I know a function $f(x)$, how can I show it's equal to a known Fourier series, using geometric series

Question

I have a function $$f(t) = \frac{1}{4}\left(\frac{3}{e^{i\,t}+3} -\frac{1}{3e^{i\,t}+2}\right)$$

I want to show that it's Fourier series is equal to

$$f(t) = \frac{1}{4}\sum^\infty_{k=-\infty}\left(\frac{-1}{3}\right)^{|k|}e^{i\,k\,t} = \frac{1}{4}-\frac{1}{6}\cos t+\frac{1}{18}\cos 2t - \frac{1}{54}\cos 3t$$

---

I'm thinking somehow geometric series is involved...

$$\sum^\infty_{k=0}z^k=\frac{1}{1-z},\,|z| < 1$$

---

Update 1

By re-writing the term $\frac{3}{e^{i\,t}+3}$, and using identifying the geometric series, we get

$$\frac{1}{1+\frac{e^{i\,t}}{3}} = \sum^\infty_{k=0}\left(\frac{-e^{i\,t}}{3}\right)^k = \sum^\infty_{k=0}\left(\frac{-1}{3}\right)^ke^{i\,k\,t}$$

This starts to look like the Fourier series we want to get to, how can I do the second term? Any idea, I'm really stuck on that one.

Update 2

It appears the textbook had a text-error at the second term. It's supposed to say $-\frac{1}{3e^{i\,t}+1}$ which makes it easy to see it's a geometric series too.

$$-\frac{1}{3e^{i\,t}+1} = \sum^\infty_{k=0}(-3)^ke^{i\,k\,t}$$

$$\implies F[t] = \frac{1}{8\pi}\int^{2\pi}_0\left (\sum^\infty_{k=0}\left(\frac{-1}{3}\right)^ke^{i\,k\,t} - \sum^\infty_{k=0}(-3)^ke^{i\,k\,t} \right)e^{-i\,k\,t}\,dt \\ = \frac{1}{8\pi}\int^{2\pi}_0\left (\sum^\infty_{k=0}\left(\frac{-1}{3}\right)^k\cdot 1 - \sum^\infty_{k=0}(-3)^k\cdot 1 \right)\,dt$$

This is where it's at now. Maybe the second term isn't a geometric series after all, considering that it diverges our integral at the moment.

Answer 1

The series representation

$$f(t)=\frac14\sum_{k=-\infty}^{\infty}\left(-\frac13\right)^{|k|}e^{ikt}$$

evaluates to

$$\begin{align}f(t)&=\frac14\left[\sum_{k=0}^{\infty}\left(-\frac{e^{it}}{3}\right)^k+\sum_{k=0}^{\infty}\left(-\frac{e^{-it}}{3}\right)^k-1\right]\\&=\frac14\left[\frac{1}{1+\frac{e^{it}}{3}}+\frac{1}{1+\frac{e^{-it}}{3}}-1\right]\\&= \frac14\left[\frac{3}{3+e^{it}}-\frac{\frac{e^{-it}}{3}}{1+\frac{e^{-it}}{3}}\right]\\&=\frac14\left[\frac{3}{3+e^{it}}-\frac{1}{1+3e^{it}}\right]\end{align}$$

which is close, but not equal, to the original function given at the top of your question. So the Fourier series of the given function is different from the series in your question.

Answer 2

You can use the fact that the Fourier Transform is unique and linear together with geometric series expansion on each term separately to see that each term in it's Fourier expansion must be the sum of two separate geometric series. If one dominates the other for each term it may look like almost one series.