# Taylor series

1. Nov 28, 2007

### buzzmath

1. The problem statement, all variables and given/known data
z is a complex number. find the taylor series expansion for g(z)=1/(z^3) about z0= 2.in what domain does the taylor series of g converge. z0 is z subscript 0

2. Relevant equations

3. The attempt at a solution

I wrote g(z)=1/(z^3) = 1/(2+(z^3)-2) = (1/2)*1/(1+(z^3 -2)/2) then i was thinking I could use the identity 1/(1+z)=summation( (z^n) / (n!) ) when |z|<1 so then the expansion would be the summation (z^3 -2)^n / 2^n+1 in the domain |z^3 - 2| <2 I think this is right but it's not the taylor series expansion because I have the z^3 in there. the domain needs to be in the form |z-2| < R and in the summation I need to have (z-2)^n not (z^3 - 2)^n How do I get rid of the z^3 and have the answer in the right form?
thanks

2. Nov 28, 2007

### Dick

You don't have to pull any tricks. The taylor series is the sum of the nth derivatives of g evaluated at 2, times (z-2)^n divided by n!. Concentrate on the first part of that. 1/2^3, -3/2^4, 3*4/2^5 etc.

3. Nov 28, 2007

### buzzmath

so I would just find the nth derivative of g which is
g^(n) (z)= (-1)^n *(n+2)!/(2*z^(n+3)) where g^(n) (z) is the nth derivative of g at z
then I evaluate these derivatives at z = 2 to and divide by n! and multiply by (z-2)^n to get g(z) = summation[ (-1)^n *(n+2)!*(z-2)^n / (n! * 2^n+4)] sorry if it looks a little messy
is there an easier way to do this or when I'm given a problem like this do I just find the nth derivatives and then plug it in like above? it seems like it could get messy sometimes. Also, how would you find the domain in which this Taylor series converges? where it needs to be in the form when g is analytic in the disk |z-z0|<R ? that cube is messing me up in this. thanks

4. Nov 28, 2007

### Dick

You want the nth derivative of g evaluated at 2. Write down a few terms (like I did) to get comfortable with the form and then figure out how to write it. To determine the radius of convergence you could apply the ratio test to the series you get. But you are expanding around z=2 and if you are doing complex analysis you will eventually learn that the radius of convergence is the distance from z=2 to the nearest singularity of g(z)=1/z^3.

5. Nov 28, 2007

### buzzmath

This is complex analysis so the only singularity is 0. I think. so the distance from 2 to 0 is 2. so to write the circle of convergence is |z-2|<2 ?
thanks

6. Nov 29, 2007

### Dick

Yessssss.

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