Complex Fourier Series & Full Fourier Series

kingwinner
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Homework Statement


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:
fourier_series.gif

where
coefficient.gif


Complex Fourier series is:
complex.gif

where
complex_coefficient.gif

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...

Homework Equations


As shown above

The Attempt at a Solution


As shown above

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

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

pde3.JPG


Is this a correct proof??

Thanks!
 
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??
 
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.
 
HallsofIvy said:
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.
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??
 
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