# Convergence of a Taylor Series

• Oct 1st 2010, 03:02 PM
eay444
Convergence of a Taylor Series
f(x)=1/(3+2x)

Problem: For which values does the Taylor Polynomial converge to the function?

I've been working on this problem for a while messing around with Legrange's Theorem for the remainder term trying to use different bounds to show how for certain intervals of x the remainder must go to zero therefore the taylor polynomial must converge at those points. This works for some intervals but then on others it doesn't work at all. I'm wondering if I'm going about this the completely wrong way and if there's something else that I can do to show convergence? Can somebody please point me in the right direction?
• Oct 1st 2010, 04:52 PM
skeeter
note ...

$\displaystyle \displaystyle \frac{1}{3+2x} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{2}{3}x\right)}$

familiar with the interval of convergence for the sum of an infinite geometric series?
• Oct 4th 2010, 06:52 AM
eay444
Thanks that helps alot.
• Oct 4th 2010, 07:56 AM
HallsofIvy
A "Taylor Series" is a power series of the form $\displaystyle \sum_{n=0}^\infty a_n(x- x_0)^n$. Both the Taylor series and its radius of convergence depend upon what $\displaystyle x_0$ you are using and you don't give that.

Skeeter assumed you meant the "MacLaurin series", the Taylor series about x= 0.

In fact, there is an easy way of answering the general question. Notice that $\displaystyle \frac{1}{3+ 2x}$ is defined and continous (and, in fact, infinitely differentiable) for all x except x= -3/2. The radius of convergence about $\displaystyle x= x_0$ will be $\displaystyle |x_0+ 3/2|$. That is, the series will converge for all x between $\displaystyle x_0$ and -3/2 and, since this is a "radius" for an equal distance on the other side of $\displaystyle x_0$. (If you posed the same question in the complex plane, the series would converge on a disk having center at $\displaystyle x_0$ and having -3/2 on its boundary so actually having r= |x_0+ 3/2| as radius.)

You could prove that, for this particular function, by writing it as $\displaystyle \frac{1}{3+ 2x}= \frac{1}{3+ 2(x-x_0)+ 2x_0}= \frac{1}{(3- 2x_0)+ 2z}$ where $\displaystyle z= x- x_0$. Now, as earboth factored a 3 out of the denominator, factor out that $\displaystyle 3- 2x_0$, leaving $\displaystyle \frac{1}{3- 2x_0}\frac{1}{1- (-\frac{2z}{3-2x_0})}$ and recognizing that as the sum of a geometric series.