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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Tue Sep 12 15:08:46 2000}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=Lab Report.cst}
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%$A_{1} \left( n\right) =\left(  \frac{1+\frac{1}{2n}}{1-\frac{1}{2n}}\right)  ^{n}\frac{120n^{2}-23}{120n^{2}-13} $
%$A_{2} \left( n\right) =\left(  \frac{1+\frac{1}{2n}}{1-\frac{1}{2n}}\right)  ^{n}\frac{67\,798\,080n^{4}-21\,422\,520n^{2}+1364\,017}{67\,798\,080n^{4}-15\,772\,680n^{2}+661\,693}\allowbreak  $
%$A_{3} \left( n\right) =\left(  \frac{2n+1}{2n-1}\right)  ^{n}\frac{675\,722\,246\,\allowbreak 828\,428\,800n^{6}-297\,926\,560\,\allowbreak 686\,363\,840n^{4}+37\,636\,542\,\allowbreak 266\,082\,480n^{2}-1194\,007\,\allowbreak 474\,028\,891}{675\,722\,246\,\allowbreak 828\,428\,800n^{6}-241\,616\,373\,\allowbreak 450\,661\,440n^{4}+23\,602\,114\,\allowbreak 762\,395\,120n^{2}-538\,818\,\allowbreak 430\,104\,349}\allowbreak  $
%$C \left( n\right) =\left(  1+\frac{1}{n}\right)  ^{n} $
%$B \left( n\right) =\left(  \frac{2n+1}{2n-1}\right)  ^{n} $
%$q_{1}=-\frac{1}{12} $
%$r_{1}=-\frac{67\,\allowbreak 583\,440\,323}{84\,\allowbreak 641\,334\,830} $
%$r_{2}=\frac{1215\,\allowbreak 052\,718\,819}{13\,407\,\allowbreak 187\,437\,072} $
%$q_{2}=\frac{7031\,839\,979}{338\,\allowbreak 565\,339\,320} $
%$r_{3}=-\frac{229\,\allowbreak 209\,775\,679}{89\,381\,\allowbreak 249\,580\,480} $
%$q_{3}=-\frac{327\,594\,\allowbreak 521\,962\,271}{337\,861\,123\,\allowbreak 414\,214\,400} $
%$a_{1}=-\frac{447\,\allowbreak 821\,309\,353}{1015\,\allowbreak 696\,017\,960} $
%$b_{1}=-\frac{363\,\allowbreak 179\,974\,523}{1015\,\allowbreak 696\,017\,960} $
%$a_{2}=\frac{7467\,\allowbreak 567\,909\,937}{134\,071\,\allowbreak 874\,370\,720} $
%$b_{2}=\frac{4682\,\allowbreak 959\,278\,253}{134\,071\,\allowbreak 874\,370\,720} $
%$a_{3}=-\frac{1194\,007\,\allowbreak 474\,028\,891}{675\,722\,246\,\allowbreak 828\,428\,800} $
%$b_{3}=-\frac{538\,818\,\allowbreak 430\,104\,349}{675\,722\,246\,\allowbreak 828\,428\,800} $
%}

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%H=36,\PARA{038<p type="texpara" tag="Body Text" >Al Lehnen}\PARA{038<p type="texpara" tag="Body Text" >Madison Area Technical College}
%F=36,\PARA{038<p type="texpara" tag="Body Text" >August 12, 2000\hfill \thepage }
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\input{tcilatex}
\begin{document}


\begin{center}
$\vspace{1pt}${\Large Use of Rational Function Factors in Approximating e}
\end{center}

\vspace{1pt}

Nearly every calculus textbook and quite a few pre-calculus ones present the
following standard limit for computing the transcendental number $%
e\allowbreak \allowbreak $ \ \ ($e\thickapprox 2.\,\allowbreak
718\,281\,828\,\allowbreak 459\,045\,235\,\allowbreak
360\,287\,471\,\allowbreak 35\allowbreak ...)$ .

\vspace{1pt}$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
e=\lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right) ^{n}$

$\vspace{1pt}\vspace{1pt}$However, students who try to calculate $e$ by this
formula are often surprised by its ''poor'' rate of convergence. In fact, on
most calculators round off error renders the computation impotent before the
calculator's decimal representation of $\exp (1)$ (calculated much more
efficiently as the sum of reciprocal factorials) is attained.

\vspace{1pt}

Recent work by John Knox and Harlan Brothers$^{(1\text{, }2\text{, }3)}$
demonstrates more efficient limit expressions for $e$ that do not involve
approximating an infinite series. This present work was inspired by theirs,
but employs a slightly different approach which results in different if not
new expressions.

\vspace{1pt}

Let the `classical' or standard approximation to $e$ be defined as follows:

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad C(n)=\left( 1+\frac{1}{n}%
\right) ^{n}$

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $=\exp \left[
n\ln (1+\frac{1}{n})\right] $

Using the Maclaurin series for $\ln (1+x)=x-\frac{x^{2}}{2}$ $+\frac{x^{3}}{3%
}-\frac{x^{4}}{4}$ $+\frac{x^{5}}{5}+...+(-1)^{j-1}\frac{x^{j}}{j}+...$

\vspace{1pt}$\ln \left[ C(n)\right] =1-\frac{1}{2n}+\frac{1}{3n^{2}}-\frac{1%
}{4n^{3}}+...(-1)^{j-1}\frac{1}{jn^{j-1}}+...$ .

Thus, $C(n)=\exp \left( 1-\frac{1}{2n}+O(\frac{1}{n^{2}})\right) =\exp
(1)\cdot \exp \left( -\frac{1}{2n}+O(\frac{1}{n^{2}})\right) =e\left( 1-%
\frac{1}{2n}+O(\frac{1}{n^{2}})\right) $.

Hence, the relative error in the standard approximation is first order in $%
\frac{1}{n}$.

\vspace{1pt}$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{C(n)-e}{e%
}=-\frac{1}{2n}+O(\frac{1}{n^{2}})$

\vspace{1pt}Thus doubling $n$ gives an approximation to $e$ only `twice as
good' as the last. This observation quantifies the poor convergence
properties of the standard approximation. To get a relatively modest 1\%
relative error requires $n$ to be as large as 50. Doubling this $n$ to 100
only approximates $e$ to about 1 part in 200, or less than two `good' digits
to the right of the decimal point.

\vspace{1pt}\pagebreak

The computational results for $C(n)$ are summarized in the following table.

