1. 1. 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

F (a, b, c; z) = ∑_{n=0}^{∞} \frac{Γ(n + a)}{Γ(a)} \frac{Γ(n + b)}{Γ(b)} \frac{Γ(c)}{Γ(n + c)} \frac{z^{n}}{n!}  (1.1)

gives

F (a, b, 1; z) = ∑_{n=0}^{∞} \frac{Γ(n + a)}{n!Γ(a)} \frac{Γ(n + b)}{n!Γ(b)} z^{n}  (1.2)

where

\frac{Γ(n + a)}{n!Γ(a)} = \binom{n + a − 1}{n}

and in particular

\frac{Γ(n + \frac{1}{2})}{n!Γ(\frac{1}{2})} = \binom{n − \frac{1}{2}}{n} = \frac{1 · 3 ⋯ (2n − 1)}{2 · 4 ⋯ 2n} = \frac{(2n)!}{2^{2n} (n!)^{2}}

\frac{Γ(n − \frac{1}{2})}{n!Γ(−\frac{1}{2})} = \binom{n − \frac{3}{2}}{n} = −\frac{1 · 3 ⋯ (2n − 3)}{2 · 4 ⋯ 2n} = −\frac{1}{2n − 1}\binom{n − \frac{1}{2}}{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:

K(k^{2}) = ∫_{0}^{π/2} \frac{dφ}{\sqrt{1 − k^{2} sin^{2} φ}}  (1.3)

and the complete elliptic integral of the second kind is

E(k^{2}) = ∫_{0}^{π/2} \sqrt{1 − k^{2} sin^{2} φ} dφ  (1.4)

From the integral representation

F (a, b, c; z) = \frac{Γ(c)}{Γ(b)Γ(c − b)} ∫_{0}^{1} t^{b−1}(1 − t)^{c−b−1}(1 − tz)^{−a}dt  (1.5)

(putting t = sin^{2} ϕ) one can derive the basic relations:

K(z) = \frac{π}{2} F (\frac{1}{2}, \frac{1}{2}, 1; z) = \frac{π}{2} [1 + ∑_{n=1}^{∞} \binom{n − \frac{1}{2}}{n}^{2}z^{n}]  (1.6)

E(z) = \frac{π}{2} F (−\frac{1}{2}, \frac{1}{2}, 1; z) = \frac{π}{2} [1 − ∑_{n=1}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{z^{n}}{2n − 1}]  (1.7)

1.1. Expansion of (K − E)/z Directly by subtraction of (1.7) from (1.6):

\frac{K − E}{z} = \frac{π}{2} ∑_{n=1}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{2n}{2n − 1} z^{n−1}  (1.8)

Alternatively, from one of the “contiguity relations”

F (\frac{1}{2}, \frac{1}{2}, 1; z) − F (−\frac{1}{2}, \frac{1}{2}, 1; z) = \frac{z}{2} F (\frac{1}{2}, \frac{3}{2}, 2; z)

so that

\frac{K − E}{z} = \frac{π}{4} F (\frac{1}{2}, \frac{3}{2}, 2; z)  (1.9)

Since

F (\frac{1}{2}, \frac{3}{2}, 2; z) = ∑_{n=0}^{∞} \frac{Γ(n + \frac{1}{2})}{n!Γ(\frac{1}{2})} \frac{Γ(n + \frac{3}{2})}{(n + 1)!Γ(\frac{3}{2})} z^{n}

and

\frac{Γ(n + \frac{3}{2})}{(n + 1)!Γ(\frac{3}{2})} = \frac{(n + \frac{1}{2})Γ(n + \frac{1}{2})}{\frac{1}{2} (n + 1) n! Γ(\frac{1}{2})} = \frac{2n + 1}{n + 1}\binom{n − \frac{1}{2}}{n}

we find

\frac{K − E}{z} = \frac{π}{2} ∑_{n=0}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{2n + 1}{2n + 2} z^{n}  (1.10)

This is equivalent to (1.8) because the sum in that equation can be rewritten as

∑_{n=0}^{∞} \binom{n + \frac{1}{2}}{n + 1}^{2} \frac{2n + 2}{2n + 1} z^{n} with \binom{n + \frac{1}{2}}{n + 1} = \binom{n − \frac{1}{2}}{n}\frac{2n + 1}{2n + 2}

