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How to get fourier transform from fourier series 
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#1
Mar3112, 06:59 AM

P: 183

How you get Fourier transform from Fourier series? Do Fourier series becomes Fourier transform as L > infinity?
http://mathworld.wolfram.com/FourierTransform.html I don't understand where discrete A sub n becomes continuous F(k)dk ( where F(k) is exactly like A sub n in fourier series)? I also have a general question which is; How to transform a discrete variable to a continuous variable in order to convert a summation to integral? 


#2
Apr412, 10:52 AM

P: 183

I think I know the answer now.
" For a function periodic in [L/2,L/2], Fourier series is [tex] f(x) = \sum_{n=\infty}^{\infty }A_{n}e^{i(2\pi nx{/}L)} \\A_{n} = 1{/}L \int_{L{/}2}^{L{/}2}f(x)e^{i(2\pi nx{/}L)}dx. [/tex]" The part that was bothering me was " The Fourier transform is a generalization of the complex Fourier series in the limit as L>infty. Replace the discrete [tex]A_{n} [/tex] with the continuous F(k)dk while letting n/L>k. Then change the sum to an integral, and the equations become [tex] f(x) = \int_{\infty}^{\infty} F(k)e^{2\pi ikx}dk \\ F(k) = \int_{\infty}^{\infty} f(x) e^{2\pi ikx}dx [/tex] " My question was how [tex]A_{n}[/tex] becomes [tex]F(k) d(k) [/tex]. Especially where the dk comes from? [tex] k=n{/}L [/tex]; then [tex]\Delta k = (n+1){/}L n{/}L =1{/}L [/tex] , when [tex]L [/tex] goes to infinity [tex]\Delta k [/tex]becomes dk. So when [tex] L \rightarrow \infty; A_{n} =F(k)dk [/tex]. Indeed! I am a happy man now :). 


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