\begin{center}
$\vspace{1pt}$%
\begin{tabular}{llll}
$n$ & $C(n)$ & Fractional Error $\frac{C(n)-e}{e}$ & $-\frac{1}{2}\cdot 
\frac{1}{n}\allowbreak $ \\ 
$1$ & $\allowbreak \allowbreak 2.0$ & $-.\,\allowbreak 264\,241\allowbreak $
& $-0.5\allowbreak $ \\ 
$2$ & $\allowbreak 2.\,\allowbreak 25\allowbreak $ & $-.\,\allowbreak
172\,271\allowbreak $ & $-0.25\allowbreak $ \\ 
$3$ & $2.\,\allowbreak 370\,370\,37\allowbreak $ & $-.\,\allowbreak
127\,989\allowbreak $ & $-0.\,\allowbreak 166\,667\allowbreak $ \\ 
$4$ & $2.\,\allowbreak 441\,406\,25$ & $\allowbreak \allowbreak
-.\,\allowbreak 101\,857\allowbreak $ & $-0.\,\allowbreak 125\allowbreak $
\\ 
$5$ & $2.\,\allowbreak 488\,32\allowbreak $ & $-8.\,\allowbreak
459\,82\times 10^{-2}$ & $-0.\,\allowbreak 1\allowbreak $ \\ 
$6$ & $\allowbreak 2.\,\allowbreak 521\,626\,372$ & $-7.\,\allowbreak
234\,55\times 10^{-2}$ & $-8.\,\allowbreak 333\,33\times 10^{-2}\allowbreak $
\\ 
$7$ & $2.\,\allowbreak 546\,499\,697\allowbreak $ & $-6.\,\allowbreak
319\,51\times 10^{-2}$ & $-7.\,\allowbreak 142\,86\times 10^{-2}$ \\ 
$8$ & $2.\,\allowbreak 565\,784\,514$ & $-5.\,\allowbreak 610\,06\times
10^{-2}$ & $-0.0\,\allowbreak 625\allowbreak $ \\ 
$9$ & $\allowbreak 2.\,\allowbreak 581\,174\,792$ & $\allowbreak
-5.\,\allowbreak 043\,89\times 10^{-2}$ & $-5.\,\allowbreak 555\,56\times
10^{-2}\allowbreak $ \\ 
$10$ & $\allowbreak 2.\,\allowbreak 593\,742\,46$ & $-4.\,\allowbreak
581\,55\times 10^{-2}\allowbreak $ & $-0.0\,\allowbreak 5\allowbreak $ \\ 
$50$ & $\allowbreak 2.\,\allowbreak 691\,588\,03\allowbreak $ & $%
-9.\,\allowbreak 820\,1\times 10^{-3}\allowbreak $ & $-0.01$ \\ 
$100$ & $\allowbreak 2.\,\allowbreak 704\,813\,83\allowbreak $ & $%
\allowbreak -4.\,\allowbreak 954\,6\times 10^{-3}\allowbreak $ & $%
\allowbreak -.00\,\allowbreak 5\allowbreak $%
\end{tabular}
\end{center}

An approximation $E(n)$ to $e$ is of order $k$ if $\ln \left[ E(n)\right]
=1+O(n^{-k})$ . For then,

$E(n)=\exp \left[ \ln \left( E(n)\right) \right] =\exp \left(
1+O(n^{-k})\right) =\exp (1)\cdot \exp \left( O(n^{-k})\right) =e\left[
1+O(n^{-k})\right] $ and the relative error, $\frac{E(n)-e}{e}$ , decreases
like $n^{-k}$ .

\vspace{1pt}

The following expression gives a second order approximation to $e$.

$\qquad \qquad \qquad \qquad \qquad B(n)=\left( \frac{2n+1}{2n-1}\right)
^{n}=\left( \frac{1+\frac{1}{2n}}{1-\frac{1}{2n}}\right) ^{n}$.

The verification that $B(n)$ is indeed second order is shown below.

\vspace{1pt}$\ln \left[ B(n)\right] =n\left[ \ln \left( 1+\frac{1}{2n}%
\right) -\ln \left( 1-\frac{1}{2n}\right) \right] $

$\qquad \qquad =n\left[ (\frac{1}{2n}-\frac{1}{2(2n)^{2}}+\frac{1}{3(2n)^{3}}%
-\frac{1}{4(2n)^{4}}+...)-(\frac{1}{-2n}-\frac{1}{2(-2n)^{2}}+\frac{1}{%
3(-2n)^{3}}-\frac{1}{4(-2n)^{4}}+...\right] $

$\qquad \qquad =2n\left[ \frac{1}{2n}+\frac{1}{3(2n)^{3}}+\frac{1}{5(2n)^{5}}%
+\frac{1}{7(2n)^{7}}+\frac{1}{9(2n)^{9}}+\frac{1}{11(2n)^{11}}+...\right] $

$\qquad \qquad =1+\frac{1}{3(2n)^{2}}+\frac{1}{5(2n)^{4}}+\frac{1}{7(2n)^{6}}%
+\frac{1}{9(2n)^{8}}+\frac{1}{11(2n)^{10}}+\frac{1}{13(2n)^{12}}+...$

So, $B(n)=e\left[ 1+\frac{1}{12n^{2}}+O(\frac{1}{n^{4}})\right] $ and $\frac{%
B(n)-e}{e}=\frac{1}{12n^{2}}+O(\frac{1}{n^{4}})$ .

\vspace{1pt}

Now, if we double $n$, the relative error of the approximation decreases by
a factor of four.

\vspace{1pt}\pagebreak

The computational results for $B(n)$ are summarized in the following table.

\vspace{1pt}

\begin{center}
\begin{tabular}{llll}
$n$ & $B(n)$ & Fractional Error $\frac{B(n)-e}{e}$ & $\frac{1}{12}\cdot 
\frac{1}{n^{2}}\allowbreak $ \\ 
$1$ & $B(1)=\allowbreak 3.0$ & $0\allowbreak .\,\allowbreak
103\,638\allowbreak $ & $\allowbreak 8.\,\allowbreak 333\,33\times
10^{-2}\allowbreak $ \\ 
$2$ & $\allowbreak B(2)\allowbreak =\allowbreak 2.\,\allowbreak
777\,777\,8\allowbreak $ & $\allowbreak 2.\,\allowbreak 188\,73\times
10^{-2}\allowbreak $ & $\allowbreak 2.\,\allowbreak 083\,33\times
10^{-2}\allowbreak $ \\ 
$3$ & $B(3)\allowbreak =\allowbreak 2.\,\allowbreak 744\allowbreak $ & $%
\allowbreak 9.\,\allowbreak 461\,19\times 10^{-3}\allowbreak $ & $%
\allowbreak 9.\,\allowbreak 259\,26\times 10^{-3}\allowbreak $ \\ 
$4$ & $B(4)\allowbreak =\allowbreak 2.\,\allowbreak 732\,611\,4\allowbreak $
& $\allowbreak \allowbreak 5.\,\allowbreak 271\,56\times 10^{-3}\allowbreak $
& $5.\,\allowbreak 208\,33\times 10^{-3}\allowbreak $ \\ 
$5$ & $B(5)=\allowbreak 2.\,\allowbreak 727\,412\,8\allowbreak $ & $%
\allowbreak 3.\,\allowbreak 359\,11\times 10^{-3}\allowbreak $ & $%
\allowbreak 3.\,\allowbreak 333\,33\times 10^{-3}\allowbreak $ \\ 
$6$ & $B(6)=\allowbreak 2.\,\allowbreak 724\,607\,8\allowbreak $ & $%
\allowbreak 2.\,\allowbreak 327\,21\times 10^{-3}\allowbreak $ & $%
\allowbreak 2.\,\allowbreak 314\,81\times 10^{-3}\allowbreak $ \\ 
$7$ & $B(7)\allowbreak =\allowbreak 2.\,\allowbreak 722\,922\,9\allowbreak $
& $\allowbreak 1.\,\allowbreak 707\,36\times 10^{-3}\allowbreak $ & $%
\allowbreak 1.\,\allowbreak 700\,68\times 10^{-3}\allowbreak $ \\ 
$8$ & $B(8)=\allowbreak 2.\,\allowbreak 721\,831\,9\allowbreak $ & $%
\allowbreak 1.30\,6\allowbreak \times 10^{-3}\allowbreak $ & $\allowbreak
1.\,\allowbreak 302\,08\times 10^{-3}\allowbreak $ \\ 
$9$ & $\allowbreak B(9)=\allowbreak 2.\,\allowbreak
721\,085\,049\,\allowbreak 4\allowbreak $ & $\allowbreak \allowbreak
1.\,\allowbreak 031\,25\times 10^{-3}\allowbreak $ & $1.\,\allowbreak
028\,81\times 10^{-3}\allowbreak $ \\ 
$10$ & $B(10)=\allowbreak 2.\,\allowbreak 720\,551\,4\allowbreak $ & $%
\allowbreak 8.\,\allowbreak 349\,34\times 10^{-4}\allowbreak $ & $%
\allowbreak 8.\,\allowbreak 333\,33\times 10^{-4}\allowbreak $ \\ 
$50$ & $\allowbreak B(50)\allowbreak =\allowbreak 2.\,\allowbreak
718\,372\,4\allowbreak $ & $\allowbreak 3.\,\allowbreak 333\,59\times
10^{-5}\allowbreak $ & $\allowbreak 3.\,\allowbreak 333\,33\times
10^{-5}\allowbreak $%
\end{tabular}
\end{center}

