Эллиптическая модульная лямбда-функция

Лямбда-звезда положительных рациональных чисел всегда являются положительными алгебраическими числами.

${\displaystyle \lambda ^{*}(x\in \mathbb {Q} ^{+})\in \mathbb {A} ^{+}}$

Следующее уравнение действительно для всех натуральных чисел n ∈ ℕ:

${\displaystyle {\sqrt {n}}=\sum _{a=1}^{n}\operatorname {dn} {\biggl \{}{\frac {2a}{n}}K{\biggl [}\lambda ^{*}{\bigl (}{\frac {1}{n}}{\bigr )}{\biggr ]};\lambda ^{*}{\bigl (}{\frac {1}{n}}{\bigr )}{\biggr \}}}$

Эллиптические функции Якоби выражаются сокращениями sn, cn и dn.

${\displaystyle \lambda ^{*}(n^{2}x)=\lambda ^{*}(x)^{n}\prod _{a=1}^{n}\operatorname {sn} \left\{{\frac {2a-1}{n}}K[\lambda ^{*}(x)];\lambda ^{*}(x)\right\}^{2}}$

Число x должно быть положительным числом, а n должно быть натуральным числом.

Эти симметричные алгебраические отношения существуют:

${\displaystyle \lambda ^{*}(x)^{2}+\lambda ^{*}(1/x)^{2}=1}$

${\displaystyle \lambda ^{*}(4x)={\frac {1-{\sqrt {1-\lambda ^{*}(x)^{2}}}}{1+{\sqrt {1-\lambda ^{*}(x)^{2}}}}}=\tan\{\arcsin[\lambda ^{*}(x)]/2\}^{2}}$

${\displaystyle \tan\{2\arctan[\lambda ^{*}(x)]\}\tan\{2\arctan[\lambda ^{*}(4/x)]\}=1}$

${\displaystyle \lambda ^{*}(x)\lambda ^{*}(4/x)+\lambda ^{*}(x)+\lambda ^{*}(4/x)=1}$

${\displaystyle \lambda ^{*}(x)-\lambda ^{*}(9x)=2\lambda ^{*}(x)^{1/4}\lambda ^{*}(9x)^{1/4}-2\lambda ^{*}(x)^{3/4}\lambda ^{*}(9x)^{3/4}}$

${\displaystyle \tan\{2\arctan[\lambda ^{*}(x)]\}-\tan\{2\arctan[\lambda ^{*}(9x)]\}=}$

${\displaystyle =2{\sqrt {2}}\tan\{2\arctan[\lambda ^{*}(x)]\}^{1/4}\tan\{2\arctan[\lambda ^{*}(9x)]\}^{1/4}+2{\sqrt {2}}\tan\{2\arctan[\lambda ^{*}(x)]\}^{3/4}\tan\{2\arctan[\lambda ^{*}(9x)]\}^{3/4}}$

${\displaystyle \tan\{2\arctan[\lambda ^{*}(x)]\}^{1/2}-\tan\{2\arctan[\lambda ^{*}(25x)]\}^{1/2}=}$

${\displaystyle =2\tan\{2\arctan[\lambda ^{*}(x)]\}^{1/12}\tan\{2\arctan[\lambda ^{*}(25x)]\}^{1/12}+2\tan\{2\arctan[\lambda ^{*}(x)]\}^{5/12}\tan\{2\arctan[\lambda ^{*}(25x)]\}^{5/12}}$

Важные константы, используемые в следующих:

имя константы	алгебраическое выражение	уравнение
Золотое сечение	${\displaystyle \Phi ={\tfrac {1}{2}}({\sqrt {5}}+1)}$	${\displaystyle \Phi ^{2}-\Phi -1=0}$
Серебряное сечение	${\displaystyle \delta _{S}={\sqrt {2}}+1}$	${\displaystyle \delta _{S}^{2}-2\delta _{S}-1=0}$
Бронзовое сечение	${\displaystyle \delta _{B}={\tfrac {1}{2}}({\sqrt {13}}+3)}$	${\displaystyle \delta _{B}^{2}-3\delta _{B}-1=0}$
Постоянная трибоначчи	${\displaystyle T_{TRI}={\tfrac {1}{3}}{\sqrt[{3}]{19+3{\sqrt {33}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{19-3{\sqrt {33}}}}+{\tfrac {1}{3}}}$	${\displaystyle T_{TRI}^{3}-T_{TRI}^{2}-T_{TRI}-1=0}$
Пластическое число	${\displaystyle \rho ={\tfrac {1}{6}}{\sqrt[{3}]{12}}({\sqrt[{3}]{9+{\sqrt {69}}}}+{\sqrt[{3}]{9-{\sqrt {69}}}})}$	${\displaystyle \rho ^{3}-\rho -1=0}$
Сверхзолотое сечение	${\displaystyle \psi ={\tfrac {1}{6}}{\sqrt[{3}]{116+12{\sqrt {93}}}}+{\tfrac {1}{6}}{\sqrt[{3}]{116-12{\sqrt {93}}}}+{\tfrac {1}{3}}}$	${\displaystyle \psi ^{3}-\psi ^{2}-1=0}$

Первые десять значений:

${\displaystyle \lambda ^{*}(1)={\tfrac {1}{2}}{\sqrt {2}}}$

${\displaystyle \lambda ^{*}(2)={\sqrt {2}}-1=\delta _{S}^{-1}}$

${\displaystyle \lambda ^{*}(3)={\tfrac {1}{4}}{\sqrt {2}}({\sqrt {3}}-1)}$

${\displaystyle \lambda ^{*}(4)=({\sqrt {2}}-1)^{2}=\delta _{S}^{-2}}$

${\displaystyle \lambda ^{*}(5)={\tfrac {1}{2}}{\sqrt {{\sqrt {5}}-1}}-{\tfrac {1}{2}}{\sqrt {3-{\sqrt {5}}}}={\tfrac {1}{2}}{\sqrt {2}}(\Phi ^{-1/2}-\Phi ^{-1})}$

${\displaystyle \lambda ^{*}(6)=(2-{\sqrt {3}})({\sqrt {3}}-{\sqrt {2}})}$

${\displaystyle \lambda ^{*}(7)={\tfrac {1}{8}}{\sqrt {2}}(3-{\sqrt {7}})}$

${\displaystyle \lambda ^{*}(8)=({\sqrt {2}}+1-{\sqrt {2{\sqrt {2}}+2}})^{2}=(\delta _{S}-{\sqrt {2\delta _{S}}})^{2}}$

${\displaystyle \lambda ^{*}(9)={\tfrac {1}{2}}({\sqrt {3}}-1)({\sqrt {2}}-{\sqrt[{4}]{3}})}$

${\displaystyle \lambda ^{*}(10)=({\sqrt {10}}-3)({\sqrt {2}}-1)^{2}}$

Дальнейшие значения нечетных чисел:

${\displaystyle \lambda ^{*}(11)={\tfrac {1}{16}}{\sqrt {2}}({\sqrt {11}}+3)({\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\tfrac {1}{3}}{\sqrt {11}}-1)^{4}={\tfrac {1}{2}}{\sqrt {1+{\tfrac {1}{2}}T_{TRI}^{-4}}}-{\tfrac {1}{2}}{\sqrt {1-{\tfrac {1}{2}}T_{TRI}^{-4}}}}$

${\displaystyle \lambda ^{*}(13)={\tfrac {1}{2}}{\sqrt {5{\sqrt {13}}-17}}-{\tfrac {1}{2}}{\sqrt {19-5{\sqrt {13}}}}={\tfrac {1}{2}}{\sqrt {1+\delta _{B}^{-3}}}-{\tfrac {1}{2}}{\sqrt {1-\delta _{B}^{-3}}}}$

${\displaystyle \lambda ^{*}(15)={\tfrac {1}{16}}{\sqrt {2}}(3-{\sqrt {5}})({\sqrt {5}}-{\sqrt {3}})(2-{\sqrt {3}})}$

${\displaystyle \lambda ^{*}(17)={\tfrac {1}{16}}{\sqrt {2}}({\sqrt {3{\sqrt {17}}+11}}-{\sqrt {5+{\sqrt {17}}}}-{\sqrt[{4}]{4{\sqrt {17}}+16}}+{\sqrt[{4}]{4{\sqrt {17}}-16}})^{2}}$

${\displaystyle \lambda ^{*}(19)={\tfrac {1}{16}}{\sqrt {2}}(3{\sqrt {19}}+13)[{\tfrac {1}{6}}({\sqrt {19}}-2+{\sqrt {3}}){\sqrt[{3}]{3{\sqrt {3}}-{\sqrt {19}}}}-{\tfrac {1}{6}}({\sqrt {19}}-2-{\sqrt {3}}){\sqrt[{3}]{3{\sqrt {3}}+{\sqrt {19}}}}-{\tfrac {1}{3}}(5-{\sqrt {19}})]^{4}}$

${\displaystyle \lambda ^{*}(21)={\tfrac {1}{8}}({\sqrt {14}}-{\sqrt {6}})[({\sqrt {3}}+1){\sqrt {2{\sqrt {7}}-4}}-4+{\sqrt {7}}-{\sqrt {3}}]}$

${\displaystyle \lambda ^{*}(23)={\tfrac {1}{32}}{\sqrt {2}}(5+{\sqrt {23}})[{\tfrac {2}{3}}+{\tfrac {1}{6}}({\sqrt {3}}+1){\sqrt[{3}]{100-12{\sqrt {69}}}}-{\tfrac {1}{6}}({\sqrt {3}}-1){\sqrt[{3}]{100+12{\sqrt {69}}}}]^{4}={\tfrac {1}{2}}{\sqrt {1+{\tfrac {1}{8}}\rho ^{-12}}}-{\tfrac {1}{2}}{\sqrt {1-{\tfrac {1}{8}}\rho ^{-12}}}}$

${\displaystyle \lambda ^{*}(25)={\tfrac {1}{2}}{\sqrt {2}}({\sqrt {5}}-2)(3-2{\sqrt[{4}]{5}})}$

${\displaystyle \lambda ^{*}(31)={\tfrac {1}{2}}{\sqrt {1+{\tfrac {1}{8}}\psi ^{-12}}}-{\tfrac {1}{2}}{\sqrt {1-{\tfrac {1}{8}}\psi ^{-12}}}}$

${\displaystyle \lambda ^{*}(37)={\tfrac {1}{2}}{\sqrt {1+({\sqrt {37}}-6)^{3}}}-{\tfrac {1}{2}}{\sqrt {1-({\sqrt {37}}-6)^{3}}}}$