놈 (수학)

3. 타원 적분과의 관계

놈(nome)은 타원 함수와 모듈러 형식을 설명하는 값으로 자주 사용된다. 사분 주기가 타원 계수 $k$의 함수이므로 함수로 간주할 수도 있다.

:$q(k)\; =\; \backslash mathrm\{e\}^\{-\backslash pi\; K\text{'}(k)/K(k)\}$.

보조 놈 $q\_1$는 다음과 같이 주어진다.

:$q\_1(k)\; =\; \backslash mathrm\{e\}^\{-\backslash pi\; K(k)/K\text{'}(k)\}.\; \backslash ,$

때때로 놈의 ''제곱''에 대해 $q=\backslash mathrm\{e\}^\; \backslash pi\; \backslash tau\}$ 표기법이 사용된다.

언급된 함수 $K$와 $K\text{'}$는 제1종 완전 타원 적분이라고 한다. 다음과 같이 정의된다.

:$K(x)\; =\; \backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{1\}\{\backslash sqrt\{1\; -\; x^2\backslash sin(\backslash varphi)^2\}\}\; \backslash ,\backslash mathrm\{d\}\backslash varphi\; =\; \backslash int\_0^1\; \backslash frac\{2\}\{\backslash sqrt\{(y^2+1)^2\; -\; 4x^2y^2\}\}\; \backslash mathrm\{d\}y$

:$K\text{'}(x)\; =\; K(\backslash sqrt\{1\; -\; x^2\})\; =\; \backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{1\}\{\backslash sqrt\{1\; -\; (1\; -\; x^2)\backslash sin(\backslash varphi)^2\}\}\; \backslash ,\backslash mathrm\{d\}\backslash varphi$

놈 함수는 제1종 및 제2종 완전 타원 적분의 정의에 사용될 수 있다.

:$K(\backslash varepsilon)\; =\; \backslash tfrac\{1\}\{2\}\backslash pi\backslash ,\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^2$

:$E(\backslash varepsilon)\; =\; 2\backslash pi\; q(\backslash varepsilon)\backslash ,\backslash vartheta\_\{00\}\text{'}[q(\backslash varepsilon)]\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^\{-3\}\; +\; \backslash tfrac\{1\}\{2\}\backslash pi(1\; -\; \backslash varepsilon^2)\backslash ,\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^2$

이 경우, 지수 위치의 대시는 소위 세타 제로 값 함수의 도함수를 나타낸다.

:$\backslash vartheta\_\{00\}\text{'}(x)\; =\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\backslash ,\backslash vartheta\_\{00\}(x)\; =\; 2\; +\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; 2(n\; +\; 1)^2\; x^\{n(n+2)\}$

4. 무한 급수 표현

놈 함수는 리하르트 데데킨트에 의해 연구되었으며, 데데킨트 에타 함수 이론의 기초를 형성한다. 또한, 카를 구스타프 야코비의 세타 함수에서 놈은 산술 기하 평균의 대수적 조합과 제1종 완전 타원 적분에 할당된 가로좌표이다. 놈 함수는 다음과 같은 무한 급수로 표현될 수 있다.^[8]

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; q(x)^\{n^2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash vartheta\_\{00\}[q(x)]\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash operatorname\{agm\}(1-x;1+x)^\{-1/2\}\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; q(x)^\{(2n-1)^2\}\; =\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{00\}[q(x)]\; -\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{01\}[q(x)]\; =\; \backslash tfrac\{1\}\{4\}(1-\backslash sqrt[4]\{1-x^2\})\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{2q(x)^\{n\}\}\{q(x)^\{2n\}\; +\; 1\}\; =\; \backslash tfrac\{1\}\{2\}\backslash vartheta\_\{00\}[q(x)]^2\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash pi^\{-1\}K(x)\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{2q(x)^\{2n-1\}\}\{q(x)^\{4n-2\}\; +\; 1\}\; =\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{00\}[q(x)]^2\; -\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{01\}[q(x)]^2\; =\; \backslash tfrac\{1\}\{2\}(1-\backslash sqrt\{1-x^2\})\backslash ,\backslash pi^\{-1\}K(x)$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; n^2\; q(x)^\{n^2\}\; =\; 2^\{-1/2\}\backslash pi^\{-5/2\}K(x)^\{3/2\}[E(x)-(1-x^2)K(x)]$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash biggl[\backslash frac\{2q(x)^\{n\}\}\{1\; +\; q(x)^\{2n\}\}\backslash biggr]^2\; =\; 2\backslash pi^\{-2\}E(x)K(x)\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash biggl[\backslash frac\{2q(x)^\{n\}\}\{1\; -\; q(x)^\{2n\}\}\backslash biggr]^2\; =\; \backslash tfrac\{2\}\{3\}\backslash pi^\{-2\}(2\; -\; x^2)K(x)^2\; -\; 2\backslash pi^\{-2\}K(x)E(x)\; +\; \backslash tfrac\{1\}\{6\}$

여기서 $\backslash vartheta\_\{00\}(x)$와 $\backslash vartheta\_\{01\}(x)$는 야코비 세타 함수를, $K(x)$는 제1종 완전 타원 적분을, $E(x)$는 제2종 완전 타원 적분을, $\backslash operatorname\{agm\}(x;y)$는 산술 기하 평균을 나타낸다.

또한, 놈 함수는 다음과 관련된 무한곱으로도 표현 가능하다.

:$\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; [1-q(x)^\{2n\}][1+q(x)^\{2n-1\}]^2\; =\; \backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; [1-q(x)^\{2n\}][1-q(x)^\{2n-1\}]^2\; \backslash sqrt[4]\{1-x^2\}\backslash ,\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

4. 1. 코테쇼베츠(Kotěšovec) 급수

코테쇼베츠는 또한 $\backslash text\{Kt\}(n+1)\; =\; \backslash frac\{8\}\{n\}\backslash sum\; \_\{m\; =\; 1\}^\{n\}\; \backslash text\{Kt\}(m)\backslash ,\backslash text\{Kn\}(n\; +\; 1\; -\; m)$ 공식을 사용하여 $\backslash text\{Kt\}(n)$ 수열을 생성했다. 여기서 $\backslash text\{Kn\}(n)$은 크네저(Kneser) 수열(OEIS A227503)이다.

코테쇼베츠 수열 구성의 예:

: $\backslash mathrm\{Kt\}(4)\; =\; \backslash frac\{8\}\{3\}\backslash sum\; \_\{m\; =\; 1\}^\{3\}\; \backslash mathrm\{Kt\}(m)\; \backslash ,\backslash mathrm\{Kn\}(4\; -\; m)\; =\; \backslash frac\{8\}\{3\}\; \backslash bigl[\backslash mathrm\{Kt\}(1)\backslash ,\backslash mathrm\{Kn\}(3)\; +\; \backslash mathrm\{Kt\}(2)\backslash ,\backslash mathrm\{Kn\}(2)\; +\; \backslash mathrm\{Kt\}(3)\backslash ,\backslash mathrm\{Kn\}(1)\; \backslash bigr]$

: $\backslash mathrm\{Kt\}(4)\; =\; \backslash frac\{8\}\{3\}\; \backslash bigl(1\; \backslash times\; 184\; +\; 8\; \backslash times\; 13\; +\; 84\; \backslash times\; 1\; \backslash bigr)\; =\; 992$

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

: $\backslash mathrm\{Kt\}(5)\; =\; \backslash frac\{8\}\{4\}\; \backslash bigl(1\; \backslash times\; 2701\; +\; 8\; \backslash times\; 184\; +\; 84\; \backslash times\; 13\; +\; 992\; \backslash times\; 1\; \backslash bigr)\; =\; 12514$

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

: $\backslash mathrm\{Kt\}(6)\; =\; \backslash frac\{8\}\{5\}\; \backslash bigl(1\; \backslash times\; 40456\; +\; 8\; \backslash times\; 2701\; +\; 84\; \backslash times\; 184\; +\; 992\; \backslash times\; 13\; +\; 12514\; \backslash times\; 1\; \backslash bigr)\; =\; 164688$

놈 함수의 매클로린 급수는 다음과 같다.

:$q(x)\; =\; \backslash frac\{1\}\{16\}x^2\; +\; \backslash frac\{8\}\{256\}x^4\; +\; \backslash frac\{84\}\{4096\}x^6\; +\; \backslash frac\{992\}\{65536\}x^8\; +\; \backslash frac\{12514\}\{1048576\}x^\{10\}\; +\; \backslash ldots$

4. 2. 슈바르츠(Schwarz) 급수

코테쇼베츠 수열 ''$\backslash text\{Kt\}(n)$''는 수정된 아페리 시퀀스 ''$\backslash text\{Ap\}(n)$''를 사용하여 생성할 수 있다. ''$\backslash text\{Kt\}(n)$''의 시작 값은 ''$\backslash text\{Kt\}(1)=1$''이며, 이후 값은 다음 두 수식으로 생성된다.

:$\backslash text\{Kt\}(1)\; =\; 1$

:$\backslash text\{Kt\}(n+1)\; =\; \backslash frac\{1\}\{n\}\; \backslash sum\_\{k\; =\; 1\}^\{n\}\; k\backslash ,\backslash text\{Kt\}(k)[16\backslash ,\backslash text\{Ap\}(n+1-k)\; -\; \backslash text\{Ap\}(n+2-k)]$

:$\backslash text\{Ap\}(n)\; =\; \backslash sum\_\{a\; =\; 0\}^\{n-1\}\; \backslash binom\{2a\}\{a\}^2\; \backslash binom\{2n-2-2a\}\{n-1-\; a\}^2$

정수 시퀀스 $Sw(n)$은 Hermann Amandus Schwarz 수를 나타낸다. 슈바르츠는 자신의 저서 ''"Formeln und Lehrsätze zum Gebrauche der elliptischen Funktionen"''의 ''"Berechnung der Grösse k"'' 장에서 이 정수 수열을 썼다. 이 슈바르츠 수열은 Karl Theodor Wilhelm Weierstrass와 Louis Melville Milne-Thomson에 의해 분석되었다. Adolf Kneser는 다음 패턴을 기반으로 이 시퀀스에 대한 합성 방법을 결정했다.

