Question

To find the power series representation for the function \(f(x) = \frac{{1 + x}}{{1 - x}}\) and determine the interval of convergence.

Short answer

The power series representation for the function \(f(x) = 1 + 2\sum\limits_{n = 1}^\infty {{{(x)}^n}} \) and the interval of convergence is \(( - 1,1)\).

Step-by-step solution

Concept of Geometric Series

Geometric Series

The sum of the geometric series with initial term\(a\)and common ratio\(r\)is\(\sum\limits_{n = 0}^\infty a {r^n} = \frac{a}{{1 - r}}\)

Calculation of the expression\(f(x) = \frac{{1 + x}}{{1 - x}}\).

Simplify the function \(f(x) = \frac{{1 + x}}{{1 - x}}\) as shown below,

\(\begin{aligned}f(x) &= \frac{{1 + x}}{{1 - x}}\\f(x) &= \frac{{2 - 1 + x}}{{1 - x}}\\f(x) &= \frac{2}{{1 - x}} - \frac{{1 - x}}{{1 - x}}\\f(x) &= \frac{2}{{1 - x}} - 1\end{aligned}\)

The resultant expression \(\frac{2}{{1 - x}} - 1\) is in the form of \(\frac{a}{{1 - r}} - 1\), where \(a = 2\) and the common ratio \(r = x\).

\(\frac{2}{{1 - x}} - 1 = \sum\limits_{n = 0}^\infty {(2)} {(x)^n} - 1,\quad \left( {\sum\limits_{n = 0}^\infty a {r^n} = \frac{a}{{1 - r}}} \right)\)

\(\begin{aligned}\frac{2}{{1 - x}} - 1 &= \left( {2 + 2x + 2{x^2} + \cdots } \right) - 1\\\frac{2}{{1 - x}} - 1 &= \left( {2 + \sum\limits_{n = 1}^\infty {(2)} {{(x)}^n}} \right) - 1\\\frac{2}{{1 - x}} - 1 &= 1 + 2\sum\limits_{n = 1}^\infty {{{(x)}^n}} \end{aligned}\)

Thus, the power series representation of the function is \(f(x) = 1 + 2\sum\limits_{n = 1}^\infty {{{(x)}^n}} \).

Calculation for the interval of convergence

We know the geometric series converges when \(|r| < 1\), it can be conclude that \(|x| < 1\).

\(|x| < 1\)

\( - 1 < x < 1\)

Therefore, the interval of convergence is \(( - 1,1)\).