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Re: Variable Assignment


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?


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