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Residue for FE P3

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Ende Jin:
I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.

JUNJING FAN:

--- Quote from: Ende Jin on December 12, 2018, 03:27:33 PM ---I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.

--- End quote ---
what i do is since it is a double-pole, i find the coefficient for the degree 2 term of the sine squared function, then i find the coefficient of the degree 1 term for the cosine function, then divide it out.

Ende Jin:

--- Quote from: JUNJING FAN on December 12, 2018, 07:02:59 PM ---
--- Quote from: Ende Jin on December 12, 2018, 03:27:33 PM ---I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$


Just asking for a check of this idea.

--- End quote ---
what i do is since it is a double-pole, i find the coefficient for the degree 2 term of the sine squared function, then i find the coefficient of the degree 1 term for the cosine function, then divide it out.

--- End quote ---
I don't think you can directly divide them out? Because the coefficient is not that simply decided?

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