Author Topic: Q6 TUT 5201  (Read 556 times)

Victor Ivrii

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Q6 TUT 5201
« on: November 17, 2018, 04:12:56 PM »
Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$
f(z)=\frac{1}{1-\cos(z)};\qquad z_0=0\quad\text{(four terms of the Laurent series)} .
$$

Shengying Yang

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Re: Q6 TUT 5201
« Reply #1 on: November 17, 2018, 04:18:55 PM »
let $h(z)=1, g(z)=1-\cos z$
as $z_0=0$, $$h(0)=1\ne0$$
$$g(0)=0, g'(0)=0, g''(0)=1\ne0$$
$$\therefore2-0=2$$, order of pole = 2
$$\therefore\frac{1}{1-\cos z}=a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots$$
$$\frac{1}{1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots}=a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots$$ Errors in the line above; correct below
$$\therefore(\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots)(a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots)=1$$
Since corresponding coefficients equal each other, we have
$$\frac{a_{-2}}{2!}=1\therefore a_{-2}=2$$
$$\frac{a_{-1}}{2!}=0\therefore a_{-1}=0 \therefore Res(f;0)=0$$
$$\frac{a_0}{2!}-\frac{a_{-2}}{4!}=0\therefore a_0=\frac{1}{6} $$
$$\frac{a_1}{2!}-\frac{a_{-1}}{4!}=0\therefore a_1=0$$
$$\therefore \frac{2}{z^2}+\frac{1}{6}+\cdots$$

« Last Edit: November 28, 2018, 05:06:48 AM by Victor Ivrii »

Qi Cui

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Re: Q6 TUT 5201
« Reply #2 on: November 17, 2018, 04:47:08 PM »
$$let\ h(z) = 0, g(z) = 1- cos(z)$$
$$As \ z _0 = 0, h(0) = 1\ne0; order = 0$$
$$g(z) = 1- cos(z)$$
$$g'(z) = sin(z), g'(0) = 0$$
$$g''(z) = cos(z), g''(0) = 1\ne0, \quad\therefore order = 2 $$
$$\quad\therefore 2-0= 2   \quad\therefore \ order\ of \ pole = 2$$
$$\quad\therefore{{1}\over1-cos(z)} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +... $$
$${{1}\over 1-(1-{{z^{2}\over {2 !}}+{z^{4}\over 4!}+...})} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...  $$
$$1= ({z^{2}\over 2!} - {z^{4}\over 4!} + {z^{6}\over {6!}}... )(a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...  ) $$
$Corresponding\ coefficients \ equal\ with\ each\ other:$
$${{a_{-2}}\over {2!}} = 1  \quad\therefore a_{-2} = 2 $$
$${{a_{-1}}\over {2!}} = 0  \quad\therefore a_{-1} = 0 $$
$${{a_{0}}\over {2!}} - {{1}\over4!}a_{-2} = 0  \quad\therefore a_{0} = {{1}\over6} $$
$${{a_{0}}\over {2!}} - {{a_{-1}}\over {4!}}  = 0 \quad\therefore a_{1} = 0 $$
$$similar\ like \ the\ above\ procedure\ we\ can\ get \ a_{2} = {{1}\over120} $$
$$\quad\therefore {{2}\over {z^2}} + {{1}\over6} + {{1}\over120} {z^2} +...,\   Res(f;0) = 0  $$
« Last Edit: November 17, 2018, 10:53:46 PM by Qi Cui »

Muyao Chen

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Re: Q6 TUT 5201
« Reply #3 on: November 17, 2018, 08:58:45 PM »
The first four terms of the Laurent series is:

$\frac{2}{z^{2}}$ + $\frac{1}{6}$ +$\frac{z^{2}}{120}$ +$\frac{z^{4}}{3024}$ +...