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 enter image description here

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. enter image description here

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?

The Wikipedia page: https://en.wikipedia.org/wiki/Lemniscatic_elliptic_function

Thank you in advance!


2 Answers 2


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}

  • $\begingroup$ 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)) $\endgroup$
    – Mas
    May 9, 2020 at 20:34
  • 1
    $\begingroup$ @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 $10$. $\endgroup$
    – Paramanand Singh
    May 10, 2020 at 0:45
  • $\begingroup$ 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 $\endgroup$
    – Mas
    May 10, 2020 at 1:02
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    May 10, 2020 at 1:05
  • $\begingroup$ Do you have any source recommendations on where one can learn how to obtain a lemniscate function from their inverse counterparts? $\endgroup$
    – 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.


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