Since I am working on a project involving square roots, I need square roots to be simplified to the max. However, some square roots expressions do not produce the disired result. Please consider checking this example:
>>> from sympy import * # just an example don't tell me that import * is obsolete
>>> x1 = simplify(factor(sqrt(3 + 2*sqrt(2))))
>>> x1 # notice that factoring doesn't work
sqrt(2*sqrt(2) + 3)
>>> x2 = sqrt(2) + 1
>>> x2
sqrt(2) + 1
>>> x1 == x2
False
>>> N(x1)
2.41421356237309
>>> N(x2)
2.41421356237309
>>> N(x1) == N(x2)
True
As you can see, the numbers are actually equal, but numpy can't recognize that because it can't factorize and simplify x1. So how do I get the simplified form of x1 so that the equality would be correct without having to convert them to float ?
Thanks in advance.
When you are working with nested sqrt expressions, sqrtdenest is a good option to try. But a great fallback to use is nsimplify which can be more useful in some situations. Since this can give an answer that is not exactly the same as the input, I like to use this "safe" function to do the simplification:
def safe_nsimplify(x):
from sympy import nsimplify
if x.is_number:
ns = nsimplify(x)
if ns != x and x.equals(ns):
return ns
return x
>>> from sympy import sqrt, sqrtdenest
>>> eq = (-sqrt(2) + sqrt(10))/(2*sqrt(sqrt(5) + 5))
>>> simplify(eq)
(-sqrt(2) + sqrt(10))/(2*sqrt(sqrt(5) + 5)) <-- no change
>>> sqrtdenest(eq)
-sqrt(2)/(2*sqrt(sqrt(5) + 5)) + sqrt(10)/(2*sqrt(sqrt(5) + 5)) <-- worse
>>> safe_nsimplify(eq)
sqrt(1 - 2*sqrt(5)/5) <-- better
On your expression
>>> safe_nsimplify(sqrt(2 * sqrt(2) + 3))
1 + sqrt(2)
And if you want to seek out such expressions wherever they occur in a larger expression you can use
>>> from sympy import bottom_up, tan
>>> bottom_up(tan(eq), safe_nsimplify)
tan(sqrt(1 - 2*sqrt(5)/5))
It might be advantageous to accept the result of sqrtdenest instead of using nsimplify as in
def safe_nsimplify(x):
from sympy import nsimplify, sqrtdenest, Pow, S
if x.is_number:
if isinstance(x, Pow) and x.exp is S.Half:
ns = sqrtdenest(x)
if ns != x:
return ns
ns = nsimplify(x)
if ns != x and x.equals(ns):
return ns
return x