:$\backslash text\{Sw\}(1)\; =\; 1$

:$\backslash text\{Sw\}(n+1)\; =\; \backslash frac\{2\}\{n\}\backslash sum\; \_\{m\; =\; 1\}^\{n\}\; \backslash text\{Sw\}(m)\backslash text\{Kn\}(n\; +\; 1\; -\; m)$

Schwarz 시퀀스 $Sw(n)$은 A002103, Kneser 시퀀스 $Kn(n)$은 A227503으로 온라인 정수열 백과사전에 등재되어 있다. Kneser 정수 시퀀스 $Kn(n)$은 다음과 같이 정의된 특수 Apéry 시퀀스 $Ap(n)$을 사용하여 구성할 수 있다.

:$\backslash text\{Ap\}(n)\; =\; \backslash sum\_\{a\; =\; 0\}^\{n-1\}\; \backslash binom\{2a\}\{a\}^2\; \backslash binom\{2n-2-2a\}\{n-1-\; a\}^2$

이러한 방식으로 모든 자연수 n에 대해 Kneser 수열을 정의할 수 있다.

:$\backslash text\{Kn\}(n\; +\; 1)\; =\; 2^\{4n\; +\; 1\}\; -\; \backslash tfrac\{1\}\{8\}\backslash text\{Ap\}(n\; +\; 2)\; -\; \backslash sum\_\{b\; =\; 1\}^\; \{n\}\; \backslash text\{Kn\}(b)\backslash text\{Ap\}(n\; +\; 2\; -\; b)$

Kneser 시퀀스는 생성 함수에 의해 다음과 같이 생성될 수도 있다.

:$\backslash frac\{\backslash pi^2\}\{8x(1\; -\; x^2)K(x)^2\}\; -\; \backslash frac\{1\}\{2x\}\; =\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{\backslash text\{Kn\}(n)\}\{2^\{4n\; -\; 2\}\}x^\{2n\; -\; 1\}$

다음 표에는 슈바르츠 수와 크네저 수, 아페리 수가 포함되어 있다.

수학자 Karl Heinrich Schellbach|카를 하인리히 쉘바흐^de는 4제곱근의 몫에 대한 매클로린 급수에서 나타나는 정수 수열을 발견했다.^[1] 이 몫은 정수함수 '''타원 놈'''을 제곱 함수로 나눈 것이다. 이 수열의 구성은 그의 저서 ''타원 적분과 세타 함수의 이론''(Die Lehre von den Elliptischen Integralen und den Thetafunktionen)에 상세히 설명되어 있다.^[2] 이 수열은 또한 헤르만 아만두스 슈바르츠에 의해 ''타원 함수의 사용을 위한 공식 및 정리''(Formeln und Lehrsätze zum Gebrauche der elliptischen Funktionen)에서 구성되었다.^[3] 이 쉘바흐 슈바르츠 수열 Sc(n)은 카를 테오도어 빌헬름 바이어슈트라스와 루이 멜빌 밀른-톰슨과 같은 수학자들에 의해서도 분석되었다. 아돌프 크네저는 다음 패턴을 기반으로 이 수열의 구성을 결정했다.

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

쉘바흐 슈바르츠 수열 Sc(n)은 온라인 정수열 백과사전(On-Line Encyclopedia of Integer Sequences)에서 번호 A002103으로 나타나고, 크네저 수열 Kn(n)은 번호 A227503으로 나타난다.

다음 표는 크네저 수와 쉘바흐 슈바르츠 수를 포함한다.

이 수열은 정확히 다음과 같은 방식으로 타원 놈의 매클로린 급수를 생성한다.^[4][5][6]

:$q(x)\; =\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{\backslash text\{Sc\}(n)\}\{2^\{4n\; -\; 3\}\}\; \backslash biggl(\backslash frac\{1\; -\; \backslash sqrt[4]\{1\; -\; x^2\}\}\{1\; +\; \backslash sqrt[4]\{1\; -\; x^2\}\}\backslash biggr)^\{4n\; -\; 3\}\; =\; x^2\backslash biggl\backslash \{\backslash frac\{1\}\{2\}\; +\; \backslash biggl[\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{\backslash text\{Sc\}(n\; +\; 1)\}\{2^\{4n\; +\; 1\}\}\; x^\{2n\}\backslash biggr]\backslash biggr\backslash \}^4$

:$q(x)\; =\; x^2\backslash bigl(\backslash frac\{1\}\{2\}\; +\; \backslash frac\{2\}\{32\}x^2\; +\; \backslash frac\{15\}\{512\}x^4\; +\; \backslash frac\{150\}\{8192\}x^6\; +\; \backslash frac\{1707\}\{131072\}x^8\; +\; \backslash ldots\backslash bigr)^4$

다음은 쉘바흐 슈바르츠 수가 어떻게 순차적으로 구성되는지에 대한 예시이다.

:$\backslash mathrm\{Sc\}(4)\; =\; \backslash frac\{2\}\{3\}\backslash sum\; \_\{m\; =\; 1\}^\{3\}\; \backslash mathrm\{Sc\}(m)\; \backslash ,\backslash mathrm\{Kn\}(4\; -\; m)\; =\; \backslash frac\{2\}\{3\}\; \backslash bigl[\backslash mathrm\{Sc\}(1)\backslash ,\backslash mathrm\{Kn\}(3)\; +\; \backslash mathrm\{Sc\}(2)\backslash ,\backslash mathrm\{Kn\}(2)\; +\; \backslash mathrm\{Sc\}(3)\backslash ,\backslash mathrm\{Kn\}(1)\; \backslash bigr]$

:$\backslash mathrm\{Sc\}(4)\; =\; \backslash frac\{2\}\{3\}\; \backslash bigl(1\; \backslash times\; 184\; +\; 2\; \backslash times\; 13\; +\; 15\; \backslash times\; 1\; \backslash bigr)\; =\; 150$

:$\backslash mathrm\{Sc\}(5)\; =\; \backslash frac\{2\}\{4\}\backslash sum\; \_\{m\; =\; 1\}^\{4\}\; \backslash mathrm\{Sc\}(m)\; \backslash ,\backslash mathrm\{Kn\}(5\; -\; m)\; =\; \backslash frac\{2\}\{4\}\; \backslash bigl[\backslash mathrm\{Sc\}(1)\backslash ,\backslash mathrm\{Kn\}(4)\; +\; \backslash mathrm\{Sc\}(2)\backslash ,\backslash mathrm\{Kn\}(3)\; +\; \backslash mathrm\{Sc\}(3)\backslash ,\backslash mathrm\{Kn\}(2)\; +\; \backslash mathrm\{Sc\}(4)\backslash ,\backslash mathrm\{Kn\}(1)\; \backslash bigr]$

:$\backslash mathrm\{Sc\}(5)\; =\; \backslash frac\{2\}\{4\}\; \backslash bigl(1\; \backslash times\; 2701\; +\; 2\; \backslash times\; 184\; +\; 15\; \backslash times\; 13\; +\; 150\; \backslash times\; 1\; \backslash bigr)\; =\; 1707$

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

:$\backslash mathrm\{Sc\}(6)\; =\; \backslash frac\{2\}\{5\}\; \backslash bigl(1\; \backslash times\; 40456\; +\; 2\; \backslash times\; 2701\; +\; 15\; \backslash times\; 184\; +\; 150\; \backslash times\; 13\; +\; 1707\; \backslash times\; 1\; \backslash bigr)\; =\; 20910$

5. 특정 값

다음은 렘니스케이트 값이다.^[1]

: $q(0)\; =\; 0$

: $q(1)\; =\; 1$

: $q(-1)\; =\; 1$

: $q(\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})\; =\; \backslash text\{e\}^\{-\backslash pi\}$

: $q[(\backslash sqrt\{2\}\; -\; 1)^2]\; =\; \backslash text\{e\}^\{-2\backslash pi\}$

: $q[2\backslash sqrt[4]\{2\}(\backslash sqrt\{2\}\; -\; 1)]\; =\; \backslash exp(-\backslash tfrac\{1\}\{2\}\backslash pi)$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{3\}\; -\; 1)(\backslash sqrt\{2\}\; -\; \backslash sqrt[4]\{3\})]\; =\; \backslash text\{e\}^\{-3\backslash pi\}$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{3\}\; -\; 1)(\backslash sqrt\{2\}\; +\; \backslash sqrt[4]\{3\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash pi)$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{10\}\; -\; 2\backslash sqrt\{2\})(3\; -\; 2\backslash sqrt[4]\{5\})]\; =\; \backslash text\{e\}^\{-5\backslash pi\}$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{10\}\; -\; 2\backslash sqrt\{2\})(3\; +\; 2\backslash sqrt[4]\{5\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{5\}\backslash pi)$

일부 비 렘니스케이트 값은 다음과 같다.^[1]

: $q[\backslash tfrac\{1\}\{4\}(\backslash sqrt\{6\}\; -\; \backslash sqrt\{2\})]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{3\}\backslash pi\}$

: $q[\backslash tfrac\{1\}\{4\}(\backslash sqrt\{6\}\; +\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash sqrt\{3\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash sin\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl(\backslash sqrt\{5\}\; -\; 2\backslash bigr)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-\backslash sqrt\{5\}\backslash pi\}$

: $q\backslash bigl\backslash \{\backslash cos\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl(\backslash sqrt\{5\}\; -\; 2\backslash bigr)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{5\}\backslash sqrt\{5\}\backslash ,\backslash pi)$

: $q[\backslash tfrac\{1\}\{8\}(3\backslash sqrt\{2\}\; -\; \backslash sqrt\{14\})]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{7\}\backslash pi\}$

: $q[\backslash tfrac\{1\}\{8\}(3\backslash sqrt\{2\}\; +\; \backslash sqrt\{14\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{7\}\backslash sqrt\{7\}\backslash ,\backslash pi)$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash bigl(\backslash sqrt\{22\}\; +\; 3\backslash sqrt\{2\}\backslash bigr)\backslash bigl(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; +\; 2\backslash sqrt\{11\}\}\; -\; \backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; -\; 2\backslash sqrt\{11\}\}\; +\; \backslash tfrac\{1\}\{3\}\backslash sqrt\{11\}\; -\; 1\backslash bigr)^4\backslash bigr]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{11\}\backslash pi\}$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash bigl(\backslash sqrt\{22\}\; -\; 3\backslash sqrt\{2\}\backslash bigr)\backslash bigl(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; +\; 2\backslash sqrt\{11\}\}\; -\; \backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; -\; 2\backslash sqrt\{11\}\}\; +\; \backslash tfrac\{1\}\{3\}\backslash sqrt\{11\}\; +\; 1\backslash bigr)^4\backslash bigr]\; =\; \backslash exp(-\backslash tfrac\{1\}\{11\}\backslash sqrt\{11\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash sin\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl(5\backslash sqrt\{13\}\; -\; 18\backslash bigr)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-\backslash sqrt\{13\}\backslash pi\}$

