In mathematics, specifically the theory of elliptic functions, the nome is a special function that belongs to the non-elementary functions. This function is of great importance in the description of the elliptic functions, especially in the description of the modular identity of the Jacobi theta function, the Hermite elliptic transcendents and the Weber modular functions, that are used for solving equations of higher degrees.
The nome function is given by
q =e-{\pi
where
K
iK'
\omega1
\omega2
0<q<1
0<q<1
q\inC
0<|q|<1
\tau
q
Notationally, the quarter periods
K
iK'
\omega1
\omega2
\omega1
\omega2
k
q(k)=e-\pi
The complementary nome
q1
q1(k)=e-\pi.
Sometimes the notation
q=e{2\rm{i
The mentioned functions
K
K'
K(x)=
\pi/2 | |
\int | |
0 |
1 | |
\sqrt{1-x2\sin(\varphi)2 |
K'(x)=K(\sqrt{1-x2})=
\pi/2 | |
\int | |
0 |
1 | |
\sqrt{1-(1-x2)\sin(\varphi)2 |
The nome solves the following equation:
|k|=
| |||||||
|
→ q(k)=e-\pi
This analogon is valid for the Pythagorean complementary modulus:
k'=\sqrt{1-k2}=
| |||||||
|
→ q(k)=e-\pi
where
\vartheta10,\theta00
K(k)
k
\vartheta00(v;w)=
infty | |
\prod | |
n=1 |
(1-w2n)[1+2\cos(2v)w2n-1+w4n-2]
\vartheta01(v;w)=
infty | |
\prod | |
n=1 |
(1-w2n)[1-2\cos(2v)w2n-1+w4n-2]
\vartheta10(v;w)=2w1/4
infty | |
\cos(v)\prod | |
n=1 |
(1-w2n)[1+2\cos(2v)w2n+w4n]
These three definition formulas are written down in the fourth edition of the book A Course in Modern Analysis written by Whittaker and Watson on the pages 469 and 470. The nome is commonly used as the starting point for the construction of Lambert series, the q-series and more generally the q-analogs. That is, the half-period ratio
\tau
q=0
q=0
\tau\toinfty
The upper half-plane (and the Poincaré disk, and the punctured disk) can thus be tiled with the fundamental domain, which is the region of values of the half-period ratio
\tau
q
K
iK'
The prototypical modular function is Klein's j-invariant. It can be written as a function of either the half-period ratio τ or as a function of the nome
q
Euler's function arises as the prototype for q-series in general.
The nome, as the
q
Every real value
x
[-1,1]
q(x)
q(x)=q(-x)
(-1,1)
q(x)
The Legendre's relation is defined that way:
KE'+EK'-KK'=\tfrac{1}{2}\pi
And as described above, the elliptic nome function
q(x)
q(x)=\exp\left[-\pi
K(\sqrt{1-x2 | |
)}{K(x)}\right] |
Furthermore, these are the derivatives of the two complete elliptic integrals:
d | |
dx |
K(x)=
1 | |
x(1-x2) |
l[E(x)-(1-x2)K(x)r]
d | |
dx |
E(x)=-
1 | |
x |
l[K(x)-E(x)r]
Therefore, the derivative of the nome function has the following expression:
d | |
dx |
q(x)=
\pi2 | |
2x(1-x2)K(x)2 |
q(x)
The second derivative can be expressed this way:
d2 | |
dx2 |
q(x)=
\pi4+2\pi2(1+x2)K(x)2-4\pi2K(x)E(x) | |
4x2(1-x2)2K(x)4 |
q(x)
And that is the third derivative:
d3 | |
dx3 |
q(x)=
\pi6+6\pi4(1+x2)K(x)2-12\pi4K(x)E(x)+8\pi2(1+x2)2K(x)4-24\pi2(1+x2)K(x)3E(x)+24\pi2K(x)2E(x)2 | |
8x3(1-x2)3K(x)6 |
q(x)
The complete elliptic integral of the second kind is defined as follows:
E(x)=
\pi/2 | |
\int | |
0 |
\sqrt{1-x2\sin(\varphi)2}d\varphi=
1 | |
2\int | |
0 |
\sqrt{(y2+1)2-4x2y2 | |
The following equation follows from these equations by eliminating the complete elliptic integral of the second kind:
3l[ | d2 |
dx2 |
q(x)r]2-2l[
d | q(x)r]l[ | |
dx |
d3 | |
dx3 |
q(x)r]=
\pi8-4\pi4(1+x2)2K(x)4 | |
16x4(1-x2)4K(x)8 |
q(x)2
Thus, the following third-order quartic differential equation is valid:
x2(1-x2)2[2q(x)2q'(x)q'''(x)-3q(x)2q''(x)2+q'(x)4]=(1+x2)2q(x)2q'(x)2
Given is the derivative of the Elliptic Nome mentioned above:
d | |
dx |
q(x)=
\pi2 | |
2x(1-x2)K(x)2 |
q(x)
The outer factor with the K-integral in the denominator shown in this equation is the derivative of the elliptic period ratio. The elliptic period ratio is the quotient of the K-integral of the Pythagorean complementary modulus divided by the K-integral of the modulus itself. And the integer number sequence in MacLaurin series of that elliptic period ratio leads to the integer sequence of the series of the elliptic nome directly.
The German mathematician Adolf Kneser researched on the integer sequence of the elliptic period ratio in his essay Neue Untersuchung einer Reihe aus der Theorie der elliptischen Funktionen and showed that the generating function of this sequence is an elliptic function. Also a further mathematician with the name Robert Fricke analyzed this integer sequence in his essay Die elliptischen Funktionen und ihre Anwendungen and described the accurate computing methods by using this mentioned sequence. The Kneser integer sequence Kn(n) can be constructed in this way:
Kn(2n)=24n-3\binom{4n}{2n}+
42n-2m\binom{4n}{2n-2m}Kn(m) | |||||||
Kn(2n+1)=24n-1\binom{4n+2}{2n+1}+
42n-2m+1\binom{4n+2}{2n-2m+1}Kn(m) |
Executed examples:
Kn(2)=2 x 6+1 x {\color{cornflowerblue}1}={\color{cornflowerblue}13} | |
Kn(3)=8 x 20+24 x {\color{cornflowerblue}1}={\color{cornflowerblue}184} | |
Kn(4)=32 x 70+448 x {\color{cornflowerblue}1}+1 x {\color{cornflowerblue}13}={\color{cornflowerblue}2701} | |
Kn(5)=128 x 252+7680 x {\color{cornflowerblue}1}+40 x {\color{cornflowerblue}13}={\color{cornflowerblue}40456} | |
Kn(6)=512 x 924+126720 x {\color{cornflowerblue}1}+1056 x {\color{cornflowerblue}13}+1 x {\color{cornflowerblue}184}={\color{cornflowerblue}613720} | |
Kn(7)=2048 x 3432+2050048 x {\color{cornflowerblue}1}+23296 x {\color{cornflowerblue}13}+56 x {\color{cornflowerblue}184}={\color{cornflowerblue}9391936} |
The Kneser sequence appears in the Taylor series of the period ratio (half period ratio):
1 | lnl( | |
4 |
16 | |
x2 |
r)-
\piK'(x) | |
4K(x) |
=
infty | |
\sum | |
n=1 |
Kn(n) | |
24nn |
x2n
{\color{limegreen} | 1 | lnl( |
4 |
16 | |
x2 |
r)-
\piK'(x) | |
4K(x) |
=
{\color{cornflowerblue | |
1}}{8}x |
2+
{\color{cornflowerblue | |
13}}{256}x |
4+
{\color{cornflowerblue | |
184}}{6144}x |
6+
{\color{cornflowerblue | |
2701}}{131072}x |
8+
{\color{cornflowerblue | |
40456}}{2621440}x |
10+\ldots}
The derivative of this equation after
x
\pi2 | |
8x(1-x2)K(x)2 |
-
1 | |
2x |
=
infty | |
\sum | |
n=1 |
Kn(n) | |
24n |
x2n
{\color{limegreen} | \pi2 |
8x(1-x2)K(x)2 |
-
1 | |
2x |
=
{\color{cornflowerblue | |
1}}{4}x |
+
{\color{cornflowerblue | |
13}}{64}x |
3+
{\color{cornflowerblue | |
184}}{1024}x |
5+
{\color{cornflowerblue | |
2701}}{16384}x |
7+
{\color{cornflowerblue | |
40456}}{262144}x |
9+\ldots}
KE'+EK'-KK'=\tfrac{1}{2}\pi
The mathematician discovered the integer number sequence that appears in the MacLaurin series of the fourth root of the quotient Elliptic Nome function divided by the square function. The construction of this sequence is detailed in his work Die Lehre von den Elliptischen Integralen und den Thetafunktionen. The sequence was also constructed by the Silesian German mathematician Hermann Amandus Schwarz in Formeln und Lehrsätze zum Gebrauche der elliptischen Funktionen (pages 54–56, chapter Berechnung der Grösse k). This Schellbach Schwarz number sequence Sc(n) was also analyzed by the mathematicians Karl Theodor Wilhelm Weierstrass and Louis Melville Milne-Thomson in the 20th century. The mathematician Adolf Kneser determined a construction for this sequence based on the following pattern:
Sc(n+1)=
2 | |
n |
\sum
n | |
m=1 |
Sc(m)Kn(n+1-m)
The Schellbach Schwarz sequence Sc(n) appears in the On-Line Encyclopedia of Integer Sequences under the number and the Kneser sequence Kn(n) appears under the number .
