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Symmetric looking equations needing a symmetric solution

  1. Jun 30, 2011 #1
    Hi all,

    I have a set of equations that look very nice and symmetric, but the only way I'm able to find solutions to them is with pages and pages of algebra! Can any members with more of a mathematical flair than myself point me in the direction of a more direct and satisfactory method of solution?

    The equations are:

    [itex]\frac{{A_1 }}{{A_1 + A_2 + A_3 }} = \frac{{b_1 c_1 }}{{b_1 c_1 + b_2 c_2 + b_3 c_3 }}[/itex]

    [itex]\frac{{A_2 }}{{A_1 + A_2 + A_3 }} = \frac{{b_2 c_2 }}{{b_1 c_1 + b_2 c_2 + b_3 c_3 }}[/itex]

    [itex]\frac{{A_3 }}{{A_1 + A_2 + A_3 }} = \frac{{b_3 c_3 }}{{b_1 c_1 + b_2 c_2 + b_3 c_3 }}[/itex]

    Given that:

    [itex]b_1 + b_2 + b_3 = 1[/itex]

    I am looking to express [itex]b_1[/itex], [itex]b_2[/itex] and [itex]b_3[/itex] in terms of [itex]A_1[/itex], [itex]A_2[/itex], [itex]A_3[/itex], [itex]c_1[/itex], [itex]c_2[/itex] and [itex]c_3[/itex].

    In general, I guess the problem would be:

    [itex]A_i \sum\limits_j {b_j c_j } = b_i c_i \sum\limits_j {A_j }[/itex]

    [itex]\sum\limits_j {b_j } = 1[/itex]

    I have solved the problem for the [itex]j = 2[/itex] case, with the neat looking solutions:

    [itex]b_1 = \frac{{A_1 c_2 }}{{A_1 c_2 + A_2 c_1 }}[/itex]

    [itex]b_2 = \frac{{A_2 c_1 }}{{A_1 c_2 + A_2 c_1 }}[/itex]

    I'm looking for similar solutions to [itex]j = 3[/itex], or more...if a general solution is obvious to someone else!

    Any ideas?

    Thanks in advance,

    FD
     
  2. jcsd
  3. Jun 30, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    I'll write your equations as:

    [Eqs. 1] [tex] \frac{A_i}{\sum_j Aj} = \frac {b_i c_i}{\sum_j b_j c_j} [/tex]


    Introduce another variable [itex] \theta [/itex] defined by

    [Eq. 2] [tex] \theta = \frac {\sum_j b_j c_j} {\sum_j A_j} [/tex]

    Multiplying the left sides of [Eqs. 1] by [itex] \frac{\theta}{\theta} [/itex] gives:

    [Eqs. 3] [tex] \frac{\theta A_i}{\sum_j b_j c_j} = \frac {b_i c_i}{\sum_j b_j c_j} [/tex]

    so

    [Eqs 4] [tex] b_i = \theta \frac{A_i}{c_i} [/tex]


    Since [itex] \sum_j b_j = 1 [/itex] we have:

    [Eq 5] [tex] \sum_j \theta \frac{A_j}{c_j} = \theta \sum_j {A_j}{c_j} = 1 [/tex]

    Sovling for [itex] \theta [/itex] gives:

    [Eq 6] [tex] \theta = \frac{1}{\sum_j \frac{A_j}{c_j}} [/tex]

    Substituting in [Eqs 4] gives

    [Eqs 7] [tex] b_i = \frac{1}{\sum_j \frac{A_j}{c_j}} \frac{A_i}{c_i} [/tex]

    Of course, you must check none of the steps involve division by zero. Also my reasoning seems suspiciously circular!
     
  4. Jul 1, 2011 #3
    Ahhhh, that works beautifully - nice one!!
     
  5. Jul 2, 2011 #4
    That was an elegant solution, I would like to add another one.
    Dividing the sides of the ith equation by the sides of the jth equation gives:
    [tex]\frac{A_i}{A_j}=\frac{b_ic_i}{b_jc_j}[/tex]
    And then:
    [tex]b_j=\frac{A_j}{c_j}\frac{c_i}{A_i}b_i[/tex]
    Thus:
    [tex]\sum_{j}b_j=\frac{c_i}{A_i}b_i\sum_{j}\frac{A_j}{c_j}[/tex]
    And since [itex]\sum_{j}b_j=1[/itex] we get:
    [tex]b_i=\frac{A_i}{c_i}\frac{1}{\sum_{j}\frac{A_j}{c_j}}[/tex]
     
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