Lemniscate elliptic functions

The lemniscate functions sl and cl can be defined as the solution to the initial value problem:[3]

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {cl} z=-\operatorname {sl} z\,{\bigl (}1+\operatorname {cl} ^{2}z{\bigr )},\ {\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {sl} z=\operatorname {cl} z\,{\bigl (}1+\operatorname {sl} ^{2}z{\bigr )},\ \operatorname {cl} (0)=1,\ \operatorname {sl} (0)=0,}$

or equivalently as the inverses of an elliptic integral, the Schwarz–Christoffel map from the complex unit disk to a square with corners ${\displaystyle {\big \{}{\tfrac {1}{2}}\varpi ,{\tfrac {1}{2}}\varpi i,-{\tfrac {1}{2}}\varpi ,-{\tfrac {1}{2}}\varpi i{\big \}}\colon }$[4]

${\displaystyle z=\int _{0}^{\operatorname {sl} z}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}=\int _{\operatorname {cl} z}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}.}$

Beyond that square, the functions can be analytically continued to the whole complex plane by a series of reflections.

By comparison, the circular sine and cosine can be defined as the solution to the initial value problem:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\cos z=-\sin z,\ {\frac {\mathrm {d} }{\mathrm {d} z}}\sin z=\cos z,\ \cos(0)=1,\ \sin(0)=0,}$

or as inverses of a map from the upper half-plane to a half-infinite strip with real part between ${\displaystyle -{\tfrac {1}{2}}\pi ,{\tfrac {1}{2}}\pi }$ and positive imaginary part:

${\displaystyle z=\int _{0}^{\sin z}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}=\int _{\cos z}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}.}$

The lemniscate sine and cosine relate the arc length of an arc of the lemniscate to the distance of one endpoint from the origin.

The trigonometric sine and cosine analogously relate the arc length of an arc of a unit-diameter circle to the distance of one endpoint from the origin.

The lemniscate of Bernoulli with half-width 1 is the locus of points in the plane such that the product of their distances from the two focal points ${\displaystyle F_{1}={\bigl (}{-{\tfrac {1}{\sqrt {2}}}},0{\bigr )}}$ and ${\displaystyle F_{2}={\bigl (}{\tfrac {1}{\sqrt {2}}},0{\bigr )}}$ is the constant ${\displaystyle {\tfrac {1}{2}}}$. This is a quartic curve satisfying the polar equation ${\displaystyle r^{2}=\cos 2\theta }$ or the Cartesian equation ${\displaystyle {\bigl (}x^{2}+y^{2}{\bigr )}{}^{2}=x^{2}-y^{2}.}$

The points on the lemniscate at distance ${\displaystyle r}$ from the origin are the intersections of the circle ${\displaystyle x^{2}+y^{2}=r^{2}}$ and the hyperbola ${\displaystyle x^{2}-y^{2}=r^{4}}$. The intersection in the positive quadrant has Cartesian coordinates:

${\displaystyle {\big (}x(r),y(r){\big )}={\biggl (}\!{\sqrt {{\tfrac {1}{2}}r^{2}{\bigl (}1+r^{2}{\bigr )}}},{\sqrt {{\tfrac {1}{2}}r^{2}{\bigl (}1-r^{2}{\bigr )}}}\,{\biggr )}.}$

Using this parametrization with ${\displaystyle r\in [0,1]}$ for a quarter of the lemniscate, the arc length from the origin to a point ${\displaystyle {\big (}x(r),y(r){\big )}}$ is:[5]

${\displaystyle {\begin{aligned}&\int _{0}^{r}{\sqrt {x'(t)^{2}+y'(t)^{2}}}\mathop {\mathrm {d} t} \\&\quad {}=\int _{0}^{r}{\sqrt {{\frac {(1+2t^{2})^{2}}{2(1+t^{2})}}+{\frac {(1-2t^{2})^{2}}{2(1-t^{2})}}}}\mathop {\mathrm {d} t} \\[6mu]&\quad {}=\int _{0}^{r}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}\\[6mu]&\quad {}=\operatorname {arcsl} r.\end{aligned}}}$

Likewise, the arc length from ${\displaystyle (1,0)}$ to ${\displaystyle {\big (}x(r),y(r){\big )}}$ is:

${\displaystyle {\begin{aligned}&\int _{r}^{1}{\sqrt {x'(t)^{2}+y'(t)^{2}}}\mathop {\mathrm {d} t} \\&\quad {}=\int _{r}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}\\[6mu]&\quad {}=\operatorname {arccl} r={\tfrac {1}{2}}\varpi -\operatorname {arcsl} r.\end{aligned}}}$

Or in the inverse direction, the lemniscate sine and cosine functions give the distance from the origin as functions of arc length from the origin and the point ${\displaystyle (1,0)}$, respectively.

