Power series The
power series expansion of the lemniscate sine at the origin is :\operatorname{sl}z=\sum_{n=0}^\infty a_n z^n=z-12\frac{z^5}{5!}+3024\frac{z^9}{9!}-4390848\frac{z^{13}}{13!}+\cdots,\quad |z| where the coefficients a_n are determined as follows: :n\not\equiv 1\pmod 4\implies a_n=0, :a_1=1,\, \forall n\in\mathbb{N}_0:\,a_{n+2}=-\frac{2}{(n+1)(n+2)}\sum_{i+j+k=n}a_ia_ja_k where i+j+k=n stands for all three-term
compositions of n. For example, to evaluate a_{13}, it can be seen that there are only six compositions of 13-2=11 that give a nonzero contribution to the sum: 11=9+1+1=1+9+1=1+1+9 and 11=5+5+1=5+1+5=1+5+5, so :a_{13}=-\tfrac{2}{12\cdot 13}(a_9a_1a_1+a_1a_9a_1+a_1a_1a_9+a_5a_5a_1+a_5a_1a_5+a_1a_5a_5)=-\tfrac{11}{15600}. The expansion can be equivalently written as :\operatorname{sl}z=\sum_{n=0}^\infty p_{2n} \frac{z^{4n+1}}{(4n+1)!},\quad \left|z\right| where :p_{n+2}=-12\sum_{j=0}^n\binom{2n+2}{2j+2}p_{n-j}\sum_{k=0}^j \binom{2j+1}{2k+1}p_k p_{j-k},\quad p_0=1,\, p_1=0. The power series expansion of \tilde{\operatorname{sl}} at the origin is :\tilde{\operatorname{sl}}\,z=\sum_{n=0}^\infty \alpha_n z^n=z-9\frac{z^3}{3!}+153\frac{z^5}{5!}-4977\frac{z^7}{7!}+\cdots,\quad \left|z\right| where \alpha_n=0 if n is even and :\tilde{\operatorname{sl}}\, z=\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}} \left(\sum_{l=0}^n 2^l \binom{2n+2}{2l+1} s_l t_{n-l}\right)\frac{z^{2n+1}}{(2n+1)!} ,\quad \left|z\right| where :s_{n+2}=3 s_{n+1} +24 \sum_{j=0}^n \binom{2n+2}{2j+2} s_{n-j} \sum_{k=0}^j \binom{2j+1}{2k+1} s_k s_{j-k},\quad s_0=1,\, s_1=3, :t_{n+2}=3 t_{n+1}+3 \sum_{j=0}^n \binom{2n+2}{2j+2} t_{n-j} \sum_{k=0}^j \binom{2j+1}{2k+1} t_k t_{j-k},\quad t_0=1,\, t_1=3. For the lemniscate cosine, :\operatorname{cl}{z}=1-\sum_{n=0}^\infty (-1)^n \left(\sum_{l=0}^n 2^l \binom{2n+2}{2l+1} q_l r_{n-l}\right) \frac{z^{2n+2}}{(2n+2)!}=1-2\frac{z^2}{2!}+12\frac{z^4}{4!}-216\frac{z^6}{6!}+\cdots ,\quad \left|z\right| :\tilde{\operatorname{cl}}\,z=\sum_{n=0}^\infty (-1)^n 2^n q_n \frac{z^{2n}}{(2n)!}=1-3\frac{z^2}{2!}+33\frac{z^4}{4!}-819\frac{z^6}{6!}+\cdots ,\quad\left|z\right| where :r_{n+2}=3 \sum_{j=0}^n \binom{2n+2}{2j+2} r_{n-j} \sum_{k=0}^j \binom{2j+1}{2k+1} r_k r_{j-k},\quad r_0=1,\, r_1=0, :q_{n+2}=\tfrac{3}{2} q_{n+1}+6 \sum_{j=0}^n \binom{2n+2}{2j+2} q_{n-j} \sum_{k=0}^j \binom{2j+1}{2k+1} q_k q_{j-k},\quad q_0=1, \,q_1=\tfrac{3}{2}.