\vspace{1pt}

To construct an approximation to $e$ that is higher than second order, we
multiply $B(n)$ by a rational function of $n$ of the form, $R_{p}(n)=\frac{%
n^{2p}+a_{1}n^{2(p-1)}+a_{2}n^{2(p-2)}+...+a_{j}n^{2(p-j)}+...+a_{p}}{%
n^{2p}+b_{1}n^{2(p-1)}+b_{2}n^{2(p-2)}+...+b_{j}n^{2(p-j)}+...+b_{p}}$.
Clearly for any choice of the $2p$ coefficients $\{a_{j},b_{j}\}_{j=1}^{p}$
, $\ \lim_{n\rightarrow \infty }R_{p}(n)=1$. Hence, $F_{p}(n)=$ $%
R_{p}(n)\cdot B(n)$ approaches $e$ for large $n$.

\vspace{1pt}$\qquad \qquad \qquad \qquad \qquad F_{p}(n)=\left( \frac{2n+1}{%
2n-1}\right) ^{n}\frac{%
n^{2p}+a_{1}n^{2(p-1)}+a_{2}n^{2(p-2)}+...+a_{j}n^{2(p-j)}+...+a_{p}}{%
n^{2p}+b_{1}n^{2(p-1)}+b_{2}n^{2(p-2)}+...+b_{j}n^{2(p-j)}+...+b_{p}}$

Now, $\ln \left[ R_{p}(n)\right] =\ln \left[ \frac{%
n^{2p}+a_{1}n^{2(p-1)}+a_{2}n^{2(p-2)}+...+a_{j}n^{2(p-j)}+...+a_{p}}{%
n^{2p}+b_{1}n^{2(p-1)}+b_{2}n^{2(p-2)}+...+b_{j}n^{2(p-j)}+...+b_{p}}\right]
=\ln \left[ \frac{1+a_{1}n^{-2}+a_{2}n^{-4}+...+a_{j}n^{-2j}+...+a_{p}n^{-2p}%
}{1+b_{1}n^{-2}+b_{2}n^{-4}+...+b_{j}n^{-2j}+...+b_{p}n^{-2p}}\right] $

\qquad \qquad \qquad \qquad $=\ln \left(
1+a_{1}n^{-2}+a_{2}n^{-4}+...+a_{j}n^{-2j}+...+a_{p}n^{-2p}\right) -\ln
\left( 1+b_{1}n^{-2}+b_{2}n^{-4}+...+b_{j}n^{-2j}+...+b_{p}n^{-2p}\right) $

\vspace{1pt}

Let $%
x=a_{1}n^{-2}+a_{2}n^{-4}+a_{3}n^{-6}+a_{4}n^{-8}+...+a_{j}n^{-2j}+...+a_{p}n^{-2p} 
$. Keeping track of the reciprocal powers of $n$ in the expansion of $\ln
(1+x)$ gives the following result.

\vspace{1pt}$\ln \left(
1+a_{1}n^{-2}+a_{2}n^{-4}+...+a_{j}n^{-2j}+...+a_{p}n^{-2p}\right) =\frac{1}{%
n^{2}}\lbrack \left( a_{1}\right) \rbrack +\frac{1}{n^{4}}\lbrack \left(
a_{2}\right) -\frac{1}{2}(a_{1}^{2})\rbrack $ $+\frac{1}{n^{6}}\lbrack
\left( a_{3}\right) -\frac{1}{2}\left( 2a_{1}a_{2}\right) +\frac{1}{3}%
(a_{1}^{3})\rbrack $

$+\frac{1}{n^{8}}\lbrack \left( a_{4}\right) -\frac{1}{2}%
(2a_{1}a_{3}+a_{2}^{2})+\frac{1}{3}(3a_{1}^{2}a_{2})-\frac{1}{4}%
(a_{1}^{4})\rbrack $

$+\frac{1}{n^{10}}\lbrack \left( a_{5}\right) -\frac{1}{2}\left(
2a_{1}a_{4}+2a_{2}a_{3}\right) +\frac{1}{3}(3a_{1}^{2}a_{3}+3a_{1}a_{2}^{2})-%
\frac{1}{4}(4a_{1}^{3}a_{2})+\frac{1}{5}(a_{1}^{5})\rbrack $

$+\frac{1}{n^{12}}\lbrack \left( a_{6}\right) -\frac{1}{2}\left(
2a_{1}a_{5}+2a_{2}a_{4}+a_{3}^{2}\right) +\frac{1}{3}%
(3a_{1}^{2}a_{4}+6a_{1}a_{2}a_{3}+a_{2}^{3})-\frac{1}{4}%
(4a_{1}^{3}a_{3}+6a_{1}^{2}a_{2}^{2})$

$+\frac{1}{5}(5a_{1}^{4}a_{2})-\frac{1}{6}(a_{1}^{6})\rbrack $

$+\frac{1}{n^{14}}\lbrack \left( a_{7}\right) -\frac{1}{2}\left(
2a_{1}a_{6}+2a_{2}a_{5}+2a_{3}a_{4}\right) +\frac{1}{3}%
(3a_{1}^{2}a_{5}+6a_{1}a_{2}a_{4}+3a_{1}a_{3}^{2}+3a_{2}^{2}a_{3})$

$-\frac{1}{4}(4a_{1}^{3}a_{4}+12a_{1}^{2}a_{2}a_{3}+4a_{1}a_{2}^{3})+\frac{1%
}{5}(5a_{1}^{4}a_{3}+10a_{1}^{3}a_{2}^{2})-\frac{1}{6}(6a_{1}^{5}a_{2})+%
\frac{1}{7}(a_{1}^{7})\rbrack $ $+$ $O(\frac{1}{n^{16}})$

\vspace{1pt}

From this equation the following combinatoric patterns emerge.

1. The coefficient of $n^{-2k}$ can be written as $\sum_{j=1}^{k}(-1)^{j+1}%
\left( \frac{1}{j}\right) S_{j}$ .

2. Each $S_{j\text{ }}$is the sum of all possible products of $j$
coefficients `$a_{m}$` where the sum of the indices $m$ is equal to $k$
,i.e., $S_{j}$ is the sum of all possible $\Pi _{l=1}^{j}a_{m_{l}}$ , with $%
\sum_{l=1}^{j}m_{l}=k$ .

\vspace{1pt}

For example, the coefficient of $n^{-10}$ has five terms $S_{1}-\frac{1}{2}%
S_{2}+\frac{1}{3}S_{3}-\frac{1}{4}S_{4}+\frac{1}{5}S_{5}$ .

