Physics 434, 2012: Lecture 17

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Back to Physics 434, 2012: Information Processing in Biology. Here we introduce the idea of Fourier series and Fourier transforms. We have discussed in the previous lecture why we need them.

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1. Consider a function $f(t)$ periodic on $-\pi \leq t\leq \pi$.
2. We would like to approximate this function as $f(t)=\sum _{k=0}^{N}a_{k}\cos kt+\sum _{k=1}^{N}b_{k}\sin kt$, takin $N\to \infty$ at some point.
3. From this expression, we can find the coefficients $a_{k},b_{k}$ self-consistently. Indeed, let's multiply the equation by $\cos mt$ and integrate from $-\pi$ to $\pi$.
4. All terms containing products $\cos kt\sin mt$ are zero.
5. For the $\cos kt\cos mt$ terms, we have $\cos kt\cos mt=1/2(\cos(k+m)t+\cos(k-m)t)$.
6. Completing the integrals, we have: $\int _{-\pi }^{\pi }dtf(t)\cos mt=\sum _{m}\int _{-\pi }^{\pi }dt{\frac {a_{k}}{2}}\left(\cos(k+m)t+\cos(k-m)t\right)=\sum _{m}\pi a_{k}(\delta _{k,m}+\delta _{k,0}\delta _{m,0})=\pi a_{m}(1+\delta _{m,0}$.
7. We can do similar to find $b_{k}$: multiply by $\cos mt$, and integrate.
8. This gives: $a_{k>0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }dtf(t)\cos k_{t}$, $a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }dtf(t)$, $b_{k>0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }dtf(t)\sin k_{t}$.