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{\bf Question}

Write down the Taylor expansion for $e^x$.  Then use Maclaurin's
 theorem to show that if $0 \leq x \leq 1$ then $\ds e^x = 1 + x+
 \frac{x^2}{2} + R(x)$, where $0 \leq R(x) \leq \frac{1}{6}e^x$.
 Hence show that $\ds 1 + x + \frac{x^2}{2}   \leq  e^x \leq
\frac{6}{5}\left(1 + x + \frac{x^2}{2}\right)$.  Use your
calculator to evaluate $e$ and check these bounds.

\vspace{.25in}

{\bf Answer}

Taylor expansion of $e^x$ around $x = 0$

$\ds e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Remainder term $\ds R_n(x) = \frac{x^n}{n!} e^{\theta x}
\hspace{.1in} 0 < \theta < 1$

Hence  $\ds e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}
e^{\theta x}$

$e^x$ is increasing $ 0 <  x < 1$\ $\ds \Rightarrow 0 \leq
\frac{x^3}{6} e^{\theta x} \leq \frac{1}{6} e^{x}$

Hence $\ds 1 + x + \frac{x^2}{2!}  \leq e^x \leq  1 + x +
\frac{x^2}{2!} + \frac{x^3}{3!} e^{x}$

Hence $\ds e^x\left(1 - \frac{x^3}{6}\right) \leq 1 + x +
\frac{x^2}{2}$

\begin{eqnarray*}
\Rightarrow e^x & \leq & \frac{1 + x + \frac{x^2}{2}}{1 -
\frac{x^3}{6}} \\ {\rm if\ }  0 < x < 1, \hspace{.2in} 1 & \geq &
1 - \frac{x^3}{6} \geq \frac{5}{6} \\ {\rm Hence \ \ \ \ } 1 &
\leq & \frac{1}{1 - \frac{x^3}{6}} \leq \frac{6}{5} \\ {\rm Hence
\ \ \ \ } e^x & \leq & \frac{1 + x + \frac{x^2}{2}}{1 -
\frac{x^3}{6}} \leq \frac{6}{5}\left(1 + x + \frac{x^2}{2}\right)
\\ \Rightarrow 1 + x + \frac{x^2}{2}  & \leq & e^x \leq
\frac{6}{5}\left(1 + x + \frac{x^2}{2}\right) \end{eqnarray*}

Putting $x=1$ gives $$1+1+\frac{1}{2}\le e\le\frac{6}{5}\left(
1+1+\frac{1}{2}\right),$$

giving $\ds \frac{5}{2}\le e\le\frac{6}{5}\times\frac{5}{2},$ i.e.
$2.5\le e \le 3,$ which is consistent with the value given on the
calculator, namely $e=2.71828\ldots$


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