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Nonlinear system

  1. Aug 24, 2015 #1
    I have this system of equation: [tex] A = \frac{\alpha + \beta + \gamma}{3}[/tex] [tex] B = \sqrt[2]{\frac{\beta \gamma + \gamma \alpha + \alpha \beta}{3}}[/tex] [tex] C = \sqrt[3]{\alpha \beta \gamma}[/tex] And I want to solve this system for α, β and γ. In other words, I want to express α, β and γ in terms of A, B and C.

    [tex] \alpha = \alpha (A,B,C)[/tex][tex] \beta = \beta (A,B,C)[/tex][tex] \gamma = \gamma (A,B,C)[/tex]
     
  2. jcsd
  3. Aug 24, 2015 #2

    mfb

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    What is your question or discussion point?
    There is an obvious way to start solving it.
     
  4. Aug 24, 2015 #3
    If α and β are the roots of the quadratic equation and A and B are the arithmetic and geometric mean, respectively, so, the quadratic formula becomes: [tex] A \pm \sqrt{A^2-B^2} [/tex]. I'm trying to solve the cubic equation in the same way...
     
  5. Aug 25, 2015 #4

    mfb

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    I would introduce new variables for B2 and C3. Solving the third equation for one variable and plugging it into another equation gives a quadratic equation for a second variable, which can be plugged into the third equation. If that has a degree of at most 4, there is an analytic solution in closed form, otherwise I doubt there is a way to solve it (because a solution would then probably allow to solve equations that are proven to have no closed analytic solution).
     
  6. Aug 26, 2015 #5

    HallsofIvy

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    So, essentially, you want to solve
    [tex]\alpha+ \beta+ \gamma= 3A[/tex]
    [tex]\alpha\beta+ \alpha\gamma+ \beta\gamma= 9B^2[/tex]
    [tex]\alpha\beta\gamma= C^3[/tex]
    and since A, B, and C are given values, so are 3A, [itex]9B^2[/itex], and [itex]C^3[/itex].

    From [itex]\alpha\beta\gamma=C^3[/itex], [itex]\gamma= \frac{C^3}{\alpha\beta}[/itex]
    so [itex]\alpha\beta+ \frac{C^3}{\beta}+ \frac{C^3}{\alpha}= 9B^2[/itex]
    Multiplying by [itex]\alpha\beta[/itex], [itex]\alpha^2\beta^2+ C^3\alpha+ C^3\beta= 9B^2\alpha\beta[/itex].

    We can write that as [itex]\alpha^2\beta^2+ C^3\beta+ (C^3\alpha- 9B^2\alpha)= 0[/itex] and use the quadratic formula to solve for [itex]\beta[/itex] in terms of [itex]\alpha[/itex], then put that into [itex]\gamma= \frac{C^3}{\alpha\beta}[/itex] to get [itex]\gamma[/itex] in terms of [itex]\alpha[/itex] only.

    Finally, put those into [itex]\alpha+ \beta+ \gamma= 3A[/itex] to get an equation in [itex]\alpha[/itex] only.
     
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