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Nome (mathematics)

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.

Definition
The nome function is given by

 * $$q

=\mathrm{e}^{-{\pi K'/K}} =\mathrm{e}^{{\rm{i}} \pi\omega_2/\omega_1} =\mathrm{e}^{{\rm{i}} \pi \tau} \, $$

where $$K$$ and $$iK'$$ are the quarter periods, and $$\omega_1$$ and $$\omega_2$$ are the fundamental pair of periods, and $\tau=\frac{iK'}{K}=\frac{\omega_2}{\omega_1}$ is the half-period ratio. The nome can be taken to be a function of any one of these quantities; conversely, any one of these quantities can be taken as functions of the nome. Each of them uniquely determines the others when $$0<q<1$$. That is, when $$0<q<1$$, the mappings between these various symbols are both 1-to-1 and onto, and so can be inverted: the quarter periods, the half-periods and the half-period ratio can be explicitly written as functions of the nome. For general $$q\in\mathbb{C}$$ with $$0<|q|<1$$, $$\tau$$ is not a single-valued function of $$q$$. Explicit expressions for the quarter periods, in terms of the nome, are given in the linked article.

Notationally, the quarter periods $$K$$ and $$iK'$$ are usually used only in the context of the Jacobian elliptic functions, whereas the half-periods $$\omega_1$$ and $$\omega_2$$ are usually used only in the context of Weierstrass elliptic functions. Some authors, notably Apostol, use $$\omega_1$$ and $$\omega_2$$ to denote whole periods rather than half-periods.

The nome is frequently used as a value with which elliptic functions and modular forms can be described; on the other hand, it can also be thought of as function, because the quarter periods are functions of the elliptic modulus $$k$$: $$q(k) = \mathrm{e}^{-\pi K'(k)/K(k)}$$.

The complementary nome $$q_1$$ is given by
 * $$q_1(k) = \mathrm{e}^{-\pi K(k)/K'(k)}. \, $$

Sometimes the notation $$q=\mathrm{e}^{{2\rm{i}} \pi \tau}$$ is used for the square of the nome.

The mentioned functions $$K$$ and $$K'$$ are called complete elliptic integrals of the first kind. They are defined as follows:

 * $$K(x) = \int_0^{\pi/2} \frac{1}{\sqrt{1 - x^2\sin(\varphi)^2}} \,\mathrm{d}\varphi = \int_0^1 \frac{2}{\sqrt{(y^2+1)^2 - 4x^2y^2}} \mathrm{d}y $$

 * $$K'(x) = K(\sqrt{1 - x^2}) = \int_0^{\pi/2} \frac{1}{\sqrt{1 - (1 - x^2)\sin(\varphi)^2}} \,\mathrm{d}\varphi $$

Applications
The nome solves the following equation:
 * $$|k| = \frac{\vartheta_{10}^2[0,q(k)]}{\vartheta_{00}^2[0,q(k)]}\rightarrow q(k) = \mathrm{e}^{-\pi K'(k)/K(k)}$$

This analogon is valid for the Pythagorean complementary modulus:
 * $$k' = \sqrt{1 - k^2} = \frac{\vartheta_{01}^2[0,q(k)]}{\vartheta_{00}^2[0,q(k)]}\rightarrow q(k) = \mathrm{e}^{-\pi K'(k)/K(k)}$$

where $$\vartheta_{10},\theta_{00}$$ are the complete Jacobi theta functions and $$K(k)$$ is the complete elliptic integral of the first kind with modulus $$k$$ shown in the formula above. For the complete theta functions these definitions introduced by Sir Edmund Taylor Whittaker and George Neville Watson are valid:

 * $$\vartheta_{00}(v;w) = \prod_{n = 1}^\infty (1-w^{2n})[1+2\cos(2v)w^{2n-1}+w^{4n-2}]$$
 * $$\vartheta_{01}(v;w) = \prod_{n = 1}^\infty (1-w^{2n})[1-2\cos(2v)w^{2n-1}+w^{4n-2}]$$
 * $$\vartheta_{10}(v;w) = 2 w^{1/4}\cos(v)\prod_{n = 1}^\infty (1-w^{2n})[1+2\cos(2v)w^{2n}+w^{4n}]$$

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 $$\tau$$ is commonly used as a coordinate on the complex upper half-plane, typically endowed with the Poincaré metric to obtain the Poincaré half-plane model. The nome then serves as a coordinate on a punctured disk of unit radius; it is punctured because $$q=0$$ is not part of the disk (or rather, $$q=0$$ corresponds to $$\tau \to \infty$$). This endows the punctured disk with the Poincaré metric.

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 $$\tau$$ (or of $$q$$, or of $$K$$ and $$iK'$$ etc.) that uniquely determine a tiling of the plane by parallelograms. The tiling is referred to as the modular symmetry given by the modular group. Some functions that are periodic on the upper half-plane are called to as modular functions; the nome, the half-periods, the quarter-periods or the half-period ratio all provide different parameterizations for these periodic functions.

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 $$q$$. The series expansion in terms of the nome or the square of the nome (the q-expansion) is famously connected to the Fisher-Griess monster by means of monstrous moonshine.

Euler's function arises as the prototype for q-series in general.

The nome, as the $$q$$ of q-series then arises in the theory of affine Lie algebras, essentially because (to put it poetically, but not factually) those algebras describe the symmetries and isometries of Riemann surfaces.

Curve sketching
Every real value $$x$$ of the interval $$[-1,1]$$ is assigned to a real number between inclusive zero and inclusive one in the nome function $$q(x)$$. The elliptic nome function is axial symmetric to the ordinate axis. Thus: $$q(x) = q(-x)$$. The functional curve of the nome passes through the origin of coordinates with the slope zero and curvature plus one eighth. For the real valued interval $$(-1,1)$$ the nome function $$q(x)$$ is strictly left-curved.

Derivatives
The Legendre's relation is defined that way:

 * $$ K\,E' + E\,K' - K\,K' = \tfrac{1}{2}\pi $$

And as described above, the elliptic nome function $$q(x)$$ has this original definition:

 * $$ q(x) = \exp\left[-\pi \,\frac{K(\sqrt{1 - x^2})}{K(x)}\right] $$

Furthermore, these are the derivatives of the two complete elliptic integrals:

 * $$ \frac{\mathrm{d}}{\mathrm{d}x} K(x) = \frac{1}{x(1 - x^2)}\bigl[E(x) - (1 - x^2)K(x)\bigr] $$
 * $$ \frac{\mathrm{d}}{\mathrm{d}x} E(x) = -\frac{1}{x}\bigl[K(x) - E(x)\bigr] $$

Therefore, the derivative of the nome function has the following expression:

 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,q(x) = \frac{\pi^2}{2x(1-x^2)K(x)^2} \,q(x) $$

The second derivative can be expressed this way:

 * $$\frac{\mathrm{d}^2}{\mathrm{d}x^2} \,q(x) = \frac{\pi^4 + 2\pi^2 (1+x^2)K(x)^2 - 4\pi^2 K(x)E(x)}{4x^2(1-x^2)^2 K(x)^4} \,q(x) $$

And that is the third derivative:

 * $$\frac{\mathrm{d}^3}{\mathrm{d}x^3} \,q(x) = \frac{\pi^6 + 6\pi^4 (1+x^2)K(x)^2 - 12\pi^4 K(x)E(x) + 8\pi^2 (1+x^2)^2 K(x)^4 - 24\pi^2 (1+x^2)K(x)^3 E(x) + 24\pi^2 K(x)^2 E(x)^2}{8x^3(1-x^2)^3 K(x)^6} \,q(x) $$

The complete elliptic integral of the second kind is defined as follows:

 * $$E(x) = \int_0^{\pi/2} \sqrt{1-x^2\sin(\varphi)^2} \,\mathrm{d}\varphi = 2\int_0^1 \frac{\sqrt{(y^2+1)^2 - 4x^2y^2}}{(y^2+1)^2} \,\mathrm{d}y $$

The following equation follows from these equations by eliminating the complete elliptic integral of the second kind:

 * $$3\biggl[\frac{\mathrm{d}^2}{\mathrm{d}x^2} q(x)\biggr]^2 - 2\biggl[\frac{\mathrm{d}}{\mathrm{d}x} q(x)\biggr]\biggl[\frac{\mathrm{d}^3}{\mathrm{d}x^3} q(x)\biggr] = \frac{\pi^8 - 4\pi^4 (1+x^2)^2 K(x)^4}{16x^4(1-x^2)^4 K(x)^8} q(x)^2 $$

Thus, the following third-order quartic differential equation is valid:
 * $$x^2 (1-x^2)^2 [2q(x)^2 q'(x)q'(x) - 3q(x)^2 q(x)^2 + q'(x)^4] = (1+x^2)^2 q(x)^2 q'(x)^2 $$

Kneser sequence
Given is the derivative of the Elliptic Nome mentioned above:

 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,q(x) = \frac{\pi^2}{2x(1-x^2)K(x)^2} \,q(x) $$

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:

 * {|class = "wikitable"

$$ \text{Kn}(2n) = 2^{4n-3} \binom{4n}{2n} + \sum_{m = 1}^{n} 4^{2n-2m} \binom{4n}{2n-2m} \text{Kn}(m) $$ $$ \text{Kn}(2n+1) = 2^{4n-1} \binom{4n+2}{2n+1} + \sum_{m = 1}^{n} 4^{2n-2m+1} \binom{4n+2}{2n-2m+1} \text{Kn}(m) $$
 * style="background-color: Yellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: GreenYellow;" |
 * }

Executed examples:

 * {|class = "wikitable"

$$ \text{Kn}(2) = 2 \times 6 + 1 \times {\color{cornflowerblue}1} = {\color{cornflowerblue}13} $$ $$ \text{Kn}(3) = 8 \times 20 + 24 \times {\color{cornflowerblue}1} = {\color{cornflowerblue}184} $$ $$ \text{Kn}(4) = 32 \times 70 + 448 \times {\color{cornflowerblue}1} + 1 \times {\color{cornflowerblue}13} = {\color{cornflowerblue}2701} $$ $$ \text{Kn}(5) = 128 \times 252 + 7680 \times {\color{cornflowerblue}1} + 40 \times {\color{cornflowerblue}13} = {\color{cornflowerblue}40456} $$ $$ \text{Kn}(6) = 512 \times 924 + 126720 \times {\color{cornflowerblue}1} + 1056 \times {\color{cornflowerblue}13} + 1 \times {\color{cornflowerblue}184} = {\color{cornflowerblue}613720} $$ $$ \text{Kn}(7) = 2048 \times 3432 + 2050048 \times {\color{cornflowerblue}1} + 23296 \times {\color{cornflowerblue}13} + 56 \times {\color{cornflowerblue}184} = {\color{cornflowerblue}9391936} $$
 * style="background-color: Yellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: Yellow;" |
 * style="background-color: Yellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: Yellow;" |
 * style="background-color: Yellow;" |
 * style="background-color: GreenYellow;" |
 * style="background-color: GreenYellow;" |
 * }

The Kneser sequence appears in the Taylor series of the period ratio (half period ratio):

 * $$\frac{1}{4}\ln\bigl(\frac{16}{x^2}\bigr) - \frac{\pi \,K'(x)}{4 \,K(x)} = \sum_{n = 1}^{\infty} \frac{\text{Kn}(n)}{2^{4n - 1}n}\,x^{2n} $$

 * $${\color{limegreen}\frac{1}{4}\ln\bigl(\frac{16}{x^2}\bigr) - \frac{\pi \,K'(x)}{4 \,K(x)} = \frac{8}x^2 + \frac{256}x^4 + \frac{6144}x^6 + \frac{131072}x^8 + \frac{2621440}x^{10} + \ldots} $$

The derivative of this equation after $$ x $$ leads to this equation that shows the generating function of the Kneser number sequence:

 * $$\frac{\pi^2}{8x(1 - x^2)K(x)^2} - \frac{1}{2x} = \sum_{n = 1}^{\infty} \frac{\text{Kn}(n)}{2^{4n - 2}}x^{2n - 1} $$

 * $${\color{limegreen}\frac{\pi^2}{8x(1 - x^2)K(x)^2} - \frac{1}{2x} = \frac{4}x + \frac{64}x^3 + \frac{1024}x^5 + \frac{16384}x^7 + \frac{262144}x^9 + \ldots} $$

This result appears because of the Legendre's relation $$ K\,E' + E\,K' - K\,K' = \tfrac{1}{2}\pi $$ in the numerator.