The following table contains the Kneser numbers and the Schellbach Schwarz numbers:
1 | 1 | 1 | |
2 | 13 | 2 | |
3 | 184 | 15 | |
4 | 2701 | 150 | |
5 | 40456 | 1707 | |
6 | 613720 | 20910 | |
7 | 9391936 | 268616 | |
8 | 144644749 | 3567400 |
And this sequence creates the MacLaurin series of the elliptic nome[1] [2] in exactly this way:
q(x)=
infty | |
\sum | |
n=1 |
Sc(n) | l( | |
24n |
1-\sqrt[4]{1-x2 | |
q(x)=
| ||||
x |
+
{\color{navy | |
2}}{32}x |
2+
{\color{navy | |
15}}{512}x |
4+
{\color{navy | |
150}}{8192}x |
6+
{\color{navy | |
1707}}{131072}x |
8+\ldots}r)4
In the following, it will be shown as an example how the Schellbach Schwarz numbers are built up successively. For this, the examples with the numbers Sc(4) = 150, Sc(5) = 1707 and Sc(6) = 20910 are used:
Sc(4)=
2 | |
3 |
\sum
3 | |
m=1 |
Sc(m)Kn(4-m)=
2 | |
3 |
l[{\color{navy}Sc(1)}{\color{cornflowerblue}Kn(3)}+{\color{navy}Sc(2)}{\color{cornflowerblue}Kn(2)}+{\color{navy}Sc(3)}{\color{cornflowerblue}Kn(1)}r]
{\color{navy}Sc(4)}=
2 | |
3 |
l({\color{navy}1} x {\color{cornflowerblue}184}+{\color{navy}2} x {\color{cornflowerblue}13}+{\color{navy}15} x {\color{cornflowerblue}1}r)={\color{navy}150}
Sc(5)=
2 | |
4 |
\sum
4 | |
m=1 |
Sc(m)Kn(5-m)=
2 | |
4 |
l[{\color{navy}Sc(1)}{\color{cornflowerblue}Kn(4)}+{\color{navy}Sc(2)}{\color{cornflowerblue}Kn(3)}+{\color{navy}Sc(3)}{\color{cornflowerblue}Kn(2)}+{\color{navy}Sc(4)}{\color{cornflowerblue}Kn(1)}r]
{\color{navy}Sc(5)}=
2 | |
4 |
l({\color{navy}1} x {\color{cornflowerblue}2701}+{\color{navy}2} x {\color{cornflowerblue}184}+{\color{navy}15} x {\color{cornflowerblue}13}+{\color{navy}150} x {\color{cornflowerblue}1}r)={\color{navy}1707}
Sc(6)=
2 | |
5 |
\sum
5 | |
m=1 |
Sc(m)Kn(6-m)=
2 | |
5 |
l[{\color{navy}Sc(1)}{\color{cornflowerblue}Kn(5)}+{\color{navy}Sc(2)}{\color{cornflowerblue}Kn(4)}+{\color{navy}Sc(3)}{\color{cornflowerblue}Kn(3)}+{\color{navy}Sc(4)}{\color{cornflowerblue}Kn(2)}+{\color{navy}Sc(5)}{\color{cornflowerblue}Kn(1)}r]
{\color{navy}Sc(6)}=
2 | |
5 |
l({\color{navy}1} x {\color{cornflowerblue}40456}+{\color{navy}2} x {\color{cornflowerblue}2701}+{\color{navy}15} x {\color{cornflowerblue}184}+{\color{navy}150} x {\color{cornflowerblue}13}+{\color{navy}1707} x {\color{cornflowerblue}1}r)={\color{navy}20910}
The MacLaurin series of the nome function
q(x)
q(x)=
infty | |
\sum | |
n=1 |
\operatorname{Kt | |
(n)}{16 |
n}x2n
And the sum with the same absolute values of the coefficients but with alternating signs generates this function:
ql[x(x2+1)-1/2r]=
infty | |
\sum | |
n=1 |
(-1)n+1\operatorname{Kt | |
(n)}{16 |
n}x2n
The radius of convergence of this Maclaurin series is 1. Here
\operatorname{Kt}(n)
\operatorname{Kt}(n)\isinN
n\isinN
\operatorname{Kt}(n)
The Kotěšovec numbers are generated in the same way as the Schellbach Schwarz numbers are constructed:
The only difference consists in the fact that this time the factor before the sum in this corresponding analogous formula is not
2 | |
n |
8 | |
n |
Kt(n+1)=
8 | |
n |
\sum
n | |
m=1 |
Kt(m)Kn(n+1-m)
Following table contains the Schellbach Schwarz numbers and the Kneser numbers and the Apéry numbers:
1 | 1 | 1 | |
2 | 13 | 8 | |
3 | 184 | 84 | |
4 | 2701 | 992 | |
5 | 40456 | 12514 | |
6 | 613720 | 164688 | |
7 | 9391936 | 2232200 | |
8 | 144644749 | 30920128 |
In the following, it will be shown as an example how the Schellbach Schwarz numbers are built up successively. For this, the examples with the numbers Kt(4) = 992, Kt(5) = 12514 and Kt(6) = 164688 are used:
Kt(4)=
8 | |
3 |
\sum
3 | |
m=1 |
Kt(m)Kn(4-m)=
8 | |
3 |
l[{\color{ForestGreen}Kt(1)}{\color{cornflowerblue}Kn(3)}+{\color{ForestGreen}Kt(2)}{\color{cornflowerblue}Kn(2)}+{\color{ForestGreen}Kt(3)}{\color{cornflowerblue}Kn(1)}r]
{\color{ForestGreen}Kt(4)}=
8 | |
3 |
l({\color{ForestGreen}1} x {\color{cornflowerblue}184}+{\color{ForestGreen}8} x {\color{cornflowerblue}13}+{\color{ForestGreen}84} x {\color{cornflowerblue}1}r)={\color{ForestGreen}992}
Kt(5)=
8 | |
4 |
\sum
4 | |
m=1 |
Kt(m)Kn(5-m)=
8 | |
4 |
l[{\color{ForestGreen}Kt(1)}{\color{cornflowerblue}Kn(4)}+{\color{ForestGreen}Kt(2)}{\color{cornflowerblue}Kn(3)}+{\color{ForestGreen}Kt(3)}{\color{cornflowerblue}Kn(2)}+{\color{ForestGreen}Kt(4)}{\color{cornflowerblue}Kn(1)}r]
{\color{ForestGreen}Kt(5)}=
8 | |
4 |
l({\color{ForestGreen}1} x {\color{cornflowerblue}2701}+{\color{ForestGreen}8} x {\color{cornflowerblue}184}+{\color{ForestGreen}84} x {\color{cornflowerblue}13}+{\color{ForestGreen}992} x {\color{cornflowerblue}1}r)={\color{ForestGreen}12514}
Kt(6)=
8 | |
5 |
\sum
5 | |
m=1 |
Kt(m)Kn(6-m)=
8 | |
5 |
l[{\color{ForestGreen}Kt(1)}{\color{cornflowerblue}Kn(5)}+{\color{ForestGreen}Kt(2)}{\color{cornflowerblue}Kn(4)}+{\color{ForestGreen}Kt(3)}{\color{cornflowerblue}Kn(3)}+{\color{ForestGreen}Kt(4)}{\color{cornflowerblue}Kn(2)}+{\color{ForestGreen}Kt(5)}{\color{cornflowerblue}Kn(1)}r]
{\color{ForestGreen}Kt(6)}=
8 | |
5 |
l({\color{ForestGreen}1} x {\color{cornflowerblue}40456}+{\color{ForestGreen}8} x {\color{cornflowerblue}2701}+{\color{ForestGreen}84} x {\color{cornflowerblue}184}+{\color{ForestGreen}992} x {\color{cornflowerblue}13}+{\color{ForestGreen}12514} x {\color{cornflowerblue}1}r)={\color{ForestGreen}164688}
So the MacLaurin series of the direct Elliptic Nome can be generated:
q(x)=
infty | |
\sum | |
n=1 |
Kt(n) | |
16n |
x2n
q(x)={\color{limegreen}
{\color{ForestGreen | |
