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 (red) and lemniscate cosine (purple) applied to a real argument, in comparison with the trigonometric sine y = sin(πx /ϖ ) (pale dashed red). 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
Definitions 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 analytically continued to the whole complex plane by a series of reflections .
By comparison, the circular sine and cosine can be defined as the solution to the initial value problem:
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 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 2ϖ i 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
Zeros, poles and symmetries 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 Curves x ² ⊕ y ² = 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 ) , {\displaystyle a\oplus b\mathrel {:=} \tan(\arctan a+\arctan b),} 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 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 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}}} 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 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 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 p i (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 )}}
Further values n {\displaystyle n} cl n ϖ {\displaystyle \operatorname {cl} n\varpi } sl n ϖ {\displaystyle \operatorname {sl} n\varpi } 3 7 {\displaystyle {\tfrac {3}{7}}} tanh { 1 2 arcoth [ 1 2 2 cos ( 3 14 π ) cot ( 1 28 π ) + cos ( 1 7 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {3}{14}}\pi )\cot({\tfrac {1}{28}}\pi )}}+\cos({\tfrac {1}{7}}\pi ){\bigr ]}{\bigr \}}} 5 12 {\displaystyle {\tfrac {5}{12}}} 1 2 8 4 [ sin ( 5 24 π ) − 3 4 sin ( 1 24 π ) ] ( 2 3 + 3 4 − 1 ) {\displaystyle {\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]{\Bigl (}{\sqrt[{4}]{2{\sqrt {3}}+3}}-1{\Bigr )}} 1 2 8 4 [ sin ( 5 24 π ) − 3 4 sin ( 1 24 π ) ] ( 2 3 + 3 4 + 1 ) {\displaystyle {\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]{\Bigl (}{\sqrt[{4}]{2{\sqrt {3}}+3}}+1{\Bigr )}} 2 5 {\displaystyle {\tfrac {2}{5}}} 1 2 ( 5 4 − 1 ) ( 5 + 2 − 1 ) {\displaystyle {\tfrac {1}{2}}({\sqrt[{4}]{5}}-1){\bigl (}{\sqrt {{\sqrt {5}}+2}}-1{\bigr )}} 2 5 − 2 4 sin ( 3 20 π ) cos ( 1 20 π ) {\displaystyle 2\,{\sqrt[{4}]{{\sqrt {5}}-2}}\,{\sqrt {\sin({\tfrac {3}{20}}\pi )\cos({\tfrac {1}{20}}\pi )}}} 3 8 {\displaystyle {\tfrac {3}{8}}} ( 2 4 − 1 ) ( 2 + 1 − 2 + 2 ) {\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\bigl (}{\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}{\bigr )}}}} ( 2 4 − 1 ) ( 2 + 1 + 2 + 2 ) {\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\bigl (}{\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}{\bigr )}}}} 5 14 {\displaystyle {\tfrac {5}{14}}} tanh { 1 2 arcoth [ 1 2 2 sin ( 1 7 π ) cot ( 3 28 π ) + sin ( 1 14 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\sin({\tfrac {1}{7}}\pi )\cot({\tfrac {3}{28}}\pi )}}+\sin({\tfrac {1}{14}}\pi ){\bigr ]}{\bigr \}}\ } 3 10 {\displaystyle {\tfrac {3}{10}}} 2 5 − 2 4 sin ( 1 20 π ) cos ( 3 20 π ) {\displaystyle 2\,{\sqrt[{4}]{{\sqrt {5}}-2}}\,{\sqrt {\sin({\tfrac {1}{20}}\pi )\cos({\tfrac {3}{20}}\pi )}}} 1 2 ( 5 4 − 1 ) ( 5 + 2 + 1 ) {\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\bigl (}{\sqrt {{\sqrt {5}}+2}}+1{\bigr )}} 2 7 {\displaystyle {\tfrac {2}{7}}} tanh { 1 2 arcoth [ 1 2 2 cos ( 1 14 π ) tan ( 5 28 π ) + sin ( 3 14 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {1}{14}}\pi )\tan({\tfrac {5}{28}}\pi )}}+\sin({\tfrac {3}{14}}\pi ){\bigr ]}{\bigr \}}\ } 3 14 {\displaystyle {\tfrac {3}{14}}} tanh { 1 2 arcoth [ 1 2 2 cos ( 1 14 π ) tan ( 5 28 π ) + sin ( 3 14 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {1}{14}}\pi )\tan({\tfrac {5}{28}}\pi )}}+\sin({\tfrac {3}{14}}\pi ){\bigr ]}{\bigr \}}\ } 1 5 {\displaystyle {\tfrac {1}{5}}} 1 2 ( 5 4 − 1 ) ( 5 + 2 + 1 ) {\displaystyle {\tfrac {1}{2}}{\bigl (}{\sqrt[{4}]{5}}-1{\bigr )}{\bigl (}{\sqrt {{\sqrt {5}}+2}}+1{\bigr )}} 2 5 − 2 4 sin ( 1 20 π ) cos ( 3 20 π ) {\displaystyle 2\,{\sqrt[{4}]{{\sqrt {5}}-2}}\,{\sqrt {\sin({\tfrac {1}{20}}\pi )\cos({\tfrac {3}{20}}\pi )}}} 1 7 {\displaystyle {\tfrac {1}{7}}} tanh { 1 2 arcoth [ 1 2 2 sin ( 1 7 π ) cot ( 3 28 π ) + sin ( 1 14 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\sin({\tfrac {1}{7}}\pi )\cot({\tfrac {3}{28}}\pi )}}+\sin({\tfrac {1}{14}}\pi ){\bigr ]}{\bigr \}}\ } 1 8 {\displaystyle {\tfrac {1}{8}}} ( 2 4 − 1 ) ( 2 + 1 + 2 + 2 ) {\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\bigl (}{\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}{\bigr )}}}} ( 2 4 − 1 ) ( 2 + 1 − 2 + 2 ) {\displaystyle {\sqrt {{\bigl (}{\sqrt[{4}]{2}}-1{\bigr )}{\bigl (}{\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}{\bigr )}}}} 1 10 {\displaystyle {\tfrac {1}{10}}} 2 5 − 2 4 sin ( 3 20 π ) cos ( 1 20 π ) {\displaystyle 2\,{\sqrt[{4}]{{\sqrt {5}}-2}}\,{\sqrt {\sin({\tfrac {3}{20}}\pi )\cos({\tfrac {1}{20}}\pi )}}} 1 2 ( 5 4 − 1 ) ( 5 + 2 − 1 ) {\displaystyle {\tfrac {1}{2}}({\sqrt[{4}]{5}}-1){\bigl (}{\sqrt {{\sqrt {5}}+2}}-1{\bigr )}} 1 12 {\displaystyle {\tfrac {1}{12}}} 1 2 8 4 [ sin ( 5 24 π ) − 3 4 sin ( 1 24 π ) ] ( 2 3 + 3 4 + 1 ) {\displaystyle {\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]{\Bigl (}{\sqrt[{4}]{2{\sqrt {3}}+3}}+1{\Bigr )}} 1 2 8 4 [ sin ( 5 24 π ) − 3 4 sin ( 1 24 π ) ] ( 2 3 + 3 4 − 1 ) {\displaystyle {\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]{\Bigl (}{\sqrt[{4}]{2{\sqrt {3}}+3}}-1{\Bigr )}} 1 14 {\displaystyle {\tfrac {1}{14}}} tanh { 1 2 arcoth [ 1 2 2 cos ( 3 14 π ) cot ( 1 28 π ) + cos ( 1 7 π ) ] } {\displaystyle \tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {3}{14}}\pi )\cot({\tfrac {1}{28}}\pi )}}+\cos({\tfrac {1}{7}}\pi ){\bigr ]}{\bigr \}}}
Relation to geometric shapes
Arc length of Bernoulli's lemniscate 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 } .
