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Thread: question about reversing equations

  1. #1
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    question about reversing equations

    Have any ways to reverse from LHS variables by RHS variables of a group of equations?
    For example:

    \begin{matrix}a^2+2bc=x<br />
\\ b^2+2ac=y<br />
\\ c^2+2ab=z<br />
\end{matrix}

    How can I reverse those equations to new equations set likes:

    \begin{matrix}f(x,y,z)=a<br />
\\ g(x,y,z)=b<br />
\\ h(x,y,z)=c<br />
\end{matrix}

    What are those functions look likes?
    Thanks
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  2. #2
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    Mathematica spits out 8 solutions, but they are extraordinarily long. I'm not even going to attempt to post them here, as they are multi-line solutions. Algebraically, though, there's nothing more complicated than powers and square roots and fractions, and such. I would try to find someone who has Mathematica, and enter the following command:

    Simplify[Solve[{a^2 + 2b c == x, b^2 + 2a c == y, c^2 + 2a b == z}, {a, b, c}]].

    That should do it.
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  3. #3
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    Quote Originally Posted by haedious View Post
    Have any ways to reverse from LHS variables by RHS variables of a group of equations?
    For example:

    \begin{matrix}a^2+2bc=x<br />
\\ b^2+2ac=y<br />
\\ c^2+2ab=z<br />
\end{matrix}

    How can I reverse those equations to new equations set likes:

    \begin{matrix}f(x,y,z)=a<br />
\\ g(x,y,z)=b<br />
\\ h(x,y,z)=c<br />
\end{matrix}

    What are those functions look likes?
    Thanks
    Let \omega = e^{2\pi i/3} be a complex cube root of unity. Then

    (a+b+c)^2 = a^2 +b^2 +c^2 +2bc+2ca+2ab = x+y+z,
    (a+\omega b+\omega^2c)^2 = a^2 +\omega^2b^2 +\omega c^2 +2bc+2\omega^2ca+2\omega ab = x+\omega^2y+\omega z,
    (a+\omega^2 b+\omega c)^2 = a^2 +\omega b^2 +\omega^2 c^2 +2bc+2\omega ca+2\omega^2 ab = x+\omega y+\omega^2 z.

    Let  X be one of the two square roots of x+y+z, and similarly let Y = \bigl(x+\omega y+\omega^2 z\bigr)^{1/2} and Z = \bigl(x+\omega^2 y+\omega z\bigr)^{1/2}. Thus there are eight possible choices for the triple of complex numbers (X,Y,Z).

    Then a+b+c=X, a+\omega^2 b+\omega c = Y and a+\omega b+\omega^2c = Z.

    Since 1+\omega+\omega^2=0, it follows that X+Y+Z = 3a, X+\omega Y + \omega^2 Z = 3b and X+\omega^2 Y + \omega Z = 3c.

    That gives the solution in the neat form

    a = \frac13(X+Y+Z),
    b = \frac13(X+\omega Y + \omega^2 Z),
    c = \frac13(X+\omega^2 Y + \omega Z).

    Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.
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    Quote Originally Posted by Ackbeet View Post
    Mathematica spits out 8 solutions, but they are extraordinarily long. I'm not even going to attempt to post them here, as they are multi-line solutions. Algebraically, though, there's nothing more complicated than powers and square roots and fractions, and such. I would try to find someone who has Mathematica, and enter the following command:

    Simplify[Solve[{a^2 + 2b c == x, b^2 + 2a c == y, c^2 + 2a b == z}, {a, b, c}]].

    That should do it.
    I'm not familiar with Mathematica but thank for you help.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    Let \omega = e^{2\pi i/3} be a complex cube root of unity. Then

    (a+b+c)^2 = a^2 +b^2 +c^2 +2bc+2ca+2ab = x+y+z,
    (a+\omega b+\omega^2c)^2 = a^2 +\omega^2b^2 +\omega c^2 +2bc+2\omega^2ca+2\omega ab = x+\omega^2y+\omega z,
    (a+\omega^2 b+\omega c)^2 = a^2 +\omega b^2 +\omega^2 c^2 +2bc+2\omega ca+2\omega^2 ab = x+\omega y+\omega^2 z.

    Let  X be one of the two square roots of x+y+z, and similarly let Y = \bigl(x+\omega y+\omega^2 z\bigr)^{1/2} and Z = \bigl(x+\omega^2 y+\omega z\bigr)^{1/2}. Thus there are eight possible choices for the triple of complex numbers (X,Y,Z).

    Then a+b+c=X, a+\omega^2 b+\omega c = Y and a+\omega b+\omega^2c = Z.

    Since 1+\omega+\omega^2=0, it follows that X+Y+Z = 3a, X+\omega Y + \omega^2 Z = 3b and X+\omega^2 Y + \omega Z = 3c.

    That gives the solution in the neat form

    a = \frac13(X+Y+Z),
    b = \frac13(X+\omega Y + \omega^2 Z),
    c = \frac13(X+\omega^2 Y + \omega Z).

    Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.
    Many thank for your works!
    It's great for having steps to the solution even I don't understand completely. And I never thought that how apply cube root of unity to solve this issue so I've difficult to digest these steps but still thanks.

    I treats above question similar as just a square root to a number consisted with {x,y,z}.
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