Schellbach Schwarz sequence
The mathematician Karl Heinrich Schellbach 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:

 * $$\text{Sc}(n+1) = \frac{2}{n}\sum _{m = 1}^{n} \text{Sc}(m)\,\text{Kn}(n + 1 - m) $$

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:

And this sequence creates the MacLaurin series of the elliptic nome  in exactly this way:

 * $$q(x) = \sum_{n = 1}^{\infty} \frac{\text{Sc}(n)}{2^{4n - 3}} \biggl(\frac{1 - \sqrt[4]{1 - x^2}}{1 + \sqrt[4]{1 - x^2}}\biggr)^{4n - 3} = x^2\biggl\{\frac{1}{2} + \biggl[\sum_{n = 1}^{\infty} \frac{\text{Sc}(n + 1)}{2^{4n + 1}} x^{2n}\biggr]\biggr\}^4$$

 * $$q(x) = x^2\bigl({\color{limegreen}\frac{2} + \frac{32}x^2 + \frac{512}x^4 + \frac{8192}x^6 + \frac{131072}x^8 + \ldots}\bigr)^4$$

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:

 * $$\mathrm{Sc}(4) = \frac{2}{3}\sum _{m = 1}^{3} \mathrm{Sc}(m) \,\mathrm{Kn}(4 - m) = \frac{2}{3} \bigl[{\color{navy}\mathrm{Sc}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{navy}\mathrm{Sc}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{navy}\mathrm{Sc}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{navy}\mathrm{Sc}(4)} = \frac{2}{3} \bigl({\color{navy}1} \times {\color{cornflowerblue}184} + {\color{navy}2} \times {\color{cornflowerblue}13} + {\color{navy}15} \times {\color{cornflowerblue}1} \bigr) = {\color{navy}150}$$

 * $$\mathrm{Sc}(5) = \frac{2}{4}\sum _{m = 1}^{4} \mathrm{Sc}(m) \,\mathrm{Kn}(5 - m) = \frac{2}{4} \bigl[{\color{navy}\mathrm{Sc}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(4)} + {\color{navy}\mathrm{Sc}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{navy}\mathrm{Sc}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{navy}\mathrm{Sc}(4)}\,{\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{navy}\mathrm{Sc}(5)} = \frac{2}{4} \bigl({\color{navy}1} \times {\color{cornflowerblue}2701} + {\color{navy}2} \times {\color{cornflowerblue}184} + {\color{navy}15} \times {\color{cornflowerblue}13} + {\color{navy}150} \times {\color{cornflowerblue}1} \bigr) = {\color{navy}1707}$$

 * $$\mathrm{Sc}(6) = \frac{2}{5}\sum _{m = 1}^{5} \mathrm{Sc}(m) \,\mathrm{Kn}(6 - m) = \frac{2}{5} \bigl[{\color{navy}\mathrm{Sc}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(5)} + {\color{navy}\mathrm{Sc}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(4)} + {\color{navy}\mathrm{Sc}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{navy}\mathrm{Sc}(4)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{navy}\mathrm{Sc}(5)}\, {\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{navy}\mathrm{Sc}(6)} = \frac{2}{5} \bigl({\color{navy}1} \times {\color{cornflowerblue}40456} + {\color{navy}2} \times {\color{cornflowerblue}2701} + {\color{navy}15} \times {\color{cornflowerblue}184} + {\color{navy}150} \times {\color{cornflowerblue}13} + {\color{navy}1707} \times {\color{cornflowerblue}1} \bigr) = {\color{navy}20910}$$

Kotěšovec sequence
The MacLaurin series of the nome function $$q(x)$$ has even exponents and positive coefficients at all positions:

 * $$ q(x) = \sum_{n = 1}^{\infty} \frac{\operatorname{Kt}(n)}{16^n}\,x^{2n} $$

And the sum with the same absolute values of the coefficients but with alternating signs generates this function:

 * $$q\bigl[x(x^2+1)^{-1/2}\bigr] = \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}\operatorname{Kt}(n)}{16^n}\,x^{2n} $$

The radius of convergence of this Maclaurin series is 1. Here $$\operatorname{Kt}(n)$$ (OEIS A005797) is a sequence of exclusively natural numbers $$\operatorname{Kt}(n) \isin \mathbb{N}$$ for all natural numbers $$n \isin \mathbb{N}$$ and this integer number sequence is not elementary. This sequence of numbers $$\operatorname{Kt}(n)$$ was researched by the Czech mathematician and fairy chess composer Václav Kotěšovec, born in 1956. Two ways of constructing this integer sequence shall be shown in the next section.

Construction method with Kneser numbers
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 $$ \frac{2}{n} $$ anymore, but $$ \frac{8}{n} $$ instead of that:

 * $$\text{Kt}(n+1) = \frac{8}{n}\sum _{m = 1}^{n} \text{Kt}(m)\,\text{Kn}(n + 1 - m) $$

Following table contains the Schellbach Schwarz numbers and the Kneser numbers and the Apéry numbers:

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:

 * $$\mathrm{Kt}(4) = \frac{8}{3}\sum _{m = 1}^{3} \mathrm{Kt}(m) \,\mathrm{Kn}(4 - m) = \frac{8}{3} \bigl[{\color{ForestGreen}\mathrm{Kt}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{ForestGreen}\mathrm{Kt}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{ForestGreen}\mathrm{Kt}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{ForestGreen}\mathrm{Kt}(4)} = \frac{8}{3} \bigl({\color{ForestGreen}1} \times {\color{cornflowerblue}184} + {\color{ForestGreen}8} \times {\color{cornflowerblue}13} + {\color{ForestGreen}84} \times {\color{cornflowerblue}1} \bigr) = {\color{ForestGreen}992}$$

 * $$\mathrm{Kt}(5) = \frac{8}{4}\sum _{m = 1}^{4} \mathrm{Kt}(m) \,\mathrm{Kn}(5 - m) = \frac{8}{4} \bigl[{\color{ForestGreen}\mathrm{Kt}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(4)} + {\color{ForestGreen}\mathrm{Kt}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{ForestGreen}\mathrm{Kt}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{ForestGreen}\mathrm{Kt}(4)}\,{\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{ForestGreen}\mathrm{Kt}(5)} = \frac{8}{4} \bigl({\color{ForestGreen}1} \times {\color{cornflowerblue}2701} + {\color{ForestGreen}8} \times {\color{cornflowerblue}184} + {\color{ForestGreen}84} \times {\color{cornflowerblue}13} + {\color{ForestGreen}992} \times {\color{cornflowerblue}1} \bigr) = {\color{ForestGreen}12514}$$

 * $$\mathrm{Kt}(6) = \frac{8}{5}\sum _{m = 1}^{5} \mathrm{Kt}(m) \,\mathrm{Kn}(6 - m) = \frac{8}{5} \bigl[{\color{ForestGreen}\mathrm{Kt}(1)}\,{\color{cornflowerblue}\mathrm{Kn}(5)} + {\color{ForestGreen}\mathrm{Kt}(2)}\,{\color{cornflowerblue}\mathrm{Kn}(4)} + {\color{ForestGreen}\mathrm{Kt}(3)}\,{\color{cornflowerblue}\mathrm{Kn}(3)} + {\color{ForestGreen}\mathrm{Kt}(4)}\,{\color{cornflowerblue}\mathrm{Kn}(2)} + {\color{ForestGreen}\mathrm{Kt}(5)}\, {\color{cornflowerblue}\mathrm{Kn}(1)} \bigr] $$

 * $${\color{ForestGreen}\mathrm{Kt}(6)} = \frac{8}{5} \bigl({\color{ForestGreen}1} \times {\color{cornflowerblue}40456} + {\color{ForestGreen}8} \times {\color{cornflowerblue}2701} + {\color{ForestGreen}84} \times {\color{cornflowerblue}184} + {\color{ForestGreen}992} \times {\color{cornflowerblue}13} + {\color{ForestGreen}12514} \times {\color{cornflowerblue}1} \bigr) = {\color{ForestGreen}164688}$$

So the MacLaurin series of the direct Elliptic Nome can be generated:

 * $$q(x) = \sum_{n = 1}^{\infty} \frac{\text{Kt}(n)}{16^{n}} \,x^{2n}$$

 * $$q(x) = {\color{limegreen}\frac{16}x^2 + \frac{256}x^4 + \frac{4096}x^6 + \frac{65536}x^8 + \frac{1048576}x^{10} + \ldots}$$

Construction method with Apéry numbers
By adding a further integer number sequence $$\operatorname{Ap}(n)$$ that denotes a specially modified Apéry sequence (OEIS A036917), the sequence of the Kotěšovec numbers $$\operatorname{Kt}(n)$$ can be generated. The starting value of the sequence $$\operatorname{Kt}(n)$$ is the value $$\operatorname{Kt}(1)=1$$ and the following values of this sequence are generated with those two formulas that are valid for all numbers $$n \isin \mathbb{N}$$:

 * $$ \operatorname{Kt}(n+1) = \frac{1}{n} \sum_{m = 1}^n m\operatorname{Kt}(m)[16\operatorname{Ap}(n+1-m) - \operatorname{Ap}(n+2-m)] $$

 * $$ \operatorname{Ap}(n) = \sum_{a = 0}^{n-1} \binom{2a}{a}^2 \binom{2n-2-2a}{n-1-a}^2 $$

This formula creates the Kotěšovec sequence too, but it only creates the sequence numbers of even indices:

 * $$ \operatorname{Kt}(2n) = \frac{1}{2} \sum_{m = 1}^{2n-1} (-1)^{2n - m + 1}16^{2n - m}\binom{2n-1}{m - 1} \operatorname{Kt}(m) $$

The Apéry sequence $$\operatorname{Ap}(n)$$ was researched especially by the mathematicians Sun Zhi-Hong and Reinhard Zumkeller. And that sequence generates the square of the complete elliptic integral of the first kind:

 * $$ 4\pi^{-2}K(x)^2 = 1 + \sum_{n = 1}^\infty \frac{\operatorname{Ap}(n + 1)x^{2n}}{16^n} $$

The first numerical values of the central binomial coefficients and the two numerical sequences described are listed in the following table:

Václav Kotěšovec wrote down the number sequence $$\operatorname{Kt}(n)$$ on the Online Encyclopedia of Integer Sequences up to the seven hundredth sequence number.

Here one example of the Kotěšovec sequence is computed:

 * {| class="wikitable"

$$ {\color{blue}1} \times {\color{blue}63504} + {\color{blue}4} \times {\color{blue}4900} + {\color{blue}36} \times {\color{blue}400} + {\color{blue}400} \times {\color{blue}36} + {\color{blue}4900} \times {\color{blue}4} + {\color{blue}63504} \times {\color{blue}1} = {\color{RoyalBlue}195008} $$

$$ \tfrac{1}{5} \times {\color{ForestGreen}1} \times (16 \times {\color{RoyalBlue}14296} - {\color{RoyalBlue}195008}) + \tfrac{2}{5} \times {\color{ForestGreen}8} \times (16 \times {\color{RoyalBlue}1088} - {\color{RoyalBlue}14296}) + \tfrac{3}{5} \times {\color{ForestGreen}84} \times (16 \times {\color{RoyalBlue}88} - {\color{RoyalBlue}1088}) + {} $$

$$ {} + \tfrac{4}{5} \times {\color{ForestGreen}992} \times (16 \times {\color{RoyalBlue}8} - {\color{RoyalBlue}88}) + \tfrac{5}{5} \times {\color{ForestGreen}12514} \times (16 \times {\color{RoyalBlue}1} - {\color{RoyalBlue}8}) = {\color{ForestGreen}164688} $$
 * }

Function values
The two following lists contain many function values of the nome function:

The first list shows pairs of values with mutually Pythagorean complementary modules:

 * $$q(\tfrac{1}{2}\sqrt{2}) = \exp(-\pi)$$
 * $$q[\tfrac{1}{4}(\sqrt{6} - \sqrt{2})] = \exp(-\sqrt{3}\,\pi)$$
 * $$q[\tfrac{1}{4}(\sqrt{6} + \sqrt{2})] = \exp(-\tfrac{1}{3}\sqrt{3}\,\pi)$$
 * $$q\bigl\{\sin\bigl[\tfrac{1}{2}\arcsin(\sqrt{5} - 2)\bigr]\bigr\} = \exp(-\sqrt{5}\,\pi)$$
 * $$q\bigl\{\cos\bigl[\tfrac{1}{2}\arcsin(\sqrt{5} - 2)\bigr]\bigr\} = \exp(-\tfrac{1}{5}\sqrt{5}\,\pi)$$
 * $$q[\tfrac{1}{8}(3\sqrt{2} - \sqrt{14})] = \exp(-\sqrt{7}\,\pi)$$
 * $$q[\tfrac{1}{8}(3\sqrt{2} + \sqrt{14})] = \exp(-\tfrac{1}{7}\sqrt{7}\,\pi)$$
 * $$q[\tfrac{1}{2}(\sqrt{3} - 1)(\sqrt{2} - \sqrt[4]{3})] = \exp(-3\pi)$$
 * $$q[\tfrac{1}{2}(\sqrt{3} - 1)(\sqrt{2} + \sqrt[4]{3})] = \exp(-\tfrac{1}{3}\pi)$$
 * $$q\bigl[\tfrac{1}{16}\bigl(\sqrt{22} + 3\sqrt{2}\bigr)\bigl(\tfrac{1}{3}\sqrt[3]{6\sqrt{3} + 2\sqrt{11}} - \tfrac{1}{3}\sqrt[3]{6\sqrt{3} - 2\sqrt{11}} + \tfrac{1}{3}\sqrt{11} - 1\bigr)^4\bigr] = \exp(-\sqrt{11}\,\pi)$$
 * $$q\bigl[\tfrac{1}{16}\bigl(\sqrt{22} - 3\sqrt{2}\bigr)\bigl(\tfrac{1}{3}\sqrt[3]{6\sqrt{3} + 2\sqrt{11}} - \tfrac{1}{3}\sqrt[3]{6\sqrt{3} - 2\sqrt{11}} + \tfrac{1}{3}\sqrt{11} + 1\bigr)^4\bigr] = \exp(-\tfrac{1}{11}\sqrt{11}\,\pi)$$
 * $$q\bigl\{\sin\bigl[\tfrac{1}{2}\arcsin(5\sqrt{13} - 18)\bigr]\bigr\} = \exp(-\sqrt{13}\,\pi)$$
 * $$q\bigl\{\cos\bigl[\tfrac{1}{2}\arcsin(5\sqrt{13} - 18)\bigr]\bigr\} = \exp(-\tfrac{1}{13}\sqrt{13}\,\pi)$$

The second list shows pairs of values with mutually tangentially complementary modules:

 * $$q(\sqrt{2} - 1) = \exp(-\sqrt{2}\,\pi)$$
 * $$q[(2 - \sqrt{3})(\sqrt{3} - \sqrt{2})] = \exp(-\sqrt{6}\,\pi)$$
 * $$q[(2 - \sqrt{3})(\sqrt{3} + \sqrt{2})] = \exp(-\tfrac{1}{3}\sqrt{6}\,\pi)$$
 * $$q[(\sqrt{10} - 3)(\sqrt{2} - 1)^2] = \exp(-\sqrt{10}\,\pi)$$
 * $$q[(\sqrt{10} - 3)(\sqrt{2} + 1)^2] = \exp(-\tfrac{1}{5}\sqrt{10}\,\pi)$$
 * $$q\bigl[\tfrac{1}{16}\sqrt{2\sqrt{2} - \sqrt{7}}\,(3\sqrt{2} - \sqrt{14})(\sqrt{2\sqrt{2} + 1} - 1)^4\bigr] = \exp(-\sqrt{14}\,\pi)$$
 * $$q\bigl[\tfrac{1}{16}\sqrt{2\sqrt{2} + \sqrt{7}}\,(3\sqrt{2} + \sqrt{14})(\sqrt{2\sqrt{2} + 1} - 1)^4\bigr] = \exp(-\tfrac{1}{7}\sqrt{14}\,\pi)$$
 * $$q[(2 - \sqrt{3})^2 (\sqrt{2} - 1)^3] = \exp(-3\sqrt{2}\,\pi)$$
 * $$q[(2 + \sqrt{3})^2 (\sqrt{2} - 1)^3] = \exp(-\tfrac{1}{3}\sqrt{2}\,\pi)$$
 * $$q[(10 - 3\sqrt{11})(3\sqrt{11} - 7\sqrt{2})] = \exp(-\sqrt{22}\,\pi)$$
 * $$q[(10 - 3\sqrt{11})(3\sqrt{11} + 7\sqrt{2})] = \exp(-\tfrac{1}{11}\sqrt{22}\,\pi)$$
 * $$q\bigl\{(\sqrt{26}+5)(\sqrt{2}-1)^2 \tan\bigl[\tfrac{1}{4}\pi-\arctan(\tfrac{1}{3}\sqrt[3]{3\sqrt{3}+\sqrt{26}}-\tfrac{1}{3}\sqrt[3]{3\sqrt{3}-\sqrt{26}}+\tfrac{1}{6}\sqrt{26}-\tfrac{1}{2}\sqrt{2})\bigr]^4\bigr\} = \exp(-\sqrt{26}\,\pi)$$
 * $$q\bigl\{(\sqrt{26}+5)(\sqrt{2}+1)^2 \tan\bigl[\arctan(\tfrac{1}{3}\sqrt[3]{3\sqrt{3}+\sqrt{26}}-\tfrac{1}{3}\sqrt[3]{3\sqrt{3}-\sqrt{26}}+\tfrac{1}{6}\sqrt{26}+\tfrac{1}{2}\sqrt{2})-\tfrac{1}{4}\pi\bigr]^4\bigr\} = \exp(-\tfrac{1}{13}\sqrt{26}\,\pi)$$

Related quartets of values are shown below:

 * {| class="wikitable"

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10} - 3)^2 (\sqrt{5} - 2)^2]\}\bigr\rangle = \exp(-\sqrt{30}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10} - 3)^2 (\sqrt{5} + 2)^2]\}\bigr\rangle = \exp(-\tfrac{1}{3}\sqrt{30}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10} + 3)^2 (\sqrt{5} - 2)^2]\}\bigr\rangle = \exp(-\tfrac{1}{5}\sqrt{30}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{10} + 3)^2 (\sqrt{5} + 2)^2]\}\bigr\rangle = \exp(-\tfrac{1}{15}\sqrt{30}\,\pi)$$ $$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7} - 3\sqrt{3})^2 (2\sqrt{2} - \sqrt{7})^2]\}\bigr\rangle = \exp(-\sqrt{42}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7} - 3\sqrt{3})^2 (2\sqrt{2} + \sqrt{7})^2]\}\bigr\rangle = \exp(-\tfrac{1}{3}\sqrt{42}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7} + 3\sqrt{3})^2 (2\sqrt{2} - \sqrt{7})^2]\}\bigr\rangle = \exp(-\tfrac{1}{7}\sqrt{42}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(2\sqrt{7} + 3\sqrt{3})^2 (2\sqrt{2} + \sqrt{7})^2]\}\bigr\rangle = \exp(-\tfrac{1}{21}\sqrt{42}\,\pi)$$ $$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5} - 2)^4 (\sqrt{2} - 1)^6]\}\bigr\rangle = \exp(-\sqrt{70}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5} - 2)^4 (\sqrt{2} + 1)^6]\}\bigr\rangle = \exp(-\tfrac{1}{5}\sqrt{70}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5} + 2)^4 (\sqrt{2} - 1)^6]\}\bigr\rangle = \exp(-\tfrac{1}{7}\sqrt{70}\,\pi)$$

$$q\bigl\langle\tan\{\tfrac{1}{2}\arctan[(\sqrt{5} + 2)^4 (\sqrt{2} + 1)^6]\}\bigr\rangle = \exp(-\tfrac{1}{35}\sqrt{70}\,\pi)$$
 * }

Sum series
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 can be described easily in terms of the elliptic nome:

 * $$\sum_{n = 1}^{\infty} q(x)^{\Box(n)} = \tfrac{1}{2}\vartheta_{00}[q(x)] - \tfrac{1}{2} = \tfrac{1}{2}\sqrt{2\pi^{-1}K(x)} - \tfrac{1}{2} = \tfrac{1}{2}\operatorname{agm}(1-x;1+x)^{-1/2} - \tfrac{1}{2} $$
 * $$\sum_{n = 1}^{\infty} q(x)^{\Box(2n-1)} = \tfrac{1}{4}\vartheta_{00}[q(x)] - \tfrac{1}{4}\vartheta_{01}[q(x)] = \tfrac{1}{4}(1-\sqrt[4]{1-x^2})\sqrt{2\pi^{-1}K(x)} $$
 * $$\sum_{n = 1}^{\infty} \frac{2q(x)^{n}}{q(x)^{2n} + 1} = \tfrac{1}{2}\vartheta_{00}[q(x)]^2 - \tfrac{1}{2} = \pi^{-1}K(x) - \tfrac{1}{2} $$
 * $$\sum_{n = 1}^{\infty} \frac{2q(x)^{2n-1}}{q(x)^{4n-2} + 1} = \tfrac{1}{4}\vartheta_{00}[q(x)]^2 - \tfrac{1}{4}\vartheta_{01}[q(x)]^2 = \tfrac{1}{2}(1-\sqrt{1-x^2})\pi^{-1}K(x) $$
 * $$\sum_{n = 1}^{\infty} \Box(n) q(x)^{\Box(n)} = 2^{-1/2}\pi^{-5/2}K(x)^{3/2}[E(x)-(1-x^2)K(x)] $$
 * $$\sum_{n = 1}^{\infty} \biggl[\frac{2q(x)^{n}}{1 + q(x)^{2n}}\biggr]^2 = 2\pi^{-2}E(x)K(x) - \tfrac{1}{2} $$
 * $$\sum_{n = 1}^{\infty} \biggl[\frac{2q(x)^{n}}{1 - q(x)^{2n}}\biggr]^2 = \tfrac{2}{3}\pi^{-2}(2 - x^2)K(x)^2 - 2\pi^{-2}K(x)E(x) + \tfrac{1}{6} $$

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: $$\Box(n)=n^2$$

The letter $$\operatorname{E}(\varepsilon)$$ describes the complete elliptic integral of the second kind, which is the quarter periphery of an ellipse in relation to the bigger half axis of the ellipse with the numerical eccentricity $$\varepsilon$$ as abscissa value.

Product series
The two most important theta functions can be defined by following product series:

 * $$\prod_{n = 1}^{\infty} [1-q(x)^{2n}][1+q(x)^{2n-1}]^2 = \vartheta_{00}[q(x)] = \sqrt{2\pi^{-1}K(x)} $$
 * $$\prod_{n = 1}^{\infty} [1-q(x)^{2n}][1-q(x)^{2n-1}]^2 = \vartheta_{01}[q(x)] = \sqrt[4]{1-x^2}\sqrt{2\pi^{-1}K(x)} $$

Furthermore, these two Pochhammer products have those two relations:

 * $$q(\varepsilon)[q(\varepsilon);q(\varepsilon)]_{\infty}^{24} = 256\,\varepsilon^2 (1 - \varepsilon^2)^4 \pi^{-{12}}K(\varepsilon)^{12} $$
 * $$\varepsilon^2 [q(\varepsilon);q(\varepsilon)^2]_{\infty}^{24} = 16\,(1 - \varepsilon^2)^2 q(\varepsilon) $$

The Pochhammer products have an important role in the pentagonal number theorem and its derivation.

Complete elliptic integrals
The nome function can be used for the definition of the complete elliptic integrals of first and second kind:

 * $$K(\varepsilon) = \tfrac{1}{2}\pi\,\vartheta_{00}[q(\varepsilon)]^2$$
 * $$E(\varepsilon) = 2\pi q(\varepsilon)\,\vartheta_{00}'[q(\varepsilon)]\vartheta_{00}[q(\varepsilon)]^{-3} + \tfrac{1}{2}\pi(1 - \varepsilon^2)\,\vartheta_{00}[q(\varepsilon)]^2$$

In this case the dash in the exponent position stands for the derivative of the so-called theta zero value function:

 * $$\vartheta_{00}'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\,\vartheta_{00}(x) = 2 + \sum_{n = 1}^{\infty} 2(n + 1)^2 x^{n(n+2)}$$

Definitions of Jacobi functions
The elliptic functions Zeta Amplitudinis and Delta Amplitudinis can be defined with the elliptic nome function easily:

 * $$\operatorname{zn}(x;k) = \sum_{n = 1}^{\infty} \frac{2\pi K(k)^{-1}\sin[\pi K(k)^{-1}x]q(k)^{2n-1}}{1-2\cos[\pi K(k)^{-1}x]q(k)^{2n-1}+q(k)^{4n-2}}$$
 * $$\operatorname{dn}(x;k) = \sqrt[4]{1-k^2}\prod_{n = 1}^{\infty} \frac{1+2\cos[\pi K(k)^{-1}x]q(k)^{2n-1}+q(k)^{4n-2}}{1-2\cos[\pi K(k)^{-1}x]q(k)^{2n-1}+q(k)^{4n-2}} $$

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:
 * $$\operatorname{sn}(x;k) = 2\sqrt[4]{k^{-2}q(k)}\,\sin[\tfrac{1}{2}\pi K(k)^{-1}x]\prod_{n = 1}^{\infty} \frac{1 - 2q(k)^{2n}\cos[\pi K(k)^{-1}x] + q(k)^{4n}}{1 - 2q(k)^{2n - 1}\cos[\pi K(k)^{-1}x] + q(k)^{4n - 2}}$$
 * $$\operatorname{cd}(x;k) = 2\sqrt[4]{k^{-2}q(k)}\,\cos[\tfrac{1}{2}\pi K(k)^{-1}x]\prod_{n = 1}^{\infty} \frac{1 + 2q(k)^{2n}\cos[\pi K(k)^{-1}x] + q(k)^{4n}}{1 + 2q(k)^{2n - 1}\cos[\pi K(k)^{-1}x] + q(k)^{4n - 2}}$$
 * $$\operatorname{cn}(x;k) = 2\sqrt[4]{k^{-2}(1 - k^2)\,q(k)}\,\cos[\tfrac{1}{2}\pi K(k)^{-1}x]\prod_{n = 1}^{\infty} \frac{1 + 2q(k)^{2n}\cos[\pi K(k)^{-1}x] + q(k)^{4n}}{1 - 2q(k)^{2n - 1}\cos[\pi K(k)^{-1}x] + q(k)^{4n - 2}}$$

These five formulas are valid for all values k from −1 until +1.

