# Homework Help: Show that ∫ dx |f(x)|^2 = ∑ |Cn|^2

1. May 12, 2016

### Poirot

1. The problem statement, all variables and given/known data
This is a question regarding Fourier series.
∫ dx |f(x)|^2 = ∑ |Cn|2 (note the integral is between -π and π, and the sum is from n= -∞ to ∞)

2. Relevant equations
Complex Fourier series: f(x) = ∑Cn einx (again between n = -∞ and ∞)

3. The attempt at a solution
So I figured the complex conjugate of the Fourier transform would be:
f*(x)= ∑ C*n e-inx

so |f(x)|2 = ∑ |Cn|2 (as the exponentials cancel)

But I don't understand how the integral comes into things? I think I'm just not overly good at manipulating summations etc, so any help would be greatly appreciated!

Thanks.

2. May 12, 2016

### ShayanJ

$|f(x)|^2=f(x) f^*(x)=[\sum_n C_n e^{-inx} ][\sum_m C^*_m e^{imx}]=\sum_n \sum_m C_n C^*_m e^{i(m-n)x}$
Now integrate the above equation!

3. May 12, 2016

### Poirot

Hi there, thanks for the help, I see what I did now, forgot to use another dummy variable for the summation!

I ran through the integral, and found that for the condition where n=m (what we want), wouldn't the exponential go to e0= 1 so the integral between -π and π be 2pi?

4. May 12, 2016

What you are proving is Parseval's theorem for the discrete case. Depending upon how you define the Fourier transform, (with a factor of 1, or $1/\sqrt{2\pi}$, or $1/2\pi$), you will pick up a different factor on the right side of your equation. When the function f(x) becomes a voltage as a function of time, it basically says that the energy that is found from an integral of V^2 ,(without an R, Power P=V^2/R), over time is the same as that of adding up the energy components in the frequency spectrum.

5. May 12, 2016

Addition to the above (post #4): The $1$,$1/\sqrt{2\pi}$, or $1/(2\pi)$ factor is for the continuous case. For the discrete case, there is no such factor, and I think you do need a $1/(2\pi)$ factor on the right side of the equation.

6. May 12, 2016

### ShayanJ

Its not that you set n=m! All the terms where n and m differ, actually integrate to zero and for the terms where n=m, you get 2π.

7. May 12, 2016

### Poirot

Thanks for all the replies,

But what I am trying to prove is that ∫ dx |f(x)|^2 = ∑ |Cn|^2, which i believe means only the case where m=n matters? And that's where I can't seem to see how to get rid of this extra factor of 2pi? I should also say that a lot of this deep math chat is a little out of my depth.

Thanks again!

8. May 12, 2016

### ShayanJ

Do you know how to solve the integral $\int_{-\pi}^\pi e^{i(m-n)x} dx$? Are you familiar with limits and L'Hôpital's rule?

That 2π depends on how you define the Fourier series. With your definition, its correct to have it. But you can always normalize your Fourier series by defining it with a $\frac{1}{\sqrt{2\pi}}$ in front of the summation.

9. May 12, 2016

### Poirot

Um, I would integrate how you would normally integrate an exponential giving the same function but over i(m-n), but I suppose that doesn't give the nice answer or 2πδm,n. And as for L'hopital, from what I remember the Lim f/g = f'/g' or something like that.

We usually use the definitions of Fourier transform with 1/root(2π), but I still can't see where that comes in this proof

10. May 12, 2016

### ShayanJ

So you know that the solution to the integral is $\frac{e^{{\large i(m-n)x}}}{i(m-n)} |_{-\pi}^{\pi}$. Expand this and see what you can learn about the case where $n\neq m$ and the limit $m \to n \Rightarrow (m-n)\to 0 \overset{s=m-n}{\Rightarrow} s\to 0$.
Well...you started with a definition of the Fourier series without $\frac{1}{\sqrt{2\pi}}$!

11. May 12, 2016

### pasmith

Are you sure the statement is not $$\frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x)|^2\,dx = \sum_{n=-\infty}^\infty |c_n|^2$$ as appears on the wikipedia page for https://en.wikipedia.org/wiki/Parseval's_identity]Parseval's[/PLAIN] [Broken] identity?

Last edited by a moderator: May 7, 2017
12. May 12, 2016

I agree with post #11. My previous statement in post #5 was almost correct. It needs the $1/(2\pi)$ on the left side of the equation.