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\(\frac{{x - \sin x}}{{{x^3}}} = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n + 1}}} \frac{{{2^{2n + 1}}}}{{(2n)!}}{x^{2n}}\) and \(f(x) = \frac{1}{6},ifx = 0\)

Step-by-step solution

Step 1: Applied the Maclaurin series

By replacing the Maclaurin series function of given function

\(\begin{aligned}{l}f(x) = \frac{{x - \sin x}}{{{x^3}}}ifx \ne 0\\\sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{1}{{(2n + 1)!}}} {x^{2n + 1}}\end{aligned}\)

Step 2: Maclaurin series function of the given function

Maclaurin series function of the given function \(\begin{aligned}{l}\frac{{x - \sin x}}{{{x^3}}} = \frac{{x - \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{1}{{(2n + 1)!}}} {x^{2n + 1}}}}{{{x^3}}}\\ = \frac{{x - x - \sum\limits_{n = 1}^2 {{{( - 1)}^n}} \frac{1}{{(2n + 1)!}}{x^{2n + 1}}}}{{{x^3}}}\\ = - \sum\limits_{n = 1}^{2\hat n} {{{( - 1)}^n}} \frac{1}{{(2n + 1)!}}{x^{2n + 1}}\end{aligned}\)

Step 3: Substitute n=n+1  in series

Substitute the n values in series

\(\begin{aligned}{l}\frac{{x - \sin x}}{{{x^3}}} = \frac{{ - \sum\limits_{n = 0}^n {{{( - 1)}^n}} \frac{1}{{(2(n + 1) + 1)!}}{x^{2(n + 1) + 1}}}}{{{x^3}}}\\ = - \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{(2n + 3)!{x^{ - 3}}}}{x^{2n + 3}}\\ = - \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{(2n + 3)!}}{x^{2n}}\\\frac{{x - \sin x}}{{{x^3}}} = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n + 1}}} \frac{{{2^{2n + 1}}}}{{(2n)!}}{x^{2n}}\end{aligned}\)