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sn(u,k) などの Lambert Series まとめ

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$$\newcommand{BA}[0]{\begin{align*}} \newcommand{BE}[0]{\begin{equation}} \newcommand{bl}[0]{\boldsymbol} \newcommand{BM}[0]{\begin{matrix}} \newcommand{cd}[0]{{\rm cd}} \newcommand{cn}[0]{{\rm cn}} \newcommand{D}[0]{\displaystyle} \newcommand{dc}[0]{{\rm dc}} \newcommand{dn}[0]{{\rm dn}} \newcommand{EA}[0]{\end{align*}} \newcommand{EE}[0]{\end{equation}} \newcommand{EM}[0]{\end{matrix}} \newcommand{h}[0]{\boldsymbol{h}} \newcommand{k}[0]{\boldsymbol{k}} \newcommand{L}[0]{\left} \newcommand{l}[0]{\boldsymbol{l}} \newcommand{m}[0]{\boldsymbol{m}} \newcommand{n}[0]{\boldsymbol{n}} \newcommand{nc}[0]{{\rm nc}} \newcommand{nd}[0]{{\rm nd}} \newcommand{ns}[0]{{\rm ns}} \newcommand{R}[0]{\right} \newcommand{sc}[0]{{\rm sc}} \newcommand{sd}[0]{{\rm sd}} \newcommand{sn}[0]{{\rm sn}} \newcommand{vep}[0]{\varepsilon} $$

文献

Evaluations of infinite series involving reciprocal hyperbolic functions /Ce Xu
Hyperbolic summations derived using the Jacobi functions dc and nc /John M. Campbell
INFINITE FAMILIES OF EXACT SUMS OF SQUARES FORMULAS, JACOBI ELLIPTIC FUNCTIONS, CONTINUED FRACTIONS, AND SCHUR FUNCTIONS /Stephen C. Milne
The best-known properties and formulas for Jacobi functions(wolfram.com)

