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3 equations with 3 unknown variables - no Show Steps in Wolfram Alpha

Posted 9 years ago

Why do equations like "x^2 - y z = 14, y^2 - x z = 28, z^2 - x y = -14" not support step-by-step solutions? I don't need answer when I don`t know how this answer was reached. Fix this, plz.

POSTED BY: ivan ivanov
8 Replies

Daniel, can you recommend a description of how GroebnerBasis works that can be understood by someone who doesn't have a Ph.D. in math?

POSTED BY: Frank Kampas

Think of row reducing a matrix. You have a pivot row and a row beneath it. You multiply bottom row by pivot coefficient, pivot row by bottom row value in pivot column, subtract, and now the resulting row has zero in the pivot position.

For linear polynomials with variables in some fixed order, a GB computation is exactly this row reduction, where columns are equivalent to variables (so col 1 has coefficients of polynomials in the top variable, col 2 has coeffs of second variable, etc.)

For nonlinear things get more complicated. The variable and monomial order give rise to an ordering of terms in polynomials. The notion of "reduction" becomes one of finding a syzygy between a pair of polynomials (forget the algebra terminology and just consider that what I now describe is related to row reduction). Say we are going to operate on a pair of polynomials. Each has a leading term. These are monomials, say 3*x*y^2 and 5*x^2*y. Find the lcm of those "power products" . It is 15*x^2*y^2. So each polynomial gets multiplied by whatever needed to cancel that lcm (poly1 gets multiplied by 5x, poly2 by 3y in this case). When we subtract the difference of the modified polyniomials the lead terms have cancelled.

Of course this is more complicated then simple row reduction. We may have increased the degree overall (since we multiplied by things that raise degree). We now do a bona fide polynomial "reduction" step, wherein polys with lead terms that divide the resulting one (up to coefficient) get used to remove high terms until no longer possible.

In row reduction we've replaced one row by another. In a GB computation, at this point we may have either a new polynomial with leading term not divisible by any polynomial in the current set, or zero. If a new polynomial then we add it to our list of polynomials in the GB, and now must consider it in similar operations involving pairs of polynomials giving rise to new ones.

The important theoretical result of Buchberger was that this process stabilizes. He also noted some extremely important efficiencies. Other important properties were proved, among them ways to use a GB to solve a system of polynomial equations.

"Ideals, Varieties, and Algorithms" by Cox, Little, and O'Shea is one of the better books on the topic, and is aimed at (advanced) undergrads.

POSTED BY: Daniel Lichtblau

Daniel, thanks for the explanation.

I decided to plot all three equations and the solutions.

Show[ContourPlot3D[{x^2 - y z == 14 , y^2 - x z == 28 , 
   z^2 - x y == -14}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, 
  ContourStyle -> Opacity[0.5], Mesh -> None], 
 Graphics3D @ {PointSize[.03], {Point[{-3, -5, 1}], 
    Point[{3, 5, -1}]}}]

enter image description here

POSTED BY: Frank Kampas

This is really funny, Simon answered the question that simple, it was not that hard though, thanks for that simon. - Paramount Song

Answered what question? (What I am suggesting is that the actual question be reread.)

POSTED BY: Daniel Lichtblau

Well Done!

POSTED BY: Simon Cadrin
Posted 9 years ago

Yes, I mean "Show steps" button :) Thats a pity that I dont see her in Wolfram Pro Subscription. If it will be added ever, it would be fine

POSTED BY: ivan ivanov

I gather you are referring to the "Show steps" capability in Wolfram|Alpha. If so, I doubt this would be considered as broken. These are not linear equations, and the underlying method used for nonlinear equations (based on a GroebnerBasis computation) does not readily lend itself to "show steps" form. Or at least not one anyone would likely want to see let alone try to make sense of.

This does not rule out the possibility that some classes of systems have "nice" by-hand approaches. I do not know if the system above would fit into such a category but it is possible that it does.

POSTED BY: Daniel Lichtblau
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