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?
The Wikipedia page:
Thank you in advance!
$\endgroup$2 Answers
$\begingroup$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}
$\endgroup$ 8 $\begingroup$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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