Q. 39 - 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.

$x \sin x$

Step-by-Step Solution

Answer

Ans: The Maclaurin series of the function $f\left(x\right)=x\sin x$ is

$0+0·x+\frac{2}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{-4}{4!}{x}^4+\dots +0·x^{2k+1}+\frac{(-1{)}^k}{(2k+1)!}·x^{2k+2}$

Step 1. Given information:

$x \sin x$

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

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

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

$\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& x\sin x& 0& 0\\ 1& x\cos x+\sin x& 0& 0\\ 2& -x\sin x+2\cos x& 2& \frac{2}{2!}\\ 3& -x\cos x-3\sin x& 0& 0\\ 4& x\sin x-4\cos x& -4& \frac{-4}{4!}\\ ·& ·& ·& ·\\ 2k+1& (-1{)}^{k}x\cos x+(-1{)}^{k}k\sin x& 0& 0\\ 2k+2& (-1{)}^{k+1}x\sin x+(-1{)}^{k}k\cos x& +(-1{)}^{k}k& \frac{(-1{)}^k}{(2k+1)!}\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=x\sin x$ is

$0+0·x+\frac{2}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{-4}{4!}{x}^4+\dots +0·x^{2k+1}+\frac{(-1{)}^k}{(2k+1)!}·x^{2k+2}$

Or, we can write as

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