solve() only gives one of infinitely many solutions.

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Arthur
Arthur 2014-11-4
评论： Sean de Wolski 2014-11-4
Hopefully the screenshot will make my question clear. From what I can tell by looking at the documentation for the solve() function, this application of solve is Supposed to give a parametrization for All solutions to x*y=1, but I only get one.
There are no additional assumptions on my variables x,y. Code in the screenshot was immediately preceded by clearing all variables.
Arthur

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回答（2 个）

Zoltán Csáti
Zoltán Csáti 2014-11-4
Your mistake is that you applied == instead of =. Do this one:
solution = solve('x*y=1','x','y');
Sean de Wolski
Sean de Wolski 2014-11-4
The problem is that you're asking for a solution for both variables. Instead, solve for one at a time:
solve(x*y==1,x)
ans =
1/y
  2 个评论
Arthur
Arthur 2014-11-4
编辑：Arthur 2014-11-4
Thanks Sean, but what I'm hoping for is a parametrization of all solutions for the variables I list in terms of a new parameter. In the example I give, a solution would look like
x = z
y = 1/z
I'm not interested so much in solving x*y==1, but this was the simplest example I could find that expressed my problem. Furthermore, I assumed that since solve() gives ALL solutions when given a system of polynomials with only finitely many solution (e.g
V = solve(x^2 + x*y + y == 3,x^2 - 4*x + 3 == 0, x, y)
) that there might be some way of parameterizing All solutions in the case where there are infinitely many?
Sean de Wolski
Sean de Wolski 2014-11-4
That's what it's doing: 1/y is the solution for all x given all y.
If you want to substitute in new values, use subs to do this.
subs(expr,old,new)

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