Re: NSolve fails where Solve succeeds!
- To: mathgroup at smc.vnet.net
- Subject: [mg43625] Re: NSolve fails where Solve succeeds!
- From: "Dana DeLouis" <delouis at bellsouth.net>
- Date: Fri, 26 Sep 2003 04:45:38 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
It appears to me that in Mathematica 5, it is setting each element of
the list equal to the right hand side.
Solve[{{2*a - 3, 4*b - 1}, {3*c - 3, 5*d - 1}} == 0]
{{a -> 3/2, b -> 1/4, c -> 1, d -> 1/5}}
It likes your example. All value are equal to zero.
Solve[{{a,b},{c,d}}==0]
However, given the above, I don't see why the following fails. Here, it
does not return all variable as equal to 1.
Solve[{{a,b},{c,d}}==1]
It says it is "Not a well-formed equation." Hmm?
It doesn't like Nsolve like you mentioned.
NSolve[{{2*a - 3, 4*b - 1},{3*c - 3, 5*d - 1}} == 0]
RowBox[{\(Solve::"elist"\), ":", "\<\"List encountered during
<snip>
It appears to like Reduce in some situations, similar to Solve:
It liked this:
Reduce[{{2*a - 3, 4*b - 1}, {3*c - 3, 5*d - 1}} == 0]
d == 1/5 && c == 1 && b == 1/4 && a == 3/2
It like this:
Reduce[{{a, b}, {c, d}} == 0]
d == 0 && c == 0 && b == 0 && a == 0
It did not like this:
Reduce[{{a, b}, {c, d}} == 1]
"....is not a quantified system of equations and inequalities. "
It appears to me that Solve and Reduce will work on each element only if
the right hand side ==0.
Suppose we have this list:
equ = {{2*a - 3, 4*b - 1}, {3*c - 3, 5*d - 1}}
Suppose we want to solve this list equal to 1. This will not work
Solve[equ == 1]
However, subtracting 1 from each side will solve for the equations:
Solve[equ - 1 == 0]
{{a -> 2, b -> 1/2, c -> 4/3, d -> 2/5}}