I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$
I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$
So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:
$$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$
But observe that I need to center it at $z_0=1$ so I rewrote it to:
$$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$
I can factor out $\frac {1}{2z}$ and use the basic identity to get :
$$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$
The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.
Can anyone help me figure this one out? What am I doing wrong?
Best Answer
$$ \begin{align} \frac1{1+z^2} &=\frac1{2+2w+w^2}\tag1\\[9pt] &=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt] &=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\ &=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4\\ &=\frac1{\sqrt2}\sum_{k=0}^\infty\left(-\frac1{\sqrt2}\right)^k\cos\left(\frac{\pi(k-1)}4\right)(z-1)^k\tag5 \end{align} $$ Explanation:
$(1)$: substitute $z=w+1$
$(2)$: partial fractions
$(3)$: use the forms $\frac1{1+aw}$ for easier series
$(4)$: write $\frac1{1+aw}$ as geometric series and undo $(1)$
$(5)$: convert to cosines