: $q\backslash bigl\backslash \{\backslash cos\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl(5\backslash sqrt\{13\}\; -\; 18\backslash bigr)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{13\}\backslash sqrt\{13\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash langle\backslash sin\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl[\backslash bigl(\backslash tfrac\{1\}\{4\}\backslash sqrt\{17\}\; +\; \backslash tfrac\{1\}\{4\}\; -\; \backslash tfrac\{1\}\{4\}\backslash sqrt\{2\backslash sqrt\{17\}\; +\; 2\}\backslash bigr)^6\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash text\{e\}^\{-\backslash sqrt\{17\}\backslash pi\}$

: $q\backslash bigl\backslash langle\backslash cos\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl[\backslash bigl(\backslash tfrac\{1\}\{4\}\backslash sqrt\{17\}\; +\; \backslash tfrac\{1\}\{4\}\; -\; \backslash tfrac\{1\}\{4\}\backslash sqrt\{2\backslash sqrt\{17\}\; +\; 2\}\backslash bigr)^6\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash exp(-\backslash tfrac\{1\}\{17\}\backslash sqrt\{17\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash langle\backslash sin\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl[(\backslash sqrt\{37\}\; -\; 6)^3\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash text\{e\}^\{-\backslash sqrt\{37\}\backslash pi\}$

: $q\backslash bigl\backslash langle\backslash cos\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl[(\backslash sqrt\{37\}\; -\; 6)^3\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash exp(-\backslash tfrac\{1\}\{37\}\backslash sqrt\{37\}\backslash ,\backslash pi)$

다음 값은 홀수의 제곱근과 함께 겔폰트 상수의 역수의 거듭제곱으로 발생한다.^[1]

: $q(\backslash sqrt\{2\}\; -\; 1)\; =\; \backslash text\{e\}^\{-\backslash sqrt\{2\}\backslash pi\}$

: $q(\backslash sqrt\{2\backslash sqrt\{2\}\; -\; 2\})\; =\; \backslash exp(-\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\}\backslash ,\backslash pi)$

: $q[(2\; -\; \backslash sqrt\{3\})(\backslash sqrt\{3\}\; -\; \backslash sqrt\{2\})]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{6\}\backslash pi\}$

: $q[(2\; -\; \backslash sqrt\{3\})(\backslash sqrt\{3\}\; +\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash sqrt\{6\}\backslash ,\backslash pi)$

: $q[(\backslash sqrt\{10\}\; -\; 3)(\backslash sqrt\{2\}\; -\; 1)^2]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{10\}\backslash pi\}$

: $q[(\backslash sqrt\{10\}\; -\; 3)(\backslash sqrt\{2\}\; +\; 1)^2]\; =\; \backslash exp(-\backslash tfrac\{1\}\{5\}\backslash sqrt\{10\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash langle\backslash tan\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arctan\backslash bigl[\backslash bigl(\backslash sqrt\{2\}\; +\; \backslash tfrac\{1\}\{2\}\; -\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{4\backslash sqrt\{2\}\; +\; 5\}\backslash bigr)^3\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash text\{e\}^\{-\backslash sqrt\{14\}\backslash pi\}$

: $q\backslash bigl\backslash langle\backslash tan\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arctan\backslash bigl[\backslash bigl(\backslash sqrt\{2\}\; +\; \backslash tfrac\{1\}\{2\}\; +\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{4\backslash sqrt\{2\}\; +\; 5\}\backslash bigr)^3\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash exp(-\backslash tfrac\{1\}\{7\}\backslash sqrt\{14\}\backslash ,\backslash pi)$

: $q[(10\; -\; 3\backslash sqrt\{11\})(3\backslash sqrt\{11\}\; -\; 7\backslash sqrt\{2\})]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{22\}\backslash pi\}$

: $q[(10\; -\; 3\backslash sqrt\{11\})(3\backslash sqrt\{11\}\; +\; 7\backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{11\}\backslash sqrt\{22\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{(\backslash sqrt\{26\}+5)(\backslash sqrt\{2\}-1)^2\; \backslash tan\backslash bigl[\backslash tfrac\{1\}\{4\}\backslash pi-\backslash arctan(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}+\backslash sqrt\{26\}\}-\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}-\backslash sqrt\{26\}\}+\backslash tfrac\{1\}\{6\}\backslash sqrt\{26\}-\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})\backslash bigr]^4\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-\backslash sqrt\{26\}\backslash pi\}$

: $q\backslash bigl\backslash \{(\backslash sqrt\{26\}+5)(\backslash sqrt\{2\}+1)^2\; \backslash tan\backslash bigl[\backslash arctan(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}+\backslash sqrt\{26\}\}-\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}-\backslash sqrt\{26\}\}+\backslash tfrac\{1\}\{6\}\backslash sqrt\{26\}+\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})-\backslash tfrac\{1\}\{4\}\backslash pi\backslash bigr]^4\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{13\}\backslash sqrt\{26\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash langle\backslash tan\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arctan\backslash bigl[\backslash bigl(\backslash tfrac\{1\}\{4\}\backslash sqrt\{17\}\; +\; \backslash tfrac\{3\}\{4\}\; -\; \backslash tfrac\{1\}\{4\}\backslash sqrt\{6\backslash sqrt\{17\}\; +\; 10\}\backslash bigr)^6\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash text\{e\}^\{-\backslash sqrt\{34\}\backslash pi\}$

: $q\backslash bigl\backslash langle\backslash tan\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash arctan\backslash bigl[\backslash bigl(\backslash tfrac\{1\}\{4\}\backslash sqrt\{17\}\; +\; \backslash tfrac\{3\}\{4\}\; +\; \backslash tfrac\{1\}\{4\}\backslash sqrt\{6\backslash sqrt\{17\}\; +\; 10\}\backslash bigr)^6\backslash bigr]\backslash bigr\backslash \}\backslash bigr\backslash rangle\; =\; \backslash exp(-\backslash tfrac\{1\}\{17\}\backslash sqrt\{34\}\backslash ,\backslash pi)$

: $q[(13\backslash sqrt\{58\}\; -\; 99)(\backslash sqrt\{2\}\; -\; 1)^6]\; =\; \backslash text\{e\}^\{-\backslash sqrt\{58\}\backslash pi\}$

: $q[(13\backslash sqrt\{58\}\; -\; 99)(\backslash sqrt\{2\}\; +\; 1)^6]\; =\; \backslash exp(-\backslash tfrac\{1\}\{29\}\backslash sqrt\{58\}\backslash ,\backslash pi)$

놈을 제곱하면 다음 값이 된다.^[1]

: $q\backslash bigl\backslash \{\backslash tan\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin\backslash bigl(\backslash sqrt\{2\}\; -\; 1\backslash bigr)\backslash bigr]^2\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-2\backslash sqrt\{2\}\backslash pi\}$

: $q[(\backslash sqrt\{3\}\; -\; \backslash sqrt\{2\})^2(\backslash sqrt\{2\}\; -\; 1)^2]\; =\; \backslash text\{e\}^\{-2\backslash sqrt\{3\}\backslash pi\}$

: $q[(\backslash sqrt\{3\}\; -\; \backslash sqrt\{2\})^2(\backslash sqrt\{2\}\; +\; 1)^2]\; =\; \backslash exp(-\backslash tfrac\{2\}\{3\}\backslash sqrt\{3\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash tan\backslash bigl[\backslash tfrac\{1\}\{4\}\backslash arcsin\backslash bigl(\backslash sqrt\{5\}\; -\; 2\backslash bigr)\backslash bigr]^2\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-2\backslash sqrt\{5\}\backslash pi\}$

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

: $q[(\backslash sqrt\{2\}\; -\; 1)^4\; (2\backslash sqrt\{2\}\; -\; \backslash sqrt\{7\})^2]\; =\; \backslash text\{e\}^\{-2\backslash sqrt\{7\}\backslash pi\}$

: $q[(\backslash sqrt\{2\}\; -\; 1)^4\; (2\backslash sqrt\{2\}\; +\; \backslash sqrt\{7\})^2]\; =\; \backslash exp(-\backslash tfrac\{2\}\{7\}\backslash sqrt\{7\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash tan\backslash bigl[\backslash tfrac\{1\}\{4\}\backslash arcsin\backslash bigl(5\backslash sqrt\{13\}\; -\; 18\backslash bigr)\backslash bigr]^2\backslash bigr\backslash \}\; =\; \backslash text\{e\}^\{-2\backslash sqrt\{13\}\backslash pi\}$

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

가치의 특정 사중주는 아래와 같다.^[1]

다음 두 목록에는 놈 함수의 많은 함수 값이 포함되어 있다.^[1]

첫 번째 목록은 서로 피타고라스 보완 모듈을 가진 값의 쌍을 보여준다.

