Question

To find the power series representation for the function \(f(x) = \frac{x}{{2{x^2} + 1}}\) and determine the interval of convergence.

Short answer

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

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\(f(x) = \frac{x}{{2{x^2} + 1}}\)

Express the function \(f(x) = \frac{x}{{2{x^2} + 1}}\) as follows,

\(\frac{x}{{2{x^2} + 1}} = \frac{x}{{1 - \left( { - 2{x^2}} \right)}}\)

Compare the equation with the geometric function as\(f(x)\)is a sum of a geometric series with initial term \(a = x\) and common ratio \(r = - 2{x^2}\).

Therefore the sum is,

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

Simplification of the expression\(\sum\limits_{n = 0}^\infty   {(x)} {\left( { - 2{x^2}} \right)^n}\)

Simplify the further terms as shown below,

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

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

Calculation for the interval convergence

We know the geometric series converges when \(|r| < 1\), it can be conclude that \(\left| { - 2{x^2}} \right| < 1\).

\(\left| {2{x^2}} \right| < 1\)

\(|x{|^2} < \frac{1}{2}\)

\(|x| < \frac{1}{{\sqrt 2 }}\)

\( - \frac{1}{{\sqrt 2 }} < x < \frac{1}{{\sqrt 2 }}\)

Therefore, the interval of convergence is \(\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)\) and the radius of convergence is \(\frac{1}{{\sqrt 2 }}\).