SU-MATH53 MAR042024 — Jemoka Knowledge Base

What if, Fourier Series, but exponential? This also motivates Discrete Fourier Transform. Also Complex Exponential. Review Recall again that if we have a periodic function, we’ve got:

\begin{equation} f(x) = \sum_{k=0}^{\infty} a_{k} \sin \left( \frac{2\pi k}{l} x\right) + b_{n} \cos \left( \frac{2\pi k x}{L}\right) \end{equation}

We note that this breaks individually into the sign and cosine series depending of the function’s oddness. Complex Fourier Series This will begin by feeling like a notation rewrite:

\begin{equation} f(x) = \sum_{-\infty}^{\infty} c_{n} e^{n \omega x i} \end{equation}

where \omega = \frac{2\pi}{L}. Why is this summing from negative to positive? Consider:

\begin{equation} \cos \left(nx\right) = \frac{e^{inx}+e^{-inx}}{2} \end{equation}

You will note that summing n \in 0 … \infty, plugging it into above, will result in summing from both n \in -\infty … \infty. Finding c_{n} Recall that complex exponentials are orthonormal + inner product over complex-valued functions Because most cancels except one thing, we get:

\begin{equation} \langle f, e^{i\omega n x} \rangle = c_{n} L \end{equation}

meaning:

\begin{equation} c_{n} = \frac{1}{L} \int_{0}^{L} f(x) e^{-i\omega n x} \dd{x} = \frac{1}{L} \int_{\frac{-L}{2}}^{\frac{L}{2}} f(x) e^{-i\omega n x} \dd{x} \end{equation}

if our function is L periodic. NOTE: this integral has a NEGATIVE power vs the series has a POSITIVE power!! Complex Exponentials with Sawtooth Consider:

\begin{equation} f(x) = x-n \end{equation}

where this function is periodic over n \leq x \leq n+1, so—

\begin{equation} c_{n} = \int_{0}^{1} x e^{-2\pi i n x} \dd{x} = -\frac{1}{2\pi i n} e^{-2 \pi i n} \end{equation}