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Re: solving for a second function

• To: mathgroup at smc.vnet.net
• Subject: [mg102344] Re: [mg102308] solving for a second function
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Thu, 6 Aug 2009 06:30:57 -0400 (EDT)
• References: <200908050942.FAA18127@smc.vnet.net>

Roger Bagula wrote:
> Yesterday I got an half Lemniscate complex function:
> a = 1; b = 1;
> f[t_] = (a*1 + b*I)/(a*1 + b*I*Exp[t]) - 1
> FullSimplify[f[t]/f[t/2]]
> gives:
> (1 + 1/(Cosh[t/2] + I*Sinh[t/2]))
> I want to solve for the other function:
> g[t]/g[t/2]=(1 + 1/(-Cosh[t/2] + I*Sinh[t/2]))
> 
> Graphics showing this:x = Re[1 + 1/(Cosh[t] + I*Sinh[t])];
> y = Im[1 + 1/(Cosh[t] + I*Sinh[t])];
> z = x + I*y;
> g1 = ParametricPlot[{x, y}, {t, -5*Pi, 5*Pi}, AspectRatio ->
> Automatic]
> (*The other half is: *)
> x = Re[1 + 1/(-Cosh[t] + I*Sinh[t])];
> y = Im[1 + 1/(-Cosh[t] + I*Sinh[t])];
> z = x + I*y;
> g0 = ParametricPlot[{x, y}, {t, -5*Pi, 5*Pi}, AspectRatio ->
> Automatic]
> Show[{g1, g0}]
> 
> Any help will be appreciated.
> Roger Bagula

Not quite in the requested form, but here goes.

Start with:

x = Re[1 + 1/(Cosh[t] + I*Sinh[t])];
y = Im[1 + 1/(Cosh[t] + I*Sinh[t])];

We translate so it is centered at the origin, then form a 
parametrization in terms of rational functions, by converting to 
exponentials and swapping Exp[t] for a new variable, r.

ratparam =
  Together[TrigToExp[
     ComplexExpand[TranslationTransform[{-1, 0}][{x, y}]]] /.
    Exp[n_.*t] -> r^n]

 From here we observe that the other parametrization is from swapping 
x<-->-x, or alternatively by a 108 degree rotation. We compute this.

otherratparam = RotationTransform[Pi][ratparam];

Now translate back to the original coordinates (where the figure is 
centered at {1,0}, then convert to exponentials.

In[359]:= otherratparamtrans =
  Simplify[TranslationTransform[{1, 0}][otherratparam] /. r -> Exp[t]]

Out[359]= {1 - (E^t + E^(3 t))/(1 + E^(4 t)), (E^t (-1 + E^(2 t)))/(
  1 + E^(4 t))}

Daniel Lichtblau
Wolfram Research

• References:
  • solving for a second function
    • From: Roger Bagula <roger.bagula@gmail.com>