\((1 + x)^n = 1 + nx + \dfrac {n(n - 1)}{2!} \: x^2 + \dfrac {n(n - 1)(n - 2)}{3!} \: x^3 + \dotso\)

\(n\) is rational and \(|x| < 1\)

Example:

Expand \((2 + x^2)^{-2}\) in ascending powers of \(x\), up to and including the term of \(x^4\), simplifying the coefficient. Find the set of values of \(x\) for which the expansion is valid.

Expansion of \((2 + x^2)^{-2}\)

\begin{equation*} \begin{split} & \left(2 \:.\: \left[1 + \dfrac 12 x^2 \right] \right)^{-2} \\\\ & 2^{-2} \:.\: \left[1 + \dfrac 12 x^2 \right]^{-2} \\\\ & \dfrac 14 \:.\: \left[1 + (-2) \:.\: \dfrac 12 x^2 + \dfrac {(-2) \:.\: (-2 - 1)}{2!} \: \left(\dfrac 12 x^2 \right)^2 \right] \\\\ & \dfrac 14 \:.\: \left[1 - x^2 + \dfrac 34 x^4 \right] \\\\ & \bbox[5px, border: 2px solid magenta] {\dfrac 14 - \dfrac 14 x^2 + \dfrac {3}{16} x^4} \end{split} \end{equation*}

The expansion is valid when \(\left| \dfrac 12 x^2 \right| < 1\)

\begin{equation*} \begin{split} & \left| \dfrac 12 x^2 \right| < 1 \\\\ & \dfrac 12 \left| x^2 \right| < 1 \\\\ & \left| x^2 \right| < 2 \\\\ & x^2 < 2 \\\\ & x^2 - 2 < 0 \\\\ & (x + \sqrt{2})(x - \sqrt{2}) < 0 \\\\ & \bbox[5px, border: 2px solid magenta] {- \sqrt{2} < x < \sqrt{2}} \end{split} \end{equation*}