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Laurent Series of z/(sin z)^2

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[SOLVED] Laurent Series of z/(sin z)^2

Homework Statement
Find the first four terms of the Laurent series of [itex]f(z) = z/(\sin z)^2[/itex] about 0.

The attempt at a solution
I know that when z = 0, f(z) is undefined so it has a singularity there. This singularity is a pole because

[tex]\lim_{z \to 0} \left|\frac{z}{(\sin z)^2}\right| = \infty[/tex]

I want to find the order of this pole, which according to my book is the order of the zero z = 0 of 1/f(z). But z = 0 is not a zero of 1/f(z) because it is undefined at z = 0.

My plan of attack is to find the order of the pole z = 0, find the form of the Laurent series of f, expand [itex]f(z)(\sin z)^2[/itex], equate it to z and get the first four terms of the Laurent series.
 

Answers and Replies

HallsofIvy
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Homework Statement
Find the first four terms of the Laurent series of [itex]f(z) = z/(\sin z)^2[/itex] about 0.

The attempt at a solution
I know that when z = 0, f(z) is undefined so it has a singularity there. This singularity is a pole because

[tex]\lim_{z \to 0} \left|\frac{z}{(\sin z)^2}\right| = \infty[/tex]

I want to find the order of this pole, which according to my book is the order of the zero z = 0 of 1/f(z). But z = 0 is not a zero of 1/f(z) because it is undefined at z = 0.
But it has a removable discontinuity at z= 0. sin(z)= z- (1/6)z3+ ... so sin2(z)= z2- (1/6)z4+ ... f(z)/z= z- (1/6)z3+... That clearly is 0 at z= 0 but its derivative is not so it has a zero of order 1 at z= 0.

My plan of attack is to find the order of the pole z = 0, find the form of the Laurent series of f, expand [itex]f(z)(\sin z)^2[/itex], equate it to z and get the first four terms of the Laurent series.
I'm not sure I am following this. What do you mean by "find the form of the Laurent series of f" if not find the Laurent series itself?
 
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But it has a removable discontinuity at z= 0.
By "it", I suspect you mean 1/f(z). That is correct. Should I redefine 1/f(z) so that it is 0 at z = 0 and work with that to find the order of z = 0?

sin(z)= z- (1/6)z3+ ... so sin2(z)= z2- (1/6)z4+ ... f(z)/z= z- (1/6)z3+... That clearly is 0 at z= 0 but its derivative is not so it has a zero of order 1 at z= 0.
This is where I'm confused. The order of a zero z0 is the least j such that the jth derivative of f evaluated at z0 is not 0. Let g(z) = 1/f(z). About z = 0, g(z) may be written as g(0) + g'(0)z + g''(0)/2 z2 + ... But g(0) is not defined. What you did seems unfair to me because you expanded the square of sin z, then divide by z (so you're assuming z is not 0), and then you evaluated at z = 0.

I'm not sure I am following this. What do you mean by "find the form of the Laurent series of f" if not find the Laurent series itself?
By the form, I mean determining from what power of z to start with in the series. You state that the order is 1 so f(z) = a-1/z + a0 + a1z + ...
 
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Forget about the Laurent series for now.

g(z) = (sin z)2/z has a singularity as z = 0. It is removable since [itex]|g(z)| \to 0[/itex] as [itex]z \to 0[/itex]. I can therefore redefine g as follows: g(z) = 0 if z = 0; otherwise g(z) = (sin z)2/z right? g is now entire and so has a power series about 0 which converges for any z:

[tex]\sum_{i=0}^\infty \frac{g^{(i)}(0)}{i!}z^i[/tex]

Right? Now g'(z) = sin(2z)/z - (sin z)2/z2 which is undefined at z = 0 so the power series can't possibly converge. Where did I go wrong?
 
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Formally, the Laurent series of f(z) about 0 is

[tex]\sum_{k=-\infty}^\infty a_k z^k[/tex]

where

[tex]a_k = \frac{1}{2\pi i} \int_{|w| = r} \frac{dw}{w^k(\sin w)^2}[/tex]

where [itex]0 < r < \pi[/itex]. I tried the case k = 0 and couldn't evaluate the integral.

I believe there must be some kind of trick to get the Laurent series for f(z). I really don't know what else to try.
 
