Wikipedia:Reference desk/Archives/Mathematics/2010 November 14

= November 14 =

System of nonlinear equations (I think?)
How would I solve a system like this? (Not homework, just curious.) I'm only in Pre-calc, but I guess this is a college-level algebra problem? $$ \begin{cases} 1/x + 3/y = 11 \\ 5/x - 2/z = 1 \\ 2/x + 5/y - 4/z = 3 \end{cases} $$ It would be easy if it weren't for those three darned variables in the denominators. 141.153.215.139 (talk) 21:51, 14 November 2010 (UTC)
 * Actually that one happens to be easy: call 1/x = a, 1/y = b, 1/z = c, you end up with 3 equations in the 3 unknowns a, b, c. Solve & substitute back. No calculus needed. In general there is no standard method for systems of nonlinear equations. Some are solveable by algebra if you have the right idea, others simply aren't. If you cannot solve them by algebra, there is always Newtons method to get you an arbitrarily exact approximate solution (but that does require calculus). 86.147.175.50 (talk) 22:59, 14 November 2010 (UTC)
 * Convergence of Newton's method is not at all guaranteed in general. My opinion is that nonlinear (perhaps differential) equations encode much of the universe's physics, at least to a good approximation, and so should be utterly horrific to solve in general. 67.158.43.41 (talk) 04:19, 15 November 2010 (UTC)
 * Substitute each of the variables as follows a = 1/x, b = 1/y and c = 1/z. Your problem becomes

\begin{cases} a + 3b = 11 \\ 5a - 2c = 1 \\ 2a + 5b - 4c = 3 \end{cases} $$
 * You can solve this using simple linear algebra. You have a = 52/29, b = 89/29 and c = 231/58. Invert to get the values of x, y and z, i.e. x = 29/52, y = 29/89 and z = 58/231. — Fly by Night  ( talk )  23:01, 14 November 2010 (UTC)

Systems of n algebraic equations in n unknowns may be solved systematically in two steps: Step 1 is exact while step 2 is approximate. So even if you didn't discover the above trick you can still solve the problem by doing some extra work. Bo Jacoby (talk) 10:29, 15 November 2010 (UTC).
 * 1) Elimination of variables leading to n equations, each in 1 unknown.
 * 2) Each equation is solved numerically by a root-finding method.
 * To Bo Jacoby: I don't believe that one. You can't generally eliminate variables from a system of equations with a symbolic method in such a way that you get equations in a single variable.  &#x2013; b_jonas 16:03, 16 November 2010 (UTC)
 * I thought the same thing, except he said "algebraic" equations, which was left a bit vague. Taken weakly enough, it's correct, but I think that would have to be a non-standard interpretation of an "algebraic equation". 67.158.43.41 (talk) 22:00, 16 November 2010 (UTC)

I should have put algebraic equations in square brackets to link to the definition. Algebraic equations generalize systems of linear equations, and exceptions regarding dependence still exist. In a system of equations like x&minus;y=0, xx&minus;xy=0 the second equation does not provide new information, so even if there are apparently two equations, there are effectively only one. A system of equations like x&minus;y=xx+yy&minus;2=0 implies xx&minus;1=yy&minus;1=0 by step 1. Bo Jacoby (talk) 10:46, 17 November 2010 (UTC).
 * For monomial terms in a single variable I see what you mean. How do you deal with mixed terms, like xy, in general? 67.158.43.41 (talk) 03:59, 18 November 2010 (UTC)
 * Scratch that. I only see what you mean in a few degenerate cases. For instance, how does one apply elimination to xy+x^2+y^3 = x^2 y^4 + y^5 = 0? 67.158.43.41 (talk) 04:34, 18 November 2010 (UTC)

Let the system of equations be P1=P2=0 where P1 and P2 are polynomials in x and y. Then x2+xy+y3=x2y4+y5=0. One obvious solution is (x,y)=(0,0). To find other solutions set P2:=P2/y4. Then x2+xy+y3=x2+y=0. To eliminate y3, set P1:=P1&minus;y2P2. Then x2+xy&minus;x2y2=x2+y=0. Set P1:=P1/x. Then x+y&minus;xy2=x2+y=0. To eliminate y2, set P1:=P1+xyP2. Then x+(x3+1)y=x2+y=0. To eliminate y from P1, set P1=P1 &minus;(x3+1)P2. Then x&minus;x5&minus;x2=x2+y=0. Set P1=&minus;P1/x. Then x4+x&minus;1=x2+y=0. Solving x4+x&minus;1=0 numerically and substituting x into y=&minus;x2 gives the real solutions (x,y)=(&minus;1.22074,&minus;1.49022), (x,y)=(0.724492,&minus;0.524889), and the complex solutions (x,y)=(0.248126+1.03398i, 1.00755&minus;0.513116i), (x,y)=(0.248126&minus;1.03398i, 1.00755+0.513116i),  in addition to the trivial solution (x,y)=(0,0). Bo Jacoby (talk) 11:55, 18 November 2010 (UTC).

PS. http://www.wolframalpha.com/input/?i=xy%2Bx^2%2By^3+%3D+x^2+y^4+%2B+y^5+%3D+0 does not identify x=y=0 as an exact solution. Perhaps wolframalpha does not eliminate variables? Bo Jacoby (talk) 10:18, 20 November 2010 (UTC).
 * I'm guessing that if there isn't a nice exact form for all solutions, it will give numerical forms for everything. Try it with a system that does have one. -- Meni Rosenfeld (talk) 16:01, 20 November 2010 (UTC)