of phi with
respect to z? It seems like this substitution is saying that
phi(z,t).diff(z) = phi(z,t)
I'm new to sympy, so I apologize if this is a stupid question.
Alex
On 05/19/2015 12:16 PM, Jonathan Lindgren wrote:
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have some
very strange behaviour with subs. The following code
from sympy.abc import phi
import sympy as sp
z=sp.Symbol('z')
t=sp.Symbol('t')
sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
