Lemniscate elliptic functions

In mathematics, the lemniscate elliptic functions are elliptic functions related to the arc length of the lemniscate of Bernoulli. They were first studied by Giulio Fagnano in 1718 and later by Leonhard Euler and Carl Friedrich Gauss, among others.

The lemniscate sine and lemniscate cosine functions, usually written with the symbols sl and cl (sometimes the symbols sinlem and coslem or sin lemn and cos lemn are used instead)^[1] are analogous to the trigonometric functions sine and cosine. While the trigonometric sine relates the arc length to the chord length in a unit-diameter circle ${\displaystyle x^{2}+y^{2}=x}$, the lemniscate sine relates the arc length to the chord length of a lemniscate ${\displaystyle {\bigl (}x^{2}+y^{2}{\bigr )}{}^{2}=x^{2}-y^{2}.}$

The lemniscate functions have periods related to a number ${\displaystyle \varpi =}$ 2.622057... called the lemniscate constant, the ratio of a lemniscate's perimeter to its diameter.

The sl and cl functions have a square period lattice (a multiple of the Gaussian integers) with fundamental periods ${\displaystyle \{(1+i)\varpi ,(1-i)\varpi \},}$^[2] and are a special case of two Jacobi elliptic functions on that lattice, ${\displaystyle \operatorname {sl} z=\operatorname {sn} (z;i),}$ ${\displaystyle \operatorname {cl} z=\operatorname {cd} (z;i)}$.

Similarly, the hyperbolic lemniscate functions slh and clh have a square period lattice with fundamental periods ${\displaystyle {\bigl \{}{\sqrt {2}}\varpi ,{\sqrt {2}}\varpi i{\bigr \}}.}$

The lemniscate functions and the hyperbolic lemniscate functions are related to the Weierstrass elliptic function ${\displaystyle \wp (z;a,0)}$.

Lemniscate sine and cosine functions[]

Definitions[]

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:

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

Arc length of Bernoulli's lemniscate[]

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]

Arc length of 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.

Lemniscate constant[]

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,^[11]

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

where Β is the beta function, Γ is the gamma function, M is the arithmetic–geometric mean and β' is the derivative of the Dirichlet beta function. Geometrically, ϖ is the ratio of the perimeter of Bernoulli's lemniscate to its diameter. The lemniscate constant was proven transcendental by Theodor Schneider in 1937.^[12] In 1975, Gregory Chudnovsky proved that π and ϖ are algebraically independent over ${\displaystyle \mathbb {Q} }$.^[13][14] The related constant ${\displaystyle G=\varpi /\pi =0.8346\ldots }$ is Gauss's constant.

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:^[15]

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

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:^[16]

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

The Wallis product for π is:

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

Analogous formulas for ϖ are:^[17]

${\displaystyle {\frac {\varpi }{2}}=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\cdot {\frac {4n^{2}}{4n^{2}+1}}\right){\frac {P}{Q}}}$

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

where

${\displaystyle P=\prod _{(n,k)\in {\mathbb {N} ^{+}}^{2}}\left({\frac {4(n+ki)^{2}}{4(n+ki)^{2}-1}}\cdot {\frac {4(n+ki)^{2}}{4(n+ki)^{2}+1}}\right),}$

${\displaystyle Q=\prod _{(n,k)\in \mathbb {N} _{0}^{2}}\left({\frac {4(n+1/2+(k+1/2)i)^{2}}{4(n+1/2+(k+1/2)i)^{2}-1}}\cdot {\frac {4(n+1/2+(k+1/2)i)^{2}}{4(n+1/2+(k+1/2)i)^{2}+1}}\right).}$

A related result 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:^[18]

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

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: ${\displaystyle {\tfrac {1}{2}}\varpi =2\operatorname {arcsl} {\tfrac {1}{2}}+\operatorname {arcsl} {\tfrac {7}{23}}.}$^[19]

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 ${\displaystyle 4}$.^[20]

Zeros, poles and symmetries[]

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:^[21]

