Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to find the MacLaurin series for the following:

$$\frac{1}{(1+2z)^2}$$

What I'm trying to do is to take the integral of that, find the Maclaurin series then derivate it, but I'm not sure that's valid for this kind of series.

Here's my attempt:

$$\int{f(z)} =\int{\frac{1}{(1+2z)^2}dz} = - \frac{1}{2+4z}$$

then I try to expand it:

$$\frac{1}{2+4z} = \frac{1}{1+(1+4z)}$$

taking $r =(-1)(1+4z)$

$$f(z) = \frac{d}{dz}(- \frac{1}{2+4z}) = -\frac{d}{dz}(\displaystyle\sum\limits_{k=0}^{+\infty} (-1)^k(1+4z)^k)$$

And derivating

I get the result

$$f(z) = \sum\limits_{k=0}^{+\infty} (-1)^{k+1}k(1+4z)^{k-1}$$

Doesn't look right, and I'm not really sure if I can do this (integrate then derivate) like I do with Laurent Series. Not homework but part of my exercises I'm using to study for an exam.

share|improve this question
1  
+1 for showing your work! –  Jyrki Lahtonen Jun 23 '11 at 14:31

1 Answer 1

up vote 3 down vote accepted

You have the right idea. Instead of $$ \frac{1}{2+4z}=\frac{1}{1+(1+4z)} $$ try $$ \frac{1}{2+4z}=\frac12\cdot\frac{1}{1+2z}, $$ and write that as a geometric series as you did in your first attempt.

share|improve this answer
    
Thank you, using your sugestion i've obtained the correct series, $\displaystyle\sum\limits_{k=0}^{+\infty}(-1)^k2^k(k+1)z^k$ –  GriffinHeart Jun 23 '11 at 15:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.