For $j=1$ , $5=5$ , so $S_{1}=a_{5}$ .

For $j=2$ , $5=1+4=4+1=2+3=3+2$, so $S_{2}=2a_{1}a_{4}+2a_{2}a_{3}$ .

For $j=3$ , $5=1+1+3=1+3+1=3+1+1=1+2+2=2+1+2=2+2+1$, so $%
S_{3}=3a_{1}^{2}a_{3}+3a_{1}a_{2}^{2}$ .

For $j=4$ , $5=1+1+1+2=1+1+2+1=1+2+1+1=2+1+1+1$, so $S_{4}=4a_{1}^{3}a_{2}$ .

\vspace{1pt}For $j=5$ , $5=1+1+1+1+1$, so $S_{5}=a_{1}^{5}$ .

\vspace{1pt}

To generate the expansion of $\ln \left(
1+b_{1}n^{-2}+b_{2}n^{-4}+...+b_{j}n^{-2j}+...+b_{p}n^{-2p}\right) $ replace 
$a$ by $b$ in the above expansion of $\ln \left(
1+a_{1}n^{-2}+a_{2}n^{-4}+...+a_{j}n^{-2j}+...+a_{p}n^{-2p}\right) $ .

\vspace{1pt}

Using the identity that $\ln \left[ F_{p}(n)\right] =\ln \left[ B(n)\right]
+\ln \left[ R_{p}(n)\right] $ \ leads to the following expansion.

$\ln \left[ F_{p}(n)\right] =1+\frac{1}{n^{2}}\lbrack \frac{1}{3\cdot 2^{2}}%
+\left( a_{1}-b_{1}\right) \rbrack +\frac{1}{n^{4}}\lbrack \frac{1}{5\cdot
2^{4}}+\left( a_{2}-b_{2}\right) -\frac{1}{2}(a_{1}^{2}-b_{1}^{2})\rbrack $

$+\frac{1}{n^{6}}\lbrack \frac{1}{7\cdot 2^{6}}+\left( a_{3}-b_{3}\right) -%
\frac{1}{2}\left( 2a_{1}a_{2}-2b_{1}b_{2}\right) +\frac{1}{3}%
(a_{1}^{3}-b_{1}^{3})\rbrack $

$+\frac{1}{n^{8}}\lbrack \frac{1}{9\cdot 2^{8}}+\left( a_{4}-b_{4}\right) -%
\frac{1}{2}(2a_{1}a_{3}-2b_{1}b_{3}+a_{2}^{2}-b_{2}^{2})+\frac{1}{3}%
(3a_{1}^{2}a_{2}-3b_{1}^{2}b_{2})-\frac{1}{4}(a_{1}^{4}-b_{1}^{4})\rbrack $

$+\frac{1}{n^{10}}\lbrack \frac{1}{11\cdot 2^{10}}+\left( a_{5}-b_{5}\right)
-\frac{1}{2}\left( 2a_{1}a_{4}-2b_{1}b_{4}+2a_{2}a_{3}-2b_{2}b_{3}\right) +%
\frac{1}{3}(3a_{1}^{2}a_{3}-3b_{1}^{2}b_{3}+3a_{1}a_{2}^{2}-3b_{1}b_{2}^{2})$

$-\frac{1}{4}(4a_{1}^{3}a_{2}-4b_{1}^{3}b_{2})+\frac{1}{5}%
(a_{1}^{5}-b_{1}^{5})\rbrack $

$+\frac{1}{n^{12}}\lbrack \frac{1}{13\cdot 2^{12}}+\left( a_{6}-b_{6}\right)
-\frac{1}{2}%
(2a_{1}a_{5}-2b_{1}b_{5}+2a_{2}a_{4}-2b_{2}b_{4}+a_{3}^{2}-b_{3}^{2})$

$+\frac{1}{3}%
(3a_{1}^{2}a_{4}-3b_{1}^{2}b_{4}+6a_{1}a_{2}a_{3}-6b_{1}b_{2}b_{3}+a_{2}^{3}-b_{2}^{3})-%
\frac{1}{4}%
(4a_{1}^{3}a_{3}-4b_{1}^{3}b_{3}+6a_{1}^{2}a_{2}^{2}-6b_{1}^{2}b_{2}^{2})$

$+\frac{1}{5}(5a_{1}^{4}a_{2}-5b_{1}^{4}b_{2})-\frac{1}{6}%
(a_{1}^{6}-b_{1}^{6})\rbrack $

$+\frac{1}{n^{14}}\lbrack \frac{1}{15\cdot 2^{14}}+\left( a_{7}-b_{7}\right)
-\frac{1}{2}\left(
2a_{1}a_{6}-2b_{1}b_{6}+2a_{2}a_{5}-2b_{2}b_{5}+2a_{3}a_{4}-2b_{3}b_{4}%
\right) +\frac{1}{3}(3a_{1}^{2}a_{5}-3b_{1}^{2}b_{5}$

$+$ $%
6a_{1}a_{2}a_{4}-6b_{1}b_{2}b_{4}+3a_{1}a_{3}^{2}-3b_{1}b_{3}^{2}+3a_{2}^{2}a_{3}-3b_{2}^{2}b_{3})-%
\frac{1}{4}(4a_{1}^{3}a_{4}-4b_{1}^{3}b_{4}+12a_{1}^{2}a_{2}a_{3}$

$-12b_{1}^{2}b_{2}b_{3}+4a_{1}a_{2}^{3}-4b_{1}b_{2}^{3})+\frac{1}{5}%
(5a_{1}^{4}a_{3}-5b_{1}^{4}b_{3}+10a_{1}^{3}a_{2}^{2}-10b_{1}^{3}b_{2}^{2})-%
\frac{1}{6}(6a_{1}^{5}a_{2}-6b_{1}^{5}b_{2})$

$+\frac{1}{7}(a_{1}^{7}-b_{1}^{7})\rbrack $ $+$ $O(\frac{1}{n^{16}})$

\vspace{1pt}

This equation demonstrates that there are infinitely many choices for the $2p
$ rational coefficients $\{a_{j},b_{j}\}_{j=1}^{p}$ which make $F_{p}(n)$ of
order $2p+2$ . For example, $F_{1}(n)=\left( \frac{2n+1}{2n-1}\right) \frac{%
n^{2}+a_{1}}{n^{2}+b_{1}}$ will exhibit fourth order convergence provided $%
\left( a_{1}-b_{1}\right) +\frac{1}{12}=0$ . Similarly, \textbf{any }$%
a_{1},b_{1},a_{2},b_{2}$ that satisfy the two constraints $\left(
a_{1}-b_{1}\right) +\frac{1}{12}=0$

and $\left( a_{2}-b_{2}\right) -\frac{1}{2}(a_{1}^{2}-b_{1}^{2})+\frac{1}{80}%
=0$ , will generate an $F_{2}(n)$ that converges to $e$ with a relative
error of $O(n^{-6})$ .

\vspace{1pt}

A second approach is to `optimize' $F_{p}(n)$ for a given value of $p$ by
choosing the $2p$ coefficients $\{a_{j},b_{j}\}_{j=1}^{p}$ so that all of
the $2p$ coefficients on $n^{-2},$ $n^{-4},...,$ $n^{-4p}$ vanish. This
gives an approximation of order $4p+2$ . The resulting optimal $F_{p}(n)$
will be designated as $A_{p}(n)$ . Below explicit expressions for $A_{1}(n)$
, $A_{2}(n)$ and $A_{3}(n)$ are developed.

