Q. 30 - 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, $R_4\left(x\right)$.

$\frac{1 + x }{1 - x}$ 

Step-by-Step Solution

Answer

The remainder is $\frac{\mathbf{2}}{{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{)}}^{\mathbf{6}}}{\mathit{x}}^{\mathbf{5}}$

Step 1 : Given Information

Given equation : $\frac{1 + x }{1 - x}$

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) $f^{\left(1\right)}\left(\frac{1+x}{1-x}\right)=2{(1-x)}^{-2}$

2) $f^{\left(2\right)}\left(\frac{1+x}{1-x}\right)=4{(1-x)}^{-3}$

3) $f^{\left(3\right)}\left(\frac{1+x}{1-x}\right)=12{(1-x)}^{-4}$

4) $f^{\left(4\right)}\left(\frac{1+x}{1-x}\right)=48{(1-x)}^{-5}$

5) $f^{\left(5\right)}\left(\frac{1+x}{1-x}\right)=240{(1-x)}^{-6}$

So,

$$\begin{gathered} R_4\left(x\right)=\frac{240{(1-c)}^{-6}}{5!}{x}^5 \\ =\boxed{\frac{\mathbf{2}}{{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{)}}^{\mathbf{6}}}{\mathit{x}}^{\mathbf{5}}} \end{gathered}$$