$\BA\D \sn(K(x)z,x)&=\frac{\pi}{xK(x)}\sum_{n=0}^\infty \frac{\sin\frac{\pi(2n+1)}{2}z}{\sinh\frac{\pi(2n+1)}{2}\cfrac{K'(x)}{K(x)}}\\ &=\L(K(x)z\R)^1-\frac{1+x^2}{3!}\L(K(x)z\R)^3+\frac{1+14x^2+x^4}{5!}\L(K(x)z\R)^5-\frac{(1+x^2)(1+134x^2+x^4)}{7!}\L(K(x)z\R)^7+\frac{1+1228x^2+5478x^4+1228x^6+x^8}{9!}\L(K(x)z\R)^9-\cdots\\ \cn(K(x)z,x)&=\frac{\pi}{xK(x)}\sum_{n=0}^\infty \frac{\cos\frac{\pi(2n+1)}{2}z}{\cosh\frac{\pi(2n+1)}{2}\cfrac{K'(x)}{K(x)}}\\ &=1-\frac{1}{2!}\L(K(x)z\R)^2+\frac{1+4x^2}{4!}\L(K(x)z\R)^4-\frac{1+44x^2+16x^4}{6!}\L(K(x)z\R)^6+\frac{1+408x^2+912x^4+64x^6}{8!}\L(K(x)z\R)^8-\cdots\\ \dn(K(x)z,x)&=\frac{\pi}{2K(x)}\sum_{n=-\infty}^\infty \frac{\cos \pi nz}{\cosh\pi n \cfrac{K'(x)}{K(x)}}\\ &=1-\frac{x^2}{2!}\L(K(x)z\R)^2+\frac{x^2(4+x^2)}{4!}\L(K(x)z\R)^4-\frac{x^2(16+44x^2+x^4)}{6!}\L(K(x)z\R)^6+\frac{x^2(64+912x^2+408x^4+x^6)}{8!}\L(K(x)z\R)^8-\cdots\\ \ns(K(x)z,x)&=\cfrac{\pi}{2K(x)}\L(\frac{1}{\sin\frac{\pi}{2}z}+4\sum_{n=0}^\infty \frac{\sin\frac{\pi(2n+1)}{2}z}{{e}^{\pi(2n+1)\frac{K'(x)}{K(x)}}-1}\R)\\ &=\frac{1}{K(x)z}+\frac{1+x^2}{6}\L(K(x)z\R)^1+\frac{7-22x^2+7x^4}{360}\L(K(x)z\R)^3+\frac{(1+x^2)(31-46x^2+31x^4)}{15120}\L(K(x)z\R)^5+\frac{127-284x^2+186x^4-284x^6+127x^8}{604800}\L(K(x)z\R)^7+\cdots\\ \nc(K(x)z,x)&=\cfrac{\pi}{2\sqrt{1-x^2}K(x)}\L(\frac{1}{\cos\frac{\pi}{2}z}-4\sum_{n=0}^\infty \frac{{(-1)}^n\cos\frac{\pi(2n+1)}{2}z}{{e}^{\pi(2n+1)\frac{K'(x)}{K(x)}}+1}\R)\\ &=1+\frac{1}{2!}(K(x)z)^2+\frac{5-4x^2}{4!}(K(x)z)^4+\frac{61-76x^2+16x^4}{6!}(K(x)z)^6+\frac{1385-2424x^2+1104x^4-64x^6}{8!}(K(x)z)^8+\cdots\\ \nd(K(x)z,x)&=\cfrac{\pi}{2\sqrt{1-x^2}K(x)}\sum_{n=-\infty}^\infty \frac{{(-1)}^n\cos \pi nz}{\cosh\pi n \cfrac{K'(x)}{K(x)}}\\ &=1+\frac{x^2}{2!}\L(K(x)z\R)^2+\frac{x^2(5x^2-4)}{4!}\L(K(x)z\R)^4+\frac{x^2(61x^2-76x^2+16)}{6!}\L(K(x)z\R)^6+\frac{x^2(1385x^6-2424x^4+1104x^2-64)}{8!}\L(K(x)z\R)^8+\cdots\\ \sd(K(x)z,x)&=\frac{2\pi}{x\sqrt{1-x^2}K(x)}\sum_{n=0}^\infty \frac{{(-1)}^n\sin\frac{\pi(2n+1)}{2}z}{\cosh\frac{\pi(2n+1)}{2}\cfrac{K'(x)}{K(x)}}\\ &=\L(K(x)z\R)^1+\frac{2x^2-1}{3!}\L(K(x)z\R)^3+\frac{16x^4-16x^2+1}{5!}\L(K(x)z\R)^5+\frac{(2x^2-1)(136x^4-136x^2+1)}{7!}\L(K(x)z\R)^7+\cdots\\ \cd(K(x)z,x)&=\frac{\pi}{xK(x)}\sum_{n=0}^\infty \frac{{(-1)}^n\cos\frac{\pi(2n+1)}{2}z}{\sinh\frac{\pi(2n+1)}{2}\frac{K'(x)}{K(x)}}\\ &=1-\frac{1-x^2}{2!}(K(x)z)^2+\frac{(1-x^2)(1-5x^2)}{4!}(K(x)z)^4-\frac{(1-x^2)(1-46x^2+61x^4)}{6!}(K(x)z)^6+\frac{(1-x^2)(1-411x^2+1731x^4-1385x^6)}{8!}(K(x)z)^8-\cdots\\ \sc(K(x)z,x)&=\frac{\pi}{2\sqrt{1-x^2}K(x)}\L(\tan\frac{\pi}{2}z+4\sum_{n=1}^\infty \frac{{(-1)}^n\sin\pi nz}{{e}^{2\pi n\frac{K'(x)}{K(x)}}-1}\R)\\ &=(K(x)z)^1+\frac{2-x^2}{3!}{(K(x)z)}^3+\frac{16-16x^2+x^4}{5!}{(K(x)z)}^5+\frac{(2-x^2)(136-136x^2+x^4)}{7!}{(K(x)z)}^7+\frac{7936-15872x^2+9168x^4-1232x^6+x^8}{9!}{(K(x)z)}^9+\cdots\\ \dc(K(x)z,x)&=\frac{\pi}{2K(x)}\L(\frac{1}{\cos \frac{\pi}{2}z}+4\sum_{n=0}^\infty \frac{{(-1)}^n\cos\frac{\pi(2n+1)}{2}z}{{e}^{\pi(2n+1)\frac{K'(x)}{K(x)}}-1}\R)\\ &=1+\frac{1-x^2}{2!}(K(x)z)^2+\frac{(1-x^2)(5-x^2)}{4!}(K(x)z)^4+\frac{(1-x^2)(61-46x^2+x^4)}{6!}(K(x)z)^6+\frac{(1-x^2)(1385-1731x^2+411x^4-x^6)}{8!}(K(x)z)^8+\cdots\\ \EA$


