$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$ Find period and Fourier expansion

Question

Given the following function: $$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$$ Find period and Fourier expansion.

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I think the period is $T=4 \pi$, observing the functions.

As for the Fourier expansion I have no clue on how to proceed. I know that $$f(t)= \pi a_0 + \sum_{k=1}^\infty a_k \cos(\frac{kt}{2})$$ $$a_k=\frac{1}{2\pi}\int_{-2\pi}^{2\pi} dt f(t)\cos(kt/2)=\frac{1}{2\pi}\int_{-2\pi}^{2\pi} dt \frac{\cos(\frac{3}{2}t) \cos(kt/2)}{1+\cos^2(\frac{t}{4})}=\frac{1}{\pi}\int_{0}^{2\pi} dt \frac{\cos(\frac{3}{2}t) \cos(kt/2)}{1+\cos^2(\frac{t}{4})}$$ Even with the residue theorem the integral looks really bad since there is that $k \in \mathbb{Z}$... Do I really have to compute this thing?

Answer 1

There is a way through this. $$\begin{align}f(t)&=\frac{4\cos\left(3\frac t2\right)}{4+4\cos^2\left(\frac t4\right)}=\frac{4\cos\left(3\frac t2\right)}{4+\left(e^{i\frac t4}+e^{-i\frac t4}\right)^2}\label{a}\tag{1}\\ &=\frac{4\cos\left(3\frac t2\right)}{e^{-i\frac t2}\left(e^{it}+6e^{i\frac t2}+1\right)}=\frac{4\cos\left(3\frac t2\right)}{z^{-1}\left(z^2+6z+1\right)}\label{b}\tag{2}\\ &=\frac{4\cos\left(3\frac t2\right)}{z^{-1}\left(z+\left(1+\sqrt2\right)^2\right)\left(z+\left(1+\sqrt2\right)^{-2}\right)}\label{c}\tag{3}\\ &=\frac{4\cos\left(3\frac t2\right)}{4\sqrt2z^{-1}}\left(\frac1{z+\left(1+\sqrt2\right)^{-2}}-\frac1{z+\left(1+\sqrt2\right)^2}\right)\label{d}\tag{4}\\ &=\frac{\cos\left(3\frac t2\right)}{\sqrt2}\left(\frac1{1+\left(1+\sqrt2\right)^{-2}z^{-1}}-\frac z{\left(1+\sqrt2\right)^2\left(1+\left(1+\sqrt2\right)^{-2}z\right)}\right)\label{e}\tag{5}\\ &=\frac{\cos\left(3\frac t2\right)}{\sqrt2}\left(\sum_{k=0}^{\infty}\frac{(-1)^kz^{-k}}{\left(1+\sqrt2\right)^{2k}}-\frac z{\left(1+\sqrt2\right)^2}\sum_{k=0}^{\infty}\frac{(-1)^kz^k}{\left(1+\sqrt2\right)^{2k}}\right)\label{f}\tag{6}\\ &=\frac{\cos\left(3\frac t2\right)}{\sqrt2}\left(1+\sum_{k=1}^{\infty}\frac{(-1)^kz^{-k}}{\left(1+\sqrt2\right)^{2k}}+\sum_{k=1}^{\infty}\frac{(-1)^kz^k}{\left(1+\sqrt2\right)^{2k}}\right)\label{g}\tag{7}\\ &=\frac{\cos\left(3\frac t2\right)}{\sqrt2}\left(1+2\sum_{k=1}^{\infty}\frac{(-1)^k\cos\left(k\frac t2\right)}{\left(1+\sqrt2\right)^{2k}}\right)\label{h}\tag{8}\\ &=\frac1{\sqrt2}\left(\cos\left(3\frac t2\right)+\sum_{k=1}^{\infty}\frac{(-1)^k}{\left(1+\sqrt2\right)^{2k}}\left(\cos\left((k+3)\frac t2\right)+\cos\left((k-3)\frac t2\right)\right)\right)\label{i}\tag{9}\\ &=\frac1{\sqrt2\left(1+\sqrt2\right)^6}\left(-1+6\cos\left(\frac t2\right)-34\cos\left(2\frac t2\right)\right)\\ &\quad-99\sqrt2\sum_{k=3}^{\infty}\frac{(-1)^k}{\left(1+\sqrt2\right)^{2k}}\cos\left(k\frac t2\right)\label{j}\tag{10}\end{align}$$ $(\ref{a})$ Convert $\cos\left(\frac t4\right)$ to complex exponential
$(\ref{b})$ Square out, substitute $z=\exp\left(i\frac t2\right)$ and isolate classic polynomial in $z$
$(\ref{c})$ Factor polynomial from equation $(\ref{b})$
$(\ref{d})$ Partial fractions decomposition
$(\ref{e})$ Ready for geometric series expansion!
$(\ref{f})$ Geometric series expansion away SIR!
$(\ref{g})$ Pull out $k=0$ term of $z^{-k}$ sum and multiply $z^k$ sum by its prefactor
$(\ref{h})$ Consolidate $z^{-k}+z^k=\exp(-ikt/2)+\exp(ikt/2)=2\cos(kt/2)$
$(\ref{i})$ Multiply through by $\cos(3t/2)$ and use $2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)$ to make it look like a Fourier series
$(\ref{j})$ Clean up Fourier series a little