\vspace{1pt}\vspace{1pt}

Case 1 : $p=1$

The system of two equations which the two coefficients $a_{1},b_{1}$must
satisfy is $
\begin{array}{l}
a_{1}-b_{1}+\frac{1}{12}=0 \\ 
-\frac{1}{2}(a_{1}^{2}-b_{1}^{2})+\frac{1}{80}=0
\end{array}
$ .

\vspace{1pt}To solve this system introduce the conjugate variables defined
by $
\begin{array}{l}
q_{1}=a_{1}-b_{1} \\ 
r_{1}=a_{1}+b_{1}
\end{array}
$ , with the inverse transformation $
\begin{array}{l}
a_{1}=\frac{r_{1}+q_{1}}{2} \\ 
b_{1}=\frac{r_{1}-q_{1}}{2}
\end{array}
$. This results in the system of two equations,\vspace{1pt}$
\begin{array}{l}
q_{1}+\frac{1}{12}=0 \\ 
-\frac{1}{2}(q_{1}r_{1})+\frac{1}{80}=0
\end{array}
$ which has the solution $
\begin{array}{l}
q_{1}=-\frac{1}{12} \\ 
r_{1}=-\frac{3}{10}
\end{array}
$ .\vspace{1pt}This gives $\allowbreak 
\begin{array}{c}
a_{1}=-\frac{23}{120} \\ 
b_{1}=-\frac{13}{120}
\end{array}
$ , so that

\vspace{1pt}

$A_{1}(n)=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{n^{2}+a_{1}}{n^{2}+b_{1}}%
=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{n^{2}-\frac{23}{120}}{n^{2}-\frac{%
13}{120}}=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{120n^{2}-23}{120n^{2}-13}
$ .

\vspace{1pt}This same result was obtained independently by John E. Beam$^{(4%
\text{, }5)}$.

The coefficient of the $\frac{1}{n^{6}}$ term in the expansion of $A_{1}(n)$
is given by

$\frac{1}{3}(a_{1}^{3}-b_{1}^{3})+\frac{1}{7\ast 2^{6}}=\allowbreak \frac{%
1121}{3628\,800}$ .

Hence,$\ A_{1}(n)=e\left[ 1+\frac{1121}{3628\,800}\cdot \frac{1}{n^{6}}%
+O\left( \frac{1}{n^{8}}\right) \right] $ .

\vspace{1pt}

The computational results for $A_{1}(n)$ are summarized in the following
table.

\begin{center}
\begin{tabular}{llll}
$n$ & $A_{1}(n)$ & Fractional Error $\frac{A_{1}(n)-e}{e}$ & $\frac{1121}{%
3628\,800}\cdot \frac{1}{n^{6}}=\allowbreak \frac{3.\,\allowbreak
089\,175\,485\times 10^{-4}}{n^{6}}\allowbreak $ \\ 
$1$ & $\allowbreak \allowbreak 2.\,\allowbreak 719\,626\,168\,\allowbreak
224\,\allowbreak $ & $\allowbreak \allowbreak 4.\,\allowbreak 945\,55\times
10^{-4}\allowbreak $ & $\allowbreak 3.\,\allowbreak 089\,18\times
10^{-4}\allowbreak $ \\ 
$2$ & $\allowbreak 2.\,\allowbreak 718\,296\,454\,\allowbreak 913\allowbreak 
$ & $\allowbreak 5.\,\allowbreak 380\,77\times 10^{-6}\allowbreak $ & $%
\allowbreak \allowbreak 4.\,\allowbreak 826\,84\times 10^{-6}\allowbreak $
\\ 
$3$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,283\,036\,\allowbreak
551\allowbreak $ & $\allowbreak 4.\,\allowbreak 444\,32\times
10^{-7}\allowbreak $ & $4.\,\allowbreak 237\,55\times 10^{-7}\allowbreak $
\\ 
$4$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,282\,039\,\allowbreak
012\allowbreak $ & $\allowbreak 7.\,\allowbreak 745\,82\times
10^{-8}\allowbreak $ & $\allowbreak 7.\,\allowbreak 541\,93\times
10^{-8}\allowbreak $ \\ 
$5$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,281\,883\,\allowbreak
1258\allowbreak $ & $\allowbreak 2.\,\allowbreak 011\,04\times
10^{-8}\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
977\,07\times 10^{-8}\allowbreak $ \\ 
$6$ & $\allowbreak 2.\,\allowbreak 718\,281\,846\,\allowbreak
671\,\allowbreak $ & $\allowbreak 6.\,\allowbreak 699\,88\times
10^{-9}\allowbreak $ & $\allowbreak 6.\,\allowbreak 621\,18\times
10^{-9}\allowbreak $ \\ 
$7$ & $\allowbreak 2.\,\allowbreak 718\,281\,835\,\allowbreak
6598\allowbreak $ & $\allowbreak \allowbreak 2.\,\allowbreak 648\,64\times
10^{-9}\allowbreak $ & $\allowbreak \allowbreak 2.\,\allowbreak
625\,76\times 10^{-9}\allowbreak $ \\ 
$8$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,281\,831\,\allowbreak
6847\allowbreak $ & $\allowbreak 1.\,\allowbreak 186\,28\times
10^{-9}\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
178\,43\times 10^{-9}\allowbreak $ \\ 
$9$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,281\,830\,\allowbreak
047\,$ & $\allowbreak 5.\,\allowbreak 843\,39\times 10^{-10}\allowbreak $ & $%
\allowbreak \allowbreak 5.\,\allowbreak 812\,83\times 10^{-10}\allowbreak $
\\ 
$10$ & $\allowbreak 2.\,\allowbreak 718\,281\,829\,\allowbreak
302\,\allowbreak $ & $\allowbreak \allowbreak 3.\,\allowbreak 102\,32\times
10^{-10}\allowbreak $ & $\allowbreak 3.\,\allowbreak 089\,18\times
10^{-10}\allowbreak $%
\end{tabular}
\end{center}

\vspace{1pt}\pagebreak

Case 2 : $p=2$

The system of four equations which the four coefficients $%
a_{1},b_{1},a_{2},b_{2}$ must satisfy is

$
\begin{array}{l}
a_{1}-b_{1}+\frac{1}{12}=0 \\ 
a_{2}-b_{2}-\frac{1}{2}(a_{1}^{2}-b_{1}^{2})+\frac{1}{80}=0 \\ 
-a_{1}a_{2}+b_{1}b_{2}+\frac{1}{3}(a_{1}^{3}-b_{1}^{3})+\frac{1}{7\ast 2^{6}}%
=0 \\ 
-\frac{1}{2}(a_{2}^{2}-b_{2}^{2})+a_{1}^{2}a_{2}-b_{1}^{2}b_{2}-\frac{1}{4}%
(a_{1}^{4}-b_{1}^{4})+\frac{1}{9\ast 2^{8}}=0
\end{array}
$ .

