# Fourier series of cycloid

What is the Fourier series representation of a cycloid? The parametric representation of the curve is as follows. $$t=\dfrac{\theta-\sin\theta}{\pi}\\ x=\dfrac{1-\cos\theta}{\pi}$$

The period is $$2$$, so the coefficients of the complex exponential Fourier series should be $$c_n=\dfrac{1}{2}\int_0^2\!x(t)e^{-jn\pi t}\,\mathrm{d}t$$ and that's where I got stuck. I performed substitution $$t=\dfrac{\theta-\sin\theta}{\pi}\implies\mathrm{d}t=\dfrac{1-\cos\theta}{\pi}\,\mathrm{d}\theta\\ c_n=\dfrac{1}{2}\int_0^{2\pi}\!\left(\dfrac{1-\cos\theta}{\pi}\right)^2e^{-jn(\theta-\sin\theta)}\,\mathrm{d}\theta$$ but don't know how to integrate this monster. Is there any other way?

The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind:

$$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$

Using the cycloid parameterization

$$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$

which results in a period of $$2\pi$$ and a maximum value of $$2$$, the Fourier series of $$y(t)$$ as a function of $$x$$, referred to as $$f(x)$$, is given by

\bbox[#f8f1ea, 0.6em, border: 0.15em solid #fd8105]{ \begin{align}f(x)&=\frac32+\sum_{n=1}^{\infty}\frac{J_{n+1}(n)-J_{n-1}(n)}{n}\cos(nx)\\&=\frac32-2\sum_{n=1}^{\infty}\frac{J'_{n}(n)}{n}\cos(nx)\end{align}}\tag{3}

where $$J'_n(x)$$ is the derivative of $$J_n(x)$$ w.r.t. $$x$$.

The figure below shows the cycloid and its Fourier series approximation according to $$(3)$$ using the first $$20$$ coefficients in the sum: Proof:

Let's consider the real-valued Fourier series. Since $$f(x)$$ is even, all sine coefficients vanish and we get

$$f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(nx)\tag{4}$$

with

$$a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx\tag{5}$$

and

$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx,\qquad n>0\tag{6}$$

With $$dx=(1-\cos t)dt$$ we obtain

$$a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}y(t)(1-\cos t)dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}(1-\cos t)^2dt=\frac32\tag{7}$$

and

\begin{align}a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}(1-\cos t)^2\cos(nt-n\sin t) dt\\&=\frac{2}{\pi}\int_{0}^{\pi}(1-\cos t)^2\cos(nt-n\sin t) dt,\qquad n>0\tag{8}\end{align}

where the last equality follows from the fact that the integrand is even.

Expanding

$$(1-\cos t)^2=1-2\cos t+\cos^2 t=\frac32-2\cos t+\frac12\cos 2t\tag{9}$$

and using $$\cos(\alpha)\cos(\beta)=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$$, the integrand in $$(8)$$ can be rewritten as

\begin{align}(1-\cos t)^2\cos(nt-n\sin t)=\frac32\cos(nt-n\sin t) -&\\\big[\cos((n-1)t-n\sin t) + \cos((n+1)t-n\sin t)\big]+\\\frac14 \big[\cos((n-2)t-n\sin t) + \cos((n+2)t-n\sin t)\big]\tag{10}\end{align}

Plugging $$(10)$$ into $$(8)$$ and using the definition of the Bessel function $$(1)$$, we can write

$$a_n=3J_n(n)-2\big[J_{n-1}(n)+J_{n+1}(n)\big]+\frac12\big[J_{n-2}(n)+J_{n+2}(n)\big],\qquad n>0\tag{11}$$

This can be further simplified using the recurrence relation (10.6.1) for $$J_n(x)$$:

$$J_{n-1}(n)+J_{n+1}(n)=\frac{2n}{x}J_n(x)\tag{12}$$

which can be used to eliminate $$J_{n-2}(n)$$ and $$J_{n+2}(n)$$ in $$(11)$$, and which finally results in

$$a_n=\frac{J_{n+1}(n)-J_{n-1}(n)}{n},\qquad n>0\tag{13}$$

Since (10.6.1)

$$J_{n-1}(x)-J_{n+1}(x)=2J'_n(x)\tag{14}$$

where $$J'_n(x)$$ denotes the derivative of $$J_n(x)$$ with respect to $$x$$, the coefficients $$a_n$$ can also be expressed as

$$a_n=-\frac{2J'_n(n)}{n},\qquad n>0\tag{15}$$

Eqs $$(7)$$, $$(13)$$, and $$(15)$$ establish the result $$(3)$$.

I haven't managed to completely evaluate the integral, but I've made some progress and perhaps someone can pick up where I leave off.

The integral you gave is $$c_n = \int_0^{2\pi}\left(\frac{1-\cos\theta}{\pi}\right)^2 e^{-jn(\theta-\sin\theta)}d\theta.$$ The first thing I did is perform a Taylor expansion on that exponential term; $$c_n = \frac{1}{\pi^2}\int_0^{2\pi}(1-\cos\theta)^2\sum_{k=0}^\infty \frac{(-jn)^k}{k!}(\theta-\sin\theta)^kd\theta.$$ According to wolfram alpha, $$1-\cos\theta=2\sin^2\theta$$. So you can write $$c_n = \frac{4}{\pi^2}\int_0^{2\pi}\sin^4\theta\sum_{k=0}^\infty \frac{(-jn)^k}{k!}(\theta-\sin\theta)^kd\theta.$$ A power of a sum like $$(\theta-\sin\theta)^k$$ can be expressed using a binomial expansion: $$(\theta-\sin\theta)^k = \sum_{m=0}^k \begin{pmatrix}k\\ m\end{pmatrix}\theta^m(-\sin\theta)^{k-m}$$ where $$\begin{pmatrix}k\\ m\end{pmatrix}$$ is the binomial coefficient.

Sums and integrals can commute, so we can write $$c_n = \frac{4}{\pi^2}\sum_{k=0}^\infty \frac{(-jn)^k}{k!}\sum_{m=0}^k \begin{pmatrix}k\\ m\end{pmatrix}(-1)^{k-m}\int_0^{2\pi} \theta^m \sin^{k-m+4} \theta d\theta$$ where I have used $$(-\sin\theta)^4=\sin^4\theta$$.

This is where I run out of skill. If anyone knows anything about the integral please feel free to add to my answer.