Prove only using Fourier Series

[Oxymoron]

Prove only using Fourier Series!

By considering the Fourier Series of f(x)=x^2 prove that

\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}

 

[Oxymoron]

Ok, so I considered the Fourier Series of f(x)=x^2 and found that

\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}

Here is how I did it

a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^2}{3} = \frac{2\pi^{2m}}{2m+1} when m = 1

a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{4}{n^2}(-1)^n

Since the function is even, b_n = 0.

Hence the Fourier Series can be written explicitly as

x^2 = \frac{\pi^2n}{2n+1} + \sum_{n=1}^{\infty}a_n \cos(nx)

And since n=1 we have

x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n\cos(nx)}{n^2}

Since x \equiv \pi we know \cos(nx)=(-1)^n, so

\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}

and rearranging

\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2)

 

[Oxymoron]

Then I tried to do the same for f(x)=x^4, for the reason that now n=2 and I will have an expression for \sum\frac{1}{n^4}. I came close, but not close enough.

I know that

\sum_{n=1}^{\infty} = \frac{\pi^4}{90} = \zeta(4)

but I am only allowed to use results from Fourier Analysis. Is there anyone who can help? Or perhaps has an idea?

 

[Kelvin]

I am not sure I am 100% correct but what I got is slightly different from what we need.
We have
\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}

If we consider the Fourier series of f(x)=x^4

x^4 = \frac{\pi}{5} + \sum_{n=1}^{\infty}(-1)^n \left( \frac{8 \pi^2}{n^2} - \frac{48}{n^4 \pi} \right) \cos(nx)

when x = \pi,

\pi^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( \frac{8 \pi^2}{n^2} - \frac{48}{n^4 \pi} \right)

rearraging,

\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{\pi}{48}=\frac{\pi^5}{90}

 

[HallsofIvy]

a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^2}{3} = \frac{2\pi^{2m}}{2m+1}

If this is a₀ WHAT is "m"?

Same question: if
x^2 = \frac{\pi^2n}{2n+1} + \sum_{n=1}^{\infty}a_n \cos(nx)

What does the "n" in the term before the sum mean?

 

[Kelvin]

I think I really made a mistake. The Fourier series of x^4 should be:

x^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( (-1)^n \frac{8(n^2\pi^2-6)}{n^4} \cos(nx) \right)

then put x = \pi, the result follows

 

[Oxymoron]

Sorry HallsovIvy, the n in the term before the sum should be "m". The m's are not summed, but the n's are - as you probably can guess.

 

[Oxymoron]

Kelvin, that's is the same Fourier Series as I got. However I can't see the connection between the two. I've played around with substituting \pi for x, but I cannot get the required result.

Is there any chance you (or someone else) could post the last remaining steps in this proof for me.

 

[Kelvin]

I think I really made a mistake. The Fourier series of x^4 should be:

x^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( (-1)^n \frac{8(n^2\pi^2-6)}{n^4} \cos(nx) \right)

then put x = \pi, the result follows

so putting x=\pi,

\pi^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( (-1)^n \frac{8(n^2\pi^2-6)}{n^4} (-1)^n \right)

\frac{4\pi^4}{5} = \sum_{n=1}^{\infty} \left( \frac{8(n^2\pi^2-6)}{n^4} \right)

\frac{4\pi^4}{5} = 8\pi^2 \sum_{n=1}^{\infty} \frac{1}{n^2} - 48 \sum_{n=1}^{\infty}\frac{1}{n^4}

using the result that

\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}

\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{1}{48}=\frac{\pi^4}{90}

 

[Oxymoron]

OMG! I forgot that I could use the result from the first part ! Thanks Kelvin.

I wonder if anyone here knows how to do it the "easy" way? You know, using the general hypergeometric function. I'd be impressed! :0

 

[OlderDan]

OMG! I forgot that I could use the result from the first part ! Thanks Kelvin.

I wonder if anyone here knows how to do it the "easy" way? You know, using the general hypergeometric function. I'd be impressed! :0

Maybe I am missing something, but I am wondering how showing the result based on the Fourier series of f(x)=x^4 is satisfying the problem you originally posed

By considering the Fourier Series of f(x)=x^2 prove that

\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}

Did you state the problem incorrectly? Is there some simple connection between the series for f(x)=x^4 and f(x)=x^2 that I am missing? OR does the original problems still need to be solved?

 

[Oxymoron]

OlderDan,

From what I can gather, the question asked to to find a Fourier series for a particular function, then notice that it looked almost like what we wanted. Then it required a little but of imagination to guess that the actual answer we want will require us to find the Fourier series of a slightly different function. Lo and behold, f(x) = x^4 does the trick because you get the \sum\frac{1}{n^4} instead of the \sum\frac{1}{n^2} you get from the Fourier series of f(x) = x^2.

I thought was that this was going to be hard to solve, since I haven't been formally introduced to the Riemann Zeta Function (which is \zeta(4) = \frac{\pi^4}{90}). But in hindsight, the Fourier analysis involved is not far above second year uni.