Lemniscate Function

The lemniscate functions arise in rectifying the Arc Length of the Lemniscate. The lemniscate functions were first studied by Jakob Bernoulli and G. Fagnano. A historical account is given by Ayoub (1984), and an extensive discussion by Siegel (1969). The lemniscate functions were the first functions defined by inversion of an integral, which was first done by Gauß.

$$L= 2a\int_0^1 (1-t^4)^{-1/2}\,dt.$$ (1)

Define the functions

$$\phi (x) \equiv {\mathop{\rm arcsinlemn}} x =\int_0^x (1-t^4)^{-1/2}\,dt$$ (2)

$$\phi'(x) \equiv {\mathop{\rm arccoslemn}} x =\int_x^1 (1-t^4)^{-1/2}\,dt,$$ (3)

where

$$\varpi \equiv{L\over a},$$ (4)

and write

$$x = \mathop{\rm sinlemn}\phi$$ (5)

$$x = \mathop{\rm coslemn}\phi'.$$ (6)

There is an identity connecting $\phi$ and $\phi'$ since

$$\phi(x)+\phi'(x)={L\over 2a} = {\textstyle{1\over 2}}\varpi,$$ (7)

so

$$\mathop{\rm sinlemn}\phi = \mathop{\rm coslemn}({\textstyle{1\over 2}}\varpi-\phi).$$ (8)

These functions can be written in terms of Jacobi Elliptic Functions,

$$u=\int_0^{{\rm sd} (u,k)} [(1-k'^2y^2)(1+k^2y^2)]^{-1/2}\,dy.$$ (9)

Now, if $k=k'=1/\sqrt{2}$, then

$$u = \int_0^{{\rm sd}(u,1/\sqrt{2})} [(1-{\textstyle{1\over 2}}y^2)(1+{\textstyle{1\over 2}}y^2)]^{-1/2}\,dy$$

$$= \int_0^{{\rm sd}(u,1/\sqrt{2})} (1-{\textstyle{1\over 4}}y^4)^{-1/2}\,dy.$$ (10)

Let $t\equiv y/\sqrt{2}$ so $dy=\sqrt{2}\,dt$,

$$u=\sqrt{2} \int_0^{{\rm sd}(u,1/\sqrt{2})/\sqrt{2}} (1-t^4)^{-1/2}\,dt$$ (11)

$${u\over \sqrt{2}}=\int_0^{{\rm sd}(u,1/\sqrt{2})/\sqrt{2}} (1-t^4)^{-1/2}\,dt$$ (12)

$$u=\int_0^{{\rm sd}(u\sqrt{2},1/\sqrt{2})/\sqrt{2}} (1-t^4)^{-1/2}\,dt,$$ (13)

and

$$\mathop{\rm sinlemn}\phi = {1\over\sqrt{2}}{\rm sd}\left({\phi\sqrt{2}, {1\over\sqrt{2}}}\right).$$ (14)

Similarly,

$$u = \int^1_{\mathop{\rm cn}\nolimits (u,k)} (1-t^2)^{-1/2}(k'^2+k^2t^2)^{-1/2}\,dt$$

$$= \int^1_{\mathop{\rm cn}\nolimits (u,1/\sqrt{2})} (1-t^2)^{-1/2}\left({{\textstyle{1\over 2}}+{\textstyle{1\over 2}}t^2}\right)^{-1/2}\,dt$$

$$= \sqrt{2} \int^1_{\mathop{\rm cn}\nolimits (u,1/\sqrt{2})} (1-t^4)^{-1/2}\,dt$$ (15)

$${u\over\sqrt{2}} = \int^1_{\mathop{\rm cn}\nolimits (u,1/\sqrt{2})} (1-t^4)^{-1/2}\,dt$$ (16)

$$u=\int^1_{\mathop{\rm cn}\nolimits (u\sqrt{2},1/\sqrt{2})} (1-t^4)^{-1/2}\,dt,$$ (17)

and

$$\mathop{\rm coslemn}\phi=\mathop{\rm cn}\nolimits \left({\phi\sqrt{2},{1\over \sqrt{2}}}\right).$$ (18)

We know

$$\mathop{\rm coslemn}({\textstyle{1\over 2}}\varpi) = \mathop... ...textstyle{1\over 2}}\varpi\sqrt{2},{1\over\sqrt{2}}}\right)=0.$$ (19)

