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Valentin Fadeev

Lemniscate functions: the lost symbols

Wednesday, 12 May 2010, 03:01
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Edited by Valentin Fadeev, Sunday, 18 Sept 2011, 23:25

The first time I encountered these weird objects of analysis was probably while surfing the book of E. Kamke "A reference of ordinary differential equations" 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.

Although these functions were already studied by Gauss (no surprise), most of the original and subsequent research concentrated on different series expansions and evaluation of particular elliptic integrals.

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 http://mathworld.wolfram.com/LemniscateFunction.html for intermediate steps):

equation sequence phi equals integral over zero under s times l times phi d times t divided by Square root of one minus t super four times phi equals integral over c times l times phi under one d times t divided by Square root of one minus t super four

(I am using the original Gaussian (and my own ;)) notation instead of the more lengthy sinlemn and coslemn)

Take the first integral and differentiate both sides by phi :

MathJax failure: TeX parse error: Extra open brace or missing close brace

equation left hand side s times l super prime times phi equals right hand side Square root of one minus s times l super four times phi

equation left hand side s times l super prime two times phi equals right hand side one minus s times l super four times phi

equation left hand side two times s times l super prime times phi times s times l super double prime times phi equals right hand side negative four times s times l cubed times phi times s times l super prime times phi

equation left hand side s times l super double prime times phi equals right hand side negative two times s times l cubed times phi

The same would hold, if we started with the second integral. Hence we obtain the differential equation for lemniscate functions:

equation left hand side y super double prime equals right hand side negative two times y cubed

Now we shall find an algebraic relation between sl and cl.

equation sequence phi equals integral over c times l times phi under one d times t divided by Square root of one minus t super four equals integral over c times l times phi under one d times t divided by Square root of one minus t squared divided by one plus t squared times open one plus t squared close

Substitute equation left hand side t squared equals right hand side one minus z squared divided by one plus z squared . Then

equation left hand side two times t times d times t equals right hand side negative two times z times open one plus z squared close minus two times z times open one minus z squared close divided by left parenthesis one plus z squared times right parenthesis squared times d times z

equation left hand side t times d times t equals right hand side negative two times z cubed times d times x divided by left parenthesis one plus z squared times right parenthesis squared

equation left hand side d times t equals right hand side negative Square root of one plus z squared divided by one minus z squared times two times z cubed times d times z divided by left parenthesis one plus z squared times right parenthesis squared

Inserting it all in the integral and observing new integration limits we obtain:

equation sequence phi equals negative integral over Square root of one minus c times l squared times phi divided by one plus c times l squared times phi under zero one divided by z times open one plus z squared close divided by two times z squared times Square root of one plus z squared divided by one minus z squared times two times z cubed times d times z divided by left parenthesis one plus z squared times right parenthesis squared equals integral over zero under Square root of one minus c times l squared times phi divided by one plus c times l squared times phi d times z divided by Square root of one minus z squared divided by one plus z squared times open one plus z squared close equals integral over zero under Square root of one minus c times l squared times phi divided by one plus c times l squared times phi d times z divided by Square root of one minus z super four equals integral over zero under Square root of one minus c times l squared times phi divided by one plus c times l squared times phi d times t divided by Square root of one minus t super four

Now comparing this with the integral defining s times l times phi and looking at the limits we conclude that:

equation left hand side s times l squared times phi equals right hand side one minus c times l squared times phi divided by one plus c times l squared times phi

Conversely:

equation left hand side c times l squared times phi equals right hand side one minus s times l squared times phi divided by one plus s times l squared times phi

Now it is easy to establish the expressions for derivatives:

equation sequence s times l super prime times phi equals Square root of one minus s times l super four times phi equals Square root of one minus s times l squared times phi divided by one plus s times l squared times phi times open one plus s times l squared times phi close equals c times l times phi times open one plus s times l squared times phi close

Formula for cl can be obtained similarly, but we can follow a different path using complimentary formula.

Rewrite the definitions in the following way:

equation sequence phi equals s times l super negative one times x equals integral over zero under x d times t divided by Square root of one minus t super four times phi macron equals c times l super negative one times x equals integral over x under one d times t divided by Square root of one minus t super four

Then:

equation sequence phi plus phi macron equals s times l super negative one times x plus c times l super negative one times x equals integral over zero under one d times t divided by Square root of one minus t super four equals c times o times n times s times t

x equals s times l times phi

x equals c times l times phi macron

So

equation left hand side s times l times phi equals right hand side c times l times open cap c minus phi close

Then we immediately obtain:

equation left hand side c times l super prime times phi equals right hand side negative s times l times phi times open one plus c times l squared times phi close

The constant value which is half the length of the "unitary" lemniscate can be evaluated substituting one minus t super four equals u in the integral:

cap c equals negative one divided by four times integral over one under zero u super negative one divided by two left parenthesis equation left hand side one minus u times right parenthesis super negative three divided by four times d times u equals right hand side one divided by four times integral over zero under one u super negative one divided by two left parenthesis equation sequence one minus u times right parenthesis super negative three divided by four times d times u equals one divided by four times cap b of one divided by two comma one divided by four equals one divided by four times normal cap gamma of one divided by two times normal cap gamma of one divided by four divided by normal cap gamma of three divided by four equals left square bracket normal cap gamma of one divided by four times right square bracket squared divided by four times Square root of two times Square root of pi

Integral of the lemniscate function is easily calculated:

equation sequence integral s times l times phi d phi equals integral t times d of s times l super negative one times t equals integral t times d times t divided by Square root of one minus t super four equals one divided by two times integral d of t squared divided by Square root of one minus open t squared times right parenthesis squared close equals arc sine of t squared divided by two equals arc sine of s times l squared times phi divided by two

Now I am only one step away from deriving the addition theorem using the same method due to Euler that works for Jacobi elliptic functions, but got somewhat lost in the algebra.. Hope to post that later on.

Tags: lemniscate function
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