1}}{16}x |
2+
{\color{ForestGreen | |
8}}{256}x |
4+
{\color{ForestGreen | |
84}}{4096}x |
6+
{\color{ForestGreen | |
992}}{65536}x |
8+
{\color{ForestGreen | |
12514}}{1048576}x |
10+\ldots}
By adding a further integer number sequence
\operatorname{Ap}(n)
\operatorname{Kt}(n)
\operatorname{Kt}(n)
\operatorname{Kt}(1)=1
n\isinN
\operatorname{Kt}(n+1)=
1 | |
n |
n | |
\sum | |
m=1 |
m\operatorname{Kt}(m)[16\operatorname{Ap}(n+1-m)-\operatorname{Ap}(n+2-m)]
\operatorname{Ap}(n)=
n-1 | |
\sum | |
a=0 |
\binom{2a}{a}2\binom{2n-2-2a}{n-1-a}2
This formula creates the Kotěšovec sequence too, but it only creates the sequence numbers of even indices:
\operatorname{Kt}(2n)=
1 | |
2 |
2n-1 | |
\sum | |
m=1 |
(-1)2n162n\binom{2n-1}{m-1}\operatorname{Kt}(m)
The Apéry sequence
\operatorname{Ap}(n)
4\pi-2K(x)2=1+
infty | |
\sum | |
n=1 |
\operatorname{Ap | |
(n |
+1)x2n
The first numerical values of the central binomial coefficients and the two numerical sequences described are listed in the following table:
Index n | Central binomial coefficient square
| Sequence number Ap(n) | Sequence number Kt(n) | |||
---|---|---|---|---|---|---|
1 | 1 | 1 | 1 | |||
2 | 4 | 8 | 8 | |||
3 | 36 | 88 | 84 | |||
4 | 400 | 1088 | 992 | |||
5 | 4900 | 14296 | 12514 | |||
6 | 63504 | 195008 | 164688 | |||
7 | 853776 | 2728384 | 2232200 | |||
8 | 11778624 | 38879744 | 30920128 | |||
9 | 165636900 | 561787864 | 435506703 | |||
10 | 2363904400 | 8206324928 | 6215660600 | |||
11 | 34134779536 | 120929313088 | 89668182220 | |||
12 | 497634306624 | 1794924383744 | 1305109502496 | |||
13 | 7312459672336 | 26802975999424 | 19138260194422 | |||
14 | 108172480360000 | 402298219288064 | 282441672732656 | |||
15 | 1609341595560000 | 6064992788397568 | 4191287776164504 | |||
16 | 24061445010950400 | 91786654611673088 | 62496081197436736 | |||
17 | 361297635242552100 | 1393772628452578264 | 935823746406530603 |
Václav Kotěšovec wrote down the number sequence
\operatorname{Kt}(n)
Here one example of the Kotěšovec sequence is computed:
{\color{blue}1} x {\color{blue}63504}+{\color{blue}4} x {\color{blue}4900}+{\color{blue}36} x {\color{blue}400}+{\color{blue}400} x {\color{blue}36}+{\color{blue}4900} x {\color{blue}4}+{\color{blue}63504} x {\color{blue}1}={\color{RoyalBlue}195008} \tfrac{1}{5} x {\color{ForestGreen}1} x (16 x {\color{RoyalBlue}14296}-{\color{RoyalBlue}195008})+\tfrac{2}{5} x {\color{ForestGreen}8} x (16 x {\color{RoyalBlue}1088}-{\color{RoyalBlue}14296})+\tfrac{3}{5} x {\color{ForestGreen}84} x (16 x {\color{RoyalBlue}88}-{\color{RoyalBlue}1088})+{} {}+\tfrac{4}{5} x {\color{ForestGreen}992} x (16 x {\color{RoyalBlue}8}-{\color{RoyalBlue}88})+\tfrac{5}{5} x {\color{ForestGreen}12514} x (16 x {\color{RoyalBlue}1}-{\color{RoyalBlue}8})={\color{ForestGreen}164688} |
The two following lists contain many function values of the nome function:
The first list shows pairs of values with mutually Pythagorean complementary modules:
q(\tfrac{1}{2}\sqrt{2})=\exp(-\pi)
q[\tfrac{1}{4}(\sqrt{6}-\sqrt{2})]=\exp(-\sqrt{3}\pi)
q[\tfrac{1}{4}(\sqrt{6}+\sqrt{2})]=\exp(-\tfrac{1}{3}\sqrt{3}\pi)
ql\{\sinl[\tfrac{1}{2}\arcsin(\sqrt{5}-2)r]r\}=\exp(-\sqrt{5}\pi)
ql\{\cosl[\tfrac{1}{2}\arcsin(\sqrt{5}-2)r]r\}=\exp(-\tfrac{1}{5}\sqrt{5}\pi)
q[\tfrac{1}{8}(3\sqrt{2}-\sqrt{14})]=\exp(-\sqrt{7}\pi)
q[\tfrac{1}{8}(3\sqrt{2}+\sqrt{14})]=\exp(-\tfrac{1}{7}\sqrt{7}\pi)
q[\tfrac{1}{2}(\sqrt{3}-1)(\sqrt{2}-\sqrt[4]{3})]=\exp(-3\pi)
q[\tfrac{1}{2}(\sqrt{3}-1)(\sqrt{2}+\sqrt[4]{3})]=\exp(-\tfrac{1}{3}\pi)
ql[\tfrac{1}{16}l(\sqrt{22}+3\sqrt{2}r)l(\tfrac{1}{3}\sqrt[3]{6\sqrt{3}+2\sqrt{11}}-\tfrac{1}{3}\sqrt[3]{6\sqrt{3}-2\sqrt{11}}+\tfrac{1}{3}\sqrt{11}-1r)4r]=\exp(-\sqrt{11}\pi)
ql[\tfrac{1}{16}l(\sqrt{22}-3\sqrt{2}r)l(\tfrac{1}{3}\sqrt[3]{6\sqrt{3}+2\sqrt{11}}-\tfrac{1}{3}\sqrt[3]{6\sqrt{3}-2\sqrt{11}}+\tfrac{1}{3}\sqrt{11}+1r)4r]=\exp(-\tfrac{1}{11}\sqrt{11}\pi)
ql\{\sinl[\tfrac{1}{2}\arcsin(5\sqrt{13}-18)r]r\}=\exp(-\sqrt{13}\pi)
ql\{\cosl[\tfrac{1}{2}\arcsin(5\sqrt{13}-18)r]r\}=\exp(-\tfrac{1}{13}\sqrt{13}\pi)
The second list shows pairs of values with mutually tangentially complementary modules:
q(\sqrt{2}-1)=\exp(-\sqrt{2}\pi)
q[(2-\sqrt{3})(\sqrt{3}-\sqrt{2})]=\exp(-\sqrt{6}\pi)
q[(2-\sqrt{3})(\sqrt{3}+\sqrt{2})]=\exp(-\tfrac{1}{3}\sqrt{6}\pi)
q[(\sqrt{10}-3)(\sqrt{2}-1)2]=\exp(-\sqrt{10}\pi)
q[(\sqrt{10}-3)(\sqrt{2}+1)2]=\exp(-\tfrac{1}{5}\sqrt{10}\pi)
ql[\tfrac{1}{16}\sqrt{2\sqrt{2}-\sqrt{7}}(3\sqrt{2}-\sqrt{14})(\sqrt{2\sqrt{2}+1}-1)4r]=\exp(-\sqrt{14}\pi)
ql[\tfrac{1}{16}\sqrt{2\sqrt{2}+\sqrt{7}}(3\sqrt{2}+\sqrt{14})(\sqrt{2\sqrt{2}+1}-1)4r]=\exp(-\tfrac{1}{7}\sqrt{14}\pi)
q[(2-\sqrt{3})2(\sqrt{2}-1)3]=\exp(-3\sqrt{2}\pi)
q[(2+\sqrt{3})2(\sqrt{2}-1)3]=\exp(-\tfrac{1}{3}\sqrt{2}\pi)
q[(10-3\sqrt{11})(3\sqrt{11}-7\sqrt{2})]=\exp(-\sqrt{22}\pi)
q[(10-3\sqrt{11})(3\sqrt{11}+7\sqrt{2})]=\exp(-\tfrac{1}{11}\sqrt{22}\pi)
ql\{(\sqrt{26}+5)(\sqrt{2}-1)2\tanl[\tfrac{1}{4}\pi-\arctan(\tfrac{1}{3}\sqrt[3]{3\sqrt{3}+\sqrt{26}}-\tfrac{1}{3}\sqrt[3]{3\sqrt{3}-\sqrt{26}}+\tfrac{1}{6}\sqrt{26}-\tfrac{1}{2}\sqrt{2})r]4r\}=\exp(-\sqrt{26}\pi)
ql\{(\sqrt{26}+5)(\sqrt{2}+1)2\tanl[\arctan(\tfrac{1}{3}\sqrt[3]{3\sqrt{3}+\sqrt{26}}-\tfrac{1}{3}\sqrt[3]{3\sqrt{3}-\sqrt{26}}+\tfrac{1}{6}\sqrt{26}+\tfrac{1}{2}\sqrt{2})-\tfrac{1}{4}\pir]4r\}=\exp(-\tfrac{1}{13}\sqrt{26}\pi)
Related quartets of values are shown below:
ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10}-3)2(\sqrt{5}-2)2]\}r\rangle=\exp(-\sqrt{30}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10}-3)2(\sqrt{5}+2)2]\}r\rangle=\exp(-\tfrac{1}{3}\sqrt{30}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10}+3)2(\sqrt{5}-2)2]\}r\rangle=\exp(-\tfrac{1}{5}\sqrt{30}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10}+3)2(\sqrt{5}+2)2]\}r\rangle=\exp(-\tfrac{1}{15}\sqrt{30}\pi) | |
ql\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7}-3\sqrt{3})2(2\sqrt{2}-\sqrt{7})2]\}r\rangle=\exp(-\sqrt{42}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7}-3\sqrt{3})2(2\sqrt{2}+\sqrt{7})2]\}r\rangle=\exp(-\tfrac{1}{3}\sqrt{42}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7}+3\sqrt{3})2(2\sqrt{2}-\sqrt{7})2]\}r\rangle=\exp(-\tfrac{1}{7}\sqrt{42}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7}+3\sqrt{3})2(2\sqrt{2}+\sqrt{7})2]\}r\rangle=\exp(-\tfrac{1}{21}\sqrt{42}\pi) | |
ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5}-2)4(\sqrt{2}-1)6]\}r\rangle=\exp(-\sqrt{70}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5}-2)4(\sqrt{2}+1)6]\}r\rangle=\exp(-\tfrac{1}{5}\sqrt{70}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5}+2)4(\sqrt{2}-1)6]\}r\rangle=\exp(-\tfrac{1}{7}\sqrt{70}\pi) ql\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5}+2)4(\sqrt{2}+1)6]\}r\rangle=\exp(-\tfrac{1}{35}\sqrt{70}\pi) |
The elliptic nome was explored by Richard Dedekind and this function is the fundament in the theory of eta functions and their related functions. The elliptic nome is the initial point of the construction of the Lambert series. In the theta function by Carl Gustav Jacobi the nome as an abscissa is assigned to algebraic combinations of the Arithmetic geometric mean and also the complete elliptic integral of the first kind. Many infinite series[3] can be described easily in terms of the elliptic nome:
infty | |
\sum | |
n=1 |
q(x)\Box(n)=\tfrac{1}{2}\vartheta00[q(x)]-\tfrac{1}{2}=\tfrac{1}{2}\sqrt{2\pi-1K(x)}-\tfrac{1}{2}=\tfrac{1}{2}\operatorname{agm}(1-x;1+x)-1/2-\tfrac{1}{2}
infty | |
\sum | |
n=1 |
q(x)\Box(2n-1)=\tfrac{1}{4}\vartheta00[q(x)]-\tfrac{1}{4}\vartheta01[q(x)]=\tfrac{1}{4}(1-\sqrt[4]{1-x2})\sqrt{2\pi-1K(x)}
infty | |
\sum | |
n=1 |
2q(x)n | |
q(x)2n+1 |
=\tfrac{1}{2}\vartheta00[q(x)]2-\tfrac{1}{2}=\pi-1K(x)-\tfrac{1}{2}
infty | |
\sum | |
n=1 |
2q(x)2n-1 | |
q(x)4n-2+1 |
=\tfrac{1}{4}\vartheta00[q(x)]2-\tfrac{1}{4}\vartheta01[q(x)]2=\tfrac{1}{2}(1-\sqrt{1-x2})\pi-1K(x)
infty | |
\sum | |
n=1 |
\Box(n)q(x)\Box(n)=2-1/2\pi-5/2K(x)3/2[E(x)-(1-x2)K(x)]
infty | ||
\sum | l[ | |
n=1 |
2q(x)n | |
1+q(x)2n |
r]2=2\pi-2E(x)K(x)-\tfrac{1}{2}
infty | ||
\sum | l[ | |
n=1 |
2q(x)n | |
1-q(x)2n |
r]2=\tfrac{2}{3}\pi-2(2-x2)K(x)2-2\pi-2K(x)E(x)+\tfrac{1}{6}
The quadrangle represents the square number of index n, because in this way of notation the two in the exponent of the exponent would appear to small. So this formula is valid:
\Box(n)=n2
The letter
\operatorname{E}(\varepsilon)
\varepsilon
The two most important theta functions can be defined by following product series:
infty | |
\prod | |
n=1 |
[1-q(x)2n][1+q(x)2n-1]2=\vartheta00[q(x)]=\sqrt{2\pi-1K(x)}
infty | |
\prod | |
n=1 |
[1-q(x)2n][1-q(x)2n-1]2=\vartheta01[q(x)]=\sqrt[4]{1-x2}\sqrt{2\pi-1K(x)}
Furthermore, these two Pochhammer products have those two relations:
24 | |
q(\varepsilon)[q(\varepsilon);q(\varepsilon)] | |
infty |
=256\varepsilon2(1-\varepsilon2)4\pi-{12
\varepsilon2
24 | |
[q(\varepsilon);q(\varepsilon) | |
infty |
=16(1-\varepsilon2)2q(\varepsilon)
The Pochhammer products have an important role in the pentagonal number theorem and its derivation.
The nome function can be used for the definition of the complete elliptic integrals of first and second kind:
K(\varepsilon)=\tfrac{1}{2}\pi\vartheta00[q(\varepsilon)]2
E(\varepsilon)=2\piq(\varepsilon)\vartheta00'[q(\varepsilon)]\vartheta00[q(\varepsilon)]-3+\tfrac{1}{2}\pi(1-
2)\vartheta | |
\varepsilon | |
00 |
[q(\varepsilon)]2
In this case the dash in the exponent position stands for the derivative of the so-called theta zero value function:
\vartheta00'(x)=
d | |
dx |
\vartheta00(x)=2+
infty | |
\sum | |
n=1 |
2(n+1)2xn(n+2)
The elliptic functions Zeta Amplitudinis and Delta Amplitudinis can be defined with the elliptic nome function[4] easily:
\operatorname{zn}(x;k)=
infty | |
\sum | |
n=1 |
2\piK(k)-1\sin[\piK(k)-1x]q(k)2n-1 | |
1-2\cos[\piK(k)-1x]q(k)2n-1+q(k)4n-2 |
\operatorname{dn}(x;k)=
infty | |
\sqrt[4]{1-k | |
n=1 |
1+2\cos[\piK(k)-1x]q(k)2n-1+q(k)4n-2 | |
1-2\cos[\piK(k)-1x]q(k)2n-1+q(k)4n-2 |
Using the fourth root of the quotient of the nome divided by the square function as it was mentioned above, following product series definitions can be set up for the Amplitude Sine, the Counter Amplitude Sine and the Amplitude Cosine in this way:
\operatorname{sn}(x;k)=2\sqrt[4]{k-2q(k)}\sin[\tfrac{1}{2}\piK(k)-1
infty | |
x]\prod | |
n=1 |
1-2q(k)2n\cos[\piK(k)-1x]+q(k)4n | |
1-2q(k)2n\cos[\piK(k)-1x]+q(k)4n |
\operatorname{cd}(x;k)=2\sqrt[4]{k-2q(k)}\cos[\tfrac{1}{2}\piK(k)-1
infty | |
x]\prod | |
n=1 |
1+2q(k)2n\cos[\piK(k)-1x]+q(k)4n | |
1+2q(k)2n\cos[\piK(k)-1x]+q(k)4n |
\operatorname{cn}(x;k)=2\sqrt[4]{k-2(1-k2)q(k)}\cos[\tfrac{1}{2}\piK(k)-1
infty | |
x]\prod | |
n=1 |
1+2q(k)2n\cos[\piK(k)-1x]+q(k)4n | |
1-2q(k)2n\cos[\piK(k)-1x]+q(k)4n |
These five formulas are valid for all values k from −1 until +1.
