# Stuck with a Taylor series

• April 16th 2012, 09:13 AM
purpleclover
Stuck with a Taylor series
So i have the f(x)=1/(x^2+5) and it says i must find the Taylor series around the point x0=0....iv tryed derivating it a couple of times hopeing i would see a pattern but it just feels wrong aproaching it this way... can someone please tell me how to aproach these type of exercices?
• April 16th 2012, 09:55 AM
Prove It
Re: Stuck with a Taylor series
Quote:

Originally Posted by purpleclover
So i have the f(x)=1/(x^2+5) and it says i must find the Taylor series around the point x0=0....iv tryed derivating it a couple of times hopeing i would see a pattern but it just feels wrong aproaching it this way... can someone please tell me how to aproach these type of exercices?

\displaystyle \begin{align*} \frac{1}{5 + x^2} &= \frac{1}{5}\left(\frac{1}{1 + \frac{x^2}{5}}\right) \\ &= \frac{1}{5}\left[\frac{1}{1 - \left(-\frac{x^2}{5}\right)}\right] \end{align*}

You should also know that \displaystyle \begin{align*} \sum_{n = 0}^{\infty}a\,r^n = \frac{a}{1 - r} \textrm{ for } |r| < 1 \end{align*}, and here \displaystyle \begin{align*} a = 1 \end{align*} and \displaystyle \begin{align*} r = -\frac{x^2}{5} \end{align*}, so the series is

\displaystyle \begin{align*} \frac{1}{5}\left[\frac{1}{1 - \left(-\frac{x^2}{5}\right)}\right] = \sum_{n = 0}^{\infty}{\left(-\frac{x^2}{5}\right)^n} \end{align*}

where

\displaystyle \begin{align*} \left|-\frac{x^2}{5}\right| &< 1 \\ \left|\frac{x^2}{5}\right| &< 1 \\ -1 < \frac{x^2}{5} &< 1 \\ -5 < x^2 &< 5 \\ x^2 &< 5 \\ |x| &< \sqrt{5} \\ -\sqrt{5} < x &< \sqrt{5} \end{align*}
• April 16th 2012, 09:59 AM
purpleclover
Re: Stuck with a Taylor series
Thanks again Prove it! ur a life savior