Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 =
                                     a
     a*(1/2*B-1/2*(B^2-4*A*C)^(1/2))/A
                                   A/a
       (1/2*B+1/2*(B^2-4*A*C)^(1/2))/a
    EDU>> [a;sol.b(2);sol.c(2);sol.d(2)]
    ans =
                                     a
     a*(1/2*B+1/2*(B^2-4*A*C)^(1/2))/A
                                   A/a
       (1/2*B-1/2*(B^2-4*A*C)^(1/2))/a
     
    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.
    [tex]Cx+\frac{A}{x}=B[/tex]
    (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:

    [tex]
    {b^2}{c^2} - Bbc + AC = 0
    [/tex]

    [tex]
    a =\frac{A}{c}
    [/tex]

    [tex]
    d = \frac{C}{b}
    [/tex]

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What kind of problem is this and what tools can we use to tackle it?
Loading...