Ramanujan's cos/cosh identity Ramanujan's famous cos/cosh identity states that if :R(s)=\frac{\pi}{\varpi\sqrt{2}}\sum_{n\in\mathbb{Z}}\frac{\cos (2n\pi s/\varpi)}{\cosh n\pi}, then :R(s)^{-2}+R(is)^{-2}=2,\quad \left|\operatorname{Re}s\right| There is a close relation between the lemniscate functions and R(s). Indeed, :\tilde{\operatorname{sl}}\,s=-\frac{\mathrm d}{\mathrm ds}R(s)\quad \left|\operatorname{Im}s\right| :\tilde{\operatorname{cl}}\,s=\frac{\mathrm d}{\mathrm ds}\sqrt{1-R(s)^2},\quad \left|\operatorname{Re}s-\frac{\varpi}{2}\right| and :R(s)=\frac{1}{\sqrt{1+\operatorname{sl}^2 s}},\quad \left|\operatorname{Im}s\right
Continued fractions For z\in\mathbb{C}\setminus\{0\}: :\int_0^\infty e^{-tz\sqrt{2}}\operatorname{cl}t\, \mathrm dt=\cfrac{1/\sqrt{2}}{z+\cfrac{a_1}{z+\cfrac{a_2}{z+\cfrac{a_3}{z+\ddots}}}},\quad a_n=\frac{n^2}{4}((-1)^{n+1}+3) :\int_0^\infty e^{-tz\sqrt{2}}\operatorname{sl}t\operatorname{cl}t \, \mathrm dt=\cfrac{1/2}{z^2+b_1-\cfrac{a_1}{z^2+b_2-\cfrac{a_2}{z^2+b_3-\ddots}}},\quad a_n=n^2(4n^2-1),\, b_n=3(2n-1)^2
Methods of computation {{quote box {{ubl |item_style=padding:0.2em 0 0 1.6em; | a_0 \leftarrow 1; b_0 \leftarrow \tfrac{1}{\sqrt2}; c_0 \leftarrow\sqrt{\tfrac12} |
for each n\ge 1
do {{ubl |item_style=padding:0.2em 0 0 1.6em; | a_n \leftarrow \tfrac12(a_{n-1}+b_{n-1}); b_n \leftarrow \sqrt{a_{n-1}b_{n-1}}; c_n \leftarrow \tfrac12(a_{n-1}-b_{n-1}) |
if c_n
then }} | \phi_N \leftarrow 2^N a_N \sqrt2x |
for each from to
do {{ubl |item_style=padding:0.2em 0 0 1.6em; | \phi_{n-1} \leftarrow \tfrac12\left(\phi_n + {\arcsin}{\left(\frac{c_n}{a_n}\sin \phi_n\right)}\right) }} |
return \frac{\sin \phi_0}{\sqrt{2-\sin^2\phi_0}} }} This is effectively using the arithmetic-geometric mean and is based on
Landen's transformations. }} Several methods of computing \operatorname{sl} x involve first making the change of variables \pi x = \varpi \tilde{x} and then computing \operatorname{sl}(\varpi \tilde{x} / \pi). A
hyperbolic series method: :\operatorname{sl}\left(\frac{\varpi}{\pi}x\right)=\frac{\pi}{\varpi}\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{\cosh (x-(n+1/2)\pi)},\quad x\in\mathbb{C} :\frac{1}{\operatorname{sl}(\varpi x/\pi)} = \frac\pi\varpi \sum_{n\in\mathbb{Z}}\frac{(-1)^n}{{\sinh} {\left(x-n\pi\right)}}=\frac\pi\varpi \sum_{n\in\mathbb{Z}}\frac{(-1)^n}{\sin (x-n\pi i)},\quad x\in\mathbb{C}
Fourier series method: :\operatorname{sl}\Bigl(\frac{\varpi}{\pi}x\Bigr)=\frac{2\pi}{\varpi}\sum_{n=0}^\infty \frac{(-1)^n\sin ((2n+1)x)}{\cosh ((n+1/2)\pi)},\quad \left|\operatorname{Im}x\right| :\operatorname{cl}\left(\frac{\varpi}{\pi}x\right)=\frac{2\pi}{\varpi}\sum_{n=0}^\infty \frac{\cos ((2n+1)x)}{\cosh ((n+1/2)\pi)},\quad\left|\operatorname{Im}x\right| :\frac{1}{\operatorname{sl}(\varpi x/\pi)}=\frac{\pi}{\varpi}\left(\frac{1}{\sin x}-4\sum_{n=0}^\infty \frac{\sin ((2n+1)x)}{e^{(2n+1)\pi}+1}\right),\quad\left|\operatorname{Im}x\right| The lemniscate