1. ## Taylor series question

The question
For the following function,
i) Find the largest open disc about the given point $z_0$ in which f is analytic.
ii) find the first 3 non-zero terms of the Taylor series of f about $z_0$
iii) Find the coefficient $a_n$ of $(z - z_0)^n$ in the Taylor series of f about $z_0$ either as an explicit function of n, or give a recursive formula for $a_n$ as appropriate.

f(z) = $\frac{1}{z^2 - z - 6}$, $z_o = 1$

My attempt
i) I expanded the function as follows:
$\frac{1}{(z - 3)(z + 2)}$
Clearly there's singularities at z = 3, -2
I draw a quick graph and noticed that the radius of convergence is |z - 1| < 2
This part I got correct. However, I get stuck for parts ii) and iii)

ii) I used partial fractions to get:
$\frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2}$
I then manipulated each fraction so I could write them in terms of known Macluarin series $\frac{1}{z \pm 1}$:
$\frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n)$ and
$\frac{-1}{15}(\sum_{n = 0}^{\infty}{(-1)^n(\frac{z - 1}{3})^n)$

Thus I got this for the series:
$\frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n) \frac{-1}{15}(\sum_{n = 0}^{\infinity}{(-1)^n(\frac{z - 1}{3})^n)$

Now, upon substituting n = 0, 1, 2 to get the first three terms, I get the following:
$-\frac{1}{5}$, $\frac{13(z - 1)}{180}$, $\frac{19(z - 1)^2}{1080}$

These are wrong, and I'm not sure why. :/

Any assistance would be appreciated.

2. ## Re: Taylor series question

\displaystyle \begin{align*} \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2} &= -\frac{1}{15}\left(\frac{1}{1 - \frac{z}{3}}\right) - \frac{1}{10}\left[\frac{1}{1 - \left(-\frac{z}{2}\right)}\right] \\ &= -\frac{1}{15}\sum_{r = 0}^{\infty}{\left(\frac{z}{3}\right)^r} - \frac{1}{10}\sum_{s = 0}^{\infty}{\left(-\frac{z}{2}\right)^s}\end{align*}

3. ## Re: Taylor series question

It needs to be around $z_0 = 1$. Does yours work for that Prove It?

4. ## Re: Taylor series question

OK I worked out what I did wrong with part ii). I thought the series was $\frac{1}{z \pm 1}$, but it's $\frac{1}{1 \pm z}$. So my solution is correct if I just add a negative to the start of my series.

However, I'm still not sure what I'm supposed to do for part iii). Thanks.

5. ## Re: Taylor series question

Originally Posted by Glitch
It needs to be around $z_0 = 1$. Does yours work for that Prove It?
I should have read the question properly :P