Q. 56 - Calculus

Question

In Exercises 49–56 find the Taylor series for the specified function and the given value of $x_0$.

56. $\sin 3x,\frac{\mathrm{\pi }}{6}$

Step-by-Step Solution

Answer

The Taylor series of the function $\sin 3x$ at $x=\frac{\mathrm{\pi }}{6}$is $p_n\left(x\right)=\sum _{k=0}^{\infty }\frac{{(-1)}^{k+1}{3}^{2k}}{\left(2k\right)!}{(x-\frac{\mathrm{\pi }}{6})}^{2k}$

Step 1. Given information.

We have given that $f\left(x\right)=\sin 3x$and $x_0=\frac{\mathrm{\pi }}{6}$.

Step 2. Table of the Taylor series

Any function $f$ with a derivative of order $n$, the Taylor series at $x=\frac{\mathrm{\pi }}{6}$ is given by,

$P_n\left(x\right)=f\left(\frac{\mathrm{\pi }}{6}\right)+f^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)(x-\frac{\mathrm{\pi }}{6})+\frac{f^{\text{'}\text{'}}\left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{2!}{(x-\frac{\mathrm{\pi }}{6})}^2+\frac{f^{\text{'}\text{'}\text{'}}\left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{3!}{(x-\frac{\mathrm{\pi }}{6})}^3+....$

So, let us find the derivatives of the given function and construct a table of the Taylor series for the function  $f\left(x\right)=\sin 3x$ at $\frac{\mathrm{\pi }}{6}$.

Step 3. Required Taylor series

Therefore, the Taylor series of the function $f\left(x\right)=\sin 3x$ at $x=\frac{\mathrm{\pi }}{6}$.

$$\begin{gathered} P_n\left(x\right)=1+0·(x-\frac{\mathrm{\pi }}{6})+\frac{-3^2}{2!}{(x-\frac{\mathrm{\pi }}{6})}^2+\frac{0}{3!}{(x-\frac{\mathrm{\pi }}{6})}^3 \\ +...+\frac{{(-1)}^k{3}^{2k}}{\left(2k\right)!}{(x-\frac{\mathrm{\pi }}{6})}^{2k}+\frac{0}{(2k+1)!}{(x-\frac{\mathrm{\pi }}{6})}^{2k+1}+... \end{gathered}$$

otherwise, it can be written as

$P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{{(-1)}^{k+1}{3}^{2k}}{\left(2k\right)!}{(x-\frac{\mathrm{\pi }}{6})}^{2k}$