$\BA\D \sn ^2\L(K(x)z,x\R)&=\frac{1}{x^2}\L(1-\frac{E(x)}{K(x)}\R)-\cfrac{\pi^2}{x^2K(x)^2}\sum_{n=1}^\infty \frac{n\cos\pi nz}{\sinh \pi n\cfrac{K'(x)}{K(x)}}\\ &=\L(K(x)z\R)^2-\frac{1+x^2}{3}\L(K(x)z\R)^4+\frac{2+13x^2+2x^4}{45}\L(K(x)z\R)^6-\frac{(1+x^2)(1+29x^2+x^4)}{315}\L(K(x)z\R)^8+\cdots\\ \ns^2(K(x)z,x)&=1-\frac{E(x)}{K(x)}+\frac{1}{\sin^2 \frac{\pi}{2}z}-\frac{2\pi^2}{K(x)^2}\sum_{n=1}^\infty \frac{n\cos\pi nz}{{e}^{\pi n\cfrac{K'(x)}{K(x)}}-1}\\ &=\frac{1}{(K(x)z)^2}+\frac{1+x^2}{3}+\frac{1-x^2+x^4}{15}(K(x)z)^2+\frac{(1+x^2)(1-2x^2)(2-x^2)}{189}(K(x)z)^4+\frac{(1-x^2+x^4)^2}{675}(K(x)z)^6+\cdots\\ \sc^2(K(x)z,x)&=\frac{1}{1-x^2}\L(-\frac{E(x)}{K(x)}+\cfrac{\pi^2}{4K(x)^2}\frac{1}{\cos^2\frac{\pi}{2}z}+\cfrac{2\pi^2}{K(x)^2}\sum_{n=1}^\infty \frac{{(-1)}^{n-1}n\cos\pi nz}{{e}^{2\pi n\cfrac{K'(x)}{K(x)}}-1}\R)\\ &={(K(x)z)}^2+\frac{2-x^2}{3}{(K(x)z)}^4+\frac{17-17x^2+2x^4}{45}{(K(x)z)}^6+\frac{(2-x^2)(31-31x^2+x^4)}{315}{(K(x)z)}^8+\frac{1382-2764x^2+1641x^4-259x^6+2x^8}{14175}{(K(x)z)}^{10}+\cdots\\ \sd^2(K(x)z,x)&=\frac{1}{x^2(1-x^2)}\L(\frac{E(x)}{K(x)}-(1-x^2)+\cfrac{\pi^2}{K(x)^2}\sum_{n=1}^\infty \frac{{(-1)}^{n-1}n\cos\pi nz}{\sinh{\pi n\cfrac{K'(x)}{K(x)}}}\R)\\ &={(K(x)z)}^2+\frac{2x^2-1}{3}{(K(x)z)}^4+\frac{17x^4-17x^2+2}{45}{(K(x)z)}^6+\frac{(2x^2-1)(31x^4-31x^2+1)}{315}{(K(x)z)}^8+\frac{1382x^8-2764x^6+1641x^4-259x^2+2}{14175}{(K(x)z)}^{10}+\cdots\\ \EA$


