Lemniscate elliptic functions

Mathematical functions

The lemniscate sine (red) and lemniscate cosine (purple) applied to a real argument, in comparison with the trigonometric sine y = sin(πx/ϖ) (pale dashed red).

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.[1]

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),[2] 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 x 2 + y 2 = x , {\displaystyle x^{2}+y^{2}=x,} [3] the lemniscate sine relates the arc length to the chord length of a lemniscate  ( x 2 + y 2 ) 2 = x 2 − y 2 . {\displaystyle {\bigl (}x^{2}+y^{2}{\bigr )}{}^{2}=x^{2}-y^{2}.}

The lemniscate functions have periods related to a number  ϖ = {\displaystyle \varpi =} 2.622057... called the lemniscate constant, the ratio of a lemniscate's perimeter to its diameter. This number is a quartic analog of the (quadratic)  π = {\displaystyle \pi =} 3.141592..., ratio of perimeter to diameter of a circle.

As complex functions, sl and cl have a square period lattice (a multiple of the Gaussian integers) with fundamental periods { ( 1 + i ) ϖ , ( 1 − i ) ϖ } , {\displaystyle \\{(1+i)\varpi ,(1-i)\varpi \\},} [4] and are a special case of two Jacobi elliptic functions on that lattice,  sl ⁡ z = sn ⁡ ( z ; i ) , {\displaystyle \operatorname {sl} z=\operatorname {sn} (z;i),} cl ⁡ z = cd ⁡ ( z ; i ) {\displaystyle \operatorname {cl} z=\operatorname {cd} (z;i)} .

Similarly, the hyperbolic lemniscate sine slh and hyperbolic lemniscate cosine clh have a square period lattice with fundamental periods  { 2 ϖ , 2 ϖ i } . {\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 ℘ ( z ; a , 0 ) {\displaystyle \wp (z;a,0)} .

## Lemniscate sine and cosine functions

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### Definitions

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The lemniscate functions sl and cl can be defined as the solution to the initial value problem:[5]

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

or equivalently as the inverses of an elliptic integral, the Schwarz–Christoffel map from the complex unit disk to a square with corners  { 1 2 ϖ , 1 2 ϖ i , − 1 2 ϖ , − 1 2 ϖ i } : {\displaystyle {\big \\{}{\tfrac {1}{2}}\varpi ,{\tfrac {1}{2}}\varpi i,-{\tfrac {1}{2}}\varpi ,-{\tfrac {1}{2}}\varpi i{\big \\}}\colon } [6]

z = ∫ 0 sl ⁡ z d t 1 − t 4 = ∫ cl ⁡ z 1 d t 1 − t 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 extended to the complex plane via analytic continuation by successive reflections.

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

d d z sin ⁡ z = cos ⁡ z ,   d d z cos ⁡ z = − sin ⁡ z ,   sin ⁡ 0 = 0 ,   cos ⁡ 0 = 1 , {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\sin z=\cos z,\ {\frac {\mathrm {d} }{\mathrm {d} z}}\cos z=-\sin z,\ \sin 0=0,\ \cos 0=1,}

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

z = ∫ 0 sin ⁡ z d t 1 − t 2 = ∫ cos ⁡ z 1 d t 1 − t 2 . {\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}}}}.}

### Relation to the lemniscate constant

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Main article: Lemniscate constant

The lemniscate sine function and hyperbolic lemniscate sine functions are defined as inverses of elliptic integrals. The complete integrals are related to the lemniscate constant ϖ.

The lemniscate functions have minimal real period 2ϖ, minimal imaginary period 2ϖi and fundamental complex periods  ( 1 + i ) ϖ {\displaystyle (1+i)\varpi } and  ( 1 − i ) ϖ {\displaystyle (1-i)\varpi } for a constant ϖ called the lemniscate constant,[7]

ϖ = 2 ∫ 0 1 d t 1 − t 4 = 2.62205 … {\displaystyle \varpi =2\int _{0}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}=2.62205\ldots }

The lemniscate functions satisfy the basic relation  cl ⁡ z = sl ( 1 2 ϖ − z ) , {\displaystyle \operatorname {cl} z={\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\varpi -z{\bigr )},} analogous to the relation  cos ⁡ z = sin ( 1 2 π − z ) . {\displaystyle \cos z={\sin }{\bigl (}{\tfrac {1}{2}}\pi -z{\bigr )}.}

The lemniscate constant ϖ is a close analog of the circle constant π, and many identities involving π have analogues involving ϖ, as identities involving the trigonometric functions have analogues involving the lemniscate functions. For example, Viète's formula for π can be written:

2 π = 1 2 ⋅ 1 2 + 1 2 1 2 ⋅ 1 2 + 1 2 1 2 + 1 2 1 2 ⋯ {\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:[8]

2 ϖ = 1 2 ⋅ 1 2 + 1 2 / 1 2 ⋅ 1 2 + 1 2 / 1 2 + 1 2 / 1 2 ⋯ {\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 Machin formula for π is  1 4 π = 4 arctan ⁡ 1 5 − arctan ⁡ 1 239 , {\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  1 4 π = arctan ⁡ 1 2 + arctan ⁡ 1 3 {\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:  1 2 ϖ = 2 arcsl ⁡ 1 2 + arcsl ⁡ 7 23 . {\displaystyle {\tfrac {1}{2}}\varpi =2\operatorname {arcsl} {\tfrac {1}{2}}+\operatorname {arcsl} {\tfrac {7}{23}}.} [9]

The lemniscate and circle constants were found by Gauss to be related to each-other by the arithmetic-geometric mean M:[10]

π ϖ = M ( 1 , 2   ) {\displaystyle {\frac {\pi }{\varpi }}=M{\left(1,{\sqrt {2}}\\!~\right)}}

## Argument identities

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### Zeros, poles and symmetries

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sl {\displaystyle \operatorname {sl} } in the complex plane.[11] In the picture, it can be seen that the fundamental periods  ( 1 + i ) ϖ {\displaystyle (1+i)\varpi } and  ( 1 − i ) ϖ {\displaystyle (1-i)\varpi } are "minimal" in the sense that they have the smallest absolute value of all periods whose real part is non-negative.

The lemniscate functions cl and sl are even and odd functions, respectively,

cl ⁡ ( − z ) = cl ⁡ z sl ⁡ ( − z ) = − sl ⁡ z {\displaystyle {\begin{aligned}\operatorname {cl} (-z)&=\operatorname {cl} z\\\\[6mu]\operatorname {sl} (-z)&=-\operatorname {sl} z\end{aligned}}}

At translations of  1 2 ϖ , {\displaystyle {\tfrac {1}{2}}\varpi ,} cl and sl are exchanged, and at translations of  1 2 i ϖ {\displaystyle {\tfrac {1}{2}}i\varpi } they are additionally rotated and reciprocated:[12]

cl ( z ± 1 2 ϖ ) = ∓ sl ⁡ z , cl ( z ± 1 2 i ϖ ) = ∓ i sl ⁡ z sl ( z ± 1 2 ϖ ) = ± cl ⁡ z , sl ( z ± 1 2 i ϖ ) = ± i cl ⁡ z {\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  ϖ {\displaystyle \varpi } (that is,  ± ϖ {\displaystyle \pm \varpi } or  ± i ϖ {\displaystyle \pm i\varpi } ), negates each function, an involution:

cl ⁡ ( z + ϖ ) = cl ⁡ ( z + i ϖ ) = − cl ⁡ z sl ⁡ ( z + ϖ ) = sl ⁡ ( z + i ϖ ) = − sl ⁡ z {\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  ϖ {\displaystyle \varpi } .[13] That is, a displacement  ( a + b i ) ϖ , {\displaystyle (a+bi)\varpi ,} with  a + b = 2 k {\displaystyle a+b=2k} for integers a, b, and k.

cl ( z + ( 1 + i ) ϖ ) = cl ( z + ( 1 − i ) ϖ ) = cl ⁡ z sl ( z + ( 1 + i ) ϖ ) = sl ( z + ( 1 − i ) ϖ ) = sl ⁡ z {\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  ( 1 + i ) ϖ {\displaystyle (1+i)\varpi } and  ( 1 − i ) ϖ {\displaystyle (1-i)\varpi } .[14] 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:

cl ⁡ z ¯ = cl ⁡ z ¯ sl ⁡ z ¯ = sl ⁡ z ¯ cl ⁡ i z = 1 cl ⁡ z sl ⁡ i z = i sl ⁡ z {\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  a ϖ + b ϖ i {\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  ( a + 1 2 ) ϖ + ( b + 1 2 ) ϖ i {\displaystyle {\bigl (}a+{\tfrac {1}{2}}{\bigr )}\varpi +{\bigl (}b+{\tfrac {1}{2}}{\bigr )}\varpi i} , with residues ( − 1 ) a − b + 1 i {\displaystyle (-1)^{a-b+1}i} . The cl function is reflected and offset from the sl function,  cl ⁡ z = sl ( 1 2 ϖ − z ) {\displaystyle \operatorname {cl} z={\operatorname {sl} }{\bigl (}{\tfrac {1}{2}}\varpi -z{\bigr )}} . It has zeros for arguments  ( a + 1 2 ) ϖ + b ϖ i {\displaystyle {\bigl (}a+{\tfrac {1}{2}}{\bigr )}\varpi +b\varpi i} and poles for arguments  a ϖ + ( b + 1 2 ) ϖ i , {\displaystyle a\varpi +{\bigl (}b+{\tfrac {1}{2}}{\bigr )}\varpi i,} with residues  ( − 1 ) a − b i . {\displaystyle (-1)^{a-b}i.}

Also

sl ⁡ z = sl ⁡ w ↔ z = ( − 1 ) m + n w + ( m + n i ) ϖ {\displaystyle \operatorname {sl} z=\operatorname {sl} w\leftrightarrow z=(-1)^{m+n}w+(m+ni)\varpi }

for some  m , n ∈ Z {\displaystyle m,n\in \mathbb {Z} } and

sl ⁡ ( ( 1 ± i ) z ) = ( 1 ± i ) sl ⁡ z sl ′ ⁡ z . {\displaystyle \operatorname {sl} ((1\pm i)z)=(1\pm i){\frac {\operatorname {sl} z}{\operatorname {sl} 'z}}.}

The last formula is a special case of complex multiplication. Analogous formulas can be given for  sl ⁡ ( ( n + m i ) z ) {\displaystyle \operatorname {sl} ((n+mi)z)} where  n + m i {\displaystyle n+mi} is any Gaussian integer – the function  sl {\displaystyle \operatorname {sl} } has complex multiplication by  Z [ i ] {\displaystyle \mathbb {Z} [i]} .[15]

There are also infinite series reflecting the distribution of the zeros and poles of sl:[16][17]

1 sl ⁡ z = ∑ ( n , k ) ∈ Z 2 ( − 1 ) n + k z + n ϖ + k ϖ i {\displaystyle {\frac {1}{\operatorname {sl} z}}=\sum _{(n,k)\in \mathbb {Z} ^{2}}{\frac {(-1)^{n+k}}{z+n\varpi +k\varpi i}}}
sl ⁡ z = − i ∑ ( n , k ) ∈ Z 2 ( − 1 ) n + k z + ( n + 1 / 2 ) ϖ + ( k + 1 / 2 ) ϖ i . {\displaystyle \operatorname {sl} z=-i\sum _{(n,k)\in \mathbb {Z} ^{2}}{\frac {(-1)^{n+k}}{z+(n+1/2)\varpi +(k+1/2)\varpi i}}.}

### Pythagorean-like identity

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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:

c l 2 ⁡ z + s l 2 ⁡ z + c l 2 ⁡ z s l 2 ⁡ z = 1 {\displaystyle \operatorname {cl^{2}} z+\operatorname {sl^{2}} z+\operatorname {cl^{2}} z\,\operatorname {sl^{2}} z=1}

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

This identity can alternately be rewritten:[18]

( 1 + c l 2 ⁡ z ) ( 1 + s l 2 ⁡ z ) = 2 {\displaystyle {\bigl (}1+\operatorname {cl^{2}} z{\bigr )}{\bigl (}1+\operatorname {sl^{2}} z{\bigr )}=2}

c l 2 ⁡ z = 1 − s l 2 ⁡ z 1 + s l 2 ⁡ z , s l 2 ⁡ z = 1 − c l 2 ⁡ z 1 + c l 2 ⁡ z {\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  a ⊕ b := tan ⁡ ( arctan ⁡ a + arctan ⁡ b ) = a + b 1 − a b , {\displaystyle a\oplus b\mathrel {:=} \tan(\arctan a+\arctan b)={\frac {a+b}{1-ab}},} gives:

c l 2 ⁡ z ⊕ s l 2 ⁡ z = 1. {\displaystyle \operatorname {cl^{2}} z\oplus \operatorname {sl^{2}} z=1.}

The functions  cl ~ {\displaystyle {\tilde {\operatorname {cl} }}} and  sl ~ {\displaystyle {\tilde {\operatorname {sl} }}} satisfy another Pythagorean-like identity:

( ∫ 0 x cl ~ t d t ) 2 + ( 1 − ∫ 0 x sl ~ t d t ) 2 = 1. {\displaystyle \left(\int _{0}^{x}{\tilde {\operatorname {cl} }}\,t\,\mathrm {d} t\right)^{2}+\left(1-\int _{0}^{x}{\tilde {\operatorname {sl} }}\,t\,\mathrm {d} t\right)^{2}=1.}

### Derivatives and integrals

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The derivatives are as follows:

d d z cl ⁡ z = c l ′ ⁡ z = − ( 1 + c l 2 ⁡ z ) sl ⁡ z = − 2 sl ⁡ z sl 2 ⁡ z + 1 c l ′ 2 ⁡ z = 1 − c l 4 ⁡ z d d z sl ⁡ z = s l ′ ⁡ z = ( 1 + s l 2 ⁡ z ) cl ⁡ z = 2 cl ⁡ z cl 2 ⁡ z + 1 s l ′ 2 ⁡ z = 1 − s l 4 ⁡ z {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {cl} z=\operatorname {cl'} z&=-{\bigl (}1+\operatorname {cl^{2}} z{\bigr )}\operatorname {sl} z=-{\frac {2\operatorname {sl} z}{\operatorname {sl} ^{2}z+1}}\\\\\operatorname {cl'^{2}} z&=1-\operatorname {cl^{4}} z\\\\[5mu]{\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {sl} z=\operatorname {sl'} z&={\bigl (}1+\operatorname {sl^{2}} z{\bigr )}\operatorname {cl} z={\frac {2\operatorname {cl} z}{\operatorname {cl} ^{2}z+1}}\\\\\operatorname {sl'^{2}} z&=1-\operatorname {sl^{4}} z\end{aligned}}}

d d z cl ~ z = − 2 sl ~ z cl ⁡ z − sl ~ z cl ⁡ z d d z sl ~ z = 2 cl ~ z cl ⁡ z − cl ~ z cl ⁡ z {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} z}}\,{\tilde {\operatorname {cl} }}\,z&=-2\,{\tilde {\operatorname {sl} }}\,z\,\operatorname {cl} z-{\frac {{\tilde {\operatorname {sl} }}\,z}{\operatorname {cl} z}}\\\\{\frac {\mathrm {d} }{\mathrm {d} z}}\,{\tilde {\operatorname {sl} }}\,z&=2\,{\tilde {\operatorname {cl} }}\,z\,\operatorname {cl} z-{\frac {{\tilde {\operatorname {cl} }}\,z}{\operatorname {cl} z}}\end{aligned}}}

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

d 2 d z 2 cl ⁡ z = − 2 c l 3 ⁡ z {\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}\operatorname {cl} z=-2\operatorname {cl^{3}} z}

d 2 d z 2 sl ⁡ z = − 2 s l 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:

∫ cl ⁡ z d z = arctan ⁡ sl ⁡ z + C ∫ sl ⁡ z d z = − arctan ⁡ cl ⁡ z + C ∫ cl ~ z d z = sl ~ z cl ⁡ z + C ∫ sl ~ z d z = − cl ~ z cl ⁡ z + C {\displaystyle {\begin{aligned}\int \operatorname {cl} z\mathop {\mathrm {d} z} &=\arctan \operatorname {sl} z+C\\\\\int \operatorname {sl} z\mathop {\mathrm {d} z} &=-\arctan \operatorname {cl} z+C\\\\\int {\tilde {\operatorname {cl} }}\,z\,\mathrm {d} z&={\frac {{\tilde {\operatorname {sl} }}\,z}{\operatorname {cl} z}}+C\\\\\int {\tilde {\operatorname {sl} }}\,z\,\mathrm {d} z&=-{\frac {{\tilde {\operatorname {cl} }}\,z}{\operatorname {cl} z}}+C\end{aligned}}}

### Argument sum and multiple identities

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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:[19]

sl ⁡ ( u + v ) = sl ⁡ u s l ′ ⁡ v + sl ⁡ v s l ′ ⁡ u 1 + s l 2 ⁡ u s l 2 ⁡ v {\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  a ⊕ b := tan ⁡ ( arctan ⁡ a + arctan ⁡ b ) {\displaystyle a\oplus b\mathrel {:=} \tan(\arctan a+\arctan b)} and tangent-difference operator  a ⊖ b := a ⊕ ( − b ) , {\displaystyle a\ominus b\mathrel {:=} a\oplus (-b),} the argument sum and difference identities can be expressed as:[20]

cl ⁡ ( u + v ) = cl ⁡ u cl ⁡ v ⊖ sl ⁡ u sl ⁡ v = cl ⁡ u cl ⁡ v − sl ⁡ u sl ⁡ v 1 + sl ⁡ u cl ⁡ u sl ⁡ v cl ⁡ v cl ⁡ ( u − v ) = cl ⁡ u cl ⁡ v ⊕ sl ⁡ u sl ⁡ v sl ⁡ ( u + v ) = sl ⁡ u cl ⁡ v ⊕ cl ⁡ u sl ⁡ v = sl ⁡ u cl ⁡ v + cl ⁡ u sl ⁡ v 1 − sl ⁡ u cl ⁡ u sl ⁡ v cl ⁡ v sl ⁡ ( u − v ) = sl ⁡ u cl ⁡ v ⊖ cl ⁡ u sl ⁡ v {\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:

cos ⁡ ( u ± v ) = cos ⁡ u cos ⁡ v ∓ sin ⁡ u sin ⁡ v sin ⁡ ( u ± v ) = sin ⁡ u cos ⁡ v ± cos ⁡ u sin ⁡ v {\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}}}

In particular, to compute the complex-valued functions in real components,

cl ⁡ ( x + i y ) = cl ⁡ x − i sl ⁡ x sl ⁡ y cl ⁡ y cl ⁡ y + i sl ⁡ x cl ⁡ x sl ⁡ y = cl ⁡ x cl ⁡ y ( 1 − sl 2 ⁡ x sl 2 ⁡ y ) cl 2 ⁡ y + sl 2 ⁡ x cl 2 ⁡ x sl 2 ⁡ y − i sl ⁡ x sl ⁡ y ( cl 2 ⁡ x + cl 2 ⁡ y ) cl 2 ⁡ y + sl 2 ⁡ x cl 2 ⁡ x sl 2 ⁡ y sl ⁡ ( x + i y ) = sl ⁡ x + i cl ⁡ x sl ⁡ y cl ⁡ y cl ⁡ y − i sl ⁡ x cl ⁡ x sl ⁡ y = sl ⁡ x cl ⁡ y ( 1 − cl 2 ⁡ x sl 2 ⁡ y ) cl 2 ⁡ y + sl 2 ⁡ x cl 2 ⁡ x sl 2 ⁡ y + i cl ⁡ x sl ⁡ y ( sl 2 ⁡ x + cl 2 ⁡ y ) cl 2 ⁡ y + sl 2 ⁡ x cl 2 ⁡ x sl 2 ⁡ y {\displaystyle {\begin{aligned}\operatorname {cl} (x+iy)&={\frac {\operatorname {cl} x-i\operatorname {sl} x\,\operatorname {sl} y\,\operatorname {cl} y}{\operatorname {cl} y+i\operatorname {sl} x\,\operatorname {cl} x\,\operatorname {sl} y}}\\\\[4mu]&={\frac {\operatorname {cl} x\,\operatorname {cl} y\left(1-\operatorname {sl} ^{2}x\,\operatorname {sl} ^{2}y\right)}{\operatorname {cl} ^{2}y+\operatorname {sl} ^{2}x\,\operatorname {cl} ^{2}x\,\operatorname {sl} ^{2}y}}-i{\frac {\operatorname {sl} x\,\operatorname {sl} y\left(\operatorname {cl} ^{2}x+\operatorname {cl} ^{2}y\right)}{\operatorname {cl} ^{2}y+\operatorname {sl} ^{2}x\,\operatorname {cl} ^{2}x\,\operatorname {sl} ^{2}y}}\\\\[12mu]\operatorname {sl} (x+iy)&={\frac {\operatorname {sl} x+i\operatorname {cl} x\,\operatorname {sl} y\,\operatorname {cl} y}{\operatorname {cl} y-i\operatorname {sl} x\,\operatorname {cl} x\,\operatorname {sl} y}}\\\\[4mu]&={\frac {\operatorname {sl} x\,\operatorname {cl} y\left(1-\operatorname {cl} ^{2}x\,\operatorname {sl} ^{2}y\right)}{\operatorname {cl} ^{2}y+\operatorname {sl} ^{2}x\,\operatorname {cl} ^{2}x\,\operatorname {sl} ^{2}y}}+i{\frac {\operatorname {cl} x\,\operatorname {sl} y\left(\operatorname {sl} ^{2}x+\operatorname {cl} ^{2}y\right)}{\operatorname {cl} ^{2}y+\operatorname {sl} ^{2}x\,\operatorname {cl} ^{2}x\,\operatorname {sl} ^{2}y}}\end{aligned}}}

Gauss discovered that

sl ⁡ ( u − v ) sl ⁡ ( u + v ) = sl ⁡ ( ( 1 + i ) u ) − sl ⁡ ( ( 1 + i ) v ) sl ⁡ ( ( 1 + i ) u ) + sl ⁡ ( ( 1 + i ) v ) {\displaystyle {\frac {\operatorname {sl} (u-v)}{\operatorname {sl} (u+v)}}={\frac {\operatorname {sl} ((1+i)u)-\operatorname {sl} ((1+i)v)}{\operatorname {sl} ((1+i)u)+\operatorname {sl} ((1+i)v)}}}

where  u , v ∈ C {\displaystyle u,v\in \mathbb {C} } such that both sides are well-defined.