Analogously, the circular sine and cosine functions relate the chord length to the arc length for the unit diameter circle with polar equation ${\displaystyle r=\cos \theta }$ or Cartesian equation ${\displaystyle x^{2}+y^{2}=x,}$ using the same argument above but with the parametrization:

${\displaystyle {\big (}x(r),y(r){\big )}={\biggl (}r^{2},{\sqrt {r^{2}{\bigl (}1-r^{2}{\bigr )}}}\,{\biggr )}.}$

The lemniscate integral and lemniscate functions satisfy an argument duplication identity discovered by Fagnano in 1718:[6]

${\displaystyle \int _{0}^{z}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}=2\int _{0}^{u}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}},\quad {\text{if }}z={\frac {2u{\sqrt {1-u^{4}}}}{1+u^{4}}}{\text{ and }}0\leq u\leq {\sqrt {{\sqrt {2}}-1}}.}$

Later mathematicians generalized this result. Analogously to the constructible polygons in the circle, the lemniscate can be divided into n sections of equal arc length using straightedge and compass whenever n is of the form ${\displaystyle n=2^{k}p_{1}p_{2}\cdots p_{m}}$ where k and m are non-negative integers and each p_i (if any) is a distinct Fermat prime.[7] This was demonstrated by Niels Abel in 1827–1828.[8]

The lemniscate sine relates the arc length to the x coordinate in the rectangular elastica.

The inverse lemniscate sine also describes the arc length s relative to the x coordinate of the rectangular elastica.[9] This curve has y coordinate and arc length:

${\displaystyle y=\int _{x}^{1}{\frac {t^{2}\mathop {\mathrm {d} t} }{\sqrt {1-t^{4}}}},\quad s=\operatorname {arcsl} x=\int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}}$

The rectangular elastica solves a problem posed by Jacob Bernoulli, in 1691, to describe the shape of an idealized flexible rod fixed in a vertical orientation at the bottom end and pulled down by a weight from the far end until it has been bent horizontal. Bernoulli's proposed solution established Euler–Bernoulli beam theory, further developed by Euler in the 18th century.

The lemniscate constant is twice the value of the complete lemniscate integral.

The lemniscate functions have minimal real period 2ϖ and fundamental complex periods ${\displaystyle (1+i)\varpi }$ and ${\displaystyle (1-i)\varpi }$ for a constant ϖ called the lemniscate constant,[10]

${\displaystyle {\begin{aligned}\varpi &=2\int _{0}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}={\sqrt {2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {1+t^{4}}}}=\int _{0}^{1}{\frac {\mathrm {d} t}{\sqrt {t-t^{3}}}}\\[6mu]&=4\int _{0}^{\infty }{\Bigl (}{\sqrt[{4}]{1+t^{4}}}-t{\Bigr )}\,\mathrm {d} t=2{\sqrt {2}}\int _{0}^{1}{\sqrt[{4}]{1-t^{4}}}\mathop {\mathrm {d} t} =3\int _{0}^{1}{\sqrt {1-t^{4}}}\,\mathrm {d} t\\[2mu]&={\tfrac {1}{2}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{2}}{\bigr )}={\frac {\Gamma {\bigl (}{\tfrac {1}{4}}{\bigr )}^{2}}{2{\sqrt {2\pi }}}}={\frac {\pi }{{\operatorname {M} }{\bigl (}1,{\sqrt {2}}{\bigr )}}}={\sqrt {\pi }}e^{\beta '(0)}\\[5mu]&=2.62205\;75542\;92119\;81046\;48395\;89891\;11941\ldots ,\end{aligned}}}$

where Β is the beta function, Γ is the gamma function, M is the arithmetic–geometric mean and β is the Dirichlet beta function. The related constant ${\displaystyle G=\varpi /\pi =0.8346\ldots }$ is Gauss's constant. The lemniscate constant was proven transcendental by Theodor Schneider in 1937.[11] In 1975, Gregory Chudnovsky proved that π and ϖ are algebraically independent over ${\displaystyle \mathbb {Q} }$.[12][13]

Geometrically, ϖ is the ratio of the perimeter of Bernoulli's lemniscate to its diameter. The lemniscate functions satisfy the basic relation ${\displaystyle \operatorname {cl} z={\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\varpi -z{\bigr )}.}$