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

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 ${\displaystyle \varpi }$.^[22] 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 ${\displaystyle (1-i)\varpi }$.^[23] 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 ${\displaystyle (-1)^{a-b}i.}$

Since the lemniscate sine is a meromorphic function, it can be written as a ratio of holomorphic functions. Gauss showed that sl has the following product expansion, reflecting the distribution of its zeros and poles:^[24]

${\displaystyle \operatorname {sl} z={\frac {M(z)}{N(z)}}}$

where ${\textstyle M(z)=z\prod _{\alpha }(1-z^{4}/\alpha ^{4})}$ and ${\textstyle N(z)=\prod _{\beta }(1-z^{4}/\beta ^{4})}$. Here, ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ denote, respectively, the zeros and poles of sl which are in the quadrant ${\displaystyle \operatorname {Re} z>0,\operatorname {Im} z\geq 0}$. Gauss conjectured that ${\displaystyle \ln N(\varpi )=\pi /2}$ (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”.^[25][26]

There is also an infinite series reflecting the distribution of the zeros of sl:^[27][28]

${\displaystyle {\frac {1}{\operatorname {sl} z}}=\sum _{(n,k)\in \mathbb {Z} ^{2}}{\frac {(-1)^{n+k}}{z-(n\varpi +k\varpi i)}}.}$

Pythagorean-like identity[]

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:^[29]

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

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

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

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

Derivatives and integrals[]

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:

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

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

Argument sum and multiple identities[]

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:^[30]

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

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:^[31]

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

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:

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

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

Duplication formulas:^[32]

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

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

Triplication formulas:^[32]

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

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

Specific values[]

Just as for the trigonometric functions, values of the lemniscate functions can be computed for divisions of the lemniscate into n parts of equal length, 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.^[33] However, the expressions become unwieldy as n grows. Below are the expressions for dividing the lemniscate ${\displaystyle (x^{2}+y^{2})^{2}=x^{2}-y^{2}}$ into n parts of equal length for some n ≤ 20. The n-division points are obtained by intersecting the lemniscate ${\displaystyle (x^{2}+y^{2})^{2}=x^{2}-y^{2}}$ with the circle ${\displaystyle x^{2}+y^{2}=\operatorname {sl} ^{2}{\tfrac {2j\varpi }{n}}}$ where ${\displaystyle j\in \{1,2,\ldots ,n\}}$.

${\displaystyle n}$	${\displaystyle \operatorname {cl} {\tfrac {2\varpi }{n}}}$	${\displaystyle \operatorname {sl} {\tfrac {2\varpi }{n}}}$
${\displaystyle 2}$	${\displaystyle -1}$	${\displaystyle 0}$
${\displaystyle 4}$	${\displaystyle 0}$	${\displaystyle 1}$
${\displaystyle 5}$	${\displaystyle {\tfrac {1}{2}}({\sqrt[{4}]{5}}-1)\left({\sqrt {{\sqrt {5}}+2}}-1\right)}$	${\displaystyle {\tfrac {1}{2{\sqrt[{4}]{2}}}}({\sqrt {5}}-1){\sqrt {{\sqrt[{4}]{20}}+{\sqrt {{\sqrt {5}}-1}}}}}$
${\displaystyle 6}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}$	${\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}$
${\displaystyle 8}$	${\displaystyle {\sqrt {{\sqrt {2}}-1}}}$	${\displaystyle {\sqrt {{\sqrt {2}}-1}}}$
${\displaystyle 10}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\Bigl (}{\sqrt {{\sqrt {5}}+2}}+1{\Bigr )}}$	${\displaystyle {\tfrac {1}{\sqrt {2}}}{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {{\sqrt {2{\bigl (}5-{\sqrt {5}}{\bigr )}}}+1-{\sqrt {5}}}}}$
${\displaystyle 12}$	${\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}$
${\displaystyle 16}$	${\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\Bigl (}{\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}{\Bigr )}}}}$	${\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\Bigl (}{\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}{\Bigr )}}}}$
${\displaystyle 20}$	${\displaystyle {\tfrac {1}{\sqrt {2}}}{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {{\sqrt {2{\bigl (}5-{\sqrt {5}}{\bigr )}}}-1+{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\Bigl (}{\sqrt {{\sqrt {5}}+2}}-1{\Bigr )}}$

