Prove only using Fourier Series

In summary: I just needed a little imagination.In summary, by using the Fourier Series of f(x)=x^4, we can find the sum of the infinite series \frac{1}{n^4} by substituting x=\pi and rearranging the resulting expression. This allows us to prove that \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} using only results from Fourier Analysis.
  • #1
Oxymoron
870
0
Prove only using Fourier Series!

By considering the Fourier Series of [itex]f(x)=x^2[/itex] prove that

[tex] \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/tex]
 
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  • #2
Ok, so I considered the Fourier Series of [itex]f(x)=x^2[/itex] and found that

[tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

Here is how I did it

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

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

Since the function is even, [itex]b_n = 0[/itex].

Hence the Fourier Series can be written explicitly as

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

And since [itex]n=1[/itex] we have

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

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

[tex]\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]

and rearranging

[tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2)[/tex]
 
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  • #3
Then I tried to do the same for [itex]f(x)=x^4[/itex], for the reason that now [itex]n=2[/itex] and I will have an expression for [itex]\sum\frac{1}{n^4}[/itex]. I came close, but not close enough.

I know that

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

but I am only allowed to use results from Fourier Analysis. Is there anyone who can help? Or perhaps has an idea?
 
  • #4
I am not sure I am 100% correct but what I got is slightly different from what we need.
We have
[tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

If we consider the Fourier series of [tex]f(x)=x^4[/tex]

[tex]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)[/tex]

when [tex]x = \pi[/tex],

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

rearraging,

[tex]\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{\pi}{48}=\frac{\pi^5}{90}[/tex]
 
  • #5
[tex]a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^2}{3} = \frac{2\pi^{2m}}{2m+1}[/tex]

If this is a0 WHAT is "m"?

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

What does the "n" in the term before the sum mean?
 
  • #6
I think I really made a mistake. The Fourier series of [tex]x^4[/tex] should be:

[tex]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)[/tex]

then put [tex]x = \pi[/tex], the result follows
 
  • #7
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.
 
  • #8
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 [itex]\pi[/itex] for [itex]x[/itex], 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.
 
  • #9
Kelvin said:
I think I really made a mistake. The Fourier series of [tex]x^4[/tex] should be:

[tex]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)[/tex]

then put [tex]x = \pi[/tex], the result follows

so putting [tex]x=\pi[/tex],

[tex]\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)[/tex]

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

[tex]\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}[/tex]

using the result that

[tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

[tex]\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{1}{48}=\frac{\pi^4}{90}[/tex]
 
  • #10
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
 
  • #11
Oxymoron said:
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 [itex]f(x)=x^4[/itex] is satisfying the problem you originally posed

Oxymoron said:
By considering the Fourier Series of [itex]f(x)=x^2[/itex] prove that

[tex] \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/tex]

Did you state the problem incorrectly? Is there some simple connection between the series for [itex]f(x)=x^4[/itex] and [itex]f(x)=x^2[/itex] that I am missing? OR does the original problems still need to be solved?
 
  • #12
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, [itex]f(x) = x^4[/itex] does the trick because you get the [itex]\sum\frac{1}{n^4}[/itex] instead of the [itex]\sum\frac{1}{n^2}[/itex] you get from the Fourier series of [itex]f(x) = x^2[/itex].

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 [itex]\zeta(4) = \frac{\pi^4}{90}[/itex]). But in hindsight, the Fourier analysis involved is not far above second year uni.
 

Related to Prove only using Fourier Series

1. What is a Fourier Series and how is it used in proofs?

A Fourier Series is a mathematical representation of a periodic function as a sum of sine and cosine functions. It is used in proofs to show that a function can be approximated by an infinite sum of simpler trigonometric functions.

2. Can all functions be proved using Fourier Series?

No, not all functions can be proved using Fourier Series. The function must be periodic and have a finite number of discontinuities in order for the Fourier Series to accurately represent it.

3. How does the Fourier Series relate to the Fourier Transform?

The Fourier Transform is a generalization of the Fourier Series for non-periodic functions. It transforms a function from the time or spatial domain to the frequency domain, while the Fourier Series only applies to periodic functions.

4. Are there any limitations to using Fourier Series in proofs?

One limitation is that the function must be continuous and piecewise smooth for the Fourier Series to converge. Additionally, the Fourier Series may not converge uniformly for certain functions, making it difficult to use in proofs.

5. How is the convergence of a Fourier Series determined?

The convergence of a Fourier Series is determined by the properties of the function being approximated. For example, if the function has a finite number of discontinuities, the Fourier Series will converge to the average of the left and right limits at each discontinuity.

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