Evaluation of Numerical series by Fourier series

[siddharth]

I have some problems which says show that
(i) $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$
and
(ii) $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}$$
And another one which says, show that for $$0<x<\pi$$
$$sin x + \frac{sin 3x}{3} + \frac{sin 5x}{5} + ... = \frac{\pi}{4}$$
The problem is that, the function I am supposed to work with ($$f(x)$$) is not given. So I want to know if there is some way to find out which function I should start out with to achieve the results.
If I know f(x), I can write it as a Fourier series and then substitute values of x to get the numerical series. But without knowing what f(x) to start with, I am stuck.

 

[fourier jr]

all you can do find the functions whose Fourier series are really close to the ones you listed. that's all i could do anyway when i had to solve those sorts of problems.

 

[Hurkyl]

Just use the formula for the coefficients of the Fourier series! But solve them for the function, instead of for the coefficients!

(I don't know if it would work, but if I had thought what you did, it seems to me to be the obvious next step)

 

[siddharth]

I got the third problem. That's the Fourier series for $$f(x)=\frac{\pi}{4}$$ in sine's. ( That was pretty obvious!)
I have noticed that if I expand $f(x)=x^2$ in $[-\pi, \pi]$ and then put $x=\pi$, I get the value of
$$\sum_{n=1}^\infty \frac{1}{n^2}$$
So, perhaps for
$$\sum_{n=1}^\infty \frac{1}{n^4}$$, I have to start with $$f(x)=x^4$$?
Finding the coeffecients in this case would require a lot of integration by parts. I'll try that and see if it works.

 

[quasar987]

There are at least 3 ways of doing this problem!

1° I guess the easiest way is the one mentioned by Hurkyl. Unfortunately, you need to know a function whose Fourier serie is just right such that when evaluated at a key number (such as 0, pi/2, 5pi, etc) you get the desired serie on one side of the equality, and a number on the other side. Personally, I don't know any function that even has 1/n^4 in its coefficient.

2° Another way is by Perseval's identity. Or at least a particular case of the more general identity perhaps. I'm talking about "If f is continuous on the circle, then

$$\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)$$

This identity is particularily useful in the case where you have developed f as an even or an odd function. Then if a_n or b_n is proportionnal to 1/n², you're in bizz. And even/odd functions with a_n or b_n proportional to 1/n², THAT I know of.

3° The third way is the most powerful of the three. With it, you can find the value of any serie of the type

$$\sum_n \frac{1}{n^i}$$

and

$$\sum_n \frac{(-1)^n}{n^i}$$

where i is even. It's quite long to write so unless someone specifies that he wants to know the trick, I won't write it out.

 

[siddharth]

2° Another way is by Perseval's identity. Or at least a particular case of the more general identity perhaps. I'm talking about "If f is continuous on the circle, then
$$\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)$$
This identity is particularily useful in the case where you have developed f as an even or an odd function. Then if a_n or b_n is proportionnal to 1/n², you're in bizz. And even/odd functions with a_n or b_n proportional to 1/n², THAT I know of.

Ok, I know that for x², the a_n's are propotional to 1/n². And that (a_n)^2 is 16/n² while the b_n's are 0. So that works! Thanks a lot!

3° The third way is the most powerful of the three. With it, you can find the value of any serie of the type
$$\sum_n \frac{1}{n^i}$$
and
$$\sum_n \frac{(-1)^n}{n^i}$$
where i is even. It's quite long to write so unless someone specifies that he wants to know the trick, I won't write it out.

If you have the time, I would be interested to know this

 

[dops]

∫_0^π▒αt d(ωt)
where α is a constant and ω is the angular velocity of a wave.


can't sove these question?

 

[dops]

The answer to the my question is απ/2 but don't know how the book got it.