functions can be computed more rapidly by :\begin{align}\operatorname{sl}\Bigl(\frac\varpi\pi x\Bigr)& = \frac{{\theta_1}{\left(x, e^{-\pi}\right)}}{{\theta_3}{\left(x, e^{-\pi}\right)}},\quad x\in\mathbb{C}\\ \operatorname{cl}\Bigl(\frac\varpi\pi x\Bigr)&=\frac{{\theta_2}{\left(x, e^{-\pi}\right)}}{{\theta_4}{\left(x, e^{-\pi}\right)}},\quad x\in\mathbb{C}\end{align} where :\begin{aligned} \theta_1(x,e^{-\pi})&=\sum_{n\in\mathbb{Z}}(-1)^{n+1}e^{-\pi (n+1/2+x/\pi)^2}=\sum_{n\in\mathbb{Z}} (-1)^n e^{-\pi (n+1/2)^2}\sin ((2n+1)x),\\ \theta_2(x,e^{-\pi})&=\sum_{n\in\mathbb{Z}}(-1)^n e^{-\pi (n+x/\pi)^2}=\sum_{n\in\mathbb{Z}} e^{-\pi (n+1/2)^2}\cos ((2n+1)x),\\ \theta_3(x,e^{-\pi})&=\sum_{n\in\mathbb{Z}}e^{-\pi (n+x/\pi)^2}=\sum_{n\in\mathbb{Z}} e^{-\pi n^2}\cos 2nx,\\ \theta_4(x,e^{-\pi})&=\sum_{n\in\mathbb{Z}}e^{-\pi (n+1/2+x/\pi)^2}=\sum_{n\in\mathbb{Z}} (-1)^n e^{-\pi n^2}\cos 2nx\end{aligned} are the
Jacobi theta functions. Fourier series for the
logarithm of the lemniscate sine: :\ln \operatorname{sl}\left(\frac\varpi\pi x\right)=\ln 2-\frac{\pi}{4}+\ln\sin x+2\sum_{n=1}^\infty \frac{(-1)^n \cos 2nx}{n(e^{n\pi}+(-1)^n)},\quad \left|\operatorname{Im}x\right| The following series identities were discovered by
Ramanujan: :\frac{\varpi ^2}{\pi ^2\operatorname{sl}^2(\varpi x/\pi)}=\frac{1}{\sin ^2x}-\frac{1}{\pi}-8\sum_{n=1}^\infty \frac{n\cos 2nx}{e^{2n\pi}-1},\quad \left|\operatorname{Im}x\right| :\arctan\operatorname{sl}\Bigl(\frac\varpi\pi x\Bigr)=2\sum_{n=0}^\infty \frac{\sin((2n+1)x)}{(2n+1)\cosh ((n+1/2)\pi)},\quad \left|\operatorname{Im}x\right| The functions \tilde{\operatorname{sl}} and \tilde{\operatorname{cl}} analogous to \sin and \cos on the unit circle have the following Fourier and hyperbolic series expansions: :\tilde{\operatorname{sl}}\,s=2\sqrt{2}\frac{\pi^2}{\varpi^2}\sum_{n=1}^\infty\frac{n\sin (2n\pi s/\varpi)}{\cosh n\pi},\quad \left|\operatorname{Im}s\right| :\tilde{\operatorname{cl}}\,s=\sqrt{2}\frac{\pi^2}{\varpi^2}\sum_{n=0}^\infty \frac{(2n+1)\cos ((2n+1)\pi s/\varpi)}{\sinh ((n+1/2)\pi)},\quad \left|\operatorname{Im}s\right| :\tilde{\operatorname{sl}}\,s=\frac{\pi^2}{\varpi^2\sqrt{2}}\sum_{n\in\mathbb{Z}}\frac{\sinh (\pi (n+s/\varpi))}{\cosh^2 (\pi (n+s/\varpi))},\quad s\in\mathbb{C} :\tilde{\operatorname{cl}}\,s=\frac{\pi^2}{\varpi^2\sqrt{2}}\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{\cosh^2 (\pi (n+s/\varpi))},\quad s\in\mathbb{C} The following identities come from product representations of the theta functions: :\mathrm{sl}\Bigl(\frac\varpi\pi x\Bigr) = 2e^{-\pi/4}\sin x\prod_{n = 1}^{\infty} \frac{1-2e^{-2n\pi}\cos 2x+e^{-4n\pi}}{1+2e^{-(2n-1)\pi}\cos 2x+e^{-(4n-2)\pi}},\quad x\in\mathbb{C} :\mathrm{cl}\Bigl(\frac\varpi\pi x\Bigr) = 2e^{-\pi/4}\cos x\prod_{n = 1}^{\infty} \frac{1+2e^{-2n\pi}\cos 2x+e^{-4n\pi}}{1-2e^{-(2n-1)\pi}\cos 2x+e^{-(4n-2)\pi}},\quad x\in\mathbb{C} A similar formula involving the \operatorname{sn} function can be given.