Then following successive definition of the other Jacobi functions is possible:

 * $$\operatorname{sn}(x;k) = \frac{2\{\operatorname{zn}(\tfrac{1}{2}x;k) + \operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}}{k^2+\{\operatorname{zn}(\tfrac{1}{2}x;k) + \operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}^2}$$
 * $$\operatorname{cd}(x;k) = \operatorname{sn}[K(k) - x;k]$$
 * $$\operatorname{cn}(x;k) = \operatorname{cd}(x;k)\operatorname{dn}(x;k)$$
 * $$\operatorname{dn}(x;k) = \frac{k^2-\{\operatorname{zn}(\tfrac{1}{2}x;k) + \operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}^2}{k^2+\{\operatorname{zn}(\tfrac{1}{2}x;k) + \operatorname{zn}[K(k)-\tfrac{1}{2}x;k]\}^2}$$

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.

Identities of Jacobi Amplitude functions
In combination with the theta functions the nome gives the values of many Jacobi amplitude function values:

 * $$\operatorname{sc}[\tfrac{2}{3}K(k);k] = \frac{\sqrt{3}\,\vartheta_{01}[q(k)^6]}{\sqrt{1 - k^2}\,\vartheta_{01}[q(k)^2]}$$
 * $$\operatorname{sn}[\tfrac{1}{3}K(k);k] = \frac{2\vartheta_{00}[q(k)]^2}{3\vartheta_{00}[q(k)^3]^2 + \vartheta_{00}[q(k)]^2} = \frac{3\vartheta_{01}[q(k)^3]^2 - \vartheta_{01}[q(k)]^2}{3\vartheta_{01}[q(k)^3]^2 + \vartheta_{01}[q(k)]^2}$$
 * $$\operatorname{cn}[\tfrac{2}{3}K(k);k] = \frac{3\vartheta_{00}[q(k)^3]^2 - \vartheta_{00}[q(k)]^2}{3\vartheta_{00}[q(k)^3]^2 + \vartheta_{00}[q(k)]^2} = \frac{2\vartheta_{01}[q(k)]^2}{3\vartheta_{01}[q(k)^3]^2 + \vartheta_{01}[q(k)]^2} $$
 * $$\operatorname{sn}[\tfrac{1}{5}K(k);k] = \biggl\{\frac{\sqrt{5}\,\vartheta_{01}[q(k)^5]}{\vartheta_{01}[q(k)]} - 1\biggr\}\biggl\{\frac{5\vartheta_{01}[q(k)^{10}]^2}{\vartheta_{01}[q(k)^2]^2} - 1\biggr\}^{-1} $$
 * $$\operatorname{sn}[\tfrac{3}{5}K(k);k] = \biggl\{\frac{\sqrt{5}\,\vartheta_{01}[q(k)^5]}{\vartheta_{01}[q(k)]} + 1\biggr\}\biggl\{\frac{5\vartheta_{01}[q(k)^{10}]^2}{\vartheta_{01}[q(k)^2]^2} - 1\biggr\}^{-1} $$
 * $$\operatorname{cn}[\tfrac{2}{5}K(k);k] = \biggl\{\frac{\sqrt{5}\,\vartheta_{00}[q(k)^5]}{\vartheta_{00}[q(k)]} + 1\biggr\}\biggl\{\frac{5\vartheta_{01}[q(k)^{10}]^2}{\vartheta_{01}[q(k)^2]^2} - 1\biggr\}^{-1} $$
 * $$\operatorname{cn}[\tfrac{4}{5}K(k);k] = \biggl\{\frac{\sqrt{5}\,\vartheta_{00}[q(k)^5]}{\vartheta_{00}[q(k)]} - 1\biggr\}\biggl\{\frac{5\vartheta_{01}[q(k)^{10}]^2}{\vartheta_{01}[q(k)^2]^2} - 1\biggr\}^{-1} $$

The abbreviation sc describes the quotient of the amplitude sine divided by the amplitude cosine.

Derivation of the nome square theorem
The law for the square of the elliptic noun involves forming the Landen daughter modulus:

 * {| class="wikitable"

 * $$q(\varepsilon)^2 = q\bigl[\varepsilon^2(1+\sqrt{1-\varepsilon^2})^{-2}\bigr] = q\bigl\{\tan\bigl[\tfrac{1}{2}\arcsin(\varepsilon)\bigr]^2\bigr\} = q\bigl\{\operatorname{tanh}\bigl[\tfrac{1}{2}\operatorname{artanh}(\varepsilon)\bigr]^2\bigr\} $$
 * }

The Landen daughter modulus is also the tangential counterpart of the Pythagorean counterpart of the mother modulus.

Examples for the nome square theorem
The Landen daughter modulus 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:

 * $$\exp(-2\sqrt{3}\,\pi) = \exp(-\sqrt{3}\,\pi)^2 = q\bigl[\sin(\tfrac{1}{12}\pi)\bigr]^2 = q\bigl[\tan(\tfrac{1}{24}\pi)^2\bigr] $$

 * $$\exp(-2\sqrt{5}\,\pi) = \exp(-\sqrt{5}\,\pi)^2 = q\bigl\{\sin\bigl[\tfrac{1}{2}\arcsin(\sqrt{5} - 2)\bigr]\bigr\}^2 = q\bigl\{\tan\bigl[\tfrac{1}{4}\arcsin(\sqrt{5} - 2)\bigr]^2\bigr\} $$

 * $$\exp(-2\sqrt{7}\,\pi) = \exp(-\sqrt{7}\,\pi)^2 = q\bigl\{\sin\bigl[\tfrac{1}{2}\arcsin(\tfrac{1}{8})\bigr]\bigr\}^2 = q\bigl\{\tan\bigl[\tfrac{1}{4}\arcsin(\tfrac{1}{8})\bigr]^2\bigr\} $$

 * $$\exp(-2\sqrt{13}\,\pi) = \exp(-\sqrt{13}\,\pi)^2 = q\bigl\{\sin\bigl[\tfrac{1}{2}\arcsin(5\sqrt{13} - 18)\bigr]\bigr\}^2 = q\bigl\{\tan\bigl[\tfrac{1}{4}\arcsin(5\sqrt{13} - 18)\bigr]^2\bigr\} $$

Hyperbolically displayed examples:

 * $$\exp(-2\sqrt{6}\,\pi) = \exp(-\sqrt{6}\,\pi)^2 =$$

 * $$= q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{2}\operatorname{arsinh}\bigl[(\sqrt{2} - 1)^2\bigr]\bigr\}\biggr\rangle^2 = q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{4}\operatorname{arsinh}\bigl[(\sqrt{2} - 1)^2\bigr]\bigr\}^2\biggr\rangle $$

 * $$\exp(-2\sqrt{10}\,\pi) = \exp(-\sqrt{10}\,\pi)^2 =$$

 * $$= q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{2}\operatorname{arsinh}\bigl[(\sqrt{5} - 2)^2\bigr]\bigr\}\biggr\rangle^2 = q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{4}\operatorname{arsinh}\bigl[(\sqrt{5} - 2)^2\bigr]\bigr\}^2\biggr\rangle $$

 * $$\exp(-2\sqrt{14}\,\pi) = \exp(-\sqrt{14}\,\pi)^2 =$$

 * $$= q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{2}\operatorname{arsinh}\bigl[(\sqrt{2} + \tfrac{1}{2} - \tfrac{1}{2}\sqrt{4\sqrt{2} + 5})^3\bigr]\bigr\}\biggr\rangle^2 = q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{4}\operatorname{arsinh}\bigl[(\sqrt{2} + \tfrac{1}{2} - \tfrac{1}{2}\sqrt{4\sqrt{2} + 5})^3\bigr]\bigr\}^2\biggr\rangle $$

 * $$\exp(-2\sqrt{22}\,\pi) = \exp(-\sqrt{22}\,\pi)^2 =$$

 * $$= q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{2}\operatorname{arsinh}\bigl[(\sqrt{2} - 1)^6\bigr]\bigr\}\biggr\rangle^2 = q\biggl\langle\operatorname{tanh}\bigl\{\tfrac{1}{4}\operatorname{arsinh}\bigl[(\sqrt{2} - 1)^6\bigr]\bigr\}^2\biggr\rangle $$

Derivation of the parametrized nome cube theorem
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.

 * {| class="wikitable"

 * $$q[u(\sqrt{u^4-u^2+1}-u^2+1)]^3 = q[u(\sqrt{u^4-u^2+1}+u^2-1)] $$
 * }

This formula was displayed exactly like this and this time it was not printed exactly after the expression $$ \varepsilon $$ with the main alignment on the mother modulus, because this formula contains a long formulation. And in the formula shown now with the parameter $$ u $$, a greatly simplified formula emerges.

Derivation of the direct nome cube theorem
On the basis of the now absolved proof a direct formula for the nome cube theorem in relation to the modulus $$ \varepsilon $$ and in combination with the Jacobi amplitude sine shall be generated:

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:

 * $$\varepsilon^2 x^4 - 2\varepsilon^2 x^3 + 2x - 1 = 0 $$

 * $$x = \text{sn}\bigl[\tfrac{1}{3}K(\varepsilon);\varepsilon\bigr] $$

Now the parametrization mentioned above is inserted into this equation:

 * $$\varepsilon = u(\sqrt{u^4-u^2+1}-u^2+1)$$

 * $$u^2(\sqrt{u^4-u^2+1}-u^2+1)^2 (x^4 - 2x^3) + 2x - 1 = 0 $$

This is the real solution of the pattern $$ \tfrac{1}{2} < x < 1 \,\cap \,x \in \R $$ of that quartic equation:

 * $$x = \frac{1}{\sqrt{u^4-u^2+1}-u^2+1} $$

Therefore, following formula is valid:

 * $$\text{sn}\bigl[\tfrac{1}{3}K(\varepsilon);\varepsilon\bigr] \bigl[\varepsilon = u(\sqrt{u^4-u^2+1}-u^2+1)\bigr] = \frac{1}{\sqrt{u^4-u^2+1}-u^2+1}$$

The parametrized nome cube formula has this mentioned form:

 * $$q\bigl[u(\sqrt{u^4-u^2+1}-u^2+1)\bigr]^3 = q\bigl[u(\sqrt{u^4-u^2+1}+u^2-1)\bigr] $$

The same formula can be designed in this alternative way:

 * $$q\bigl[u(\sqrt{u^4-u^2+1}-u^2+1)\bigr]^3 = q\bigl\{\bigl[u(\sqrt{u^4-u^2+1}-u^2+1)\bigr]^3 \bigl(\sqrt{u^4-u^2+1}-u^2+1\bigr)^{-4} \bigr\} $$

So this result appears as the direct nome cube theorem:

 * $$q(\varepsilon)^3 = q\bigl\{\varepsilon^3 \text{sn}\bigl[\tfrac{1}{3}K(\varepsilon);\varepsilon\bigr]^4 \bigr\}$$

Examples for the nome cube theorem
Alternatively, this formula can be set up:

 * {| class="wikitable"

 * $$q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(t^3)\bigr]\bigr\}^3 = q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(t^3)\bigr]^3 \tan\bigl[\arctan\bigl(\,\sqrt{2\sqrt{t^4 - t^2 + 1} - t^2 + 2} + \sqrt{t^2 + 1}\,\bigr) - \tfrac{1}{4}\pi\bigr]^4\bigr\} $$
 * }

The now presented formula is used for simplified computations, because the given elliptical modulus can be used to determine the value $$t $$ in an easy way. The value $$t $$ can be evoked by taking the tangent duplication of the modulus and then taking the cube root of that in order to get the parameterization value $$t $$ directly.