: $q(\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})\; =\; \backslash exp(-\backslash pi)$

: $q[\backslash tfrac\{1\}\{4\}(\backslash sqrt\{6\}\; -\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash sqrt\{3\}\backslash ,\backslash pi)$

: $q[\backslash tfrac\{1\}\{4\}(\backslash sqrt\{6\}\; +\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash sqrt\{3\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash sin\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin(\backslash sqrt\{5\}\; -\; 2)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash sqrt\{5\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash cos\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin(\backslash sqrt\{5\}\; -\; 2)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{5\}\backslash sqrt\{5\}\backslash ,\backslash pi)$

: $q[\backslash tfrac\{1\}\{8\}(3\backslash sqrt\{2\}\; -\; \backslash sqrt\{14\})]\; =\; \backslash exp(-\backslash sqrt\{7\}\backslash ,\backslash pi)$

: $q[\backslash tfrac\{1\}\{8\}(3\backslash sqrt\{2\}\; +\; \backslash sqrt\{14\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{7\}\backslash sqrt\{7\}\backslash ,\backslash pi)$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{3\}\; -\; 1)(\backslash sqrt\{2\}\; -\; \backslash sqrt[4]\{3\})]\; =\; \backslash exp(-3\backslash pi)$

: $q[\backslash tfrac\{1\}\{2\}(\backslash sqrt\{3\}\; -\; 1)(\backslash sqrt\{2\}\; +\; \backslash sqrt[4]\{3\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash pi)$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash bigl(\backslash sqrt\{22\}\; +\; 3\backslash sqrt\{2\}\backslash bigr)\backslash bigl(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; +\; 2\backslash sqrt\{11\}\}\; -\; \backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; -\; 2\backslash sqrt\{11\}\}\; +\; \backslash tfrac\{1\}\{3\}\backslash sqrt\{11\}\; -\; 1\backslash bigr)^4\backslash bigr]\; =\; \backslash exp(-\backslash sqrt\{11\}\backslash ,\backslash pi)$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash bigl(\backslash sqrt\{22\}\; -\; 3\backslash sqrt\{2\}\backslash bigr)\backslash bigl(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; +\; 2\backslash sqrt\{11\}\}\; -\; \backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{6\backslash sqrt\{3\}\; -\; 2\backslash sqrt\{11\}\}\; +\; \backslash tfrac\{1\}\{3\}\backslash sqrt\{11\}\; +\; 1\backslash bigr)^4\backslash bigr]\; =\; \backslash exp(-\backslash tfrac\{1\}\{11\}\backslash sqrt\{11\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash sin\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin(5\backslash sqrt\{13\}\; -\; 18)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash sqrt\{13\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{\backslash cos\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin(5\backslash sqrt\{13\}\; -\; 18)\backslash bigr]\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{13\}\backslash sqrt\{13\}\backslash ,\backslash pi)$

두 번째 목록은 서로 접선 보완 모듈을 가진 값의 쌍을 보여준다.

: $q(\backslash sqrt\{2\}\; -\; 1)\; =\; \backslash exp(-\backslash sqrt\{2\}\backslash ,\backslash pi)$

: $q[(2\; -\; \backslash sqrt\{3\})(\backslash sqrt\{3\}\; -\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash sqrt\{6\}\backslash ,\backslash pi)$

: $q[(2\; -\; \backslash sqrt\{3\})(\backslash sqrt\{3\}\; +\; \backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash sqrt\{6\}\backslash ,\backslash pi)$

: $q[(\backslash sqrt\{10\}\; -\; 3)(\backslash sqrt\{2\}\; -\; 1)^2]\; =\; \backslash exp(-\backslash sqrt\{10\}\backslash ,\backslash pi)$

: $q[(\backslash sqrt\{10\}\; -\; 3)(\backslash sqrt\{2\}\; +\; 1)^2]\; =\; \backslash exp(-\backslash tfrac\{1\}\{5\}\backslash sqrt\{10\}\backslash ,\backslash pi)$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash sqrt\{2\backslash sqrt\{2\}\; -\; \backslash sqrt\{7\}\}\backslash ,(3\backslash sqrt\{2\}\; -\; \backslash sqrt\{14\})(\backslash sqrt\{2\backslash sqrt\{2\}\; +\; 1\}\; -\; 1)^4\backslash bigr]\; =\; \backslash exp(-\backslash sqrt\{14\}\backslash ,\backslash pi)$

: $q\backslash bigl[\backslash tfrac\{1\}\{16\}\backslash sqrt\{2\backslash sqrt\{2\}\; +\; \backslash sqrt\{7\}\}\backslash ,(3\backslash sqrt\{2\}\; +\; \backslash sqrt\{14\})(\backslash sqrt\{2\backslash sqrt\{2\}\; +\; 1\}\; -\; 1)^4\backslash bigr]\; =\; \backslash exp(-\backslash tfrac\{1\}\{7\}\backslash sqrt\{14\}\backslash ,\backslash pi)$

: $q[(2\; -\; \backslash sqrt\{3\})^2\; (\backslash sqrt\{2\}\; -\; 1)^3]\; =\; \backslash exp(-3\backslash sqrt\{2\}\backslash ,\backslash pi)$

: $q[(2\; +\; \backslash sqrt\{3\})^2\; (\backslash sqrt\{2\}\; -\; 1)^3]\; =\; \backslash exp(-\backslash tfrac\{1\}\{3\}\backslash sqrt\{2\}\backslash ,\backslash pi)$

: $q[(10\; -\; 3\backslash sqrt\{11\})(3\backslash sqrt\{11\}\; -\; 7\backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash sqrt\{22\}\backslash ,\backslash pi)$

: $q[(10\; -\; 3\backslash sqrt\{11\})(3\backslash sqrt\{11\}\; +\; 7\backslash sqrt\{2\})]\; =\; \backslash exp(-\backslash tfrac\{1\}\{11\}\backslash sqrt\{22\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{(\backslash sqrt\{26\}+5)(\backslash sqrt\{2\}-1)^2\; \backslash tan\backslash bigl[\backslash tfrac\{1\}\{4\}\backslash pi-\backslash arctan(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}+\backslash sqrt\{26\}\}-\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}-\backslash sqrt\{26\}\}+\backslash tfrac\{1\}\{6\}\backslash sqrt\{26\}-\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})\backslash bigr]^4\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash sqrt\{26\}\backslash ,\backslash pi)$

: $q\backslash bigl\backslash \{(\backslash sqrt\{26\}+5)(\backslash sqrt\{2\}+1)^2\; \backslash tan\backslash bigl[\backslash arctan(\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}+\backslash sqrt\{26\}\}-\backslash tfrac\{1\}\{3\}\backslash sqrt[3]\{3\backslash sqrt\{3\}-\backslash sqrt\{26\}\}+\backslash tfrac\{1\}\{6\}\backslash sqrt\{26\}+\backslash tfrac\{1\}\{2\}\backslash sqrt\{2\})-\backslash tfrac\{1\}\{4\}\backslash pi\backslash bigr]^4\backslash bigr\backslash \}\; =\; \backslash exp(-\backslash tfrac\{1\}\{13\}\backslash sqrt\{26\}\backslash ,\backslash pi)$

관련된 값의 사중주는 다음과 같다.^[1]

6. 지수 법칙

대수 숫자의 놈을 밑수로 하고 양의 유리수를 지수로 사용하는 모든 거듭제곱은 대수 숫자의 놈이 된다.^[11][12]

:$q(\backslash varepsilon\_1\; \backslash in\; \backslash mathbb\{A\}^\{+\})^\{w\; \backslash in\; \backslash mathbb\{Q^\{+\}\}\}\; =\; q(\backslash varepsilon\_2\; \backslash in\; \backslash mathbb\{A\}^\{+\})$

여기서 $\backslash operatorname\{sn\}$은 야코비 타원 함수의 진폭 사인(amplitude sine)을 나타내며, ''K''는 제1종 완전 타원 적분, ε는 타원 모듈러스, ''n''은 자연수이다. 실수 구간 $[-1,1]$에서 대수적인 $x$ 값에 대해, 표시된 진폭 사인 표현식은 항상 대수적이다.

일반적인 거듭제곱 정리는 다음과 같다.

:$q(\backslash varepsilon)^\{2n\}\; =\; q\backslash biggl\backslash \{\backslash varepsilon^\{2n\}\backslash prod\_\{k\; =\; 1\}^\{n\}\backslash operatorname\{sn\}\backslash bigl[\backslash tfrac\{2k-1\}\{2n\}K(\backslash varepsilon);\backslash varepsilon\backslash bigr]^4\backslash biggr\backslash \}$

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

이 정리는 모든 자연수 ''n''에 대해 유효하며, 다음과 같은 예시들을 포함한다.

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

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

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

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

:$q(\backslash varepsilon)^6\; =\; q\backslash \{\backslash varepsilon^6\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{6\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{2\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{5\}\{6\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash \}$

:$q(\backslash varepsilon)^7\; =\; q\backslash \{\backslash varepsilon^7\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{7\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{3\}\{7\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{5\}\{7\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash \}$

:$q(\backslash varepsilon)^8\; =\; q\backslash \{\backslash varepsilon^8\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{8\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{3\}\{8\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{5\}\{8\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{7\}\{8\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash \}$

:$q(\backslash varepsilon)^9\; =\; q\backslash \{\backslash varepsilon^9\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{9\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{3\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{5\}\{9\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{7\}\{9\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash \}$
계산 힌트:야코비 진폭 사인 표현식은 다음 방정식들을 풀 수 있다.^[17][18][19]

다섯 제곱 정리 예시::$q(\backslash varepsilon)^5\; =\; q\backslash \{\backslash varepsilon^5\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{5\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash operatorname\{sn\}[\backslash tfrac\{3\}\{5\}K(\backslash varepsilon);\backslash varepsilon]^4\backslash \}$

• 렘니스케이트 예시:

• 일반 예시:

6. 1. 제곱 법칙

랜든 딸 계수는 또한 어미 계수의 피타고라스 짝의 접선 대응이다.

위 방정식에서 대체와 미분을 통해 다음 표현을 유도할 수 있다.

:$(1\; +\; \backslash sqrt\{1\; -\; \backslash varepsilon^2\})K(\backslash varepsilon)\; =\; 2K\backslash bigl[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}\backslash bigr]$

:$K\backslash bigl[2\backslash sqrt[4]\{1\; -\; \backslash varepsilon^2\}(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-1\}\backslash bigr]\; =\; (1\; +\; \backslash sqrt\{1\; -\; \backslash varepsilon^2\})K(\backslash sqrt\{1-\backslash varepsilon^2\})$

위 두 공식을 결합하면 주기 비율을 나타내는 몫 방정식을 얻는다.

:$2\backslash ,\backslash frac\{K(\backslash sqrt\{1-\backslash varepsilon^2\})\}\{K(\backslash varepsilon)\}\; =\; \backslash frac\{K[2\backslash sqrt[4]\{1\; -\; \backslash varepsilon^2\}(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-1\}]\}\{K[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}]\}$

타원 놈은 음의 원 숫자에 실제 주기 비율을 곱한 지수 함수로 정의되며, 실제 주기 비율은 피타고라스 보완 계수의 K 적분을 계수 자체의 K 적분으로 나눈 몫으로 정의된다.

