Question

How do you find a power series representation for $\dfrac{{{x^2}}}{{{{(1 - 2x)}^2}}}$?

Answer 1

Hint: We are going to solve this question by first relating the expansion of the given term to the expansion of the terms we already know. Then next we will use a substitution method in order to simplify the equation.

Complete step by step answer:
First we will start off by relating the given expression to the known power series expansion.
$\dfrac{1}{{1 - x}} = \sum\limits_{n = 1}^\infty {{x^n}} $
Now in the beginning we will just disregard the term ${x^2}$ and we will consider a function as
$f(x) = {x^2}\dfrac{1}{{{{(1 - 2x)}^2}}}$
Now we will take the integral of the term $\dfrac{1}{{{{(1 - 2x)}^2}}}$.
$\int {\dfrac{{dx}}{{{{(1 - 2x)}^2}}}} $
Now, here we will be using substitution.
$\begin{array}{*{20}{c}}
  u& = &{1 - 2x} \\
  {\,\,du}& = &{ - 2dx} \\
  { - \dfrac{1}{2}du}& = &{dx}
\end{array}$
\[ - \dfrac{1}{2}\int {{u^{ - 2}}du = \dfrac{1}{{2u}}} = \dfrac{1}{{2(1 - 2x)}}\]
Now, here we know that if we differentiate this integrated expression returns the original term which is $\dfrac{1}{{{{(1 - 2x)}^2}}}$. So, hence we can say,
\[f(x) = {x^2}\dfrac{d}{{dx}}\left( {\dfrac{1}{2}.\dfrac{1}{{1 - 2x}}} \right)\]
So, we can easily relate the terms \[\dfrac{1}{{1 - 2x}}\] to \[\dfrac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \].
\[\dfrac{1}{{1 - 2x}} = \sum\limits_{n = 0}^\infty {{{(2x)}^n}} = \sum\limits_{n = 0}^\infty {{2^n}{x^n}} \]
Hence, we can write,
\[f(x) = {x^2}\dfrac{d}{{dx}}\dfrac{1}{2}\sum\limits_{n = 0}^\infty {{2^n}{x^n}} \]
Now if we get the term $\dfrac{1}{2}$ in the interval.
\[f(x) = {x^2}\dfrac{d}{{dx}}\sum\limits_{n = 0}^\infty {{2^{n - 1}}{x^n}} \]
Now we will differentiate the summation with respect to $x$, recalling that differentiating the summation causes the index to shift up by $1$.
\[
  f(x) = {x^2}\dfrac{d}{{dx}}\sum\limits_{n = 0}^\infty {{2^{n - 1}}{x^n}} \\
  f(x) = {x^2}\sum\limits_{n = 0}^\infty {{2^{n - 1}}n{x^{n - 1}}} \\
 \]
Multiply in the
\[
  f(x) = {x^2}\sum\limits_{n = 0}^\infty {{2^{n - 1}}n{x^{n - 1 + 2}}} \\
  f(x) = {x^2}\sum\limits_{n = 0}^\infty {{2^{n - 1}}n{x^{n + 1}}} \\
 \]

Hence, the power series representation of the term is $\dfrac{{{x^2}}}{{{{(1 - 2x)}^2}}}$ is \[f(x) = {x^2}\sum\limits_{n = 0}^\infty {{2^{n - 1}}n{x^{n + 1}}} \].

Note: While relating such terms, make sure you relate along with the powers and the respective signs. While substituting any terms, substitute such that the integral becomes easy to solve. Also, remember that the differentiation of ${x^2}$ is $2x$.