Then following successive definition of the other Jacobi functions is possible:
\operatorname{sn}(x;k)=
2\{\operatorname{zn | |
(\tfrac{1}{2}x;k) |
+\operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}}{k2+\{\operatorname{zn}(\tfrac{1}{2}x;k)+\operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}2}
\operatorname{cd}(x;k)=\operatorname{sn}[K(k)-x;k]
\operatorname{cn}(x;k)=\operatorname{cd}(x;k)\operatorname{dn}(x;k)
\operatorname{dn}(x;k)=
k2-\{\operatorname{zn | |
(\tfrac{1}{2}x;k) |
+\operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}2}{k2+\{\operatorname{zn}(\tfrac{1}{2}x;k)+\operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}2}
The product definition of the amplitude sine was written down in the essay π and the AGM by the Borwein brothers on page 60 and this formula is based on the theta function definition of Whittaker und Watson.
In combination with the theta functions the nome gives the values of many Jacobi amplitude function values:
\operatorname{sc}[\tfrac{2}{3}K(k);k]=
\sqrt{3 | |
\vartheta |
01[q(k)6]}{\sqrt{1-
2}\vartheta | |
k | |
01 |
[q(k)2]}
\operatorname{sn}[\tfrac{1}{3}K(k);k]=
2\vartheta00[q(k)]2 | |
3\vartheta00[q(k)3]2+\vartheta00[q(k)]2 |
=
3\vartheta01[q(k)3]2-\vartheta01[q(k)]2 | |
3\vartheta01[q(k)3]2+\vartheta01[q(k)]2 |
\operatorname{cn}[\tfrac{2}{3}K(k);k]=
3\vartheta00[q(k)3]2-\vartheta00[q(k)]2 | |
3\vartheta00[q(k)3]2+\vartheta00[q(k)]2 |
=
2\vartheta01[q(k)]2 | |
3\vartheta01[q(k)3]2+\vartheta01[q(k)]2 |
\operatorname{sn}[\tfrac{1}{5}K(k);k]=l\{
\sqrt{5 | |
\vartheta |
01
5]}{\vartheta | |
[q(k) | |
01 |
[q(k)]}-1r\}l\{
5\vartheta01[q(k)10]2 | |
\vartheta01[q(k)2]2 |
-1r\}-1
\operatorname{sn}[\tfrac{3}{5}K(k);k]=l\{
\sqrt{5 | |
\vartheta |
01
5]}{\vartheta | |
[q(k) | |
01 |
[q(k)]}+1r\}l\{
5\vartheta01[q(k)10]2 | |
\vartheta01[q(k)2]2 |
-1r\}-1
\operatorname{cn}[\tfrac{2}{5}K(k);k]=l\{
\sqrt{5 | |
\vartheta |
00
5]}{\vartheta | |
[q(k) | |
00 |
[q(k)]}+1r\}l\{
5\vartheta01[q(k)10]2 | |
\vartheta01[q(k)2]2 |
-1r\}-1
\operatorname{cn}[\tfrac{4}{5}K(k);k]=l\{
\sqrt{5 | |
\vartheta |
00
5]}{\vartheta | |
[q(k) | |
00 |
[q(k)]}-1r\}l\{
5\vartheta01[q(k)10]2 | |
\vartheta01[q(k)2]2 |
-1r\}-1
The abbreviation sc describes the quotient of the amplitude sine divided by the amplitude cosine.
The law for the square of the elliptic noun involves forming the Landen daughter modulus:
q(\varepsilon)2=ql[\varepsilon2(1+\sqrt{1-\varepsilon2})-2r]=ql\{\tanl[\tfrac{1}{2}\arcsin(\varepsilon)r]2r\}=ql\{\operatorname{tanh}l[\tfrac{1}{2}\operatorname{artanh}(\varepsilon)r]2r\} |
The Landen daughter modulus is also the tangential counterpart of the Pythagorean counterpart of the mother modulus.
This formula results as a combination of the following equations: (1+\sqrt{1-\varepsilon2})Fl[\arctan(x);\varepsilonr]=Fl[\arctan(x)+\arctan(\sqrt{1-\varepsilon2}x);\varepsilon2(1+\sqrt{1-\varepsilon2})-2r] w The next equation follows directly from the previous equation: (1+\sqrt{1-\varepsilon2})K(\varepsilon)=2Kl[\varepsilon2(1+\sqrt{1-\varepsilon2})-2r] By changing the substitution this expression is generated: Kl[2\sqrt[4]{1-\varepsilon2}(1+\sqrt{1-\varepsilon2})-1r]=(1+\sqrt{1-\varepsilon2})K(\sqrt{1-\varepsilon2}) The combination of both formulas leads to that quotient equation:
={\color{green}
2})-1]}{K[\varepsilon2(1+\sqrt{1-\varepsilon2})-2]}} Both sides of this equation scale show period ratios. For on both sides of this balance the modulus in the numerator is Pythagorean complementary to the modulus in the denominator. The elliptic nome is defined as an exponential function from the negative circle number times the real period ratio. And the real period ratio is defined as the quotient of the K integral of the Pythagorean complementary modulus divided by the K integral of the modulus itself. This is the consequence: q(\varepsilon)2=\expl[-\pi
r]2=\expl[-\pi
2=\expl\{-\pil[{\color{blue}2
= =\expl\{-\pi{\color{green}
2})-1]}{K[\varepsilon2(1+\sqrt{1-\varepsilon2})-2]}}r\}=\expl\{-\pi
-2]}{K[\varepsilon2(1+\sqrt{1-\varepsilon2})-2]}r\}=ql[\varepsilon2(1+\sqrt{1-\varepsilon2})-2r] |
The Landen daughter modulus[5] is identical to the tangential opposite of the Pythagorean opposite of the mother modulus.
Three examples shall be shown in the following:
Trigonometrically displayed examples:
\exp(-2\sqrt{3}\pi)=\exp(-\sqrt{3}\pi)2=ql[\sin(\tfrac{1}{12}\pi)r]2=ql[\tan(\tfrac{1}{24}\pi)2r]
\exp(-2\sqrt{5}\pi)=\exp(-\sqrt{5}\pi)2=ql\{\sinl[\tfrac{1}{2}\arcsin(\sqrt{5}-2)r]r\}2=ql\{\tanl[\tfrac{1}{4}\arcsin(\sqrt{5}-2)r]2r\}
\exp(-2\sqrt{7}\pi)=\exp(-\sqrt{7}\pi)2=ql\{\sinl[\tfrac{1}{2}\arcsin(\tfrac{1}{8})r]r\}2=ql\{\tanl[\tfrac{1}{4}\arcsin(\tfrac{1}{8})r]2r\}
\exp(-2\sqrt{13}\pi)=\exp(-\sqrt{13}\pi)2=ql\{\sinl[\tfrac{1}{2}\arcsin(5\sqrt{13}-18)r]r\}2=ql\{\tanl[\tfrac{1}{4}\arcsin(5\sqrt{13}-18)r]2r\}
Hyperbolically displayed examples:
\exp(-2\sqrt{6}\pi)=\exp(-\sqrt{6}\pi)2=
=ql\langle\operatorname{tanh}l\{\tfrac{1}{2}\operatorname{arsinh}l[(\sqrt{2}-1)2r]r\}r\rangle2=ql\langle\operatorname{tanh}l\{\tfrac{1}{4}\operatorname{arsinh}l[(\sqrt{2}-1)2r]r\}2r\rangle
\exp(-2\sqrt{10}\pi)=\exp(-\sqrt{10}\pi)2=
=ql\langle\operatorname{tanh}l\{\tfrac{1}{2}\operatorname{arsinh}l[(\sqrt{5}-2)2r]r\}r\rangle2=ql\langle\operatorname{tanh}l\{\tfrac{1}{4}\operatorname{arsinh}l[(\sqrt{5}-2)2r]r\}2r\rangle
\exp(-2\sqrt{14}\pi)=\exp(-\sqrt{14}\pi)2=
=ql\langle\operatorname{tanh}l\{\tfrac{1}{2}\operatorname{arsinh}l[(\sqrt{2}+\tfrac{1}{2}-\tfrac{1}{2}\sqrt{4\sqrt{2}+5})3r]r\}r\rangle2=ql\langle\operatorname{tanh}l\{\tfrac{1}{4}\operatorname{arsinh}l[(\sqrt{2}+\tfrac{1}{2}-\tfrac{1}{2}\sqrt{4\sqrt{2}+5})3r]r\}2r\rangle
\exp(-2\sqrt{22}\pi)=\exp(-\sqrt{22}\pi)2=
=ql\langle\operatorname{tanh}l\{\tfrac{1}{2}\operatorname{arsinh}l[(\sqrt{2}-1)6r]r\}r\rangle2=ql\langle\operatorname{tanh}l\{\tfrac{1}{4}\operatorname{arsinh}l[(\sqrt{2}-1)6r]r\}2r\rangle
Not only the law for the square but also the law for the cube of the elliptic nome leads to an elementary modulus transformation.This parameterized formula for the cube of the elliptic noun is valid for all values −1 < u < 1.