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:[30]
∫ 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[31]
( 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 lemniscate integral and lemniscate functions satisfy an argument duplication identity discovered by Fagnano in 1718:[32]
∫ 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 p i (if any) is a distinct Fermat prime .[33] 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.[34] 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; the theorem refers to constructing the division points only.
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 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 .[35] 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 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
Power series The power series expansion of the lemniscate sine at the origin is[36]
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[37]
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[38]
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[39]
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,[40]
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 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[38]
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,[38] [41]
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 For z ∈ C ∖ { 0 } {\displaystyle z\in \mathbb {C} \setminus \{0\}} :[42]
∫ 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 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:[45] [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:[38] [41] [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} } Two other fast computation methods use the following sum and product series:
This product is described in the textbook A Course of Modern Analysis :[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} } The brothers Peter and Jonathan Borwein described a similar formula in their work π and the AGM on page 60 ff by mentioning the Jacobi elliptic function general case.
In the same pattern following sum formulas can be set up with the help of the tangent duplication theorem:
sl ( ϖ π x ) = f ( 4 π ϖ sin x ∑ n = 1 ∞ cosh [ ( 2 n − 1 ) π ] cosh 2 [ ( 2 n − 1 ) π ] − cos 2 x ) {\displaystyle {\text{sl}}{\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=f{\biggl (}{\frac {4\pi }{\varpi }}\sin x\sum _{n=1}^{\infty }{\frac {\cosh[(2n-1)\pi ]}{\cosh ^{2}[(2n-1)\pi ]-\cos ^{2}x}}{\biggr )}} cl ( ϖ π x ) = f ( 4 π ϖ cos x ∑ n = 1 ∞ cosh [ ( 2 n − 1 ) π ] cosh 2 [ ( 2 n − 1 ) π ] − sin 2 x ) {\displaystyle {\text{cl}}{\Bigl (}{\frac {\varpi }{\pi }}x{\Bigr )}=f{\biggl (}{\frac {4\pi }{\varpi }}\cos x\sum _{n=1}^{\infty }{\frac {\cosh[(2n-1)\pi ]}{\cosh ^{2}[(2n-1)\pi ]-\sin ^{2}x}}{\biggr )}} where f ( x ) = tan ( 2 arctan x ) = 2 x / ( 1 − x 2 ) . {\displaystyle f(x)=\tan(2\arctan x)=2x/(1-x^{2}).}
The lemniscate functions as a ratio of entire functions Since the lemniscate sine is a meromorphic function in the whole complex plane, it can be written as a ratio of entire functions . Gauss showed that sl has the following product expansion, reflecting the distribution of its zeros and poles:[53]
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.[53] [54]
Proof of the infinite product for the lemniscate sine
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 }
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”.[55] Gauss expanded the products for M {\displaystyle M} and N {\displaystyle N} as infinite series. 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}.} Since the functions M {\displaystyle M} and N {\displaystyle N} are entire, their power series expansions converge everywhere in the complex plane:[56] [57] [58]
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} .} 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[59]
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,
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 ) = N ( z ) 4 − 2 N ( z ) 2 M ( z ) 2 − M ( z ) 4 , {\displaystyle S(2z)=N(z)^{4}-2N(z)^{2}M(z)^{2}-M(z)^{4},} T ( 2 z ) = N ( z ) 4 + 2 N ( z ) 2 M ( z ) 2 − M ( z ) 4 {\displaystyle T(2z)=N(z)^{4}+2N(z)^{2}M(z)^{2}-M(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}} for z ∈ C {\displaystyle z\in \mathbb {C} } .
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 theta functions and the above functions are not equivalent).
