2 Babby and to be quite honest I am #triggered at the fact that for most sums we can at best prove they converge but have no fucking idea to what.

When will this pain end? Will numerical analysis save me from the nothing I've become next year or will I have to wait until my junior year so that real analysis teach me how not to be a faggot?

Or is real analysis not even enough? Fuck, do I need a PhD in infinite series or what?

>>8484041
>>8483886

>>8484236

Use the TEX icon.

>>8484048

I guess it's more [math] x \mapsto \pi - x [/math] on [math] [0,2\pi] [/math]
For the other values, you can or use cotan, or the Bernouilli Polynoms.

>>8484041

Any continuous function can be developped as a Fourier series on any compact.

>>8484041

?_?

Formula using Bernoulli numbers ? You mean Euler-Maclaurin formula ?

>>8484237

Lol.

Most of differential equations can not be solved.

Most real numbers are not "computable".

We can even not find the roots of polynoms, or compute power series.

Maths is a hard world. Deal with it.

>>8484335
If this is all true then tbqh I think it is a time for a new axiom.

>>8483819
Consider the function [math]\displaystyle f(z) = \frac{2\pi i}{z^6(e^{2\pi i z} - 1)}[/math]. f has a pole of order 1 at each nonzero integer n with residue [math]\frac{1}{n^6}[/math].
Now, consider the square [math]K_N = \mathrm{conv}\left\{\left(N+\frac{1}{2}\right)(\epsilon_1 + i \epsilon_2), (\epsilon_1, \epsilon_2)\in \{-1,1\}^2\right\}[/math].
The boundary [math]\partial K_N[/math] does not contain any pole of f, hence by the residue theorem [math]\displaystyle \int_{\partial K_N} f(z)dz = \sum_{n = -N}^N Res_n(f) = Res_0(f) + 2\sum_{n=1}^N \frac{1}{n^6}[/math].
Now, it is easily seen that [math]\int_{\partial K_N} f(z)dz = O\left(\frac{1}{N^5}\right)[/math].
Finally, we get [math]\displaystyle \sum_{n = 1}^N \frac{1}{n} = -\frac{1}{2} Res_0(f) + O(1/N)[/math]. Passing to the limit, we get [math]\displaystyle \sum_{n \ge 1} \frac{1}{n^6} = -\frac{1}{2}Res_0(f)[/math]
Now, we have by definition of Bernoulli numbers, for z sufficiently small [math]\displaystyle \frac{2\pi iz}{e^{2\pi i z} - 1} = \sum_{n \ge 0} \frac{B_n}{n!}(2\pi i z)^n[/math]
Now, for z sufficiently small and nonzero, we get [math]\displaystyle f(z) = \sum_{n \ge 0} \frac{B_n}{n!}(2\pi i)^n z^{n-7}[/math] and hence [math]\displaystyle Res_0(f) = \frac{B_6}{6!} (2\pi i)^6[/math] and hence [math]\displaystyle \sum_{n \ge 1}\frac{1}{n^6} = \frac{B_62^5\pi^6}{720}[/math]
Now let's compute B_6: Remember the definining formal series equality [math]\displaystyle X = (e^X-1)\sum_{n \ge 0} \frac{B_n}{n!} X^n = \sum_{n \ge 1} \left(\sum_{k=1}^n {n \choose k} B_{n-k}\right)\frac{X^n}{n!}[/math]
We then get [math]B_0 = 1[/math] and [math]\sum_{k = 1}^n {n \choose k} B_{n-k} = 0[/math] for each [math] n \ge 2[/math]. Plugging in, we get [math]B_6 = 1/42[/math] and [math]\displaystyle \sum_{n \ge 1}\frac{1}{n^6} = \frac{\pi^6}{945}[/math]

>>8484584
Unnecessarily complicated.

>>8484927
Go ahead and provide a simple one if you can - that one is pretty nice.

>>8484984

Calculate the Fourier (sine) series of x, x^2, and x^3. This pretty much just means calculate 3 simple integrals.

Use basic algebraic manipulation to get the result of for the 1/n^2 series from the Fourier series of x.

