Re: Variable Assignment
- To: mathgroup at smc.vnet.net
- Subject: [mg14012] Re: [mg14007] Variable Assignment
- From: BobHanlon at aol.com
- Date: Mon, 14 Sep 1998 02:57:54 -0400
- Sender: owner-wri-mathgroup at wolfram.com
Clear[x];
x = x /. FindRoot[Exp[x] == x^2, {x, -0.5}]
-0.7034674295409823
Exp[x] - x^2
-1.339360800045596*^-8
Clear[x];
x = x /. Solve[Exp[x] == x^2, {x}]
InverseFunction::ifun:
Warning: Inverse functions are being used. Values may be
lost for multivalued inverses.
InverseFunction::ifun:
Warning: Inverse functions are being used. Values may be
lost for multivalued inverses.
Solve::ifun:
Inverse functions are being used by Solve, so some
solutions may not be found.
{-2*ProductLog[-1/2], -2*ProductLog[1/2]}
% // N
{1.588047264689378 - 1.540223501020758*I,
-0.7034674224983916}
If you are only interested in real solutions:
x = Select[x, Im[#] == 0&]
{-2*ProductLog[1/2]}
x = x[[1]]
-2*ProductLog[1/2]
Exp[x] - x^2 // FullSimplify
0
?ProductLog
ProductLog[z] gives the principal solution for w in z = w
e^w. ProductLog[k, z] gives the kth solution.
Bob Hanlon
In a message dated 9/13/98 3:18:11 AM, wax at waxer.com wrote:
>FindRoot[Exp[x] == x^2, {x, -0.5}]
>
>{x --> -0.703467}
>
>How do I assign the result [-0.7033467] to a variable?