Relation to other functions
Relation to Weierstrass and Jacobi elliptic functions The lemniscate functions are closely related to the Weierstrass elliptic function ℘ ( z ; 1 , 0 ) {\displaystyle \wp (z;1,0)} (the "lemniscatic case"), with invariants g 2 = 1 and g 3 = 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 g 2 = a , g 3 = 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".[60]
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 g 2 and g 3 . The lemniscate sine is a rational function in the Weierstrass elliptic function and its derivative:[61]
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.[62] [63]
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 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 The inverse function of the lemniscate sine is the lemniscate arcsine, defined as
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}\left({\tfrac {1}{2}},{\tfrac {1}{4}};{\tfrac {5}{4}};x^{4}\right).} 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:
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:
aslh ( x ) = ∫ 0 x 1 y 4 + 1 d y = 1 2 F [ 2 arctan ( x ) ; 1 2 2 ] {\displaystyle \operatorname {aslh} (x)=\int _{0}^{x}{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y={\frac {1}{2}}F{\bigl [}2\arctan(x);{\frac {1}{2}}{\sqrt {2}}\,{\bigr ]}} aclh ( x ) = ∫ x ∞ 1 y 4 + 1 d y = 1 2 F [ 2 arccot ( x ) ; 1 2 2 ] {\displaystyle \operatorname {aclh} (x)=\int _{x}^{\infty }{\frac {1}{\sqrt {y^{4}+1}}}\mathrm {d} y={\frac {1}{2}}F{\bigl [}2\operatorname {arccot}(x);{\frac {1}{2}}{\sqrt {2}}\,{\bigr ]}} aclh ( x ) = ϖ 2 − aslh ( x ) {\displaystyle \operatorname {aclh} (x)={\frac {\varpi }{\sqrt {2}}}-\operatorname {aslh} (x)} aslh ( x ) = 2 arcsl [ x ( x 4 + 1 + 1 ) − 1 / 2 ] {\displaystyle \operatorname {aslh} (x)={\sqrt {2}}\operatorname {arcsl} {\bigl [}x({\sqrt {x^{4}+1}}+1)^{-1/2}{\bigr ]}} arcsl ( x ) = 2 aslh [ x ( 1 + 1 − x 4 ) − 1 / 2 ] {\displaystyle \operatorname {arcsl} (x)={\sqrt {2}}\operatorname {aslh} {\bigl [}x(1+{\sqrt {1-x^{4}}})^{-1/2}{\bigr ]}}
Expression using elliptic integrals The lemniscate arcsine and the lemniscate arccosine can also be expressed by the Legendre-Form:
These functions can be displayed directly by using the incomplete elliptic integral of the first kind:
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):
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:
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 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
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 } [64] ) appears in entry 34e of chapter 11 of Ramanujan's second notebook.[65]
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
Hyperbolic Lemniscate Tangent and Cotangent 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} .[66] 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:
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:
Derivation of the Hyperbolic Lemniscate functions 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.[67] 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 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}}}}
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 This list shows the values of the Hyperbolic Lemniscate Sine accurately:
s l h ( ϖ 2 2 ) = 1 {\displaystyle \mathrm {slh} \,\left({\frac {\varpi }{2{\sqrt {2}}}}\right)=1} s l h ( ϖ 3 2 ) = 1 3 4 2 3 − 3 4 {\displaystyle \mathrm {slh} \,\left({\frac {\varpi }{3{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{3}}}{\sqrt[{4}]{2{\sqrt {3}}-3}}} s l h ( 2 ϖ 3 2 ) = 2 3 + 3 4 {\displaystyle \mathrm {slh} \,\left({\frac {2\varpi }{3{\sqrt {2}}}}\right)={\sqrt[{4}]{2{\sqrt {3}}+3}}} s l h ( ϖ 4 2 ) = 1 2 4 ( 2 + 1 − 1 ) {\displaystyle \mathrm {slh} \,\left({\frac {\varpi }{4{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{2}}}({\sqrt {{\sqrt {2}}+1}}-1)} s l h ( 3 ϖ 4 2 ) = 1 2 4 ( 2 + 1 + 1 ) {\displaystyle \mathrm {slh} \,\left({\frac {3\varpi }{4{\sqrt {2}}}}\right)={\frac {1}{\sqrt[{4}]{2}}}({\sqrt {{\sqrt {2}}+1}}+1)} s l h ( 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 \mathrm {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 )}}} s l h ( 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 \mathrm {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 )}}} s l h ( 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 \mathrm {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 )}}} s l h ( 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 \mathrm {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 )}}} s l h ( ϖ 6 2 ) = 1 2 ( 2 3 + 3 + 1 ) ( 1 − 2 3 − 3 4 ) {\displaystyle \mathrm {slh} \,\left({\frac {\varpi }{6{\sqrt {2}}}}\right)={\frac {1}{2}}({\sqrt {2{\sqrt {3}}+3}}+1)(1-{\sqrt[{4}]{2{\sqrt {3}}-3}})} s l h ( 5 ϖ 6 2 ) = 1 2 ( 2 3 + 3 + 1 ) ( 1 + 2 3 − 3 4 ) {\displaystyle \mathrm {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} 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 In combination with the Hyperbolic Lemniscate Areasine , 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}}}} The square of the Hyperbolic Lemniscate Tangent is the Pythagorean counterpart of the square of the Hyperbolic Lemniscate cotangent because the sum of the fourth powers of tlh {\displaystyle \operatorname {tlh} } and ctlh {\displaystyle \operatorname {ctlh} } is always equal to the value one.
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 )}} 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}}
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 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] [68] 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} .[68]
Hurwitz numbers 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[69] [70]
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} .[71] This yields[69]
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[72]
denom H 4 n = 2 ∏ ( p − 1 ) | 4 n p {\displaystyle \operatorname {denom} \mathrm {H} _{4n}=2\prod _{(p-1)|4n}p} where p ∈ P {\displaystyle p\in \mathbb {P} } such that p ≡ 1 ( mod 4 ) , {\displaystyle p\equiv 1\,({\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 states that
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} , a = a p {\displaystyle a=a_{p}} is uniquely determined and[69]
H 4 n − 1 2 − ∑ ( p − 1 ) | 4 n ( 2 a p ) 4 n / ( p − 1 ) p = def G n ∈ Z , n ≥ 1 , {\displaystyle \mathrm {H} _{4n}-{\frac {1}{2}}-\sum _{(p-1)|4n}{\frac {(2a_{p})^{4n/(p-1)}}{p}}\mathrel {\overset {\text{def}}{=}} \mathrm {G} _{n}\in \mathbb {Z} ,\quad n\geq 1,} sl z = ∑ n = 0 ∞ k 4 n + 1 z 4 n + 1 ( 4 n + 1 ) ! , | z | < ϖ 2 ⟹ k p ≡ 2 a p ( mod p ) . {\displaystyle \operatorname {sl} z=\sum _{n=0}^{\infty }k_{4n+1}{\frac {z^{4n+1}}{(4n+1)!}},\quad \left|z\right|<{\frac {\varpi }{\sqrt {2}}}\implies k_{p}\equiv 2a_{p}\,({\text{mod}}\,p).} The sequence of the integers G n {\displaystyle \mathrm {G} _{n}} starts with 0 , − 1 , 5 , 253 , … . {\displaystyle 0,-1,5,253,\ldots .} [69]
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)} .[73]
Some authors instead define the Hurwitz numbers as H n ′ = H 4 n {\displaystyle \mathrm {H} _{n}'=\mathrm {H} _{4n}} .
Appearances in Laurent series The Hurwitz numbers appear in several Laurent series expansions related to the lemniscate functions:[74]
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 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[75] for these numbers) such that[76]
( 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 .
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