Improve factoring for square roots

[fredrik-johansson]

Composing matrix exponentials and logarithms is a good way to find expressions that don't simplify:

>>> ca_mat([[1,-1],[-1,-1]]).exp().log()
ca_mat of size 2 x 2
[1.00000 {(-b*d*f+b*e*f)/(2*c) where a = 1.41421 [Log(4.11325 {(c+d+e)/2})], b = -1.41421 [Log(0.243117 {(-c+d+e)/2})], c = 3.87013 [Sqrt(14.9779 {d^2+e^2-2})], d = 4.11325 [Exp(1.41421 {f})], e = 0.243117 [Exp(-1.41421 {-f})], f = 1.41421 [f^2-2=0]}, -1.00000 {(b*d^2+b*e^2-2*b)/(c*d*f-c*e*f) where a = 1.41421 [Log(4.11325 {(c+d+e)/2})], b = -1.41421 [Log(0.243117 {(-c+d+e)/2})], c = 3.87013 [Sqrt(14.9779 {d^2+e^2-2})], d = 4.11325 [Exp(1.41421 {f})], e = 0.243117 [Exp(-1.41421 {-f})], f = 1.41421 [f^2-2=0]}]
[-1.00000 {(b*d*f-b*e*f)/(2*c) where a = 1.41421 [Log(4.11325 {(c+d+e)/2})], b = -1.41421 [Log(0.243117 {(-c+d+e)/2})], c = 3.87013 [Sqrt(14.9779 {d^2+e^2-2})], d = 4.11325 [Exp(1.41421 {f})], e = 0.243117 [Exp(-1.41421 {-f})], f = 1.41421 [f^2-2=0]},             -1.00000 {(b*d*f-b*e*f)/(2*c) where a = 1.41421 [Log(4.11325 {(c+d+e)/2})], b = -1.41421 [Log(0.243117 {(-c+d+e)/2})], c = 3.87013 [Sqrt(14.9779 {d^2+e^2-2})], d = 4.11325 [Exp(1.41421 {f})], e = 0.243117 [Exp(-1.41421 {-f})], f = 1.41421 [f^2-2=0]}]

This is the problem:

>>> a = exp(2*sqrt(2))
>>> b = exp(-2*sqrt(2))
>>> sqrt(a+b-2); sqrt(a+1/a-2)
3.87013 {a where a = 3.87013 [Sqrt(14.9779 {b+c-2})], b = 16.9188 [Exp(2.82843 {2*d})], c = 0.0591057 [Exp(-2.82843 {-2*d})], d = 1.41421 [d^2-2=0]}
3.87013 {(a*b-a)/(b) where a = 4.11325 [Sqrt(16.9188 {b})], b = 16.9188 [Exp(2.82843 {2*c})], c = 1.41421 [c^2-2=0]}
>>> 
>>> sqrt(a+b-2) - (a-1)/sqrt(a)
0e-1125 {(a*b-c+1)/(b) where a = 3.87013 [Sqrt(14.9779 {c+d-2})], b = 4.11325 [Sqrt(16.9188 {c})], c = 16.9188 [Exp(2.82843 {2*e})], d = 0.0591057 [Exp(-2.82843 {-2*e})], e = 1.41421 [e^2-2=0]}
>>> 
>>> sqrt(a+1/a-2) - (a-1)/sqrt(a)
0

a+b-2 does not factor, though the equivalent rational function a+1/a-2 does.

The factoring algorithm for field elements somehow needs to be made aware of this. A possibility is to look specifically for exponentials to eliminate, but perhaps there's a more general approach.

[fredrik-johansson]

Another useful example of (lack of) rational function simplification:

>>> (pi-1) / (sqrt(pi) - 1); sqrt(pi) + 1
2.77245 {(b-1)/(a-1) where a = 1.77245 [Sqrt(3.14159 {b})], b = 3.14159 [Pi]}
2.77245 {a+1 where a = 1.77245 [Sqrt(3.14159 {b})], b = 3.14159 [Pi]}