# How to use Lemniscate sine and Lemniscate cosine elliptic integrals?

I’ve been reading up on lemniscate sine and cosine functions and found that they are essentially the lemniscate analogues of circular sine and cosine functions. The following wiki page elaborates a bit more on this I’ve highlighted the parts of the article I didn’t quite understand. For example, what do variables s and c represent? What does t represent?

Below I’ve attached a visualization of the mental boundary I’m trying to overcome. Is it that using the lemniscate sine and lemniscate cosine function give the lengths of the y and x components of the hypotenuse drawn from the origin (0,0) to endpoint of the lemniscate arc length?

Just take a point $$P$$ in first quadrant which lies on lemniscate. Traverse the curve from origin to $$P$$ in first quadrant. Let the length of this traversed part of curve be $$l$$ then line segment $$OP$$ has a length equal to $$\operatorname {sl} (l)$$.

On the other hand if $$l'$$ is the length of curve traversed from point $$(1,0)$$ to $$P$$ (see yellow part in your image) in first quadrant then we have $$OP=\operatorname{cl} (l')$$.

Further note that neither $$x$$ nor $$y$$ of your image but the length $$OP =\sqrt{x^2+y^2}$$ is given by these lemniscatic functions.

Do you now understand what Wikipedia says? Let me know if some more clarification is needed.

The polar equation of lemniscate is $$r^2=\cos 2\theta$$ and hence the arc-length $$l'$$ from $$(1,0)$$ to $$P=(\rho\cos\phi, \rho\sin\phi)$$ is given by $$l'=\int_{0}^{\phi}\sqrt{r^2+\left(\frac{dr}{d\theta}\right) ^2}\,d\theta =\int_{0}^{\phi}\frac{d\theta}{\sqrt{\cos 2\theta}}$$ Putting $$t^2=\cos 2\theta$$ and noting that $$\rho=\sqrt{\cos 2\phi}$$ we get $$l'=\int_{\rho} ^{1}\frac {dt} {\sqrt{1-t^4}}$$ Note that $$\rho=OP=\operatorname{cl} (l')$$ so we have $$l'=\int_{\operatorname {cl} (l')} ^{1}\frac{dt}{\sqrt{1-t^4}}$$ and similarly $$l=\int_{0}^{\operatorname {sl} (l)} \frac{dt} {\sqrt{1-t^4}}$$ and you get the formulas in Wikipedia (Wikipedia article replaces both $$l, l'$$ by $$r$$ and writes $$s=\operatorname {sl} (r), c=\operatorname {cl} (r)$$).

The functions defined above can be expressed in terms of Jacobian elliptic functions of modulus $$k=1/\sqrt {2}$$ (and corresponding nome $$q=e^{-\pi}$$) as $$\operatorname {sl} (u) =\frac{1}{\sqrt{2}}\operatorname {sd} (\sqrt{2}u,k),\operatorname {cl} (u) =\operatorname{cn} (\sqrt{2}u,k)$$ These elliptic functions can then be evaluated using their Fourier series given below: \begin{align} \operatorname {sd} (u, k) &=\frac{2\pi}{Kkk'}\sum_{n=0}^{\infty} (-1)^n\cdot\frac{q^{n+(1/2)}\sin((2n+1)\pi u/(2K))}{1+q^{2n+1}}\notag\\ \operatorname {cn} (u, k) &=\frac{2\pi}{Kk}\sum_{n=0}^{\infty} \frac{q^{n+(1/2)}\cos((2n+1)\pi u/(2K))} {1+q^{2n+1}} \notag \end{align}

• This makes a lot more sense! I now understand why the upper and lower limits differ between the two functions. I think I mostly understand the derivation and was wondering how it would be put into application. For example, say I had a lemniscate with arclength 10 cm if I started this traverse from origin O (0,0). What would the length of line segment OP (which we justified to be = sl(arclength) in this case) be? (My approach was to first find a way to integrate the integral, then isolate for sl(arclength) after inputting both sl(arclength) and 0 into t, where I got (tan r)^0.5 = sl(r))
– Mas
May 9, 2020 at 20:34
• @Thunderbolt114: you want to find $\operatorname {sl} (10)$. Then note that this function is periodic with period $4\omega$ where $$\omega=2\int_{0}^{1}\frac{dt}{\sqrt{1-t^4}}$$ (when you have traversed all the 4 quadrants things begin to repeat, $\omega$ is the arc-length for one quadrant). Using a numerical value of $\omega$ we can essentially figure out the quadrant in which $P$ lies when arc-length is $10$. May 10, 2020 at 0:45
• Thank you for the clarification! If you don’t mind, I was wondering what the steps would look like if one was to solve for sl(10). That way, I can better visualize how the original Wikipedia formula could be used, and how extensions of this kind of application can be evaluated
– Mas
May 10, 2020 at 1:02
• @Thunderbolt114: well the integrals give you the inverse of lemniscatic sine function. To get the lemniscatic function from these integrals is not easy. You need to study elliptic functions in general. May 10, 2020 at 1:05
• Do you have any source recommendations on where one can learn how to obtain a lemniscate function from their inverse counterparts?
– Mas
May 10, 2020 at 1:24

These integrals occur when integrating the quartic potential well:

$$T = m \dot{x}^2 / 2$$, $$V = m k^2 x^4 / 2$$

thus $$L = - m \ddot{x} + 2 m k^2 x^3$$

which gives an equation of motion $$\ddot{x} = 2 k^2 x^3$$

The equation $$\ddot{x} = F(x)$$ can be solved by multiplying both sides by $$\dot{x}$$ and integrating by parts: $$\frac{d}{dt} \left( \frac{\dot{x}^2}{2} \right) = \ddot{x} \dot{x} = 2 k^2 \dot{x} x^3 = 2 k^2 \frac{d}{dt} \left( \frac{x^4}{4} \right)$$

After some manipulation you obtain Gauss's elliptic integral:

$$k(t - t_0 ) = \int_0^x \frac{ds}{\sqrt{a^4 - s^4}}$$

whose inverse gives the solution $$x(t) = a\text{sl}(ak(t-t_0 ))$$.

The integration constant $$a$$ arises from the first integral and $$t_0$$ from the second.