Answer
${p_5}\left( {\frac{1}{2}} \right) \approx 1.6487$
Work Step by Step
From Example 2, we obtain the $n$th Maclaurin polynomial for ${{\rm{e}}^x}$:
${p_n}\left( x \right) = \mathop \sum \limits_{k = 0}^n \dfrac{{{x^k}}}{{k!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \cdot\cdot\cdot + \dfrac{{{x^n}}}{{n!}}$
Thus,
${{\rm{e}}^{1/2}} \approx {p_n}\left( {\dfrac{1}{2}} \right) = 1 + \dfrac{1}{2} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{{2!}} + \cdot\cdot\cdot + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^n}}}{{n!}}$
According to Theorem 9.7.4, the polynomial approximation for $\sqrt {\rm{e}} $ is to achieve four decimal-place accuracy, we must choose $n$ so that the absolute value of the $n$th remainder at $x = \frac{1}{2}$ satisfies
$\left| {{R_n}\left( {\dfrac{1}{2}} \right)} \right| \le 0.00005$
We use $f\left( x \right) = {{\rm{e}}^x}$, $x = \dfrac{1}{2}$, ${x_0} = 0$, and the interval $\left[ {0,\dfrac{1}{2}} \right]$. By Theorem 9.7.4,
$|{R_n}\left( {\dfrac{1}{2}} \right)\left| { \le \dfrac{M}{{\left( {n + 1} \right)!}}\cdot } \right|\dfrac{1}{2} - 0{\bigg|^{n + 1}}$
$\left| {{R_n}\left( {\dfrac{1}{2}} \right)} \right| \le \dfrac{M}{{{2^{n + 1}}\left( {n + 1} \right)!}}$
Since ${{\rm{e}}^x}$ is increasing, its maximum value on the interval $\left[ {0,\dfrac{1}{2}} \right]$ occurs at $x = \dfrac{1}{2}$. Thus, we can take $M = {{\rm{e}}^{1/2}}$. So then
$\left| {{R_n}\left( {\dfrac{1}{2}} \right)} \right| \le \dfrac{{\sqrt {\rm{e}} }}{{{2^{n + 1}}\left( {n + 1} \right)!}}$
$\dfrac{{\sqrt {\rm{e}} }}{{{2^{n + 1}}\left( {n + 1} \right)!}} \le 0.00005$
To find the solution, we use a computing utility to compute the following and list them in a table:
$\begin{array}{*{20}{c}}
n&{}&{\dfrac{{\sqrt {\rm{e}} }}{{{2^{n + 1}}\left( {n + 1} \right)!}}}\\
4&{}&{0.0004294}\\
5&{}&{0.0000358}\\
6&{}&{0.0000026}
\end{array}$
From the result, we choose $n=5$. Thus, the required polynomial to achieve four decimal-place accuracy is
${p_5}\left( {\dfrac{1}{2}} \right) = 1 + \dfrac{1}{2} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^4}}}{{4!}} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^5}}}{{5!}} \approx 1.6487$
As a verification, we compute using a calculator and obtain $\sqrt e \approx 1.64872$. Hence, it is verified:
$\left| {{R_n}\left( {\dfrac{1}{2}} \right)} \right| = \left| {\sqrt e - {p_5}\left( {\dfrac{1}{2}} \right)} \right| = 0.00002 \le 0.00005$