Furthermore, ϖ is related to the area under the curve ${\displaystyle x^{4}+y^{4}=1}$. Defining ${\displaystyle \pi _{n}\mathrel {:=} \mathrm {B} {\bigl (}{\tfrac {1}{n}},{\tfrac {1}{n}}{\bigr )}}$, twice the area in the positive quadrant under the curve ${\displaystyle x^{n}+y^{n}=1}$ is ${\textstyle 2\int _{0}^{1}{\sqrt[{n}]{1-x^{n}}}\mathop {\mathrm {d} x} ={\tfrac {1}{n}}\pi _{n}.}$ In the quartic case, ${\displaystyle {\tfrac {1}{4}}\pi _{4}={\tfrac {1}{\sqrt {2}}}\varpi .}$

Euler discovered in 1738 that for the rectangular elastica:[14]

${\displaystyle {\textrm {arc}}\ {\textrm {length}}\cdot {\textrm {height}}=\int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt {1-x^{4}}}}\cdot \int _{0}^{1}{\frac {x^{2}\mathop {\mathrm {d} x} }{\sqrt {1-x^{4}}}}={\frac {\varpi }{2}}\cdot {\frac {\pi }{2\varpi }}={\frac {\pi }{4}}}$

Viète's formula for π can be written:

${\displaystyle {\frac {2}{\pi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}}}\cdots }$

An analogous formula for ϖ is:[15]

${\displaystyle {\frac {2}{\varpi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\bigg /}\!{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\Bigg /}\!{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\bigg /}\!{\sqrt {\frac {1}{2}}}}}}}\cdots }$

The Wallis product for π is:

${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\biggl (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\biggr )}{\biggl (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\biggr )}{\biggl (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\biggr )}\cdots }$

An analogous formula for ϖ is:[16]

${\displaystyle {\frac {\varpi }{2}}=\prod _{n=1}^{\infty }\left({\frac {4n-1}{4n-2}}\cdot {\frac {4n}{4n+1}}\right)={\biggl (}{\frac {3}{2}}\cdot {\frac {4}{5}}{\biggr )}{\biggl (}{\frac {7}{6}}\cdot {\frac {8}{9}}{\biggr )}{\biggl (}{\frac {11}{10}}\cdot {\frac {12}{13}}{\biggr )}\cdots }$

The ratio of these two is:

${\displaystyle {\frac {\varpi }{\pi }}=\prod _{n=1}^{\infty }\left({\frac {4n-1}{4n}}\cdot {\frac {4n+2}{4n+1}}\right)={\biggl (}{\frac {3}{4}}\cdot {\frac {6}{5}}{\biggr )}{\biggl (}{\frac {7}{8}}\cdot {\frac {10}{9}}{\biggr )}{\biggl (}{\frac {11}{12}}\cdot {\frac {14}{13}}{\biggr )}\cdots }$

An infinite series for ${\displaystyle \varpi /\pi }$ discovered by Gauss is:[17]

${\displaystyle {\frac {\varpi }{\pi }}=\sum _{n=0}^{\infty }(-1)^{n}\prod _{k=1}^{n}{\frac {(2k-1)^{2}}{(2k)^{2}}}=1-{\frac {1^{2}}{2^{2}}}+{\frac {1^{2}\cdot 3^{2}}{2^{2}\cdot 4^{2}}}-{\frac {1^{2}\cdot 3^{2}\cdot 5^{2}}{2^{2}\cdot 4^{2}\cdot 6^{2}}}+\cdots }$

The Machin formula for π is ${\textstyle {\tfrac {1}{4}}\pi =4\arctan {\tfrac {1}{5}}-\arctan {\tfrac {1}{239}},}$ and several similar formulas for π can be developed using trigonometric angle sum identities, e.g. Euler's formula ${\textstyle {\tfrac {1}{4}}\pi =\arctan {\tfrac {1}{2}}+\arctan {\tfrac {1}{3}}}$. Analogous formulas can be developed for ϖ, including the following found by Gauss: ${\displaystyle {\tfrac {1}{2}}\varpi =2\operatorname {arcsl} {\tfrac {1}{2}}+\operatorname {arcsl} {\tfrac {7}{23}}.}$[18]

In a spirit similar to that of the Basel problem,

${\displaystyle \sum _{z\in \mathbf {Z} [i]\setminus \{0\}}{\frac {1}{z^{4}}}=G_{4}(i)={\frac {\varpi ^{4}}{15}}}$

where ${\displaystyle \mathbf {Z} [i]}$ are the Gaussian integers and ${\displaystyle G_{4}(\tau )}$ is the Eisenstein series of weight ${\displaystyle 4}$.[19]