The lemniscate functions as a ratio of entire functions Since the lemniscate sine is a meromorphic function in the whole complex plane, it can be written as a ratio of
entire functions. Gauss showed that has the following product expansion, reflecting the distribution of its zeros and poles: :\operatorname{sl}z=\frac{M(z)}{N(z)} where :M(z)=z\prod_{\alpha}\left(1-\frac{z^4}{\alpha^4}\right),\quad N(z)=\prod_{\beta}\left(1-\frac{z^4}{\beta^4}\right). Here, \alpha and \beta denote, respectively, the zeros and poles of which are in the quadrant \operatorname{Re}z>0,\operatorname{Im}z\ge 0. A proof can be found in. Importantly, the infinite products converge to the same value for all possible orders in which their terms can be multiplied, as a consequence of
uniform convergence.
Proof by logarithmic differentiation It can be easily seen (using uniform and
absolute convergence arguments to justify
interchanging of limiting operations) that :\frac{M'(z)}{M(z)}=-\sum_{n=0}^\infty 2^{4n}\mathrm{H}_{4n}\frac{z^{4n-1}}{(4n)!},\quad \left|z\right| (where \mathrm{H}_n are the Hurwitz numbers defined in
Lemniscate elliptic functions § Hurwitz numbers) and :\frac{N'(z)}{N(z)}=(1+i)\frac{M'((1+i)z)}{M((1+i)z)}-\frac{M'(z)}{M(z)}. Therefore :\frac{N'(z)}{N(z)}=\sum_{n=0}^\infty 2^{4n}(1-(-1)^n 2^{2n})\mathrm{H}_{4n}\frac{z^{4n-1}}{(4n)!},\quad \left|z\right| It is known that :\frac{1}{\operatorname{sl}^2z}=\sum_{n=0}^\infty 2^{4n}(4n-1)\mathrm{H}_{4n}\frac{z^{4n-2}}{(4n)!},\quad \left|z\right| Then from :\frac{\mathrm d}{\mathrm dz}\frac{\operatorname{sl}'z}{\operatorname{sl}z}=-\frac{1}{\operatorname{sl}^2z}-\operatorname{sl}^2z and :\operatorname{sl}^2z=\frac{1}{\operatorname{sl}^2z}-\frac{(1+i)^2}{\operatorname{sl}^2((1+i)z)} we get :\frac{\operatorname{sl}'z}{\operatorname{sl}z}=-\sum_{n=0}^\infty 2^{4n}(2-(-1)^n 2^{2n})\mathrm{H}_{4n}\frac{z^{4n-1}}{(4n)!},\quad \left|z\right| Hence :\frac{\operatorname{sl}'z}{\operatorname{sl}z}=\frac{M'(z)}{M(z)}-\frac{N'(z)}{N(z)},\quad \left|z\right| Therefore :\operatorname{sl}z=C\frac{M(z)}{N(z)} for some constant C for \left|z\right| but this result holds for all z\in\mathbb{C} by analytic continuation. Using :\lim_{z\to 0}\frac{\operatorname{sl}z}{z}=1 gives C=1 which completes the proof. \blacksquare '''Proof by Liouville's theorem''' Let :f(z)=\frac{M(z)}{N(z)}=\frac{(1+i)M(z)^2}{M((1+i)z)}, with patches at removable singularities. The shifting formulas :M(z+2\varpi)=e^{2\frac{\pi}{\varpi}(z+\varpi)}M(z),\quad M(z+2\varpi i)=e^{-2\frac{\pi}{\varpi}(iz-\varpi)}M(z) imply that f is an elliptic function with periods 2\varpi and 2\varpi i, just as \operatorname{sl}. It follows that the function g defined by :g(z)=\frac{\operatorname{sl}z}{f(z)}, when patched, is an elliptic function without poles. By
