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 \(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{1}{{{2^{2n}}(2n)!}}} {x^{4n + 1}}\)

Step-by-step solution

Step 1: Applied the Maclaurin series

By replacing the Maclaurin series function of given function

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

Replaced in maclaurin series

Step 2: Maclaurin series function of the given function

Maclaurin series function of the given function

\(\begin{aligned}{l}x\cos \left( {\frac{1}{2}{x^2}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{{\left( {\frac{1}{2}{x^2}} \right)}^{2n}}}}{{(2n)!}}} \\x\cos \left( {\frac{1}{2}{x^2}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{1}{{{2^{2n}}(2n)!}}} {x^{4n + 1}}\end{aligned}\)