이러한 정의를 바탕으로 다음 등식이 성립한다.

:$q(\backslash varepsilon)^2\; =\; \backslash exp\backslash biggl[-\backslash pi\; \backslash ,\backslash frac\{K\text{'}(\backslash varepsilon)\}\{K(\backslash varepsilon)\}\backslash biggr]^2\; =\; \backslash exp\backslash biggl[-\backslash pi\; \backslash ,\backslash frac\{K(\backslash sqrt\{1\; -\; \backslash varepsilon^2\})\}\{K(\backslash varepsilon)\}\backslash biggr]^2\; =\; \backslash exp\backslash biggl[-2\backslash pi\; \backslash ,\backslash frac\{K(\backslash sqrt\{1\; -\; \backslash varepsilon^2\})\}\{K(\backslash varepsilon)\}\backslash biggr]\; =$

:$=\; \backslash exp\backslash biggl\backslash \{-\backslash pi\; \backslash ,\backslash frac\{K[2\backslash sqrt[4]\{1\; -\; \backslash varepsilon^2\}(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-1\}]\}\{K[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}]\}\backslash biggr\backslash \}\; =\; \backslash exp\backslash biggl\backslash \{-\backslash pi\; \backslash ,\backslash frac\{K\text{'}[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}]\}\{K[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}]\}\backslash biggr\backslash \}\; =\; q\backslash bigl[\backslash varepsilon^2(1+\backslash sqrt\{1-\backslash varepsilon^2\})^\{-2\}\backslash bigr]$

다음은 제곱 법칙을 보여주는 몇 가지 예시이다.

삼각법으로 표시된 예시:

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

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

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

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

쌍곡선으로 표시된 예시:

:$\backslash exp(-2\backslash sqrt\{6\}\backslash ,\backslash pi)\; =\; \backslash exp(-\backslash sqrt\{6\}\backslash ,\backslash pi)^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; -\; 1)^2\backslash bigr]\backslash bigr\backslash \}\backslash biggr\backslash rangle^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{4\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; -\; 1)^2\backslash bigr]\backslash bigr\backslash \}^2\backslash biggr\backslash rangle$

:$\backslash exp(-2\backslash sqrt\{10\}\backslash ,\backslash pi)\; =\; \backslash exp(-\backslash sqrt\{10\}\backslash ,\backslash pi)^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{5\}\; -\; 2)^2\backslash bigr]\backslash bigr\backslash \}\backslash biggr\backslash rangle^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{4\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{5\}\; -\; 2)^2\backslash bigr]\backslash bigr\backslash \}^2\backslash biggr\backslash rangle$

:$\backslash exp(-2\backslash sqrt\{14\}\backslash ,\backslash pi)\; =\; \backslash exp(-\backslash sqrt\{14\}\backslash ,\backslash pi)^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; +\; \backslash tfrac\{1\}\{2\}\; -\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{4\backslash sqrt\{2\}\; +\; 5\})^3\backslash bigr]\backslash bigr\backslash \}\backslash biggr\backslash rangle^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{4\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; +\; \backslash tfrac\{1\}\{2\}\; -\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{4\backslash sqrt\{2\}\; +\; 5\})^3\backslash bigr]\backslash bigr\backslash \}^2\backslash biggr\backslash rangle$

:$\backslash exp(-2\backslash sqrt\{22\}\backslash ,\backslash pi)\; =\; \backslash exp(-\backslash sqrt\{22\}\backslash ,\backslash pi)^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{2\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; -\; 1)^6\backslash bigr]\backslash bigr\backslash \}\backslash biggr\backslash rangle^2\; =\; q\backslash biggl\backslash langle\backslash operatorname\{tanh\}\backslash bigl\backslash \{\backslash tfrac\{1\}\{4\}\backslash operatorname\{arsinh\}\backslash bigl[(\backslash sqrt\{2\}\; -\; 1)^6\backslash bigr]\backslash bigr\backslash \}^2\backslash biggr\backslash rangle$

6. 2. 세제곱 법칙

예시:

• t = 1일 때:

• t = Φ^-2 = (3 - √5)/2 (Φ는 황금비)일 때:

7. 미분

타원 놈 함수 $q(x)$의 미분은 다음과 같이 표현된다.^[1]

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; q(x)\; =\; \backslash frac\{\backslash pi^2\}\{2x(1-x^2)K(x)^2\}\; q(x)$

여기서 $K(x)$는 제1종 완전 타원 적분이다.

이계 도함수와 삼계 도함수는 다음과 같다.^[1]

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

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

여기서 $E(x)$는 제2종 완전 타원 적분이다.

제2종 완전 타원 적분을 소거하면 다음과 같은 방정식을 얻을 수 있다.^[1]

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

따라서, 다음 삼차 사차 미분 방정식이 성립한다.^[1]

:$x^2\; (1-x^2)^2\; [2q(x)^2\; q\text{'}(x)q\text{'}\text{'}\text{'}(x)\; -\; 3q(x)^2\; q\text{'}\text{'}(x)^2\; +\; q\text{'}(x)^4]\; =\; (1+x^2)^2\; q(x)^2\; q\text{'}(x)^2$

자코비 세타 함수 중 주 세타 함수의 첫 번째 도함수는 연쇄 법칙과 타원 놈의 유도 공식을 사용하여 유도할 수 있다.^[1]

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash vartheta\_\{00\}(x)\; =\; \backslash vartheta\_\{00\}(x)\backslash bigl[\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\backslash bigr]\backslash biggl\backslash \{\backslash frac\{1\}\{2\backslash pi\; x\}E\backslash biggl[\backslash frac\{\backslash vartheta\_\{00\}(x)^2-\backslash vartheta\_\{01\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\}\backslash biggr]\; -\; \backslash frac\{\backslash vartheta\_\{01\}(x)^2\}\{4x\}\backslash biggr\backslash \}$

마찬가지로, 세타 함수의 다른 1차 도함수와 그 조합은 다음과 같이 유도할 수 있다.^[1]

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash vartheta\_\{01\}(x)\; =\; \backslash vartheta\_\{01\}(x)\backslash bigl[\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\backslash bigr]\backslash biggl\backslash \{\backslash frac\{1\}\{2\backslash pi\; x\}E\backslash biggl[\backslash frac\{\backslash vartheta\_\{00\}(x)^2-\backslash vartheta\_\{01\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\}\backslash biggr]\; -\; \backslash frac\{\backslash vartheta\_\{00\}(x)^2\}\{4x\}\backslash biggr\backslash \}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash vartheta\_\{10\}(x)\; =\; \backslash frac\{1\}\{2\backslash pi\; x\}\; \backslash vartheta\_\{10\}(x)\backslash vartheta\_\{00\}(x)^2\; E\backslash biggl[\backslash frac\{\backslash vartheta\_\{10\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2\}\backslash biggr]$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash vartheta\_\{00\}(x)\}\{\backslash vartheta\_\{01\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{00\}(x)^5\; -\; \backslash vartheta\_\{00\}(x)\backslash vartheta\_\{01\}(x)^4\}\{4x\backslash vartheta\_\{01\}(x)\}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash vartheta\_\{10\}(x)\}\{\backslash vartheta\_\{00\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{10\}(x)\backslash vartheta\_\{01\}(x)^4\}\{4x\backslash vartheta\_\{00\}(x)\}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash vartheta\_\{10\}(x)\}\{\backslash vartheta\_\{01\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{10\}(x)\backslash vartheta\_\{00\}(x)^4\}\{4x\backslash vartheta\_\{01\}(x)\}$

8. 세타 함수와의 관계

타원 놈는 리하르트 데데킨트에 의해 연구되었으며, 이는 그의 에타 함수 이론의 기초를 형성한다. 타원형 놈는 카를 구스타프 야코비의 테타 함수에서 가로 좌표로 지정되며, 산술 기하 평균의 대수적 조합에 대한 람베르트 급수 구성의 시작점을 형성한다.

일반적으로 많은 급수 전개는 타원 놈로 설명된다:

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; q(x)^\{n^2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash vartheta\_\{00\}[q(x)]\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash tfrac\{1\}\{2\}\backslash operatorname\{agm\}(1-x;1+x)^\{-1/2\}\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; q(x)^\{(2n-1)^2\}\; =\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{00\}[q(x)]\; -\; \backslash tfrac\{1\}\{4\}\backslash vartheta\_\{01\}[q(x)]\; =\; \backslash tfrac\{1\}\{4\}(1-\backslash sqrt[4]\{1-x^2\})\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{2q(x)^\{n\}\}\{q(x)^\{2n\}\; +\; 1\}\; =\; \backslash tfrac\{1\}\{2\}\backslash vartheta\_\{00\}[q(x)]^2\; -\; \backslash tfrac\{1\}\{2\}\; =\; \backslash pi^\{-1\}K(x)\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{2q(x)^\{2n-1\}\}\{q(x)^\{4n-2\}\; +\; 1\}\; =\; \backslash tfrac\{1\}\{2\}(1-\backslash sqrt\{1-x^2\})\backslash ,\backslash pi^\{-1\}K(x)$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; n^2\; q(x)^\{n^2\}\; =\; 2^\{-1/2\}\backslash pi^\{-5/2\}K(x)^\{3/2\}[E(x)-(1-x^2)K(x)]$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash biggl[\backslash frac\{2q(x)^\{n\}\}\{1\; +\; q(x)^\{2n\}\}\backslash biggr]^2\; =\; 2\backslash pi^\{-2\}E(x)K(x)\; -\; \backslash tfrac\{1\}\{2\}$

:$\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash biggl[\backslash frac\{2q(x)^\{n\}\}\{1\; -\; q(x)^\{2n\}\}\backslash biggr]^2\; =\; \backslash tfrac\{2\}\{3\}\backslash pi^\{-2\}(2\; -\; x^2)K(x)^2\; -\; 2\backslash pi^\{-2\}K(x)E(x)\; +\; \backslash tfrac\{1\}\{6\}$

:$\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; [1-q(x)^\{2n\}][1+q(x)^\{2n-1\}]^2\; =\; \backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; [1-q(x)^\{2n\}][1-q(x)^\{2n-1\}]^2\; =\; \backslash sqrt[4]\{1-x^2\}\backslash ,\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

세 가지 주요 테타 함수는 다음과 같이 정의된다:

:$\backslash vartheta\_\{00\}(y)\; =\; 1\; +\; 2\backslash biggl(\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; y^\{n^2\}\backslash biggr)$

:$\backslash vartheta\_\{01\}(y)\; =\; 1\; +\; 2\backslash biggl[\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; (-1)^\{n\}y^\{n^2\}\backslash biggr]$

:$\backslash vartheta\_\{10\}(y)\; =\; 2\backslash biggl[\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; y^\{(n\; -\; 1/2)^2\}\backslash biggr]$

야코비 항등식은 이 세 가지 함수를 수학적 조합으로 제공한다:

:$\backslash vartheta\_\{10\}(y)\; =\; \backslash sqrt[4]\{\backslash vartheta\_\{00\}(y)^4\; -\; \backslash vartheta\_\{01\}(y)^4\}$

테타 함수와 타원형 놈 함수는 서로 다음과 같은 관계를 갖는다:

:$\backslash vartheta\_\{00\}[q(x)]\; =\; \backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash vartheta\_\{01\}[q(x)]\; =\; \backslash sqrt[4]\{1-x^2\}\backslash ,\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash vartheta\_\{00\}[q(x)^2]\; =\; \backslash cos\backslash bigl[\backslash tfrac\{1\}\{2\}\backslash arcsin(x)\backslash bigr]\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

:$\backslash vartheta\_\{01\}[q(x)^2]\; =\; \backslash sqrt[8]\{1\; -\; x^2\}\backslash ,\backslash sqrt\{2\backslash pi^\{-1\}K(x)\}$

놈 함수는 제1종 및 제2종 완전 타원 적분의 정의에 사용될 수 있다.

:$K(\backslash varepsilon)\; =\; \backslash tfrac\{1\}\{2\}\backslash pi\backslash ,\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^2$

:$E(\backslash varepsilon)\; =\; 2\backslash pi\; q(\backslash varepsilon)\backslash ,\backslash vartheta\_\{00\}\text{'}[q(\backslash varepsilon)]\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^\{-3\}\; +\; \backslash tfrac\{1\}\{2\}\backslash pi(1\; -\; \backslash varepsilon^2)\backslash ,\backslash vartheta\_\{00\}[q(\backslash varepsilon)]^2$

여기서 지수 위치의 대시는 소위 세타 제로 값 함수의 도함수를 나타낸다.

:$\backslash vartheta\_\{00\}\text{'}(x)\; =\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\backslash ,\backslash vartheta\_\{00\}(x)\; =\; 2\; +\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; 2(n\; +\; 1)^2\; x^\{n(n+2)\}$

타원 함수 진폭의 제타와 델타 진폭은 타원 노메 함수^[9]를 사용하여 쉽게 정의할 수 있다.

:$\backslash operatorname\{zn\}(x;k)\; =\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{2\backslash pi\; K(k)^\{-1\}\backslash sin[\backslash pi\; K(k)^\{-1\}x]q(k)^\{2n-1\}\}\{1-2\backslash cos[\backslash pi\; K(k)^\{-1\}x]q(k)^\{2n-1\}+q(k)^\{4n-2\}\}$

:$\backslash operatorname\{dn\}(x;k)\; =\; \backslash sqrt[4]\{1-k^2\}\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{1+2\backslash cos[\backslash pi\; K(k)^\{-1\}x]q(k)^\{2n-1\}+q(k)^\{4n-2\}\}\{1-2\backslash cos[\backslash pi\; K(k)^\{-1\}x]q(k)^\{2n-1\}+q(k)^\{4n-2\}\}$

앞서 언급했듯이, 노메를 제곱 함수로 나눈 몫의 네제곱근을 사용하여 진폭 사인, 반대 진폭 사인 및 진폭 코사인에 대한 곱 급수 정의^[10]를 설정할 수 있다.

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

:$\backslash operatorname\{cd\}(x;k)\; =\; 2\backslash sqrt[4]\{k^\{-2\}q(k)\}\backslash ,\backslash cos[\backslash tfrac\{1\}\{2\}\backslash pi\; K(k)^\{-1\}x]\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{1\; +\; 2q(k)^\{2n\}\backslash cos[\backslash pi\; K(k)^\{-1\}x]\; +\; q(k)^\{4n\}\}\{1\; +\; 2q(k)^\{2n\; -\; 1\}\backslash cos[\backslash pi\; K(k)^\{-1\}x]\; +\; q(k)^\{4n\; -\; 2\}\}$

:$\backslash operatorname\{cn\}(x;k)\; =\; 2\backslash sqrt[4]\{k^\{-2\}(1\; -\; k^2)\backslash ,q(k)\}\backslash ,\backslash cos[\backslash tfrac\{1\}\{2\}\backslash pi\; K(k)^\{-1\}x]\backslash prod\_\{n\; =\; 1\}^\{\backslash infty\}\; \backslash frac\{1\; +\; 2q(k)^\{2n\}\backslash cos[\backslash pi\; K(k)^\{-1\}x]\; +\; q(k)^\{4n\}\}\{1\; -\; 2q(k)^\{2n\; -\; 1\}\backslash cos[\backslash pi\; K(k)^\{-1\}x]\; +\; q(k)^\{4n\; -\; 2\}\}$

이 공식들은 −1부터 +1까지의 모든 k 값에 유효하다.

다른 야코비 함수의 연속 정의는 다음과 같다.

:$\backslash operatorname\{sn\}(x;k)\; =\; \backslash frac\{2\backslash \{\backslash operatorname\{zn\}(\backslash tfrac\{1\}\{2\}x;k)\; +\; \backslash operatorname\{zn\}[K(k)-\backslash tfrac\{1\}\{2\}x;k]\backslash \}\}\{k^2+\backslash \{\backslash operatorname\{zn\}(\backslash tfrac\{1\}\{2\}x;k)\; +\; \backslash operatorname\{zn\}[K(k)-\backslash tfrac\{1\}\{2\}x;k]\backslash \}^2\}$

:$\backslash operatorname\{cd\}(x;k)\; =\; \backslash operatorname\{sn\}[K(k)\; -\; x;k]$

:$\backslash operatorname\{cn\}(x;k)\; =\; \backslash operatorname\{cd\}(x;k)\backslash operatorname\{dn\}(x;k)$

:$\backslash operatorname\{dn\}(x;k)\; =\; \backslash frac\{k^2-\backslash \{\backslash operatorname\{zn\}(\backslash tfrac\{1\}\{2\}x;k)\; +\; \backslash operatorname\{zn\}[K(k)-\backslash tfrac\{1\}\{2\}x;k]\backslash \}^2\}\{k^2+\backslash \{\backslash operatorname\{zn\}(\backslash tfrac\{1\}\{2\}x;k)\; +\; \backslash operatorname\{zn\}[K(k)-\backslash tfrac\{1\}\{2\}x;k]\backslash \}^2\}$

진폭 사인의 곱 정의는 Borwein 형제가 쓴 ''π와 AGM''이라는 에세이 60페이지에 적혀 있으며, 이 공식은 Whittaker와 Watson의 세타 함수 정의를 기반으로 한다.

세타 함수와 결합하면 놈은 많은 야코비 진폭 함수 값들을 제공한다.

:$\backslash operatorname\{sc\}[\backslash tfrac\{2\}\{3\}K(k);k]\; =\; \backslash frac\{\backslash sqrt\{3\}\backslash ,\backslash vartheta\_\{01\}[q(k)^6]\}\{\backslash sqrt\{1\; -\; k^2\}\backslash ,\backslash vartheta\_\{01\}[q(k)^2]\}$

:$\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{3\}K(k);k]\; =\; \backslash frac\{2\backslash vartheta\_\{00\}[q(k)]^2\}\{3\backslash vartheta\_\{00\}[q(k)^3]^2\; +\; \backslash vartheta\_\{00\}[q(k)]^2\}\; =\; \backslash frac\{3\backslash vartheta\_\{01\}[q(k)^3]^2\; -\; \backslash vartheta\_\{01\}[q(k)]^2\}\{3\backslash vartheta\_\{01\}[q(k)^3]^2\; +\; \backslash vartheta\_\{01\}[q(k)]^2\}$

:$\backslash operatorname\{cn\}[\backslash tfrac\{2\}\{3\}K(k);k]\; =\; \backslash frac\{3\backslash vartheta\_\{00\}[q(k)^3]^2\; -\; \backslash vartheta\_\{00\}[q(k)]^2\}\{3\backslash vartheta\_\{00\}[q(k)^3]^2\; +\; \backslash vartheta\_\{00\}[q(k)]^2\}\; =\; \backslash frac\{2\backslash vartheta\_\{01\}[q(k)]^2\}\{3\backslash vartheta\_\{01\}[q(k)^3]^2\; +\; \backslash vartheta\_\{01\}[q(k)]^2\}$

:$\backslash operatorname\{sn\}[\backslash tfrac\{1\}\{5\}K(k);k]\; =\; \backslash biggl\backslash \{\backslash frac\{\backslash sqrt\{5\}\backslash ,\backslash vartheta\_\{01\}[q(k)^5]\}\{\backslash vartheta\_\{01\}[q(k)]\}\; -\; 1\backslash biggr\backslash \}\backslash biggl\backslash \{\backslash frac\{5\backslash vartheta\_\{01\}[q(k)^\{10\}]^2\}\{\backslash vartheta\_\{01\}[q(k)^2]^2\}\; -\; 1\backslash biggr\backslash \}^\{-1\}$

:$\backslash operatorname\{sn\}[\backslash tfrac\{3\}\{5\}K(k);k]\; =\; \backslash biggl\backslash \{\backslash frac\{\backslash sqrt\{5\}\backslash ,\backslash vartheta\_\{01\}[q(k)^5]\}\{\backslash vartheta\_\{01\}[q(k)]\}\; +\; 1\backslash biggr\backslash \}\backslash biggl\backslash \{\backslash frac\{5\backslash vartheta\_\{01\}[q(k)^\{10\}]^2\}\{\backslash vartheta\_\{01\}[q(k)^2]^2\}\; -\; 1\backslash biggr\backslash \}^\{-1\}$

:$\backslash operatorname\{cn\}[\backslash tfrac\{2\}\{5\}K(k);k]\; =\; \backslash biggl\backslash \{\backslash frac\{\backslash sqrt\{5\}\backslash ,\backslash vartheta\_\{00\}[q(k)^5]\}\{\backslash vartheta\_\{00\}[q(k)]\}\; +\; 1\backslash biggr\backslash \}\backslash biggl\backslash \{\backslash frac\{5\backslash vartheta\_\{01\}[q(k)^\{10\}]^2\}\{\backslash vartheta\_\{01\}[q(k)^2]^2\}\; -\; 1\backslash biggr\backslash \}^\{-1\}$

:$\backslash operatorname\{cn\}[\backslash tfrac\{4\}\{5\}K(k);k]\; =\; \backslash biggl\backslash \{\backslash frac\{\backslash sqrt\{5\}\backslash ,\backslash vartheta\_\{00\}[q(k)^5]\}\{\backslash vartheta\_\{00\}[q(k)]\}\; -\; 1\backslash biggr\backslash \}\backslash biggl\backslash \{\backslash frac\{5\backslash vartheta\_\{01\}[q(k)^\{10\}]^2\}\{\backslash vartheta\_\{01\}[q(k)^2]^2\}\; -\; 1\backslash biggr\backslash \}^\{-1\}$

약어 sc는 진폭 코사인이 나뉜 진폭 사인의 몫을 설명한다.

