Question

To find the power series representation for the function for the function \(f(x) = \frac{{{x^2} + x}}{{{{(1 - x)}^3}}}\) and determine the radius of convergence.

Short answer

The power series radius of convergence is \(R = 1\) and the power series representation is \(\frac{{\left( {{x^2} + x} \right)}}{{{{(1 - x)}^3}}} = \sum\limits_{n = 1}^\infty {{n^2}} {x^n}\).

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\(f(x) = \frac{{{x^2} + x}}{{{{(1 - x)}^3}}}\)

The power series representation of \(\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \)

Let

\(f(x) = \frac{{{x^2} + x}}{{{{(1 - x)}^3}}}\)

Then, by the power series

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

Double differentiate for \(\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \) on both sides with respect to \(x\)

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{{1 - x}}} \right]\)

\(\frac{{dy}}{{dx}} = \frac{d}{{d(1 - x)}}\left[ {\frac{1}{{1 - x}}} \right] \times \frac{{d(1 - x)}}{d}\)

\(\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(1 - x)}^2}}} \times ( - 1)\)

\(\frac{{dy}}{{dx}} = \frac{1}{{{{(1 - x)}^2}}}\)

Differentiate the series as shown below:

\(\frac{{dy}}{{dx}} = {(x - 1)^{ - 2}}\frac{{{d^2}y}}{{d{x^2}}}\)

\(\frac{{dy}}{{dx}} = - 2{(x - 1)^{ - 3}}\)

\(\frac{{dy}}{{dx}} = - \frac{2}{{{{(x - 1)}^3}}}\)

\(\frac{{dy}}{{dx}} = \frac{2}{{{{(1 - x)}^3}}}\)

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

Simplification of the expression\(\frac{2}{{{{(1 - x)}^3}}} = \sum\limits_{n = 2}^\infty  n (n - 1){x^{n - 2}}\)

Multiply both sides by \(\frac{{{x^2} + x}}{2}\) for simplification.

\(\frac{{2\left( {{x^2} + x} \right)}}{{2{{(1 - x)}^3}}} = \sum\limits_{n = 2}^\infty n (n - 1){x^{n - 2}} \times \frac{{{x^2} + x}}{2}\frac{{\left( {{x^2} + x} \right)}}{{{{(1 - x)}^3}}}\)

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

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

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

Replace \(n\) by \(n - 1\) in the second part of the summation.

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

When \(n = 1\) in the first part of the summation, \(\sum\limits_{n = 2}^\infty {\frac{{n(n - 1)}}{2}} {x^{n - 1}} = \sum\limits_{n = 1}^\infty {\frac{{(n + 1)n}}{2}} {x^n}\)

Expansion of the series\(f(x) = \frac{{{x^2} + x}}{{{{(1 - x)}^3}}}\)

Expand the series as follows:

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

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

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

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

Calculation of the expression\(f\left( x \right) = {n^2}{x^n}\) by ratio test

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

By using the ratio test,

Let

\({a_n} = {n^2}{x^n}\).

Then,

\({a_{n + 1}} = {(n + 1)^2}{x^{n + 1}}\).

Now for\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{{{(n + 1)}^2}{x^{n + 1}}}}{{{n^2}{x^n}}}} \right|\)

Take \(\mathop {\lim }\limits_{n \to \infty } \) on both sides.