Calculating Circle of Convergence for f(z) = (3z+1)/(15+2z-z^{2}) at z=1

[elimenohpee]

Homework Statement

Expand f(z) = (3z+1)/(15+2z-z^{2}) at z=1 and find the circle of convergence.

Homework Equations

The Attempt at a Solution

I think this is pretty straight forward, but I want to make sure I'm doing everything correctly. I used a power series representation to find the circle of convergence, and expanded the function using the taylor series evaluated at z=1.

f(z) = (3z+1)/(15+2z-z^{2}) <br /> = -1/3 (1 / (1 + z/3)) + 2/5(1 / (1 - (z/5)) <br /> = -1/3\sum(-1)^{n} (z/3)^{n} + 2/5\sum (z/5)^{n} , |z|&lt; 3

circle of convergence is then |z| < 3

I worked out the taylor series, and it matched with wolfram alpha here:
http://www.wolframalpha.com/input/?i=taylor+series+%283z%2B1%29%2F%2815%2B2z-z^2%29++at+z%3D1

Does everything look correct?

 

[Dick]

If you expand the function around z=0 then, yes, the circle of convergence is |z|<3. But they want you to expand around z=1. Don't you want an expression like |z-1|<R?

 

[elimenohpee]

If you expand the function around z=0 then, yes, the circle of convergence is |z|<3. But they want you to expand around z=1. Don't you want an expression like |z-1|<R?

Wow, I have no idea why I didn't see that.. DOH

Basically, what I need to do is find the series representation of f(z) evaluated at z=1, then I can use a ratio/root test to easily find the circle of convergence right? If I understand correctly, that test is needed to find the region of convergence.

When looking at the expanded series about z=1, I think I have the series representation, but I'm confused on how to formulate the '3' entering the expansion for odd values in the summation. Looks like the series representation would look like this (without the 3 entering for odd numbers):

\sum(z-1)^{n}/4^{n+1}

Once I know the series representation, I think I can handle the rest of it.

 

[Dick]

Here's a trick. Put u=z-1. Now express the function in terms of u. The radius of convergence of f(u) around u=0 is the same as the radius of convergence of f(z) around z=1.

 

[elimenohpee]

Clever trick! Looks as if the circle of convergence would be |z-1| < 3

Its not needed to answer the question, but how exactly would you account for the '3' term entering the taylors expansion for odd values of n? Its bugging me!

Thank you again for the help!

 

[Dick]

Clever trick! Looks as if the circle of convergence would be |z-1| < 3

Its not needed to answer the question, but how exactly would you account for the '3' term entering the taylors expansion for odd values of n? Its bugging me!

Thank you again for the help!

It's not |z-1|<3. The radius of convergence is larger than that. Like I said, try and express the function in terms of u.

 

[elimenohpee]

It's not |z-1|<3. The radius of convergence is larger than that. Like I said, try and express the function in terms of u.

Let me show my work I've done up to this point and see where I am failing to understand.

f(z) = \frac{3z+1}{15+2z-z^{2}}

perform a partial fraction expansion to get the expression in a easily recognized series:

= \frac{-1}{3}(\frac{1}{1+\frac{z}{3}}) + \frac{2}{5}(\frac{1}{1-\frac{z}{5}})

= \frac{-1}{3}\sum^{\infty}_{n=0} (-1)^{n} (\frac{z}{3})^{n} + \frac{2}{5}\sum^{\infty}_{n=0} (\frac{z}{5})^{n}, (|z|&lt; 3, |z| &lt; 5)

This is all about the point z=0.

If I make the substitution u = z-1

= \frac{-1}{3}\sum^{\infty}_{n=0} (-1)^{n} (\frac{u}{3})^{n} + \frac{2}{5}\sum^{\infty}_{n=0} (\frac{u}{5})^{n}, (|u|&lt; 3, |u| &lt; 5)

|u| &lt; 3 == |z-1|&lt; 3 , |u|&lt; 5 == |z-1|&lt; 5

Am I correct the previous steps? Isn't |z-1|<3 a stronger case than |z-1|<5? Or am I going about this the wrong way.

 

[Dick]

You are still expanding around z=0 and then trying to patch things up. Substitute u=z-1 by replacing z with u+1 in f(z) BEFORE you power expand. Express it in terms of u. THEN power expand in u.

 

[elimenohpee]

ahhhhhh..ok I got it now. Thank you! I will do that and repost to see if I get the right circle of convergence. Thanks :)

 

[elimenohpee]

Before I type all of my work up, I get the convergence to be |z-1|< 4 ... if that's not right I'm lost lol.

 

[Dick]

Before I type all of my work up, I get the convergence to be |z-1|< 4 ... if that's not right I'm lost lol.

It's right.

 

[elimenohpee]

Thank you for the help once again.