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Solving equation in terms of symbols

  1. Nov 13, 2012 #1
    Hi friends
    I wanna solve an equation which is function of x,y,z,m but i want to find answer like this m=... i want my answer(m) to be a function of x, y, z
    but when i solve it in matlab or mathematica none of them give me the answer
    I know that my equation must have 4 answer for "m" in terms of x , y , z

    in code of MATLAB my equation is:

    -36 + m* (-64 - 24 *m + m^3) - 16* (2 + m)^2 *cos(x) + 8* cos(2*x) - 28* cos(x - y) + 2 *cos(2 *(x - y)) - 64 *cos(y) - 8* (2* m *(4 + m) + (1 + m)* (7 + m)* cos(x)) *cos(y) + 8 *cos(2*y) - 28* cos(x + y) + 2 *cos(2 *(x + y)) - 64 *cos(z) - 8* (2* m* (4 + m) + (14 + m *(8 + m) + 2 *cos(2*x)) *cos(y) + cos(x) *(14 + m *(8 + m) + 8 *(3 + m) *cos(y) + 2 *cos(2*y))) *cos(z) + 8 *(cos(x) - cos(y))^2 *cos(2*z)==0



    i want to solve it and find "m" which is in terms of x,y,z

    thank for your guidence
     
  2. jcsd
  3. Nov 13, 2012 #2

    Simon Bridge

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    But surely you can do it by hand?

    $$
    -36 + m(-64 - 24 m + m^3) - 16(2 + m)^2 \cos(x) + 8\cos(2x) - 28\cos(x - y)
    + 2\cos(2(x - y)) - 64\cos(y)\\ - 8(2m(4 + m) + (1 + m)(7 + m)\cos(x))\cos(y) + 8\cos(2y) - 28\cos(x + y) + 2\cos(2(x + y)) - 64\cos(z)\\ - 8(2m(4 + m) + (14 + m(8 + m) + 2\cos(2x))\cos(y) + (14 + m(8 + m) + 8(3 + m)\cos(y) + 2\cos(2y))\cos(x))\cos(z)\\ + 8(\cos(x) - \cos(y))^2\cos(2z)=0$$... is basically finding the roots of a quartic in m:
    http://en.wikipedia.org/wiki/Quartic_function

    ... expand the brackets and group the terms.
    I know it's painful: use a whiteboard or a large window.
     
  4. Nov 14, 2012 #3
    Mathematica finds the four exact solutions of this in a few seconds.

    m/.Solve[-36 + m*(-64 - 24*m + m^3) - 16*(2 + m)^2*Cos[x] + 8*Cos[2*x] - 28*Cos[x - y] + 2*Cos[2*(x - y)] - 64*Cos[y] - 8*(2*m*(4 + m) + (1 + m)*(7 + m)*Cos[x])*Cos[y] + 8*Cos[2*y] - 28*Cos[x + y] + 2*Cos[2*(x + y)] - 64*Cos[z] - 8*(2*m*(4 + m) + (14 + m*(8 + m) + 2*Cos[2*x])*Cos[y] + Cos[x]*(14 + m*(8 + m) + 8*(3 + m)*Cos[y] + 2*Cos[2*y]))*Cos[z] + 8*(Cos[x] - Cos[y])^2*Cos[2*z] == 0, m]

    Each is huge and contains a forest of Cos[], but they are the exact solutions.

    Just curious, what did you give Mathematica that did not find the solutions?
     
  5. Nov 14, 2012 #4
    thanks for your help
    I used "solve[]" too
    and also now I test the sentence that you write above but mathematica said:
    "a very larg output was generated"
    maybe its reason is that the version of my mathematica is older than yours.

    however finally I could reach to my result with other version of Matlab and it quickly gave answer

    thank you friend
     
  6. Nov 14, 2012 #5
    I have never seen that error before. What version are you using?

    $Version

    will tell you this.
     
  7. Nov 14, 2012 #6
    Mathematica 7

    thanks for your help
     
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