Also

sl ⁡ ( u + v ) sl ⁡ ( u − v ) = sl 2 ⁡ u − sl 2 ⁡ v 1 + sl 2 ⁡ u sl 2 ⁡ v {\displaystyle \operatorname {sl} (u+v)\operatorname {sl} (u-v)={\frac {\operatorname {sl} ^{2}u-\operatorname {sl} ^{2}v}{1+\operatorname {sl} ^{2}u\operatorname {sl} ^{2}v}}}

where  u , v ∈ C {\displaystyle u,v\in \mathbb {C} } such that both sides are well-defined; this resembles the trigonometric analog

sin ⁡ ( u + v ) sin ⁡ ( u − v ) = sin 2 ⁡ u − sin 2 ⁡ v . {\displaystyle \sin(u+v)\sin(u-v)=\sin ^{2}u-\sin ^{2}v.}

Bisection formulas:

cl 2 ⁡ 1 2 x = 1 + cl ⁡ x 1 + sl 2 ⁡ x 1 + sl 2 ⁡ x + 1 {\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}}}

sl 2 ⁡ 1 2 x = 1 − cl ⁡ x 1 + sl 2 ⁡ x 1 + 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:[21]

cl ⁡ 2 x = − 1 + 2 cl 2 ⁡ x + cl 4 ⁡ x 1 + 2 cl 2 ⁡ x − cl 4 ⁡ x {\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}}}

sl ⁡ 2 x = 2 sl ⁡ x cl ⁡ x 1 + sl 2 ⁡ x 1 + sl 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:[21]

cl ⁡ 3 x = − 3 cl ⁡ x + 6 cl 5 ⁡ x + cl 9 ⁡ x 1 + 6 cl 4 ⁡ x − 3 cl 8 ⁡ x {\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}}}

sl ⁡ 3 x = 3 sl ⁡ x − 6 sl 5 ⁡ x − 1 sl 9 ⁡ x 1 + 6 sl 4 ⁡ x − 3 sl 8 ⁡ x {\displaystyle \operatorname {sl} 3x={\frac {\color {red}{3}\,\color {black}{\operatorname {sl} x-\,}\color {green}{6}\,\color {black}{\operatorname {sl} ^{5}x-\,}\color {blue}{1}\,\color {black}{\operatorname {sl} ^{9}x}}{\color {blue}{1}\,\color {black}{+\,}\,\color {green}{6}\,\color {black}{\operatorname {sl} ^{4}x-\,}\color {red}{3}\,\color {black}{\operatorname {sl} ^{8}x}}}}

Note the "reverse symmetry" of the coefficients of numerator and denominator of  sl ⁡ 3 x {\displaystyle \operatorname {sl} 3x} . This phenomenon can be observed in multiplication formulas for  sl ⁡ β x {\displaystyle \operatorname {sl} \beta x} where  β = m + n i {\displaystyle \beta =m+ni} whenever  m , n ∈ Z {\displaystyle m,n\in \mathbb {Z} } and  m + n {\displaystyle m+n} is odd.[15]

### Lemnatomic polynomials

[edit]

Let  L {\displaystyle L} be the lattice

L = Z ( 1 + i ) ϖ + Z ( 1 − i ) ϖ . {\displaystyle L=\mathbb {Z} (1+i)\varpi +\mathbb {Z} (1-i)\varpi .}

Furthermore, let  K = Q ( i ) {\displaystyle K=\mathbb {Q} (i)} ,  O = Z [ i ] {\displaystyle {\mathcal {O}}=\mathbb {Z} [i]} ,  z ∈ C {\displaystyle z\in \mathbb {C} } ,  β = m + i n {\displaystyle \beta =m+in} ,  γ = m ′ + i n ′ {\displaystyle \gamma =m'+in'} (where  m , n , m ′ , n ′ ∈ Z {\displaystyle m,n,m',n'\in \mathbb {Z} } ),  m + n {\displaystyle m+n} be odd,  m ′ + n ′ {\displaystyle m'+n'} be odd,  γ ≡ 1 mod 2 ( 1 + i ) {\displaystyle \gamma \equiv 1\,\operatorname {mod} \,2(1+i)} and  sl ⁡ β z = M β ( sl ⁡ z ) {\displaystyle \operatorname {sl} \beta z=M_{\beta }(\operatorname {sl} z)} . Then

M β ( x ) = i ε x P β ( x 4 ) Q β ( x 4 ) {\displaystyle M_{\beta }(x)=i^{\varepsilon }x{\frac {P_{\beta }(x^{4})}{Q_{\beta }(x^{4})}}}

for some coprime polynomials  P β ( x ) , Q β ( x ) ∈ O [ x ] {\displaystyle P_{\beta }(x),Q_{\beta }(x)\in {\mathcal {O}}[x]} and some  ε ∈ { 0 , 1 , 2 , 3 } {\displaystyle \varepsilon \in \\{0,1,2,3\\}} [22] where

x P β ( x 4 ) = ∏ γ | β Λ γ ( x ) {\displaystyle xP_{\beta }(x^{4})=\prod _{\gamma |\beta }\Lambda _{\gamma }(x)}

and

Λ β ( x ) = ∏ [ α ] ∈ ( O / β O ) × ( x − sl ⁡ α δ β ) {\displaystyle \Lambda _{\beta }(x)=\prod _{[\alpha ]\in ({\mathcal {O}}/\beta {\mathcal {O}})^{\times }}(x-\operatorname {sl} \alpha \delta _{\beta })}

where  δ β {\displaystyle \delta _{\beta }} is any  β {\displaystyle \beta } -torsion generator (i.e.  δ β ∈ ( 1 / β ) L {\displaystyle \delta _{\beta }\in (1/\beta )L} and  [ δ β ] ∈ ( 1 / β ) L / L {\displaystyle [\delta _{\beta }]\in (1/\beta )L/L} generates  ( 1 / β ) L / L {\displaystyle (1/\beta )L/L} as an  O {\displaystyle {\mathcal {O}}} -module). Examples of  β {\displaystyle \beta } -torsion generators include  2 ϖ / β {\displaystyle 2\varpi /\beta } and  ( 1 + i ) ϖ / β {\displaystyle (1+i)\varpi /\beta } . The polynomial  Λ β ( x ) ∈ O [ x ] {\displaystyle \Lambda _{\beta }(x)\in {\mathcal {O}}[x]} is called the  β {\displaystyle \beta } -th lemnatomic polynomial. It is monic and is irreducible over  K {\displaystyle K} . The lemnatomic polynomials are the "lemniscate analogs" of the cyclotomic polynomials,[23]

Φ k ( x ) = ∏ [ a ] ∈ ( Z / k Z ) × ( x − ζ k a ) . {\displaystyle \Phi _{k}(x)=\prod _{[a]\in (\mathbb {Z} /k\mathbb {Z} )^{\times }}(x-\zeta _{k}^{a}).}

The  β {\displaystyle \beta } -th lemnatomic polynomial  Λ β ( x ) {\displaystyle \Lambda _{\beta }(x)} is the minimal polynomial of  sl ⁡ δ β {\displaystyle \operatorname {sl} \delta _{\beta }} in  K [ x ] {\displaystyle K[x]} . For convenience, let  ω β = sl ⁡ ( 2 ϖ / β ) {\displaystyle \omega _{\beta }=\operatorname {sl} (2\varpi /\beta )} and  ω ~ β = sl ⁡ ( ( 1 + i ) ϖ / β ) {\displaystyle {\tilde {\omega }}_{\beta }=\operatorname {sl} ((1+i)\varpi /\beta )} . So for example, the minimal polynomial of  ω 5 {\displaystyle \omega _{5}} (and also of  ω ~ 5 {\displaystyle {\tilde {\omega }}_{5}} ) in  K [ x ] {\displaystyle K[x]} is

Λ 5 ( x ) = x 16 + 52 x 12 − 26 x 8 − 12 x 4 + 1 , {\displaystyle \Lambda _{5}(x)=x^{16}+52x^{12}-26x^{8}-12x^{4}+1,}

and[24]

ω 5 = − 13 + 6 5 + 2 85 − 38 5 4 {\displaystyle \omega _{5}={\sqrt[{4}]{-13+6{\sqrt {5}}+2{\sqrt {85-38{\sqrt {5}}}}}}}
ω ~ 5 = − 13 − 6 5 + 2 85 + 38 5 4 {\displaystyle {\tilde {\omega }}_{5}={\sqrt[{4}]{-13-6{\sqrt {5}}+2{\sqrt {85+38{\sqrt {5}}}}}}} [25]

(an equivalent expression is given in the table below). Another example is[23]

Λ − 1 + 2 i ( x ) = x 4 − 1 + 2 i {\displaystyle \Lambda _{-1+2i}(x)=x^{4}-1+2i}

which is the minimal polynomial of  ω − 1 + 2 i {\displaystyle \omega _{-1+2i}} (and also of  ω ~ − 1 + 2 i {\displaystyle {\tilde {\omega }}_{-1+2i}} ) in  K [ x ] . {\displaystyle K[x].}

If  p {\displaystyle p} is prime and  β {\displaystyle \beta } is positive and odd,[26] then[27]

deg ⁡ Λ β = β 2 ∏ p | β ( 1 − 1 p ) ( 1 − ( − 1 ) ( p − 1 ) / 2 p ) {\displaystyle \operatorname {deg} \Lambda _{\beta }=\beta ^{2}\prod _{p|\beta }\left(1-{\frac {1}{p}}\right)\left(1-{\frac {(-1)^{(p-1)/2}}{p}}\right)}

which can be compared to the cyclotomic analog

deg ⁡ Φ k = k ∏ p | k ( 1 − 1 p ) . {\displaystyle \operatorname {deg} \Phi _{k}=k\prod _{p|k}\left(1-{\frac {1}{p}}\right).}

### Specific values

[edit]

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, if and only if n is of the form  n = 2 k p 1 p 2 ⋯ p m {\displaystyle n=2^{k}p_{1}p_{2}\cdots p_{m}} where k is a non-negative integer and each pi (if any) is a distinct Fermat prime.[28]

n {\displaystyle n} |  cl ⁡ n ϖ {\displaystyle \operatorname {cl} n\varpi } |  sl ⁡ n ϖ {\displaystyle \operatorname {sl} n\varpi }

|
|

1 {\displaystyle 1} |  − 1 {\displaystyle -1} |  0 {\displaystyle 0}
5 6 {\displaystyle {\tfrac {5}{6}}} |  − 2 3 − 3 4 {\displaystyle -{\sqrt[{4}]{2{\sqrt {3}}-3}}} |  1 2 ( 3 + 1 − 12 4 ) {\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}
3 4 {\displaystyle {\tfrac {3}{4}}} |  − 2 − 1 {\displaystyle -{\sqrt {{\sqrt {2}}-1}}} |  2 − 1 {\displaystyle {\sqrt {{\sqrt {2}}-1}}}
2 3 {\displaystyle {\tfrac {2}{3}}} |  − 1 2 ( 3 + 1 − 12 4 ) {\displaystyle -{\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}} |  2 3 − 3 4 {\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}
1 2 {\displaystyle {\tfrac {1}{2}}} |  0 {\displaystyle 0} |  1 {\displaystyle 1}
1 3 {\displaystyle {\tfrac {1}{3}}} |  1 2 ( 3 + 1 − 12 4 ) {\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}} |  2 3 − 3 4 {\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}}
1 4 {\displaystyle {\tfrac {1}{4}}} |  2 − 1 {\displaystyle {\sqrt {{\sqrt {2}}-1}}} |  2 − 1 {\displaystyle {\sqrt {{\sqrt {2}}-1}}}
1 6 {\displaystyle {\tfrac {1}{6}}} |  2 3 − 3 4 {\displaystyle {\sqrt[{4}]{2{\sqrt {3}}-3}}} |  1 2 ( 3 + 1 − 12 4 ) {\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt {3}}+1-{\sqrt[{4}]{12}}{\bigr )}}

## Relation to geometric shapes

[edit]

### Arc length of Bernoulli's lemniscate

[edit]

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.

L {\displaystyle {\mathcal {L}}} , the lemniscate of Bernoulli with unit distance from its center to its furthest point (i.e. with unit "half-width"), is essential in the theory of the lemniscate elliptic functions. It can be characterized in at least three ways:

Angular characterization: Given two points  A {\displaystyle A} and  B {\displaystyle B} which are unit distance apart, let  B ′ {\displaystyle B'} be the reflection of  B {\displaystyle B} about  A {\displaystyle A} . Then  L {\displaystyle {\mathcal {L}}} is the closure of the locus of the points  P {\displaystyle P} such that  | A P B − A P B ′ | {\displaystyle |APB-APB'|} is a right angle.[29]

Focal characterization: L {\displaystyle {\mathcal {L}}} is the locus of points in the plane such that the product of their distances from the two focal points  F 1 = ( − 1 2 , 0 ) {\displaystyle F_{1}={\bigl (}{-{\tfrac {1}{\sqrt {2}}}},0{\bigr )}} and  F 2 = ( 1 2 , 0 ) {\displaystyle F_{2}={\bigl (}{\tfrac {1}{\sqrt {2}}},0{\bigr )}} is the constant  1 2 {\displaystyle {\tfrac {1}{2}}} .

Explicit coordinate characterization: L {\displaystyle {\mathcal {L}}} is a quartic curve satisfying the polar equation  r 2 = cos ⁡ 2 θ {\displaystyle r^{2}=\cos 2\theta } or the Cartesian equation  ( x 2 + y 2 ) 2 = x 2 − y 2 . {\displaystyle {\bigl (}x^{2}+y^{2}{\bigr )}{}^{2}=x^{2}-y^{2}.}

The perimeter of  L {\displaystyle {\mathcal {L}}} is  2 ϖ {\displaystyle 2\varpi } .[30]

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

( x ( r ) , y ( r ) ) = ( 1 2 r 2 ( 1 + r 2 ) , 1 2 r 2 ( 1 − r 2 ) ) . {\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  r ∈ [ 0 , 1 ] {\displaystyle r\in [0,1]} for a quarter of  L {\displaystyle {\mathcal {L}}} , the arc length from the origin to a point  ( x ( r ) , y ( r ) ) {\displaystyle {\big (}x(r),y(r){\big )}} is:[31]

∫ 0 r x ′ ( t ) 2 + y ′ ( t ) 2 d t = ∫ 0 r ( 1 + 2 t 2 ) 2 2 ( 1 + t 2 ) + ( 1 − 2 t 2 ) 2 2 ( 1 − t 2 ) d t = ∫ 0 r d t 1 − t 4 = arcsl ⁡ r . {\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  ( 1 , 0 ) {\displaystyle (1,0)} to  ( x ( r ) , y ( r ) ) {\displaystyle {\big (}x(r),y(r){\big )}} is:

∫ r 1 x ′ ( t ) 2 + y ′ ( t ) 2 d t = ∫ r 1 d t 1 − t 4 = arccl ⁡ r = 1 2 ϖ − arcsl ⁡ r . {\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  ( 1 , 0 ) {\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  r = cos ⁡ θ {\displaystyle r=\cos \theta } or Cartesian equation  x 2 + y 2 = x , {\displaystyle x^{2}+y^{2}=x,} using the same argument above but with the parametrization:

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

Alternatively, just as the unit circle x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1} is parametrized in terms of the arc length  s {\displaystyle s} from the point  ( 1 , 0 ) {\displaystyle (1,0)} by

( x ( s ) , y ( s ) ) = ( cos ⁡ s , sin ⁡ s ) , {\displaystyle (x(s),y(s))=(\cos s,\sin s),}

L {\displaystyle {\mathcal {L}}} is parametrized in terms of the arc length  s {\displaystyle s} from the point  ( 1 , 0 ) {\displaystyle (1,0)} by[32]

( x ( s ) , y ( s ) ) = ( cl ⁡ s 1 + sl 2 ⁡ s , sl ⁡ s cl ⁡ s 1 + sl 2 ⁡ s ) = ( cl ~ s , sl ~ s ) . {\displaystyle (x(s),y(s))=\left({\frac {\operatorname {cl} s}{\sqrt {1+\operatorname {sl} ^{2}s}}},{\frac {\operatorname {sl} s\operatorname {cl} s}{\sqrt {1+\operatorname {sl} ^{2}s}}}\right)=\left({\tilde {\operatorname {cl} }}\,s,{\tilde {\operatorname {sl} }}\,s\right).}

The notation  cl ~ , sl ~ {\displaystyle {\tilde {\operatorname {cl} }},\,{\tilde {\operatorname {sl} }}} is used solely for the purposes of this article; in references, notation for general Jacobi elliptic functions is used instead.

The lemniscate integral and lemniscate functions satisfy an argument duplication identity discovered by Fagnano in 1718:[33]

∫ 0 z d t 1 − t 4 = 2 ∫ 0 u d t 1 − t 4 , if  z = 2 u 1 − u 4 1 + u 4  and  0 ≤ u ≤ 2 − 1 . {\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}}.}
A lemniscate divided into 15 sections of equal arclength (red curves). Because the prime factors of 15 (3 and 5) are both Fermat primes, this polygon (in black) is constructible using a straightedge and compass.

