Question

How do you find a power series representation for $\dfrac{1+x}{1-x}$ and what is the radius of convergence?

Answer 1

We recall the power series and radius of convergence. We find the power series of the given function using the power series of $\dfrac{1}{1-x}=1+x+{{x}^{2}}+...=\sum\limits_{n=1}^{\infty }{{{x}^{n}}}$. We do require simplification to get the power series. We use the fact that $\sum\limits_{n=1}^{\infty }{{{x}^{n}}}$ is geometric progression (GP) series which is always convergent if common ratio $\left| r \right| < 1$.

Complete step by step answer:
We know that power series are given with infinite terms
$\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{\left( x-c \right)}^{n}}}={{a}_{0}}+{{a}_{1}}\left( x-c \right)+{{a}_{2}}{{\left( x-c \right)}^{2}}+...$
Here $n$ is the power and $c$ is called centre. We say is series is convergent if we can find an $R$ such that series converges if around the centre $c$ if $\left| x-c \right| < R$ and divergent if $\left| x-c \right|>R$.We can obtain $R$ from the bounds of
$L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|$
We also know that we can approximate any infinitely differentiable function $f\left( x \right)$ as power series with centre $x=a$using Taylor’s approximation formula as written below
$f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( a \right)\left( x-a \right)}{n!}}$
We know the Taylor’s series approximation of the function $\dfrac{1}{1-x}$ with the centre $c=0$ is given by

& \dfrac{1}{1-x}=1+x+{{x}^{2}}+... \\\ & \Rightarrow \dfrac{1}{1-x}=\sum\limits_{n=0}^{\infty }{{{x}^{n}}}........\left( 1 \right) \\\ \end{aligned}$$ We are asked to find the power series for $\dfrac{1+x}{1-x}$ . Let us consider $$\dfrac{1+x}{1-x}=\dfrac{1}{1-x}+\dfrac{x}{1-x}.......\left( 2 \right)$$ We multiply $x$ on both sides of the power series (1) to have; $$\begin{aligned} & \Rightarrow \dfrac{x}{1-x}=\sum\limits_{n=0}^{\infty }{{{x}^{n+1}}} \\\ & \Rightarrow \dfrac{x}{1-x}=\sum\limits_{n=1}^{\infty }{{{x}^{n}}}......\left( 3 \right) \\\ \end{aligned}$$ We can write the power series s(1) as $$\begin{aligned} & \dfrac{1}{1-x}=\sum\limits_{n=0}^{\infty }{{{x}^{n}}}= \\\ & \Rightarrow \dfrac{1}{1-x}={{x}^{0}}+\sum\limits_{n=1}^{\infty }{{{x}^{n}}} \\\ & \Rightarrow \dfrac{1}{1-x}=1+\sum\limits_{n=1}^{\infty }{{{x}^{n}}}...\left( 4 \right) \\\ \end{aligned}$$ We put (3) and (4) in equation (2) to have; $$\begin{aligned} & \dfrac{1+x}{1-x}=1+\sum\limits_{n=1}^{\infty }{{{x}^{n}}}+\sum\limits_{n=1}^{\infty }{{{x}^{n}}} \\\ & \Rightarrow \dfrac{1+x}{1-x}=1+2\sum\limits_{n=1}^{\infty }{{{x}^{n}}} \\\ \end{aligned}$$ The above obtained power series is the required power series. The convergence of above series depends upon $\sum\limits_{n=1}^{\infty }{{{x}^{n}}}$ which is a GP series. We know that a GP series is convergent if is common ratio $\left| r \right| < 1$. Here the common ratio is $\dfrac{x}{1}=\dfrac{{{x}^{2}}}{x}=x$. The power series is convergent if $\left| x \right| < 1\Rightarrow \left| x-0 \right| < 1$.So the radius of the convergence is 1. **Note:** We note that the Taylor’s series with centre 0 is called a Mclaurin’s series. We can similarly take derivatives to find the power series of $\dfrac{1}{{{\left( 1-x \right)}^{2}}}$. We note that the GP series diverges for $\left| r \right|\ge 1$. The sum of converging GP series is given by $\dfrac{a}{1-r}$ where $a$ is the first term of the series. We can also power series $\dfrac{1}{1+x}=\dfrac{1}{1-\left( -x \right)}=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}{{x}^{n}}}$ .