Two examples are to be treated exemplarily:

In the first example, the value $$t = 1 $$ is inserted:
 * $${\color{blue}\exp(-3\sqrt{2}\,\pi)} = \exp(-\sqrt{2}\,\pi)^3 = q(\sqrt{2} - 1)^3 = q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(1)\bigr]\bigr\}^3 = $$

 * $$= q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(1)\bigr]^3 \tan\bigl[\arctan(\sqrt{3} + \sqrt{2}) - \tfrac{1}{4}\pi\bigr]^4\bigr\} = {\color{blue}q\bigl[(\sqrt{2} - 1)^3 (\tfrac{1}{2}\sqrt{6} - \tfrac{1}{2}\sqrt{2})^4\bigr]} $$

In the second example, the value $$t = \Phi^{-2} = \tfrac{1}{2}(3 - \sqrt{5})$$ is inserted:
 * $${\color{blue}\exp(-3\sqrt{10}\,\pi)} = \exp(-\sqrt{10}\,\pi)^3 = q\bigl[(\sqrt{10} - 3)(\sqrt{2} - 1)^2\bigr]^3 = q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(\Phi^{-6})\bigr]\bigr\}^3 = $$

 * $$= q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(\Phi^{-6})\bigr]^3 \tan\bigl[\arctan\bigl(\sqrt{2\sqrt{\Phi^{-8} - \Phi^{-4} + 1} - \Phi^{-4} + 2} + \sqrt{\Phi^{-4} + 1}\bigr) - \tfrac{1}{4}\pi\bigr]^4\bigr\} = $$

 * $$= {\color{blue}q\bigl\{(\sqrt{10} - 3)^3(\sqrt{2} - 1)^6 \tan\bigl[\arctan\bigl(\sqrt{2\sqrt{\Phi^{-8} - \Phi^{-4} + 1} - \Phi^{-4} + 2} + \sqrt{\Phi^{-4} + 1}\bigr) - \tfrac{1}{4}\pi\bigr]^4\bigr\}} $$

The constant $$ \Phi $$ represents the Golden ratio number $$ \Phi = \tfrac{1}{2}(\sqrt{5} + 1)$$ exactly. Indeed, the formula for the cube of the nome involves a modulus transformation that really contains elementary cube roots because it involves the solution of a regular quartic equation. However the laws for the fifth power and the seventh power of the elliptic nome do not lead to an elementary nome transformation, but to a non elementary transformation. This was proven by the Abel–Ruffini theorem and by the Galois theory too.

Exponentiation theorems with Jacobi amplitude functions
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:

 * $$q(\varepsilon_1 \in \mathbb{A}^{+})^{w \in \mathbb{Q^{+}}} = q(\varepsilon_2 \in \mathbb{A}^{+}) $$

These are the most important examples of the general exponentiation theorem:

 * $$q(\varepsilon)^2 = q\{\varepsilon^2\operatorname{sn}[\tfrac{1}{2}K(\varepsilon);\varepsilon]^4\} = q[\varepsilon^2(1+\sqrt{1-\varepsilon^2})^{-2}] $$
 * $$q(\varepsilon)^3 = q\{\varepsilon^3\operatorname{sn}[\tfrac{1}{3}K(\varepsilon);\varepsilon]^4\} $$
 * $$q(\varepsilon)^4 = q\{\varepsilon^4\operatorname{sn}[\tfrac{1}{4}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{4}K(\varepsilon);\varepsilon]^4\} = q[(1-\sqrt[4]{1-\varepsilon^2})^{2}(1+\sqrt[4]{1-\varepsilon^2})^{-2}] $$
 * $$q(\varepsilon)^5 = q\{\varepsilon^5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]^4\} $$
 * $$q(\varepsilon)^6 = q\{\varepsilon^6\operatorname{sn}[\tfrac{1}{6}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{1}{2}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{5}{6}K(\varepsilon);\varepsilon]^4\} $$
 * $$q(\varepsilon)^7 = q\{\varepsilon^7\operatorname{sn}[\tfrac{1}{7}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{7}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{5}{7}K(\varepsilon);\varepsilon]^4\} $$
 * $$q(\varepsilon)^8 = q\{\varepsilon^8\operatorname{sn}[\tfrac{1}{8}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{8}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{5}{8}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{7}{8}K(\varepsilon);\varepsilon]^4\} $$
 * $$q(\varepsilon)^9 = q\{\varepsilon^9\operatorname{sn}[\tfrac{1}{9}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{1}{3}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{5}{9}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{7}{9}K(\varepsilon);\varepsilon]^4\} $$

The abbreviation $$\operatorname{sn}$$ stands for the Jacobi elliptic function amplitude sine.

For algebraic $$x$$ values in the real interval $$[-1,1]$$ the shown amplitude sine expressions are always algebraic.

This are the general exponentiation theorems:

 * $$q(\varepsilon)^{2n} = q\biggl\{\varepsilon^{2n}\prod_{k = 1}^{n}\operatorname{sn}\bigl[\tfrac{2k-1}{2n}K(\varepsilon);\varepsilon\bigr]^4\biggr\} $$
 * $$q(\varepsilon)^{2n+1} = q\biggl\{\varepsilon^{2n+1}\prod_{k = 1}^{n}\operatorname{sn}\bigl[\tfrac{2k-1}{2n+1}K(\varepsilon);\varepsilon\bigr]^4\biggr\} $$

That theorem is valid for all natural numbers n.

Important computation clues:

The following Jacobi amplitude sine expressions solve the subsequent equations:

Examples for the exponentiation theorems
For these nome power theorems important examples shall be formulated:

Given is the fifth power theorem:

 * $$q(\varepsilon)^5 = q\{\varepsilon^5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]^4\} $$

Lemniscatic example for the fifth power theorem:

 * {|class = "wikitable"

 * $$ \tfrac{1}{8} x^6 - \tfrac{1}{2} x^5 + \tfrac{5}{4} x^4 - \tfrac{5}{2} x^2 + 4x - 1 = 0 $$

$$x = \text{sn}[\tfrac{1}{5}K(k);k]\text{sn}[\tfrac{3}{5}K(k);k](k = \tfrac{1}{2}\sqrt{2}) = \tfrac{1}{2}(\sqrt{5} - 1)(\sqrt[4]{5} - 1) $$

$${\color{blue}\exp(-5\pi)} = \exp(-\pi)^5 = q(\tfrac{1}{2}\sqrt{2})^5 = q\{\varepsilon^5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]^4\} (\varepsilon = \tfrac{1}{2}\sqrt{2}) = $$

$$= q\bigl\{\tfrac{1}{8}\sqrt{2}\bigl[\tfrac{1}{2}(\sqrt{5} - 1)(\sqrt[4]{5} - 1)\bigr]^4\bigr\} = {\color{blue}q\bigl[\tfrac{1}{2}(\sqrt{10} - 2\sqrt{2})(3 - 2\sqrt[4]{5})\bigr]}$$
 * }

A next example for the fifth power theorem:

 * {|class = "wikitable"

 * $$(\sqrt{2} - 1)^6 x^6 - 4(\sqrt{2} - 1)^6 x^5 + 5(\sqrt{2} - 1)^4 x^4 - 5(\sqrt{2} - 1)^2 x^2 + 4x - 1 = 0 $$

$$x = \text{sn}[\tfrac{1}{5}K(k);k]\text{sn}[\tfrac{3}{5}K(k);k](k = \sqrt{2} - 1) = (\sqrt{2} + 1)\tan\bigl[\arctan\bigl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}\,\bigr)-\tfrac{1}{8}\pi\bigr] $$

$${\color{blue}\exp(-5\sqrt{2}\,\pi)} = \exp(-\sqrt{2}\,\pi)^5 = q(\sqrt{2} - 1)^5 = $$

$$= q\{\varepsilon^5\operatorname{sn}[\tfrac{1}{5}K(\varepsilon);\varepsilon]^4\operatorname{sn}[\tfrac{3}{5}K(\varepsilon);\varepsilon]^4\} (\varepsilon = \sqrt{2} - 1) = $$

$$= q\biggl\langle (\sqrt{2} - 1)^5\bigl\{(\sqrt{2} + 1)\tan\bigl[\arctan\bigl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}\,\bigr)-\tfrac{1}{8}\pi\bigr]\bigr\}^4\biggr\rangle = $$

$$= {\color{blue}q\bigl\{(\sqrt{2}-1) \tan\bigl[\arctan\bigl(\tfrac{1}{3}\sqrt{5}-\tfrac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\tfrac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}}\,\bigr)-\tfrac{1}{8}\pi\bigr]^4\bigr\}}$$
 * }

Reflection theorems
If two positive numbers $$a$$ and $$b$$ are Pythagorean opposites to each other and thus the equation $$a^2+b^2=1$$ is valid, then this relation is valid:
 * $$\ln[\operatorname{q}(a)]\ln[\operatorname{q}(b)] = \pi^2$$

If two positive numbers $$c$$ and $$d$$ are tangential opposites to each other and thus the equation $$(c+1)(d+1)=2$$ is valid, then that relation is valid:
 * $$\ln[\operatorname{q}(c)]\ln[\operatorname{q}(d)] = 2\pi^2$$

Therefore, these representations have validity for all real numbers x:

Pythagorean opposites:

 * $$\ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi - \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi + \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle = \pi^2 $$
 * $$\ln\bigl\{q\bigl[\tfrac{1}{2}\sqrt{2-2x(x^2+1)^{-1/2}}\bigr] \bigr\} \ln\bigl\{q\bigl[\tfrac{1}{2} \sqrt{2+2x(x^2+1)^{-1/2}}\bigr]\bigr\} = \pi^2 $$

Tangential opposites:

 * $$\ln\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{8}\pi - \tfrac{1}{4}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln \biggl\langle q \bigl\{\tan\bigl[\tfrac{1}{8}\pi + \tfrac{1}{4}\arctan(x)\bigr] \bigr\} \biggr\rangle = 2\pi^2 $$
 * $$\ln\bigl\{ q\bigl[\sqrt{(\sqrt{x^2+1}+x)^2+1}-\sqrt{x^2+1}-x\bigr]\bigr\} \ln\bigl\{ q \bigl[ \sqrt{(\sqrt{x^2+1}-x)^2+1}-\sqrt{x^2+1}+x\bigr] \bigr\} = 2\pi^2 $$

Direct results of mentioned theorems
The following examples should be used to determine the nouns:

Example 1: Given is the formula of the Pythagorean counterparts:

 * {| class="wikitable"

 * $$\ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi - \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi + \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle = \pi^2 $$
 * }

For x = 0, this formula gives this equation:
 * $$\ln\bigl\{q\bigl[\sin(\tfrac{1}{4}\pi)\bigr]\bigr\}^2 = \pi^2 $$
 * $$q\bigl[\sin(\tfrac{1}{4}\pi)\bigr] = \exp(-\pi) $$

Example 2:

Given is the formula of the tangential counterparts:

 * {| class="wikitable"

 * $$\ln\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{8}\pi - \tfrac{1}{4}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln \biggl\langle q \bigl\{\tan\bigl[\tfrac{1}{8}\pi + \tfrac{1}{4}\arctan(x)\bigr] \bigr\} \biggr\rangle = 2\pi^2 $$
 * }

For x = 0, the formula for the tangential counterparts gives the following equation:
 * $$\ln\bigl\{q\bigl[\tan(\tfrac{1}{8}\pi)\bigr]\bigr\}^2 = 2\pi^2 $$
 * $$q\bigl[\tan(\tfrac{1}{8}\pi)\bigr] = \exp(-\sqrt{2}\,\pi) $$

Combinations of two theorems each
Example 1: Equianharmonic case

The formula of the Pythagorean counterparts is used again:

 * {| class="wikitable"

 * $$\ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi - \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln\biggl\langle q\bigl\{\sin\bigl[\tfrac{1}{4}\pi + \tfrac{1}{2}\arctan(x)\bigr]\bigr\}\biggr\rangle = \pi^2 $$
 * }

For $$x = \sqrt{3}$$, this equation results from this formula:

 * $$\ln\bigl\{q\bigl[\sin(\tfrac{1}{12}\pi)\bigr]\bigr\}\ln\bigl\{q\bigl[\sin(\tfrac {5}{12}\pi)\bigr]\bigr\} = \pi^2 $$