\vspace{1pt}

Using the conjugate variables $
\begin{array}{l}
q_{1}=a_{1}-b_{1} \\ 
r_{1}=a_{1}+b_{1} \\ 
q_{2}=a_{2}-b_{2} \\ 
r_{2}=a_{2}+b_{2}
\end{array}
$ and the inverse transformation $
\begin{array}{l}
a_{1}=\frac{r_{1}+q_{1}}{2} \\ 
b_{1}=\frac{r_{1}-q_{1}}{2} \\ 
a_{2}=\frac{r_{2}+q_{2}}{2} \\ 
b_{2}=\frac{r_{2}-q_{2}}{2}
\end{array}
$ results in the system of four equations \vspace{1pt}$
\begin{array}{l}
q_{1}+\frac{1}{12}=0 \\ 
q_{2}-\frac{1}{2}r_{1}q_{1}+\frac{1}{80}=0 \\ 
-\frac{1}{2}r_{1}q_{2}-\frac{1}{2}q_{1}r_{2}+\frac{1}{4}r_{1}^{2}q_{1}+\frac{%
1}{12}q_{1}^{3}+\frac{1}{448}=0 \\ 
\frac{1}{2304}-\frac{1}{8}r_{1}^{3}q_{1}-\frac{1}{2}r_{2}q_{2}-\frac{1}{8}%
r_{1}q_{1}^{3}+\frac{1}{4}q_{1}^{2}q_{2}+\allowbreak \frac{1}{4}%
r_{1}^{2}q_{2}+\frac{1}{2}r_{1}q_{1}r_{2}=0
\end{array}
$ ,

which has the solution\vspace{1pt} $
\begin{array}{l}
q_{1}=-\frac{1}{12} \\ 
r_{1}=-\frac{615}{1121} \\ 
q_{2}=\frac{929}{89\,680} \\ 
r_{2}=\frac{202\,571}{6779\,808}
\end{array}
$ . This gives $\allowbreak 
\begin{array}{l}
a_{1}=-\frac{8501}{26\,904} \\ 
b_{1}=-\frac{6259}{26\,904} \\ 
a_{2}=\frac{1364\,017}{67\,798\,080} \\ 
b_{2}=\frac{661\,693}{67\,798\,080}
\end{array}
$ , so that

\vspace{1pt}$A_{2}(n)=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{%
n^{4}+a_{1}n^{2}+a_{2}}{n^{4}+b_{1}n^{2}+b_{2}}=\left( \frac{2n+1}{2n-1}%
\right) ^{n}\frac{n^{4}-\frac{8501}{26\,904}n^{2}+\frac{1364\,017}{%
67\,798\,080}}{n^{4}-\frac{6259}{26\,904}n^{2}+\frac{661\,693}{67\,798\,080}}%
=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{67\,798\,080n^{4}-21\,422%
\,520n^{2}+1364\,017}{67\,798\,080n^{4}-15\,772\,680n^{2}+661\,693}$ .

\vspace{1pt}

The coefficient of the $\frac{1}{n^{10}}$ term in the expansion of $A_{2}(n)$
is given by

$\frac{1}{11\cdot 2^{10}}%
+(a_{1}a_{2}^{2}-b_{1}b_{2}^{2})-(a_{1}^{3}a_{2}-b_{1}^{3}b_{2})+\frac{1}{5}%
(a_{1}^{5}-b_{1}^{5})=\allowbreak \frac{651\,087\,191}{541\,256\,\allowbreak
479\,948\,800}$ .

\vspace{1pt}

Hence, $A_{2}(n)=e\left[ 1+\frac{651\,087\,191}{541\,256\,\allowbreak
479\,948\,800}\cdot \frac{1}{n^{10}}+O(\frac{1}{n^{12}})\right] $

\pagebreak

The computational results for $A_{2}(n)$ are summarized in the following
table.

\begin{center}
\begin{tabular}{llll}
$n$ & $A_{2}(n)$ & Fract Error $\frac{A_{2}(n)-e}{e}$ & $\frac{651\,087\,191%
}{541\,256\,\allowbreak 479\,948\,800}\cdot \frac{1}{n^{10}}\allowbreak
\allowbreak \allowbreak =\allowbreak \frac{1.\,\allowbreak 202\,918\,1\times
10^{-6}}{n^{10}}\allowbreak $ \\ 
$1$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,288\,727\,\allowbreak
753\,494\allowbreak $ & $\allowbreak 2.\,\allowbreak 538\,11\times
10^{-6}\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
202\,92\times 10^{-6}\allowbreak $ \\ 
$2$ & $\allowbreak 2.\,\allowbreak 718\,281\,832\,\allowbreak
253\,27\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
395\,82\times 10^{-9}\allowbreak $ & $\allowbreak \allowbreak
1.\,\allowbreak 174\,72\times 10^{-9}\allowbreak $ \\ 
$3$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
518\,771\allowbreak $ & $\allowbreak 2.\,\allowbreak 197\,19\times
10^{-11}\allowbreak $ & $\allowbreak 2.\,\allowbreak 037\,15\times
10^{-11}\allowbreak $ \\ 
$4$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
462\,299\allowbreak $ & $\allowbreak 1.\,\allowbreak 196\,81\times
10^{-12}\allowbreak $ & $\allowbreak 1.\,\allowbreak 147\,19\times
10^{-12}\allowbreak $ \\ 
$5$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,389\allowbreak $ & $\allowbreak 1.\,\allowbreak 265\,55\times
10^{-13}\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
231\,79\times 10^{-13}\allowbreak $ \\ 
$6$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,1\allowbreak $ & $\allowbreak 2.\,\allowbreak 027\,08\times
10^{-14}\allowbreak $ & $\allowbreak 1.\,\allowbreak 989\,41\times
10^{-14}\allowbreak $ \\ 
$7$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,057\allowbreak $ & $\allowbreak \allowbreak 4.\,\allowbreak
317\,54\times 10^{-15}\allowbreak $ & $\allowbreak \allowbreak
4.\,\allowbreak 258\,49\times 10^{-15}\allowbreak $ \\ 
$8$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,048\allowbreak $ & $\allowbreak \allowbreak 1.\,\allowbreak
132\,17\times 10^{-15}\allowbreak $ & $\allowbreak \allowbreak
1.\,\allowbreak 120\,3\times 10^{-15}\allowbreak $ \\ 
$9$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,046\allowbreak $ & $\allowbreak 3.\,\allowbreak 478\,78\times
10^{-16}\allowbreak $ & $\allowbreak 3.\,\allowbreak 449\,94\times
10^{-16}\allowbreak $ \\ 
$10$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,046\allowbreak $ & $\allowbreak 1.\,\allowbreak 211\,06\times
10^{-16}\allowbreak $ & $\allowbreak 1.\,\allowbreak 202\,92\times
10^{-16}\allowbreak $%
\end{tabular}
\end{center}

\vspace{1pt}

Case 3 : $p=3$

The system of six equations which the six coefficients $%
a_{1},b_{1},a_{2},b_{2},a_{3},b_{3}$ must satisfy is