Liouville's theorem, it is a constant. By using \operatorname{sl}z=z+\operatorname{O}(z^5), M(z)=z+\operatorname{O}(z^5) and N(z)=1+\operatorname{O}(z^4), this constant is 1, which proves the theorem. \blacksquare Gauss conjectured that \ln N(\varpi)=\pi/2 (this later turned out to be true) and commented that this “is most remarkable and a proof of this property promises the most serious increase in analysis”. Gauss expanded the products for M and N as infinite series (see below). He also discovered several identities involving the functions M and N, such as :N(z)=\frac{M((1+i)z)}{(1+i)M(z)},\quad z\notin \varpi\mathbb{Z}[i] and :N(2z)=M(z)^4+N(z)^4. Thanks to a certain theorem on splitting limits, we are allowed to multiply out the infinite products and collect like powers of z. Doing so gives the following power series expansions that are convergent everywhere in the complex plane: :M(z)=z-2\frac{z^5}{5!}-36\frac{z^9}{9!}+552\frac{z^{13}}{13!}+\cdots,\quad z\in\mathbb{C} :N(z)=1+2\frac{z^4}{4!}-4\frac{z^8}{8!}+408\frac{z^{12}}{12!}+\cdots,\quad z\in\mathbb{C}. This can be contrasted with the power series of \operatorname{sl} which has only finite radius of convergence (because it is not entire). We define S and T by :S(z)=N\left(\frac{z}{1+i}\right)^2-iM\left(\frac{z}{1+i}\right)^2,\quad T(z)=S(iz). Then the lemniscate cosine can be written as :\operatorname{cl}z=\frac{S(z)}{T(z)} where :S(z)=1-\frac{z^2}{2!}-\frac{z^4}{4!}-3\frac{z^6}{6!}+17\frac{z^8}{8!}-9\frac{z^{10}}{10!}+111\frac{z^{12}}{12!}+\cdots,\quad z\in\mathbb{C} :T(z)=1+\frac{z^2}{2!}-\frac{z^4}{4!}+3\frac{z^6}{6!}+17\frac{z^8}{8!}+9\frac{z^{10}}{10!}+111\frac{z^{12}}{12!}+\cdots,\quad z\in\mathbb{C}. Furthermore, the identities :M(2z)=2 M(z) N(z) S(z) T(z), :S(2z)=S(z)^4-2M(z)^4, :T(2z)=T(z)^4-2M(z)^4 and the Pythagorean-like identities :M(z)^2+S(z)^2=N(z)^2, :M(z)^2+N(z)^2=T(z)^2 hold for all z\in\mathbb{C}. The quasi-addition formulas :M(z+w)M(z-w)=M(z)^2N(w)^2-N(z)^2M(w)^2, :N(z+w)N(z-w)=N(z)^2N(w)^2+M(z)^2M(w)^2 (where z,w\in\mathbb{C}) imply further multiplication formulas for M and N by recursion. Gauss' M and N satisfy the following system of differential equations: :M(z)M''(z)=M'(z)^2-N(z)^2, :N(z)N''(z)=N'(z)^2+M(z)^2 where z\in\mathbb{C}. Both M and N satisfy the differential equation :X(z)X''''(z)=4X'(z)X'
(z)-3X(z)^2+2X(z)^2,\quad z\in\mathbb{C}. The functions can be also expressed by integrals involving elliptic functions: :M(z)=z\exp\left(-\int_0^z\int_0^w \left(\frac{1}{\operatorname{sl}^2v}-\frac{1}{v^2}\right)\, \mathrm dv\,\mathrm dw\right), :N(z)=\exp\left(\int_0^z\int_0^w \operatorname{sl}^2v\,\mathrm dv\,\mathrm dw\right) where the contours do not cross the poles; while the innermost integrals are path-independent, the outermost ones are path-dependent; however, the path dependence cancels out with the non-injectivity of the complex
exponential function. An alternative way of expressing the lemniscate functions as a ratio of entire functions involves the theta functions (see
Lemniscate elliptic functions § Methods of computation); the relation between M,N and \theta_1,\theta_3 is :M(z)=2^{-1/4}e^{\pi z^2/(2\varpi^2)}\sqrt{\frac{\pi}{\varpi}}\theta_1\left(\frac{\pi z}{\varpi},e^{-\pi}\right), :N(z)=2^{-1/4}e^{\pi z^2/(2\varpi^2)}\sqrt{\frac{\pi}{\varpi}}\theta_3\left(\frac{\pi z}{\varpi},e^{-\pi}\right) where z\in\mathbb{C}. == Relation to other functions ==