Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

$\mathit{s}\mathit{i}\mathit{n}{\mathit{x}}^{\mathbf{2}}$

Short answer

$\sin x^2=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2(2n+1)}}{(2n+1)!}$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Find Maclaurin series

Consider the function$\sin x^2 ---\left(1\right)$

We have Maclaurin series:

$$\begin{gathered} \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n}{(2n+1)!}{x}^{2n+1} ---\left(2\right) \end{gathered}$$

Replace$x$ by$x^2$ in (2) we get,

$\sin x^2=x^2-\frac{{\left(x^2\right)}^3}{3!}+\frac{{\left(x^2\right)}^5}{5!}-\frac{{\left(x^2\right)}^7}{7!}+\dots$

Thus the Maclaurin series is $\sin x^2=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2(2n+1)}}{(2n+1)!}$.