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\begin{document}

\section{\textbf{Hypergeometric functions and elliptic integrals}}

For the calculation of the fields of loops and coils, and of their
capacitance and inductance, we need certain combinations of the complete
elliptic integrals $K$ and $E$, which are related to hypergeometric
functions of the type $F(a,b,m;z)$, where $a$ and $b$ are half-integers and $%
m$ is an integer. The general formula for the hypergeometric series 
\begin{equation}
F(a,b,c;z)=\sum_{n=0}^\infty \frac{\Gamma (n+a)}{\Gamma (a)}\frac{\Gamma
(n+b)}{\Gamma (b)}\frac{\Gamma (c)}{\Gamma (n+c)}\frac{z^n}{n!}  \label{Hs}
\end{equation}
gives 
\begin{equation}
F(a,b,1;z)=\sum_{n=0}^\infty \frac{\Gamma (n+a)}{n!\Gamma (a)}\frac{\Gamma
(n+b)}{n!\Gamma (b)}\;z^n  \label{H1}
\end{equation}
where 
\[
\frac{\Gamma (n+a)}{n!\Gamma (a)}=\binom{n+a-1}n 
\]
and in particular 
\[
\frac{\Gamma (n+\frac 12)}{n!\Gamma (\frac 12)}=\binom{n-\frac 12}n=\frac{%
1\cdot 3\,\cdots \,(2n-1)}{2\cdot 4\,\cdots \,2n}=\frac{\left( 2n\right) !}{%
2^{2n}\left( n!\right) ^2} 
\]
\[
\frac{\Gamma (n-\frac 12)}{n!\Gamma (-\frac 12)}=\binom{n-\frac 32}n=-\frac{%
1\cdot 3\,\cdots \,(2n-3)}{2\cdot 4\,\cdots \,2n}=-\frac 1{2n-1}\binom{%
n-\frac 12}n 
\]
for $n>0$ (The $n=0$ term in the series for $F$ is equal to $1$).

The complete elliptic integral of the first kind is by definition: 
\begin{equation}
K(k^2)=\int_0^{\pi /2}\frac{d\varphi }{\sqrt{1-k^2\sin ^2\varphi }}
\label{K}
\end{equation}
and the complete elliptic integral of the second kind is 
\begin{equation}
E(k^2)=\int_0^{\pi /2}\sqrt{1-k^2\sin ^2\varphi }\,d\varphi  \label{E}
\end{equation}
From the integral representation 
\begin{equation}
F(a,b,c;z)=\frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}%
\int_0^1t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}dt  \label{Hi}
\end{equation}
(putting $t=\sin ^2\phi $) one can derive the basic relations: 
\begin{equation}
K(z)=\frac \pi 2\,F(\frac 12,\frac 12,1;z)=\frac \pi 2\left[
1+\sum_{n=1}^\infty \binom{n-\frac 12}n^2z^n\right]  \label{Kh}
\end{equation}
\begin{equation}
E(z)=\frac \pi 2\,F(-\frac 12,\frac 12,1;z)=\frac \pi 2\left[
1-\sum_{n=1}^\infty \binom{n-\frac 12}n^2\frac{z^n}{2n-1}\right]  \label{Eh}
\end{equation}