But it is true that

$$\mathop{\rm cn}\nolimits (K,k) = 0,$$ (20)

so

$$K\left({1\over\sqrt{2}}\right)= {\textstyle{1\over 2}}\sqrt{2}\,\varpi = {1\over\sqrt{2}}\varpi$$ (21)

$${\Gamma^2({\textstyle{1\over 4}})\over 4\sqrt{\pi}} = {1\over \sqrt{2}}\varpi$$ (22)

$$L=a\varpi = a\sqrt{2}\,{\Gamma^2({\textstyle{1\over 4}})\ove... ... {\Gamma^2({\textstyle{1\over 4}})\over 2^{3/2}\sqrt{\pi}}\,a.$$ (23)

By expanding $(1-t^4)^{-1/2}$ in a Binomial Series and integrating term by term, the arcsinlemn function can be written

$$\phi(x)=\int_0^v {dt\over\sqrt{1-t^4}} = \sum_{n=0}^\infty {({\textstyle{1\over 2}})_n x^{4n+1}\over n!(4n+1)},$$ (24)

where $(a)_n$ is the Rising Factorial (Berndt 1994). Ramanujan gave the following inversion Formula for $\phi(x)$. If

$${\theta\mu\over\sqrt{2}}=\sum_{n=0}^\infty {({\textstyle{1\over 2}})_n x^{4n+1}\over n!(4n+1)},$$ (25)

where

$$\mu={\Gamma^2({\textstyle{1\over 4}})\over 2\pi^{3/2}}$$ (26)

is the constant obtained by letting $x=1$ and $\theta=\pi/2$, and

$$v=2^{-1/2} \mathop{\rm sd}\nolimits (\mu\theta),$$ (27)

then

$${\mu^2\over 2x^2}=\csc^2\theta-{1\over\pi}-8\sum_{n=1}^\infty {n\cos(2n\theta)\over e^{2\pi n}-1}$$ (28)

(Berndt 1994). Ramanujan also showed that if $0<\theta<\pi/2$, then

$$-{\mu\over\sqrt{2}} \sum_{n=0}^\infty {({\textstyle{1\over 2... ...er\pi} +4\sum_{n=1}^\infty {\sin(2n\theta)\over 2^{2\pi n}-1},$$ (29)

$$\ln v+{\textstyle{1\over 6}}\pi-{\textstyle{1\over 2}}\ln 2+... ...\pi}-2\sum_{n=1}^\infty {\cos(2n\theta)\over n(e^{2\pi n}-1)},$$ (30)

$${\textstyle{1\over 2}}\tan^{-1} v=\sum_{n=0}^\infty {\sin[(2n+1)\theta]\over (2n+1)\cosh[{\textstyle{1\over 2}}(2n+1)\pi]},$$ (31)

$${\textstyle{1\over 4}}\cos^{-1}(v^2)=\sum_{n=0}^\infty {(-1)... ...+1)\theta]\over (2n+1)\cosh[{\textstyle{1\over 2}}(2n+1)\pi]},$$ (32)

and

$${\sqrt{2}\over 4\mu}\sum_{n=0}^\infty {2^{2n}(n!)^2\over (2n... ...+1)\theta]\over(2n+1)^2\cosh[{\textstyle{1\over 2}}(2n+1)\pi]}$$ (33)

(Berndt 1994).

A generalized version of the lemniscate function can be defined by letting $0\leq\theta\leq\pi/2$ and $0\leq v\leq 1$. Write

$${\textstyle{2\over 3}}\theta\mu=\int_0^v {dt\over\sqrt{1-t^6}},$$ (34)

where $\mu$ is the constant obtained by setting $\theta=\pi/2$ and $v=1$. Then

$$\mu={\sqrt{\pi}\over\Gamma({\textstyle{2\over 3}})\Gamma({\textstyle{5\over 6}})},$$ (35)

and Ramanujan showed

$${4\mu^2\over 9v^2}=\csc^2\theta-{2\over\pi\sqrt{3}}+8\sum_{n... ...nfty {(-1)^{n-1}n\cos(2n\theta)\over e^{\pi n\sqrt{3}}-(-1)^n}$$ (36)

(Berndt 1994).

See also Hyperbolic Lemniscate Function

References

Ayoub, R. ``The Lemniscate and Fagnano's Contributions to Elliptic Integrals.'' Arch. Hist. Exact Sci. 29, 131-149, 1984.

Berndt, B. C. Ramanujan's Notebooks, Part IV. New York: Springer-Verlag, pp. 245, and 247-255, 258-260, 1994.

Siegel, C. L. Topics in Complex Function Theory, Vol. 1. New York: Wiley, 1969.