On an intuitively motivated proof of Fourier series of a function?

Question

Consider a function such as

$sin^3 (x) cosx$

How would one find the Fourier series of this?

I have read of Fourier trick from a physics book "electricity and magnetism by Griffith" where he goes over solving laplace equation using a Fourier trick. And there it seems that the principle works due to ideas related to solutions of differential equations. However how would one apply this to some periodic function?

I have seen the formula of Fourier series,however I can not understand the idea behind each term. IF someone cone could provide a simple intuitive explanation of it, it would be awesome

Answer 1

As Don Antonio as spoken about, Fourier series exist for periodic functions. However, if you are only trying to represent a function $f$ on a finite interval $[a,b]$, you can just pretend that $f$ is periodic and $[a,b]$ is contained in one period.

For simplicity, assume that $f$ has period $2\pi$, and you want to represent it on $[-\pi, \pi]$. If $f$ is representable as a Fourier series, then $$f(\theta) = \sum_{n=0}^\infty A_n\cos n\theta + \sum_{n=1}^\infty B_n\sin n\theta$$

Now there are some useful trigonometric integral formulas for integer $m,n$: $$\int_{-\pi}^\pi\cos n\theta\,d\theta = \begin{cases}0& n \ne 0\\2\pi & n = 0\end{cases}$$ $$\int_{-\pi}^\pi\sin n\theta\,d\theta = 0$$ $$\int_{-\pi}^\pi\cos n\theta\cos m\theta\,d\theta =\int_{-\pi}^\pi\sin n\theta\sin m\theta\,d\theta = \begin{cases}0& n\ne m\\\pi & n = m\ne 0\end{cases}$$ $$\int_{-\pi}^\pi\cos n\theta\sin m\theta\,d\theta = 0$$ (Because sine and cosine are periodic, these actually hold for any interval of integration of width $2\pi$.) This happens because other than the exceptional cases, the integrands are all sinusoidal with $y=0$ as the midline, and symmetry matches the area below the midline to the area above, so it all cancels out.

So if we integrate $$\begin{align}\int_{-\pi}^\pi f(\theta)\,d\theta &= \sum_{n=0}^\infty A_n\int_{-\pi}^\pi\cos n\theta\,d\theta + \sum_{n=1}^\infty B_n\int_{-\pi}^\pi\sin n\theta\,d\theta \\&= 2\pi A_0 + 0\end{align}$$ and for $m > 0$, $$\begin{align}\int_{-\pi}^\pi f(\theta)\cos m\theta\,d\theta &= \sum_{n=0}^\infty A_n\int_{-\pi}^\pi\cos n\theta\cos m\theta\,d\theta + \sum_{n=1}^\infty B_n\int_{-\pi}^\pi\sin n\theta\cos m\theta\,d\theta\\&=A_m\pi + 0\end{align}$$ $$\begin{align}\int_{-\pi}^\pi f(\theta)\sin m\theta\,d\theta &= \sum_{n=0}^\infty A_n\int_{-\pi}^\pi\cos n\theta\sin m\theta\,d\theta + \sum_{n=1}^\infty B_n\int_{-\pi}^\pi\sin n\theta\sin m\theta\,d\theta\\&=0 + B_m\pi\end{align}$$

So you can find the coefficients by $$A_0 = \frac 1{2\pi} \int_{-\pi}^\pi f(\theta)\,d\theta\\ A_m = \frac 1\pi \int_{-\pi}^\pi f(\theta)\cos m\theta\,d\theta, \quad m > 0$$ and $$B_m = \frac 1\pi \int_{-\pi}^\pi f(\theta)\sin m\theta\,d\theta, \quad m > 0$$

For functions with different periods, you just have to introduce a scaling factor.

Answer 2

Computing a Fourier series is about expanding a function into linear combination of exponentials (or sines and cosines) with frequencies which are multiple of a dominant one.

But your function is already in this form, just a little in disguise perhaps:

\begin{align} \sin(x)^3 \cos(x) &= \left ( \frac{e^{ix} - e^{-ix}}{2i} \right)^3 \left ( \frac{e^{ix} + e^{-ix}}{2}\right) \\ &= \frac{1}{8} i e^{-2 i x}-\frac{1}{8} i e^{2 i x}-\frac{1}{16} i e^{-4 i x}+\frac{1}{16} i e^{4 i x} \\ &= \frac{1}{4} \sin (2 x)-\frac{1}{8} \sin (4 x) \end{align}

Added

I see that this was not exactly what you were asking, which is a bit vague. You mention a trick used to solve differential equations with Fourier methods.

One trick that is commonly used in this context is the following, perhaps better explained with Fourier integral transform. It boils down to the following fact. Given

$$ \mathcal{F}(f)(k) = \int_{-\infty}^\infty dx \frac{e^{-ikx}}{\sqrt{2\pi}} f(x) $$

Then for sufficiently nice $f$ (i.e. $f\in\mathscr{S}$)

\begin{align} \left (i \frac{d}{dk} \right )^n \mathcal{F}(f) &= \mathcal{F}( x^n f(x)) \\ \mathcal{F} \left ( \left ( -i \frac{d}{dx} \right )^n f \right)(k) &= k^n \mathcal{F}(f)(k) \end{align}

With these formulae one can transform a (partial) differential equation to an algebraic equation. This is how Fourier solved the heat equation.