\subsection{Expansion of $(K-E)/z$ and of $A_\phi $ for a loop}

Directly by subtraction of (\ref{Eh}) from (\ref{Kh}): 
\begin{equation}
\frac{K-E}z=\frac \pi 2\sum_{n=1}^\infty \binom{n-\frac 12}n^2\frac{2n}{2n-1}%
\;z^{n-1}  \label{KE}
\end{equation}
Alternatively, from one of the ``contiguity relations'' 
\[
F(\frac 12,\frac 12,1;z)-F(-\frac 12,\frac 12,1;z)=\frac z2\,F(\frac
12,\frac 32,2;z)
\]
so that 
\begin{equation}
\frac{K-E}z=\frac \pi 4\,F(\frac 12,\frac 32,2;z)  \label{KE1}
\end{equation}
Since 
\[
F(\frac 12,\frac 32,2;z)=\sum_{n=0}^\infty \frac{\Gamma (n+\frac 12)}{%
n!\Gamma (\frac 12)}\frac{\Gamma (n+\frac 32)}{\left( n+1\right) !\Gamma
(\frac 32)}\;z^n
\]
and 
\[
\frac{\Gamma (n+\frac 32)}{\left( n+1\right) !\Gamma (\frac 32)}=\frac{%
(n+\frac 12)\Gamma (n+\frac 12)}{\frac 12\left( n+1\right) n!\,\Gamma (\frac
12)}=\frac{2n+1}{n+1}\binom{n-\frac 12}n
\]
we find 
\begin{equation}
\frac{K-E}z=\frac \pi 2\sum_{n=0}^\infty \binom{n-\frac 12}n^2\frac{2n+1}{%
2n+2}\;z^n  \label{KE2}
\end{equation}
This is equivalent to (\ref{KE}) because the sum in that equation can be
rewritten as 
\[
\sum_{n=0}^\infty \binom{n+\frac 12}{n+1}^2\frac{2n+2}{2n+1}\;z^n\,\text{%
\quad \quad with\quad \quad }\binom{n+\frac 12}{n+1}=\binom{n-\frac 12}n%
\frac{2n+1}{2n+2}
\]
The vector potential of a loop is
\[
A_\phi (r,\theta )=\frac{4Ia}{c\sqrt{a^2+r^2+2ar\sin \theta }}\left[ 2\,%
\frac{K(z)-E(z)}z-K(z)\right] 
\]
with 
\[
z=\frac{4ar\sin \theta }{a^2+r^2+2ar\sin \theta }
\]
Using eqs. (\ref{KE2}) and (\ref{KH}) we easily find
\[
2\,\frac{K(z)-E(z)}z-K(z)=\frac \pi 2\sum_{n=1}^\infty \binom{n-\frac 12}%
n^2\frac n{n+1}\,z^n
\]

\subsection{Expansion of $E/(1-z)$}

Using 
\[
F(a,b,1;z)=(1-z)^{1-a-b}F(1-a,1-b,1;z) 
\]
with $a=-\frac 12$ and $b=\frac 12$ , we find 
\begin{equation}
\frac E{1-z}=\frac \pi 2\,F(\frac 32,\frac 12,1;z)  \label{E1}
\end{equation}
and, using 
\[
\frac{\Gamma (n+\frac 32)}{n!\Gamma (\frac 32)}=\binom{n+\frac 12}n=\frac{%
3\cdot 5\,\cdots \,(2n+1)}{2\cdot 4\,\cdots \,2n}=(2n+1)\binom{n-\frac 12}n, 
\]
we obtain 
\begin{equation}
\frac E{1-z}=\frac \pi 2\sum_{n=0}^\infty \binom{n-\frac 12}n^2\left(
2n+1\right) \;z^n\,  \label{E2}
\end{equation}

\subsection{Expansion of the inductance}

Combining the results (\ref{KE2}) and (\ref{E2}): 
\[
\frac{K-E}z+\frac E{1-z}=\frac \pi 2\sum_{n=0}^\infty \binom{n-\frac 12}n^2%
\frac{\left( 2n+1\right) \left( 2n+3\right) }{2n+2}\;z^n 
\]
Equivalently, (\ref{KE1}) and (\ref{E1}) give: 
\[
\frac{K-E}z+\frac E{1-z}=\frac \pi 2\left[ F(\frac 32,\frac 12,2;z)+\frac
12\,F(\frac 32,\frac 12,1;z)\right] 
\]
Using AS , this formula can be rewritten as 
\[
\frac{K-E}z+\frac E{1-z}=\frac{3\pi }4\,F(\frac 52,\frac 12,2;z) 
\]
with 
\[
F(\frac 52,\frac 12,2;z)=\sum_{n=0}^\infty \frac{\Gamma (n+\frac 52)}{\left(
n+1\right) !\Gamma (\frac 52)}\frac{\Gamma (n+\frac 12)}{n!\Gamma (\frac 12)}%
\;z^n=\sum_{n=0}^\infty \binom{n-\frac 12}n^2\frac{\left( 2n+3\right) \left(
2n+1\right) }{3\left( n+1\right) }\;z^n 
\]
Using the transformation formula 
\[
F(a,b,c;z)=(1-z)^{-b}F\left( b,c-a,c;\frac z{z-1}\right) 
\]
and putting 
\[
z=\frac{x^2}{1+x^2}\quad \text{so that \quad }\frac z{z-1}=-x^2\text{,} 
\]
where $x=2a/l$ (ratio of solenoid diameter to length), we have finally 
\begin{eqnarray*}
\frac 4{3\pi }\sqrt{1-z}\,\left( \frac{K-E}z+\frac E{1-z}\right)
&=&\,F(\frac 12,-\frac 12,2;-x^2) \\
&=&1-\sum_{n=1}^\infty \binom{n-\frac 12}n^2\frac{\left( -x^2\right) ^n}{%
(n+1)(2n-1)}
\end{eqnarray*}
As a check, the leading terms are: 
\[
1+\frac 12\left( \frac x2\right) ^2-\frac 14\left( \frac x2\right) ^4+\frac
5{16}\left( \frac x2\right) ^6-\frac{35}{64}\left( \frac x2\right) ^8+\frac{%
147}{128}\left( \frac x2\right) ^{10}+\cdots 
\]
Then the inductance of a finite solenoid is $Lf,$ where $L$ is computed
neglecting end effects ($L=(4\pi /c^2)N^2\pi a^2/l$) and 
\[
f=F(\frac 12,-\frac 12,2;-x^2)-\frac{4x}{3\pi } 
\]
\[
F(\left( \frac x2\right) )=1-\frac{8\left( \frac x2\right) }{3\pi }+\frac
12\left( \frac x2\right) ^2-\frac 14\left( \frac x2\right) ^4+\frac
5{16}\left( \frac x2\right) ^6-\frac{35}{64}\left( \frac x2\right) ^8+\frac{%
147}{128}\left( \frac x2\right) ^{10} 
\]
$-\sum_{n=6}^{10}\binom{n-\frac 12}n^2\frac{\left( -4x^2\right) ^n}{%
(n+1)(2n-1)}=-\frac{693}{256}x^{12}+\frac{14157}{2048}x^{14}-\frac{306735}{%
16384}x^{16}+\frac{1738165}{32768}x^{18}+O\left( x^{20}\right) $