Power series[]

The power series expansion of the lemniscate sine at the origin is^[34]

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

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

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

${\displaystyle a_{1}=1,\,\forall n\in \mathbb {N} _{0}:\,a_{n+2}=-{\frac {2}{(n+1)(n+2)}}\sum _{i+j+k=n}a_{i}a_{j}a_{k}}$

where ${\displaystyle i+j+k=n}$ stands for all three-term compositions of ${\displaystyle n}$. 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

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

Relation to Weierstrass and Jacobi elliptic functions[]

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 periods ${\displaystyle \omega _{1}={\sqrt {2}}\varpi ,}$ 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".^[35]

The square of the lemniscate sine can be represented as

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

where the second and third argument of ${\displaystyle \wp }$ denote the lattice invariants g₂ and g₃. Another representation is

${\displaystyle \operatorname {sl} ^{2}z={\frac {\varpi ^{2}}{\wp (z/\varpi ,i)}}}$

where the second argument of ${\displaystyle \wp }$ denotes the period ratio ${\displaystyle \tau }$. The lemniscate sine is a rational function in the Weierstrass elliptic function and its derivative:^[36]

${\displaystyle \operatorname {sl} z=-2{\frac {\wp ((1+i)z;1/4,0)}{\wp '((1+i)z;1/4,0)}}}$

where the second and third argument of ${\displaystyle \wp }$ denote the lattice invariants g₂ and g₃. In terms of the period ratio ${\displaystyle \tau }$, this becomes

${\displaystyle \operatorname {sl} z=-2{\frac {\wp ((1+i)z/(2\varpi ),i)}{\wp '((1+i)z/(2\varpi ),i)}}.}$

The lemniscate functions can also be written in terms of Jacobi elliptic functions. The Jacobi elliptic functions ${\displaystyle \operatorname {sn} }$ and ${\displaystyle \operatorname {cd} }$ with positive real elliptic modulus have an "upright" rectangular lattice aligned with real and imaginary axes. Alternately, the functions ${\displaystyle \operatorname {sn} }$ and ${\displaystyle \operatorname {cd} }$ with modulus i (and ${\displaystyle \operatorname {sd} }$ and ${\displaystyle \operatorname {cn} }$ with modulus ${\displaystyle 1/{\sqrt {2}}}$) have a square period lattice rotated 1/8 turn.^[37]

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

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

where the second arguments denote the elliptic modulus.

Relation to the modular lambda function[]

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:

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

Methods of computation[]

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:^{[40][27][41][42]}

${\displaystyle \operatorname {sl} \left({\frac {\varpi }{\pi }}x\right)={\frac {\pi }{\varpi }}\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{\cosh(x-(n+1/2)\pi )}}}$

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

Fourier series method:^[43]

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

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

${\displaystyle {\begin{aligned}\theta _{1}(x,q)&=\sum _{n\in \mathbb {Z} }(-1)^{n}q^{(n+1/2)^{2}}\sin((2n+1)x),\\\theta _{3}(x,q)&=\sum _{n\in \mathbb {Z} }q^{n^{2}}\cos 2nx\end{aligned}}}$

are the Jacobi theta functions.^[44]

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:^[45]

${\displaystyle {\frac {\varpi ^{2}}{\pi ^{2}\operatorname {sl} ^{2}(\varpi x/\pi )}}={\frac {1}{\sin ^{2}x}}-{\frac {1}{\pi }}-8\sum _{n=1}^{\infty }{\frac {n\cos 2nx}{e^{2n\pi }-1}}}$

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

Inverse functions[]

The inverse function of the lemniscate sine is the lemniscate arcsine, defined as:

${\displaystyle \operatorname {arcsl} x=\int _{0}^{x}{\frac {1}{\sqrt {1-t^{4}}}}\mathrm {d} t}$