In a previous section this theorem was stated:

 * {| class="wikitable"

 * $$q[u(\sqrt{u^4-u^2+1}-u^2+1)]^3 = q[u(\sqrt{u^4-u^2+1}+ u^2-1)] $$
 * }

From this theorem for cubing, the following equation results for $$u = 1/\sqrt{2}$$:

 * $$q\bigl[\sin(\tfrac{5}{12}\pi)\bigr]^3 = q\bigl[\sin(\tfrac{1}{12}\pi)\bigr] $$

The solution to the system of equations with two unknowns then reads as follows:

 * $$q\bigl[\sin(\tfrac{1}{12}\pi)\bigr] = \exp(-\sqrt{3}\,\pi) $$

 * $$q\bigl[\sin(\tfrac{5}{12}\pi)\bigr] = \exp(-\tfrac{1}{3}\sqrt{3}\,\pi) $$

Example 2: A further case with the cube formula

The formula of the tangential counterparts is used again:

 * {| class="wikitable"

 * $$\ln\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{8}\pi - \tfrac{1}{4}\arctan(x)\bigr]\bigr\}\biggr\rangle \ln \biggl\langle q \bigl\{\tan\bigl[\tfrac{1}{8}\pi + \tfrac{1}{4}\arctan(x)\bigr] \bigr\} \biggr\rangle = 2\pi^2 $$
 * }

For $$x = \sqrt{8}$$ this formula results in the following equation:

 * $$\ln\bigl\{q\bigl[(2 - \sqrt{3})(\sqrt{3} - \sqrt{2})\bigr]\bigr\}\ln\bigl\{q \bigl[(2 - \sqrt{3})(\sqrt{3} + \sqrt{2})\bigr]\bigr\} = 2\pi^2 $$

The theorem for cubing is also used here:

 * {| class="wikitable"

 * $$q[u(\sqrt{u^4-u^2+1}-u^2+1)]^3 = q[u(\sqrt{u^4-u^2+1}+ u^2-1)] $$
 * }

From the previously mentioned theorem for cubing, the following equation results for $$u = (\sqrt{3} - 1)/\sqrt{2}$$:

 * $$q\bigl[(2 - \sqrt{3})(\sqrt{3} + \sqrt{2})\bigr]^3 = q\bigl[(2 - \sqrt{3})( \sqrt{3} - \sqrt{2})\bigr] $$

The solution to the system of equations with two unknowns then reads as follows:

 * $$q\bigl[(2 - \sqrt{3})(\sqrt{3} - \sqrt{2})\bigr] = \exp(-\sqrt{6}\,\pi) $$
 * $$q\bigl[(2 - \sqrt{3})(\sqrt{3} + \sqrt{2})\bigr] = \exp(-\tfrac{1}{3}\sqrt{6 }\,\pi) $$

Investigations about incomplete integrals
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:

 * {| class="wikitable"

 * $$F\biggl[2\arctan\biggl(\frac{x^3 + \sqrt{3}\,x}{\sqrt{3}\,x^2 + 1}\biggr);\frac {1}{4}(\sqrt{6} + \sqrt{2})\biggr] = \sqrt{3} \,F\biggl[2\arctan(x);\frac{1}{4}( \sqrt{6} - \sqrt{2})\biggr] $$

The correctness of this formula can be proved by computing the differential quotient after the variable $$x$$ on both sides of the balance of equation.

Using the value $$x = 1$$ gives this result:

$$K\bigl[\frac{1}{4}(\sqrt{6} + \sqrt{2})\bigr] = \sqrt{3} \,K\bigl[\frac{1}{4 }(\sqrt{6} - \sqrt{2})\bigr] $$

The following two results emerge:

$$q\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2})\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac{ 1}{4}(\sqrt{6} + \sqrt{2})\bigr]\div K\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2})\bigr ]\bigr\} = \exp(-\sqrt{3}\,\pi) $$

$$q\bigl[\frac{1}{4}(\sqrt{6} + \sqrt{2})\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac{ 1}{4}(\sqrt{6} - \sqrt{2})\bigr]\div K\bigl[\frac{1}{4}(\sqrt{6} + \sqrt{2})\bigr ]\bigr\} = \exp\bigl(- \frac{1}{3}\sqrt{3}\,\pi\bigr) $$
 * }

Second example:
 * {| class="wikitable"

 * $$F\biggl[2\arctan\biggl(\frac{x^5 + 2\sqrt{5}\,\Phi^{-1/2}\,x^3 + \sqrt{5}\,x}{\sqrt{5}\,x^4 + 2\sqrt{5}\,\Phi^{-1/2}\,x^2 + 1}\biggr);\frac{1}{ 2}\sqrt{2}\bigl(\Phi^{-1/2} + \Phi^{-1}\bigr)\biggr] = \sqrt{5} \,F\biggl[2\arctan(x );\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} - \Phi^{-1}\bigr)\biggr] $$

The correctness of this formula can be proved by differentiating both sides of the equation balance.

$$K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} + \Phi^{-1}\bigr)\bigr] = \sqrt{ 5} \,K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} - \Phi^{-1}\bigr)\bigr] $$

The following two results emerge:

$$q\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} - \Phi^{-1}\bigr)\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} + \Phi^{-1}\bigr)\bigr]\div K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} - \Phi^{-1}\bigr)\bigr]\bigr\} = \exp(- \sqrt{5}\,\pi) $$

$$q\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} + \Phi^{-1}\bigr)\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} - \Phi^{-1}\bigr)\bigr]\div K\bigl[\frac{1}{2}\sqrt{2}\bigl(\Phi^{-1/2} + \Phi^{-1}\bigr)\bigr]\bigr\} = \exp\bigl(-\frac{1}{5}\sqrt{5}\,\pi\bigr) $$
 * }

Third example:
 * {| class="wikitable"

 * $$F\biggl[2\arctan\biggl(\frac{x^5 + 2\,\sqrt{7}\,x^5 + 7x^3 + \sqrt{7}\,x}{\sqrt{7}\,x^6 + 7x^4 + 2\,\sqrt{7}\,x^2 + 1}\biggr);\frac{1}{8}(3\sqrt{2} + \sqrt{14})\biggr] = \sqrt{7} \,F\biggl[2\arctan(x);\frac{1}{8}(3\sqrt{2} - \sqrt{14}) \biggr] $$

The correctness of this formula can be proved by differentiating both sides of the equation balance.

Using the value $$x = 1$$ gives this result:

$$K\bigl[\frac{1}{8}(3\sqrt{2} + \sqrt{14})\bigr] = \sqrt{7} \,K\bigl[\frac{1}{ 8}(3\sqrt{2} - \sqrt{14})\bigr] $$

The following two results emerge:

$$q\bigl[\frac{1}{8}(3\sqrt{2} - \sqrt{14})\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac {1}{8}(3\sqrt{2} + \sqrt{14})\bigr]\div K\bigl[\frac{1}{8}(3\sqrt{2} - \sqrt{14} )\bigr]\bigr\} = \exp(- \sqrt{7}\,\pi) $$

$$q\bigl[\frac{1}{8}(3\sqrt{2} + \sqrt{14})\bigr] = \exp\bigl\{- \pi \,K\bigl[\frac {1}{8}(3\sqrt{2} - \sqrt{14})\bigr]\div K\bigl[\frac{1}{8}(3\sqrt{2} + \sqrt{14} )\bigr]\bigr\} = \exp\bigl(- \frac{1}{7}\sqrt{7}\,\pi\bigr) $$
 * }

Derivation of the derivative
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:

 * $$\frac{\pi^2}{2\varepsilon(1-\varepsilon^2)K(\varepsilon)^2} \,q(\varepsilon)\,\biggl\{\frac{\mathrm{d}}{\mathrm{d}\,q(\varepsilon)}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr]\biggr\} = \biggl[\frac{\mathrm{d}}{\mathrm{d}\varepsilon} \,q(\varepsilon)\biggr]\biggl\{\frac{\mathrm{d}}{\mathrm{d}\,q(\varepsilon)}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr]\biggr\} = \frac{\mathrm{d}}{\mathrm{d}\varepsilon}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr] = \frac{\mathrm{d}}{\mathrm{d}\varepsilon}\,\sqrt{2\pi^{-1}K(\varepsilon)} = $$

 * $$= \frac{1}{2}\sqrt{2}\,\pi^{-1/2}\,K(\varepsilon)^{-1/2}\biggl[\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\,K(\varepsilon)\biggr] = \frac{1}{2}\sqrt{2}\,\pi^{-1/2}\,K(\varepsilon)^{-1/2}\,\frac{E(\varepsilon) - (1 - \varepsilon^2)K(\varepsilon)}{\varepsilon(1 - \varepsilon^2)} $$

For the now mentioned derivation part this identity is the fundament:

 * $$\vartheta_{00}[q(\varepsilon)] = \sqrt{2\pi^{-1} K(\varepsilon)}$$

Therefore, this equation results:

 * $$\frac{\mathrm{d}}{\mathrm{d}\,q(\varepsilon)}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr] = \sqrt{2}\,\pi^{-5/2}\,q(\varepsilon)^{-1}\,K(\varepsilon)^{3/2}\bigl[E(\varepsilon) - (1 - \varepsilon^2)K(\varepsilon)\bigr] $$

The complete elliptic integrals of the second kind have that identity:

 * $$(1 + \sqrt{1 - \varepsilon^2})\,E\left(\frac{1 - \sqrt{1 - \varepsilon^2}}{1 + \sqrt{1 - \varepsilon^2}}\right) = E(\varepsilon) + \sqrt{1 - \varepsilon^2}\,K(\varepsilon) $$

Along with this modular identity, following formula transformation can be made:

 * $$\frac{\mathrm{d}}{\mathrm{d}\,q(\varepsilon)}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr] = \sqrt{2}\,\pi^{-5/2}\,q(\varepsilon)^{-1}\,K(\varepsilon)^{3/2}(1 + \sqrt{1 - \varepsilon^2})\left[E\left(\frac{1 - \sqrt{1 - \varepsilon^2}}{1 + \sqrt{1 - \varepsilon^2}}\right) - \sqrt{1 - \varepsilon^2}\,K(\varepsilon)\right] $$

Furthermore, this identity is valid:

 * $$\vartheta_{01}[q(\varepsilon)] = \sqrt[4]{1 - \varepsilon^2}\sqrt{2\pi^{-1} K(\varepsilon)}$$

By using the theta function expressions ϑ00(x) and ϑ01(x) following representation is possible:

 * $$\frac{\mathrm{d}}{\mathrm{d}\,q(\varepsilon)}\,\vartheta_{00}\bigl[q(\varepsilon)\bigr] = \frac{1}{2\pi}\,q(\varepsilon)^{-1}\vartheta_{00}[q(\varepsilon)]\bigl\{\vartheta_{00}[q(\varepsilon)]^2 + \vartheta_{01}[q(\varepsilon)]^2\bigr\}\biggl\langle E\biggl\{\frac{\vartheta_{00}[q(\varepsilon)]^2 - \vartheta_{01}[q(\varepsilon)]^2}{\vartheta_{00}[q(\varepsilon)]^2 + \vartheta_{01}[q(\varepsilon)]^2}\biggr\} - \frac{\pi}{2}\,\vartheta_{01}\bigl[q(\varepsilon)\bigr]^2\biggr\rangle $$

This is the final result:
 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\vartheta_{00}(x) = \vartheta_{00}(x)\bigl[\vartheta_{00}(x)^2+\vartheta_{01}(x)^2\bigr]\biggl\{\frac{1}{2\pi x}E\biggl[\frac{\vartheta_{00}(x)^2-\vartheta_{01}(x)^2}{\vartheta_{00}(x)^2+\vartheta_{01}(x)^2}\biggr] - \frac{\vartheta_{01}(x)^2}{4x}\biggr\}$$

Related first derivatives
In a similar way following other first derivatives of theta functions and their combinations can also be derived:

 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\vartheta_{01}(x) = \vartheta_{01}(x)\bigl[\vartheta_{00}(x)^2+\vartheta_{01}(x)^2\bigr]\biggl\{\frac{1}{2\pi x}E\biggl[\frac{\vartheta_{00}(x)^2-\vartheta_{01}(x)^2}{\vartheta_{00}(x)^2+\vartheta_{01}(x)^2}\biggr] - \frac{\vartheta_{00}(x)^2}{4x}\biggr\}$$
 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\vartheta_{10}(x) = \frac{1}{2\pi x} \vartheta_{10}(x)\vartheta_{00}(x)^2 E\biggl[\frac{\vartheta_{10}(x)^2}{\vartheta_{00}(x)^2}\biggr]$$
 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\vartheta_{00}(x)}{\vartheta_{01}(x)} = \frac{\vartheta_{00}(x)^5 - \vartheta_{00}(x)\vartheta_{01}(x)^4}{4x\,\vartheta_{01}(x)}$$
 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\vartheta_{10}(x)}{\vartheta_{00}(x)} = \frac{\vartheta_{10}(x)\vartheta_{01}(x)^4}{4x\,\vartheta_{00}(x)}$$
 * $$\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\vartheta_{10}(x)}{\vartheta_{01}(x)} = \frac{\vartheta_{10}(x)\vartheta_{00}(x)^4}{4x\,\vartheta_{01}(x)}$$

Important definition:
 * $$\vartheta_{10}(x) = 2x^{1/4} + 2x^{1/4}\sum_{n = 1}^{\infty} x^{2\bigtriangleup(n)} $$
 * $$\bigtriangleup(n) = \tfrac{1}{2}n(n + 1) $$

title: Nome (mathematics)


text:
Juchitán de Zaragoza

Juchitán de Zaragoza (Spanish name; Isthmus Zapotec: Xabizende ) is an indigenous town in the southeast of the Mexican state of Oaxaca. It is part of the Juchitán District in the west of the Istmo de Tehuantepec region. With a 2020 census population of 88,280, it is the third-largest city in the state. The majority of the indigenous inhabitants are Zapotecs and Huaves. The town also serves as the municipal seat for the surrounding municipality, with which it shares a name. The municipality has an area of 414.64 km2 (160.1 sq mi) and a population of 113,570, the state's third-largest in population.