$\frac 12x^2-\frac 14x^4+\frac 5{16}x^6-\frac{35}{64}x^8+\frac{147}{128}%
x^{10}-\frac{693}{256}x^{12}+\frac{14157}{2048}x^{14}-\frac{306735}{16384}%
x^{16}+\frac{1738165}{32768}x^{18}+O\left( x^{20}\right) $

$16384/256=64$ and $32768/2048=16$

$\binom{n-\frac 12}n=\frac{\left( 2n\right) !}{2^{2n}\left( n!\right) ^2}$
and $\binom{7-\frac 12}7=\frac{\left( 14\right) !}{2^{14}\left( 7!\right) ^2}
$ is true $\frac{150!}{2^{150}\left( 75!\right) ^2}=\frac{%
580\,16293\,58544\,29935\,61866\,13429\,53782\,01212\,57345}{%
8920\,29807\,94122\,49256\,61428\,73090\,59344\,60239\,21664}$ and

$8920\,29807\,94122\,49256\,61428\,73090\,59344\,60239\,21664/1024^{15}=%
\frac 1{16}$ and

$580\,16293\,58544\,29935\,61866\,13429\,53782\,0121257345/3^2=$

$64\,46254\,84282\,69992\,84651\,79269\,94864\,66801\,39705/5=$

$12\,89250\,96856\,53998\,56930\,35853\,98972\,93360\,27941/7=$

$1\,84178\,70979\,50571\,22418\,62264\,85567\,56194\,32563/11=$

$16743\,51907\,22779\,20219\,87478\,62324\,32381\,30233/13=$

$1287\,96300\,55598\,40016\,91344\,50948\,02490\,86941/19=$

$67\,78752\,66084\,12632\,46912\,86892\,00131\,09839/41=$

$1\,65335\,43075\,22259\,32851\,53338\,82930\,02679/43=$

$3845\,01001\,74936\,26345\,38449\,74021\,62853/47=$

$81\,80872\,37764\,60135\,00818\,07957\,90699/79=$

$1\,03555\,34655\,24811\,83554\,65923\,51781/83=$

$1247\,65477\,77407\,37151\,26095\,46407/89=$

$14\,01859\,30083\,22889\,34001\,07263/97=14452\,15774\,05390\,61175%
\,26879/101=$

$143\,09067\,06984\,06546\,28979/103=1\,38922\,98126\,05888\,79893/107=$

$1298\,34561\,92578\,39999/109=11\,91142\,76996\,13211/113=$

$10541\,08645\,98347/7=1505\,86949\,42621/11=136\,89722\,67511/127=$

$1\,07793\,09193/131=82284803/137=600619/139=4321/149=29$

\end{document}