# Maclaurian series

## Homework Statement

find the maclaurian series for f(z) = z^3 / (2 - z)^2

and determine the radius of convergence

## The Attempt at a Solution

i found the first 8 terms of the series using the maclaurian formula:

z^3/(2-z)^2 = 0 + 0 + 0 + 1/4 * z^3 + 1/4 * z^4 + 3/16 * z^5 + 1/8 * z^6 + 5/64 * z^7

the problem i'm having is comming up wtih a formula for the coefficients so that I can write it out in compact series form.

another way to write it would be:

3/2 * z^3/3! + 6 * z^4/4! + 45/2 * z^5/5! + 90 * z^6/6! * 1575/4 * z^7/7!

so I need a formula for 3/2, 6, 45/2, 90, 1575/4,...

the series starting at n = 3 [ (some formula) * z^(n)/n! ]

could someone please direct me on how to determine a formula for the coefficients.

Thanks

## Answers and Replies

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When finding the Maclaurin series you should avoid using the McLaurin formula. Instead, you should do manipulations using the known series expansions of standard functions. In fact, if you want to find the nth derivative fo some complicated function at some point for large n, it is usually easier to find the series expansion first and then extract the derivative using McLaurin's formula for the series expansion.

In this caseyou can use that:

1/(1-x) = 1 + x + x^2 + .....

Take the derivative:

-1/(1-x)^2 = 1 + 2 x + 3 x^2 + 4x^3 + ....

Therefore:

1/(2-z)^2 = 1/4 1/[1-(z/2)]^2 =

-1/4 sum from n = 0 to infinity of (n+1)/2^n z^n

Ah. Thanks, that was very helpful!

i found that the maclarian series is

sum from 0 to inf [ 1/4 * (n+1)/2^n * z^(n+3) ]

i calculated the first 10 terms and they all match. I pretty much just used your last formula and multiplied it by z^3.

using the ratio test I found that the radius of convergence is 2. i.e. the series converges for |z| < 2 and diverges for |z| > 2.

is this right? thanks! Also, since this is on the complex plane do I write the radius of convergence any differently? for example, should I say that the series converges inside the disk centered at 0 with radius 2?

Thanks