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 only straightedge and compass if and only if n is of the form  n = 2 k p 1 p 2 ⋯ p m {\displaystyle n=2^{k}p_{1}p_{2}\cdots p_{m}} where k is a non-negative integer and each pi (if any) is a distinct Fermat prime.[34] The "if" part of the theorem was proved by Niels Abel in 1827–1828, and the "only if" part was proved by Michael Rosen in 1981.[35] Equivalently, the lemniscate can be divided into n sections of equal arc length using only straightedge and compass if and only if  φ ( n ) {\displaystyle \varphi (n)} is a power of two (where  φ {\displaystyle \varphi } is Euler's totient function). The lemniscate is not assumed to be already drawn, as that would go against the rules of straightedge and compass constructions; instead, it is assumed that we are given only two points by which the lemniscate is defined, such as its center and radial point (one of the two points on the lemniscate such that their distance from the center is maximal) or its two foci.

Let  r j = sl ⁡ 2 j ϖ n {\displaystyle r_{j}=\operatorname {sl} {\dfrac {2j\varpi }{n}}} . Then the n-division points for  L {\displaystyle {\mathcal {L}}} are the points

( r j 1 2 ( 1 + r j 2 ) ,   ( − 1 ) ⌊ 4 j / n ⌋ 1 2 r j 2 ( 1 − r j 2 ) ) , j ∈ { 1 , 2 , … , n } {\displaystyle \left(r_{j}{\sqrt {{\tfrac {1}{2}}{\bigl (}1+r_{j}^{2}{\bigr )}}},\ (-1)^{\left\lfloor 4j/n\right\rfloor }{\sqrt {{\tfrac {1}{2}}r_{j}^{2}{\bigl (}1-r_{j}^{2}{\bigr )}}}\right),\quad j\in \\{1,2,\ldots ,n\\}}

where  ⌊ ⋅ ⌋ {\displaystyle \lfloor \cdot \rfloor } is the floor function. See below for some specific values of  sl ⁡ 2 ϖ n {\displaystyle \operatorname {sl} {\dfrac {2\varpi }{n}}} .

### Arc length of rectangular elastica

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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.[36] This curve has y coordinate and arc length:

y = ∫ x 1 t 2 d t 1 − t 4 , s = arcsl ⁡ x = ∫ 0 x d t 1 − t 4 {\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.

### Elliptic characterization

[edit]

The lemniscate elliptic functions and an ellipse

Let  C {\displaystyle C} be a point on the ellipse  x 2 + 2 y 2 = 1 {\displaystyle x^{2}+2y^{2}=1} in the first quadrant and let  D {\displaystyle D} be the projection of  C {\displaystyle C} on the unit circle  x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1} . The distance  r {\displaystyle r} between the origin  A {\displaystyle A} and the point  C {\displaystyle C} is a function of  φ {\displaystyle \varphi } (the angle  B A C {\displaystyle BAC} where  B = ( 1 , 0 ) {\displaystyle B=(1,0)} ; equivalently the length of the circular arc  B D {\displaystyle BD} ). The parameter  u {\displaystyle u} is given by

u = ∫ 0 φ r ( θ ) d θ = ∫ 0 φ d θ 1 + sin 2 ⁡ θ . {\displaystyle u=\int _{0}^{\varphi }r(\theta )\,\mathrm {d} \theta =\int _{0}^{\varphi }{\frac {\mathrm {d} \theta }{\sqrt {1+\sin ^{2}\theta }}}.}

If  E {\displaystyle E} is the projection of  D {\displaystyle D} on the x-axis and if  F {\displaystyle F} is the projection of  C {\displaystyle C} on the x-axis, then the lemniscate elliptic functions are given by

cl ⁡ u = A F ¯ , sl ⁡ u = D E ¯ , {\displaystyle \operatorname {cl} u={\overline {AF}},\quad \operatorname {sl} u={\overline {DE}},}
cl ~ u = A F ¯ A C ¯ , sl ~ u = A F ¯ F C ¯ . {\displaystyle {\tilde {\operatorname {cl} }}\,u={\overline {AF}}{\overline {AC}},\quad {\tilde {\operatorname {sl} }}\,u={\overline {AF}}{\overline {FC}}.}

## Series Identities

[edit]

### Power series

[edit]

The power series expansion of the lemniscate sine at the origin is[37]

sl ⁡ z = ∑ n = 0 ∞ a n z n = z − 12 z 5 5 ! + 3024 z 9 9 ! − 4390848 z 13 13 ! + ⋯ , | z | < ϖ 2 {\displaystyle \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|<{\tfrac {\varpi }{\sqrt {2}}}}

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

n ≢ 1 ( mod 4 ) ⟹ a n = 0 , {\displaystyle n\not \equiv 1{\pmod {4}}\implies a_{n}=0,}
a 1 = 1 , ∀ n ∈ N 0 : a n + 2 = − 2 ( n + 1 ) ( n + 2 ) ∑ i + j + k = n a i a j a k {\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  i + j + k = n {\displaystyle i+j+k=n} stands for all three-term compositions of  n {\displaystyle n} . For example, to evaluate  a 13 {\displaystyle a_{13}} , it can be seen that there are only six compositions of  13 − 2 = 11 {\displaystyle 13-2=11} that give a nonzero contribution to the sum:  11 = 9 + 1 + 1 = 1 + 9 + 1 = 1 + 1 + 9 {\displaystyle 11=9+1+1=1+9+1=1+1+9} and  11 = 5 + 5 + 1 = 5 + 1 + 5 = 1 + 5 + 5 {\displaystyle 11=5+5+1=5+1+5=1+5+5} , so

a 13 = − 2 12 ⋅ 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 ) = − 11 15600 . {\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 expansion can be equivalently written as[38]

sl ⁡ z = ∑ n = 0 ∞ p 2 n z 4 n + 1 ( 4 n + 1 ) ! , | z | < ϖ 2 {\displaystyle \operatorname {sl} z=\sum _{n=0}^{\infty }p_{2n}{\frac {z^{4n+1}}{(4n+1)!}},\quad \left|z\right|<{\frac {\varpi }{\sqrt {2}}}}

where

p n + 2 = − 12 ∑ j = 0 n ( 2 n + 2 2 j + 2 ) p n − j ∑ k = 0 j ( 2 j + 1 2 k + 1 ) p k p j − k , p 0 = 1 , p 1 = 0. {\displaystyle 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  sl ~ {\displaystyle {\tilde {\operatorname {sl} }}} at the origin is

sl ~ z = ∑ n = 0 ∞ α n z n = z − 9 z 3 3 ! + 153 z 5 5 ! − 4977 z 7 7 ! + ⋯ , | z | < ϖ 2 {\displaystyle {\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|<{\frac {\varpi }{2}}}

where  α n = 0 {\displaystyle \alpha _{n}=0} if  n {\displaystyle n} is even and[39]

α n = 2 π ϖ ( − 1 ) ( n − 1 ) / 2 n ! ∑ k = 1 ∞ ( 2 k π / ϖ ) n + 1 cosh ⁡ k π , | α n | ∼ 2 n + 5 / 2 n + 1 ϖ n + 2 {\displaystyle \alpha _{n}={\sqrt {2}}{\frac {\pi }{\varpi }}{\frac {(-1)^{(n-1)/2}}{n!}}\sum _{k=1}^{\infty }{\frac {(2k\pi /\varpi )^{n+1}}{\cosh k\pi }},\quad \left|\alpha _{n}\right|\sim 2^{n+5/2}{\frac {n+1}{\varpi ^{n+2}}}}

if  n {\displaystyle n} is odd.

The expansion can be equivalently written as[40]

sl ~ z = ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 ( ∑ l = 0 n 2 l ( 2 n + 2 2 l + 1 ) s l t n − l ) z 2 n + 1 ( 2 n + 1 ) ! , | z | < ϖ 2 {\displaystyle {\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|<{\frac {\varpi }{2}}}

where

s n + 2 = 3 s n + 1 + 24 ∑ j = 0 n ( 2 n + 2 2 j + 2 ) s n − j ∑ k = 0 j ( 2 j + 1 2 k + 1 ) s k s j − k , s 0 = 1 , s 1 = 3 , {\displaystyle s_{n+2}=3s_{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 ∑ j = 0 n ( 2 n + 2 2 j + 2 ) t n − j ∑ k = 0 j ( 2 j + 1 2 k + 1 ) t k t j − k , t 0 = 1 , t 1 = 3. {\displaystyle t_{n+2}=3t_{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,[41]

cl ⁡ z = 1 − ∑ n = 0 ∞ ( − 1 ) n ( ∑ l = 0 n 2 l ( 2 n + 2 2 l + 1 ) q l r n − l ) z 2 n + 2 ( 2 n + 2 ) ! = 1 − 2 z 2 2 ! + 12 z 4 4 ! − 216 z 6 6 ! + ⋯ , | z | < ϖ 2 , {\displaystyle \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|<{\frac {\varpi }{2}},}
cl ~ z = ∑ n = 0 ∞ ( − 1 ) n 2 n q n z 2 n ( 2 n ) ! = 1 − 3 z 2 2 ! + 33 z 4 4 ! − 819 z 6 6 ! + ⋯ , | z | < ϖ 2 {\displaystyle {\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|<{\frac {\varpi }{2}}}

where

r n + 2 = 3 ∑ j = 0 n ( 2 n + 2 2 j + 2 ) r n − j ∑ k = 0 j ( 2 j + 1 2 k + 1 ) r k r j − k , r 0 = 1 , r 1 = 0 , {\displaystyle 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 = 3 2 q n + 1 + 6 ∑ j = 0 n ( 2 n + 2 2 j + 2 ) q n − j ∑ k = 0 j ( 2 j + 1 2 k + 1 ) q k q j − k , q 0 = 1 , q 1 = 3 2 . {\displaystyle 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

[edit]

Ramanujan's famous cos/cosh identity states that if

R ( s ) = π ϖ 2 ∑ n ∈ Z cos ⁡ ( 2 n π s / ϖ ) cosh ⁡ n π , {\displaystyle R(s)={\frac {\pi }{\varpi {\sqrt {2}}}}\sum _{n\in \mathbb {Z} }{\frac {\cos(2n\pi s/\varpi )}{\cosh n\pi }},}

then[39]

R ( s ) − 2 + R ( i s ) − 2 = 2 , | Re ⁡ s | < ϖ 2 , | Im ⁡ s | < ϖ 2 . {\displaystyle R(s)^{-2}+R(is)^{-2}=2,\quad \left|\operatorname {Re} s\right|<{\frac {\varpi }{2}},\left|\operatorname {Im} s\right|<{\frac {\varpi }{2}}.}

There is a close relation between the lemniscate functions and  R ( s ) {\displaystyle R(s)} . Indeed,[39][42]

sl ~ s = − d d s R ( s ) | Im ⁡ s | < ϖ 2 {\displaystyle {\tilde {\operatorname {sl} }}\,s=-{\frac {\mathrm {d} }{\mathrm {d} s}}R(s)\quad \left|\operatorname {Im} s\right|<{\frac {\varpi }{2}}}
cl ~ s = d d s 1 − R ( s ) 2 , | Re ⁡ s − ϖ 2 | < ϖ 2 , | Im ⁡ s | < ϖ 2 {\displaystyle {\tilde {\operatorname {cl} }}\,s={\frac {\mathrm {d} }{\mathrm {d} s}}{\sqrt {1-R(s)^{2}}},\quad \left|\operatorname {Re} s-{\frac {\varpi }{2}}\right|<{\frac {\varpi }{2}},\,\left|\operatorname {Im} s\right|<{\frac {\varpi }{2}}}

and

R ( s ) = 1 1 + sl 2 ⁡ s , | Im ⁡ s | < ϖ 2 . {\displaystyle R(s)={\frac {1}{\sqrt {1+\operatorname {sl} ^{2}s}}},\quad \left|\operatorname {Im} s\right|<{\frac {\varpi }{2}}.}

### Continued fractions

[edit]

For  z ∈ C ∖ { 0 } {\displaystyle z\in \mathbb {C} \setminus \\{0\\}} :[43]

∫ 0 ∞ e − t z 2 cl ⁡ t d t = 1 / 2 z + a 1 z + a 2 z + a 3 z + ⋱ , a n = n 2 4 ( ( − 1 ) n + 1 + 3 ) {\displaystyle \int _{0}^{\infty }e^{-tz{\sqrt {2}}}\operatorname {cl} t\,\mathrm {d} t={\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)}
∫ 0 ∞ e − t z 2 sl ⁡ t cl ⁡ t d t = 1 / 2 z 2 + b 1 − a 1 z 2 + b 2 − a 2 z 2 + b 3 − ⋱ , a n = n 2 ( 4 n 2 − 1 ) , b n = 3 ( 2 n − 1 ) 2 {\displaystyle \int _{0}^{\infty }e^{-tz{\sqrt {2}}}\operatorname {sl} t\operatorname {cl} t\,\mathrm {d} t={\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

[edit]

> A fast algorithm, returning approximations to  sl ⁡ x {\displaystyle \operatorname {sl} x} (which get closer to  sl ⁡ x {\displaystyle \operatorname {sl} x} with increasing  N {\displaystyle N} ), is the following:[44]

>   * a 0 ← 1 ; {\displaystyle a_{0}\leftarrow 1;} b 0 ← 1 2 ; {\displaystyle b_{0}\leftarrow {\tfrac {1}{\sqrt {2}}};} c 0 ← 1 2 {\displaystyle c_{0}\leftarrow {\sqrt {\tfrac {1}{2}}}}
>   * for each n ≥ 1 {\displaystyle n\geq 1} do

>     * a n ← 1 2 ( a n − 1 + b n − 1 ) ; {\displaystyle a_{n}\leftarrow {\tfrac {1}{2}}(a_{n-1}+b_{n-1});} b n ← a n − 1 b n − 1 ; {\displaystyle b_{n}\leftarrow {\sqrt {a_{n-1}b_{n-1}}};} c n ← 1 2 ( a n − 1 − b n − 1 ) {\displaystyle c_{n}\leftarrow {\tfrac {1}{2}}(a_{n-1}-b_{n-1})}
>     * if c n < tolerance {\displaystyle c_{n}<{\textrm {tolerance}}} then

>       * N ← n ; {\displaystyle N\leftarrow n;} break

>   * ϕ N ← 2 N a N 2 x {\displaystyle \phi _{N}\leftarrow 2^{N}a_{N}{\sqrt {2}}x}
>   * for each n from N to 0 do

>     * ϕ n − 1 ← 1 2 ( ϕ n + arcsin ( c n a n sin ⁡ ϕ n ) ) {\displaystyle \phi _{n-1}\leftarrow {\tfrac {1}{2}}\left(\phi _{n}+{\arcsin }{\left({\frac {c_{n}}{a_{n}}}\sin \phi _{n}\right)}\right)}

>   * return sin ⁡ ϕ 0 2 − sin 2 ⁡ ϕ 0 {\displaystyle {\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.[45]

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

A hyperbolic series method:[46][47]

sl ⁡ ( ϖ π x ) = π ϖ ∑ n ∈ Z ( − 1 ) n cosh ⁡ ( x − ( n + 1 / 2 ) π ) , x ∈ C {\displaystyle \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} }

1 sl ⁡ ( ϖ x / π ) = π ϖ ∑ n ∈ Z ( − 1 ) n sinh ( x − n π ) = π ϖ ∑ n ∈ Z ( − 1 ) n sin ⁡ ( x − n π i ) , x ∈ C {\displaystyle {\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:[48]

sl ⁡ ( ϖ π x ) = 2 π ϖ ∑ n = 0 ∞ ( − 1 ) n sin ⁡ ( ( 2 n + 1 ) x ) cosh ⁡ ( ( n + 1 / 2 ) π ) , | Im ⁡ x | < π 2 {\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}}}
cl ⁡ ( ϖ π x ) = 2 π ϖ ∑ n = 0 ∞ cos ⁡ ( ( 2 n + 1 ) x ) cosh ⁡ ( ( n + 1 / 2 ) π ) , | Im ⁡ x | < π 2 {\displaystyle \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 {\pi }{2}}}
1 sl ⁡ ( ϖ x / π ) = π ϖ ( 1 sin ⁡ x − 4 ∑ n = 0 ∞ sin ⁡ ( ( 2 n + 1 ) x ) e ( 2 n + 1 ) π + 1 ) , | Im ⁡ x | < π {\displaystyle {\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|<\pi }

The lemniscate functions can be computed more rapidly by

sl ⁡ ( ϖ π x ) = θ 1 ( x , e − π ) θ 3 ( x , e − π ) , x ∈ C cl ⁡ ( ϖ π x ) = θ 2 ( x , e − π ) θ 4 ( x , e − π ) , x ∈ C {\displaystyle {\begin{aligned}\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{aligned}}}

where

θ 1 ( x , e − π ) = ∑ n ∈ Z ( − 1 ) n + 1 e − π ( n + 1 / 2 + x / π ) 2 = ∑ n ∈ Z ( − 1 ) n e − π ( n + 1 / 2 ) 2 sin ⁡ ( ( 2 n + 1 ) x ) , θ 2 ( x , e − π ) = ∑ n ∈ Z ( − 1 ) n e − π ( n + x / π ) 2 = ∑ n ∈ Z e − π ( n + 1 / 2 ) 2 cos ⁡ ( ( 2 n + 1 ) x ) , θ 3 ( x , e − π ) = ∑ n ∈ Z e − π ( n + x / π ) 2 = ∑ n ∈ Z e − π n 2 cos ⁡ 2 n x , θ 4 ( x , e − π ) = ∑ n ∈ Z e − π ( n + 1 / 2 + x / π ) 2 = ∑ n ∈ Z ( − 1 ) n e − π n 2 cos ⁡ 2 n x {\displaystyle {\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.[49]

Fourier series for the logarithm of the lemniscate sine:

ln ⁡ sl ⁡ ( ϖ π x ) = ln ⁡ 2 − π 4 + ln ⁡ sin ⁡ x + 2 ∑ n = 1 ∞ ( − 1 ) n cos ⁡ 2 n x n ( e n π + ( − 1 ) n ) , | Im ⁡ x | < π 2 {\displaystyle \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|<{\frac {\pi }{2}}}

The following series identities were discovered by Ramanujan:[50]

ϖ 2 π 2 sl 2 ⁡ ( ϖ x / π ) = 1 sin 2 ⁡ x − 1 π − 8 ∑ n = 1 ∞ n cos ⁡ 2 n x e 2 n π − 1 , | Im ⁡ x | < π {\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}},\quad \left|\operatorname {Im} x\right|<\pi }
arctan ⁡ sl ⁡ ( ϖ π x ) = 2 ∑ n = 0 ∞ sin ⁡ ( ( 2 n + 1 ) x ) ( 2 n + 1 ) cosh ⁡ ( ( n + 1 / 2 ) π ) , | Im ⁡ x | < π 2 {\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 )}},\quad \left|\operatorname {Im} x\right|<{\frac {\pi }{2}}}

