# Homework Help: Power series

1. Jan 30, 2012

### fauboca

$$z\in\mathbb{Z}$$

$$\frac{1}{1-(-z)}=\sum_{n=0}^{\infty}(-z)^n$$

$$\frac{1}{(z+2)^2}=\frac{d}{dz} \frac{-1}{1-(1-z)} = \frac{d}{dz} (1 + (1-z) + (1-z)^2+\cdots = 0 -1 -2(1-z)-3(1-z)^2 - \cdots = \sum_{n=0}^{\infty} ???$$

Not to sure about the second one.

Last edited: Jan 30, 2012
2. Jan 31, 2012

### susskind_leon

$$\frac{-1}{1-(1-z)}=\frac{-1}{z}$$
$$\frac{d}{dz}\frac{-1}{z}=\frac{1}{z^2}\neq\frac{1}{(z+2)^2}$$

3. Jan 31, 2012

### genericusrnme

There's a pretty well known power series for $\frac{1}{1-x}$
If you let x=-z, what do you get?

4. Jan 31, 2012

### susskind_leon

What you can do is $$\frac{d}{dx} \frac{1}{(1+x)} = -\frac{1}{(1+x)^2}$$
If you just define y=x-1, you get
$$\frac{1}{(2+y)^2}=-\frac{d}{dy}\sum_{n=0}^\infty (-(y+1)^n)$$
However, you still need to make sure that you are inside the radius of convergence, so |y+1|<1