The same technique may be used IIRC to establish for the radius of a Kepler orbit $$r=\frac{a\left(1-e^2\right)}{1+e\cos\theta}=a\sqrt{1-e^2}\left(1+2\sum_{k=1}^{\infty}\frac{(-1)^k}{\left(\frac{1+\sqrt{1-e^2}}e\right)^k}\cos(k\theta)\right)$$ EDIT: Just to illustrate the disconnect between the line of the original question that asked for $$f(t)=a_0+\sum_{k=1}^{\infty}a_k\cos\left(\frac{kt}2\right)$$ And the next line that computed the coefficients of $$\bar f(t)=\bar a_0+\sum_{k=1}^{\infty}\left(\bar a_k\cos\left(kt\right)+\bar b_k\sin\left(kt\right)\right)$$ We present a Matlab program that plots the original function on $[0,2\pi]$ along with the Fourier cosine series that is the periodic extension of $\bar f$ as an even function and the total Fourier series for $\bar f$, Gibbs phenomenon and all.

% fourier.m

clear all;
close all;
g = @(t) cos(3/2*t)./(1+cos(t/4).^2);
t= linspace(0,2*pi,300);
nf = 100;
for k = 1:nf,
    h = @(t) g(t).*cos(k*t);
    a(k) = integral(h,0,2*pi)/pi;
    j = @(t) g(t).*sin(k*t);
    b(k) = integral(j,0,2*pi)/pi;
end
a0 = integral(g,0,2*pi)/(2*pi);
f = g(t);
for n = 1:length(t),
    e(n)= a0+cos([1:nf]*t(n))*a';
    d(n) = e(n)+sin([1:nf]*t(n))*b';
end
plot(t,f,'k-',t,e,'b-',t,d,'r-');
title('Fourier Series for Problem');
xlabel('t');
ylabel('y');
legend('f(t)','Cosine series','Total series','Location','Best');

Answer 2

I will still show the evaluation of the integrals, since it's not always possible to use tricks, sometimes we need the general methods as well. (For what it's worth, in this case the integrals are very simple).

Edited because the period of the function is $4 \pi$ the correct expression should be:

$$a_k=\frac{1}{\pi}\int_{0}^{2\pi} \frac{\cos(\frac{3}{2}t) \cos \left( \frac{kt}{2} \right)}{1+\cos^2(\frac{t}{4})}dt=\frac{4}{\pi }\int_{0}^{\pi} \frac{\cos(3x) \cos(kx)}{3+\cos x}dx=$$

Now the residue theorem is the most convenient way here, in my opinion, so we transform the integral the following way:

$$=\frac{2}{\pi } \Re \int_{-\pi}^{\pi} \frac{\cos(3x) e^{ikx}}{3+\cos x}dx=\frac{2}{\pi } \Re \int_{-\pi}^{\pi} \frac{\left(e^{3ix}+e^{-3ix} \right) e^{ikx}}{6+e^{ix}+e^{-ix}}dx=$$

Setting $z=e^{ix}$ and choosing the unit circle as the contour, we have:

$$=\frac{2}{\pi } \Re \left( -i \oint \frac{\left(z^3+z^{-3} \right) z^{k-1}}{6+z+z^{-1}}dz \right)=\frac{2}{\pi } \Re \left( -i \oint \frac{\left(z^6+1 \right) z^{k-3}}{6z+z^2+1}dz \right)=$$

We assume $k>2$, the cases $k=0,1,2$ can be dealt with separately (there will be an additional pole at $z=0$ of order $1,2,3$). For other cases, we need to consider only the poles coming from the denominator:

$$z^2+6z+1=0$$

$$z= \pm 2 \sqrt{2}-3$$

Of the roots, only one lies inside the unit circle ($z= 2 \sqrt{2}-3$), so we only need to compute one residue:

$$=\frac{2}{\pi } \Re \left( 2 \pi \frac{\left((2 \sqrt{2}-3)^6+1 \right) (2 \sqrt{2}-3)^{k-3}}{2 \sqrt{2}-3+2 \sqrt{2}+3} \right)=\frac{(2 \sqrt{2}-3)^{k-3}+(2 \sqrt{2}-3)^{k+3}}{\sqrt{2}} $$

So, finally, for $k>2$, $k \in \mathbb{Z}$ we have:

$$\frac{1}{\pi}\int_{0}^{2\pi} \frac{\cos(\frac{3}{2}t) \cos \left( \frac{kt}{2} \right)}{1+\cos^2(\frac{t}{4})}dt=\frac{(2 \sqrt{2}-3)^{k-3}+(2 \sqrt{2}-3)^{k+3}}{\sqrt{2}} $$

Note that:

$$2 \sqrt{2}-3=\frac{1-\sqrt{2}}{1+\sqrt{2}}=-\frac{1}{(1+\sqrt{2})^2}$$