$\BA\D \frac{\sn(K(x)z,x)\,\cn(K(x)z,x)}{\dn(K(x)z,x)}&=\frac{2\pi}{x^2K(x)}\sum_{n=0}^\infty \frac{\sin\pi(2n+1)z}{\sinh\pi(2n+1)\cfrac{K'(x)}{K(x)}}\\ &=\L(K(x)z\R)^1+\frac{x^2-2}{3}\L(K(x)z\R)^3+\frac{2(x^4-x^2+1)}{15}\L(K(x)z\R)^5+\frac{(x^2-2)(17x^4-2x^2+2)}{315}\L(K(x)z\R)^7+\cdots\\ \frac{\sn(K(x)z,x)\,\dn(K(x)z,x)}{\cn(K(x)z,x)}&=\frac{\pi}{2K(x)}\L(\tan\frac{\pi}{2}z+4\sum_{n=1}^\infty \frac{\sin\pi nz}{{e}^{\pi n\cfrac{K'(x)}{K(x)}}+{(-1)}^n}\R)\\ &=\L(K(x)z\R)^1+\frac{1-2x^2}{3}\L(K(x)z\R)^3+\frac{2(1-x^2+x^4)}{15}\L(K(x)z\R)^5+\frac{(1-2x^2)(17-2x^2+2x^4)}{315}\L(K(x)z\R)^7+\cdots\\ \frac{\sn(K(x)z,x)}{\cn(K(x)z,x)\,\dn(K(x)z,x)}&=\frac{\pi}{2(1-x^2)K(x)}\L(\tan\frac{\pi}{2}z+4\sum_{n=1}^\infty \frac{{(-1)}^n\sin\pi nz}{{e}^{\pi n\cfrac{K'(x)}{K(x)}}+1}\R)\\ &=\L(K(x)z\R)^1+\frac{1+x^2}{3}\L(K(x)z\R)^3+\frac{2(1-x^2+x^4)}{15}\L(K(x)z\R)^5+\frac{(1+x^2)(17-32x^2+17x^4)}{315}\L(K(x)z\R)^7+\cdots\\ \frac{\sn^2(K(x)z,x)\,\cn^2(K(x)z,x)}{\dn^2(K(x)z,x)}&=\frac{1}{x^4}\L(2-x^2-\frac{2E(x)}{K(x)}-\cfrac{4\pi^2}{K(x)^2}\sum_{n=1}^\infty \frac{n\cos2\pi nz}{\sinh2\pi n\cfrac{K'(x)}{K(x)}}\R)\\ &=\L(K(x)z\R)^2+\frac{2(x^2-2)}{3}\L(K(x)z\R)^4+\frac{17x^4-32x^2+32}{45}\L(K(x)z\R)^6+\frac{2(x^2-2)(31x^4-16x^2+16)}{315}\L(K(x)z\R)^8+\cdots\\ \frac{\sn^2(K(x)z,x)\,\dn^2(K(x)z,x)}{\cn^2(K(x)z,x)}&=1-\frac{2E(x)}{K(x)}+\cfrac{\pi^2}{4K(x)^2}\frac{1}{\cos^2\frac{\pi}{2}z}-\cfrac{2\pi^2}{K(x)^2}\sum_{n=1}^\infty \frac{n\cos\pi nz}{{e}^{\pi n\cfrac{K'(x)}{K(x)}}-{(-1)}^n}\\ &=\L(K(x)z\R)^2+\frac{2(1-2x^2)}{3}\L(K(x)z\R)^4+\frac{17-32x^2+32x^4}{45}\L(K(x)z\R)^6+\frac{(1-2x^2)(31-16x^2+16x^4)}{315}\L(K(x)z\R)^8+\cdots\\ \EA$

投稿日:30日前

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