# How do you find a power series representation for f(x)= 1/(1+4x^2) and what is the radius of convergence?

Sep 28, 2015

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n}$ with radius of convergence $\frac{1}{2}$

#### Explanation:

Consider the power series:

${\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n} = 1 - 4 {x}^{2} + 16 {x}^{4} - 64 {x}^{6} + \ldots$

Then:

$\left(1 + 4 {x}^{2}\right) \left({\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n}\right)$

$= {\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n} + 4 {x}^{2} {\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n}$

$= {\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n} - {\sum}_{n = 1}^{\infty} {\left(- 4 {x}^{2}\right)}^{n}$

$= {\left(- 4 {x}^{2}\right)}^{0} = 1$

provided the sum ${\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n}$ converges.

So

${\sum}_{n = 0}^{\infty} {\left(- 4 {x}^{2}\right)}^{n} = \frac{1}{1 + 4 {x}^{2}} = f \left(x\right)$

This is a geometric sequence, so will converge if the common ratio has absolute value $< 1$.

That is:

$\left\mid - 4 {x}^{2} \right\mid < 1$, so ${x}^{2} < \frac{1}{4}$, so $\left\mid x \right\mid < \frac{1}{2}$

In general

$\frac{1}{1 + a} = {\sum}_{n = 0}^{\infty} {\left(- a\right)}^{n}$

which converges if $\left\mid a \right\mid < 1$.