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.