$
\begin{array}{l}
a_{1}-b_{1}+\frac{1}{12}=0 \\ 
a_{2}-b_{2}-\frac{1}{2}(a_{1}^{2}-b_{1}^{2})+\frac{1}{80}=0 \\ 
(a_{3}-b_{3})-a_{1}a_{2}+b_{1}b_{2}+\frac{1}{3}(a_{1}^{3}-b_{1}^{3})+\frac{1%
}{7\ast 2^{6}}=0 \\ 
-\frac{1}{2}%
(2a_{1}a_{3}-2b_{1}b_{3}+a_{2}^{2}-b_{2}^{2})+a_{1}^{2}a_{2}-b_{1}^{2}b_{2}-%
\frac{1}{4}(a_{1}^{4}-b_{1}^{4})+\frac{1}{9\ast 2^{8}}=0 \\ 
-\left( a_{2}a_{3}-b_{2}b_{3}\right)
+(a_{1}^{2}a_{3}-b_{1}^{2}b_{3}+a_{1}a_{2}^{2}-b_{1}b_{2}^{2})-\left(
a_{1}^{3}a_{2}-b_{1}^{3}b_{2}\right) +\frac{1}{5}\left(
a_{1}^{5}-b_{1}^{5}\right) +\frac{1}{11\cdot 2^{10}}=0 \\ 
-\frac{1}{2}\left( a_{3}^{2}-b_{3}^{2}\right) +\frac{1}{3}%
(6a_{1}a_{2}a_{3}-6b_{1}b_{2}b_{3}+a_{2}^{3}-b_{2}^{3})-\frac{1}{4}\left(
4a_{1}^{3}a_{3}-4b_{1}^{3}b_{3}+6a_{1}^{2}a_{2}^{2}-6b_{1}^{2}b_{2}^{2}%
\right) +\left( a_{1}^{4}a_{2}-b_{1}^{4}b_{2}\right) -\frac{1}{6}\left(
a_{1}^{6}-b_{1}^{6}\right) +\frac{1}{13\cdot 2^{12}}=0
\end{array}
$

\vspace{1pt}Using the conjugate variables $
\begin{array}{l}
q_{1}=a_{1}-b_{1} \\ 
r_{1}=a_{1}+b_{1} \\ 
q_{2}=a_{2}-b_{2} \\ 
r_{2}=a_{2}+b_{2} \\ 
q_{3}=a_{3}-b_{3} \\ 
r_{3}=a_{3}+b_{3}
\end{array}
$ and the inverse transformation $
\begin{array}{l}
a_{1}=\frac{r_{1}+q_{1}}{2} \\ 
b_{1}=\frac{r_{1}-q_{1}}{2} \\ 
a_{2}=\frac{r_{2}+q_{2}}{2} \\ 
b_{2}=\frac{r_{2}-q_{2}}{2} \\ 
a_{3}=\frac{r_{3}+q_{3}}{2} \\ 
b_{3}=\frac{r_{3}-q_{3}}{2}
\end{array}
$ results in the following system of six equations\vspace{1pt}.

$\allowbreak $

$
\begin{array}{l}
q_{1}+\frac{1}{12}=0 \\ 
\allowbreak q_{2}-\frac{1}{2}r_{1}q_{1}+\frac{1}{80}=0 \\ 
\allowbreak q_{3}-\frac{1}{2}r_{1}q_{2}-\frac{1}{2}q_{1}r_{2}+\frac{1}{4}%
r_{1}^{2}q_{1}+\frac{1}{12}q_{1}^{3}+\frac{1}{448}=0 \\ 
-\frac{1}{2}r_{1}q_{3}-\frac{1}{2}q_{1}r_{3}-\frac{1}{2}r_{2}q_{2}+\frac{1}{4%
}r_{1}^{2}q_{2}+\frac{1}{4}q_{1}^{2}\allowbreak q_{2}-\frac{1}{8}%
r_{1}^{3}q_{1}-\frac{1}{8}r_{1}q_{1}^{3}+\frac{1}{2}r_{1}q_{1}r_{2}+\frac{1}{%
2304}=0 \\ 
\frac{1}{2}r_{1}r_{2}q_{2}+\frac{1}{2}r_{1}q_{1}r_{3}-\frac{3}{8}%
r_{1}q_{1}^{2}q_{2}-\frac{3}{8}r_{1}^{2}q_{1}\allowbreak r_{2}-\frac{1}{2}%
r_{2}q_{3}-\frac{1}{2}q_{2}r_{3}+\frac{1}{4}r_{1}^{2}q_{3}+\frac{1}{4}%
q_{1}^{2}q_{3}+\frac{1}{4}q_{1}\allowbreak r_{2}^{2}+\frac{1}{4}%
q_{1}q_{2}^{2}-\frac{1}{8}r_{1}^{3}q_{2}-\frac{1}{8}q_{1}^{3}r_{2}+\frac{1}{%
16}r_{1}^{4}q_{1}+\allowbreak \frac{1}{8}r_{1}^{2}q_{1}^{3}+\frac{1}{11\,264}%
+\frac{1}{80}q_{1}^{5}=0 \\ 
-\frac{3}{8}r_{1}q_{1}^{2}q_{3}-\frac{3}{8}r_{1}^{2}q_{1}r_{3}+\frac{1}{2}%
q_{1}q_{2}q_{3}+\frac{1}{12}q_{2}^{3}+\allowbreak \frac{1}{4}r_{2}^{2}q_{2}-%
\frac{1}{2}r_{3}q_{3}+\frac{1}{53\,248}-\frac{3}{8}r_{1}^{2}r_{2}q_{2}+\frac{%
1}{2}q_{1}\allowbreak r_{2}r_{3}+\frac{1}{2}r_{1}q_{2}r_{3}+\frac{1}{2}%
r_{1}r_{2}q_{3}+\frac{1}{16}r_{1}^{4}q_{2}+\frac{1}{4}r_{1}^{3}\allowbreak
q_{1}r_{2}+\frac{3}{8}r_{1}^{2}q_{1}^{2}q_{2}+\frac{1}{4}r_{1}q_{1}^{3}r_{2}-%
\frac{3}{8}r_{1}q_{1}r_{2}^{2}-\allowbreak \frac{3}{8}r_{1}q_{1}q_{2}^{2}-%
\frac{3}{8}q_{1}^{2}r_{2}q_{2}-\frac{1}{8}q_{1}^{3}r_{3}-\frac{1}{8}%
r_{1}^{3}q_{3}+\allowbreak \frac{1}{16}q_{1}^{4}q_{2}-\frac{1}{32}%
r_{1}^{5}q_{1}-\frac{5}{48}r_{1}^{3}q_{1}^{3}-\frac{1}{32}%
r_{1}q_{1}^{5}\allowbreak =0
\end{array}
$, $\allowbreak $

\vspace{1pt}

which has the solution\vspace{1pt} $
\begin{array}{l}
q_{1}=-\frac{1}{12} \\ 
r_{1}=-\frac{67\,\allowbreak 583\,440\,323}{84\,\allowbreak 641\,334\,830}
\\ 
q_{2}=\frac{7031\,839\,979}{338\,\allowbreak 565\,339\,320} \\ 
r_{2}=\frac{1215\,\allowbreak 052\,718\,819}{13\,407\,\allowbreak
187\,437\,072} \\ 
q_{3}=-\frac{327\,594\,\allowbreak 521\,962\,271}{337\,861\,123\,\allowbreak
414\,214\,400} \\ 
r_{3}=-\frac{229\,\allowbreak 209\,775\,679}{89\,381\,\allowbreak
249\,580\,480}
\end{array}
$ . \vspace{1pt}

This gives $\ \allowbreak 
\begin{array}{c}
a_{1}=-\frac{447\,\allowbreak 821\,309\,353}{1015\,\allowbreak 696\,017\,960}
\\ 
b_{1}=-\frac{363\,\allowbreak 179\,974\,523}{1015\,\allowbreak 696\,017\,960}
\\ 
a_{2}=\frac{7467\,\allowbreak 567\,909\,937}{134\,071\,\allowbreak
874\,370\,720} \\ 
b_{2}=\frac{4682\,\allowbreak 959\,278\,253}{134\,071\,\allowbreak
874\,370\,720} \\ 
a_{3}=-\frac{1194\,007\,\allowbreak 474\,028\,891}{675\,722\,246\,%
\allowbreak 828\,428\,800} \\ 
b_{3}=-\frac{538\,818\,\allowbreak 430\,104\,349}{675\,722\,246\,\allowbreak
828\,428\,800}
\end{array}
\allowbreak $ , so that\vspace{1pt}

