Fourier series of cycloid

Question

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?

Answer 1

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 $(3)$.

Answer 2

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.