Editing Appendix/Ramblings/PowerSeriesExpressions (section)

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> &#8212; that is, in the case of an infinitesimally thin, axisymmetric disk &#8212; the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{c}{a}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
  </td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
  <td align="right">
<math>
\sin^{-1} e
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
  </td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>A_1</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{A_3}{2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
  </td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2 
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6 
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ \cdots
\biggr\}
</math>
&nbsp; &nbsp; &nbsp; for, &nbsp; &nbsp; &nbsp;
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, . 
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the  [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2 
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6 
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3 
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7 
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5 
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7 
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 +  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9 
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9  
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) -  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 -  \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 -  \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7 
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 -  \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9 

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9  
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
  </td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) -  \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 -  \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 -  \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7 
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 -  \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9 

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9  
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) 
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr] 
+  \frac{1}{2} 
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+  \biggl[\frac{1}{2^2\cdot 3}\biggr] 
+ \biggl[\frac{3}{2^3 }\biggr] 
\biggr\}\biggl(\frac{c}{a}\biggr)^5 

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr] 
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+  \biggl[\frac{1}{2^4}\biggr]\ 
+ \biggl[\frac{5}{2^4 }\biggr] 
\biggr\}\biggl(\frac{c}{a}\biggr)^7 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr] 
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr] 
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+  \biggl[\frac{5}{2^5\cdot 3}\biggr] 
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\}  \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) 
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5 

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7 

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \,  .
</math>
  </td>
</tr>
</table>

</td></tr></table>


Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
  </td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathcal{F}_2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8 
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10} 
- \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8 
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4 
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6 
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6 
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{1}{2\cdot 3} 
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4 
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} 
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 } 
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4 
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{ 93}{5 \cdot 7} 
\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
  </td>
</tr>
</table>
Once again from the binomial theorem, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
  </td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4 
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{ 93}{5 \cdot 7} 
\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- 2\biggl(\frac{c}{a}\biggr)^2 
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4 
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[\frac{ 2^7}{5 \cdot 7} 
\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
  </td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
- \biggl(\frac{c}{a}\biggr)^2 
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4 
+ 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+ 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
- \biggl[ \frac{ 93}{5 \cdot 7} 
\biggr]\biggl(\frac{c}{a}\biggr)^8 
+ 
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr) 
+ 2\biggl(\frac{c}{a}\biggr)^2 
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3 
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4 
- 
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5 
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
- 
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7 
+ \biggl[ \frac{ 2^7}{5 \cdot 7} 
\biggr]\biggl(\frac{c}{a}\biggr)^8 
- 
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9 
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
  </td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.