Question

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

Short answer

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

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

Simplify the function \(f(x) = \frac{{{x^2}}}{{{a^3} - {x^3}}}\) as shown below,

\(\frac{{{x^2}}}{{{a^3} - {x^3}}} = \frac{{{x^2}}}{{{a^3}\left( {1 - \left( {\frac{{{x^3}}}{{{a^3}}}} \right)} \right)}}\)

The resultant expression \(\frac{{{x^2}}}{{{a^3}\left( {1 - \left( {\frac{{{x^3}}}{{{a^3}}}} \right)} \right)}}\) is in the form of \(\frac{a}{{1 - r}}\), where \(a = \frac{{{x^2}}}{{{a^3}}}\) and the common ratio \(r = \frac{{{x^3}}}{{{a^3}}}\).

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

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

Simplification of the expression\(f(x) = \sum\limits_{n = 0}^\infty   {\frac{{{x^{3n + 2}}}}{{{a^{3n + 3}}}}} \)

Let,

\({a_n} = \frac{{{x^{3n + 2}}}}{{{a^{3n + 3}}}}\).

Then,

\(\begin{aligned}{a_{n + 1}} &= \frac{{{x^{3(n + 1) + 2}}}}{{{a^{3(n + 1) + 3}}}}\\{a_{n + 1}} &= \frac{{{x^{3n + 5}}}}{{{a^{3n + 6}}}}\end{aligned}\)

Now on solving,\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\).

We get,

\(\begin{aligned}\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mid \frac{{\frac{{{x^{3n + 5}}}}{{{a^{3n + 6}}}}}}{{\frac{{{x^{3n + 2}}}}{{{a^{3n + 3}}}}\mid }}\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{{x^{3n + 5}}}}{{{a^{3n + 6}}}} \cdot \frac{{{a^{3n + 3}}}}{{{x^{3n + 2}}}}} \right|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{{x^{3n + 2}} \cdot {x^3}}}{{{a^{3n + 3}} \cdot {a^3}}} \cdot \frac{{{a^{3n + 3}}}}{{{x^{3n + 2}}}}} \right|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{{x^3}}}{{{a^3}}}} \right|\end{aligned}\)

Calculation for the interval of convergence

Ratio test:

If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\), then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is absolutely convergent.

Apply the limit,

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{{x^3}}}{{{a^3}}}} \right|\).

Thus, by the Ratio test,

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

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

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

\(|x| < a\)

Here, the series is divergent at the end points \(x = - a\) and \(x = a\) by the test for divergent.

That is, \( - a < x < a\).

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