Re: Taylor series question

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

Re: Taylor series question

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

Re: Taylor series question

OK I worked out what I did wrong with part ii). I thought the series was $\displaystyle \frac{1}{z \pm 1}$, but it's $\displaystyle \frac{1}{1 \pm z}$. 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.

Re: Taylor series question

Quote:

Originally Posted by

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

I should have read the question properly :P