MathGroup Archive 2014

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Inverse function solution


Am 28.04.2014 03:45, schrieb Narasimham:
> Solve[ {x == Cos[u], y == Cos[u + v] }, {u, v} ]
>
> Its closed/analytic solution is not possible, even numerically.
>
> The known solutions are ellipses from sine waves with a phase difference, having x^2, x y and y^2 terms, as also sketched in Lissajous curves:
>
> ParametricPlot[{Cos[u], Cos[u + v]}, {u, -Pi, Pi}, {v, -Pi, Pi}]
>
> Can there be a work around?
>

Yes, replace trigonometric functions by rationals of exponentials.

{x ==  Cos[u], y == Cos[u + v]} // TrigExpand // TrigToExp

{x == E^(-I u)/2 + E^(I u)/2,
  y == 1/2 E^(-I u - I v) + 1/2 E^(I u + I v)}

Solve[ {2 x ==  q + 1/q, 2 y == p q + 1/(p q)}, {q, p}]

q-> Exp[I u], p-> Exp[I v]

-- 

Roland Franzius



  • Prev by Date: Re: Inverse function solution
  • Next by Date: Differential Equation
  • Previous by thread: Re: Inverse function solution
  • Next by thread: Re: Inverse function solution