Taylor series $\frac{z^2+z}{(z-1)^2}$ at $z=-1$

Question

Taylor series expansion for $\frac{z^2+z}{(z-1)^2}$ at $z=-1$. Also find its radius of convergence. One way is to expand with the general formula and calculate derivative of all orders but that is very tedious. Is there a way to find the series without calculating every term?

Here's the answer from WA if it helps: $-\frac14(z + 1) + \frac1{16} (z + 1)^3 + \frac1{16} (z + 1)^4 +\frac3{64} (z + 1)^5 + \frac1{32} (z + 1)^6 + O((z + 1)^7) $

Answer 1

The brute force strategy is pretty straight-forward: Find the series for $\frac1{z-1} = \frac{1}{(z+1)-2}$, square it, and finally multiply by $z^2 + z = (z+1)^2 - (z+1)$.

For simplicity, I will set $x = z+1$ and expand about $x = 0$ so I can actually see what's going on.

So, without further ado: $$ \frac{1}{x-2} = -\frac12\cdot\frac{1}{1-\frac x2} = -\frac12\left(1+\frac12 x +\frac14x^2 + \frac18x^3 + \cdots \right) $$ Squaring this yields $$ \frac{1}{(x-2)^2} = \frac14\left(1 + \frac 22x + \frac34x^2 + \frac48x^3 + \cdots\right) $$ (where I've kept the fractions unsimplified to make the pattern stand out more).

Finally, we multiply this by $x^2 - x$ to get $$ \frac{x^2 - x}{(x-2)^2} = -\frac14x + \frac14\left(1 - \frac 22\right)x^2 + \frac14\left(\frac22 - \frac34\right)x^3 + \frac14\left(\frac34 - \frac48\right)x^4 + \cdots $$ Reinserting $x = z+1$ gives the result you're after.

Edit: I just realized I haven't addressed the second part of the question: the radius of convergence. The radius of convergence is equal to the distance to the nearest singularity of the function. That happens at $z=1$, obviously, so the radius of convergence is $2$.

Answer 2

Note that\begin{align}\frac{x^2+x}{(x-1)^2}&=\frac{x^2+x-2}{(x-1)^2}+\frac2{(x-1)^2}\\&=\frac{(x-1)(x+2)}{(x-1)^2}+2\left(\frac1{1-x}\right)'\\&=1-3\frac1{1-x}+2\left(\frac1{1-x}\right)'.\end{align}Now, compute the Taylor series of $\frac3{1-x}$ and of its derivative centered around $-1$. For that, use the fact that $1-x=2-(x+1)$.

Answer 3

Hint:

Set $u=z+1$, i.e. $z=u-1$, and rewrite the fraction: $$ \frac{z^2+z}{(z-1)^2}=\frac{u^2-u}{(u-2)^2}=1+\frac{3u-4}{(u-2)^2}= 1+\frac14\frac{3u-4}{(\frac u 2-1)^2}= 1+\frac14\frac{6(\frac u2-1)-1}{(\frac u 2-1)^2}, $$ so, setting $t=\frac u2$, we ultimately obtain \begin{align} \frac{z^2+z}{(z-1)^2}&=1+\frac32\frac{(\frac u2-1)}{(\frac u 2-1)^2}-\frac 14\frac1{(\frac u 2-1)^2} = 1-\frac32\frac{1}{1-t}-\frac14\frac1{(1-t)^2} \\[1ex] &= 1-\frac32\frac{1}{1-t}+\frac14\biggl(\frac1{1-t}\biggr)^{\negmedspace\prime}. \end{align} Now expand with the standard formula, and replace $t$ with $\;\frac12(z+1)$.