HallsofIvy
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Forget about the Laurent series for now.

g(z) = (sin z)2/z has a singularity as z = 0. It is removable since [itex]|g(z)| \to 0[/itex] as [itex]z \to 0[/itex]. I can therefore redefine g as follows: g(z) = 0 if z = 0; otherwise g(z) = (sin z)2/z right? g is now entire and so has a power series about 0 which converges for any z:

[tex]\sum_{i=0}^\infty \frac{g^{(i)}(0)}{i!}z^i[/tex]

Right? Now g'(z) = sin(2z)/z - (sin z)2/z2 which is undefined at z = 0 so the power series can't possibly converge. Where did I go wrong?
???
[tex]g'= \frac{2zsin(z)cos(z)- sin^2(z)}{z^2}= 2(\frac{sin(z)}{z})cos(z)-
(\frac{sin(z)}{z})^2)[/tex] which goes to 2-1= 1 as z goes to 0. How did you get the "sin(2z)"?
 
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2(sin z)(cos z) = sin 2z.

Why are you taking the limit?
 
Avodyne
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Now g'(z) = sin(2z)/z - (sin z)2/z2 which is undefined at z = 0 so the power series can't possibly converge. Where did I go wrong?
The singularity at z=0 is removable, since the limit of g'(z) as z->0 is finite (specifically, as noted by HallsofIvy, the limit equals one).

The fast way to compute the Taylor series for g(z) is to first write it as g(z)=(1-cos(2z))/(2z), then use the known series for the cosine to get 1-cos(2z), and finally divide by 2z, term by term.

You should find a series of the form a1z + a2z2+... . To get the series for f(z)=1/g(z), write the series for g(z) as a1z[1 + (a2/a1)z + ...], and take the reciprocal by standard methods.

By the way, you can see that f(z) has a pole of order one at z=0 without doing any real calculating: near z=0 we have sin(z) ~ z, so z/(sin z)2 ~ 1/z.
 
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The fast way to compute the Taylor series for g(z) is to first write it as g(z)=(1-cos(2z))/(2z), then use the known series for the cosine to get 1-cos(2z), and finally divide by 2z, term by term.
True. But what bugs me is that it can't get the power series through the standard method, i.e by computing the coefficients in the Taylor expansion.

You should find a series of the form a1z + a2z2+... . To get the series for f(z)=1/g(z), write the series for g(z) as a1z[1 + (a2/a1)z + ...], and take the reciprocal by standard methods.
What standard methods? I'm not familiar how to compute the reciprocal of a power series. Perhaps you mean by using partial fractions?

By the way, you can see that f(z) has a pole of order one at z=0 without doing any real calculating: near z=0 we have sin(z) ~ z, so z/(sin z)2 ~ 1/z.
True. I wanted to show that formally and got stuck.
 
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Since I know the f(z) has a pole of order 1 at 0, then f(z) = b-1z-1 + b0 + b1z+ b2z2 + ...

f(z)(sin z)2 = z and so computing the power series of the product on the left and equating the coefficients with those on the right should be enough to compute the Laurent series of f right?
 
Avodyne
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But what bugs me is that it can't get the power series through the standard method, i.e by computing the coefficients in the Taylor expansion.
You can. You just have to define the derivatives of g(z) at z=0 by a limit process (because of the removable singularity there).
What standard methods? I'm not familiar how to compute the reciprocal of a power series.
Consider S = 1 + c1x + c2x2 + ... . Let y = S-1. Then 1/S = 1 - y + y2 - ... = 1 - (c1x + c2x2 + ...) + (c1x + c2x2 + ...)2 + ... . Then expand out the powers, and retain terms up to the order you need.
f(z)(sin z)2 = z and so computing the power series of the product on the left and equating the coefficients with those on the right should be enough to compute the Laurent series of f right?
Yes, that works as well.
 
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You can. You just have to define the derivatives of g(z) at z=0 by a limit process (because of the removable singularity there).
Aha. I didn't know that. The book I'm using is scarce on details.

Consider S = 1 + c1x + c2x2 + ... . Let y = S-1. Then 1/S = 1 - y + y2 - ... = 1 - (c1x + c2x2 + ...) + (c1x + c2x2 + ...)2 + ... . Then expand out the powers, and retain terms up to the order you need.
Interesting. I don't see why that works though. Would you elaborate?

Thanks a lot for the help.
 
Avodyne
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OK, let me try again. Let y = c1x + c2x2 + ... , so that S = 1+y. Then 1/(1+y) = 1 - y + y2 - ... is a standard Taylor expansion. (It's good to know a simple library of expansions like this.) Now just substitute in the expression for y in terms of x, and expand in powers of x.
 

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