3
$\begingroup$

So I have the function $$\frac{z^2}{(z+i)(z-i)^2}.$$

I want to determine the power series around $z=0$ of this function.

I know that the power series is $\sum_{n=0}^\infty a_n(z-a)^n$, where $a_n=\frac{f^{(n)}(a)}{n!}$. But this gives me coefficients, how can I find a series for this?

Edit: maybe we can use partial fractions?

$\endgroup$
  • $\begingroup$ Would this help? $$\frac{z^2}{(z+i)(z-i)^2}\equiv \frac{z^2(z+i)}{(z^2+1)^2}$$ $\endgroup$ – Kenny Lau Apr 17 '16 at 13:18
  • $\begingroup$ @KennyLau I'm not sure, how would that help? $\endgroup$ – Peter Bond Apr 17 '16 at 13:19
  • $\begingroup$ I'll answer this question, then. $\endgroup$ – Kenny Lau Apr 17 '16 at 13:20
  • $\begingroup$ @KennyLau Thankyou very much $\endgroup$ – Peter Bond Apr 17 '16 at 13:22
0
$\begingroup$

Note that: $$\frac{z^2}{(z+i)(z-i)^2}\equiv \frac{z^2(z+i)}{(z^2+1)^2}$$

This means, that I would need to only find the power series of $\displaystyle\frac1{(z^2+1)^2}$.

We have: $$\frac1{1-x}\equiv\sum_{n\mathop=0}^\infty x^n$$

Taking derivative of both sides: $$\frac{-1}{(1-x)^2}\equiv\sum_{n\mathop=0}^\infty nx^{n-1}\equiv\sum_{n\mathop=0}^\infty (n+1)x^n$$

Substituting $x=-z^2$: $$\frac{-1}{(1+z^2)^2}\equiv\sum_{n\mathop=0}^\infty (n+1)(-z^2)^n\equiv\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n}$$

Multiplying both sides by $-z^2(z+i)$: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv(z+i)\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n+2}\equiv(z+i)\sum_{n\mathop=1}^\infty n(iz)^{2n}$$

Distributing: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty i^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty i^{2n+1}nz^{2n}$$

Combining: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty (-i)^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty (-i)^{2n-1}nz^{2n}$$

$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\frac{n+1}2z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$

$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$

$\displaystyle\equiv\sum_{n\mathop=0}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}$

$\displaystyle\equiv\sum_{n\mathop=2}^\infty (-i)^{n-1}\left\lceil\frac n2\right\rceil z^n$

$\endgroup$
  • $\begingroup$ Look, maybe you could skip the final steps. $\endgroup$ – Kenny Lau Apr 17 '16 at 13:37
1
$\begingroup$

It's also convenient to perform a partial fraction decomposition followed by a binomial series expansion.

We obtain by partial fraction decomposition \begin{align*} \frac{z^2}{(z+i)(z-i)^2}&=\frac{1}{4(z+i)}+\frac{3}{4(z-i)}+\frac{i}{2(z-i)^2}\\ &=-\frac{i}{4}\cdot\frac{1}{1-iz}+\frac{3i}{4}\cdot\frac{1}{1+iz}-\frac{i}{2}\cdot\frac{1}{(1+iz)^2}\tag{1}\\ &=-\frac{i}{4}\sum_{n\geq 0}\left(iz\right)^n+\frac{3i}{4}\sum_{n\geq 0}\left(-iz\right)^n -\frac{i}{2}\sum_{n\geq 0}\binom{-2}{n}\left(iz\right)^n\tag{2}\\ &=\sum_{n\geq 0}\left(-\frac{1}{4}+\frac{3}{4}(-1)^n-\frac{1}{2}(n+1)(-1)^n\right)i^nz^n\tag{3}\\ &=\sum_{n\geq 0}\left(\frac{2n-1}{4}(-1)^{n+1}-\frac{1}{4}\right)i^{n+1}z^n \end{align*}

Comment:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.