자코비 세타 함수 중 주 세타 함수의 첫 번째 도함수는 연쇄 법칙과 타원 놈의 유도 공식을 사용하여 유도할 수 있다.

:$\backslash frac\{\backslash pi^2\}\{2\backslash varepsilon(1-\backslash varepsilon^2)K(\backslash varepsilon)^2\}\; \backslash ,q(\backslash varepsilon)\backslash ,\backslash biggl\backslash \{\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}\backslash ,q(\backslash varepsilon)\}\backslash ,\backslash vartheta\_\{00\}\backslash bigl[q(\backslash varepsilon)\backslash bigr]\backslash biggr\backslash \}\; =\; \backslash biggl[\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}\backslash varepsilon\}\; \backslash ,q(\backslash varepsilon)\backslash biggr]\backslash biggl\backslash \{\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}\backslash ,q(\backslash varepsilon)\}\backslash ,\backslash vartheta\_\{00\}\backslash bigl[q(\backslash varepsilon)\backslash bigr]\backslash biggr\backslash \}\; =\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}\backslash varepsilon\}\backslash ,\backslash vartheta\_\{00\}\backslash bigl[q(\backslash varepsilon)\backslash bigr]\; =\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}\backslash varepsilon\}\backslash ,\backslash sqrt\{2\backslash pi^\{-1\}K(\backslash varepsilon)\}\; =$

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

지금 언급한 유도 부분에서 이 항등식이 기초가 된다:

:$\backslash vartheta\_\{00\}[q(\backslash varepsilon)]\; =\; \backslash sqrt\{2\backslash pi^\{-1\}\; K(\backslash varepsilon)\}$

그러므로 이 방정식의 결과는 다음과 같다:

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

제2종 완전 타원 적분은 다음 항등식을 가진다:

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

이 모듈식 항등식과 함께 다음 공식 변환을 수행할 수 있다:

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

더욱이, 다음 항등식이 유효하다:

:$\backslash vartheta\_\{01\}[q(\backslash varepsilon)]\; =\; \backslash sqrt[4]\{1\; -\; \backslash varepsilon^2\}\backslash sqrt\{2\backslash pi^\{-1\}\; K(\backslash varepsilon)\}$

세타 함수 표현 ₀₀(x)과 ₀₁(x)를 사용하여 다음 표현이 가능하다:

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

이것이 최종 결과이다:

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash vartheta\_\{00\}(x)\; =\; \backslash vartheta\_\{00\}(x)\backslash bigl[\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\backslash bigr]\backslash biggl\backslash \{\backslash frac\{1\}\{2\backslash pi\; x\}E\backslash biggl[\backslash frac\{\backslash vartheta\_\{00\}(x)^2-\backslash vartheta\_\{01\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\}\backslash biggr]\; -\; \backslash frac\{\backslash vartheta\_\{01\}(x)^2\}\{4x\}\backslash biggr\backslash \}$

유사한 방식으로, 세타 함수의 다른 1차 도함수와 이들의 조합도 다음과 같이 유도할 수 있다.

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash vartheta\_\{01\}(x)\; =\; \backslash vartheta\_\{01\}(x)\backslash bigl[\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\backslash bigr]\backslash biggl\backslash \{\backslash frac\{1\}\{2\backslash pi\; x\}E\backslash biggl[\backslash frac\{\backslash vartheta\_\{00\}(x)^2-\backslash vartheta\_\{01\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2+\backslash vartheta\_\{01\}(x)^2\}\backslash biggr]\; -\; \backslash frac\{\backslash vartheta\_\{00\}(x)^2\}\{4x\}\backslash biggr\backslash \}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash vartheta\_\{10\}(x)\; =\; \backslash frac\{1\}\{2\backslash pi\; x\}\; \backslash vartheta\_\{10\}(x)\backslash vartheta\_\{00\}(x)^2\; E\backslash biggl[\backslash frac\{\backslash vartheta\_\{10\}(x)^2\}\{\backslash vartheta\_\{00\}(x)^2\}\backslash biggr]$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash frac\{\backslash vartheta\_\{00\}(x)\}\{\backslash vartheta\_\{01\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{00\}(x)^5\; -\; \backslash vartheta\_\{00\}(x)\backslash vartheta\_\{01\}(x)^4\}\{4x\backslash ,\backslash vartheta\_\{01\}(x)\}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash frac\{\backslash vartheta\_\{10\}(x)\}\{\backslash vartheta\_\{00\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{10\}(x)\backslash vartheta\_\{01\}(x)^4\}\{4x\backslash ,\backslash vartheta\_\{00\}(x)\}$

:$\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash ,\backslash frac\{\backslash vartheta\_\{10\}(x)\}\{\backslash vartheta\_\{01\}(x)\}\; =\; \backslash frac\{\backslash vartheta\_\{10\}(x)\backslash vartheta\_\{00\}(x)^4\}\{4x\backslash ,\backslash vartheta\_\{01\}(x)\}$

중요한 정의:

:$\backslash vartheta\_\{10\}(x)\; =\; 2x^\{1/4\}\; +\; 2x^\{1/4\}\backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; x^\{2\backslash bigtriangleup(n)\}$

:$\backslash bigtriangleup(n)\; =\; \backslash tfrac\{1\}\{2\}n(n\; +\; 1)$

9. 5차 방정식의 해

아벨-루피니 정리에 따르면 5차 방정식의 일반적인 경우는 기본적으로 풀 수 없다. 그러나 타원 명사와 세타 함수의 조합으로 모든 오차 방정식을 풀 수 있다. 다음은 브링-제라드 정규형의 5차 다항식에 대해 언급된 타원 함수가 있는 실제 해를 나타낸 것이다.^[1]

:$x^5\; +\; 5\backslash ,x\; =\; 4\backslash ,c$

:$Q\; =\; q\backslash bigl[\backslash bigl(2\backslash ,c^2\; +\; 2\; +\; 2\backslash sqrt\{c^4\; +\; 1\}\backslash bigr)^\{-1/2\}\backslash bigl(\backslash sqrt\{\backslash sqrt\{c^4\; +\; 1\}\; +\; 1\}\; +\; c\backslash bigr)\backslash bigr]$

:$x\; =\; \backslash frac\{\backslash bigl[\backslash vartheta\_\{00\}(Q^\{1/5\})^2\; -\; 5\backslash ,\backslash vartheta\_\{00\}(Q^5)^2\backslash bigr]\backslash sqrt\{\backslash vartheta\_\{00\}(Q^\{1/5\})^2\; +\; 5\backslash ,\backslash vartheta\_\{00\}(Q^5)^2\; -\; 4\backslash ,\backslash vartheta\_\{00\}(Q)^2\; -\; 2\backslash ,\backslash vartheta\_\{00\}(Q^\{1/5\})\backslash ,\backslash vartheta\_\{00\}(Q^5)\}\}\{4\backslash ,\backslash vartheta\_\{10\}(Q)\backslash ,\backslash vartheta\_\{01\}(Q)\backslash ,\backslash vartheta\_\{00\}(Q)\}$

기본 함수로 풀 수 있는 방정식의 예시는 다음과 같다.
첫 번째 계산 예:

두 번째 계산 예:

타원 함수로만 풀 수 있는 계산 예:

:$x^5\; +\; 5\backslash ,x\; =\; 2\backslash sqrt[4]\{2\}$

:$Q\; =\; q\backslash bigl[\backslash bigl(2\backslash ,c^2\; +\; 2\; +\; 2\backslash sqrt\{c^4\; +\; 1\}\backslash bigr)^\{-1/2\}\backslash bigl(\backslash sqrt\{\backslash sqrt\{c^4\; +\; 1\}\; +\; 1\}\; +\; c\backslash bigr)\backslash bigr](c\; =\; 2^\{-3/4\})\; =\; q\backslash bigl[\backslash cos\backslash bigl(\backslash tfrac\{1\}\{8\}\backslash pi\backslash bigr)\backslash bigr]$

:$Q\; \backslash approx\; 0.11785793531185771914155254110648923544545944879394196130445$

:$x\; =\; \backslash frac\{\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^\{1/5\}\backslash bigr\backslash \}^2\; -\; 5\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^5\backslash bigr\backslash \}^2\}\{4\backslash ,\backslash vartheta\_\{10\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]\backslash bigr\backslash \}\backslash ,\backslash vartheta\_\{01\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]\backslash bigr\backslash \}\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]\backslash bigr\backslash \}\}\; \backslash times$

:$\backslash times\backslash sqrt\{\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^\{1/5\}\backslash bigr\backslash \}^2\; +\; 5\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^5\backslash bigr\backslash \}^2\; -\; 4\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]\backslash bigr\backslash \}^2\; -\; 2\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^\{1/5\}\backslash bigr\backslash \}\backslash ,\backslash vartheta\_\{00\}\backslash bigl\backslash \{q\backslash bigl[\backslash cos(\backslash tfrac\{1\}\{8\}\backslash pi)\backslash bigr]^5\backslash bigr\backslash \}\}$

:$x\; \backslash approx\; 0.4710447387949811740242434591671230417409496435087081512857$

솔루션이 타원 함수의 도움으로만 표현될 수 있는 두 번째 예:

:$x^5\; +\; 5\backslash ,x\; =\; 4$

:$Q\; =\; q\backslash bigl[\backslash bigl(2c^2\; +\; 2\; +\; 2\backslash sqrt\{c^4\; +\; 1\}\backslash bigr)^\{-1/2\}\backslash bigl(\backslash sqrt\{\backslash sqrt\{c^4\; +\; 1\}\; +\; 1\}\; +\; c\backslash bigr)\backslash bigr]\backslash bigl(c\; =\; 1\backslash bigr)\; =\; q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\}\backslash bigr)\backslash bigr]$

:$Q\; \backslash approx\; 0.18520287008030014142515182307361246060360377625$

:$x\; =\; \backslash frac\{\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\})\backslash bigr]^\{1/5\}\backslash \}^2\; -\; 5\backslash ,\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\})\backslash bigr]^5\backslash \}^2\}\{4\backslash ,\backslash vartheta\_\{10\}\backslash \{q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\})\backslash bigr]\backslash \}\backslash ,\backslash vartheta\_\{01\}\backslash \{q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\})\backslash bigr]\backslash \}\backslash ,\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash frac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash frac\{\backslash pi\}\{8\})\backslash bigr]\backslash \}\}\; \backslash times$

:$\backslash times\; \backslash sqrt\{\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash tfrac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash tfrac\{\backslash pi\}\{8\})\backslash bigr]^\{1/5\}\backslash \}^2\; +\; 5\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash tfrac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash tfrac\{\backslash pi\}\{8\})\backslash bigr]^5\backslash \}^2\; -\; 4\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash tfrac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash tfrac\{\backslash pi\}\{8\})\backslash bigr]\backslash \}^2\; -\; 2\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash tfrac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash tfrac\{\backslash pi\}\{8\})\backslash bigr]^\{1/5\}\backslash \}\backslash ,\backslash vartheta\_\{00\}\backslash \{q\backslash bigl[\backslash tfrac\{\backslash sqrt[4]\{2\}\}\{2\}\; +\; \backslash sin(\backslash tfrac\{\backslash pi\}\{8\})\backslash bigr]^5\backslash \}\}$

:$x\; \backslash approx\; 0.75192639869405948026865366345020738740978383913$

모듈러스 $\backslash varepsilon$에 관한 놈 세제곱 정리의 직접적인 공식과 야코비 진폭 사인을 조합하여 생성한다.^[1]

Johansson의 저서 ''대수 방정식의 해석적 해''와 Bagis의 ''5차 타원 특이 모듈의 평가''는 인용된 저서에서 완전 제1종 적분 K의 3분의 1에 대한 야코비 진폭 사인이 다음 4차 방정식을 푼다는 것을 보여주었다.^[1]

:$\backslash varepsilon^2\; x^4\; -\; 2\backslash varepsilon^2\; x^3\; +\; 2x\; -\; 1\; =\; 0$

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

위에 언급된 매개변수화를 이 방정식에 삽입한다.

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

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

이것은 패턴 에 대한 4차 방정식의 실수 해이다.

:$x\; =\; \backslash frac\{1\}\{\backslash sqrt\{u^4-u^2+1\}-u^2+1\}$

따라서 다음 공식이 유효하다.

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

매개변수화된 놈 세제곱 공식은 다음과 같다.

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

같은 공식을 다음과 같은 대안적인 방식으로 설계할 수 있다.

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

따라서 이 결과는 직접 놈 세제곱 정리로 나타난다.

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

10. 응용

놈은 타원 함수와 모듈러 형식을 설명하는 값으로 자주 사용된다.

놈은 람베르트 급수, q-급수, 더 일반적으로 q-아날로그를 구성하는 시작점으로 사용된다. 즉, 반 주기 비 \(\tau\)는 일반적으로 복소수 상반평면의 좌표로 사용되며, 푸앵카레 메트릭이 부여되어 푸앵카레 상반평면 모형을 얻는다. 그런 다음 놈은 반지름이 1인 구멍 뚫린 원반의 좌표 역할을 한다. 구멍이 뚫린 이유는 \(q=0\)이 원반의 일부가 아니기 때문이다. \(q=0\)은 \(\tau \to \infty\)에 해당한다. 이렇게 하면 구멍 뚫린 원반에 푸앵카레 메트릭이 부여된다.

상반평면(및 푸앵카레 원반 및 구멍 뚫린 원반)은 기본 영역으로 타일링할 수 있으며, 이는 평행 사변형으로 평면 타일링을 고유하게 결정하는 반주기 비율 \(\tau\)(또는 \(q\) 또는 \(K\) 및 \(iK'\) 등의 값)의 영역이다. 타일링은 모듈 그룹에 의해 주어진 모듈 대칭이라고 한다. 상반평면에서 주기적인 일부 함수는 모듈 함수라고 한다. 놈, 반주기, 사분기 또는 반주기 비율은 모두 이러한 주기 함수에 대한 다양한 매개변수화를 제공한다.

전형적인 모듈 함수는 클라인의 j-불변량이다. 반주기 비율 \(\tau\) 또는 놈 \(q\)의 함수로 작성할 수 있다. 놈 또는 놈의 제곱(''q''-전개)에 따른 급수 전개는 피셔-그라이스 괴물과 괴물 달빛에 의해 유명하게 연결된다.

오일러 함수는 일반적으로 ''q''-급수의 원형으로 나타난다.

놈은 ''q''-급수의 \(q\)이므로 본질적으로 아핀 리 대수 이론에서 나타난다.

11. 참고 문헌

• Milton Abramowitz and Irene A. Stegun, ''Handbook of Mathematical Functions'' (1964), Dover Publications, New York. 섹션 16.27.4 및 17.3.17 참조. 1972년 판:
• Tom M. Apostol, ''Modular Functions and Dirichlet Series in Number Theory, Second Edition'' (1990), Springer, New York
• Folkmar Bornemann, Dirk Laurie, Stan Wagon and Jörg Waldvogel, ''Vom Lösen numerischer Probleme'', 275쪽
• Edmund Taylor Whittaker and George Neville Watson: ''A Course in Modern Analysis, 4th ed.'' Cambridge, England: Cambridge University Press, 1990. 469–470쪽.
• Toshio Fukushima: ''Fast Computation of Complete Elliptic Integrals and Jacobian Elliptic Functions''. 2012, National Astronomical Observatory of Japan (国立天文台)
• Lowan, Blanch and Horenstein: ''On the Inversion of the q-Series Associated with Jacobian Elliptic Functions''. Bull. Amer. Math. Soc. 48, 1942
• H. Ferguson, D. E. Nielsen, G. Cook: ''A partition formula for the integer coefficients of the theta function nome''. Mathematics of computation, Volume 29, number 131, Juli 1975
• J. D. Fenton and R. S. Gardiner-Garden: ''Rapidly-convergent methods for evaluating elliptic integrals and theta and elliptic functions''. J. Austral. Math. Soc. (Series B) 24, 1982, 57쪽
• Charles Hermite: ''Sur la résolution de l'Équation du cinquiéme degré Comptes rendus''. Acad. Sci. Paris, Nr. 11, 1858
• Nikolaos Bagis: ''On the solution of the general quintic using the Rogers-Ramanujan continued fraction''. Pella, Makedonien, Griechenland, 2015
• Nikolaos Bagis: ''Solution of Polynomial Equations with Nested Radicals''. Pella, Makedonien, Griechenland, 2020
• Viktor Prasolov (Прасолов) und Yuri Solovyev (Соловьёв): ''Elliptic Functions and Elliptic Integrals''. Volume 170, Rhode Island, 1991. 149 – 159쪽
• Sun Zhi-Hong: ''New congruences involving Apery-like numbers''. Huaiyin Normal University, Huaian (淮安), China, 2020. 2쪽
• Robert Fricke: ''Die elliptischen Funktionen und ihre Anwendungen: Dritter Teil''. Springer-Verlag Berlin Heidelberg, 2012.
• Adolf Kneser: ''Neue Untersuchung einer Reihe aus der Theorie der elliptischen Funktionen''. J. reine u. angew. Math. 157, 1927. 209 – 218쪽
• G. P. Young: ''Solution of Solvable Irreducible Quintic Equations, Without the Aid of a Resolvent Sextic''. In: Amer. J. Math. Band 7, 170–177쪽, 1885.
• C. Runge: ''Über die auflösbaren Gleichungen von der Form x 5 + u x + v = 0 {\displaystyle x^{5}+ux+v=0} x^{5}+ux+v=0''. In: Acta Math. Band 7, 173–186쪽, 1885, doi:10.1007/BF02402200.
• Edward Neuman: ''Two-sided inequalitites for the lemniscate functions.'' Volume 1, Southern Illinois University Carbondale, USA, 2014.
• Ji-en Deng und Chao-ping Chen: ''Sharp Shafer-Fink type inequalities for Gauss lemniscate functions.'' Universität Henan (河南大学), China, 2014.
• Jun-Ling Sun und Chao-ping Chen: ''Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions.'' Universität Henan, China, 2016.
• Minjie Wei, Yue He and Gendi Wang: ''Shafer–Fink type inequalities for arc lemniscate functions''. Zhejiang Sci-Tech University, Hangzhou, China, 2019

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_[3] 간행물 Neue Untersuchung einer Reihe aus der Theorie der elliptischen Funktionen. https://eudml.org/do[...] 2023-06-11
_[4] 간행물 Application of theta functions for numerical evaluation of complete elliptic integrals of the first and second kinds https://www.osti.gov[...] Oak Ridge National Lab. (ORNL) 2023-06-11
_[5] 웹사이트 A002103 - OEIS https://oeis.org/A00[...] 2023-05-28
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_[11] 학술지 Arithmetisch geometrisches Mittel, Werke, Bd. 3 1876
_[12] 문서
_[13] 서적 Riflessioni intorno alla soluzione delle equazioni algebraiche generali opuscolo del cav. dott. Paolo Ruffini ... https://books.google[...] presso la Societa Tipografica 1813
_[14] 서적 Œuvres Complètes de Niels Henrik Abel Grøndahl & Søn Forlag
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_[16] 서적 Galois' Theory of Algebraic Equations https://archive.org/[...] World Scientific
_[17] 간행물 Elliptic Functions and Elliptic Integrals https://staff.math.s[...] 2023-06-24
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