Answer
(a) Five terms are needed.
(b) Using a computer algebra system, we get:
${F_4}\left( 1 \right) \approx 0.747487$, ${\ \ \ \ \ }$ $F\left( 1 \right) \approx 0.7468244$
Hence, it is verified that $\left| {F\left( 1 \right) - {F_4}\left( 1 \right)} \right| \approx 0.00066264 \lt 0.001$.
Work Step by Step
From Table 2:
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $ - {t^2}$ for $x$ in the series above gives
${{\rm{e}}^{ - {t^2}}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - {t^2}} \right)}^n}}}{{n!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{n!}} = 1 - {t^2} + \dfrac{{{t^4}}}{{2!}} - \dfrac{{{t^6}}}{{3!}} + \dfrac{{{t^8}}}{{4!}} + \cdot\cdot\cdot$,
converges for all $t$.
Next, we express the function $F\left( x \right) = \mathop \smallint \limits_0^x {{\rm{e}}^{ - {t^2}}}{\rm{d}}t$ using the series above:
$F\left( x \right) = \mathop \smallint \limits_0^x {{\rm{e}}^{ - {t^2}}}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^x \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{n!}}{\rm{d}}t$
$ = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 1} \right)n!}}} \right]_0^x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)n!}}$
Thus, $F\left( x \right) = \mathop \smallint \limits_0^x {{\rm{e}}^{ - {t^2}}}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)n!}}$.
(a) For $x=1$, we have $F\left( 1 \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)n!}}$, where the positive terms are ${b_n} = \dfrac{1}{{\left( {2n + 1} \right)n!}}$.
By Eq. (2) in Section 11.4:
$\left| {F\left( 1 \right) - {F_n}\left( 1 \right)} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {F\left( 1 \right) - {F_n}\left( 1 \right)} \right| \lt {b_{n + 1}} \lt {10^{ - 3}}$
$\dfrac{1}{{\left( {2n + 3} \right)\left( {n + 1} \right)!}} \lt {10^{ - 3}}$
$\left( {2n + 3} \right)\left( {n + 1} \right)! \gt {10^3}$
We evaluate and list some values for $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{\left( {2n + 3} \right)\left( {n + 1} \right)!}\\
2&{}&{42}\\
3&{}&{216}\\
4&{}&{1320}\\
5&{}&{9360}
\end{array}$
Using the results in the table above, we choose $N=4$ such that the approximation of the integral for $x=1$ is within an error of at most $0.001$.
(b) Using a computer algebra system, we compute:
${F_4}\left( 1 \right) \approx 0.747487$, ${\ \ \ \ \ }$ $F\left( 1 \right) \approx 0.7468244$
Hence, it is verified that $\left| {F\left( 1 \right) - {F_4}\left( 1 \right)} \right| \approx 0.00066264 \lt 0.001$.