q[u(\sqrt{u4-u2+1}-u2+1)]3=q[u(\sqrt{u4-u2+1}+u2-1)] |
This formula was displayed exactly like this and this time it was not printed exactly after the expression
\varepsilon
u
This formula results as a combination of the following equations: (2\sqrt{u4-u2+1}-2u2+1)Fl[\arctan(w);u(\sqrt{u4-u2+1}-u2+1)r]= =Fl\{\arctan(w)+2\arctanl[(\sqrt{u4-u2+1}-u2)wr];u(\sqrt{u4-u2+1}+u2-1)r\} w The next equation follows directly from the previous equation: (2\sqrt{u4-u2+1}-2u2+1)Kl[u(\sqrt{u4-u2+1}-u2+1)r]=3Kl[u(\sqrt{u4-u2+1}+u2-1)r] By changing the substitution this expression is generated: Kl[\sqrt{1-u2}(\sqrt{u4-u2+1}+u2)r]=(2\sqrt{u4-u2+1}-2u2+1)Kl[\sqrt{1-u2}(\sqrt{u4-u2+1}-u2)r] The combination of both formulas leads to that quotient equation:
4-u2+1}-u2)]}{K[u(\sqrt{u4-u2+1}-u2+1)]}}={\color{green}
4-u2+1}+u2)]}{K[u(\sqrt{u4-u2+1}+u2-1)]}} Both sides of this equation scale show period ratios. For on both sides of this balance the modulus in the numerator is Pythagorean complementary to the modulus in the denominator. The elliptic nome is defined as an exponential function from the negative circle number times the real period ratio. And the real period ratio is defined as the quotient of the K integral of the Pythagorean complementary modulus divided by the K integral of the modulus itself. This is the consequence: ql[u(\sqrt{u4-u2+1}-u2+1)r]3=\expl\{-\pi
2+1)]}{K[u(\sqrt{u4-u2+1}-u2+1)]}r\}3= =\expl\{-\pi
4-u2+1}-u2)]}{K[u(\sqrt{u4-u2+1}-u2+1)]}r\}3=\expl\langle-\pil\{{\color{blue}3
4-u2+1}-u2)]}{K[u(\sqrt{u4-u2+1}-u2+1)]}}r\}r\rangle= =\expl\{-\pi{\color{green}
4-u2+1}+u2)]}{K[u(\sqrt{u4-u2+1}+u2-1)]}}r\}=\expl\{-\pi
2-1)]}{K[u(\sqrt{u4-u2+1}+u2-1)]}r\}=ql[u(\sqrt{u4-u2+1}+u2-1)r] |
On the basis of the now absolved proof a direct formula for the nome cube theorem in relation to the modulus
\varepsilon
The works Analytic Solutions to Algebraic Equations by Johansson and Evaluation of Fifth Degree Elliptic Singular Moduli by Bagis showed in their quotated works that the Jacobi amplitude sine of the third part of the complete first kind integral K solves following quartic equation:
\varepsilon2x4-2\varepsilon2x3+2x-1=0
x=snl[\tfrac{1}{3}K(\varepsilon);\varepsilonr]
Now the parametrization mentioned above is inserted into this equation:
\varepsilon=u(\sqrt{u4-u2+1}-u2+1)
u2(\sqrt{u4-u2+1}-u2+1)2(x4-2x3)+2x-1=0
This is the real solution of the pattern
\tfrac{1}{2}<x<1\capx\in\R
x=
1 | |
\sqrt{u4-u2+1 |
-u2+1}
Therefore, following formula is valid:
snl[\tfrac{1}{3}K(\varepsilon);\varepsilonr]l[\varepsilon=u(\sqrt{u4-u2+1}-u2+1)r]=
1 | |
\sqrt{u4-u2+1 |
-u2+1}
The parametrized nome cube formula has this mentioned form:
ql[u(\sqrt{u4-u2+1}-u2+1)r]3=ql[u(\sqrt{u4-u2+1}+u2-1)r]
The same formula can be designed in this alternative way:
ql[u(\sqrt{u4-u2+1}-u2+1)r]3=ql\{l[u(\sqrt{u4-u2+1}-u2+1)r]3l(\sqrt{u4-u2+1}-u2+1r)-4r\}
So this result appears as the direct nome cube theorem:
q(\varepsilon)3=ql\{\varepsilon3snl[\tfrac{1}{3}K(\varepsilon);\varepsilonr]4r\}
Alternatively, this formula can be set up:
ql\{\tanl[\tfrac{1}{2}\arctan(t3)r]r\}3=ql\{\tanl[\tfrac{1}{2}\arctan(t3)r]3\tanl[\arctanl(\sqrt{2\sqrt{t4-t2+1}-t2+2}+\sqrt{t2+1}r)-\tfrac{1}{4}\pir]4r\} |
The now presented formula is used for simplified computations, because the given elliptical modulus can be used to determine the value
t
t
t
Two examples are to be treated exemplarily:
In the first example, the value
t=1
{\color{blue}\exp(-3\sqrt{2}\pi)}=\exp(-\sqrt{2}\pi)3=q(\sqrt{2}-1)3=ql\{\tanl[\tfrac{1}{2}\arctan(1)r]r\}3=
=ql\{\tanl[\tfrac{1}{2}\arctan(1)r]3\tanl[\arctan(\sqrt{3}+\sqrt{2})-\tfrac{1}{4}\pir]4r\}={\color{blue}ql[(\sqrt{2}-1)3(\tfrac{1}{2}\sqrt{6}-\tfrac{1}{2}\sqrt{2})4r]}
In the second example, the value
t=\Phi-2=\tfrac{1}{2}(3-\sqrt{5})
{\color{blue}\exp(-3\sqrt{10}\pi)}=\exp(-\sqrt{10}\pi)3=ql[(\sqrt{10}-3)(\sqrt{2}-1)2r]3=ql\{\tanl[\tfrac{1}{2}\arctan(\Phi-6)r]r\}3=
=ql\{\tanl[\tfrac{1}{2}\arctan(\Phi-6)r]3\tanl[\arctanl(\sqrt{2\sqrt{\Phi-8-\Phi-4+1}-\Phi-4+2}+\sqrt{\Phi-4+1}r)-\tfrac{1}{4}\pir]4r\}=
={\color{blue}ql\{(\sqrt{10}-3)3(\sqrt{2}-1)6\tanl[\arctanl(\sqrt{2\sqrt{\Phi-8-\Phi-4+1}-\Phi-4+2}+\sqrt{\Phi-4+1}r)-\tfrac{1}{4}\pir]4r\}}
The constant
\Phi
\Phi=\tfrac{1}{2}(\sqrt{5}+1)
Every power of a nome of a positive algebraic number as base and a positive rational number as exponent is equal to a nome value of a positive algebraic number:
q(\varepsilon1\inA+
| ||||
) |
=q(\varepsilon2\inA+)
These are the most important examples of the general exponentiation theorem:
q(\varepsilon)2=q\{\varepsilon2\operatorname{sn}[\tfrac{1}{2}K(\varepsilon);\varepsilon]4\}=q[\varepsilon2(1+\sqrt{1-\varepsilon2})-2]
q(\varepsilon)3=q\{\varepsilon3\operatorname{sn}[\tfrac{1}{3}K(\varepsilon);\varepsilon]4\}
q(\varepsilon)4=q\{\varepsilon4\operatorname{sn}[\tfrac{1}{4}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{4}K(\varepsilon);\varepsilon]4\}=q[(1-\sqrt[4]{1-\varepsilon2})2(1+\sqrt[4]{1-\varepsilon2})-2]
q(\varepsilon)5=q\{\varepsilon5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]4\}
q(\varepsilon)6=q\{\varepsilon6\operatorname{sn}[\tfrac{1}{6}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{1}{2}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{5}{6}K(\varepsilon);\varepsilon]4\}
q(\varepsilon)7=q\{\varepsilon7\operatorname{sn}[\tfrac{1}{7}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{7}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{5}{7}K(\varepsilon);\varepsilon]4\}
q(\varepsilon)8=q\{\varepsilon8\operatorname{sn}[\tfrac{1}{8}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{8}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{5}{8}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{7}{8}K(\varepsilon);\varepsilon]4\}
q(\varepsilon)9=q\{\varepsilon9\operatorname{sn}[\tfrac{1}{9}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{1}{3}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{5}{9}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{7}{9}K(\varepsilon);\varepsilon]4\}
The abbreviation
\operatorname{sn}
For algebraic
x
[-1,1]
This are the general exponentiation theorems:
q(\varepsilon)2n=ql\{\varepsilon2n
n | |
\prod | |
k=1 |
\operatorname{sn}l[\tfrac{2k-1}{2n}K(\varepsilon);\varepsilonr]4r\}
q(\varepsilon)2n+1=ql\{\varepsilon2n+1
n | |
\prod | |
k=1 |
\operatorname{sn}l[\tfrac{2k-1}{2n+1}K(\varepsilon);\varepsilonr]4r\}
That theorem is valid for all natural numbers n.
