Question

Find a power series representation for the function and determine the interval of convergence (check my work please :D ).

Answer 1

\[f(x) = \frac{ 1+x }{ 1-x }\]

Answer 2

can u help me plz asap

Answer 3

\[= \frac{ 1 }{ 1-x }+\frac{ x }{ 1-x }\]\[\frac{ 1 }{ 1-x }=\sum_{n=0}^{\infty}x^n\]\[\frac{ x }{ 1-x }=x \sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}x^{n+1}\]\[\sum_{}^{}x^n+\sum_{}^{}x^{n+1}=\sum_{}^{}(x^n+x^{n+1})=\sum_{n=0}^{\infty}(1+x)x^n\]

Is the last step correct? Otherwise, I know how to do the problem.

Answer 4

http://math.stackexchange.com/questions/363650/power-series-representation-of-1x-1-x

Answer 5

It depends on what you consider a power series. The usual definition is a series of the form \(\sum\limits_n a_nx^n\), where \(a_n\) depends only on \(n\) and the only way \(x\) occurs in the summand is as a base to the \(n\)th power.

What you have is mathematically correct, but the last line isn't what you should be doing. Instead, for the second series, you should be shifting the index.
\[\sum_{n\ge0}x^{n+1}=x+x^2+x^2+\cdots=\sum_{n\ge1}x^n\]Now if you add and subtract \(1\), you can write
\[-1+1+\sum_{n\ge1}x^n=-1+\left(1+x+x^2+\cdots\right)=-1+\sum_{n\ge0}x^n\]This means a series representation for your function should be
\[\frac{1+x}{1-x}=-1+2\sum_{n\ge0}x^n\]But this isn't *the* power series representation, because the series above maintains that \(a_n=2\) for all \(n\ge0\), yet the first term (the constant term) is actually \(2x^0-1=1\). So to fix that, take out the first term of the summation and combine the constants to get
\[-1+2\sum_{n\ge0}x^n=-1+2+2\sum_{n\ge1}x^n=1+2\sum_{n\ge1}x^n\]and this is your power series. The coefficients are determined by the sequence
\[a_n=\begin{cases}1&\text{for }n=0\\[1ex]2&\text{for }n\ge1\end{cases}\]

Answer 6

As for the interval of convergence, you can easily find that via the ratio test:
\[\lim_{n\to\infty}\left|\frac{2x^{n+1}}{2x^n}\right|=|x|\]For the series to converge, this limit must be less than \(1\). The conclusion follows.