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What kind of problem is this and what tools can we use to tackle it?

  1. Mar 16, 2010 #1
    Basically I have the following system:

    [tex] ac = A [/tex]

    [tex] ad + bc = B [/tex]

    [tex] bd = C [/tex]

    where [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex] are constants.

    Solving for [tex]a, b, c,[/tex] and [tex]d[/tex], what kind of problem/system am I encountering and what appropriate tools (vectors and/or numerical methods perhaps[?]) would help to find the set of solutions for [tex]a, b, c,[/tex] and [tex]d[/tex]?

    I know that the system factors a trinomial but I don't know what kind of problem is presented by the system itself.

    Thanks in advance for any insight.
  2. jcsd
  3. Mar 16, 2010 #2
    Inserting the first and third equation into the second gives you a quadratic equation (altghough for a product of two unknowns).
    You will still end up with one unknown variable as you do not have enough equations.
  4. Mar 16, 2010 #3
    Could there still be numerical methods for finding the solutions?

    Thanks for your reply.
  5. Mar 16, 2010 #4
    My messing around on MATLAB. Not sure how useful it is...

    Code (Text):

    EDU>> syms a b c A B C
    EDU>> sol = solve('a*c = A','a*d + b*c = B','b*d = C');
    EDU>> [a;sol.b(1);sol.c(1);sol.d(1)]
    ans =
    EDU>> [a;sol.b(2);sol.c(2);sol.d(2)]
    ans =
    I guess that's two sets of solutions?

    edit - I generally prefer linear algebra to sets of equations where variables are multiplied. It's kind of a nightmare actually - I can see a uniqueness/existence problem if a = 0 right away, or B^2-4*A*C < 0? What a mess...
    Last edited: Mar 16, 2010
  6. Mar 16, 2010 #5
    I recommend substituting the top and bottom into the middle.
    (where [itex]x=\frac{a}{b}[/itex]) and solve for x.

    The variable b is arbitrary (but not equal 0) and [itex]a=xb[/itex]
    Use top and bottom equation to find rest.

    EDIT: just as bebel said :D
  7. Mar 16, 2010 #6
    Thanks for the output MikeyW, your MATLAB solution looks consistent with what I have found. I have found that the system can be simplified to a function of [tex]b, c[/tex], and the constants. Letting one of the variables assume any value, I can solve for the other with the quadratic formula or newtons method, and then find the other two variables.

    After simplifying:

    {b^2}{c^2} - Bbc + AC = 0

    a =\frac{A}{c}

    d = \frac{C}{b}

    I don't know if this would necessarily work for higher degree polynomials (especially an odd number of variables), and wonder if there is a solution using vectors/linear algebra. My linear algebra knowledge is still elementary and haven't had to encounter multiple variables before so I don't know if there are tools that can help.
  8. Mar 16, 2010 #7
    A little more complex would be the following system which factors a quintic when [tex]a, b, c, d, e, f, g, h, i [/tex] and [tex]j[/tex] are solved for with [tex]A, B, C, D, E,[/tex] and [tex]F[/tex] being constants.

    [tex] acegi = A[/tex]

    [tex] acegj + acehi + acfgi + egadi + egbci = B [/tex]

    [tex] acehj + acfgj + egadj + egbcj + acfhi + adehi + adfgi + bcehi + bcfgi + bdegi = C [/tex]

    [tex] acfhj + adehj + bcehj + bcfgj + bdegj + adfhi + bcfhi + ehbdi + fgbdi + adfgj = D [/tex]

    [tex] adfhj + bcfhj + ehbdj + fgbdj + bdfhi = E [/tex]

    [tex] bdfhj = F [/tex]

    Is there any way to solve for the above ten variables with a method more streamlined than the "plug and chug" method used previously for the quadratic?

    *I am not as interested in any solution as I am in any algorithm ("plug and chug" or other)

    Thanks again for any info. It is greatly appreciated.
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