Show that$\frac{\mathbf{2}}{\sqrt{\mathbf{4}\mathbf{-}\mathbf{x}}}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathit{x}$with an error less than$\frac{\mathbf{1}}{\mathbf{32}}$for 0<x<1. Hint: Let x=4y , and use the theorem (14.4) .

$$\begin{gathered} \mathrm{The}\mathrm{function}\mathrm{is} \\ \frac{2}{\sqrt{4-\mathrm{x}}}=1+\frac{1}{8}\mathrm{x}. \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{formula}\mathrm{states}\mathrm{that}: \\ \frac{1}{\sqrt{1+\mathrm{x}}}=1-\frac{1}{2}\mathrm{x}+\frac{3}{8}{\mathrm{x}}^2-\frac{5}{16}{\mathrm{x}}^3+\frac{35}{128}{\mathrm{x}}^4+\dots \dots \\ \mathrm{Substitute}\mathrm{x}\mathrm{by}-\mathrm{y}\mathrm{as}: \\ \frac{1}{\sqrt{1-\mathrm{y}}}=1+\frac{1}{2}\mathrm{y}+\frac{3}{8}{\mathrm{y}}^2+\dots \dots . \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{theorem}\mathrm{states}\mathrm{that}: \\ \mathrm{If}\mathrm{S}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{coverages}\mathrm{for}\left|\mathrm{x}\right|<1,\mathrm{and}\mathrm{if}\left|{\mathrm{a}}_{\mathrm{n}+1}\right|<\left|{\mathrm{a}}_{\mathrm{n}}\right|\mathrm{for}\mathrm{n}>\mathrm{N},\mathrm{then} \\ \left|\mathrm{S}-\sum _{\mathrm{n}=0}^{\mathrm{N}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right|<\frac{\left|{\mathrm{a}}_{\mathrm{N}-1}{\mathrm{x}}^{\mathrm{N}-1}\right|}{\left(1-\left|\mathrm{x}\right|\right)} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{theorem},\mathrm{and}\mathrm{we}\mathrm{get}, \end{gathered}$$

The calculated series is Underestimate if the Error is less than zero. The calculated series is Overestimate if the Error is more than zero.

The function is:

$\frac{2}{\sqrt{4-x}}=1+\frac{1}{8}x$

The formula states that:

$\frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+{\frac{3}{8}}^2-\frac{5}{16}{x}^3+\frac{35}{128}{x}^4+$ ……

$$\begin{gathered} \mathrm{The}\mathrm{theorem}\mathrm{states}\mathrm{that}: \\ \mathrm{If}\mathrm{S}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{coverages}\mathrm{for}\left|\mathrm{x}\right|<1,\mathrm{and}\mathrm{if}\left|{\mathrm{a}}_{\mathrm{n}+1}\right|<\left|{\mathrm{a}}_{\mathrm{n}}\right|\mathrm{for}\mathrm{n}>\mathrm{N},\mathrm{then} \\ \left|\mathrm{S}-\sum _{\mathrm{n}=0}^{\mathrm{N}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right|<\frac{\left|{\mathrm{a}}_{\mathrm{N}-1}{\mathrm{x}}^{\mathrm{N}-1}\right|}{\left(1-\left|\mathrm{x}\right|\right)} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{theorem},\mathrm{and}\mathrm{we}\mathrm{get}, \end{gathered}$$

$$\begin{gathered} \left|S-\sum _{n=0}^1{a}_n{x}^n\right|<\frac{\left|a_2{x}^2\right|}{\left(1-\left|x\right|\right)} \\ \left|\frac{1}{\sqrt{1-y}}-\left(1+\frac{y}{2}\right)\right|<\frac{\left|a_2{y}^2\right|}{\left(1-\left|y\right|\right)} \\ \left|\frac{1}{\sqrt{1-y}}-\left(1+\frac{y}{2}\right)\right|<\frac{\left|{\displaystyle \frac{-3}{8}}\times {\displaystyle \frac{1}{4^2}}\right|}{1-\left|{\displaystyle \frac{1}{4}}\right|} \\ \left|\frac{1}{\sqrt{1-y}}-\left(1+\frac{y}{2}\right)\right|<\frac{0.0234}{0.75} \end{gathered}$$

$$\begin{gathered} \mathrm{Solve}\mathrm{further},\mathrm{and}\mathrm{we}\mathrm{get}, \\ \left|\frac{1}{\sqrt{1-\mathrm{y}}}-\left(1+\frac{\mathrm{y}}{2}\right)\right|<\frac{0.0234}{0.75} \\ \left|\frac{1}{\sqrt{1-\mathrm{y}}}-\left(1+\frac{\mathrm{y}}{2}\right)\right|<0.0312 \\ \mathrm{Hence}\mathrm{the}\mathrm{Error}\mathrm{is}\mathrm{less}\mathrm{than}\frac{1}{32}. \\ \mathrm{The}\mathrm{statement}\mathrm{has}\mathrm{been}\mathrm{verified}. \end{gathered}$$