Oregon Curriculum Network
School of Tomorrow
The sympy simplifier, on its own, just makes baroque expressions, but if confronted with $\sqrt{2}$ and $\sqrt{3}$ as equalities, as suggested on StackOverFlow, the identities are confirmed.
import sympy as sym
ϕ = sym.Symbol('ϕ')
This will be $\sqrt{2}$:
e = sym.sqrt(ϕ + ϕ**-2)
e
This will be $\sqrt{3}$:
e2 = sym.sqrt(ϕ**2 + ϕ**-2)
e2
ϕ = (sym.sqrt(5)+1)/2 # real value time!
ϕ.evalf(50)
e = sym.sqrt(ϕ + ϕ**-2)
e
e.evalf(50)
sym.N(sym.sqrt(2), 50)
e.equals(sym.sqrt(2))
True
e2 = sym.sqrt(ϕ**2 + ϕ**-2)
e2
e2.evalf(50)
sym.N(sym.sqrt(3), 50)
e2.equals(sym.sqrt(3))
True