# 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.

### Main Lecture

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