It is located 26 km northeast of the city of Tehuantepec. Its Palacio Municipal dates back to the middle of the 19th century and perhaps is the widest "palace" in Mexico with 31 arches in its front portal. Its main church is the Parroquia de San Vicente Ferrer (Parish of San Vicente Ferrer) which dates from the 17th century. To the west of the Palacio is a large market where local products can be seen and a local variant of the Zapotec language can be heard.

History
The people of Juchitán have led some local revolts over time: In 1834, "Che Gorio Melendre", a native of Juchitán, directed a revolt against the government of Oaxaca, demanding the control of salt mines on the coast located at the southwest of Juchitán and for local autonomy of the county. The revolt was interrupted by the Mexican–American War in 1847. Irregular troops commanded by Melendre joined the resistance against the invasion. After the invasion by the United States, the governor of Oaxaca, Benito Juárez responded to the local demands of Che Gorio Melendre on May 19, 1850, by sending troops to burn the city of Juchitán and to assassinate their leader Melendre.

On September 5, 1866, during the French intervention in Mexico, the indigenous people of Juchitán, Unión Hidalgo, San Blas Atempa, and Ixtaltepec defeated the Royal French Army stationed in Tehuantepec. Most of the army of Porfirio Díaz, later President of Mexico, were natives of Juchitán. José Fructuoso Gómez, nicknamed Che Gómez directed a 1910 revolt in support of the Mexican Revolution, allied with Zapata and Villa. In the 1970s, a group of left wing students, workers and farmers organized with the intent of taking control of the local county through elections, instead of by force. In February 2001, Juchitán municipality received the caravan of Zapatista Army of National Liberation (EZLN).

The violent history of Juchitán involves the strategic geopolitical location of the area, which is located on the Isthmus of Tehuantepec, the thin part of Mexico between the Pacific Ocean and the Gulf of Mexico. The zone has been coveted by many countries since the McLane–Ocampo Treaty, which was signed in December 1859. Under the treaty, President Benito Juárez received a loan in exchange for the use of the isthmus of Tehuantepec by the United States. In the 1970s an attempt to resurrect the treaty, called the Alfa–Omega project, was aborted. In 2000, the project was finally approved as the Plan Puebla Panama.

Gamesa and Iberdrola are currently making important investments in Juchitán, to create a big wind power eolic park -called Proyecto La Venta II- able to produce at least 88 megawatts of power. The project will make Juchitán the center of the alternative energies in Mexico becoming an example to the rest of Latin America as the eolic park would be the largest in all the region. This project has been criticized because of the lack of information given by Gamesa, Iberdrola and the Mexican Government about its possible ecological, political and cultural consequences on a region where its culture is based on the property of the land used by the Proyecto La Venta II.

In 2006, it was renamed as "Heroica Ciudad de Juchitán de Zaragoza" (Heroic City of Juchitán de Zaragoza) by the State Congress for its inhabitants' defense against the French invasion.

Between June and October 2020, Juchitán reported 241 COVID-19 pandemic-related deaths; space in local cemeteries was at a premium.

In February 2021 the National Institute of Indigenous Peoples (INPI) and Comisión Federal de Electricidad (CFE) reached an agreement with Santa María del Mar agency to allow the operation of 540 kV solar power plant, providing electricity for the first time in ten years.

2017 Chiapas earthquake
Shortly before midnight on 7 September 2017, an earthquake struck off the coast of Chiapas, registered at either 8.1 or 8.2. The historic earthquake was said to have been the strongest in Mexico in a century.

Juchitán de Zaragoza, on the Oaxacan coast, was one of the cities most damaged by the earthquake. Entire streets were destroyed, its 1860 municipal palace suffered notable destruction, with a large part of the building completely collapsed. In the aftermath of the quake, a resident retrieved the national flag of Mexico and placed it on top of the rubble - the image quickly went viral and became a symbol of patriotism and national unity in the disaster-stricken nation.

Culture
Citizens of Juchitán have also made contributions in the arts, such as painting, poetry, music, folk dance, and sculpture.

In May, residents celebrate the Fiesta de las Velas (Festival of the Candles) in honor of its patron saint San Vicente Ferrer, with a large procession.

The Isthmus Zapotec, an indigenous people who comprise about 70 percent of the population of Juchitan, a city in the south of Mexico, practice a melding of Catholic and indigenous spiritual traditions. In a slide-illustrated lecture and book, anthropologist Anya Peterson Royce shows how the Zapotec use flowers, processions and prayer in rituals that protect and guide spirits on their journey of dying. She also describes the Day of the Dead and Holy Week rituals and the role of the community healer.

The municipality also has a reputation as a significant oasis of tolerance and respect for LGBTQ people in Mexico, due to the traditional Zapotec culture of muxes.

The municipality

As municipal seat, Juchitán has governing jurisdiction over the following communities:

15 de Agosto, 5 de Septiembre, Álvaro Obregón, Cerro Cristo, Charis, Chicapa de Castro, Colonia 19 de Marzo (Colonia San José), Colonia de la Amistad, Colonia Jesús González Cortaza, Colonia José Yola, Colonia Mariano Montero, Colonia San Vicente, Colonia Santa Rosita, Dos Arbolitos, Dos Hermanos, El Caballero Burro, El Chamizal (Toledo Cueto), El Chaparral, El Porvenir, El Tamarindo, Emiliano Zapata, Esquipulas 1, Esquipulas 2, Estero Guiee, Gaspar Torres Urbieta, Huanacastal, La Esperanza, La Estancia (Santa Cecilia), La Guadalupana, La Liebre (Paraje la Liebre), La Negrita, La Providencia, La Venta, La Ventosa 1, La Ventosa 2, Los Aguacates, Los Cocos Los Ordaz II, Los Vicentes  Minerva, Onésimas (Argelino Solórzano), Parada San Vicente, Pepe y Lolita, Piedra Larga, Playa San Vicente, Primera Curva, Rancho Adelma, Rancho Babel López Sánchez, Rancho Chuvalessa, Rancho de los Vásquez (Colonia Palomar), Rancho Domitilo Marquez, Rancho Don Cutberto, Rancho el Ángel, Rancho Esquipulas, Rancho Francisco  Rancho Guadalupe, Rancho Juanita, Rancho Lucita, Rancho Martín Vicente, Rancho Nemesio Valdivieso López, Rancho San Antonio, Rancho San Luis, Rancho Torres, Río Viejo (Mojonera), Salinas Santa Cruz, San Antonio, San Isidro, San José, Santa Clara, Santa Fe 1, Santa Fe 2, Santa Lucía, Santa María del Mar, and Santa Rita

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Juchitán de Zaragoza


text:
Bryant, Seattle

Bryant is a residential neighborhood in northeast Seattle, Washington. According to the City of Seattle's neighborhood maps (as pictured), it is bounded by 35th Avenue NE and NE 45th Place on the west, beyond which is Ravenna; Sand Point Way NE and 45th Ave NE on the east, beyond which are Laurelhurst and Windermere; and NE 75th Street and NE 65th Street on the north, beyond which are View Ridge and Wedgwood.

The neighborhood is sometimes known as Ravenna-Bryant, due to its proximity to Ravenna Park. The Burke–Gilman Trail runs along the southern and eastern margins of the neighborhood, paralleling Blakeley Street, Union Bay Place, 45th Street, and Sand Point Way. Bryant Park is located on NE 65th Street at 40th Avenue NE. In late 2012 the Ravenna-Bryant association incorporated a small neighborhood bounded by 35th Avenue NE on the west, 40th Avenue NE on the east, NE 75th Street on the north, and NE 65th Street on the south. The area had previously been a 'donut hole' between the neighborhood belonging to the Ravenna-Bryant association and those of Wedgwood and View Ridge.

The North East branch of the Seattle Public Library, is located at 35th Avenue NE and NE 68th Street. Bryant is also the home to Assumption Catholic Church, University Unitarian Church, Ravenna Methodist, which houses a local preschool cooperative, and three synagogues, Congregation Beth Shalom (Conservative) at 35th Avenue NE and NE 68th Street, Emmanuel Congregation (Modern Orthodox) on NE 65th Street and 35th Avenue NE, and Congregation Shaarei Tefilah–Lubavitch (Orthodox) on NE 65th Street and 43rd Avenue NE.

Bryant Elementary School, the Nathan Eckstein Middle School, and the North East Library all have status as Seattle city landmarks, as does Fire Station #38 at 5503 33rd Avenue N.E.

Education
Bryant Elementary School is located at 33rd Avenue NE and NE 60th Street. The school was named after William Cullen Bryant. Bryant School has an event called the Bryant Blast, now in the fall, that offers many games and food, including bouncy toys, cake walks, raffles, and more. On the 4th of July, an annual block party takes place along the two streets west of the elementary school. The neighborhood is also home to the Nathan Eckstein Middle School.

Parks
Bryant Playground is a 3.1 acre (13,000 m2) playground and park located at N.E. 65th Street and 40th Avenue N.E. This places it at the very northern end of the Hawthorne Hills/Bryant neighborhood or the very southern end of the View Ridge neighborhood. It was acquired by the city in 1976 from Brook Haven Baptist Church for $155,000. The park and playground were developed two years later at a cost of $415,000.

Commercial Establishments
There are commercial establishments along NE 55th Street between 25th and 35th Avenues NE. The west end is home to the original Kidd Valley burger restaurant. The Duchess Bar on 55th west of 30th Avenue NE is a popular place for patrons to watch University of Washington football games, which are played just south of the neighborhood at Husky Stadium. Across the street is Queen Mary's, a teahouse, and Pair, restaurant specializing in dishes that may be shared. A coffee shop, Ventoux, a Spanish restaurant, Guadi, an Italian café, MioPosto, and several shops, top the hill. The neighborhood is also home to two grocery stores: Puget Consumer Co-op (PCC), a local cooperative chain, at 40th Avenue NE and NE 65th St, and Metropolitan Market, a former Albertsons, on 40th Avenue NE at NE 55th Street, at the southwest corner of the subneighborhood of Hawthorne Hills.

title: Bryant, Seattle


text:
Athens 98.4 FM

Athens 98.4 FM (Αθήνα 9.84), officially Municipality of Athens Broadcasting Corporation "Athina"  is the first non-state radio station to begin broadcasting in Greece in 1987. The station is the forerunner of the municipal radio sector in Greece.

History
The station went on air on May 31, 1987, without a license but backed by then mayor of Athens Miltiadis Evert. The station initially faced many challenges, as a legal framework for private broadcasting was not finalized in Greece until 1989.

The station used to broadcast on 98.4 FM, but moved to 98.3 FM in March 2001 after frequencies in the city of Athens were reassigned. The station also webcasts online and broadcasts via satellite.

The station also operates its website of news in both Greek and English language.

In 2004, it established the foreign-language station Athens International Radio as a spin-off project.

title: Athens 98.4 FM


text:
The Body Politic

The Body Politic was a Canadian monthly magazine, which was published from 1971 to 1987. It was one of Canada's first significant gay publications, and played a prominent role in the development of the LGBT community in Canada.

