1. ## Find Taylor Series

Hi, I could use some help on part (b) of the following question from my Complex Variables, book. I didn't know if this belonged in the Calculus forum or not.

Question 3.2.5
(a) Use the identity
$\frac1z = \frac1{(z+1)-1}$

to establish
$\frac1z = - \sum_{n=0}^\infty (z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

(b) Use the above identity to also establish
$\frac1{z^2} = \sum_{n=0}^\infty (n+1)(z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

Check the answer by differentiating the series from (a)

Part (a) was easy:
$\frac1z = \frac{1}{(z+1)-1} = \frac{-1}{1-(z-1)} = -\sum_{n=0}^\infty(z+1)^n$

The check in part (b) is easy, too. But I can't do part (b) using the identity given. I tried squaring the identity, but this is as far as I've gotten (essentially, I just eliminated the negative sign):
$\frac1{z^2} = \frac1{z*z} = \frac1{(z+1)-1}* \frac1{(z+1)-1} = \left(- \sum_{n=0}^\infty (z+1)^n \right)* \left(- \sum_{n=0}^\infty (z+1)^n \right) = \left( \sum_{n=0}^\infty (z+1)^n \right)^2$

I've also got this (but I don't know if it's worth anything, or if it is even correct):

$\ \sum_{n=0}^\infty (z+1)^n \right) * \left( \sum_{n=0}^\infty (z+1)^n$

$= \left[1 + (z+1) + (z+1)^2 + ... \right] * \sum_{n=0}^\infty (z+1)^n$

$= 1 * \sum_{n=0}^\infty (z+1)^n + (z+1) * \sum_{n=0}^\infty (z+1)^n + (z+1)^2 * \sum_{n=0}^\infty (z+1)^n + ...$

$= \sum_{n=0}^\infty (z+1)^n + \sum_{n=0}^\infty (z+1)^{n+1} + \sum_{n=0}^\infty (z+1)^{n+2} + ...$

$= \sum_{n=0}^\infty \sum_{m=0}^\infty (z+1)^{m+n}$

I have answered part (a) and the check in part (b) of the following question. But I cannot figure out part (b)

2. Originally Posted by MSUMathStdnt
Hi, I could use some help on part (b) of the following question from my Complex Variables, book. I didn't know if this belonged in the Calculus forum or not.

Question 3.2.5
(a) Use the identity
$\frac1z = \frac1{(z+1)-1}$

to establish
$\frac1z = - \sum_{n=0}^\infty (z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

(b) Use the above identity to also establish
$\frac1{z^2} = \sum_{n=0}^\infty (n+1)(z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

Check the answer by differentiating the series from (a)
$\displaystyle \dfrac{1}{z^2}=\left[ \sum_{k=0}^{\infty}(z+1)^k\right]\times \left[ \sum_{k=0}^{\infty}(z+1)^k\right]$

Now consider how we obtain the coefficient of the term in $(z+1)^n$ when multiplying out the right hand side. We get $1$ from $(z+1)^0\times (z+1)^n$, and another $1$ from $(z+1)\times (z+1)^{n-1}$ ... and $1$ form $(z+1)^n\times (z+1)^0$. Which total to $(n+1)$. Hence the coefficient of $(z+1)^n$ in the expansion is $(n+1)$ which is what you need.

CB

3. Originally Posted by CaptainBlack
$\displaystyle \dfrac{1}{z^2}=\left[ \sum_{k=0}^{\infty}(z+1)^k\right]\times \left[ \sum_{k=0}^{\infty}(z+1)^k\right]$

Now consider how we obtain the coefficient of the term in $(z+1)^n$ when multiplying out the right hand side. We get $1$ from $(z+1)^0\times (z+1)^n$, and another $1$ from $(z+1)\times (z+1)^{n-1}$ ... and $1$ form $(z+1)^n\times (z+1)^0$. Which total to $(n+1)$. Hence the coefficient of $(z+1)^n$ in the expansion is $(n+1)$ which is what you need.

CB
Thanks, CB. Your reply was a little difficult for me to understand. But I think you're trying to say something similar to what I found, anyway:

$1/(z^2)=\left(-\sum(z+1)^n \right) \cdot \left(-\sum(z+1)^n \right)=\sum(z+1)^n \cdot \sum(z+1)^n=$

Expand the left summation (but not the right one):
$\left[1+(z+1)+(z+1)^2+(z+1)^3+... \right] \cdot \sum(z+1)^n=$

Distribute the right summation:
$1 \cdot \sum(z+1)^n+(z+1) \cdot \sum(z+1)^n+(z+1)^2 \cdot \sum(z+1)^n+(z+1)^3 \cdot \sum(z+1)^n+... =$

$\sum(z+1)^n +\sum(z+1)^{n+1}+\sum(z+1)^{n+2}+\sum(z+1)^{n+3}+. .. =$

Expand each summation on a row (lining them up nice would make the following step easier to see, too):
$\sum(z+1)^{n} \hspace{4mm} = 1+(z+1)+(z+1)^2+(z+1)^3+(z+1)^4...$
$\sum(z+1)^{n+1} = \hspace{7mm} (z+1)+(z+1)^2+(z+1)^3+(z+1)^4+...$
$\sum(z+1)^{n+2} = \hspace{24mm} (z+1)^2+(z+1)^3+(z+1)^4+...$
$\sum(z+1)^{n+3} = \hspace{43mm} (z+1)^3+(z+1)^4+...$

Now, draw a line underneath and add like terms vertically (you know, like you did when you first learned addition back in the first grade) and you get:
$1+2(z+1)+3(z+1)^2+4(z+1)^3 +...= \sum_{n=0}^\infty (n+1)(z+1)^n$

So the coefficient of each term is one more than the exponent, giving the summation on the right. Which is what you said.