The functions  sl ~ {\displaystyle {\tilde {\operatorname {sl} }}} and  cl ~ {\displaystyle {\tilde {\operatorname {cl} }}} analogous to  sin {\displaystyle \sin } and  cos {\displaystyle \cos } on the unit circle have the following Fourier and hyperbolic series expansions:[39][42][51]

sl ~ s = 2 2 π 2 ϖ 2 ∑ n = 1 ∞ n sin ⁡ ( 2 n π s / ϖ ) cosh ⁡ n π , | Im ⁡ s | < ϖ 2 {\displaystyle {\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|<{\frac {\varpi }{2}}}
cl ~ s = 2 π 2 ϖ 2 ∑ n = 0 ∞ ( 2 n + 1 ) cos ⁡ ( ( 2 n + 1 ) π s / ϖ ) sinh ⁡ ( ( n + 1 / 2 ) π ) , | Im ⁡ s | < ϖ 2 {\displaystyle {\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|<{\frac {\varpi }{2}}}
sl ~ s = π 2 ϖ 2 2 ∑ n ∈ Z sinh ⁡ ( π ( n + s / ϖ ) ) cosh 2 ⁡ ( π ( n + s / ϖ ) ) , s ∈ C {\displaystyle {\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} }
cl ~ s = π 2 ϖ 2 2 ∑ n ∈ Z ( − 1 ) n cosh 2 ⁡ ( π ( n + s / ϖ ) ) , s ∈ C {\displaystyle {\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:[52]

s l ( ϖ π x ) = 2 e − π / 4 sin ⁡ x ∏ n = 1 ∞ 1 − 2 e − 2 n π cos ⁡ 2 x + e − 4 n π 1 + 2 e − ( 2 n − 1 ) π cos ⁡ 2 x + e − ( 4 n − 2 ) π , x ∈ C {\displaystyle \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} }
c l ( ϖ π x ) = 2 e − π / 4 cos ⁡ x ∏ n = 1 ∞ 1 + 2 e − 2 n π cos ⁡ 2 x + e − 4 n π 1 − 2 e − ( 2 n − 1 ) π cos ⁡ 2 x + e − ( 4 n − 2 ) π , x ∈ C {\displaystyle \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  sn {\displaystyle \operatorname {sn} } function can be given.[53]

### The lemniscate functions as a ratio of entire functions

[edit]

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 sl has the following product expansion, reflecting the distribution of its zeros and poles:[54]

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

where

M ( z ) = z ∏ α ( 1 − z 4 α 4 ) , N ( z ) = ∏ β ( 1 − z 4 β 4 ) . {\displaystyle 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,  α {\displaystyle \alpha } and  β {\displaystyle \beta } denote, respectively, the zeros and poles of sl which are in the quadrant  Re ⁡ z > 0 , Im ⁡ z ≥ 0 {\displaystyle \operatorname {Re} z>0,\operatorname {Im} z\geq 0} . A proof can be found in.[54][55] 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.[56]

Proof of the infinite product for the lemniscate sine

Proof by logarithmic differentiation

It can be easily seen (using uniform and absolute convergence arguments to justify interchanging of limiting operations) that

M ′ ( z ) M ( z ) = − ∑ n = 0 ∞ 2 4 n H 4 n z 4 n − 1 ( 4 n ) ! , | z | < ϖ {\displaystyle {\frac {M'(z)}{M(z)}}=-\sum _{n=0}^{\infty }2^{4n}\mathrm {H} _{4n}{\frac {z^{4n-1}}{(4n)!}},\quad \left|z\right|<\varpi }

(where  H n {\displaystyle \mathrm {H} _{n}} are the Hurwitz numbers defined in Lemniscate elliptic functions § Hurwitz numbers) and

N ′ ( z ) N ( z ) = ( 1 + i ) M ′ ( ( 1 + i ) z ) M ( ( 1 + i ) z ) − M ′ ( z ) M ( z ) . {\displaystyle {\frac {N'(z)}{N(z)}}=(1+i){\frac {M'((1+i)z)}{M((1+i)z)}}-{\frac {M'(z)}{M(z)}}.}

Therefore

N ′ ( z ) N ( z ) = ∑ n = 0 ∞ 2 4 n ( 1 − ( − 1 ) n 2 2 n ) H 4 n z 4 n − 1 ( 4 n ) ! , | z | < ϖ 2 . {\displaystyle {\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|<{\frac {\varpi }{\sqrt {2}}}.}

It is known that

1 sl 2 ⁡ z = ∑ n = 0 ∞ 2 4 n ( 4 n − 1 ) H 4 n z 4 n − 2 ( 4 n ) ! , | z | < ϖ . {\displaystyle {\frac {1}{\operatorname {sl} ^{2}z}}=\sum _{n=0}^{\infty }2^{4n}(4n-1)\mathrm {H} _{4n}{\frac {z^{4n-2}}{(4n)!}},\quad \left|z\right|<\varpi .}

Then from

d d z sl ′ ⁡ z sl ⁡ z = − 1 sl 2 ⁡ z − sl 2 ⁡ z {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}{\frac {\operatorname {sl} 'z}{\operatorname {sl} z}}=-{\frac {1}{\operatorname {sl} ^{2}z}}-\operatorname {sl} ^{2}z}

and

sl 2 ⁡ z = 1 sl 2 ⁡ z − ( 1 + i ) 2 sl 2 ⁡ ( ( 1 + i ) z ) {\displaystyle \operatorname {sl} ^{2}z={\frac {1}{\operatorname {sl} ^{2}z}}-{\frac {(1+i)^{2}}{\operatorname {sl} ^{2}((1+i)z)}}}

we get

sl ′ ⁡ z sl ⁡ z = − ∑ n = 0 ∞ 2 4 n ( 2 − ( − 1 ) n 2 2 n ) H 4 n z 4 n − 1 ( 4 n ) ! , | z | < ϖ 2 . {\displaystyle {\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|<{\frac {\varpi }{\sqrt {2}}}.}

Hence

sl ′ ⁡ z sl ⁡ z = M ′ ( z ) M ( z ) − N ′ ( z ) N ( z ) , | z | < ϖ 2 . {\displaystyle {\frac {\operatorname {sl} 'z}{\operatorname {sl} z}}={\frac {M'(z)}{M(z)}}-{\frac {N'(z)}{N(z)}},\quad \left|z\right|<{\frac {\varpi }{\sqrt {2}}}.}

Therefore

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

for some constant  C {\displaystyle C} for  | z | < ϖ / 2 {\displaystyle \left|z\right|<\varpi /{\sqrt {2}}} but this result holds for all  z ∈ C {\displaystyle z\in \mathbb {C} } by analytic continuation. Using

lim z → 0 sl ⁡ z z = 1 {\displaystyle \lim _{z\to 0}{\frac {\operatorname {sl} z}{z}}=1}

gives  C = 1 {\displaystyle C=1} which completes the proof.  ◼ {\displaystyle \blacksquare }

Proof by Liouville's theorem

Let

f ( z ) = M ( z ) N ( z ) = ( 1 + i ) M ( z ) 2 M ( ( 1 + i ) z ) , {\displaystyle 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 ϖ ) = e 2 π ϖ ( z + ϖ ) M ( z ) , M ( z + 2 ϖ i ) = e − 2 π ϖ ( i z − ϖ ) M ( z ) {\displaystyle 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 {\displaystyle f} is an elliptic function with periods  2 ϖ {\displaystyle 2\varpi } and  2 ϖ i {\displaystyle 2\varpi i} , just as  sl {\displaystyle \operatorname {sl} } . It follows that the function  g {\displaystyle g} defined by

g ( z ) = sl ⁡ z f ( z ) , {\displaystyle 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  sl ⁡ z = z + O ⁡ ( z 5 ) {\displaystyle \operatorname {sl} z=z+\operatorname {O} (z^{5})} ,  M ( z ) = z + O ⁡ ( z 5 ) {\displaystyle M(z)=z+\operatorname {O} (z^{5})} and  N ( z ) = 1 + O ⁡ ( z 4 ) {\displaystyle N(z)=1+\operatorname {O} (z^{4})} , this constant is  1 {\displaystyle 1} , which proves the theorem.  ◼ {\displaystyle \blacksquare }

Gauss conjectured that  ln ⁡ N ( ϖ ) = π / 2 {\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”.[57] Gauss expanded the products for  M {\displaystyle M} and  N {\displaystyle N} as infinite series (see below). He also discovered several identities involving the functions  M {\displaystyle M} and  N {\displaystyle N} , such as

The  M {\displaystyle M} function in the complex plane. The complex argument is represented by varying hue. The  N {\displaystyle N} function in the complex plane. The complex argument is represented by varying hue.

N ( z ) = M ( ( 1 + i ) z ) ( 1 + i ) M ( z ) , z ∉ ϖ Z [ i ] {\displaystyle N(z)={\frac {M((1+i)z)}{(1+i)M(z)}},\quad z\notin \varpi \mathbb {Z} [i]}

and

N ( 2 z ) = M ( z ) 4 + N ( z ) 4 . {\displaystyle N(2z)=M(z)^{4}+N(z)^{4}.}

Thanks to a certain theorem[58] on splitting limits, we are allowed to multiply out the infinite products and collect like powers of  z {\displaystyle z} . Doing so gives the following power series expansions that are convergent everywhere in the complex plane:[59][60][61][62][63]

M ( z ) = z − 2 z 5 5 ! − 36 z 9 9 ! + 552 z 13 13 ! + ⋯ , z ∈ C {\displaystyle 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 z 4 4 ! − 4 z 8 8 ! + 408 z 12 12 ! + ⋯ , z ∈ C . {\displaystyle 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  sl {\displaystyle \operatorname {sl} } which has only finite radius of convergence (because it is not entire).

We define  S {\displaystyle S} and  T {\displaystyle T} by

S ( z ) = N ( z 1 + i ) 2 − i M ( z 1 + i ) 2 , T ( z ) = S ( i z ) . {\displaystyle 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

cl ⁡ z = S ( z ) T ( z ) {\displaystyle \operatorname {cl} z={\frac {S(z)}{T(z)}}}

where[64]

S ( z ) = 1 − z 2 2 ! − z 4 4 ! − 3 z 6 6 ! + 17 z 8 8 ! − 9 z 10 10 ! + 111 z 12 12 ! + ⋯ , z ∈ C {\displaystyle 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 + z 2 2 ! − z 4 4 ! + 3 z 6 6 ! + 17 z 8 8 ! + 9 z 10 10 ! + 111 z 12 12 ! + ⋯ , z ∈ C . {\displaystyle 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 ( 2 z ) = 2 M ( z ) N ( z ) S ( z ) T ( z ) , {\displaystyle M(2z)=2M(z)N(z)S(z)T(z),}
S ( 2 z ) = S ( z ) 4 − 2 M ( z ) 4 , {\displaystyle S(2z)=S(z)^{4}-2M(z)^{4},}
T ( 2 z ) = T ( z ) 4 − 2 M ( z ) 4 {\displaystyle T(2z)=T(z)^{4}-2M(z)^{4}}

and the Pythagorean-like identities

M ( z ) 2 + S ( z ) 2 = N ( z ) 2 , {\displaystyle M(z)^{2}+S(z)^{2}=N(z)^{2},}
M ( z ) 2 + N ( z ) 2 = T ( z ) 2 {\displaystyle M(z)^{2}+N(z)^{2}=T(z)^{2}}

hold for all  z ∈ C {\displaystyle z\in \mathbb {C} } .

The quasi-addition formulas

M ( z + w ) M ( z − w ) = M ( z ) 2 N ( w ) 2 − N ( z ) 2 M ( w ) 2 , {\displaystyle M(z+w)M(z-w)=M(z)^{2}N(w)^{2}-N(z)^{2}M(w)^{2},}
N ( z + w ) N ( z − w ) = N ( z ) 2 N ( w ) 2 + M ( z ) 2 M ( w ) 2 {\displaystyle N(z+w)N(z-w)=N(z)^{2}N(w)^{2}+M(z)^{2}M(w)^{2}}

(where  z , w ∈ C {\displaystyle z,w\in \mathbb {C} } ) imply further multiplication formulas for  M {\displaystyle M} and  N {\displaystyle N} by recursion.[65]

Gauss'  M {\displaystyle M} and  N {\displaystyle N} satisfy the following system of differential equations:

M ( z ) M ″ ( z ) = M ′ ( z ) 2 − N ( z ) 2 , {\displaystyle M(z)M''(z)=M'(z)^{2}-N(z)^{2},}
N ( z ) N ″ ( z ) = N ′ ( z ) 2 + M ( z ) 2 {\displaystyle N(z)N''(z)=N'(z)^{2}+M(z)^{2}}

where  z ∈ C {\displaystyle z\in \mathbb {C} } . Both  M {\displaystyle M} and  N {\displaystyle N} satisfy the differential equation[66]

X ( z ) X ⁗ ( z ) = 4 X ′ ( z ) X ‴ ( z ) − 3 X ″ ( z ) 2 + 2 X ( z ) 2 , z ∈ C . {\displaystyle 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 ⁡ ( − ∫ 0 z ∫ 0 w ( 1 sl 2 ⁡ v − 1 v 2 ) d v d w ) , {\displaystyle M(z)=z\exp \left(-\int _{0}^{z}\int _{0}^{w}\left({\frac {1}{\operatorname {sl} ^{2}v}}-{\frac {1}{v^{2}}}\right)\,\mathrm {d} v\,\mathrm {d} w\right),}
N ( z ) = exp ⁡ ( ∫ 0 z ∫ 0 w sl 2 ⁡ v d v d w ) {\displaystyle N(z)=\exp \left(\int _{0}^{z}\int _{0}^{w}\operatorname {sl} ^{2}v\,\mathrm {d} v\,\mathrm {d} w\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 {\displaystyle M,N} and  θ 1 , θ 3 {\displaystyle \theta _{1},\theta _{3}} is

M ( z ) = 2 − 1 / 4 e π z 2 / ( 2 ϖ 2 ) π ϖ θ 1 ( π z ϖ , e − π ) , {\displaystyle 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 π z 2 / ( 2 ϖ 2 ) π ϖ θ 3 ( π z ϖ , e − π ) {\displaystyle 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 ∈ C {\displaystyle z\in \mathbb {C} } .

## Relation to other functions

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### Relation to Weierstrass and Jacobi elliptic functions

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The lemniscate functions are closely related to the Weierstrass elliptic function ℘ ( z ; 1 , 0 ) {\displaystyle \wp (z;1,0)} (the "lemniscatic case"), with invariants g2 = 1 and g3 = 0. This lattice has fundamental periods  ω 1 = 2 ϖ , {\displaystyle \omega _{1}={\sqrt {2}}\varpi ,} and  ω 2 = i ω 1 {\displaystyle \omega _{2}=i\omega _{1}} . The associated constants of the Weierstrass function are  e 1 = 1 2 ,   e 2 = 0 ,   e 3 = − 1 2 . {\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  ℘ ( z ; − 1 , 0 ) {\displaystyle \wp (z;-1,0)} is called the "pseudolemniscatic case".[67]

The square of the lemniscate sine can be represented as

sl 2 ⁡ z = 1 ℘ ( z ; 4 , 0 ) = i 2 ℘ ( ( 1 − i ) z ; − 1 , 0 ) = − 2 ℘ ( 2 z + ( i − 1 ) ϖ 2 ; 1 , 0 ) {\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. The lemniscate sine is a rational function in the Weierstrass elliptic function and its derivative:[68]

sl ⁡ z = − 2 ℘ ( z ; − 1 , 0 ) ℘ ′ ( z ; − 1 , 0 ) . {\displaystyle \operatorname {sl} z=-2{\frac {\wp (z;-1,0)}{\wp '(z;-1,0)}}.}

The lemniscate functions can also be written in terms of Jacobi elliptic functions. The Jacobi elliptic functions  sn {\displaystyle \operatorname {sn} } and  cd {\displaystyle \operatorname {cd} } with positive real elliptic modulus have an "upright" rectangular lattice aligned with real and imaginary axes. Alternately, the functions  sn {\displaystyle \operatorname {sn} } and  cd {\displaystyle \operatorname {cd} } with modulus i (and  sd {\displaystyle \operatorname {sd} } and  cn {\displaystyle \operatorname {cn} } with modulus  1 / 2 {\displaystyle 1/{\sqrt {2}}} ) have a square period lattice rotated 1/8 turn.[69][70]

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

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

where the second arguments denote the elliptic modulus  k {\displaystyle k} .

The functions  sl ~ {\displaystyle {\tilde {\operatorname {sl} }}} and  cl ~ {\displaystyle {\tilde {\operatorname {cl} }}} can also be expressed in terms of Jacobi elliptic functions:

sl ~ z = cd ⁡ ( z ; i ) sd ⁡ ( z ; i ) = dn ⁡ ( z ; 2 ) sn ⁡ ( z ; 2 ) = 1 2 cn ⁡ ( 2 z ; 1 2 ) sn ⁡ ( 2 z ; 1 2 ) , {\displaystyle {\tilde {\operatorname {sl} }}\,z=\operatorname {cd} (z;i)\operatorname {sd} (z;i)=\operatorname {dn} (z;{\sqrt {2}})\operatorname {sn} (z;{\sqrt {2}})={\tfrac {1}{\sqrt {2}}}\operatorname {cn} \left({\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}\right)\operatorname {sn} \left({\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}\right),}
cl ~ z = cd ⁡ ( z ; i ) nd ⁡ ( z ; i ) = dn ⁡ ( z ; 2 ) cn ⁡ ( z ; 2 ) = cn ⁡ ( 2 z ; 1 2 ) dn ⁡ ( 2 z ; 1 2 ) . {\displaystyle {\tilde {\operatorname {cl} }}\,z=\operatorname {cd} (z;i)\operatorname {nd} (z;i)=\operatorname {dn} (z;{\sqrt {2}})\operatorname {cn} (z;{\sqrt {2}})=\operatorname {cn} \left({\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}\right)\operatorname {dn} \left({\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}\right).}

### Relation to the modular lambda function

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

∏ k = 1 n sl ( 2 k − 1 2 n + 1 ϖ 2 ) = λ ( ( 2 n + 1 ) i ) 1 − λ ( ( 2 n + 1 ) i ) 8 {\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:

sl ( 1 14 ϖ ) sl ( 3 14 ϖ ) sl ( 5 14 ϖ ) = λ ( 7 i ) 1 − λ ( 7 i ) 8 = tan ( 1 2 arccsc ( 1 2 8 7 + 21 + 1 2 7 + 1 ) ) = 2 2 + 7 + 21 + 8 7 + 2 14 + 6 7 + 455 + 172 7 sl ( 1 18 ϖ ) sl ( 3 18 ϖ ) sl ( 5 18 ϖ ) sl ( 7 18 ϖ ) = λ ( 9 i ) 1 − λ ( 9 i ) 8 = tan ( π 4 − arctan ( 2 2 3 − 2 3 − 2 2 − 3 3 + 3 − 1 12 4 ) ) {\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 )}\\\\[7mu]&\quad {}={\frac {2}{2+{\sqrt {7}}+{\sqrt {21+8{\sqrt {7}}}}+{\sqrt {2{14+6{\sqrt {7}}+{\sqrt {455+172{\sqrt {7}}}}}}}}}\\\\[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}}}

## Inverse functions

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The inverse function of the lemniscate sine is the lemniscate arcsine, defined as[71]

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

It can also be represented by the hypergeometric function:

arcsl ⁡ x = x 2 F 1 ( 1 2 , 1 4 ; 5 4 ; x 4 ) {\displaystyle \operatorname {arcsl} x=x\,{}_{2}F_{1}{\bigl (}{\tfrac {1}{2}},{\tfrac {1}{4}};{\tfrac {5}{4}};x^{4}{\bigr )}}

which can be easily seen by using the binomial series.