The Body Politic was a queer, activism-based Canadian monthly magazine that published from 1971 to 1987. It was one of the first significant gay publications in Canada, and played a large role in amplifying the Gay Liberation Movement and creating a space for queer issues and voices to be heard. The Body Politic focused on issues related to Gay and Lesbian sex, Queer culture, and other issues relating to the LGBT community at the time. Although The Body Politic was often criticized for talking about taboo topics such as queer sex and relations, queer media continued to grow. Xtra!, an LGBTQ focused magazine expanded in 1993 by launching its sister editions, ''Xtra! West, based in Vancouver, and Capital Xtra!'', based in Ottawa. All three of these publications remained in print until 2015, when Pink Triangle Press discontinued the print publications of the magazines, however, Daily Xtra continues to publish as an online magazine. The Body Politic was ranked as the 17th most influential magazine in Canadian publishing history by Masthead, and in May 2016 the Canadian actor and playwright Nick Green premiered his historical play, Body Politic. This play discussed the role of the magazine in the early Gay Liberation Movement. In 2017, the play was awarded with the Dora Mavor Moore Award for Outstanding New Play.

History
The magazine was first published on November 1, 1971, by an informal collective, operating out of the home of Glad Day Bookshop owner Jearld Moldenhauer. Many members of the collective had been associated with the underground publication Guerilla, which had been relatively gay-friendly but alienated some of its LGBT contributors when it altered Moldenhauer's article about the We Demand protest of August 28, 1971. At a September meeting of the Toronto Gay Alliance, Moldenhauer first proposed the idea of launching a gay-focused publication. Names considered for the magazine included Mandala and Radical Pervert.

In 1973, the publication ran into difficulty with the Toronto Star, which refused to print an advertisement for the magazine because of its policy of refusing to accept ads relating to sexual activity. While the magazine won an Ontario Press Council ruling that the Star ' s refusal of the ad had been discriminatory, the Star retaliated by discontinuing The Body Politic ' s printing contract with its commercial printing subsidiary Newsweb Enterprises.

The magazine's editorial collective also created the Canadian Lesbian and Gay Archives in 1973.

The collective was incorporated as Pink Triangle Press in 1975. In that same year, the magazine ran into minor legal difficulties when an installment of cartoonist Rand Holmes' Harold Hedd strip depicted an act of fellatio.

The Body Politic was twice charged with publishing obscene material, in 1977 for Gerald Hannon's article "Men Loving Boys Loving Men", and in 1982 for "Lust with a Very Proper Stranger", an article on fisting. The 1977 raid sparked international protests, especially to the police's seizure of the magazine's subscriber list—Harvey Milk was one of the organizers of a protest at the Canadian consulate in San Francisco against the action. The magazine was ultimately acquitted in both trials, although materials seized by police in the Hannon trial were not returned to the magazine until 1985.

In 1977 and 1978, the magazine was involved in the production of Gay News and Views, an LGBT magazine series on Toronto cable.

In 1982, Toronto City Councillor Joe Piccininni failed in an attempt to have the magazine barred from the city council's press gallery, following a cover story on the Sisters of Perpetual Indulgence which Piccininni deemed disrespectful to Roman Catholicism. In this era, the magazine also became noted for its coverage of the emergence of HIV and AIDS.

The magazine ceased publication in 1987, following Pink Triangle Press's launch of the tabloid Xtra! in 1984.

Legacy
Xtra! expanded in 1993 to launch sister editions Xtra! West in Vancouver and Capital Xtra! in Ottawa. All three publications remained in print until 2015, when Pink Triangle Press discontinued the print publications, but remain in operation today as the online magazine Daily Xtra.

In 2008, The Body Politic was ranked as the 17th most influential magazine in Canadian publishing history by Masthead, the trade magazine of the Canadian magazine publishing industry.

Body Politic, a historical play by Nick Green about the magazine and its role in the early gay liberation movement, premiered at Buddies in Bad Times in May 2016, and won the Dora Mavor Moore Award for Outstanding New Play in 2017. As well, actor and writer Diane Flacks was nominated for a Dora award for outstanding performance in the play Body Politic.

Contributors
Writers associated with the magazine included Gerald Hannon, Rick Bébout, Chris Bearchell, Hugh Brewster, Stan Persky, Michael Lynch, Stephen O. Murray, John Greyson, David Rayside, Herbert Spiers, Ian Young, Ed Jackson, Sue Golding, Robin Hardy, Richard Summerbell, Thomas Waugh, John Alan Lee and Gary Kinsman.

title: The Body Politic


text:
Tamie Sheffield

Tamie Michelle Sheffield (born July 27, 1970) is an American film actress, model and former professional wrestler.

Career
Sheffield was born in Mechanicsburg, Pennsylvania and graduated from Mechanicsburg Senior High School in 1988. She graduated cum laude from West Chester University with a degree in Elementary Education. She was also a cheerleader from junior high through college. Sheffield gave campus tours and was in the Phi Sigma Sigma sorority.

Sheffield became a professional wrestler in the California-based Women of Wrestling (WOW) promotion. Using the name Sandy, Sheffield was given the gimmick of a Baywatch-like lifeguard along with a tag team partner named Summer. The duo was collectively known as The Beach Patrol.

Sheffield started her acting career in 1999 in the movie Wildflower and appeared in Slammed, Cheerleader Massacre, The Deviants and Black Tie Nights. She also hosted two Lingerie Bowls. In 2003, Sheffield was a contestant on the third season of NBC's Fear Factor, becoming the winner of her episode. She returned later in the season for the semi-final, where she was unsuccessful.

title: Tamie Sheffield


text:
Par, Cornwall

Par ( An Porth, meaning creek or harbour ) is a village and fishing port with a harbour on the south coast of Cornwall, England, United Kingdom. The village is situated in the civil parish of Tywardreath and Par, although West Par and the docks lie in the parish of St Blaise.

Par is approximately 3.5 mi east of St Austell. Par has a population of around 1,600 (in 2012).

It became developed in the second quarter of the 19th century when the harbour was developed, to serve copper mines and other mineral sites in and surrounding the Luxulyan Valley; china clay later became the dominant traffic as copper working declined, and the harbour and the china clay dries remain as distinctive features of the industrial heritage; however the mineral activity is much reduced.

Par Harbour and the beach at Par Sands are south of the village, and the latter includes a large static caravan holiday park; another small beach is at Spit Point west of the harbour. Between these two beaches the South West Coast Path takes an inland diversion through the village.

Par lies in a triangle of streets which form a one-way traffic system. There is a variety of shops, a post office, a public house and other businesses.

Churches
The Anglican church of St Mary the Virgin at Biscovey was completed in 1849. It was built mainly from the local reddish coloured Biscovey slates. The parish of Par was formed out of parts of St Blazey and Tywardreath parishes in 1846. In the churchyard is an inscribed cross shaft removed from the highroad in 1896. This stone is a sepulchral monument to a son of Ullicus erected by Alroron. The church was the first to be designed by the notable architect G. E. Street. The design is an original and subtle adaptation of the Early English style. The chancel, nave and south aisle are well proportioned and the steeple is placed at the west end of the south aisle. The church is a Grade II* listed building. The Church of the Good Shepherd at Par Green was designed by E. H. Sedding and built of granite with Polyphant stone dressings in 1896. It is in the Early English style; the sanctuary was embellished by Stephen Dykes Bower in 1948.

Joseph Treffry
Before 1800 the village was a small group of houses below the cliff overlooking the mouth of the River Par; the river was crossed by a ferry. During the first years of the nineteenth century small scale workings of china stone, china clay, copper and granite were developed.

Joseph Austen, born 1782, was an important Fowey businessman; he later changed his name to Joseph Treffry, and as that name is much better known it is used here. He acquired an interest on many mines and pits, and he re-opened the dormant Lanescot copper mine on the hill overlooking Par, and developed it further. With adjacent workings it became the rich and highly productive Fowey Consols mine. Treffry sought to build a tramway connection to Fowey Harbour from his workings, but was unable to acquire the necessary land, and instead he decided to develop a harbour at Par. He purchased the ferry and replaced it with a bridge in 1824, and started improvement of the harbour in 1829; it was completed in 1840.

Par Canal
To bring the copper ore to Par, Treffry built a canal from Pontsmill to Par by canalising the river; he constructed a tramway on an inclined plane from Fowey Consols down to Pontsmill, so that Par harbour became a key location in the transport chain. The harbour development led to the expansion of Par, and the community was detached from the parish of St Blaise (later St Blazey) in the mid 19th century.

Treffry later built a new tramway up the Luxulyan Valley to Molinnis, and extended it down from Pontsmill to Par, by-passing the canal; this further developed the importance of Par. However copper was exported to Swansea for smelting and coal for powering mine engines were imported from there; this involved a difficult sea passage around Land's End, and Treffry announced his intention to continue his tramway to Newquay, on the north coast of Cornwall. This was not achieved in his lifetime.

China clay
In the fourth quarter of the nineteenth century, the importance of copper working had seriously diminished, due to exhaustion and the availability of cheaper supplies of the mineral elsewhere in the world. At the same time, china clay (kaolinite) became ever more important, and industrialisation of the extraction and processing work took place. This mineral became the dominant outward traffic at Par, and clay dries were erected in the vicinity (at Par Moors and elsewhere), together with further expansion of the harbour.

The opening of the Cornwall Railway from Plymouth in 1859 encouraged further expansion of Par north-eastwards towards Tywardreath. The boundaries between the three settlements are now somewhat indistinct.

In 1858 15,154 tons of china clay were shipped out of Par. By 1885 86,325 tons were being handled at Par, but by this time Fowey had a railway connection and handled 114,403 tons. In 1987 Par handled 700,000 tons, by 2002 the port served 284 vessels per year which were loaded with 318455 MT of china clay, and 107 vessels loaded with 136970 MT of secondary aggregates for the building trade.

The harbour developed a range of industrial facilities including a lead smelter with a 248 ft high chimney known as Par Stack. This was used as a navigation aid by shipping until it was demolished in 1907.

Present-day Harbour
A 450 ft breakwater encloses 35 acre of water which is tidal with only 16 ft depth of water and, unlike nearby Fowey, it cannot accommodate large ocean-going ships. The harbour is operated by the French mineral extraction company Imerys.

Today china clay is piped to the harbour in slurry form; most is dried in large sheds before exporting either from Par or Fowey, the two being linked by a private road. One berth at Par can also load clay slurry into coasting vessels. The harbour also has a rail link that is used to carry away dried clay loaded in rail vans.

A major reduction in china clay operations, announced on 4 July 2006, included proposals to close Par to commercial shipping and to close some of the clay dryers. The closures took effect in 2007. There were plans to re-develop the docks as part of the St Austell and Clay Country Eco-town. This would include a new marina and 500–700 homes.

Railways
The first railway in Par was the southwards extension of Treffry Tramways, a horse-operated mineral railway that connected Molinnis and pits and quarries in the Luxulyan Valley with Par. This opened in 1855, replacing the canal.

The Cornwall Railway opened from Plymouth to Truro on 4 May 1859 when Par railway station was opened to the north-east of Par, giving the village a connection to London Paddington. This involved the building of Par Viaduct to cross the old tramway. A siding was opened down to the harbour from the west end of the viaduct on 13 February 1860. The Cornwall Railway was built to the broad gauge of 7 ft but Treffry's line was to the, so that through running was not possible. The Cornwall Railway line was converted to standard gauge on 21 May 1892, from which time ordinary interworking became possible.

Treffry's tramway was taken over by the Cornwall Minerals Railway (CMR). It was upgraded for locomotive operation and extended to form a link throughout to Newquay and Fowey. It reopened in this form on 1 June 1874; passenger operation started in 1876.

Par harbour continued in importance in the 20th century; the restricted railway facilities on the site led to dedicated shunting locomotives, such as Bagnall 0-4-0ST "Alfred" and "Judy", being built specially for the work.

The railway line from Par to Fowey closed on 1 July 1968, and was converted to a private haul road linking the two harbours; it is now owned by Imerys. Par railway station is still open on the Cornish Main Line from Plymouth to Penzance. It is the junction for the Atlantic Coast Line local passenger train service to Newquay.

Today it is served by Great Western Railway services from London Paddington to Penzance as well as the Newquay line, while CrossCountry services operate north to Manchester and Glasgow.

Cornish wrestling
Cornish wrestling tournaments took place in the following locations all around Par Green over the last 200 years:
 * A field at the back of Par Green
 * Field at the back of the Welcome Home Inn which was a field behind the old foundry
 * Field behind the Par Green Post Office

title: Par, Cornwall