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Coefficient in a Laurent Expansion in terms of an Integral

shen07

Member
Aug 14, 2013
54
Hi guys, i need your help to go about his question,

Question:

$$\text{Show that the coefficient }C_n \text{in the Laurent expansion of }$$
$$f(z)=(z+\frac{1}{z}) \text{ about z=0 is given by}$$
$$C_n=\frac{1}{2\pi}\int^{2\pi}_0 \cos(2cos(\theta))cos(n\theta)\, d\theta ,n\in\mathbb{z}$$
 

shen07

Member
Aug 14, 2013
54
Hi guys, i need your help to go about his question,

Question:

$$\text{Show that the coefficient }C_n \text{in the Laurent expansion of }$$
$$f(z)=(z+\frac{1}{z}) \text{ about z=0 is given by}$$
$$C_n=\frac{1}{2\pi}\int^{2\pi}_0 \cos(2cos(\theta))cos(n\theta)\, d\theta ,n\in\mathbb{z}$$
the function is actually $$f(z)=cos(z+z^{-1})$$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
the function is actually $$f(z)=cos(z+z^{-1})$$
Hi shen07,

Perhaps you can start by filling in your function into the formula for the Laurent series coefficients?
The formula is:
$$C_n=\frac{1}{2\pi i} \oint_\gamma \frac{f(z)\,\mathrm{d}z}{(z-c)^{n+1}}$$
where the function f(z) is expanded around c and $\gamma$ is a closed counter clockwise curve around c.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Let $\gamma$ be the unit circle.

Then since $\gamma$ lies in a annulus in which $ \displaystyle \cos \left( z + \frac{1}{z} \right)$ is analytic,

$ \displaystyle C_{n} = \frac{1}{2\pi i} \int_{\gamma} \frac{\cos (z + \frac{1}{z})}{z^{n+1}} \ dz = \frac{1}{4 \pi i} \int_{\gamma} \frac{e^{i(z + \frac{1}{z})} + e^{-i(z+ \frac{1}{z})}}{z^{n+1}} \ dz $

Parametrize the contour by letting $z = e^{i \theta}$.

$\displaystyle = \frac{1}{4 \pi i} \int_{-\pi}^{\pi} \frac{e^{i (e^{i\theta}+e^{-i\theta})} + e^{i(e^{i\theta}+e^{-i\theta})}}{e^{i(n+1)\theta}} i e^{i\theta} \ d \theta = \frac{1}{4 \pi} \int_{- \pi}^{\pi} \frac{e^{i (2 \cos \theta)} + e^{-i(2 \cos \theta)}}{e^{i n \theta}} \ d \theta$

$ = \displaystyle \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos(2 \cos \theta) e^{-i n \theta} \ d \theta = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos(2 \cos \theta) \cos(n \theta)$ since the imaginary part of the integrand is odd

$ \displaystyle = \frac{(-1)^{n}}{2 \pi} \int_{0}^{2 \pi} \cos(2 \cos u) \cos(n u) \ du$

But since the integral is zero when $n$ is odd,

$ \displaystyle C_{n} = \frac{1}{2 \pi} \int_{0}^{2 \pi} \cos(2 \cos u) \cos(n u) \ \ du$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$ \displaystyle = \frac{(-1)^{n}}{2 \pi} \int_{0}^{2 \pi} \cos(2 \cos u) \cos(n u) \ du$

But since the integral is zero when $n$ is odd,
How did you deduce that the integral is zero ?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Here's an indirect argument.


Let $n$ be an odd integer.


$ \displaystyle C_{n} = \frac{1}{2 \pi i} \int_{|z|=1} \frac{\cos (z+ \frac{1}{z})}{z^{n+1}} \ dx = \frac{1}{2 \pi i} \text{Re} \int_{|z|=1} \frac{e^{i(z+\frac{1}{z})}}{z^{n+1}} \ dz = \text{Re} \ \text{Res}\Big[ \frac{e^{i(z+\frac{1}{z})}}{z^{n+1}},0 \Big]$


$ \displaystyle \frac{e^{i(z+\frac{1}{z})}}{z^{n+1}} = \frac{1}{z^{n+1}} \sum_{k=0}^{\infty} \frac{[i(z+\frac{1}{z})]^{k}}{k!}= \frac{1}{z^{n+1}} \sum_{k=0}^{\infty} \frac{i^{k}}{k!} \sum_{m=0}^{k} \binom{k}{m} z^{m} \left(\frac{1}{z} \right)^{k-m} $

$ \displaystyle = \sum_{k=0}^{\infty} \frac{i^{k}}{k!} \sum_{m=0}^{k} \binom{k}{m} z^{2m-k-n-1} $


We want to the know the coefficient of the $\displaystyle \frac{1}{z}$ term.

So we're only interested when $ \displaystyle m = \frac{k+n}{2}$.

But since $n$ is assumed to be odd, $k$ must also be odd.

But if $k$ can only assume odd values, then every term of the series is imaginary. So $C_{n}$ must be zero when $n$ is odd.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Another way would be using

\(\displaystyle \cos(z+z^{-1}) = \cos(z) \cos\left(\frac{1}{z}\right)-\sin(z) \sin \left( \frac{1}{z}\right)=\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots \right)\left(1-\frac{1}{2!\,z^2}+\frac{1}{4!\,z^4}+\cdots \right)-\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}\cdots \right)\left(\frac{1}{z}-\frac{1}{3!\,z^3}+\frac{1}{5!\,z^5}+\cdots \right)\)

Evidently all the terms are even . Another way would be using Cauchy product formula.