The inverse function of the lemniscate cosine is the lemniscate arccosine. This function is defined by following expression:

${\displaystyle \operatorname {arccl} x=\int _{x}^{1}{\frac {1}{\sqrt {1-t^{4}}}}\mathrm {d} t={\tfrac {1}{2}}\varpi -\operatorname {arcsl} x}$

For x in the interval ${\displaystyle -1\leq x\leq 1}$, ${\displaystyle \operatorname {sl} (\operatorname {arcsl} x)=x}$ and ${\displaystyle \operatorname {cl} (\operatorname {arccl} x)=x}$

For the halving of the lemniscate arc length these formulas are valid:

${\displaystyle {\begin{aligned}{\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\operatorname {arcsl} x{\bigr )}&={\sin }{\bigl (}{\tfrac {1}{2}}\arcsin x{\bigr )}\,{\operatorname {sech} }{\bigl (}{\tfrac {1}{2}}\operatorname {arsinh} x{\bigr )}\\{\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\operatorname {arcsl} x{\bigr )}^{2}&={\tan }{\bigl (}{\tfrac {1}{4}}\arcsin x^{2}{\bigr )}\end{aligned}}}$

Expression using elliptic integrals[]

The lemniscate arcsine and the lemniscate arccosine can also be expressed by the Legendre-Form:

These functions can be displayed directly by using the incomplete elliptic integral of the first kind:

${\displaystyle \operatorname {arcsl} x={\frac {1}{\sqrt {2}}}F\left[{\arcsin }{\left({\frac {{\sqrt {2}}x}{\sqrt {1+x^{2}}}}\right)};{\frac {1}{\sqrt {2}}}\right]}$

${\displaystyle \operatorname {arcsl} x=2({\sqrt {2}}-1)F\left[{\arcsin }{\left[{\frac {({\sqrt {2}}+1)x}{{\sqrt {1+x^{2}}}+1}}\right]};({\sqrt {2}}-1)^{2}\right]}$

The arc lengths of the lemniscate can also be expressed by only using the arc lengths of ellipses (calculated by elliptic integrals of the second kind):

${\displaystyle {\begin{aligned}\operatorname {arcsl} x={}&{\frac {2+{\sqrt {2}}}{2}}E\left[{\arcsin }{\left[{\frac {({\sqrt {2}}+1)x}{{\sqrt {1+x^{2}}}+1}}\right]};({\sqrt {2}}-1)^{2}\right]\\[5mu]&\ \ -E\left[{\arcsin }{\left({\frac {{\sqrt {2}}x}{\sqrt {1+x^{2}}}}\right)};{\frac {1}{\sqrt {2}}}\right]+{\frac {x{\sqrt {1-x^{2}}}}{{\sqrt {2}}(1+x^{2}+{\sqrt {1+x^{2}}})}}\end{aligned}}}$

The lemniscate arccosine has this expression:

${\displaystyle \operatorname {arccl} x={\frac {1}{\sqrt {2}}}F\left[\arccos x;{\frac {1}{\sqrt {2}}}\right]}$

Use in integration[]

The lemniscate can be used to integrate many functions. Here is a list of important integrals (the constants of integration are omitted):

${\displaystyle \int {\frac {1}{\sqrt {1-x^{4}}}}\,\mathrm {d} x=\operatorname {arcsl} x}$

