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Mathematica Equation Solving Symbolic Computations

Solving equations with multiple unique solutions

Maya Greene

Posted 10 years ago

Hi All, I am new to Mathematica and I apologize if this is a very basic question. I am attempting to use the solve function to derive multiple solutions for a variable, however it seems as though I can only have the same number of equations as I do unknowns, which then leads to only one solution per unknown. E.g.: Solve[{m == ab, n == ac, o == b*c}, {a, b, c}] What I am trying to solve for is this: Solve[{m == ab, n == ac, o == ad, p == bc, q == bd, r == cd}, {a, b, c, d}] This indicates that each unknown (a,b,c,d) will have more than one unique solution. However, this statement leads to an empty solution. I can try to break up the over-identified (determined?) statement and eventually produce all unique solutions, however this could get complicated and tiresome with a larger model. I was wondering, is there a way to get Mathematica to produce all unique solutions in a situation like this?

POSTED BY: Maya Greene

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

(1) `ab` is not the same as`ab`. (2) The second system is overdetermined. Might instead use `Reduce` for it. Reduce[{m == ab, n == ac, o == ad, p == bc, q == bd, r == cd}, {a, b, c, d}] ( (o == 0 && n == 0 && m == 0 && a == 0 && r != 0 && (b == -((Sqrt[p] Sqrt[q])/Sqrt[r]) \|\| b == (Sqrt[p] Sqrt[q])/Sqrt[r]) && b != 0 && c == p/b && d == q/b) \|\| (q == 0 && p == 0 && m == 0 && r != 0 && (a == -((Sqrt[n] Sqrt[o])/Sqrt[r]) \|\| a == (Sqrt[n] Sqrt[o])/Sqrt[r]) && b == 0 && a != 0 && c == n/a && d == o/a) \|\| (r == 0 && p == 0 && n == 0 && q != 0 && (a == -((Sqrt[m] Sqrt[o])/Sqrt[q]) \|\| a == (Sqrt[m] Sqrt[o])/Sqrt[q]) && a != 0 && b == m/a && c == 0 && d == o/a) \|\| (r == 0 && q == 0 && o == 0 && p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) \|\| a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a && c == n/a && d == 0) \|\| (r == 0 && q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b == 0 && c == 0) \|\| (r == 0 && q == 0 && p == 0 && o == 0 && n == 0 && a != 0 && b == m/a && c == 0 && d == 0) \|\| (q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b == 0 && c != 0 && d == r/c && r != 0) \|\| (r == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && c == 0 && b != 0 && d == q/b && q != 0) \|\| (r == 0 && q == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b != 0 && c == p/b && d == 0 && p != 0) \|\| (r == 0 && q == 0 && p == 0 && n == 0 && m == 0 && b == 0 && c == 0 && a != 0 && d == o/a && o != 0) \|\| (r == 0 && q == 0 && p == 0 && o == 0 && m == 0 && b == 0 && a != 0 && c == n/a && d == 0 && n != 0) \|\| (r == 0 && q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b == 0 && c != 0 && d == 0) \|\| (r == 0 && q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b != 0 && c == 0 && d == 0) \|\| (q != 0 && n == (o p)/q && r != 0 && m == (o p)/r && p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) \|\| a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a && c == n/a && d == o/a) *)

(1) ab is not the same asa*b.

(2) The second system is overdetermined. Might instead use Reduce for it.

Reduce[{m == a*b, n == a*c, o == a*d, p == b*c, q == b*d,  r == c*d}, {a, b, c, d}]

(* (o == 0 && n == 0 && m == 0 && a == 0 && 
   r != 0 && (b == -((Sqrt[p] Sqrt[q])/Sqrt[r]) || 
     b == (Sqrt[p] Sqrt[q])/Sqrt[r]) && b != 0 && c == p/b && 
   d == q/b) || (q == 0 && p == 0 && m == 0 && 
   r != 0 && (a == -((Sqrt[n] Sqrt[o])/Sqrt[r]) || 
     a == (Sqrt[n] Sqrt[o])/Sqrt[r]) && b == 0 && a != 0 && c == n/a &&
    d == o/a) || (r == 0 && p == 0 && n == 0 && 
   q != 0 && (a == -((Sqrt[m] Sqrt[o])/Sqrt[q]) || 
     a == (Sqrt[m] Sqrt[o])/Sqrt[q]) && a != 0 && b == m/a && c == 0 &&
    d == o/a) || (r == 0 && q == 0 && o == 0 && 
   p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) || 
     a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a && 
   c == n/a && d == 0) || (r == 0 && q == 0 && p == 0 && o == 0 && 
   n == 0 && m == 0 && a == 0 && b == 0 && c == 0) || (r == 0 && 
   q == 0 && p == 0 && o == 0 && n == 0 && a != 0 && b == m/a && 
   c == 0 && d == 0) || (q == 0 && p == 0 && o == 0 && n == 0 && 
   m == 0 && a == 0 && b == 0 && c != 0 && d == r/c && 
   r != 0) || (r == 0 && p == 0 && o == 0 && n == 0 && m == 0 && 
   a == 0 && c == 0 && b != 0 && d == q/b && q != 0) || (r == 0 && 
   q == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b != 0 && 
   c == p/b && d == 0 && p != 0) || (r == 0 && q == 0 && p == 0 && 
   n == 0 && m == 0 && b == 0 && c == 0 && a != 0 && d == o/a && 
   o != 0) || (r == 0 && q == 0 && p == 0 && o == 0 && m == 0 && 
   b == 0 && a != 0 && c == n/a && d == 0 && n != 0) || (r == 0 && 
   q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 && 
   b == 0 && c != 0 && d == 0) || (r == 0 && q == 0 && p == 0 && 
   o == 0 && n == 0 && m == 0 && a == 0 && b != 0 && c == 0 && 
   d == 0) || (q != 0 && n == (o p)/q && r != 0 && m == (o p)/r && 
   p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) || 
     a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a && 
   c == n/a && d == o/a) *)

POSTED BY: Daniel Lichtblau

Maya Greene

Posted 10 years ago

Thank you, Daniel. I think this will work for my purposes! Looks like I can just replace the first unknown in the second phrase to get the multiple solutions for it.

POSTED BY: Maya Greene

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