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

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

For x in the interval  − 1 ≤ x ≤ 1 {\displaystyle -1\leq x\leq 1} ,  sl ⁡ arcsl ⁡ x = x {\displaystyle \operatorname {sl} \operatorname {arcsl} x=x} and  cl ⁡ arccl ⁡ x = x {\displaystyle \operatorname {cl} \operatorname {arccl} x=x}

For the halving of the lemniscate arc length these formulas are valid:[citation needed]

sl ( 1 2 arcsl ⁡ x ) = sin ( 1 2 arcsin ⁡ x ) sech ( 1 2 arsinh ⁡ x ) sl ( 1 2 arcsl ⁡ x ) 2 = tan ( 1 4 arcsin ⁡ x 2 ) {\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}}}

Furthermore there are the so called Hyperbolic lemniscate area functions:[citation needed]

aslh ⁡ ( x ) = ∫ 0 x 1 y 4 + 1 d y = 1 2 F ( 2 arctan ⁡ x ; 1 2 ) {\displaystyle \operatorname {aslh} (x)=\int _{0}^{x}{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y={\tfrac {1}{2}}F\left(2\arctan x;{\tfrac {1}{\sqrt {2}}}\right)}

aclh ⁡ ( x ) = ∫ x ∞ 1 y 4 + 1 d y = 1 2 F ( 2 arccot ⁡ x ; 1 2 ) {\displaystyle \operatorname {aclh} (x)=\int _{x}^{\infty }{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y={\tfrac {1}{2}}F\left(2\operatorname {arccot} x;{\tfrac {1}{\sqrt {2}}}\right)}

aclh ⁡ ( x ) = ϖ 2 − aslh ⁡ ( x ) {\displaystyle \operatorname {aclh} (x)={\frac {\varpi }{\sqrt {2}}}-\operatorname {aslh} (x)}

aslh ⁡ ( x ) = 2 arcsl ⁡ ( x / 1 + x 4 + 1 ) {\displaystyle \operatorname {aslh} (x)={\sqrt {2}}\operatorname {arcsl} \left(x{\Big /}{\sqrt {\textstyle 1+{\sqrt {x^{4}+1}}}}\right)}

arcsl ⁡ ( x ) = 2 aslh ⁡ ( x / 1 + 1 − x 4 ) {\displaystyle \operatorname {arcsl} (x)={\sqrt {2}}\operatorname {aslh} \left(x{\Big /}{\sqrt {\textstyle 1+{\sqrt {1-x^{4}}}}}\right)}

### Expression using elliptic integrals

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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:[citation needed]

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

arcsl ⁡ x = 2 ( 2 − 1 ) F ( arcsin ( 2 + 1 ) x 1 + x 2 + 1 ; ( 2 − 1 ) 2 ) {\displaystyle \operatorname {arcsl} x=2({\sqrt {2}}-1)F\left({\arcsin }{\frac {({\sqrt {2}}+1)x}{{\sqrt {1+x^{2}}}+1}};({\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):[citation needed]

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

arccl ⁡ x = 1 2 F ( arccos ⁡ x ; 1 2 ) {\displaystyle \operatorname {arccl} x={\frac {1}{\sqrt {2}}}F\left(\arccos x;{\frac {1}{\sqrt {2}}}\right)}

### Use in integration

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

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

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

∫ 1 x 4 + 6 x 2 + 1 d x = arcsl 2 x x 4 + 6 x 2 + 1 + x 2 + 1 {\displaystyle \int {\frac {1}{\sqrt {x^{4}+6x^{2}+1}}}\,\mathrm {d} x={\operatorname {arcsl} }{\frac {{\sqrt {2}}x}{\sqrt {{\sqrt {x^{4}+6x^{2}+1}}+x^{2}+1}}}}

∫ 1 x 4 + 1 d x = 2 arcsl x x 4 + 1 + 1 {\displaystyle \int {\frac {1}{\sqrt {x^{4}+1}}}\,\mathrm {d} x={{\sqrt {2}}\operatorname {arcsl} }{\frac {x}{\sqrt {{\sqrt {x^{4}+1}}+1}}}}

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

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

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

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

∫ 1 ( a x 2 + b x + c ) 3 4 d x = 2 2 4 a 2 c − a b 2 4 arcsl 2 a x + b 4 a ( a x 2 + b x + c ) + 4 a c − b 2 {\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} }{\frac {2ax+b}{{\sqrt {4a(ax^{2}+bx+c)}}+{\sqrt {4ac-b^{2}}}}}}

∫ sech ⁡ x d x = 2 arcsl tanh ⁡ 1 2 x {\displaystyle \int {\sqrt {\operatorname {sech} x}}\,\mathrm {d} x={2\operatorname {arcsl} }\tanh {\tfrac {1}{2}}x}

∫ sec ⁡ x d x = 2 arcsl tan ⁡ 1 2 x {\displaystyle \int {\sqrt {\sec x}}\,\mathrm {d} x={2\operatorname {arcsl} }\tan {\tfrac {1}{2}}x}

## Hyperbolic lemniscate functions

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### Fundamental information

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The hyperbolic lemniscate sine (red) and hyperbolic lemniscate cosine (purple) applied to a real argument, in comparison with the trigonometric tangent (pale dashed red). The hyperbolic lemniscate sine in the complex plane. Dark areas represent zeros and bright areas represent poles. The complex argument is represented by varying hue.

For convenience, let  σ = 2 ϖ {\displaystyle \sigma ={\sqrt {2}}\varpi } .  σ {\displaystyle \sigma } is the "squircular" analog of  π {\displaystyle \pi } (see below). The decimal expansion of  σ {\displaystyle \sigma } (i.e.  3.7081 … {\displaystyle 3.7081\ldots } [72]) appears in entry 34e of chapter 11 of Ramanujan's second notebook.[73]

The hyperbolic lemniscate sine (slh) and cosine (clh) can be defined as inverses of elliptic integrals as follows:

z = ∗ ∫ 0 slh ⁡ z d t 1 + t 4 = ∫ clh ⁡ z ∞ d t 1 + t 4 {\displaystyle z\mathrel {\overset {*}{=}} \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}}}}}

where in  ( ∗ ) {\displaystyle (*)} ,  z {\displaystyle z} is in the square with corners  { σ / 2 , σ i / 2 , − σ / 2 , − σ i / 2 } {\displaystyle \\{\sigma /2,\sigma i/2,-\sigma /2,-\sigma i/2\\}} . Beyond that square, the functions can be analytically continued to meromorphic functions in the whole complex plane.

The complete integral has the value:

∫ 0 ∞ d t t 4 + 1 = 1 4 B ( 1 4 , 1 4 ) = σ 2 = 1.85407 46773 01371 … {\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 )}={\frac {\sigma }{2}}=1.85407\;46773\;01371\ldots }

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

slh ⁡ z = clh ( σ 2 − z ) {\displaystyle \operatorname {slh} z={\operatorname {clh} }{{\Bigl (}{\frac {\sigma }{2}}-z{\Bigr )}}}

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

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

The functions  slh {\displaystyle \operatorname {slh} } and  clh {\displaystyle \operatorname {clh} } have a square period lattice with fundamental periods  { σ , σ i } {\displaystyle \\{\sigma ,\sigma i\\}} .

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

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

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

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

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

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

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

slh ⁡ z = 1 − i 2 sl ⁡ ( 1 + i 2 z ) = sl ⁡ ( − 1 4 z ) − 1 4 {\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:

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

### Relation to quartic Fermat curve

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#### Hyperbolic Lemniscate Tangent and Cotangent

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This image shows the standardized superelliptic Fermat squircle curve of the fourth degree:

Superellipse with the relation  x 4 + y 4 = 1 {\displaystyle x^{4}+y^{4}=1}

In a quartic Fermat curve x 4 + y 4 = 1 {\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  x 2 + y 2 = 1 {\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  x = 1 {\displaystyle x=1} .[74] Just as  π {\displaystyle \pi } is the area enclosed by the circle  x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1} , the area enclosed by the squircle  x 4 + y 4 = 1 {\displaystyle x^{4}+y^{4}=1} is  σ {\displaystyle \sigma } . Moreover,

M ( 1 , 1 / 2 ) = π σ {\displaystyle M(1,1/{\sqrt {2}})={\frac {\pi }{\sigma }}}

where  M {\displaystyle M} is the arithmetic–geometric mean.

The hyperbolic lemniscate sine satisfies the argument addition identity:

slh ⁡ ( a + b ) = slh ⁡ a slh ′ ⁡ b + slh ⁡ b slh ′ ⁡ a 1 − slh 2 ⁡ a slh 2 ⁡ b {\displaystyle \operatorname {slh} (a+b)={\frac {\operatorname {slh} a\operatorname {slh} 'b+\operatorname {slh} b\operatorname {slh} 'a}{1-\operatorname {slh} ^{2}a\,\operatorname {slh} ^{2}b}}}

When  u {\displaystyle u} is real, the derivative and the original antiderivative of  slh {\displaystyle \operatorname {slh} } and  clh {\displaystyle \operatorname {clh} } can be expressed in this way:

d d u slh ⁡ ( u ) = 1 + slh ⁡ ( u ) 4 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}\operatorname {slh} (u)={\sqrt {1+\operatorname {slh} (u)^{4}}}}

d d u clh ⁡ ( u ) = − 1 + clh ⁡ ( u ) 4 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}\operatorname {clh} (u)=-{\sqrt {1+\operatorname {clh} (u)^{4}}}}

d d u 1 2 arsinh ⁡ [ slh ⁡ ( u ) 2 ] = slh ⁡ ( u ) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}\,{\frac {1}{2}}\operatorname {arsinh} {\bigl [}\operatorname {slh} (u)^{2}{\bigr ]}=\operatorname {slh} (u)}

d d u − 1 2 arsinh ⁡ [ clh ⁡ ( u ) 2 ] = clh ⁡ ( u ) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}-\,{\frac {1}{2}}\operatorname {arsinh} {\bigl [}\operatorname {clh} (u)^{2}{\bigr ]}=\operatorname {clh} (u)}

There are also the Hyperbolic lemniscate tangent and the Hyperbolic lemniscate coangent als further functions:

The functions tlh and ctlh fulfill the identities described in the differential equation mentioned:

tlh ( 2 u ) = sin 4 ⁡ ( 2 u ) = sl ⁡ ( u ) cl 2 ⁡ u + 1 sl 2 ⁡ u + cl 2 ⁡ u {\displaystyle {\text{tlh}}({\sqrt {2}}\,u)=\sin _{4}({\sqrt {2}}\,u)=\operatorname {sl} (u){\sqrt {\frac {\operatorname {cl} ^{2}u+1}{\operatorname {sl} ^{2}u+\operatorname {cl} ^{2}u}}}}
ctlh ( 2 u ) = cos 4 ⁡ ( 2 u ) = cl ⁡ ( u ) sl 2 ⁡ u + 1 sl 2 ⁡ u + cl 2 ⁡ u {\displaystyle {\text{ctlh}}({\sqrt {2}}\,u)=\cos _{4}({\sqrt {2}}\,u)=\operatorname {cl} (u){\sqrt {\frac {\operatorname {sl} ^{2}u+1}{\operatorname {sl} ^{2}u+\operatorname {cl} ^{2}u}}}}

The functional designation sl stands for the lemniscatic sine and the designation cl stands for the lemniscatic cosine. In addition, those relations to the Jacobi elliptic functions are valid:

tlh ( u ) = sn ( u ; 1 2 2 ) cd ( u ; 1 2 2 ) 4 + sn ( u ; 1 2 2 ) 4 4 {\displaystyle {\text{tlh}}(u)={\frac {{\text{sn}}(u;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(u;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(u;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}}
ctlh ( u ) = cd ( u ; 1 2 2 ) cd ( u ; 1 2 2 ) 4 + sn ( u ; 1 2 2 ) 4 4 {\displaystyle {\text{ctlh}}(u)={\frac {{\text{cd}}(u;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(u;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(u;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}}

When  u {\displaystyle u} is real, the derivative and quarter period integral of  tlh {\displaystyle \operatorname {tlh} } and  ctlh {\displaystyle \operatorname {ctlh} } can be expressed in this way:

d d u tlh ⁡ ( u ) = ctlh ⁡ ( u ) 3 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}\operatorname {tlh} (u)=\operatorname {ctlh} (u)^{3}}

d d u ctlh ⁡ ( u ) = − tlh ⁡ ( u ) 3 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} u}}\operatorname {ctlh} (u)=-\operatorname {tlh} (u)^{3}}

∫ 0 ϖ / 2 tlh ⁡ ( u ) d u = ϖ 2 {\displaystyle \int _{0}^{\varpi /{\sqrt {2}}}\operatorname {tlh} (u)\,\mathrm {d} u={\frac {\varpi }{2}}}

∫ 0 ϖ / 2 ctlh ⁡ ( u ) d u = ϖ 2 {\displaystyle \int _{0}^{\varpi /{\sqrt {2}}}\operatorname {ctlh} (u)\,\mathrm {d} u={\frac {\varpi }{2}}}

#### Derivation of the Hyperbolic Lemniscate functions

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With respect to the quartic Fermat curve  x 4 + y 4 = 1 {\displaystyle x^{4}+y^{4}=1} , the hyperbolic lemniscate sine is analogous to the trigonometric tangent function. Unlike  slh {\displaystyle \operatorname {slh} } and  clh {\displaystyle \operatorname {clh} } , the functions  sin 4 {\displaystyle \sin _{4}} and  cos 4 {\displaystyle \cos _{4}} cannot be analytically extended to meromorphic functions in the whole complex plane.[75]

The horizontal and vertical coordinates of this superellipse are dependent on twice the enclosed area w = 2A, so the following conditions must be met:

x ( w ) 4 + y ( w ) 4 = 1 {\displaystyle x(w)^{4}+y(w)^{4}=1}
d d w x ( w ) = − y ( w ) 3 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} w}}x(w)=-y(w)^{3}}
d d w y ( w ) = x ( w ) 3 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} w}}y(w)=x(w)^{3}}
x ( w = 0 ) = 1 {\displaystyle x(w=0)=1}
y ( w = 0 ) = 0 {\displaystyle y(w=0)=0}

The solutions to this system of equations are as follows:

x ( w ) = cl ⁡ ( 1 2 2 w ) [ sl ⁡ ( 1 2 2 w ) 2 + 1 ] 1 / 2 [ sl ⁡ ( 1 2 2 w ) 2 + cl ⁡ ( 1 2 2 w ) 2 ] − 1 / 2 {\displaystyle x(w)=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)[\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+1]^{1/2}[\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}]^{-1/2}}
y ( w ) = sl ⁡ ( 1 2 2 w ) [ cl ⁡ ( 1 2 2 w ) 2 + 1 ] 1 / 2 [ sl ⁡ ( 1 2 2 w ) 2 + cl ⁡ ( 1 2 2 w ) 2 ] − 1 / 2 {\displaystyle y(w)=\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)[\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+1]^{1/2}[\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}]^{-1/2}}

The following therefore applies to the quotient:

y ( w ) x ( w ) = sl ⁡ ( 1 2 2 w ) [ cl ⁡ ( 1 2 2 w ) 2 + 1 ] 1 / 2 cl ⁡ ( 1 2 2 w ) [ sl ⁡ ( 1 2 2 w ) 2 + 1 ] 1 / 2 = slh ⁡ ( w ) {\displaystyle {\frac {y(w)}{x(w)}}={\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)[\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+1]^{1/2}}{\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}w)[\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}w)^{2}+1]^{1/2}}}=\operatorname {slh} (w)}

The functions x(w) and y(w) are called cotangent hyperbolic lemniscatus and hyperbolic tangent.

x ( w ) = ctlh ( w ) {\displaystyle x(w)={\text{ctlh}}(w)}
y ( w ) = tlh ( w ) {\displaystyle y(w)={\text{tlh}}(w)}

The sketch also shows the fact that the derivation of the Areasinus hyperbolic lemniscatus function is equal to the reciprocal of the square root of the successor of the fourth power function.

#### First proof: comparison with the derivative of the arctangent

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There is a black diagonal on the sketch shown on the right. The length of the segment that runs perpendicularly from the intersection of this black diagonal with the red vertical axis to the point (1|0) should be called s. And the length of the section of the black diagonal from the coordinate origin point to the point of intersection of this diagonal with the cyan curved line of the superellipse has the following value depending on the slh value:

D ( s ) = ( 1 s 4 + 1 4 ) 2 + ( s s 4 + 1 4 ) 2 = s 2 + 1 s 4 + 1 4 {\displaystyle D(s)={\sqrt {{\biggl (}{\frac {1}{\sqrt[{4}]{s^{4}+1}}}{\biggr )}^{2}+{\biggl (}{\frac {s}{\sqrt[{4}]{s^{4}+1}}}{\biggr )}^{2}}}={\frac {\sqrt {s^{2}+1}}{\sqrt[{4}]{s^{4}+1}}}}

This connection is described by the Pythagorean theorem.

An analogous unit circle results in the arctangent of the circle trigonometric with the described area allocation.

The following derivation applies to this:

d d s arctan ⁡ ( s ) = 1 s 2 + 1 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} s}}\arctan(s)={\frac {1}{s^{2}+1}}}

To determine the derivation of the areasinus lemniscatus hyperbolicus, the comparison of the infinitesimally small triangular areas for the same diagonal in the superellipse and the unit circle is set up below. Because the summation of the infinitesimally small triangular areas describes the area dimensions. In the case of the superellipse in the picture, half of the area concerned is shown in green. Because of the quadratic ratio of the areas to the lengths of triangles with the same infinitesimally small angle at the origin of the coordinates, the following formula applies:

d d s aslh ( s ) = [ d d s arctan ⁡ ( s ) ] D ( s ) 2 = 1 s 2 + 1 D ( s ) 2 = 1 s 2 + 1 ( s 2 + 1 s 4 + 1 4 ) 2 = 1 s 4 + 1 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} s}}{\text{aslh}}(s)={\biggl [}{\frac {\mathrm {d} }{\mathrm {d} s}}\arctan(s){\biggr ]}D(s)^{2}={\frac {1}{s^{2}+1}}D(s)^{2}={\frac {1}{s^{2}+1}}{\biggl (}{\frac {\sqrt {s^{2}+1}}{\sqrt[{4}]{s^{4}+1}}}{\biggr )}^{2}={\frac {1}{\sqrt {s^{4}+1}}}}

#### Second proof: integral formation and area subtraction

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In the picture shown, the area tangent lemniscatus hyperbolicus assigns the height of the intersection of the diagonal and the curved line to twice the green area. The green area itself is created as the difference integral of the superellipse function from zero to the relevant height value minus the area of the adjacent triangle:

atlh ( v ) = 2 ( ∫ 0 v 1 − w 4 4 d w ) − v 1 − v 4 4 {\displaystyle {\text{atlh}}(v)=2{\biggl (}\int _{0}^{v}{\sqrt[{4}]{1-w^{4}}}\mathrm {d} w{\biggr )}-v{\sqrt[{4}]{1-v^{4}}}}
d d v atlh ( v ) = 2 1 − v 4 4 − ( d d v v 1 − v 4 4 ) = 1 ( 1 − v 4 ) 3 / 4 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} v}}{\text{atlh}}(v)=2{\sqrt[{4}]{1-v^{4}}}-{\biggl (}{\frac {\mathrm {d} }{\mathrm {d} v}}v{\sqrt[{4}]{1-v^{4}}}{\biggr )}={\frac {1}{(1-v^{4})^{3/4}}}}

The following transformation applies:

aslh ( x ) = atlh ( x x 4 + 1 4 ) {\displaystyle {\text{aslh}}(x)={\text{atlh}}{\biggl (}{\frac {x}{\sqrt[{4}]{x^{4}+1}}}{\biggr )}}

And so, according to the chain rule, this derivation holds:

d d x aslh ( x ) = d d x atlh ( x x 4 + 1 4 ) = ( d d x x x 4 + 1 4 ) [ 1 − ( x x 4 + 1 4 ) 4 ] − 3 / 4 = {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\text{aslh}}(x)={\frac {\mathrm {d} }{\mathrm {d} x}}{\text{atlh}}{\biggl (}{\frac {x}{\sqrt[{4}]{x^{4}+1}}}{\biggr )}={\biggl (}{\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {x}{\sqrt[{4}]{x^{4}+1}}}{\biggr )}{\biggl [}1-{\biggl (}{\frac {x}{\sqrt[{4}]{x^{4}+1}}}{\biggr )}^{4}{\biggr ]}^{-3/4}=}
= 1 ( x 4 + 1 ) 5 / 4 [ 1 − ( x x 4 + 1 4 ) 4 ] − 3 / 4 = 1 ( x 4 + 1 ) 5 / 4 ( 1 x 4 + 1 ) − 3 / 4 = 1 x 4 + 1 {\displaystyle ={\frac {1}{(x^{4}+1)^{5/4}}}{\biggl [}1-{\biggl (}{\frac {x}{\sqrt[{4}]{x^{4}+1}}}{\biggr )}^{4}{\biggr ]}^{-3/4}={\frac {1}{(x^{4}+1)^{5/4}}}{\biggl (}{\frac {1}{x^{4}+1}}{\biggr )}^{-3/4}={\frac {1}{\sqrt {x^{4}+1}}}}

