Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function

Short answer

The Maclaurin series for the given function is \(\sum\limits_{n = 1}^\infty {{{( - 1)}^{n + 1}}} \frac{{{2^{2n - 1}}}}{{(2n)!}}{x^{2n}}\)

Step-by-step solution

Step 1: Applied the Maclaurin series

By replacing the Maclaurin series function of given function

\(\begin{aligned}{l}f(x) = {\sin ^2}x\\{\sin ^2}x = \frac{1}{2}(1 - \cos 2x)\\\cos x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{1}{{(2n)!}}}

Replaced x by 2x in Maclaurin series

Step 2: Maclaurin series function of the given function

Maclaurin series function of the given function \(\begin{aligned}{l}\cos (2x) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{(2n)!}}{(2x)^{2n}}\\ = 1 - \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} \frac{1}{{(2n)!}}{(2)^{2n}}{(x)^{2n}}\\\frac{1}{2}(1 - \cos 2x) = \frac{1}{2}\left( {1 - 1 - \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{2^{2n}}}}{{(2n)!}}{x^{2n}}} \right)\\ = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n + 1}}} \frac{{{2^{2n - 1}}}}{{(2n)!}}{x^{2n}}\end{aligned}\)