At translations of ${\displaystyle {\tfrac {1}{2}}\varpi }$ the lemniscate functions cl and sl are exchanged, and at translations of ${\displaystyle {\tfrac {1}{2}}i\varpi }$ they are additionally rotated and reciprocated:

${\displaystyle {\begin{aligned}{\operatorname {cl} }{\bigl (}z\pm {\tfrac {1}{2}}\varpi {\bigr )}&=\mp \operatorname {sl} z,&{\operatorname {cl} }{\bigl (}z\pm {\tfrac {1}{2}}i\varpi {\bigr )}&={\frac {\mp i}{\operatorname {sl} z}}\\[6mu]{\operatorname {sl} }{\bigl (}z\pm {\tfrac {1}{2}}\varpi {\bigr )}&=\pm \operatorname {cl} z,&{\operatorname {sl} }{\bigl (}z\pm {\tfrac {1}{2}}i\varpi {\bigr )}&={\frac {\pm i}{\operatorname {cl} z}}\end{aligned}}}$

Doubling these to translations by a unit-Gaussian-integer multiple of ${\displaystyle \varpi }$ (that is, ${\displaystyle \pm \varpi }$ or ${\displaystyle \pm i\varpi }$), negates each function, an involution:

${\displaystyle {\begin{aligned}\operatorname {cl} (z+\varpi )&=\operatorname {cl} (z+i\varpi )=-\operatorname {cl} z\\[4mu]\operatorname {sl} (z+\varpi )&=\operatorname {sl} (z+i\varpi )=-\operatorname {sl} z\end{aligned}}}$

As a result, both functions are invariant under translation by an even-Gaussian-integer multiple of ${\displaystyle \varpi }$.[20] That is, a displacement ${\displaystyle (a+bi)\varpi ,}$ with ${\displaystyle a+b=2k}$ for integers a, b, and k.

${\displaystyle {\begin{aligned}{\operatorname {cl} }{\bigl (}z+(1+i)\varpi {\bigr )}&={\operatorname {cl} }{\bigl (}z+(1-i)\varpi {\bigr )}=\operatorname {cl} z\\[4mu]{\operatorname {sl} }{\bigl (}z+(1+i)\varpi {\bigr )}&={\operatorname {sl} }{\bigl (}z+(1-i)\varpi {\bigr )}=\operatorname {sl} z\end{aligned}}}$

This makes them elliptic functions (doubly periodic meromorphic functions in the complex plane) with a diagonal square period lattice of fundamental periods ${\displaystyle (1+i)\varpi }$ and ${\displaystyle (1-i)\varpi }$.[21] Elliptic functions with a square period lattice are more symmetrical than arbitrary elliptic functions, following the symmetries of the square.

Reflections and quarter-turn rotations of lemniscate function arguments have simple expressions:

${\displaystyle {\begin{aligned}\operatorname {cl} {\bar {z}}&={\overline {\operatorname {cl} z}}\\[6mu]\operatorname {sl} {\bar {z}}&={\overline {\operatorname {sl} z}}\\[4mu]\operatorname {cl} iz&={\frac {1}{\operatorname {cl} z}}\\[6mu]\operatorname {sl} iz&=i\operatorname {sl} z\end{aligned}}}$

The sl function has simple zeros at Gaussian integer multiples of ϖ, complex numbers of the form ${\displaystyle a\varpi +b\varpi i}$ for integers a and b. It has simple poles at Gaussian half-integer multiples of ϖ, complex numbers of the form ${\displaystyle {\bigl (}a+{\tfrac {1}{2}}{\bigr )}\varpi +{\bigl (}b+{\tfrac {1}{2}}{\bigr )}\varpi i}$, with residues ${\displaystyle (-1)^{a-b+1}i}$. The cl function is reflected and offset from the sl function, ${\displaystyle \operatorname {cl} z={\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\varpi -z{\bigr )}}$. It has zeros for arguments ${\displaystyle {\bigl (}a+{\tfrac {1}{2}}{\bigr )}\varpi +b\varpi i}$ and poles for arguments ${\displaystyle a\varpi +{\bigl (}b+{\tfrac {1}{2}}{\bigr )}\varpi i,}$ with residues ${\displaystyle (-1)^{a-b}i.}$

Curves x² ⊕ y² = a for various values of a. Negative a in green, positive a in blue, a = ±1 in red, a = ∞ in black.