### Specific values

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This list shows the values of the Hyperbolic Lemniscate Sine accurately. Recall that,

∫ 0 ∞ d ⁡ t t 4 + 1 = 1 4 B ( 1 4 , 1 4 ) = ϖ 2 = σ 2 = 1.85407 … {\displaystyle \int _{0}^{\infty }{\frac {\operatorname {d} t}{\sqrt {t^{4}+1}}}={\tfrac {1}{4}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{4}}{\bigr )}={\frac {\varpi }{\sqrt {2}}}={\frac {\sigma }{2}}=1.85407\ldots }

whereas  1 2 B ( 1 2 , 1 2 ) = π 2 , {\displaystyle {\tfrac {1}{2}}\mathrm {B} {\bigl (}{\tfrac {1}{2}},{\tfrac {1}{2}}{\bigr )}={\tfrac {\pi }{2}},} so the values below such as  slh ( ϖ 2 2 ) = slh ( σ 4 ) = 1 {\displaystyle {\operatorname {slh} }{\bigl (}{\tfrac {\varpi }{2{\sqrt {2}}}}{\bigr )}={\operatorname {slh} }{\bigl (}{\tfrac {\sigma }{4}}{\bigr )}=1} are analogous to the trigonometric  sin ( π 2 ) = 1 {\displaystyle {\sin }{\bigl (}{\tfrac {\pi }{2}}{\bigr )}=1} .

slh ( ϖ 2 2 ) = 1 {\displaystyle \operatorname {slh} \,\left({\frac {\varpi }{2{\sqrt {2}}}}\right)=1}
slh ( ϖ 3 2 ) = 1 3 4 2 3 − 3 4 {\displaystyle \operatorname {slh} \,\left({\frac {\varpi }{3{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{3}}}{\sqrt[{4}]{2{\sqrt {3}}-3}}}
slh ( 2 ϖ 3 2 ) = 2 3 + 3 4 {\displaystyle \operatorname {slh} \,\left({\frac {2\varpi }{3{\sqrt {2}}}}\right)={\sqrt[{4}]{2{\sqrt {3}}+3}}}
slh ( ϖ 4 2 ) = 1 2 4 ( 2 + 1 − 1 ) {\displaystyle \operatorname {slh} \,\left({\frac {\varpi }{4{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{2}}}({\sqrt {{\sqrt {2}}+1}}-1)}
slh ( 3 ϖ 4 2 ) = 1 2 4 ( 2 + 1 + 1 ) {\displaystyle \operatorname {slh} \,\left({\frac {3\varpi }{4{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{2}}}({\sqrt {{\sqrt {2}}+1}}+1)}
slh ( ϖ 5 2 ) = 1 8 4 5 − 1 20 4 − 5 + 1 = 2 5 − 2 4 sin ⁡ ( 1 20 π ) sin ⁡ ( 3 20 π ) {\displaystyle \operatorname {slh} \,\left({\frac {\varpi }{5{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{8}}}{\sqrt {{\sqrt {5}}-1}}{\sqrt {{\sqrt[{4}]{20}}-{\sqrt {{\sqrt {5}}+1}}}}=2{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {\sin({\tfrac {1}{20}}\pi )\sin({\tfrac {3}{20}}\pi )}}}
slh ( 2 ϖ 5 2 ) = 1 2 2 4 ( 5 + 1 ) 20 4 − 5 + 1 = 2 5 + 2 4 sin ⁡ ( 1 20 π ) sin ⁡ ( 3 20 π ) {\displaystyle \operatorname {slh} \,\left({\frac {2\varpi }{5{\sqrt {2}}}}\right)={\frac {1}{2{\sqrt[{4}]{2}}}}({\sqrt {5}}+1){\sqrt {{\sqrt[{4}]{20}}-{\sqrt {{\sqrt {5}}+1}}}}=2{\sqrt[{4}]{{\sqrt {5}}+2}}{\sqrt {\sin({\tfrac {1}{20}}\pi )\sin({\tfrac {3}{20}}\pi )}}}
slh ( 3 ϖ 5 2 ) = 1 8 4 5 − 1 20 4 + 5 + 1 = 2 5 − 2 4 cos ⁡ ( 1 20 π ) cos ⁡ ( 3 20 π ) {\displaystyle \operatorname {slh} \,\left({\frac {3\varpi }{5{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{8}}}{\sqrt {{\sqrt {5}}-1}}{\sqrt {{\sqrt[{4}]{20}}+{\sqrt {{\sqrt {5}}+1}}}}=2{\sqrt[{4}]{{\sqrt {5}}-2}}{\sqrt {\cos({\tfrac {1}{20}}\pi )\cos({\tfrac {3}{20}}\pi )}}}
slh ( 4 ϖ 5 2 ) = 1 2 2 4 ( 5 + 1 ) 20 4 + 5 + 1 = 2 5 + 2 4 cos ⁡ ( 1 20 π ) cos ⁡ ( 3 20 π ) {\displaystyle \operatorname {slh} \,\left({\frac {4\varpi }{5{\sqrt {2}}}}\right)={\frac {1}{2{\sqrt[{4}]{2}}}}({\sqrt {5}}+1){\sqrt {{\sqrt[{4}]{20}}+{\sqrt {{\sqrt {5}}+1}}}}=2{\sqrt[{4}]{{\sqrt {5}}+2}}{\sqrt {\cos({\tfrac {1}{20}}\pi )\cos({\tfrac {3}{20}}\pi )}}}
slh ( ϖ 6 2 ) = 1 2 ( 2 3 + 3 + 1 ) ( 1 − 2 3 − 3 4 ) {\displaystyle \operatorname {slh} \,\left({\frac {\varpi }{6{\sqrt {2}}}}\right)={\frac {1}{2}}({\sqrt {2{\sqrt {3}}+3}}+1)(1-{\sqrt[{4}]{2{\sqrt {3}}-3}})}
slh ( 5 ϖ 6 2 ) = 1 2 ( 2 3 + 3 + 1 ) ( 1 + 2 3 − 3 4 ) {\displaystyle \operatorname {slh} \,\left({\frac {5\varpi }{6{\sqrt {2}}}}\right)={\frac {1}{2}}({\sqrt {2{\sqrt {3}}+3}}+1)(1+{\sqrt[{4}]{2{\sqrt {3}}-3}})}

That table shows the most important values of the Hyperbolic Lemniscate Tangent and Cotangent functions:

z {\displaystyle z} |  clh ⁡ z {\displaystyle \operatorname {clh} z} |  slh ⁡ z {\displaystyle \operatorname {slh} z} |  ctlh ⁡ z = cos 4 ⁡ z {\displaystyle \operatorname {ctlh} z=\cos _{4}z} |  tlh ⁡ z = sin 4 ⁡ z {\displaystyle \operatorname {tlh} z=\sin _{4}z}

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0 {\displaystyle 0} |  ∞ {\displaystyle \infty } |  0 {\displaystyle 0} |  1 {\displaystyle 1} |  0 {\displaystyle 0}
1 4 σ {\displaystyle {\tfrac {1}{4}}\sigma } |  1 {\displaystyle 1} |  1 {\displaystyle 1} |  1 / 2 4 {\displaystyle 1{\big /}{\sqrt[{4}]{2}}} |  1 / 2 4 {\displaystyle 1{\big /}{\sqrt[{4}]{2}}}
1 2 σ {\displaystyle {\tfrac {1}{2}}\sigma } |  0 {\displaystyle 0} |  ∞ {\displaystyle \infty } |  0 {\displaystyle 0} |  1 {\displaystyle 1}
3 4 σ {\displaystyle {\tfrac {3}{4}}\sigma } |  − 1 {\displaystyle -1} |  − 1 {\displaystyle -1} |  − 1 / 2 4 {\displaystyle -1{\big /}{\sqrt[{4}]{2}}} |  1 / 2 4 {\displaystyle 1{\big /}{\sqrt[{4}]{2}}}
σ {\displaystyle \sigma } |  ∞ {\displaystyle \infty } |  0 {\displaystyle 0} |  − 1 {\displaystyle -1} |  0 {\displaystyle 0}

### Combination and halving theorems

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Given the hyperbolic lemniscate tangent ( tlh {\displaystyle \operatorname {tlh} } ) and hyperbolic lemniscate cotangent ( ctlh {\displaystyle \operatorname {ctlh} } ). Recall the hyperbolic lemniscate area functions from the section on inverse functions,

aslh ⁡ ( x ) = ∫ 0 x 1 y 4 + 1 d y {\displaystyle \operatorname {aslh} (x)=\int _{0}^{x}{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y}
aclh ⁡ ( x ) = ∫ x ∞ 1 y 4 + 1 d y {\displaystyle \operatorname {aclh} (x)=\int _{x}^{\infty }{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y}

Then the following identities can be established,

tlh [ aslh ( x ) ] = ctlh [ aclh ( x ) ] = x x 4 + 1 4 {\displaystyle {\text{tlh}}{\bigl [}{\text{aslh}}(x){\bigr ]}={\text{ctlh}}{\bigl [}{\text{aclh}}(x){\bigr ]}={\frac {x}{\sqrt[{4}]{x^{4}+1}}}}
ctlh [ aslh ( x ) ] = tlh [ aclh ( x ) ] = 1 x 4 + 1 4 {\displaystyle {\text{ctlh}}{\bigl [}{\text{aslh}}(x){\bigr ]}={\text{tlh}}{\bigl [}{\text{aclh}}(x){\bigr ]}={\frac {1}{\sqrt[{4}]{x^{4}+1}}}}

hence the 4th power of  tlh {\displaystyle \operatorname {tlh} } and  ctlh {\displaystyle \operatorname {ctlh} } for these arguments is equal to one,

tlh [ aslh ( x ) ] 4 + ctlh [ aslh ( x ) ] 4 = 1 {\displaystyle {\text{tlh}}{\bigl [}{\text{aslh}}(x){\bigr ]}^{4}+{\text{ctlh}}{\bigl [}{\text{aslh}}(x){\bigr ]}^{4}=1}
tlh [ aclh ( x ) ] 4 + ctlh [ aclh ( x ) ] 4 = 1 {\displaystyle {\text{tlh}}{\bigl [}{\text{aclh}}(x){\bigr ]}^{4}+{\text{ctlh}}{\bigl [}{\text{aclh}}(x){\bigr ]}^{4}=1}

so a 4th power version of the Pythagorean theorem. The bisection theorem of the hyperbolic sinus lemniscatus reads as follows:

slh [ 1 2 aslh ( x ) ] = 2 x x 2 + 1 + x 4 + 1 + x 4 + 1 − x 2 + 1 {\displaystyle {\text{slh}}{\bigl [}{\tfrac {1}{2}}{\text{aslh}}(x){\bigr ]}={\frac {{\sqrt {2}}x}{{\sqrt {x^{2}+1+{\sqrt {x^{4}+1}}}}+{\sqrt {{\sqrt {x^{4}+1}}-x^{2}+1}}}}}

This formula can be revealed as a combination of the following two formulas:

a s l h ( x ) = 2 arcsl [ x ( x 4 + 1 + 1 ) − 1 / 2 ] {\displaystyle \mathrm {aslh} (x)={\sqrt {2}}\,{\text{arcsl}}{\bigl [}x({\sqrt {x^{4}+1}}+1)^{-1/2}{\bigr ]}}
arcsl ( x ) = 2 aslh ( 2 x 1 + x 2 + 1 − x 2 ) {\displaystyle {\text{arcsl}}(x)={\sqrt {2}}\,{\text{aslh}}{\bigl (}{\frac {{\sqrt {2}}x}{{\sqrt {1+x^{2}}}+{\sqrt {1-x^{2}}}}}{\bigr )}}

In addition, the following formulas are valid for all real values  x ∈ R {\displaystyle x\in \mathbb {R} } :

slh [ 1 2 aclh ( x ) ] = x 4 + 1 + x 2 − 2 x x 4 + 1 + x 2 = ( x 4 + 1 − x 2 + 1 ) − 1 / 2 ( x 4 + 1 + 1 − x ) {\displaystyle {\text{slh}}{\bigl [}{\tfrac {1}{2}}{\text{aclh}}(x){\bigr ]}={\sqrt {{\sqrt {x^{4}+1}}+x^{2}-{\sqrt {2}}x{\sqrt {{\sqrt {x^{4}+1}}+x^{2}}}}}={\bigl (}{\sqrt {x^{4}+1}}-x^{2}+1{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {x^{4}+1}}+1}}-x{\bigr )}}
clh [ 1 2 aclh ( x ) ] = x 4 + 1 + x 2 + 2 x x 4 + 1 + x 2 = ( x 4 + 1 − x 2 + 1 ) − 1 / 2 ( x 4 + 1 + 1 + x ) {\displaystyle {\text{clh}}{\bigl [}{\tfrac {1}{2}}{\text{aclh}}(x){\bigr ]}={\sqrt {{\sqrt {x^{4}+1}}+x^{2}+{\sqrt {2}}x{\sqrt {{\sqrt {x^{4}+1}}+x^{2}}}}}={\bigl (}{\sqrt {x^{4}+1}}-x^{2}+1{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {x^{4}+1}}+1}}+x{\bigr )}}

These identities follow from the last-mentioned formula:

tlh [ 1 2 aclh ( x ) ] 2 = 1 2 2 − 2 2 x x 4 + 1 − x 2 = ( 2 x 2 + 2 + 2 x 4 + 1 ) − 1 / 2 ( x 4 + 1 + 1 − x ) {\displaystyle {\text{tlh}}[{\tfrac {1}{2}}{\text{aclh}}(x)]^{2}={\tfrac {1}{2}}{\sqrt {2-2{\sqrt {2}}\,x{\sqrt {{\sqrt {x^{4}+1}}-x^{2}}}}}={\bigl (}2x^{2}+2+2{\sqrt {x^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {x^{4}+1}}+1}}-x{\bigr )}}
ctlh [ 1 2 aclh ( x ) ] 2 = 1 2 2 + 2 2 x x 4 + 1 − x 2 = ( 2 x 2 + 2 + 2 x 4 + 1 ) − 1 / 2 ( x 4 + 1 + 1 + x ) {\displaystyle {\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}(x)]^{2}={\tfrac {1}{2}}{\sqrt {2+2{\sqrt {2}}\,x{\sqrt {{\sqrt {x^{4}+1}}-x^{2}}}}}={\bigl (}2x^{2}+2+2{\sqrt {x^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {x^{4}+1}}+1}}+x{\bigr )}}

Hence, their 4th powers again equal one,

tlh [ 1 2 aclh ( x ) ] 4 + ctlh [ 1 2 aclh ( x ) ] 4 = 1 {\displaystyle {\text{tlh}}{\bigl [}{\tfrac {1}{2}}{\text{aclh}}(x){\bigr ]}^{4}+{\text{ctlh}}{\bigl [}{\tfrac {1}{2}}{\text{aclh}}(x){\bigr ]}^{4}=1}

The following formulas for the lemniscatic sine and lemniscatic cosine are closely related:

sl [ 1 2 2 aclh ( x ) ] = cl [ 1 2 2 aslh ( x ) ] = x 4 + 1 − x 2 {\displaystyle {\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}(x)]={\text{cl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aslh}}(x)]={\sqrt {{\sqrt {x^{4}+1}}-x^{2}}}}
sl [ 1 2 2 aslh ( x ) ] = cl [ 1 2 2 aclh ( x ) ] = x ( x 4 + 1 + 1 ) − 1 / 2 {\displaystyle {\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aslh}}(x)]={\text{cl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}(x)]=x{\bigl (}{\sqrt {x^{4}+1}}+1{\bigr )}^{-1/2}}

### Coordinate Transformations

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Analogous to the determination of the improper integral in the Gaussian bell curve function, the coordinate transformation of a general cylinder can be used to calculate the integral from 0 to the positive infinity in the function  f ( x ) = exp ⁡ ( − x 4 ) {\displaystyle f(x)=\exp(-x^{4})} integrated in relation to x. In the following, the proofs of both integrals are given in a parallel way of displaying.

This is the cylindrical coordinate transformation in the Gaussian bell curve function:

[ ∫ 0 ∞ exp ⁡ ( − x 2 ) d x ] 2 = ∫ 0 ∞ ∫ 0 ∞ exp ⁡ ( − y 2 − z 2 ) d y d z = {\displaystyle {\biggl [}\int _{0}^{\infty }\exp(-x^{2})\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{\infty }\int _{0}^{\infty }\exp(-y^{2}-z^{2})\,\mathrm {d} y\,\mathrm {d} z=}
= ∫ 0 π / 2 ∫ 0 ∞ det [ ∂ / ∂ r r cos ⁡ ( ϕ ) ∂ / ∂ ϕ r cos ⁡ ( ϕ ) ∂ / ∂ r r sin ⁡ ( ϕ ) ∂ / ∂ ϕ r sin ⁡ ( ϕ ) ] exp ⁡ { − [ r cos ⁡ ( ϕ ) ] 2 − [ r sin ⁡ ( ϕ ) ] 2 } d r d ϕ = {\displaystyle =\int _{0}^{\pi /2}\int _{0}^{\infty }\det {\begin{bmatrix}\partial /\partial r\,\,r\cos(\phi )&\partial /\partial \phi \,\,r\cos(\phi )\\\\\partial /\partial r\,\,r\sin(\phi )&\partial /\partial \phi \,\,r\sin(\phi )\end{bmatrix}}\exp {\bigl \\{}-{\bigl [}r\cos(\phi ){\bigr ]}^{2}-{\bigl [}r\sin(\phi ){\bigr ]}^{2}{\bigr \\}}\,\mathrm {d} r\,\mathrm {d} \phi =}
= ∫ 0 π / 2 ∫ 0 ∞ r exp ⁡ ( − r 2 ) d r d ϕ = ∫ 0 π / 2 1 2 d ϕ = π 4 {\displaystyle =\int _{0}^{\pi /2}\int _{0}^{\infty }r\exp(-r^{2})\,\mathrm {d} r\,\mathrm {d} \phi =\int _{0}^{\pi /2}{\frac {1}{2}}\,\mathrm {d} \phi ={\frac {\pi }{4}}}

And this is the analogous coordinate transformation for the lemniscatory case:

[ ∫ 0 ∞ exp ⁡ ( − x 4 ) d x ] 2 = ∫ 0 ∞ ∫ 0 ∞ exp ⁡ ( − y 4 − z 4 ) d y d z = {\displaystyle {\biggl [}\int _{0}^{\infty }\exp(-x^{4})\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{\infty }\int _{0}^{\infty }\exp(-y^{4}-z^{4})\,\mathrm {d} y\,\mathrm {d} z=}
= ∫ 0 ϖ / 2 ∫ 0 ∞ det [ ∂ / ∂ r r ctlh ( ϕ ) ∂ / ∂ ϕ r ctlh ( ϕ ) ∂ / ∂ r r tlh ( ϕ ) ∂ / ∂ ϕ r tlh ( ϕ ) ] exp ⁡ { − [ r ctlh ( ϕ ) ] 4 − [ r tlh ( ϕ ) ] 4 } d r d ϕ = {\displaystyle =\int _{0}^{\varpi /{\sqrt {2}}}\int _{0}^{\infty }\det {\begin{bmatrix}\partial /\partial r\,\,r\,{\text{ctlh}}(\phi )&\partial /\partial \phi \,\,r\,{\text{ctlh}}(\phi )\\\\\partial /\partial r\,\,r\,{\text{tlh}}(\phi )&\partial /\partial \phi \,\,r\,{\text{tlh}}(\phi )\end{bmatrix}}\exp {\bigl \\{}-{\bigl [}r\,{\text{ctlh}}(\phi ){\bigr ]}^{4}-{\bigl [}r\,{\text{tlh}}(\phi ){\bigr ]}^{4}{\bigr \\}}\,\mathrm {d} r\,\mathrm {d} \phi =}
= ∫ 0 ϖ / 2 ∫ 0 ∞ r exp ⁡ ( − r 4 ) d r d ϕ = ∫ 0 ϖ / 2 π 4 d ϕ = ϖ π 4 2 {\displaystyle =\int _{0}^{\varpi /{\sqrt {2}}}\int _{0}^{\infty }r\exp(-r^{4})\,\mathrm {d} r\,\mathrm {d} \phi =\int _{0}^{\varpi /{\sqrt {2}}}{\frac {\sqrt {\pi }}{4}}\,\mathrm {d} \phi ={\frac {\varpi {\sqrt {\pi }}}{4{\sqrt {2}}}}}

In the last line of this elliptically analogous equation chain there is again the original Gauss bell curve integrated with the square function as the inner substitution according to the Chain rule of infinitesimal analytics (analysis).

