# 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?

• According to wolfram alpha, $1−cosθ=2sin2θ$ Here is a mistake: $1−cos(2θ) = 2*(sinθ)^2$ Commented Apr 17 at 14:25

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 $$J_n(x)$$:

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

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.