Logarithms in sympy
Question:
Please tell me how you can change log(e) to 1. And in general is there a way in sympy to get a float-type answer?
For example, instead of log(2), get 0.69314718056
from math import *
from sympy import *
x1=symbols('x1')
x2=symbols('x2')
Func = e**(2*x1)*(x1+x2**2+2*x2)
print(Func.diff(x2))
Here I already get log(e). In other calculations, the same situation
Answers:
The problem is that you are mixing the math module and sympy. This leads to "inconsistent" results. Note that in your code, e comes from the math module, and it is just the number 2.71828182845905. SymPy also exposes the Euler’s number, but it is called E, which is a symbolic entity.
Hence, when working with SymPy, try to use as many SymPy objects as possible. Your expression becomes:
Func = E**(2*x1)*(x1+x2**2+2*x2)
By using SymPy objects, canonical simplifications are possible: log(E) gives 1.
If you want to convert "symbolic numbers" to floating point, you can use the n() method. For example log(2).n() gives 0.693147180559945.
Please tell me how you can change log(e) to 1. And in general is there a way in sympy to get a float-type answer?
For example, instead of log(2), get 0.69314718056
from math import *
from sympy import *
x1=symbols('x1')
x2=symbols('x2')
Func = e**(2*x1)*(x1+x2**2+2*x2)
print(Func.diff(x2))
Here I already get log(e). In other calculations, the same situation
The problem is that you are mixing the math module and sympy. This leads to "inconsistent" results. Note that in your code, e comes from the math module, and it is just the number 2.71828182845905. SymPy also exposes the Euler’s number, but it is called E, which is a symbolic entity.
Hence, when working with SymPy, try to use as many SymPy objects as possible. Your expression becomes:
Func = E**(2*x1)*(x1+x2**2+2*x2)
By using SymPy objects, canonical simplifications are possible: log(E) gives 1.
If you want to convert "symbolic numbers" to floating point, you can use the n() method. For example log(2).n() gives 0.693147180559945.