Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around $z_0 = 1$?
I know that, centered around 0 we can manipulate the geometric series to obtain $ \frac{1}{1+z^2} = \sum_{k \geq 0} (-1)^k z^{2k} $. However, I want to write it on the form $ \sum_{k \geq 0} c_k (z-1)^k $.
 A: Usually, when I want a series about a point $z_0 \ne 0$, I change variables and write $w = z-z_0, z = w+z_0$ and do the series in powers of $w$.
Example
$\frac{1}{1+z^2}$ centered at $z=1$.  Write $w=z-1, z=w+1$, so
$$
\frac{1}{1+z^2} = \frac{1}{1+(w+1)^2} = \frac{1}{2+2w+w^2}
$$
For this I could use partial fractions
$$
\frac{1}{2+2w+w^2} = \frac{i/2}{w+(1+i)} - \frac{i/2}{w+(1-i)}
$$
and each of these is a geometric series.  It turns out (as expected) that
the imaginary parts all cancel.  We can see a pattern:
$$
\frac{1}{2} -\frac{1}{2}(z-1) 
+\frac{1}{4}(z-1)^2 
-\frac{1}{8}(z-1)^4 +\frac{1}{8}(z-1)^5 
-\frac{1}{16}(z-1)^6 
+\frac{1}{32}(z-1)^8 -\frac{1}{32}(z-1)^9 
+\frac{1}{64}(z-1)^{10} \dots
$$
A: Just manipulate the fraction
$$
\dfrac{1}{1-z} = \dfrac{1}{1-z+1/2-1/2} = \dfrac{1}{1/2-z+1/2} = \dfrac{1}{1/2-(z-1/2)} = \dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)}.
$$
And
$$
\dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)} = 2 \sum_{k=0}^{\infty} \Big(2 \, (z - 1/2) \Big)^k = \sum_{k=0}^{\infty} 2^{k+1} \Big(z - \dfrac{1}{2}\Big)^k,
$$
with
$$
| 2 (z-1/2) | < 1, \\
|(z-1/2) | < 1/2.
$$
Since I like GEdgar's answer, I will write most of the details of his answer.
You want to manipulate something like this
$$
\dfrac{1}{1-z},
$$
not this
$$
\dfrac{1}{1+z^2}.
$$
Easy, use partial fractions
$$
\dfrac{1}{1+z^2} = \dfrac{-i}{2}\dfrac{1}{(z-i)} + \dfrac{i}{2}\dfrac{1}{(z+i)},
$$
and manipulate
$$
\dfrac{-i}{2}\dfrac{1}{(z-i)} = \dfrac{-i}{2}\dfrac{1}{(z-i + 1 -1)} = \dfrac{-i}{2(-1-i)} \dfrac{1}{ \Bigg[ 1- \dfrac{z-1}{-1-i} \Bigg]} \\
= \dfrac{-i}{2(-1-i)} \sum_{k=0}^{\infty} \Big(\dfrac{z-1}{-1-i}\Big)^k.
$$
Using the polar form of $1/(-1-i) = e^{i\, 3\pi/4}/ \sqrt{2} $
$$
\dfrac{-i}{2(-1-i)} \sum_{k=0}^{\infty} \Big(\dfrac{z-1}{-1-i}\Big)^k = \dfrac{-i}{2} \dfrac{1}{\sqrt 2} e^{i\, 3\pi/4} \sum_{k=0}^{\infty}  \Bigg(\dfrac{1}{\sqrt 2} e^{i\, 3\pi/4} \Bigg)^{\!k} (z-1)^k .
$$
Doing some calculations, the other fraction is
$$
\dfrac{i}{2}\dfrac{1}{(z+i)} = \dfrac{i}{2}  \dfrac{1}{\sqrt 2} e^{-i\, 3\pi/4} \sum_{k=0}^{\infty}  \Bigg(\dfrac{1}{\sqrt 2} e^{-i\, 3\pi/4} \Bigg)^{\!k} (z-1)^k .
$$
Joining everything, the power series of $1/(1+z^2)$ around $z_0=1$ is
$$
\dfrac{1}{1+z^2} = \dfrac{i}{2 \sqrt 2} \sum_{k=0}^{\infty}  \bigg(\dfrac{1}{\sqrt 2}\bigg)^{\!k} \bigg(\exp \Big(\!-i\, \dfrac{3\pi(k+1)}{4} \Big) - \exp \Big(i\, \dfrac{3\pi(k+1)}{4} \Big) \bigg) (z-1)^k .
$$
Simplifying both exponentials
$$
\exp \Big(\!-i\, \dfrac{3\pi(k+1)}{4} \Big) - \exp \Big(i\, \dfrac{3\pi(k+1)}{4} \Big) = -2\, i \sin \Big(\dfrac{3\pi(k+1)}{4} \Big)
$$
The result is
$$
\dfrac{1}{1+z^2} = \sum_{k=0}^{\infty}  \bigg(\dfrac{1}{\sqrt 2}\bigg)^{\!k+1}\! \sin \Big(\dfrac{3\pi(k+1)}{4} \Big) (z-1)^k,
$$
with
$$
|z-1| < \dfrac{1}{\sqrt 2}.
$$
