# Finding the Fourier series of $\frac{4-2\cos x}{5-4\cos x}$

I am required to find the Fourier Series of the following function, on $$[-\pi,\pi]$$:

$$f(x)=\frac{4-2\cos(x)}{5-4\cos(x)}$$

This question seemed simple, at first. I managed to compute $$a_0$$ (which wasn't too easy) and noticed that $$b_n=0$$ for all $$n\in\mathbb{N}$$ (since the given $$f$$ is even). However, computing the coefficients $$a_n$$ seemed like an impossible mission. So I cheated a little, and looked at the final answer, which is:

$$f(x)\simeq1+\sum_{n=1}^{\infty}\frac{1}{2^n}\cos(nx)\triangleq\mathfrak{F}(x)$$

The problem is $$a_n$$ - I don't know how to compute them, and therefore I couldn't find the Fourier Series by myself. However, after finding out what the final answer is, I thought of a creative way to show the final answer is actually the Fourier Series of $$f(x)$$. This is kind of cheating, but I'll show my solution anyway:

Let $$z=\frac{\cos(x)+i\sin(x)}{2}, z\in\mathbb{C}.$$ Using the fact that $$z^n=\frac{\cos(nx)+i\sin(nx)}{2^n}$$, and that $$|z|=\frac 12<1$$, we receive:

$$\mathfrak{F}(x)=\sum_{n=0}^{\infty}\Re{(z^n)}=\Re{\left(\sum_{n=0}^{\infty}z^n\right)}=\Re{\left(\frac{1}{1-z}\right)}$$$$=\Re{\left(\frac{1-\bar{z}}{|1-z|^2}\right)}=\Re{\left(\frac{1-0.5\cos(x)+0.5i\sin(x)}{(1-0.5\cos(x))^2+(0.5\sin(x))^2}\right)}$$$$=\frac{1-0.5\cos(x)}{1-\cos(x)+0.25\cos^2(x)+0.25\sin^2(x)}=\frac{4-2\cos(x)}{5-4\cos(x)}\equiv f(x)$$

We know that $$f(x)$$ is differentiable everywhere. Therefore, it has only one Fourier Series, and it has to be $$\mathfrak{F}(x)$$, since it is a Fourier Series that converges to $$f(x)$$. This might not be a true statement (so please correct me if so), but either way, I would like to hear your thoughts about a more legit solution to the problem (without knowing the Fourier Series in advance).

• Do you know the residue theorem in complex analysis? If so you can define $z=e^{ix}$ and proceed automatically from there (you get poles at $0$ and $\frac12$). – John B Dec 23 '19 at 21:23
• @JohnB Yes I do know the theorem, but I've never seen it in the context of finding Fourier Series. I can see why defining $z=e^{ix}$ would help, but only if I know the series in advance. Unfortunately I can't see how would it help to find the Fourier Series of $f(x)$ – Amit Zach Dec 23 '19 at 21:26
• Just use convolution theorem backwards. – mathreadler Dec 23 '19 at 21:41

Write $$2\cos(x) = z+z^{-1}$$ for some $$\lvert z \rvert = 1$$. Then $$\frac{4-2\cos(x)}{5-4\cos(x)} = \frac{4-z-z^{-1}}{5-2z-2z^{-1}}=\frac12 + \frac1{2-z} + \frac1{4z-2}.$$ Since $$\lvert z \rvert = 1$$ this expands as $$\frac12 + \frac12 \sum_{n=0}^{\infty}\frac{z^n}{2^n} + \frac12 \sum_{n=1}^{\infty}\frac{z^{-n}}{2^n} = 1 + \frac12 \sum_{n=1}^{\infty} \frac{z^n+z^{-n}}{2^n}.$$ The Fourier series is apparent from this expansion.

• Brilliant! Thank you very much! – Amit Zach Dec 24 '19 at 7:26

Notice that $$\frac1\pi\int_{-\pi}^\pi \frac{4-2\cos x}{5-4\cos x}\cos(nx)\,dx=-\frac i\pi\int_\gamma g,$$ where $$\gamma\colon[-\pi,\pi]\to\mathbb C$$ is defined by $$\gamma(t)=e^{it}$$ and where $$g(z)=\frac{(z^2-4 z+1) (z^{2 n}+1)}{2 z^{n+1}(2 z^2-5 z+2)}.$$ This is obtained from the identity $$\cos x=(e^{ix}+e^{-ix})/2$$ (remember the term $$\gamma'(t)$$ in the integral). The rest comes from applying the residue theorem more or less automatically.

Here is a noncomplex way to do the integral. Multiply top and bottom by $$5+4\cos x$$:

$$\frac{20+6\cos x +8\cos^2x}{25-16\cos^2 x} = -\frac{1}{2} +\frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x}$$

$$I = \int_{-\pi}^\pi -\frac{1}{2} + \frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x} \:dx = \pi +\frac{1}{2}\arctan\left(\frac{4}{3}\sin x \right)\Biggr|_{-\pi}^\pi + \frac{65}{2}\int_{-\pi}^\pi \frac{\sec^2 x}{9 + 25\tan^2 x} \:dx$$

$$= \pi + 0 + \frac{13}{6}\arctan\left(\frac{5}{3}\tan x \right)\Biggr|_{-\pi}^\pi$$

That last limit is not simple to evaluate because of the singularities, but just take the limits separately to each one. Each interval of $$\pi$$ picks up a value of $$\pi$$ so we have

$$\pi +\frac{13\pi}{3} = \frac{16\pi}{3}$$

First write

$$f(x)=\frac{4-2\cos x}{5-4\cos x}=\frac12+\frac{3}{10-8\cos x}$$

then use the generalization

$$\frac{1}{a+b\cos x}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nx)},\ a>b$$

set $$a=10$$ and $$b=-8$$ to get the Fourier series of $$f(x)$$