The lemniscate functions satisfy a Pythagorean-like identity:

${\displaystyle \operatorname {cl^{2}} z+\operatorname {sl^{2}} z+\operatorname {cl^{2}} z\,\operatorname {sl^{2}} z=1}$

As a result, the parametric equation ${\displaystyle (x,y)=(\operatorname {cl} t,\operatorname {sl} t)}$ parametrizes the quartic curve ${\displaystyle x^{2}+y^{2}+x^{2}y^{2}=1.}$

This identity can alternately be rewritten:[22]

${\displaystyle {\bigl (}1+\operatorname {cl^{2}} z{\bigr )}{\bigl (}1+\operatorname {sl^{2}} z{\bigr )}=2}$

${\displaystyle \operatorname {cl^{2}} z={\frac {1-\operatorname {sl^{2}} z}{1+\operatorname {sl^{2}} z}},\quad \operatorname {sl^{2}} z={\frac {1-\operatorname {cl^{2}} z}{1+\operatorname {cl^{2}} z}}}$

Defining a tangent-sum operator as ${\displaystyle a\oplus b\mathrel {:=} \tan(\arctan a+\arctan b),}$ gives:

${\displaystyle \operatorname {cl^{2}} z\oplus \operatorname {sl^{2}} z=1}$

The derivatives are as follows:

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {cl} z=\operatorname {cl'} z&=-\operatorname {sl} z\,{\bigl (}1+\operatorname {cl^{2}} z{\bigr )}\\\operatorname {cl'^{2}} z&=1-\operatorname {cl^{4}} z\\[5mu]{\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {sl} z=\operatorname {sl'} z&=\operatorname {cl} z\,{\bigl (}1+\operatorname {sl^{2}} z{\bigr )}\\\operatorname {sl'^{2}} z&=1-\operatorname {sl^{4}} z\end{aligned}}}$

The second derivatives of lemniscate sine and lemniscate cosine are their negative duplicated cubes:

${\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}\operatorname {cl} z=-2\operatorname {cl^{3}} z}$

${\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}\operatorname {sl} z=-2\operatorname {sl^{3}} z}$

The lemniscate functions can be integrated using the inverse tangent function:

${\displaystyle \int \operatorname {cl} z\mathop {\mathrm {d} z} =\arctan(\operatorname {sl} z)+C}$

${\displaystyle \int \operatorname {sl} z\mathop {\mathrm {d} z} =-\arctan(\operatorname {cl} z)+C}$

Like the trigonometric functions, the lemniscate functions satisfy argument sum and difference identities. The original identity used by Fagnano for bisection of the lemniscate was:[23]

${\displaystyle \operatorname {sl} (u+v)={\frac {\operatorname {sl} u\,\operatorname {sl'} v+\operatorname {sl} v\,\operatorname {sl'} u}{1+\operatorname {sl^{2}} u\,\operatorname {sl^{2}} v}}}$

The derivative and Pythagorean-like identities can be used to rework the identity used by Fagano in terms of sl and cl. Defining a tangent-sum operator ${\displaystyle a\oplus b\mathrel {:=} \tan(\arctan a+\arctan b)}$ and tangent-difference operator ${\displaystyle a\ominus b\mathrel {:=} a\oplus (-b),}$ the argument sum and difference identities can be expressed as:[24]

${\displaystyle {\begin{aligned}\operatorname {cl} (u+v)&=\operatorname {cl} u\,\operatorname {cl} v\ominus \operatorname {sl} u\,\operatorname {sl} v={\frac {\operatorname {cl} u\,\operatorname {cl} v-\operatorname {sl} u\,\operatorname {sl} v}{1+\operatorname {sl} u\,\operatorname {cl} u\,\operatorname {sl} v\,\operatorname {cl} v}}\\[2mu]\operatorname {cl} (u-v)&=\operatorname {cl} u\,\operatorname {cl} v\oplus \operatorname {sl} u\,\operatorname {sl} v\\[2mu]\operatorname {sl} (u+v)&=\operatorname {sl} u\,\operatorname {cl} v\oplus \operatorname {cl} u\,\operatorname {sl} v={\frac {\operatorname {sl} u\,\operatorname {cl} v+\operatorname {cl} u\,\operatorname {sl} v}{1-\operatorname {sl} u\,\operatorname {cl} u\,\operatorname {sl} v\,\operatorname {cl} v}}\\[2mu]\operatorname {sl} (u-v)&=\operatorname {sl} u\,\operatorname {cl} v\ominus \operatorname {cl} u\,\operatorname {sl} v\end{aligned}}}$

These resemble their trigonometric analogs:

${\displaystyle {\begin{aligned}\cos(u\pm v)&=\cos u\,\cos v\mp \sin u\,\sin v\\[6mu]\sin(u\pm v)&=\sin u\,\cos v\pm \cos u\,\sin v\end{aligned}}}$

Bisection formulas:

${\displaystyle \operatorname {cl} ^{2}{\bigl (}{\tfrac {1}{2}}x{\bigr )}={\frac {1+\operatorname {cl} x{\sqrt {1+\operatorname {sl} ^{2}x}}}{{\sqrt {1+\operatorname {sl} ^{2}x}}+1}}}$

${\displaystyle \operatorname {sl} ^{2}{\bigl (}{\tfrac {1}{2}}x{\bigr )}={\frac {1-\operatorname {cl} x{\sqrt {1+\operatorname {sl} ^{2}x}}}{{\sqrt {1+\operatorname {sl} ^{2}x}}+1}}}$

Duplication formulas:[25]

${\displaystyle \operatorname {cl} (2x)={\frac {-1+2\,\operatorname {cl} ^{2}x+\operatorname {cl} ^{4}x}{1+2\,\operatorname {cl} ^{2}x-\operatorname {cl} ^{4}x}}}$

${\displaystyle \operatorname {sl} (2x)=2\,\operatorname {sl} x\,\operatorname {cl} x{\frac {1+\operatorname {sl} ^{2}x}{1+\operatorname {sl} ^{4}x}}}$

Triplication formulas:[25]

${\displaystyle \operatorname {cl} (3x)={\frac {-3\,\operatorname {cl} x+6\,\operatorname {cl} ^{5}x+\operatorname {cl} ^{9}x}{1+6\,\operatorname {cl} ^{4}x-3\,\operatorname {cl} ^{8}x}}}$

${\displaystyle \operatorname {sl} (3x)={\frac {3\,\operatorname {sl} x-6\,\operatorname {sl} ^{5}x-\operatorname {sl} ^{9}x}{1+6\,\operatorname {sl} ^{4}x-3\,\operatorname {sl} ^{8}x}}}$

Just as for the trigonometric functions, values of the lemniscate functions can be computed for divisions of the lemniscate into n parts, using only basic arithmetic and square roots, whenever n is of the form ${\displaystyle n=2^{k}p_{1}p_{2}\cdots p_{m}}$ where k and m are non-negative integers and each p_i (if any) is a distinct Fermat prime.[26] However, the expressions become unwieldy as n grows. Below are the expressions for some n ≤ 10:

${\displaystyle x}$	${\displaystyle \operatorname {cl} x}$	${\displaystyle \operatorname {sl} x}$
${\displaystyle 0}$	${\displaystyle 1}$	${\displaystyle 0}$
${\displaystyle \varpi }$	${\displaystyle -1}$	${\displaystyle 0}$
${\displaystyle {\frac {\varpi }{2}}}$	${\displaystyle 0}$	${\displaystyle 1}$
${\displaystyle {\frac {\varpi }{3}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}$	${\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}$
${\displaystyle {\frac {\varpi }{4}}}$	${\displaystyle {\sqrt {{\sqrt {2}}-1}}}$	${\displaystyle {\sqrt {{\sqrt {2}}-1}}}$
${\displaystyle {\frac {\varpi }{5}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\Bigl (}{\sqrt {{\sqrt {5}}+2}}+1{\Bigr )}}$	${\displaystyle {\tfrac {1}{\sqrt {2}}}{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {{\sqrt {2{\bigl (}5-{\sqrt {5}}{\bigr )}}}+1-{\sqrt {5}}}}}$
${\displaystyle {\frac {\varpi }{6}}}$	${\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}$
${\displaystyle {\frac {\varpi }{8}}}$	${\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\Bigl (}{\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}{\Bigr )}}}}$	${\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\Bigl (}{\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}{\Bigr )}}}}$
${\displaystyle {\frac {\varpi }{10}}}$	${\displaystyle {\tfrac {1}{\sqrt {2}}}{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {{\sqrt {2{\bigl (}5-{\sqrt {5}}{\bigr )}}}-1+{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\Bigl (}{\sqrt {{\sqrt {5}}+2}}-1{\Bigr )}}$

The power series expansion of the lemniscate sine at the origin is[27]

${\displaystyle \operatorname {sl} z=\sum _{n=0}^{\infty }a_{n}z^{n}=z-{\tfrac {1}{10}}z^{5}+{\tfrac {1}{120}}z^{9}-{\tfrac {11}{15600}}z^{13}+\cdots ,\quad z\in \mathbb {C} ,\,|z|<{\tfrac {\varpi }{\sqrt {2}}}}$

where the coefficients ${\displaystyle a_{n}}$ are determined as follows:

${\displaystyle a_{1}=1,\,n\not \equiv 1{\pmod {4}}\implies a_{n}=0,}$

${\displaystyle a_{n+2}=-{\frac {2}{(n+1)(n+2)}}\sum _{i+j+k=n}a_{i}a_{j}a_{k}.}$

For example, to evaluate ${\displaystyle a_{13}}$, it can be seen that there are only six compositions of ${\displaystyle 13-2=11}$ that give a nonzero contribution to the sum: ${\displaystyle 11=9+1+1=1+9+1=1+1+9}$ and ${\displaystyle 11=5+5+1=5+1+5=1+5+5}$, so

${\displaystyle a_{13}=-{\tfrac {2}{12\cdot 13}}(a_{9}a_{1}a_{1}+a_{1}a_{9}a_{1}+a_{1}a_{1}a_{9}+a_{5}a_{5}a_{1}+a_{5}a_{1}a_{5}+a_{1}a_{5}a_{5})=-{\tfrac {11}{15600}}.}$

The lemniscate functions are closely related to the Weierstrass elliptic function ${\displaystyle \wp (z;1,0)}$ (the "lemniscatic case"), with invariants g₂ = 1 and g₃ = 0. This lattice has fundamental half periods ${\displaystyle \omega _{1}=\varpi /{\sqrt {2}},}$ and ${\displaystyle \omega _{2}=i\omega _{1},}$. The associated constants of the Weierstrass function are ${\displaystyle e_{1}={\tfrac {1}{2}},\ e_{2}=0,\ e_{3}=-{\tfrac {1}{2}}.}$

The related case of a Weierstrass elliptic function with g₂ = a, g₃ = 0 may be handled by a scaling transformation. However, this may involve complex numbers. If it is desired to remain within real numbers, there are two cases to consider: a > 0 and a < 0. The period parallelogram is either a square or a rhombus. The Weierstrass elliptic function ${\displaystyle \wp (z;-1,0)}$ is called the "pseudolemniscatic case".[28] The square of the lemniscate sine can be represented as

${\displaystyle \operatorname {sl} ^{2}z={\frac {1}{\wp (z;4,0)}}={\frac {i}{2\wp ((1-i)z;-1,0)}}={-2\wp }{\left({\sqrt {2}}z+(i-1)\varpi /{\sqrt {2}};1,0\right)}}$

where the second and third argument of ${\displaystyle \wp }$ denote the lattice invariants.

The lemniscate functions can also be written in terms of Jacobi elliptic functions. The Jacobi functions with positive real elliptic modulus have an "upright" rectangular lattice aligned with real and imaginary axes. Alternately, the Jacobi elliptic function with modulus i has a square period lattice rotated 1/8 turn.[29]

${\displaystyle \operatorname {sl} z=\operatorname {sn} (z;i)={{\tfrac {1}{\sqrt {2}}}\operatorname {sd} }{\bigl (}{\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}{\bigr )}}$

${\displaystyle \operatorname {cl} z=\operatorname {cd} (z;i)={\operatorname {cn} }{\bigl (}{\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}{\bigr )}}$

where the second arguments denote the elliptic modulus.

The lemniscate sine can be used for the computation of values of the modular lambda function:

${\displaystyle \prod _{k=1}^{n}\;{\operatorname {sl} }{\left({\frac {2k-1}{2n+1}}{\frac {\varpi }{2}}\right)}={\sqrt[{8}]{\frac {\lambda ((2n+1)i)}{1-\lambda ((2n+1)i)}}}}$

For example:

