Question

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Find a power series representation for the function. Determine the interval of convergence. (Give your power series representation centered at x=0)
f(x)=1/6+x

Answer 1

Answer:

The series representation is $\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}$ and the interval of convergence is (-6,6)

Step-by-step explanation:

We want to find a series, such that f(x) = \sum_{n=0}{\infty}a_n(x-a)^{n}[/tex], were a is the value that we are using to center the series expansion. In our case, a=0.

We will use the geometric series formula as follows. For |r|<1 then

$\frac{1}{1-r}=\sum_{n=0}^{\infty} r^n$

In our case, with some algebreaic manipulation we have that

$f(x) = \frac{1}{6+x} = \frac{1}{6}\frac{1}{1-(\frac{-x}{6})}$

Taking $r = \frac{-x}{6}$ we get that

$f(x) = \frac{1}{6}\sum_{n=0}^{\infty} (\frac{-x}{6})^n = \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}$

This representation is valid (that means that the series converges to the value of f(x)) only for |r|<1. That is

$\left|\frac{-x}{6}\right|= \left|\frac{x}{6}\right|$

which implies that |x|<6. So the interval of convergence is (-6,6).