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Re: Inverse function solution

• To: mathgroup at smc.vnet.net
• Subject: [mg132654] Re: Inverse function solution
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Tue, 29 Apr 2014 01:33:51 -0400 (EDT)
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• References: <20140428014444.9FBD16A4E@smc.vnet.net>

In Mathematica 9.0.1, I'm getting:

      Solve[{x == Cos[u], y == Cos[u + v]}, {u, v}, Reals]
    {{u -> ConditionalExpression[-ArcCos[x] +
     2 \[Pi] C[1], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1],
  v -> ConditionalExpression[
    ArcCos[x] - ArcCos[y] - 2 \[Pi] C[1] +
     2 \[Pi] C[2], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1]}, {u ->
   ConditionalExpression[-ArcCos[x] +
     2 \[Pi] C[1], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1],
  v -> ConditionalExpression[
    ArcCos[x] + ArcCos[y] - 2 \[Pi] C[1] +
     2 \[Pi] C[2], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1]}, {u ->
   ConditionalExpression[
    ArcCos[x] +
     2 \[Pi] C[1], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1],
  v -> ConditionalExpression[-ArcCos[x] - ArcCos[y] - 2 \[Pi] C[1] +
     2 \[Pi] C[2], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1]}, {u ->
   ConditionalExpression[
    ArcCos[x] +
     2 \[Pi] C[1], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1],
  v -> ConditionalExpression[-ArcCos[x] + ArcCos[y] - 2 \[Pi] C[1] +
     2 \[Pi] C[2], (C[1] | C[2]) \[Element] Integers && -1 <= x <=
      1 && -1 <= y <= 1]}}

Or, a bit simpler:

        Reduce[{x == Cos[u], y == Cos[u + v]}, {u, v}, Reals]
    (C[1] | C[2]) \[Element] Integers && -1 <= x <= 1 && -1 <= y <=
  1 && ((u == -ArcCos[x] +
       2 \[Pi] C[1] && (v ==
        ArcCos[x] - ArcCos[y] - 2 \[Pi] C[1] + 2 \[Pi] C[2] ||
       v == ArcCos[x] + ArcCos[y] - 2 \[Pi] C[1] +
         2 \[Pi] C[2])) || (u ==
      ArcCos[x] +
       2 \[Pi] C[1] && (v == -ArcCos[x] - ArcCos[y] - 2 \[Pi] C[1] +
         2 \[Pi] C[2] ||
       v == -ArcCos[x] + ArcCos[y] - 2 \[Pi] C[1] + 2 \[Pi] C[2])))
On Apr 27, 2014, at 9:44 PM, Narasimham <mathma18 at gmail.com> wrote:

> 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?
>
> Narasimham
>

Murray Eisenberg                                murray at math.umass.edu
Mathematics & Statistics Dept.      
Lederle Graduate Research Tower      phone 240 246-7240 (H)
University of Massachusetts               
710 North Pleasant Street                
Amherst, MA 01003-9305

• References:
  • Inverse function solution
    • From: Narasimham <mathma18@gmail.com>