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I need to approximate a function f, but I cannot do so with frequencies that exceed 1kHz

What is the best approximation I can get? Is taking the Fourier transform then zeroing any term above 1kHz the best approximation? Or can I fiddle with the lower order terms and get a better fit?

This is a made up scenario, but I have to prove the same concept with Walsh transforms. I am fairly certain that the lower order terms form the best approximation from random twiddling and hill climbing searches, but I need proof.

I believe the proof is something very similar to a least squares regression proof, but I can't get it. Has this problem been solved before? At least in the Fourier domain?

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I think this belongs on digital signal processing...if someone can migrate it there? –  Fixed Point Mar 20 '13 at 7:06
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1 Answer

The Fourier transform is unitary. Therefore the best fit in the $L^2$ norm in the frequency domain is also the best fit in the time domain. This means that suppressing all frequencies outside the allowed band indeed gives the best approximation.

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Is there a proof or a deeper explanation of this? I do not understand it. I read about Unitary. That means the matrix has orthogonal rows/columns. I don't understand what $L^2$ norm is and how that relates across the domains. I can see intuitively how it is an orthogonal decomposition and therefore probably is the best approximation. –  SwimBikeRun Mar 21 '13 at 4:00
    
Ah $L^2$ is just the norm of the squares. So the key factor is that it is unitary? –  SwimBikeRun Mar 21 '13 at 6:41
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