| Author |
Comment/Response |
yehuda
|
 01/31/13 08:58am
Sure you can
You need to do the following steps
1. Expand the expression inside the square root
2. Factor it
you do this using pattern matching + relevant functions
expr = 1.01792*10^-40 \[Sqrt](2.6336*10^72 + 4.80538*10^77 kx^2 +
8.76809*10^82 kx^4 + 4.80538*10^77 ky^2 +
3.65593*10^83 kx^2 ky^2 + 8.76809*10^82 ky^4 -
9.61076*10^77 kz^2 + 3.65593*10^83 kx^2 kz^2 +
3.65593*10^83 ky^2 kz^2 + 8.76809*10^82 kz^4)
expr /. Sqrt[s_] :> Sqrt[Factor[Expand[s]]]
Notice that kz has the highest degree (4) so its coefficient will be normalized to 1, so the result is
30.1416 \[Sqrt](3.00362*10^-11 + 5.48053*10^-6 kx^2 + 1. kx^4 +
5.48053*10^-6 ky^2 + 4.16959 kx^2 ky^2 + 1. ky^4 -
0.0000109611 kz^2 + 4.16959 kx^2 kz^2 + 4.16959 ky^2 kz^2 +
1. kz^4)
yehuda
URL: , |
|