${\displaystyle \int {\frac {1}{\sqrt {(x^{2}+1)(2x^{2}+1)}}}\,\mathrm {d} x={\operatorname {arcsl} }{\left({\frac {x}{\sqrt {x^{2}+1}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt {x^{4}+6x^{2}+1}}}\,\mathrm {d} x={\operatorname {arcsl} }{\left({\frac {{\sqrt {2}}x}{\sqrt {{\sqrt {x^{4}+6x^{2}+1}}+x^{2}+1}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt {x^{4}+1}}}\,\mathrm {d} x={{\sqrt {2}}\operatorname {arcsl} }{\left({\frac {x}{\sqrt {{\sqrt {x^{4}+1}}+1}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt[{4}]{(1-x^{4})^{3}}}}\,\mathrm {d} x={{\sqrt {2}}\operatorname {arcsl} }{\left({\frac {x}{\sqrt {1+{\sqrt {1-x^{4}}}}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt[{4}]{(x^{4}+1)^{3}}}}\,\mathrm {d} x={\operatorname {arcsl} }{\left({\frac {x}{\sqrt[{4}]{x^{4}+1}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt[{4}]{(1-x^{2})^{3}}}}\,\mathrm {d} x={2\operatorname {arcsl} }{\left({\frac {x}{1+{\sqrt {1-x^{2}}}}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt[{4}]{(x^{2}+1)^{3}}}}\,\mathrm {d} x={2\operatorname {arcsl} }{\left({\frac {x}{{\sqrt {x^{2}+1}}+1}}\right)}}$

${\displaystyle \int {\frac {1}{\sqrt[{4}]{(ax^{2}+bx+c)^{3}}}}\,\mathrm {d} x={{\frac {2{\sqrt {2}}}{\sqrt[{4}]{4a^{2}c-ab^{2}}}}\operatorname {arcsl} }{\left({\frac {2ax+b}{{\sqrt {4a(ax^{2}+bx+c)}}+{\sqrt {4ac-b^{2}}}}}\right)}}$

${\displaystyle \int {\sqrt {\operatorname {sech} x}}\,\mathrm {d} x={2\operatorname {arcsl} }{\bigl (}\tanh {\tfrac {1}{2}}x{\bigr )}}$

${\displaystyle \int {\sqrt {\sec x}}\,\mathrm {d} x={2\operatorname {arcsl} }{\bigl (}\tan {\tfrac {1}{2}}x{\bigr )}}$

Hyperbolic lemniscate functions[]

The hyperbolic lemniscate sine (slh) and cosine (clh) can be defined by their inverse functions as follows:

${\displaystyle z=\int _{0}^{\operatorname {slh} z}{\frac {\mathrm {d} t}{\sqrt {1+t^{4}}}}=\int _{\operatorname {clh} z}^{\infty }{\frac {\mathrm {d} t}{\sqrt {1+t^{4}}}}}$

The complete integral has the value:

${\displaystyle \int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {t^{4}+1}}}={\tfrac {1}{4}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{4}}{\bigr )}={\tfrac {1}{\sqrt {2}}}\varpi =1.85407\;46773\;01371\ldots }$

Therefore, the two defined functions have following relation to each other:

${\displaystyle \operatorname {slh} z={\operatorname {clh} }{{\Bigl (}{\tfrac {1}{\sqrt {2}}}\varpi -z{\Bigr )}}}$

The product of hyperbolic lemniscate sine and hyperbolic lemniscate cosine is equal to one:

${\displaystyle \operatorname {slh} z\,\operatorname {clh} z=1}$

The hyperbolic lemniscate functions can be expressed in terms of lemniscate sine and lemniscate cosine:

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

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

But there is also a relation to the Jacobi elliptic functions with the elliptic modulus one by square root of two:

${\displaystyle \operatorname {slh} x={\frac {\operatorname {sn} (x;1/{\sqrt {2}})}{\operatorname {cd} (x;1/{\sqrt {2}})}}}$

${\displaystyle \operatorname {clh} x={\frac {\operatorname {cd} (x;1/{\sqrt {2}})}{\operatorname {sn} (x;1/{\sqrt {2}})}}}$

The hyperbolic lemniscate sine has following imaginary relation to the lemniscate sine:

${\displaystyle \operatorname {slh} z={\frac {1-i}{\sqrt {2}}}\operatorname {sl} \left({\frac {1+i}{\sqrt {2}}}z\right)={\frac {\operatorname {sl} \left({\sqrt[{4}]{-1}}z\right)}{\sqrt[{4}]{-1}}}}$

This is analogous to the relationship between hyperbolic and trigonometric sine:

${\displaystyle \sinh z=-i\sin(iz)={\frac {\sin \left({\sqrt[{2}]{-1}}z\right)}{\sqrt[{2}]{-1}}}}$

In a quartic Fermat curve ${\displaystyle x^{4}+y^{4}=1}$ (sometimes called a squircle) the hyperbolic lemniscate sine and cosine are analogous to the tangent and cotangent functions in a unit circle ${\displaystyle x^{2}+y^{2}=1}$ (the quadratic Fermat curve). If the origin and a point on the curve are connected to each other by a line L, the hyperbolic lemniscate sine of twice the enclosed area between this line and the x-axis is the y-coordinate of the intersection of L with the line ${\displaystyle x=1}$.^[46]

The hyperbolic lemniscate sine satisfies the argument addition identity:

${\displaystyle \operatorname {slh} (a+b)={\frac {\operatorname {slh} a{\sqrt {1+\operatorname {slh} ^{4}b}}+\operatorname {slh} b{\sqrt {1+\operatorname {slh} ^{4}a}}}{1-\operatorname {slh} ^{2}a\,\operatorname {slh} ^{2}b}}}$

The derivative can be expressed in this way:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {slh} x={\sqrt {1+\operatorname {slh} ^{4}x}}}$

Furthermore the function ${\displaystyle \sin _{4}(t)}$ is called Hyperbolic Lemniscate Tangent and the function ${\displaystyle \cot _{4}(t)}$ is called Hyperbolic Lemniscate Cotangent.

${\displaystyle {\text{tlh}}(t)=\sin _{4}(t)}$

${\displaystyle {\text{ctlh}}(t)=\cos _{4}(t)}$

Number theory[]

In algebraic number theory, every abelian extension of the Gaussian rationals ${\displaystyle \mathbf {Q} (i)}$ is obtained by adjoining ${\displaystyle \operatorname {sl} z}$, where ${\displaystyle z}$ is a solution of the equation ${\displaystyle \operatorname {sl} mz=0}$ over the complexes and ${\displaystyle m}$ is a Gaussian integer.^[47] This is analogous to the Kronecker–Weber theorem for the rational numbers ${\displaystyle \mathbf {Q} }$ which is based on division of the circle. Both are special cases of Kronecker's Jugendtraum, which became Hilbert's twelfth problem.

The field ${\displaystyle \mathbf {Q} (i,\operatorname {sl} (\varpi /n))}$ (for positive odd ${\displaystyle n}$) is the extension of ${\displaystyle \mathbf {Q} (i)}$ generated by the ${\displaystyle x}$- and ${\displaystyle y}$-coordinates of the ${\displaystyle (1+i)n}$-torsion points on the elliptic curve ${\displaystyle y^{2}=4x^{3}+x}$.^[48]

World map projections[]

The Peirce quincuncial projection, designed by Charles Sanders Peirce of the US Coast Survey in the 1870s, is a world map projection based on the inverse lemniscate sine of stereographically projected points (treated as complex numbers).^[49]

When lines of constant real or imaginary part are projected onto the complex plane via the hyperbolic lemniscate sine, and thence stereographically projected onto the sphere (see Riemann sphere), the resulting curves are spherical conics, the spherical analog of planar ellipses and hyperbolas.^[50] Thus the lemniscate functions (and more generally, the Jacobi elliptic functions) provide a parametrization for spherical conics.

A conformal map projection from the globe onto the 6 square faces of a cube can also be defined using the lemniscate functions.^[51] Because many partial differential equations can be effectively solved by conformal mapping, this map from sphere to cube is convenient for atmospheric modeling.^[52]

See also[]

• Elliptic function
  • Abel elliptic functions
  • Dixon elliptic functions
  • Jacobi elliptic functions
  • Weierstrass elliptic function
• Elliptic Gauss sum
• Gauss's constant
• Peirce quincuncial projection
• Schwarz–Christoffel mapping

External links[]

• Parker, Matt (2021). "What is the area of a Squircle?". Stand-up Maths. YouTube. Archived from the original on 2021-12-19.

Notes[]

References[]