Find the order of the pole of a function

[docnet]

Homework Statement

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Relevant Equations

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I learned that $f$ has another singular point at $z=1.715$, but i don't think this would be related to the pole at $z=0$

I tried substitutine $u=2\cos z-2+z^2$

and $$f(u)=\frac{1}{u^2}$$ has a pole of order 2 at $u=0$ which happens i.f.f. $z=0$ or $z=1.715$.

so $f$ has a pole of second order at $u=0$ equivalent to $z=0$.

 

[docnet]

to be honest i feel ilke my logic is shaky at the conclusion.. I'm not sure how to draw the connection between the the orders of the pole at u=0 and the pole at z=0. like they are the equivalent pole of $f$ so its order stays the same during the $u$ substitution?

 

[fresh_42]

Have you considered using the series expansion of the cosine?

 

[docnet]

Have you considered using the series expansion of the cosine?

trying the series expansion of the cosine:

$$f=\frac{1}{(2\cos z-2+z^2)^2}=\frac{1}{(2(1-\frac{z^2}{2!}+\cdots)-2+z^2)^2}=\frac{1}{(\frac{2z^4}{4!}-\cdots)^2}$$
$$\frac{1}{z^8(\frac{2}{4!}-\cdots)^2}$$

$$f=\frac{g}{z^8}\quad\text{with}\quad g=\frac{1}{(\frac{2}{4!}-\cdots)^2}$$

$$g(0)=144\neq 0$$. $z=0$ is an 8th order pole of $f$!

 

[Office_Shredder]

That looks good to me.

To make a somewhat silly counterexample to your first attempt, $f(z)=1/(z^2)^2$, let $u=z^2$, then $f(u)=1/u^2$. The problem is the $u$ that you picked has a zero of order 4 at $z=0$.

 

[fresh_42]

trying the series expansion of the cosine:

$$f=\frac{1}{(2\cos z-2+z^2)^2}=\frac{1}{(2(1-\frac{z^2}{2!}+\cdots)-2+z^2)^2}=\frac{1}{(\frac{2z^4}{4!}-\cdots)^2}$$
$$\frac{1}{z^8(\frac{2}{4!}-\cdots)^2}$$

$$f=\frac{g}{z^8}\quad\text{with}\quad g=\frac{1}{(\frac{2}{4!}-\cdots)^2}$$

$$g(0)=144\neq 0$$. $z=0$ is an 8th order pole of $f$!

You can check it with WolframAlpha:
https://www.wolframalpha.com/input/?i=f(z)=z^7/(2cos+z+-2++z^2)^2
https://www.wolframalpha.com/input/?i=f(z)=z^8/(2cos+z+-2++z^2)^2

 

[docnet]

That looks good to me.

To make a somewhat silly counterexample to your first attempt, $f(z)=1/(z^2)^2$, let $u=z^2$, then $f(u)=1/u^2$. The problem is the $u$ that you picked has a zero of order 4 at $z=0$.

would you mind please explaining this more?

 

[Office_Shredder]

would you mind please explaining this more?

When you reduced f to $f(u)=1/u^2$, your $u(z)$ was a function that looked like $z^4$, so you couldn't use the order of the pole of $f(u)$ to compute the order of the pole of $f(z)$, because $u(z)$ had a high degree zero at 0, i e. Looked like $z^4$.

If it's still not clear I wouldn't worry too much about it, as long as you understand the technique in your first post doesn't work and you have to do a series expansion to get the answer.

 

[docnet]

When you reduced f to $f(u)=1/u^2$, your $u(z)$ was a function that looked like $z^4$, so you couldn't use the order of the pole of $f(u)$ to compute the order of the pole of $f(z)$, because $u(z)$ had a high degree zero at 0, i e. Looked like $z^4$.

If it's still not clear I wouldn't worry too much about it, as long as you understand the technique in your first post doesn't work and you have to do a series expansion to get the answer.

Sorry I'm still not understanding. how does one compute the degree of the sine funciton? does $u=2\sin z - 2 +z^2$ have a degree of 2? I'm confused and I don't know how to find a degree of a zero

 

[Office_Shredder]

Sorry I'm still not understanding. how does one compute the degree of the sine funciton? does $u=2\sin z - 2 +z^2$ have a degree of 2? I'm confused and I don't know how to find a degree of a zero

It's just the same way you computed the order of the pole. Although you flipped cosine with sine.

$2\cos(z) -2+z^2 = z^4g(z)$ where $g(0)$ is not zero, so this is a zero of order 4. I was calling it a zero of degree 4 but I think that's not actually standard terminology, sorry.