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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?

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  • $\begingroup$ Looks like you made a mistake on the beginning of your last equation. When you factored out $2z$, apparently you missed a minus sign, correct? $\endgroup$ – Pedro A Dec 10 '15 at 12:36
  • $\begingroup$ Yes I did miss a minus sign thanks :) hopefully its corrected. $\endgroup$ – Symphonized Dec 10 '15 at 12:37
  • $\begingroup$ @Element118 - my wolframalpha input is 1/(1+z^2) not 1/(1-z^2) $\endgroup$ – Symphonized Dec 10 '15 at 12:38
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    $\begingroup$ I don't think you are allowed to keep $z$, except as part of $z-1$. $\endgroup$ – Empy2 Dec 10 '15 at 12:38
  • $\begingroup$ @Michael - is that true? :O.. do you have any pointers what method might be used in this case? $\endgroup$ – Symphonized Dec 10 '15 at 12:40
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$$\frac1{1+z^2}=\frac1{(z-1)^2+2(z-1)+2}\\ =\frac12\left(1+(z-1)+\frac12(z-1)^2\right)^{-1}\\ =\frac12\left(1-((z-1)+\frac12(z-1)^2)+((z-1)+\frac12(z-1)^2)^2-...\right)\\ =\frac12-\frac12(z-1)+\frac14(z-1)^2+0(z-1)^3+...$$

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  • $\begingroup$ I understand what you are doing up to 90% but I don't see what identity you are applying to get the series.. I'm used to writing them in the form 1/(1-z) $\endgroup$ – Symphonized Dec 10 '15 at 13:01
  • $\begingroup$ I expanded the series. The next term is $-((z-1)+\frac12(z-1)^2)^3$, which has $(z-1)^3,(z-1)^4,(z-1)^5$ and $(z-1)^6$ terms in it. So I picked out the $(z-1)^3$ term from there, and the $(z-1)^3$ term in the middle of $((z-1)+\frac12(z-1)^2)^2$. $\endgroup$ – Empy2 Dec 10 '15 at 13:07
  • $\begingroup$ Is there an easy way to write this with the sum (sigma) notation without having to expand the series? $\endgroup$ – Symphonized Dec 10 '15 at 13:17
  • $\begingroup$ Thanks I finally understood what is happening. I also found the correct answer. $\endgroup$ – Symphonized Dec 10 '15 at 16:23
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$$ \begin{align} \frac1{1+z^2} &=\frac1{2+2w+w^2}\tag1\\[9pt] &=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt] &=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\ &=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4\\ &=\frac1{\sqrt2}\sum_{k=0}^\infty\left(-\frac1{\sqrt2}\right)^k\cos\left(\frac{\pi(k-1)}4\right)(z-1)^k\tag5 \end{align} $$ 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

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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.

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