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I want to compute the (even) fourier coefficients for $\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.

Namely, I want to compute the integral: $$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$

Integrating this in mathematica, it seems like $b_n = 0$ if $n$ is odd and $b_n = \left(\frac 1 2 \right)^{n/2}$ for $n$ even.

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    The $b_n=0$ when $n$ is odd is easy to see, becauseyour function has period $\pi$, which means that $f(\theta+\pi)=f(\theta)$, but when $n$ is odd, $\cos(n(\theta+\pi))=-\cos n\theta$. – 2017-02-01
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    Not sure how to do it directly, but it is true that the value is $\frac{1}{2^{n/2}}$ when $n$ is even. I can show this using complex Fourier coefficients... – 2017-02-01

2 Answers 2

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Complex analysis approach:

Let $$f(z)=\frac{3}{5-2(z^2+z^{-2})}=\frac{-3z^2}{2z^4-5z^2+2}=\frac{1}{2z^2-1}-\frac{2}{z^2-2}$$ Then $f(e^{i\theta})=\frac{3}{5-4\cos 2\theta}$.

Now, for $|z|<\sqrt{2}$, we have:

$$\frac{-2}{z^2-2}=\frac{1}{1-z^2/2}=\sum_{k=0}^{\infty} \frac{1}{2^k}z^{2k}$$

and for $|z|>\frac{1}{\sqrt{2}}$ you have:

$$\frac{1}{2z^2-1}=\frac{1}{2z^2}\frac{1}{1-\frac{1}{2z^2}}=\sum_{k=1}^{\infty}\frac{1}{2^k}z^{-2k}$$

Since $z=e^{i\theta}$ is in this range, this means that:

$$\frac{3}{5-4\cos 2\theta} = \sum_{k=\infty}^\infty \frac{1}{2^{|k|}}e^{i2k\theta}$$

Combining $k$ and $-k$ terms, this gives:

$$\frac{3}{5-4\cos2\theta} = 1 + \sum_{k=1}^{\infty} \frac{1}{2^{k-1}}\cos 2k\theta$$

I might have missed something in this computation, but this is close to your answer - the coefficients are $0$ if $n$ is odd and $\frac{1}{2^{k-1}}$ if $n=2k$.

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    Hi Thomas. Thanks so much for your help! Your final expression of $1 + \sum_{k=1}^\infty \frac{1}{2^{k-1}}\cos(2k\theta)$ is indeed correct. The missing factor of 1/2 comes from the actual integration. Namely, when we now integrate $\int_0^{2\pi} \frac{1}{2\pi} \cos(n\theta) (1+\sum_{k=1} \frac{1}{2^{k-1}}\cos(2k\theta)) d\theta = \int \frac{1}{2}^{\frac{n}{2}} \frac{1}{\pi}\cos(n\theta)^2d\theta = \frac{1}{2}^{\frac{n}{2}}$. – 2017-02-01
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Hint: Observe \begin{align} \frac{3}{5-4\cos 2\theta} = \frac{3/5}{1-\frac{4}{5}\cos 2\theta} = \frac{3}{5}\sum^\infty_{n=0} \left(\frac{4}{5}\right)^n \cos^n 2\theta \end{align} and \begin{align} \cos^n 2\theta = \left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)^n = \frac{1}{2^n}\sum^n_{k=0}\binom{n}{k} e^{2ik\theta}e^{-2i(n-k)\theta}. \end{align} Thus \begin{align} \int^{2\pi}_0 e^{2ik\theta}e^{-2i(n-k)\theta} \frac{e^{in\theta}+e^{-in\theta}}{2}\ dx = \text{ something }. \end{align}

Of course, you could have just consider the integral \begin{align} \int^{2\pi}_0 \cos(n\theta) \cos^{n} 2\theta\ \frac{d\theta}{2\pi}. \end{align}

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    Hi Jacky, thanks for your help. I tried expanding the series using the binomial expansion and I wasn't getting nice cancellations, as I got two ($k=\frac{n}{4}$ and $k=\frac{3n}{4}$) terms which looked a bit messy. The complex analysis approach below seemed a little bit easier. – 2017-02-01