Q. 31 - Calculus

Question

In Exercises 21–30 in Section 8.2 you were asked to find the fourth Maclaurin polynomial $P_4\left(x\right)$ for the specified function. In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_4\left(x\right)$.

$x \sin x$ 

Step-by-Step Solution

Answer

The remainder is $\frac{\mathbf{c}\mathbf{ }\mathbf{cos}\mathbf{c}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{5}\mathbf{sin}\mathbf{c}}{\mathbf{120}}{\mathit{x}}^{\mathbf{5}}$

Step 1 : Given Information

Given equation : $x \sin x$

Theory used : For $n\mathit{>}\mathit{0}, \mathrm{if} \mathit{\left|}{\mathit{f}}^{\mathit{(}\mathit{n}\mathit{+}\mathit{1}\mathit{)}}\mathit{\left(}\mathit{c}\mathit{\right)}\mathit{\right|}\mathit{\le }\mathit{1}$for every value of$x$, then using the Lagrange's form for the remainder, we have 

$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$

So, the Lagrange's form for the remainder, $R_4\left(x\right) \mathrm{is}$

$R_4\left(x\right)=\frac{f^{\left(5\right)}\left(c\right)}{5!}{x}^5$

Step 2 : Calculating Lagrange’s form for the corresponding remainder

1) $f^{\left(1\right)}(x \sin x)=x\cos x + \sin x$

2) $f^{\left(2\right)}(x \sin x)=-x\sin x + 2\cos x$

3) $f^{\left(3\right)}(x \sin x)=-x\cos x - 3\sin x$

4) $f^{\left(4\right)}(x \sin x)=x\sin x - 4\cos x$

5) $f^{\left(5\right)}(x \sin x)=x\cos x + 5\sin x$

So, 

$$\begin{gathered} R_4\left(x\right)=\frac{c \cos c + 5\sin c}{5!}{x}^5 \\ =\boxed{\frac{\mathbf{c}\mathbf{ }\mathbf{cos}\mathbf{c}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{5}\mathbf{sin}\mathbf{c}}{\mathbf{120}}{\mathit{x}}^{\mathbf{5}}} \end{gathered}$$