## Saturday, 14 January 2012

### Lemniscate functions

The first time I encountered these weird objects of analysis was probably while surfing the book of E. Kamke which in turn gave a reference to Whittakker and Watson. I was at that time delving into the methods of analytic geometry and only new that lemniscate was an 8-shaped algebraic curve of the 4th order, a particular case of Cassini ovals. So I was pretty shocked to find out that it could give rise to some "trigonometric" system.
Being fresh from the first course on calculus, I attempted an investigation of the properties of lemniscate functions by means of only the very basic techniques used derive similar results for circular functions. I wanted to derive formulas for derivatives and primitives, addition theorems, complementary formulas, etc.
However, looking back at that paper I see that in the most crucial steps I just took the known relations for Jacobi elliptic functions (from W&W book) and then reduced them to the particular case of lemniscate functions.
I think that lemniscate functions can be used for changing variable when evaluating certain integrals, or converting differential equations into manageable forms. Although in my rather limited practice I have never encountered the cases where it would be appropriate, I still want to make a small, but independent account of these devices and keep them in my arsenal waiting for the right moment to come.
So, enough of the words, let's get down to business.
Lemniscate functions arise when rectifying the length of lemniscate and are defined by inversion of an integral (see Wolfram article for intermediate steps)
$$\phi=\int_0^{\operatorname{sl}\phi}\frac{dt}{\sqrt{1-t^4}}\qquad \phi=\int_{\operatorname{cl}\phi}^1\frac{dt}{\sqrt{1-t^4}}$$

I am using the Gaussian notation instead of the more lengthy sinlemn and coslemn.
Take the first integral and differentiate both sides by $$\phi$$:

$$1=\operatorname{sl}'\phi\frac{1}{\sqrt{1-\operatorname{sl}^4\phi}}$$
$$\operatorname{sl}'\phi=\sqrt{1-\operatorname{sl}^4\phi}$$
$$\operatorname{sl}'^2\phi=1-\operatorname{sl}^4\phi$$
$$2\operatorname{sl}'\phi \operatorname{sl}''\phi=-4\operatorname{sl}^3\phi \operatorname{sl}'\phi$$
$$\operatorname{sl}''\phi=-2\operatorname{sl}^3\phi$$

The same would hold, if we started with the second integral. Hence we obtain the differential equation for lemniscate functions:
$$y''=-2y^3$$

Now we shall find an algebraic relation between sl and cl.
$$\phi=\int_{\operatorname{cl}\phi}^1\frac{dt}{\sqrt{1-t^4}}=\int_{\operatorname{cl}\phi}^1\frac{dt}{\sqrt{\frac{1-t^2}{1+t^2}}(1+t^2)}$$
Substitute $$t^2=\frac{1-z^2}{1+z^2}$$. Then
$$2tdt=\frac{-2z(1+z^2)-2z(1-z^2)}{(1+z^2)^2}dz$$
$$tdt=-\frac{2z^3dx}{(1+z^2)^2}$$
$$dt=-\sqrt{\frac{1+z^2}{1-z^2}}\frac{2z^3dz}{(1+z^2)^2}$$

Inserting it all in the integral and observing new integration limits we obtain:
$$\phi=-\int_{\sqrt{\frac{1-\operatorname{cl}^2\phi}{1+\operatorname{cl}^2\phi}}}^0\frac{1}{z}\frac{(1+z^2)}{2z^2}\sqrt{\frac{1+z^2}{1-z^2}}\frac{2z^3dz}{(1+z^2)^2}=\int_0^{\sqrt{\frac{1-\operatorname{cl}^2\phi}{1+\operatorname{cl}^2\phi}}}\frac{dz}{\sqrt{\frac{1-z^2}{1+z^2}}(1+z^2)}=$$
$$=\int_0^{\sqrt{\frac{1-\operatorname{cl}^2\phi}{1+\operatorname{cl}^2\phi}}}\frac{dz}{\sqrt{1-z^4}}=\int_0^{\sqrt{\frac{1-\operatorname{cl}^2\phi}{1+\operatorname{cl}^2\phi}}}\frac{dt}{\sqrt{1-t^4}}$$

Now comparing this with the integral defining $$sl\phi$$ and looking at the limits we conclude that:
$$\operatorname{sl}^2\phi=\frac{1-\operatorname{cl}^2\phi}{1+\operatorname{cl}^2\phi}$$
Conversely:
$$\operatorname{cl}^2\phi=\frac{1-\operatorname{sl}^2\phi}{1+\operatorname{sl}^2\phi}$$
Now it is easy to establish the expressions for derivatives:
$$sl'\phi=\sqrt{1-\operatorname{sl}^4\phi}=\sqrt{\frac{1-\operatorname{sl}^2\phi}{1+\operatorname{sl}^2\phi}}(1+\operatorname{sl}^2\phi)=\operatorname{cl}\phi(1+\operatorname{sl}^2\phi)$$

Formula for cl can be obtained similarly, but we can follow a different path using complimentary formula.
Rewrite the definitions in the following way:
$$\phi=\operatorname{sl}^{-1}x=\int_0^x\frac{dt}{\sqrt{1-t^4}}\qquad \overline\phi=\operatorname{cl}^{-1}x=\int_x^1\frac{dt}{\sqrt{1-t^4}}$$
Then
$$\phi+\overline\phi=\operatorname{sl}^{-1}x+\operatorname{cl}^{-1}x=\int_0^1\frac{dt}{\sqrt{1-t^4}}=const$$
$$x=\operatorname{sl}\phi$$
$$x=\operatorname{cl}\bar{\phi}$$

So
$$\operatorname{sl}\phi=\operatorname{cl}(C-\phi)$$
Then we immediately obtain:
$$\operatorname{cl}'\phi=-\operatorname{sl}\phi(1+\operatorname{cl}^2\phi)$$
The constant value which is half the length of the "unitary" lemniscate can be evaluated substituting $$u=1-t^4$$ in the integral:
$$C=-\frac{1}{4}\int_1^0u^{-\frac{1}{2}}(1-u)^{-\frac{3}{4}}du=\frac{1}{4}\int_0^1u^{-\frac{1}{2}}(1-u)^{-\frac{3}{4}}du=\frac{1}{4}B(\frac{1}{2},\frac{1}{4})=$$
$$=\frac{1}{4}\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=\frac{[\Gamma(\frac{1}{4})]^2}{4\sqrt{2}\sqrt{\pi}}$$

Integral of the lemniscate function can be easily calculated:
$$\int \operatorname{sl}\phi d\phi=\int td(\operatorname{sl}^{-1}t)=\int\frac{tdt}{\sqrt{1-t^4}}=\frac{1}{2}\int\frac{d(t^2)}{\sqrt{1-(t^2)^2)}}=$$
$$=\frac{\arcsin{t^2}}{2}+C=\frac{\arcsin{\operatorname{sl}^2\phi}}{2}+C$$