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nemenman>Ilya at 13:21, 1 November 2012

2012-11-01T13:21:58Z

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{{PHYS434-2012}}
Here we introduce the idea of Fourier series and Fourier transforms. We have discussed in the previous lecture why we need them.

===Main Lecture===
# Consider a function <math>f(t)</math> periodic on <math>-\pi\le t\le \pi</math>.
# We would like to approximate this function as <math>f(t)=\sum_{k=0}^N a_k \cos kt + \sum_{k=1}^N b_k \sin kt</math>, takin <math>N\to\infty</math> at some point.
# From this expression, we can find the coefficients <math> a_k, b_k</math> self-consistently. Indeed, let's multiply the equation by <math>\cos mt</math> and integrate from <math>-\pi</math> to <math>\pi</math>.
#All terms containing products <math>\cos kt\sin mt</math> are zero.
#For the <math>\cos kt\cos mt</math> terms, we have <math>\cos kt\cos mt=1/2 ( \cos (k+m)t +\cos(k-m)t)</math>.
#Completing the integrals, we have: <math>\int_{-\pi}^{\pi}dt f(t) \cos mt= \sum_m \int_{-\pi}^{\pi} dt \frac{a_k}{2} \left( \cos (k+m)t +\cos(k-m)t\right) = \sum_m \pi a_k (\delta_{k,m}+\delta_{k,0}\delta_{m,0})=\pi a_m (1+\delta_{m,0}</math>.
#We can do similar to find <math>b_k</math>: multiply by <math>\cos mt</math>, and integrate.
#This gives: <math>a_{k>0}= \frac{1}{\pi}\int_{-\pi}^{\pi}dt f(t)\cos k_t</math>, <math>a_{0}= \frac{1}{2\pi}\int_{-\pi}^{\pi}dt f(t)</math>, <math>b_{k>0}= \frac{1}{\pi}\int_{-\pi}^{\pi}dt f(t)\sin k_t</math>.

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Physics 434, 2012: Lecture 17 - Revision history

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nemenman>Ilya at 13:21, 1 November 2012