${\displaystyle {\begin{aligned}&{\operatorname {sl} }{\bigl (}{\tfrac {1}{14}}\varpi {\bigr )}\,{\operatorname {sl} }{\bigl (}{\tfrac {3}{14}}\varpi {\bigr )}\,{\operatorname {sl} }{\bigl (}{\tfrac {5}{14}}\varpi {\bigr )}\\[7mu]&\quad {}={\sqrt[{8}]{\frac {\lambda (7i)}{1-\lambda (7i)}}}={\tan }{\Bigl (}{{\tfrac {1}{2}}\operatorname {arccsc} }{\Bigl (}{\tfrac {1}{2}}{\sqrt {8{\sqrt {7}}+21}}+{\tfrac {1}{2}}{\sqrt {7}}+1{\Bigr )}{\Bigr )}\\[18mu]&{\operatorname {sl} }{\bigl (}{\tfrac {1}{18}}\varpi {\bigr )}\,{\operatorname {sl} }{\bigl (}{\tfrac {3}{18}}\varpi {\bigr )}\,{\operatorname {sl} }{\bigl (}{\tfrac {5}{18}}\varpi {\bigr )}\,{\operatorname {sl} }{\bigl (}{\tfrac {7}{18}}\varpi {\bigr )}\\[-3mu]&\quad {}={\sqrt[{8}]{\frac {\lambda (9i)}{1-\lambda (9i)}}}={\tan }{\Biggl (}{\frac {\pi }{4}}-{\arctan }{\Biggl (}{\frac {2{\sqrt[{3}]{2{\sqrt {3}}-2}}-2{\sqrt[{3}]{2-{\sqrt {3}}}}+{\sqrt {3}}-1}{\sqrt[{4}]{12}}}{\Biggr )}{\Biggr )}\end{aligned}}}$

Several methods of computing ${\displaystyle \operatorname {sl} x}$ involve first making the change of variables ${\displaystyle \pi x=\varpi {\tilde {x}}}$ and then computing ${\displaystyle \operatorname {sl} (\varpi {\tilde {x}}/\pi ).}$

A hyperbolic series method:[30]

${\displaystyle \operatorname {sl} {\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}={\frac {\varpi }{\pi }}{\Bigg /}\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{{\sinh }{\left(x-n\pi \right)}}}}$

Fourier series method:[31]

${\displaystyle \operatorname {sl} {\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}={\frac {2\pi }{\varpi }}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin((2n+1)x)}{\cosh((n+1/2)\pi )}},\quad |\operatorname {Im} x|<{\frac {\pi }{2}}}$

The lemniscate sine can be computed more rapidly by

${\displaystyle \operatorname {sl} {\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}={\frac {{\theta _{1}}{\left(x,e^{-\pi }\right)}}{{\theta _{3}}{\left(x,e^{-\pi }\right)}}}}$

where

${\displaystyle {\begin{aligned}\theta _{1}(x,q)&=\sum _{n\in \mathbb {Z} }q^{(n+1/2)^{2}}\exp \left((2n+1)ix+{\bigl (}n-{\tfrac {1}{2}}{\bigr )}i\pi \right)\\&=2\sum _{n=0}^{\infty }(-1)^{n}q^{(n+1/2)^{2}}\sin((2n+1)x),\end{aligned}}}$

${\displaystyle \theta _{3}(x,q)=\sum _{n\in \mathbb {Z} }q^{n^{2}}e^{2nix}=1+2\sum _{n=1}^{\infty }q^{n^{2}}\cos(2nx)}$

are the Jacobi theta functions.[32]

Another fast algorithm for computing ${\displaystyle \operatorname {sl} x}$ is the following:[33]

This is effectively using the arithmetic-geometric mean and is based on Landen's transformations.[34]

Two other fast computation methods use the following sum and product series:

${\displaystyle {\text{sl}}{\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=f{\biggl (}{\frac {4\pi }{\varpi }}\sin(x)\sum _{n=1}^{\infty }{\frac {\cosh[(2n-1)\pi ]}{\cosh[(2n-1)\pi ]^{2}-\cos(x)^{2}}}{\biggr )}}$

${\displaystyle {\text{cl}}{\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=f{\biggl (}{\frac {4\pi }{\varpi }}\cos(x)\sum _{n=1}^{\infty }{\frac {\cosh[(2n-1)\pi ]}{\cosh[(2n-1)\pi ]^{2}-\sin(x)^{2}}}{\biggr )}}$

${\displaystyle \mathrm {sl} {\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=2\exp {\bigl (}{-{\tfrac {1}{4}}\pi }{\bigr )}\sin(x)\prod _{n=1}^{\infty }{\frac {1-2\cos(2x)\exp(-2n\pi )+\exp(-4n\pi )}{1+2\cos(2x)\exp[-(2n-1)\pi ]+\exp[-(4n-2)\pi ]}}}$

${\displaystyle \mathrm {cl} {\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=2\exp {\bigl (}{-{\tfrac {1}{4}}\pi }{\bigr )}\cos(x)\prod _{n=1}^{\infty }{\frac {1+2\cos(2x)\exp(-2n\pi )+\exp(-4n\pi )}{1-2\cos(2x)\exp[-(2n-1)\pi ]+\exp[-(4n-2)\pi ]}}}$

where ${\displaystyle f(x)=\tan(2\arctan x)=2x/(1-x^{2}).}$

The following series identities were discovered by Ramanujan:[35]