In both cases, the determinant of the Jacobi matrix is multiplied to the original function in the integration domain.

The resulting new functions in the integration area are then integrated according to the new parameters.

## Number theory

[edit]

In algebraic number theory, every finite abelian extension of the Gaussian rationals Q ( i ) {\displaystyle \mathbb {Q} (i)} is a subfield of  Q ( i , ω n ) {\displaystyle \mathbb {Q} (i,\omega _{n})} for some positive integer  n {\displaystyle n} .[23][76] This is analogous to the Kronecker–Weber theorem for the rational numbers  Q {\displaystyle \mathbb {Q} } which is based on division of the circle – in particular, every finite abelian extension of  Q {\displaystyle \mathbb {Q} } is a subfield of  Q ( ζ n ) {\displaystyle \mathbb {Q} (\zeta _{n})} for some positive integer  n {\displaystyle n} . Both are special cases of Kronecker's Jugendtraum, which became Hilbert's twelfth problem.

The field Q ( i , sl ⁡ ( ϖ / n ) ) {\displaystyle \mathbb {Q} (i,\operatorname {sl} (\varpi /n))} (for positive odd  n {\displaystyle n} ) is the extension of  Q ( i ) {\displaystyle \mathbb {Q} (i)} generated by the  x {\displaystyle x} \- and  y {\displaystyle y} -coordinates of the  ( 1 + i ) n {\displaystyle (1+i)n} -torsion points on the elliptic curve y 2 = 4 x 3 + x {\displaystyle y^{2}=4x^{3}+x} .[76]

### Hurwitz numbers

[edit]

The Bernoulli numbers B n {\displaystyle \mathrm {B} _{n}} can be defined by

B n = lim z → 0 d n d z n z e z − 1 , n ≥ 0 {\displaystyle \mathrm {B} _{n}=\lim _{z\to 0}{\frac {\mathrm {d} ^{n}}{\mathrm {d} z^{n}}}{\frac {z}{e^{z}-1}},\quad n\geq 0}

and appear in

∑ k ∈ Z ∖ { 0 } 1 k 2 n = ( − 1 ) n − 1 B 2 n ( 2 π ) 2 n ( 2 n ) ! = 2 ζ ( 2 n ) , n ≥ 1 {\displaystyle \sum _{k\in \mathbb {Z} \setminus \\{0\\}}{\frac {1}{k^{2n}}}=(-1)^{n-1}\mathrm {B} _{2n}{\frac {(2\pi )^{2n}}{(2n)!}}=2\zeta (2n),\quad n\geq 1}

where  ζ {\displaystyle \zeta } is the Riemann zeta function.

The Hurwitz numbers H n , {\displaystyle \mathrm {H} _{n},} named after Adolf Hurwitz, are the "lemniscate analogs" of the Bernoulli numbers. They can be defined by[77][78]

H n = − lim z → 0 d n d z n z ζ ( z ; 1 / 4 , 0 ) , n ≥ 0 {\displaystyle \mathrm {H} _{n}=-\lim _{z\to 0}{\frac {\mathrm {d} ^{n}}{\mathrm {d} z^{n}}}z\zeta (z;1/4,0),\quad n\geq 0}

where  ζ ( ⋅ ; 1 / 4 , 0 ) {\displaystyle \zeta (\cdot ;1/4,0)} is the Weierstrass zeta function with lattice invariants  1 / 4 {\displaystyle 1/4} and  0 {\displaystyle 0} . They appear in

∑ z ∈ Z [ i ] ∖ { 0 } 1 z 4 n = H 4 n ( 2 ϖ ) 4 n ( 4 n ) ! = G 4 n ( i ) , n ≥ 1 {\displaystyle \sum _{z\in \mathbb {Z} [i]\setminus \\{0\\}}{\frac {1}{z^{4n}}}=\mathrm {H} _{4n}{\frac {(2\varpi )^{4n}}{(4n)!}}=G_{4n}(i),\quad n\geq 1}

where  Z [ i ] {\displaystyle \mathbb {Z} [i]} are the Gaussian integers and  G 4 n {\displaystyle G_{4n}} are the Eisenstein series of weight  4 n {\displaystyle 4n} , and in

∑ n = 1 ∞ n k e 2 π n − 1 = { 1 24 − 1 8 π if   k = 1 B k + 1 2 k + 2 if   k ≡ 1 ( m o d 4 )   and   k ≥ 5 B k + 1 2 k + 2 + H k + 1 2 k + 2 ( ϖ π ) k + 1 if   k ≡ 3 ( m o d 4 )   and   k ≥ 3. {\displaystyle \displaystyle {\begin{array}{ll}\displaystyle \sum _{n=1}^{\infty }{\dfrac {n^{k}}{e^{2\pi n}-1}}={\begin{cases}{\dfrac {1}{24}}-{\dfrac {1}{8\pi }}&{\text{if}}\ k=1\\\\{\dfrac {\mathrm {B} _{k+1}}{2k+2}}&{\text{if}}\ k\equiv 1\,(\mathrm {mod} \,4)\ {\text{and}}\ k\geq 5\\\\{\dfrac {\mathrm {B} _{k+1}}{2k+2}}+{\dfrac {\mathrm {H} _{k+1}}{2k+2}}\left({\dfrac {\varpi }{\pi }}\right)^{k+1}&{\text{if}}\ k\equiv 3\,(\mathrm {mod} \,4)\ {\text{and}}\ k\geq 3.\\\\\end{cases}}\end{array}}}

The Hurwitz numbers can also be determined as follows:  H 4 = 1 / 10 {\displaystyle \mathrm {H} _{4}=1/10} ,

H 4 n = 3 ( 2 n − 3 ) ( 16 n 2 − 1 ) ∑ k = 1 n − 1 ( 4 n 4 k ) ( 4 k − 1 ) ( 4 ( n − k ) − 1 ) H 4 k H 4 ( n − k ) , n ≥ 2 {\displaystyle \mathrm {H} _{4n}={\frac {3}{(2n-3)(16n^{2}-1)}}\sum _{k=1}^{n-1}{\binom {4n}{4k}}(4k-1)(4(n-k)-1)\mathrm {H} _{4k}\mathrm {H} _{4(n-k)},\quad n\geq 2}

and  H n = 0 {\displaystyle \mathrm {H} _{n}=0} if  n {\displaystyle n} is not a multiple of  4 {\displaystyle 4} .[79] This yields[77]

H 8 = 3 10 , H 12 = 567 130 , H 16 = 43 659 170 , … {\displaystyle \mathrm {H} _{8}={\frac {3}{10}},\,\mathrm {H} _{12}={\frac {567}{130}},\,\mathrm {H} _{16}={\frac {43\,659}{170}},\,\ldots }

Also[80]

denom ⁡ H 4 n = ∏ ( p − 1 ) | 4 n p {\displaystyle \operatorname {denom} \mathrm {H} _{4n}=\prod _{(p-1)|4n}p}

where  p ∈ P {\displaystyle p\in \mathbb {P} } such that  p ≢ 3 ( mod 4 ) , {\displaystyle p\not \equiv 3\,({\text{mod}}\,4),} just as

denom ⁡ B 2 n = ∏ ( p − 1 ) | 2 n p {\displaystyle \operatorname {denom} \mathrm {B} _{2n}=\prod _{(p-1)|2n}p}

where  p ∈ P {\displaystyle p\in \mathbb {P} } (by the von Staudt–Clausen theorem).

In fact, the von Staudt–Clausen theorem determines the fractional part of the Bernoulli numbers:

B 2 n + ∑ ( p − 1 ) | 2 n 1 p ∈ Z , n ≥ 1 {\displaystyle \mathrm {B} _{2n}+\sum _{(p-1)|2n}{\frac {1}{p}}\in \mathbb {Z} ,\quad n\geq 1}

(sequence A000146 in the OEIS) where  p {\displaystyle p} is any prime, and an analogous theorem holds for the Hurwitz numbers: suppose that  a ∈ Z {\displaystyle a\in \mathbb {Z} } is odd,  b ∈ Z {\displaystyle b\in \mathbb {Z} } is even,  p {\displaystyle p} is a prime such that  p ≡ 1 ( m o d 4 ) {\displaystyle p\equiv 1\,(\mathrm {mod} \,4)} ,  p = a 2 + b 2 {\displaystyle p=a^{2}+b^{2}} (see Fermat's theorem on sums of two squares) and  a ≡ b + 1 ( m o d 4 ) {\displaystyle a\equiv b+1\,(\mathrm {mod} \,4)} . Then for any given  p {\displaystyle p} ,  2 a = ν ( p ) {\displaystyle 2a=\nu (p)} is uniquely determined; equivalently  ν ( p ) = p − N p {\displaystyle \nu (p)=p-{\mathcal {N}}_{p}} where  N p {\displaystyle {\mathcal {N}}_{p}} is the number of solutions of the congruence  X 3 − X ≡ Y 2 ( mod ⁡ p ) {\displaystyle X^{3}-X\equiv Y^{2}\,(\operatorname {mod} p)} in variables  X , Y {\displaystyle X,Y} that are non-negative integers.[81] The Hurwitz theorem then determines the fractional part of the Hurwitz numbers:[77]

H 4 n − 1 2 − ∑ ( p − 1 ) | 4 n ν ( p ) 4 n / ( p − 1 ) p = def G n ∈ Z , n ≥ 1. {\displaystyle \mathrm {H} _{4n}-{\frac {1}{2}}-\sum _{(p-1)|4n}{\frac {\nu (p)^{4n/(p-1)}}{p}}\mathrel {\overset {\text{def}}{=}} \mathrm {G} _{n}\in \mathbb {Z} ,\quad n\geq 1.}

The sequence of the integers  G n {\displaystyle \mathrm {G} _{n}} starts with  0 , − 1 , 5 , 253 , … . {\displaystyle 0,-1,5,253,\ldots .} [77]

Let  n ≥ 2 {\displaystyle n\geq 2} . If  4 n + 1 {\displaystyle 4n+1} is a prime, then  G n ≡ 1 ( m o d 4 ) {\displaystyle \mathrm {G} _{n}\equiv 1\,(\mathrm {mod} \,4)} . If  4 n + 1 {\displaystyle 4n+1} is not a prime, then  G n ≡ 3 ( m o d 4 ) {\displaystyle \mathrm {G} _{n}\equiv 3\,(\mathrm {mod} \,4)} .[82]

Some authors instead define the Hurwitz numbers as  H n ′ = H 4 n {\displaystyle \mathrm {H} _{n}'=\mathrm {H} _{4n}} .

#### Appearances in Laurent series

[edit]

The Hurwitz numbers appear in several Laurent series expansions related to the lemniscate functions:[83]

sl 2 ⁡ z = ∑ n = 1 ∞ 2 4 n ( 1 − ( − 1 ) n 2 2 n ) H 4 n 4 n z 4 n − 2 ( 4 n − 2 ) ! , | z | < ϖ 2 sl ′ ⁡ z sl ⁡ z = 1 z − ∑ n = 1 ∞ 2 4 n ( 2 − ( − 1 ) n 2 2 n ) H 4 n 4 n z 4 n − 1 ( 4 n − 1 ) ! , | z | < ϖ 2 1 sl ⁡ z = 1 z − ∑ n = 1 ∞ 2 2 n ( ( − 1 ) n 2 − 2 2 n ) H 4 n 4 n z 4 n − 1 ( 4 n − 1 ) ! , | z | < ϖ 1 sl 2 ⁡ z = 1 z 2 + ∑ n = 1 ∞ 2 4 n H 4 n 4 n z 4 n − 2 ( 4 n − 2 ) ! , | z | < ϖ {\displaystyle {\begin{aligned}\operatorname {sl} ^{2}z&=\sum _{n=1}^{\infty }{\frac {2^{4n}(1-(-1)^{n}2^{2n})\mathrm {H} _{4n}}{4n}}{\frac {z^{4n-2}}{(4n-2)!}},\quad \left|z\right|<{\frac {\varpi }{\sqrt {2}}}\\\\{\frac {\operatorname {sl} 'z}{\operatorname {sl} {z}}}&={\frac {1}{z}}-\sum _{n=1}^{\infty }{\frac {2^{4n}(2-(-1)^{n}2^{2n})\mathrm {H} _{4n}}{4n}}{\frac {z^{4n-1}}{(4n-1)!}},\quad \left|z\right|<{\frac {\varpi }{\sqrt {2}}}\\\\{\frac {1}{\operatorname {sl} z}}&={\frac {1}{z}}-\sum _{n=1}^{\infty }{\frac {2^{2n}((-1)^{n}2-2^{2n})\mathrm {H} _{4n}}{4n}}{\frac {z^{4n-1}}{(4n-1)!}},\quad \left|z\right|<\varpi \\\\{\frac {1}{\operatorname {sl} ^{2}z}}&={\frac {1}{z^{2}}}+\sum _{n=1}^{\infty }{\frac {2^{4n}\mathrm {H} _{4n}}{4n}}{\frac {z^{4n-2}}{(4n-2)!}},\quad \left|z\right|<\varpi \end{aligned}}}

Analogously, in terms of the Bernoulli numbers:

1 sinh 2 ⁡ z = 1 z 2 − ∑ n = 1 ∞ 2 2 n B 2 n 2 n z 2 n − 2 ( 2 n − 2 ) ! , | z | < π . {\displaystyle {\frac {1}{\sinh ^{2}z}}={\frac {1}{z^{2}}}-\sum _{n=1}^{\infty }{\frac {2^{2n}\mathrm {B} _{2n}}{2n}}{\frac {z^{2n-2}}{(2n-2)!}},\quad \left|z\right|<\pi .}

### A quartic analog of the Legendre symbol

[edit]

Let  p {\displaystyle p} be a prime such that  p ≡ 1 ( mod 4 ) {\displaystyle p\equiv 1\,({\text{mod}}\,4)} . A quartic residue (mod  p {\displaystyle p} ) is any number congruent to the fourth power of an integer. Define  ( a p ) 4 {\displaystyle \left({\tfrac {a}{p}}\right)_{4}} to be  1 {\displaystyle 1} if  a {\displaystyle a} is a quartic residue (mod  p {\displaystyle p} ) and define it to be  − 1 {\displaystyle -1} if  a {\displaystyle a} is not a quartic residue (mod  p {\displaystyle p} ).

If  a {\displaystyle a} and  p {\displaystyle p} are coprime, then there exist numbers  p ′ ∈ Z [ i ] {\displaystyle p'\in \mathbb {Z} [i]} (see[84] for these numbers) such that[85]

( a p ) 4 = ∏ p ′ sl ⁡ ( 2 ϖ a p ′ / p ) sl ⁡ ( 2 ϖ p ′ / p ) . {\displaystyle \left({\frac {a}{p}}\right)_{4}=\prod _{p'}{\frac {\operatorname {sl} (2\varpi ap'/p)}{\operatorname {sl} (2\varpi p'/p)}}.}

This theorem is analogous to

( a p ) = ∏ n = 1 p − 1 2 sin ⁡ ( 2 π a n / p ) sin ⁡ ( 2 π n / p ) {\displaystyle \left({\frac {a}{p}}\right)=\prod _{n=1}^{\frac {p-1}{2}}{\frac {\sin(2\pi an/p)}{\sin(2\pi n/p)}}}

where  ( ⋅ ⋅ ) {\displaystyle \left({\tfrac {\cdot }{\cdot }}\right)} is the Legendre symbol.

## World map projections

[edit]

"The World on a Quincuncial Projection", from Peirce (1879).

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).[86]

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.[87] 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.[88] Because many partial differential equations can be effectively solved by conformal mapping, this map from sphere to cube is convenient for atmospheric modeling.[89]

## See also

[edit]

* Elliptic function
* Abel elliptic functions
* Dixon elliptic functions
* Jacobi elliptic functions
* Weierstrass elliptic function
* Elliptic Gauss sum
* Lemniscate constant
* Peirce quincuncial projection
* Schwarz–Christoffel mapping

## Notes

[edit]

