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Re: How to find more solutions for an periodical equation with infinity solutions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122923] Re: How to find more solutions for an periodical equation with infinity solutions
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 16 Nov 2011 04:46:14 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201111151050.FAA23758@smc.vnet.net>

On 15 Nov 2011, at 11:50, Gy Peng wrote:

> Dear all,
>
> I have a matrix defined as:
> Matrix[{\[Alpha]_, \[Beta]_, \[Gamma]_}, \[Theta]_] := {{\[Alpha]^2 \
> (1 - Cos[\[Theta]]) +
>    Cos[\[Theta]], \[Alpha] \[Beta] (1 -
>       Cos[\[Theta]]) - \[Gamma] Sin[\[Theta]], \[A lpha] \[Gamma] (1 -
>        Cos[\[Theta]]) + \[Beta] Sin[\[Theta]]}, {\[Alpha] \[Beta] (1 \
> - Cos[\[Theta]]) + \[Gamma] Sin[\[Theta]], \[Beta]^2 (1 -
>       Cos[\[Theta]]) +
>    Cos[\[Theta]], \[Beta] \[Gamma] (1 -
>       Cos[\[Theta]]) - \[Alpha] Sin[\[Theta]]}, {\[Alpha] \[Gamma] \
> (1 - Cos[\[Theta]]) - \[Beta] Sin[\[Theta]], \[Beta] \[Gamma] (1 -
>       Cos[\[Theta]]) + \[Alpha] Sin[\[Theta]], \[Gamma]^2 (1 -
>       Cos[\[Theta]]) + Cos[\[Theta]]}}
>
> Solve[{x, y, z} == Matrix[{0, 0, 1}, \[Theta]].{x, y, z}, \[Theta]]
>
> I know this equation is periodical and has infinity solutions. So, Mathematica only gave me only one solution: {{\[Theta] -> 0}} and show me the message:
>
> Solve::ifun: Inverse functions are being used by Solve, so some solutions
> may not be found; use Reduce for complete solution information. >>
>
> My question is, how could I see for example first 6 solution, because I
> know the first 5 or 6 solutions should be different and then
> repeat themselves periodically. What should I do to find the first 5 or 6
> solutions?
>
> Thank you very much!


Mathematica told you precisely what to do so why don't you do that?

Reduce[{x, y, z} ==
  Matrix[{0, 0, 1}, \[Theta]] . {x, y, z}, \[Theta]]

(Element[C[1], Integers] && (\[Theta] == 2*Pi*C[1] ||
         (x == 0 && y == 0 && \[Theta] == 2*Pi*C[1] + Pi))) ||
   (NotElement[(\[Theta] - Pi)/(2*Pi), Integers] && x == 0 && y == 0)

If you know that either x or y is not zero then the only solutions integer multiples of 2Pi

Simplify[%, x != 0]

Element[C[1], Integers] && \[Theta] == 2*Pi*C[1]

Andrzej Kozlowski






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