Don's question at Yahoo! Answers (Taylor series)

[Fernando Revilla]

Here is the question:

is there a short way to work this out?

Here is a link to the question:

How to Find the Taylor series for the function f(x)=(1)/(x) centered at a=-3? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Don,

Denoting $t=x+3$ and using the geometric series:

$$\frac{1}{x}=\frac{1}{t-3}=-\frac{1}{3}\cdot\frac{1}{1-\frac{t}{3}}=-\frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{t}{3}\right)^n \qquad \left(\;\left|\frac{t}{3}\right|<1\;\right)$$ Hence, $f(x)=-\displaystyle\sum_{n=0}^{\infty}\frac{(x+3)^n}{3^{n+1}}$, valid expasion for $|x+3|<3$, or equivalently for $x\in (-6,0).$