Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$

Question

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?

Answer 1

$$\frac1{1+z^2}=\frac1{(z-1)^2+2(z-1)+2}\\ =\frac12\left(1+(z-1)+\frac12(z-1)^2\right)^{-1}\\ =\frac12\left(1-((z-1)+\frac12(z-1)^2)+((z-1)+\frac12(z-1)^2)^2-...\right)\\ =\frac12-\frac12(z-1)+\frac14(z-1)^2+0(z-1)^3+...$$

Answer 2

Unfortunately (irresponsibly?), the "answer" was accepted, and newer questions about the same series are closed as duplicates (by people having more privileges than judgement). Just in case some naive people follow the link to this old turkey, let's post something deserving the name "answer" (hopefully).

If you look for a power series, you don't need just the first three terms without much of an explanation, you need an algorithm how to compute as many coefficients as you need, or, if possible, an explicit formula.

So let's have a look at $$\frac1{1+x^2}=\frac1{2+2\,(x-1)+(x-1)^2}=\sum^\infty_{n=0}a_n\,(x-1)^n:$$ Multiplying both sides by the denominator and comparing equal powers of $x-1$, we get $$2\,a_0=1,$$ $$2\,a_1+2\,a_0=0$$ and $$2\,a_n+2\,a_{n-1}+a_{n-2}=0$$ for $n\ge2$. That means $$a_0=1/2,$$ $$a_1=-1/2$$ and $$a_n=-a_{n-1}-a_{n-2}/2\tag{Recursion}$$ for $n\ge2$, allowing to calculate as many coefficients as needed. Of course, this gives the same values $a_2=1/4$ and $a_3=0$, but you can continue that ad nauseam, without much effort.

If you want an explicit formula for $a_n$, there are several ways, each opening another can of worms. You could solve (Recursion), that's not so hard because of the constant coefficients, but boring. Or you could recognize the generating function of Chebyshev polynomials of second kind, $$\frac1{1-2\,t\,y+t^2}=\sum^\infty_{n=0}U_n(y)\,t^n.$$ Since $$\frac1{2+2\,(x-1)+(x-1)^2}=\frac12\,\frac1{1-2\cdot\frac{-1}{\sqrt{2}}\cdot\frac{x-\,1}{\sqrt{2}}+\left(\frac{x-1}{\sqrt{2}}\right)^2},$$ we see that $$\frac1{1+x^2}=\frac12\,\sum^\infty_{n=0}\,U_n\left(\frac{-1}{\sqrt{2}}\right)\,\left(\frac{x-1}{\sqrt{2}}\right)^n.$$ From the trigonometric definition, we have $$U_n(\cos\theta)=\frac{\sin (n+1)\theta}{\sin\theta},$$ and since $$\frac{-1}{\sqrt{2}}=\cos\frac{3\pi}4,$$ we have $$U_n\left(\frac{-1}{\sqrt{2}}\right)=\sqrt{2}\,\sin\frac{3(n+1)\pi}4,$$ i.e. $$a_n=\frac{\sin\frac{3(n+1)\pi}4}{2^{\frac{n+1}2}}.$$ Those infos about Chebyshev polynomials can be found here: https://en.wikipedia.org/wiki/Chebyshev_polynomials

Yet another possibility would be to use $$\frac1{1+x^2}=\frac1{2i}\left(\frac1{x-i}-\frac1{x+i}\right)$$ and the good old geometric series, but that (if you know complex numbers) would give the same explicit formula for $a_n$, naturally.

Answer 3

$$ \begin{align} \frac1{1+z^2} &=\frac1{2+2w+w^2}\tag1\\[9pt] &=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt] &=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\ &=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4\\ &=\frac1{\sqrt2}\sum_{k=0}^\infty\left(-\frac1{\sqrt2}\right)^k\cos\left(\frac{\pi(k-1)}4\right)(z-1)^k\tag5 \end{align} $$ 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