I want to take the derivative of a multivariable function using SymPy and then for a) the symbolic result to be printed and then b) the result of the derivative at a point to be printed. I'm using the following code
import math as m
import numpy
import scipy
#define constants
lambdasq = 0.09
Ca = 3
qOsq = 2
def f1(a,b,NN,ktsq,x):
return NN*x**(-a)*ktsq**b*m.exp(m.sqrt(16*Ca/9*m.log(1/x)*m.log((m.log(ktsq/lambdasq))/m.log(qOsq/lambdasq))))
from sympy import *
x = symbols('x')
def f2(NN,a,b,x,ktsq):
return -x*diff(m.log(f1),x)
This runs but I can't find a way to get the symbolic result to be printed and when I try to evaluate at a point, say e.g adding in print(f2(0.3,0.1,-0.2,0.1,3))
I get an error
TypeError: must be real number, not function
When I replace f1
with its symbolic representation, I get instead the error
ValueError:
Can't calculate 1st derivative wrt 0.100000000000000.
So I can summarise my question as follows
a) How to print out a symbolic derivative and its value at a point when I call diff(m.log(f1),x)
(i.e without having to replace f1
by its actual representation)
b) If I have to use the symbolic representation in the differentiation (i.e use diff(m.log(NN*x**(-a)*ktsq**b*m.exp(m.sqrt(16*Ca/9*m.log(1/x)*m.log((m.log(ktsq\
/lambdasq))/m.log(qOsq/lambdasq))))),x)
then how to print out the symbolic derivative and its value at a point?
New to Python so hopefully there is a relatively simple fix. Thanks!
First, math
functions are numeric, they cannot work with SymPy's symbols. Use the corresponding functions from SymPy (exp, log, sqrt) which you already imported with from sympy import *
:
def f1(a, b, NN, ktsq, x):
return NN*x**(-a)*ktsq**b*exp(sqrt(16*Ca/9*log(1/x)*log((log(ktsq/lambdasq))/log(qOsq/lambdasq))))
Second, within f2 you are trying to differentiate f1
. But f1 is a callable Python function, not a SymPy expression. You need to pass in some arguments to get a SymPy expression, which can then be differentiated.
def f2(NN, a, b, x0, ktsq):
return (-x*diff(log(f1(a, b, NN, ktsq, x)), x)).subs(x, x0)
Here the numeric arguments, except the value x0, are passed to f1, resulting in a SymPy expression containing x. That is a thing to be differentiated. After that, the numeric value x0 is substituted for x.
print(f2(0.3,0.1,-0.2,0.1,3)) # 0.366748952743614
A take-away point is that SymPy differentiates expressions, not functions. There is no concept of f'
in SymPy, only f'(x)
.