1.2. 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{1}{2} and b = \frac{1}{2} , we find

\frac{E}{1 − z} = \frac{π}{2} F (\frac{3}{2}, \frac{1}{2}, 1; z)  (1.11)

and, using

\frac{Γ(n + \frac{3}{2})}{n!Γ(\frac{3}{2})} = \binom{n + \frac{1}{2}}{n} = \frac{3 · 5 ⋯ (2n + 1)}{2 · 4 ⋯ 2n} = (2n + 1)\binom{n − \frac{1}{2}}{n},

we obtain

\frac{E}{1 − z} = \frac{π}{2} ∑_{n=0}^{∞} \binom{n − \frac{1}{2}}{n}^{2} (2n + 1) z^{n}  (1.12)

1.3. Expansion of the inductance Combining the results (1.10) and (1.12):

\frac{K − E}{z} + \frac{E}{1 − z} = \frac{π}{2} ∑_{n=0}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{(2n + 1) (2n + 3)}{2n + 2} z^{n}

Equivalently, (1.9) and (1.11) give:

\frac{K − E}{z} + \frac{E}{1 − z} = \frac{π}{2} [F (\frac{3}{2}, \frac{1}{2}, 2; z) + \frac{1}{2} F (\frac{3}{2}, \frac{1}{2}, 1; z)]

Using AS , this formula can be rewritten as

\frac{K − E}{z} + \frac{E}{1 − z} = \frac{3π}{4} F (\frac{5}{2}, \frac{1}{2}, 2; z)

with

F (\frac{5}{2}, \frac{1}{2}, 2; z) = ∑_{n=0}^{∞} \frac{Γ(n + \frac{5}{2})}{(n + 1)!Γ(\frac{5}{2})} \frac{Γ(n + \frac{1}{2})}{n!Γ(\frac{1}{2})} z^{n} = ∑_{n=0}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{(2n + 3) (2n + 1)}{3 (n + 1)} z^{n}

Using the transformation formula

F (a, b, c; z) = (1 − z)^{−b}F (b, c − a, c; \frac{z}{z − 1})

and putting

z = \frac{x^{2}}{1 + x^{2}} so that \frac{z}{z − 1} = −x^{2},

where x = 2a/l (ratio of solenoid diameter to length), we have finally

\frac{4}{3π} \sqrt{1 − z} (\frac{K − E}{z} + \frac{E}{1 − z} ) = F (\frac{1}{2}, −\frac{1}{2}, 2; −x^{2})
= 1 − ∑_{n=1}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{(−x^{2})^{n}}{(n + 1)(2n − 1)}

As a check, the leading terms are:

1 + \frac{1}{2} (\frac{x}{2})^{2} − \frac{1}{4} (\frac{x}{2})^{4} + \frac{5}{16} (\frac{x}{2})^{6} − \frac{35}{64} (\frac{x}{2})^{8} + \frac{147}{128} (\frac{x}{2})^{10} + ⋯

Then the inductance of a finite solenoid is Lf, where L is computed neglecting end effects (L = (4π/c^{2})N^{2}πa^{2}/l) and

f = F (\frac{1}{2}, −\frac{1}{2}, 2; −x^{2}) − \frac{4x}{3π}

F ((\frac{x}{2})) = 1 − \frac{8 (\frac{x}{2})}{3π} + \frac{1}{2} (\frac{x}{2})^{2} − \frac{1}{4} (\frac{x}{2})^{4} + \frac{5}{16} (\frac{x}{2})^{6} − \frac{35}{64} (\frac{x}{2})^{8} + \frac{147}{128} (\frac{x}{2})^{10}

− ∑_{n=6}^{10} \binom{n− \frac{1}{2}}{n}^{2}^{\frac{(−4x^{2})^{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 (x^{20}) \frac{1}{2}x^{2} − \frac{1}{4}x^{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 (x^{20}) 16384/256 = 64 and 32768/2048 = 16 \binom{n− \frac{1}{2}}{n} = \frac{(2n)!}{2^{2n}(n!)^{2}} and \binom{7− \frac{1}{2}}{7} = \frac{(14)!}{2^{14}(7!)^{2}} is true \frac{150!}{2^{150}(75!)^{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