Q. 36 - Calculus

Question

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

$\sin 3x$

Step-by-Step Solution

Answer

Ans: The Maclaurin series of the function $f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k{3}^{2k+1}}{(2k+1)!}{x}^{2k+1}$

Step 1. Given information:

$f\left(x\right)=\sin 3x$

Step 2. Finding the general form Maclaurin series:

Since for any function $f$ with derivatives of all orders at the point $x_0=0$, then the Maclurin series is

$f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}}\left(0\right)}{3!}{x}^3+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{4!}{x}^4+\dots$

Or, we can write the general form Maclurin series of the function $f$ is

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function $f\left(x\right)=\sin 3x$

$\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& \sin 3x& 0& 0\\ 1& 3\cos 3x& 3& 3\\ 2& -9\sin 3x& 0& 0\\ 3& -27\cos 3x& -27& -\frac{27}{3!}\\ ·& *& ·& ·\\ 2k& (-9{)}^k\sin 3x& 0& ·\\ 2k+1& (-1{)}^k{3}^{2k+1}\cos 3x& (-1{)}^k{3}^{2k+1}& (-1{)}^k\frac{3^{2k+1}}{(2k+1)!}\\ ·& ·& ·& ·\\ ·& ·& ·& ·\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=\sin 3x$ is $0+3·x+\frac{0}{2!}{x}^2+\frac{(-27)}{3!}{x}^3+\frac{0}{4!}{x}^4+\dots$

Or, we can write as

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