Important computation clues:
The following Jacobi amplitude sine expressions solve the subsequent equations:
Thirds of the K: x=sn[\tfrac{1}{3}K(k);k] k2x4-2k2x3+2x-1=0 | Fifths of the K: x=sn[\tfrac{1}{5}K(k);k]sn[\tfrac{3}{5}K(k);k] k6x6-4k6x5+5k4x4-5k2x2+4x-1=0 | |
Sevenths of the K: x=sn[\tfrac{1}{7}K(k);k]sn[\tfrac{3}{7}K(k);k]sn[\tfrac{5}{7}K(k);k] k12x8-8k12x7+28k10x6-56k8x5+70k6x4-56k4x3+28k2x2-8x+1=0 (1-k2x)8=(1-k2)(1-k14x8) | ||
Elevenths of the K: x=sn[\tfrac{1}{11}K(k);k]sn[\tfrac{3}{11}K(k);k]sn[\tfrac{5}{11}K(k);k]sn[\tfrac{7}{11}K(k);k]sn[\tfrac{9}{11}K(k);k] k30x12+(-32k30+22k28)x11+44k26x10-(88k24+22k22)x9+165k20x8-132k18x7+(-44k16+44k14)x6+ +132k12x5-165k10x4+(22k8+88k6)x3-44k4x2+(-22k2+32)x-1=0 |
For these nome power theorems important examples shall be formulated:
Given is the fifth power theorem:
q(\varepsilon)5=q\{\varepsilon5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]4\}
Lemniscatic example for the fifth power theorem:
\tfrac{1}{8}x6-\tfrac{1}{2}x5+\tfrac{5}{4}x4-\tfrac{5}{2}x2+4x-1=0 x=sn[\tfrac{1}{5}K(k);k]sn[\tfrac{3}{5}K(k);k](k=\tfrac{1}{2}\sqrt{2})=\tfrac{1}{2}(\sqrt{5}-1)(\sqrt[4]{5}-1) {\color{blue}\exp(-5\pi)}=\exp(-\pi)5=q(\tfrac{1}{2}\sqrt{2})5=q\{\varepsilon5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]4\}(\varepsilon=\tfrac{1}{2}\sqrt{2})= =ql\{\tfrac{1}{8}\sqrt{2}l[\tfrac{1}{2}(\sqrt{5}-1)(\sqrt[4]{5}-1)r]4r\}={\color{blue}ql[\tfrac{1}{2}(\sqrt{10}-2\sqrt{2})(3-2\sqrt[4]{5})r]} |
A next example for the fifth power theorem:
(\sqrt{2}-1)6x6-4(\sqrt{2}-1)6x5+5(\sqrt{2}-1)4x4-5(\sqrt{2}-1)2x2+4x-1=0 x=sn[\tfrac{1}{5}K(k);k]sn[\tfrac{3}{5}K(k);k](k=\sqrt{2}-1)=(\sqrt{2}+1)\tanl[\arctanl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}r)-\tfrac{1}{8}\pir] {\color{blue}\exp(-5\sqrt{2}\pi)}=\exp(-\sqrt{2}\pi)5=q(\sqrt{2}-1)5= =q\{\varepsilon5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]4\}(\varepsilon=\sqrt{2}-1)= =ql\langle(\sqrt{2}-1)5l\{(\sqrt{2}+1)\tanl[\arctanl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}r)-\tfrac{1}{8}\pir]r\}4r\rangle= ={\color{blue}ql\{(\sqrt{2}-1)\tanl[\arctanl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}r)-\tfrac{1}{8}\pir]4r\}} |
If two positive numbers
a
b
a2+b2=1
ln[\operatorname{q}(a)]ln[\operatorname{q}(b)]=\pi2
If two positive numbers
c
d
(c+1)(d+1)=2
ln[\operatorname{q}(c)]ln[\operatorname{q}(d)]=2\pi2
Therefore, these representations have validity for all real numbers x:
Pythagorean opposites:
lnl\langleql\{\sinl[\tfrac{1}{4}\pi-\tfrac{1}{2}\arctan(x)r]r\}r\ranglelnl\langleql\{\sinl[\tfrac{1}{4}\pi+\tfrac{1}{2}\arctan(x)r]r\}r\rangle=\pi2
lnl\{ql[\tfrac{1}{2}\sqrt{2-2x(x2+1)-1/2
Tangential opposites:
lnl\langleql\{\tanl[\tfrac{1}{8}\pi-\tfrac{1}{4}\arctan(x)r]r\}r\ranglelnl\langleql\{\tanl[\tfrac{1}{8}\pi+\tfrac{1}{4}\arctan(x)r]r\}r\rangle=2\pi2
lnl\{ql[\sqrt{(\sqrt{x2+1}+x)2+1}-\sqrt{x2+1}-xr]r\}lnl\{ql[\sqrt{(\sqrt{x2+1}-x)2+1}-\sqrt{x2+1}+xr]r\}=2\pi2
The following examples should be used to determine the nouns:
Example 1:Given is the formula of the Pythagorean counterparts:
lnl\langleql\{\sinl[\tfrac{1}{4}\pi-\tfrac{1}{2}\arctan(x)r]r\}r\ranglelnl\langleql\{\sinl[\tfrac{1}{4}\pi+\tfrac{1}{2}\arctan(x)r]r\}r\rangle=\pi2 |
For x = 0, this formula gives this equation:
lnl\{ql[\sin(\tfrac{1}{4}\pi)r]r\}2=\pi2
ql[\sin(\tfrac{1}{4}\pi)r]=\exp(-\pi)
Given is the formula of the tangential counterparts:
lnl\langleql\{\tanl[\tfrac{1}{8}\pi-\tfrac{1}{4}\arctan(x)r]r\}r\ranglelnl\langleql\{\tanl[\tfrac{1}{8}\pi+\tfrac{1}{4}\arctan(x)r]r\}r\rangle=2\pi2 |
For x = 0, the formula for the tangential counterparts gives the following equation:
lnl\{ql[\tan(\tfrac{1}{8}\pi)r]r\}2=2\pi2
ql[\tan(\tfrac{1}{8}\pi)r]=\exp(-\sqrt{2}\pi)
Example 1: Equianharmonic case
The formula of the Pythagorean counterparts is used again:
lnl\langleql\{\sinl[\tfrac{1}{4}\pi-\tfrac{1}{2}\arctan(x)r]r\}r\ranglelnl\langleql\{\sinl[\tfrac{1}{4}\pi+\tfrac{1}{2}\arctan(x)r]r\}r\rangle=\pi2 |
For
x=\sqrt{3}
lnl\{ql[\sin(\tfrac{1}{12}\pi)r]r\}lnl\{ql[\sin(\tfrac{5}{12}\pi)r]r\}=\pi2
In a previous section this theorem was stated:
q[u(\sqrt{u4-u2+1}-u2+1)]3=q[u(\sqrt{u4-u2+1}+u2-1)] |
From this theorem for cubing, the following equation results for
u=1/\sqrt{2}
ql[\sin(\tfrac{5}{12}\pi)r]3=ql[\sin(\tfrac{1}{12}\pi)r]
The solution to the system of equations with two unknowns then reads as follows:
ql[\sin(\tfrac{1}{12}\pi)r]=\exp(-\sqrt{3}\pi)
ql[\sin(\tfrac{5}{12}\pi)r]=\exp(-\tfrac{1}{3}\sqrt{3}\pi)
Example 2: A further case with the cube formula
The formula of the tangential counterparts is used again:
lnl\langleql\{\tanl[\tfrac{1}{8}\pi-\tfrac{1}{4}\arctan(x)r]r\}r\ranglelnl\langleql\{\tanl[\tfrac{1}{8}\pi+\tfrac{1}{4}\arctan(x)r]r\}r\rangle=2\pi2 |
For
x=\sqrt{8}
lnl\{ql[(2-\sqrt{3})(\sqrt{3}-\sqrt{2})r]r\}lnl\{ql[(2-\sqrt{3})(\sqrt{3}+\sqrt{2})r]r\}=2\pi2
q[u(\sqrt{u4-u2+1}-u2+1)]3=q[u(\sqrt{u4-u2+1}+u2-1)] |
From the previously mentioned theorem for cubing, the following equation results for
u=(\sqrt{3}-1)/\sqrt{2}
ql[(2-\sqrt{3})(\sqrt{3}+\sqrt{2})r]3=ql[(2-\sqrt{3})(\sqrt{3}-\sqrt{2})r]
The solution to the system of equations with two unknowns then reads as follows:
ql[(2-\sqrt{3})(\sqrt{3}-\sqrt{2})r]=\exp(-\sqrt{6}\pi)
ql[(2-\sqrt{3})(\sqrt{3}+\sqrt{2})r]=\exp(-\tfrac{1}{3}\sqrt{6}\pi)
With the incomplete elliptic integrals of the first kind, the values of the elliptic noun function can be derived directly.
