Question

To find the power series representation for the function \(f(x) = \frac{2}{{3 - x}}\) and determine the interval of convergence.

Short answer

The power series representation for the function \(f(x) = 2\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{3^{n + 1}}}}} \) and the interval of convergence is\(( - 3,3)\).

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\(f(x) = \frac{2}{{3 - x}}\)

Consider the function \(f(x) = \frac{2}{{3 - x}}\)

Divide the numerator and denominator by\(3\).

\(\begin{aligned}\frac{2}{{1 - x}} &= \frac{{\frac{2}{3}}}{{\frac{{1 - x}}{3}}}\\\frac{2}{{1 - x}} &= \frac{2}{3} \cdot \frac{1}{{1 - \left( {\frac{x}{3}} \right)}}\end{aligned}\)

\(\frac{1}{{1 - x}} = 1 + x + {x^2} + {x^3} + \cdots = \sum\limits_{n = 0}^\infty {{x^n}} \)

Substitute \(\frac{x}{3}\) for \(x\) in the above equation,

\(\begin{aligned}\frac{1}{{1 - x}} &= \frac{1}{{1 - \left( {\frac{x}{3}} \right)}}\\\frac{1}{{1 - x}} &= 1 + \left( {\frac{x}{3}} \right) + {\left( {\frac{x}{3}} \right)^2} + {\left( {\frac{x}{3}} \right)^3} + \ldots &= \sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{3}} \right)}^n}} \end{aligned}\)

Therefore, \(\frac{2}{{1 - x}} = \frac{2}{3} \cdot \sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{3}} \right)}^n}} \).

Simplification of the expression\(\frac{2}{{1 - x}}\)

Simplify the terms further as shown below,

\(\begin{aligned}\frac{2}{{1 - x}} = \frac{2}{3}\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{3^n}}}} \\\frac{2}{{1 - x}} = 2\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{3 \cdot {3^n}}}} \\\frac{2}{{1 - x}} = 2\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{3^{n + 1}}}}} \end{aligned}\)

Thus, the power series representation of \(f(x) = 2\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{3^{n + 1}}}}} \).

Calculation for the interval of convergence

We know the series is converges if \(|r| < 1,\left| {\frac{x}{3}} \right| < 1\).

\(\left| {\frac{x}{3}} \right| < 1\)

\(|x| < 3\)

\( - 3 < x < 3\)

Therefore, the interval of convergence is \(( - 3,3)\) and the radius of convergence is\(3\).