# Complex Fourier Series & Full Fourier Series

1. Dec 9, 2009

### kingwinner

1. The problem statement, all variables and given/known data
Claim: If f(x) is a REAL-valued function on x E [-L,L], then the full Fourier series is exactly equivalent to the complex Fourier series.

This is a claim stated in my textbook, but without any proof. I also searched some other textbooks, but still I have no luck of finding the proof.
I've already spent an hour thinking about how to show that this is true, but still I am not having much progress. Here is what I've got so far:

Full Fourier series is:

where

Complex Fourier series is:

where

And now I am having trouble with this...how can I use the last part to show that if f(x) is REAL-valued, the complex Fourier series can be reduced to the full Fourier series. Can someone please show me how to continue from here? I also don't see how a sum from negative infinity to infinity (for complex Fourier series) can possibly be reduced to a sum from 0 to infinity (for full Fourier series). It seems like I have no hope...

2. Relevant equations
As shown above

3. The attempt at a solution
As shown above

I am really frustrated now and any help is very much appreciated! :)

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Last edited: Dec 9, 2009
2. Dec 9, 2009

### HallsofIvy

Staff Emeritus
An obvious first step is to use the fact that $e^{in\pi x/L}= cos(n\pi x/L)+ i sin(n\pi x/L)$. Multiply it out and use the fact that cos(-x)= cos(x), sin(-x)= sin(x).

3. Dec 9, 2009

### kingwinner

OK, the following is what I got.
(for simplicity I am taking the interval to be from -pi to pi)

Is this a correct proof??

Thanks!

4. Dec 9, 2009

### kingwinner

The claim is
"If f(x) is a REAL-valued function on x E [-L,L], then the full Fourier series is exactly equivalent to the complex Fourier series."

But nowhere in the proof have I assumed f(x) is real-valued. Is it absolutely necessary for f(x) to be REAL-valued in order to prove that the full Fourier series is exactly equivalent to the complex Fourier series??

5. Dec 10, 2009

### HallsofIvy

Staff Emeritus
That's because the equality of those sums does not depend upon real or complex numbers. We require that F be real valued in order to have the coefficients real numbers.

6. Dec 10, 2009

### kingwinner

But looking at my proof above, I believe that the full Fourier series and the complex Fourier series are equivalent in general, even when f(x) is complex-valued. Right??