\vspace{1pt}$A_{3}(n)=\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{%
n^{6}+a_{1}n^{4}+a_{2}n^{2}+a_{3}}{n^{6}+b_{1}n^{4}+b_{2}n^{2}+b_{3}}%
\allowbreak \allowbreak $

$\qquad \ \ =\left( \frac{2n+1}{2n-1}\right) ^{n}\frac{675\,722\,246\,%
\allowbreak 828\,428\,800n^{6}-297\,926\,560\,\allowbreak
686\,363\,840n^{4}+37\,636\,542\,\allowbreak
266\,082\,480n^{2}-1194\,007\,\allowbreak 474\,028\,891}{675\,722\,246\,%
\allowbreak 828\,428\,800n^{6}-241\,616\,373\,\allowbreak
450\,661\,440n^{4}+23\,602\,114\,\allowbreak
762\,395\,120n^{2}-538\,818\,\allowbreak 430\,104\,349}\allowbreak $ .

\vspace{1pt}The coefficient of the $\frac{1}{n^{14}}$ term in the expansion
of $A_{3}(n)$ is given by\vspace{1pt}$\frac{1}{15\cdot 2^{14}}%
+(a_{1}a_{3}^{2}-b_{1}b_{3}^{2}+a_{2}^{2}a_{3}-b_{2}^{2}b_{3})-(3a_{1}^{2}a_{2}a_{3}-3b_{1}^{2}b_{2}b_{3}+a_{1}a_{2}^{3}-b_{1}b_{2}^{3})+ 
$

$%
(a_{1}^{4}a_{3}-b_{1}^{4}b_{3}+2a_{1}^{3}a_{2}^{2}-2b_{1}^{3}b_{2}^{2})-(a_{1}^{5}a_{2}-b_{1}^{5}b_{2})+%
\frac{1}{7}(a_{1}^{7}-b_{1}^{7})\allowbreak =\allowbreak \frac{%
39\,\allowbreak 521\,551\,119\,\allowbreak 193\,251\,271}{%
8415\,472\,972\,\allowbreak 046\,720\,337\,\allowbreak 838\,080\,000}$

$=\allowbreak 4.\,\allowbreak 696\,3\times 10^{-9}\allowbreak $

\vspace{1pt}

Hence, $\ A_{3}(n)=e\left[ 1+\frac{39\,\allowbreak
521\,551\,119\,\allowbreak 193\,251\,271}{8415\,472\,972\,\allowbreak
046\,720\,337\,\allowbreak 838\,080\,000}\cdot \frac{1}{n^{14}}+O(\frac{1}{%
n^{16}})\right] $

\vspace{1pt}\pagebreak

The computational results for $A_{3}(n)$ are summarized in the following
table.

\begin{center}
\begin{tabular}{llll}
$n$ & $A_{3}(n)$ & Fract Error $\frac{A_{3}(n)-e}{e}$ & $\allowbreak
\allowbreak 4.\,\allowbreak 696\,3\times 10^{-9}\allowbreak \cdot \frac{1}{%
n^{14}}\allowbreak \allowbreak \allowbreak \allowbreak \allowbreak $ \\ 
$1$ & $\allowbreak \allowbreak 2.\,\allowbreak 718\,281\,863\,\allowbreak
999\,051\,753\,\allowbreak 898\,2\allowbreak $ & $\allowbreak
1.\,\allowbreak 307\,44\times 10^{-8}\allowbreak $ & $\allowbreak
4.\,\allowbreak 696\,3\times 10^{-9}\allowbreak $ \\ 
$2$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
460\,032\,228\,\allowbreak 737\,1\allowbreak $ & $\allowbreak
3.\,\allowbreak 630\,95\times 10^{-13}\allowbreak $ & $\allowbreak
2.\,\allowbreak 866\,39\times 10^{-13}\allowbreak $ \\ 
$3$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,048\,195\,\allowbreak 978\,2\allowbreak $ & $\allowbreak
1.\,\allowbreak 089\,15\times 10^{-15}\allowbreak $ & $\allowbreak
9.\,\allowbreak 818\,8\times 10^{-16}\allowbreak $ \\ 
$4$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,285\,\allowbreak 759\,4\allowbreak $ & $\allowbreak
1.\,\allowbreak 854\,08\times 10^{-17}\allowbreak $ & $\allowbreak
1.\,\allowbreak 749\,51\times 10^{-17}\allowbreak $ \\ 
$5$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,237\,\allowbreak 530\,8\allowbreak $ & $\allowbreak \allowbreak
7.\,\allowbreak 985\,05\times 10^{-19}\allowbreak $ & $\allowbreak
7.\,\allowbreak 694\,42\times 10^{-19}\allowbreak $ \\ 
$6$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 527\,4\allowbreak $ & $\allowbreak
6.\,\allowbreak 149\,04\times 10^{-20}\allowbreak $ & $\allowbreak
\allowbreak \allowbreak 5.\,\allowbreak 992\,92\times 10^{-20}\allowbreak $
\\ 
$7$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 379\,5\allowbreak $ & $\allowbreak
7.\,\allowbreak 056\,4\times 10^{-21}\allowbreak $ & $\allowbreak
6.\,\allowbreak 924\,42\times 10^{-21}\allowbreak $ \\ 
$8$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 363\,2\allowbreak $ & $\allowbreak
1.\,\allowbreak 083\,36\times 10^{-21}\allowbreak $ & $\allowbreak
1.\,\allowbreak 067\,81\times 10^{-21}\allowbreak $ \\ 
$9$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 360\,9\allowbreak $ & $\allowbreak
2.\,\allowbreak 076\,43\times 10^{-22}\allowbreak $ & $\allowbreak
\allowbreak \allowbreak 2.\,\allowbreak 052\,87\times 10^{-22}\allowbreak $
\\ 
$10$ & $\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 360\,4\allowbreak $ & $\allowbreak
4.\,\allowbreak 739\,91\times 10^{-23}\allowbreak $ & $\allowbreak
\allowbreak \allowbreak 4.\,\allowbreak 696\,3\times 10^{-23}\allowbreak $%
\end{tabular}
\end{center}

\vspace{1pt}

While it is possible to extend these results to larger values of $p$ and
even faster convergence, the computational complexity in the determination
of the required coefficients makes this scheme rather unwieldy. It is hoped,
however, that instructors or possibly even texts could introduce the
computational advantages of $B(n)$ and $A_{1}(n).$

\vspace{1pt}

References:

\begin{quotation}
\vspace{1pt}$(1)$ H.J. Brothers and J.A. Knox, \textit{Math Intelligencer} 
\textbf{20}, 25 (1998).

$(2)$ J.A. Knox and H.J. Brothers, \textit{College Math. J.} \textbf{30},
269 (1999).

$(3)$ Web site article, \textit{Serendipit-e}, at \textbf{%
http://www.giss.nasa.gov/research/intro/knox.03/}

\vspace{1pt}\vspace{1pt}$(4)$ \textit{On Wisconsin} (UW Madison Alumni
Magazine)\textbf{100}, 5 (Summer 1999).

\vspace{1pt}$(5)$ Private email correspondence with the author.
\end{quotation}

\vspace{1pt}

\end{document}