With two accurate examples, these direct derivations are to be carried out in the following:
First example:
2+1}r);
(\sqrt{6}+\sqrt{2})r]=\sqrt{3}Fl[2\arctan(x);
(\sqrt{6}-\sqrt{2})r] The correctness of this formula can be proved by computing the differential quotient after the variable x Using the value x=1
(\sqrt{6}+\sqrt{2})r]=\sqrt{3}Kl[
(\sqrt{6}-\sqrt{2})r] The following two results emerge:
(\sqrt{6}-\sqrt{2})r]=\expl\{-\piKl[
(\sqrt{6}+\sqrt{2})r] ÷ Kl[
(\sqrt{6}-\sqrt{2})r]r\}=\exp(-\sqrt{3}\pi)
(\sqrt{6}+\sqrt{2})r]=\expl\{-\piKl[
(\sqrt{6}-\sqrt{2})r] ÷ Kl[
(\sqrt{6}+\sqrt{2})r]r\}=\expl(-
\sqrt{3}\pir) |
Second example:
-1/2x3+\sqrt{5}x}{\sqrt{5}x4+2\sqrt{5}\Phi-1/2x2+1}r);
\sqrt{2}l(\Phi-1/2+\Phi-1r)r]=\sqrt{5}Fl[2\arctan(x);
\sqrt{2}l(\Phi-1/2-\Phi-1r)r] The correctness of this formula can be proved by differentiating both sides of the equation balance.
\sqrt{2}l(\Phi-1/2+\Phi-1r)r]=\sqrt{5}Kl[
\sqrt{2}l(\Phi-1/2-\Phi-1r)r] The following two results emerge:
\sqrt{2}l(\Phi-1/2-\Phi-1r)r]=\expl\{-\piKl[
\sqrt{2}l(\Phi-1/2+\Phi-1r)r] ÷ Kl[
\sqrt{2}l(\Phi-1/2-\Phi-1r)r]r\}=\exp(-\sqrt{5}\pi)
\sqrt{2}l(\Phi-1/2+\Phi-1r)r]=\expl\{-\piKl[
\sqrt{2}l(\Phi-1/2-\Phi-1r)r] ÷ Kl[
\sqrt{2}l(\Phi-1/2+\Phi-1r)r]r\}=\expl(-
\sqrt{5}\pir) |
Third example:
5+7x3+\sqrt{7}x}{\sqrt{7}x6+7x4+2\sqrt{7}x2+1}r);
(3\sqrt{2}+\sqrt{14})r]=\sqrt{7}Fl[2\arctan(x);
(3\sqrt{2}-\sqrt{14})r] The correctness of this formula can be proved by differentiating both sides of the equation balance. Using the value x=1
(3\sqrt{2}+\sqrt{14})r]=\sqrt{7}Kl[
(3\sqrt{2}-\sqrt{14})r] The following two results emerge:
(3\sqrt{2}-\sqrt{14})r]=\expl\{-\piKl[
(3\sqrt{2}+\sqrt{14})r] ÷ Kl[
(3\sqrt{2}-\sqrt{14})r]r\}=\exp(-\sqrt{7}\pi)
(3\sqrt{2}+\sqrt{14})r]=\expl\{-\piKl[
(3\sqrt{2}-\sqrt{14})r] ÷ Kl[
(3\sqrt{2}+\sqrt{14})r]r\}=\expl(-
\sqrt{7}\pir) |
The first derivative of the principal theta function among the Jacobi theta functions can be derived in the following way using the chain rule and the derivation formula of the elliptic nome:
\pi2 | q(\varepsilon)l\{ | |
2\varepsilon(1-\varepsilon2)K(\varepsilon)2 |
d | |
dq(\varepsilon) |
\vartheta00l[q(\varepsilon)r]r\}=l[
d | q(\varepsilon)r]l\{ | |
d\varepsilon |
d | |
dq(\varepsilon) |
\vartheta00l[q(\varepsilon)r]r\}=
d | |
d\varepsilon |
\vartheta00l[q(\varepsilon)r]=
d | |
d\varepsilon |
\sqrt{2\pi-1K(\varepsilon)}=
=
1 | |
2 |
\sqrt{2}\pi-1/2K(\varepsilon)-1/2l[
d | |
d\varepsilon |
K(\varepsilon)r]=
1 | |
2 |
\sqrt{2}\pi-1/2K(\varepsilon)-1/2
E(\varepsilon)-(1-\varepsilon2)K(\varepsilon) | |
\varepsilon(1-\varepsilon2) |
For the now mentioned derivation part this identity is the fundament:
\vartheta00[q(\varepsilon)]=\sqrt{2\pi-1K(\varepsilon)}
Therefore, this equation results:
d | |
dq(\varepsilon) |
\vartheta00l[q(\varepsilon)r]=\sqrt{2}\pi-5/2q(\varepsilon)-1K(\varepsilon)3/2l[E(\varepsilon)-(1-\varepsilon2)K(\varepsilon)r]
The complete elliptic integrals of the second kind have that identity:
(1+\sqrt{1-
| ||||
\varepsilon |
Along with this modular identity, following formula transformation can be made:
d | |
dq(\varepsilon) |
\vartheta00l[q(\varepsilon)r]=\sqrt{2}\pi-5/2q(\varepsilon)-1K(\varepsilon)3/2(1+\sqrt{1-
| ||||
\varepsilon |
Furthermore, this identity is valid:
\vartheta01[q(\varepsilon)]=\sqrt[4]{1-\varepsilon2}\sqrt{2\pi-1K(\varepsilon)}
By using the theta function expressions 00(x) and 01(x) following representation is possible:
d | |
dq(\varepsilon) |
\vartheta00l[q(\varepsilon)r]=
1 | |
2\pi |
q(\varepsilon)-1\vartheta00[q(\varepsilon)]l\{\vartheta00[q(\varepsilon)]2+\vartheta01[q(\varepsilon)]2r\}l\langleEl\{
\vartheta00[q(\varepsilon)]2-\vartheta01[q(\varepsilon)]2 | |
\vartheta00[q(\varepsilon)]2+\vartheta01[q(\varepsilon)]2 |
r\}-
\pi | |
2 |
\vartheta01l[q(\varepsilon)r]2r\rangle
This is the final result:
d | |
dx |
\vartheta00(x)=\vartheta00(x)l[\vartheta00
2+\vartheta | |
(x) | |
01 |
| |||||
(x) | El[ |
| |||||||||||||
|
r]-
\vartheta01(x)2 | |
4x |
r\}
In a similar way following other first derivatives of theta functions and their combinations can also be derived:
d | |
dx |
\vartheta01(x)=\vartheta01(x)l[\vartheta00
2+\vartheta | |
(x) | |
01 |
| |||||
(x) | El[ |
| |||||||||||||
|
r]-
\vartheta00(x)2 | |
4x |
r\}
d | |
dx |
\vartheta10(x)=
1 | |
2\pix |
\vartheta10(x)\vartheta00(x)2El[
\vartheta10(x)2 | |
\vartheta00(x)2 |
r]
d | |
dx |
\vartheta00(x) | |
\vartheta01(x) |
=
\vartheta00(x)5-\vartheta00(x)\vartheta01(x)4 | |
4x\vartheta01(x) |
d | |
dx |
\vartheta10(x) | |
\vartheta00(x) |
=
\vartheta10(x)\vartheta01(x)4 | |
4x\vartheta00(x) |
d | |
dx |
\vartheta10(x) | |
\vartheta01(x) |
=
\vartheta10(x)\vartheta00(x)4 | |
4x\vartheta01(x) |
Important definition:
\vartheta10(x)=2x1/4+2x1/4
infty | |
\sum | |
n=1 |
x2triangleup(n)
triangleup(n)=\tfrac{1}{2}n(n+1)