1. ^ Fagnano (1718–1723); Euler (1761); Gauss (1917)
2. ^ Gauss (1917) p. 199 used the symbols sl and cl for the lemniscate sine and cosine, respectively, and this notation is most common today: see e.g. Cox (1984) p. 316, Eymard & Lafon (2004) p. 204, and Lemmermeyer (2000) p. 240. Ayoub (1984) uses sinlem and coslem. Whittaker & Watson (1920) use the symbols sin lemn and cos lemn. Some sources use the generic letters s and c. Prasolov & Solovyev (1997) use the letter φ for the lemniscate sine and φ′ for its derivative.
3. ^ The circle  x 2 + y 2 = x {\displaystyle x^{2}+y^{2}=x} is the unit-diameter circle centered at  ( 1 2 , 0 ) {\textstyle {\bigl (}{\tfrac {1}{2}},0{\bigr )}} with polar equation  r = cos ⁡ θ , {\displaystyle r=\cos \theta ,} the degree-2 clover under the definition from Cox & Shurman (2005). This is not the unit-radius circle  x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1} centered at the origin. Notice that the lemniscate  ( x 2 + y 2 ) 2 = x 2 − y 2 {\displaystyle {\bigl (}x^{2}+y^{2}{\bigr )}{}^{2}=x^{2}-y^{2}} is the degree-4 clover.
4. ^ The fundamental periods  ( 1 + i ) ϖ {\displaystyle (1+i)\varpi } and  ( 1 − i ) ϖ {\displaystyle (1-i)\varpi } are "minimal" in the sense that they have the smallest absolute value of all periods whose real part is non-negative.
5. ^ Robinson (2019a) starts from this definition and thence derives other properties of the lemniscate functions.
6. ^ This map was the first ever picture of a Schwarz–Christoffel mapping, in Schwarz (1869) p. 113.
7. ^ Schappacher (1997). OEIS sequence A062539 lists the lemniscate constant's decimal digits.
8. ^ Levin (2006)
9. ^ Todd (1975)
10. ^ Cox (1984)
11. ^ Dark areas represent zeros, and bright areas represent poles. As the argument of  sl ⁡ z {\displaystyle \operatorname {sl} z} changes from  − π {\displaystyle -\pi } (excluding  − π {\displaystyle -\pi } ) to  π {\displaystyle \pi } , the colors go through cyan, blue  ( Arg ≈ − π / 2 ) {\displaystyle (\operatorname {Arg} \approx -\pi /2)} , magneta, red  ( Arg ≈ 0 ) {\displaystyle (\operatorname {Arg} \approx 0)} , orange, yellow  ( Arg ≈ π / 2 ) {\displaystyle (\operatorname {Arg} \approx \pi /2)} , green, and back to cyan  ( Arg ≈ π ) {\displaystyle (\operatorname {Arg} \approx \pi )} .
12. ^ Combining the first and fourth identity gives  sl ⁡ z = − i / sl ⁡ ( z − ( 1 + i ) ϖ / 2 ) {\displaystyle \operatorname {sl} z=-i/\operatorname {sl} (z-(1+i)\varpi /2)} . This identity is (incorrectly) given in Eymard & Lafon (2004) p. 226, without the minus sign at the front of the right-hand side.
13. ^ The even Gaussian integers are the residue class of 0, modulo 1 + i, the black squares on a checkerboard.
14. ^ Prasolov & Solovyev (1997); Robinson (2019a)
15. ^ a b Cox (2012)
16. ^ Reinhardt & Walker (2010a) §22.12.6, §22.12.12
17. ^ Analogously,  1 sin ⁡ z = ∑ n ∈ Z ( − 1 ) n z + n π . {\displaystyle {\frac {1}{\sin z}}=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n}}{z+n\pi }}.}
18. ^ Lindqvist & Peetre (2001) generalizes the first of these forms.
19. ^ Ayoub (1984); Prasolov & Solovyev (1997)
20. ^ Euler (1761) §44 p. 79, §47 pp. 80–81
21. ^ a b Euler (1761) §46 p. 80
22. ^ In fact,  i ε = sl ⁡ β ϖ 2 {\displaystyle i^{\varepsilon }=\operatorname {sl} {\tfrac {\beta \varpi }{2}}} .
23. ^ a b c Cox & Hyde (2014)
24. ^ Gómez-Molleda & Lario (2019)
25. ^ The fourth root with the least positive principal argument is chosen.
26. ^ The restriction to positive and odd  β {\displaystyle \beta } can be dropped in  deg ⁡ Λ β = | ( O / β O ) × | {\displaystyle \operatorname {deg} \Lambda _{\beta }=\left|({\mathcal {O}}/\beta {\mathcal {O}})^{\times }\right|} .
27. ^ Cox (2013) p. 142, Example 7.29(c)
28. ^ Rosen (1981)
29. ^ Eymard & Lafon (2004) p. 200
30. ^ And the area enclosed by  L {\displaystyle {\mathcal {L}}} is  1 {\displaystyle 1} , which stands in stark contrast to the unit circle (whose enclosed area is a non-constructible number).
31. ^ Euler (1761); Siegel (1969). Prasolov & Solovyev (1997) use the polar-coordinate representation of the Lemniscate to derive differential arc length, but the result is the same.
32. ^ Reinhardt & Walker (2010a) §22.18.E6
33. ^ Siegel (1969); Schappacher (1997)
34. ^ Such numbers are OEIS sequence A003401.
35. ^ Abel (1827–1828); Rosen (1981); Prasolov & Solovyev (1997)
36. ^ Euler (1786); Sridharan (2004); Levien (2008)
37. ^ "A104203". The On-Line Encyclopedia of Integer Sequences.
38. ^ Lomont, J.S.; Brillhart, John (2001). Elliptic Polynomials. CRC Press. pp. 12, 44. ISBN 1-58488-210-7.
39. ^ a b c d "A193543 - Oeis".
40. ^ Lomont, J.S.; Brillhart, John (2001). Elliptic Polynomials. CRC Press. ISBN 1-58488-210-7. p. 79, eq. 5.36
41. ^ Lomont, J.S.; Brillhart, John (2001). Elliptic Polynomials. CRC Press. ISBN 1-58488-210-7. p. 79, eq. 5. 36 and p. 78, eq. 5.33
42. ^ a b "A289695 - Oeis".
43. ^ Wall, H. S. (1948). Analytic Theory of Continued Fractions. Chelsea Publishing Company. pp. 374–375.
44. ^ Reinhardt & Walker (2010a) §22.20(ii)
45. ^ Carlson (2010) §19.8
46. ^ Reinhardt & Walker (2010a) §22.12.12
47. ^ In general,  sinh ⁡ ( x − n π ) {\displaystyle \sinh(x-n\pi )} and  sin ⁡ ( x − n π i ) = − i sinh ⁡ ( i x + n π ) {\displaystyle \sin(x-n\pi i)=-i\sinh(ix+n\pi )} are not equivalent, but the resulting infinite sum is the same.
48. ^ Reinhardt & Walker (2010a) §22.11
49. ^ Reinhardt & Walker (2010a) §22.2.E7
50. ^ Berndt (1994) p. 247, 248, 253
51. ^ Reinhardt & Walker (2010a) §22.11.E1
52. ^ Whittaker & Watson (1927)
53. ^ Borwein & Borwein (1987)
54. ^ a b Eymard & Lafon (2004) p. 227.
55. ^ Cartan, H. (1961). Théorie élémentaire des fonctions analytiques d'une ou plusieurs variables complexes (in French). Hermann. pp. 160–164.
56. ^ More precisely, suppose  { a n } {\displaystyle \\{a_{n}\\}} is a sequence of bounded complex functions on a set  S {\displaystyle S} , such that  ∑ | a n ( z ) | {\textstyle \sum \left|a_{n}(z)\right|} converges uniformly on  S {\displaystyle S} . If  { n 1 , n 2 , n 3 , … } {\displaystyle \\{n_{1},n_{2},n_{3},\ldots \\}} is any permutation of  { 1 , 2 , 3 , … } {\displaystyle \\{1,2,3,\ldots \\}} , then  ∏ n = 1 ∞ ( 1 + a n ( z ) ) = ∏ k = 1 ∞ ( 1 + a n k ( z ) ) {\textstyle \prod _{n=1}^{\infty }(1+a_{n}(z))=\prod _{k=1}^{\infty }(1+a_{n_{k}}(z))} for all  z ∈ S {\displaystyle z\in S} . The theorem in question then follows from the fact that there exists a bijection between the natural numbers and  α {\displaystyle \alpha } 's (resp.  β {\displaystyle \beta } 's).
57. ^ Bottazzini & Gray (2013) p. 58
58. ^ More precisely, if for each  k {\displaystyle k} ,  lim n → ∞ a k ( n ) {\textstyle \lim _{n\to \infty }a_{k}(n)} exists and there is a convergent series  ∑ k = 1 ∞ M k {\textstyle \sum _{k=1}^{\infty }M_{k}} of nonnegative real numbers such that  | a k ( n ) | ≤ M k {\displaystyle \left|a_{k}(n)\right|\leq M_{k}} for all  n ∈ N {\displaystyle n\in \mathbb {N} } and  1 ≤ k ≤ n {\displaystyle 1\leq k\leq n} , then

lim n → ∞ ∑ k = 1 n a k ( n ) = ∑ k = 1 ∞ lim n → ∞ a k ( n ) . {\displaystyle \lim _{n\to \infty }\sum _{k=1}^{n}a_{k}(n)=\sum _{k=1}^{\infty }\lim _{n\to \infty }a_{k}(n).}
59. ^ Alternatively, it can be inferred that these expansions exist just from the analyticity of  M {\displaystyle M} and  N {\displaystyle N} . However, establishing the connection to "multiplying out and collecting like powers" reveals identities between sums of reciprocals and the coefficients of the power series, like  ∑ α 1 α 4 = − the coefficient of z 5 {\textstyle \sum _{\alpha }{\frac {1}{\alpha ^{4}}}=-\,{\text{the coefficient of}}\,z^{5}} in the  M {\displaystyle M} series, and infinitely many others.
60. ^ Gauss, C. F. (1866). Werke (Band III) (in Latin and German). Herausgegeben der Königlichen Gesellschaft der Wissenschaften zu Göttingen. p. 405; there's an error on the page: the coefficient of  φ 17 {\displaystyle \varphi ^{17}} should be  107 7 410 154 752 000 {\displaystyle {\tfrac {107}{7\,410\,154\,752\,000}}} , not  107 207 484 333 056 000 {\displaystyle {\tfrac {107}{207\,484\,333\,056\,000}}} .
61. ^ If  M ( z ) = ∑ n = 0 ∞ a n z n + 1 {\textstyle M(z)=\sum _{n=0}^{\infty }a_{n}z^{n+1}} , then the coefficients  a n {\displaystyle a_{n}} are given by the recurrence  a n + 1 = − 1 n + 1 ∑ k = 0 n 2 n − k + 1 a k H n − k + 1 ( n − k + 1 ) ! {\textstyle a_{n+1}=-{\frac {1}{n+1}}\sum _{k=0}^{n}2^{n-k+1}a_{k}{\frac {\mathrm {H} _{n-k+1}}{(n-k+1)!}}} with  a 0 = 1 {\displaystyle a_{0}=1} where  H n {\displaystyle \mathrm {H} _{n}} are the Hurwitz numbers defined in Lemniscate elliptic functions § Hurwitz numbers.
62. ^ The power series expansions of  M {\displaystyle M} and  N {\displaystyle N} are useful for finding a  β {\displaystyle \beta } -division polynomial for the  β {\displaystyle \beta } -division of the lemniscate  L {\displaystyle {\mathcal {L}}} (where  β = m + n i {\displaystyle \beta =m+ni} where  m , n ∈ Z {\displaystyle m,n\in \mathbb {Z} } such that  m + n {\displaystyle m+n} is odd). For example, suppose we want to find a  3 {\displaystyle 3} -division polynomial. Given that

M ( 3 z ) = d 9 M ( z ) 9 + d 5 M ( z ) 5 N ( z ) 4 + d 1 M ( z ) N ( z ) 8 {\displaystyle M(3z)=d_{9}M(z)^{9}+d_{5}M(z)^{5}N(z)^{4}+d_{1}M(z)N(z)^{8}}
for some constants  d 1 , d 5 , d 9 {\displaystyle d_{1},d_{5},d_{9}} , from

3 z − 2 ( 3 z ) 5 5 ! − 36 ( 3 z ) 9 9 ! + O ⁡ ( z 13 ) = d 9 x 9 + d 5 x 5 y 4 + d 1 x y 8 , {\displaystyle 3z-2{\frac {(3z)^{5}}{5!}}-36{\frac {(3z)^{9}}{9!}}+\operatorname {O} (z^{13})=d_{9}x^{9}+d_{5}x^{5}y^{4}+d_{1}xy^{8},}
where

x = z − 2 z 5 5 ! − 36 z 9 9 ! + O ⁡ ( z 13 ) , y = 1 + 2 z 4 4 ! − 4 z 8 8 ! + O ⁡ ( z 12 ) , {\displaystyle x=z-2{\frac {z^{5}}{5!}}-36{\frac {z^{9}}{9!}}+\operatorname {O} (z^{13}),\quad y=1+2{\frac {z^{4}}{4!}}-4{\frac {z^{8}}{8!}}+\operatorname {O} (z^{12}),}
we have

{ d 1 , d 5 , d 9 } = { 3 , − 6 , − 1 } . {\displaystyle \\{d_{1},d_{5},d_{9}\\}=\\{3,-6,-1\\}.}
Therefore, a  3 {\displaystyle 3} -division polynomial is

− X 9 − 6 X 5 + 3 X {\displaystyle -X^{9}-6X^{5}+3X}
(meaning one of its roots is  sl ⁡ ( 2 ϖ / 3 ) {\displaystyle \operatorname {sl} (2\varpi /3)} ). The equations arrived at by this process are the lemniscate analogs of

X n = 1 {\displaystyle X^{n}=1}
(so that  e 2 π i / n {\displaystyle e^{2\pi i/n}} is one of the solutions) which comes up when dividing the unit circle into  n {\displaystyle n} arcs of equal length. In the following note, the first few coefficients of the monic normalization of such  β {\displaystyle \beta } -division polynomials are described symbolically in terms of  β {\displaystyle \beta } .
63. ^ By utilizing the power series expansion of the  N {\displaystyle N} function, it can be proved that a polynomial having  sl ⁡ ( 2 ϖ / β ) {\displaystyle \operatorname {sl} (2\varpi /\beta )} as one of its roots (with  β {\displaystyle \beta } from the previous note) is

∑ n = 0 ( β β ¯ − 1 ) / 4 a 4 n + 1 ( β ) X β β ¯ − 4 n {\displaystyle \sum _{n=0}^{(\beta {\overline {\beta }}-1)/4}a_{4n+1}(\beta )X^{\beta {\overline {\beta }}-4n}}
where

a 1 ( β ) = 1 , a 5 ( β ) = β 4 − β β ¯ 12 , a 9 ( β ) = − β 8 − 70 β 5 β ¯ + 336 β 4 + 35 β 2 β ¯ 2 − 300 β β ¯ 10080 {\displaystyle {\begin{aligned}a_{1}(\beta )&=1,\\\a_{5}(\beta )&={\frac {\beta ^{4}-\beta {\overline {\beta }}}{12}},\\\a_{9}(\beta )&={\frac {-\beta ^{8}-70\beta ^{5}{\overline {\beta }}+336\beta ^{4}+35\beta ^{2}{\overline {\beta }}^{2}-300\beta {\overline {\beta }}}{10080}}\end{aligned}}}
and so on.
64. ^ Zhuravskiy, A. M. (1941). Spravochnik po ellipticheskim funktsiyam (in Russian). Izd. Akad. Nauk. U.S.S.R.
65. ^ For example, by the quasi-addition formulas, the duplication formulas and the Pythagorean-like identities, we have

M ( 3 z ) = − M ( z ) 9 − 6 M ( z ) 5 N ( z ) 4 + 3 M ( z ) N ( z ) 8 , {\displaystyle M(3z)=-M(z)^{9}-6M(z)^{5}N(z)^{4}+3M(z)N(z)^{8},}
N ( 3 z ) = N ( z ) 9 + 6 M ( z ) 4 N ( z ) 5 − 3 M ( z ) 8 N ( z ) , {\displaystyle N(3z)=N(z)^{9}+6M(z)^{4}N(z)^{5}-3M(z)^{8}N(z),}

sl ⁡ 3 z = − M ( z ) 9 − 6 M ( z ) 5 N ( z ) 4 + 3 M ( z ) N ( z ) 8 N ( z ) 9 + 6 M ( z ) 4 N ( z ) 5 − 3 M ( z ) 8 N ( z ) . {\displaystyle \operatorname {sl} 3z={\frac {-M(z)^{9}-6M(z)^{5}N(z)^{4}+3M(z)N(z)^{8}}{N(z)^{9}+6M(z)^{4}N(z)^{5}-3M(z)^{8}N(z)}}.}
On dividing the numerator and the denominator by  N ( z ) 9 {\displaystyle N(z)^{9}} , we obtain the triplication formula for  sl {\displaystyle \operatorname {sl} } :

sl ⁡ 3 z = − sl 9 ⁡ z − 6 sl 5 ⁡ z + 3 sl ⁡ z 1 + 6 sl 4 ⁡ z − 3 sl 8 ⁡ z . {\displaystyle \operatorname {sl} 3z={\frac {-\operatorname {sl} ^{9}z-6\operatorname {sl} ^{5}z+3\operatorname {sl} z}{1+6\operatorname {sl} ^{4}z-3\operatorname {sl} ^{8}z}}.}
66. ^ Gauss (1866), p. 408
67. ^ Robinson (2019a)
68. ^ Eymard & Lafon (2004) p. 234
69. ^ Armitage, J. V.; Eberlein, W. F. (2006). Elliptic Functions. Cambridge University Press. p. 49\. ISBN 978-0-521-78563-1.
70. ^ The identity  cl ⁡ z = cn ( 2 z ; 1 2 ) {\displaystyle \operatorname {cl} z={\operatorname {cn} }\left({\sqrt {2}}z;{\tfrac {1}{\sqrt {2}}}\right)} can be found in Greenhill (1892) p. 33.
71. ^ Siegel (1969)
72. ^ Sloane, N. J. A. (ed.). "Sequence A175576". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
73. ^ Berndt, Bruce C. (1989). Ramanujan's Notebooks Part II. Springer. ISBN 978-1-4612-4530-8. p. 96
74. ^ Levin (2006); Robinson (2019b)
75. ^ Levin (2006) p. 515
76. ^ a b Cox (2012) p. 508, 509
77. ^ a b c d Arakawa, Tsuneo; Ibukiyama, Tomoyoshi; Kaneko, Masanobu (2014). Bernoulli Numbers and Zeta Functions. Springer. ISBN 978-4-431-54918-5. p. 203—206
78. ^ Equivalently,  H n = − lim z → 0 d n d z n ( ( 1 + i ) z / 2 sl ⁡ ( ( 1 + i ) z / 2 ) + z 2 E ( z 2 ; i ) ) {\displaystyle \mathrm {H} _{n}=-\lim _{z\to 0}{\frac {\mathrm {d} ^{n}}{\mathrm {d} z^{n}}}\left({\frac {(1+i)z/2}{\operatorname {sl} ((1+i)z/2)}}+{\frac {z}{2}}{\mathcal {E}}\left({\frac {z}{2}};i\right)\right)} where  n ≥ 4 {\displaystyle n\geq 4} and  E ( ⋅ ; i ) {\displaystyle {\mathcal {E}}(\cdot ;i)} is the Jacobi epsilon function with modulus  i {\displaystyle i} .
79. ^ The Bernoulli numbers can be determined by an analogous recurrence:  B 2 n = − 1 2 n + 1 ∑ k = 1 n − 1 ( 2 n 2 k ) B 2 k B 2 ( n − k ) {\displaystyle \mathrm {B} _{2n}=-{\frac {1}{2n+1}}\sum _{k=1}^{n-1}{\binom {2n}{2k}}\mathrm {B} _{2k}\mathrm {B} _{2(n-k)}} where  n ≥ 2 {\displaystyle n\geq 2} and  B 2 = 1 / 6 {\displaystyle \mathrm {B} _{2}=1/6} .
80. ^ Katz, Nicholas M. (1975). "The congruences of Clausen — von Staudt and Kummer for Bernoulli-Hurwitz numbers". Mathematische Annalen. 216 (1): 1–4\. doi:10.1007/BF02547966. See eq. (9)
81. ^ For more on the  ν {\displaystyle \nu } function, see Lemniscate constant.
82. ^ Hurwitz, Adolf (1963). Mathematische Werke: Band II (in German). Springer Basel AG. p. 370
83. ^ Arakawa et al. (2014) define  H 4 n {\displaystyle \mathrm {H} _{4n}} by the expansion of  1 / sl 2 . {\displaystyle 1/\operatorname {sl} ^{2}.}
84. ^ Eisenstein, G. (1846). "Beiträge zur Theorie der elliptischen Functionen". Journal für die reine und angewandte Mathematik (in German). 30. Eisenstein uses  φ = sl {\displaystyle \varphi =\operatorname {sl} } and  ω = 2 ϖ {\displaystyle \omega =2\varpi } .
85. ^ Ogawa (2005)
86. ^ Peirce (1879). Guyou (1887) and Adams (1925) introduced transverse and oblique aspects of the same projection, respectively. Also see Lee (1976). These authors write their projection formulas in terms of Jacobi elliptic functions, with a square lattice.
87. ^ Adams (1925)
88. ^ Adams (1925); Lee (1976).
89. ^ Rančić, Purser & Mesinger (1996); McGregor (2005).

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## External links

[edit]

* Parker, Matt (2021). "What is the area of a Squircle?". Stand-up Maths. YouTube.

* Ramalingam, Muthu Veerappan (2023). "Bernoulli Lemniscate and the Squircle  | | {\displaystyle ||} A remarkable Geometric fun fact!!?". Act of Learning. YouTube. Relation shown in the video amounts to  cl ⁡ ( 2 t ) = cos 4 2 ⁡ ( t ) − sin 4 2 ⁡ ( t ) cos 4 2 ⁡ ( t ) + sin 4 2 ⁡ ( t ) {\displaystyle \operatorname {cl} ({\sqrt {2}}t)={\frac {\cos _{4}^{2}(t)-\sin _{4}^{2}(t)}{\cos _{4}^{2}(t)+\sin _{4}^{2}(t)}}}