Q. 29 - 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, $R_4\left(x\right)$.

$\sqrt{1 - x}$

Step-by-Step Solution

Answer

The remainder is $-\frac{\mathbf{7}}{\mathbf{256}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{)}}^{\frac{\mathbf{9}}{\mathbf{2}}}}{\mathit{x}}^{\mathbf{5}}$.

Step 1 : Given Information

Given equation : $\sqrt{1 - x}$

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

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

3) $f^{\left(3\right)}\left(\sqrt{1-x}\right)=-\frac{3}{8}{(1-x)}^{-\frac{5}{2}}$

4) $f^{\left(4\right)}\left(\sqrt{1-x}\right)=-\frac{15}{16}{(1-x)}^{-\frac{7}{2}}$

5) $f^{\left(5\right)}\left(\sqrt{1-x}\right)=-\frac{105}{32}{(1-x)}^{-\frac{9}{2}}$

So,

$$\begin{gathered} R_4\left(x\right)=\frac{-\frac{105}{32}{(1-c)}^{-\frac{9}{2}}}{5!}{x}^5 \\ =\boxed{-\frac{\mathbf{7}}{\mathbf{256}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{)}}^{\frac{\mathbf{9}}{\mathbf{2}}}}{\mathit{x}}^{\mathbf{5}}} \end{gathered}$$