Use this result + the Fourier series of x^2 to get the sum of the series 1/n^4.

Use this result + the Fourier series of x^3 to get the sum of the series 1/n^6.

>>8485013
That proof would be no simpler than the previous one - it just involves a different method. You not being familiar with the basics of complex analysis, or Bernoulli numbers, does not mean the proof is "unnecessarily complicated."

>>8485013
do it

>>8485299
I did it on my HW like 2 weeks ago.

>>8485314
I'm trying to do it but I get nonsense, I'm trying to do it fourier series method with pasrval

[eqn]\int sin(x^3)[/eqn]
help

>>8485299
>>8485325
It's not difficult, it's just a bunch of computation.
You want to use Parseval's theorem, which is
[eqn]
\frac{1}{2}a_0^2 + \sum_{n\geq1}(a_n^2 + b_n^2) = \frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 \text{d}x
[/eqn]
If you haven't worked with Fourier series before, you're going to want to look up how odd/even functions are evaluated.
While the Fourier series for [math]x[/math] and [math]x^2[/math] are simple to use, [math]x^3[/math] is a little more of a pain in the ass.

>>8485325
Follow this pattern.

- Calculate coefficients and you find...
[math]x = \sum\limits_{n = 1}^\infty {\left( {\frac{{2\ell {{\left( { - 1} \right)}^{n + 1}}}}{{\pi n}}} \right)} \sin \left( {n\pi x/\ell } \right)[/math]

- Apply Parseval
[math]\left\| x \right\|_{{L^2}}^2 = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell {{\left( { - 1} \right)}^{n + 1}}}}{{\pi n}}} \right|}^2}} \left\| {\sin \left( {n\pi x/\ell } \right)} \right\|_{{L^2}}^2[/math]

=>

[math]\int\limits_0^\ell {{x^2}\operatorname{dx} } = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell }}{{\pi n}}} \right|}^2}} \int\limits_0^\ell {{{\sin }^2}\left( {n\pi x/\ell } \right)} \operatorname{dx} [/math]

=>

[math]\frac{{{\ell ^3}}}{3} = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell }}{{\pi n}}} \right|}^2}} \frac{\ell }{2}[/math]

=>

[math]\frac{{{\pi ^2}}}{6} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} [/math]

>>8484002
Kek

>>8484927
It looks complicated but it is very robust (and actually involves very little computation). You just have to replace 6 with 2k everywhere to compute zeta(2k), there is literally nothing to change.
Besides, the method adapts readily to other series: If you have a sum of the form [math] \sum_{n\ge 1} \frac{1}{g(n)}[/math] with f an entire function, then you can try doing this with the function [math]\displaystyle f(z) = \frac{2\pi i}{g(z)(e^{2\pi i z} - 1}[/math] (maybe the contour will have to be more complicated though)

>>8483892
Top kek

The answer: Fourier analysis + Parseval's identity.

>>8483892
Your answer is logically incoherent because it assumes existence of dark numbers greater than 10^200 and that meta number like π can be treated as regular number

File: 1471673920166.jpg (34KB, 645x773px) Image search: [Google]

34KB, 645x773px

>>8483819

Speaking of infinite series
the limit as n goes infinity of (2^n +1)/(2^(n+1))

I'hops rule doesn't seem to be helping, so I have no idea how to determine that the limit is 1/2.

It might be as simple as the power of 2^(∞+1) expands greater than 2^∞. the Wolfram engine disagrees as it claims that both are still ∞ and that it would lead to ∞/∞

>tfw this brainlet can't grasp the mechanics of infinity

>>8486289
Replace (2^n +1) with (2^n) because that constant becomes insignificant. (2^n)/(2^(n+1)) = 1/2 for all n

File: 1479661058377.png (339KB, 670x503px) Image search: [Google]

339KB, 670x503px

>>8486301

That should have been obvious to me.

>>8486289
an intuitive way could have been splitting it into (2^n)/(2^(n+1)) + 1/(2^(n+1))

do the limit thing then 1/2 + 0 = 1/2