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When I type
import sympy as sp
x = sp.Symbol('x')
sp.simplify(sp.log(sp.exp(x)))
I obtain
log(e^x)
Instead of x. I know that "there are no guarantees" on this function.
Question. Is there some specific simplification (through series expansion or whatsoever) to convert logarithm of exponent into identity function?
935
With the help of sympy. log() function, we can simplify the principal branch of the natural logarithm. Logarithms are taken with the natural base, e. To get a logarithm of a different base b, use log(x, y), which is essentially short-hand for log(x) / log(y).
simplify() method, we can simplify any mathematical expression. Parameters: expression – It is the mathematical expression which needs to be simplified. Returns: Returns a simplified mathematical expression corresponding to the input expression.
Note that by default in SymPy the base of the natural logarithm is E (capital E ). That is, exp(x) is the same as E**x .
You have to set x to real type and your code will work:
import sympy as sp
x = sp.Symbol('x', real=True)
print(sp.simplify(sp.log(sp.exp(x))))
Output: x.
For complex x result of this formula is not always is equal to x. Example is here.
82
If you want to force the simplification, expand can help because it offers the force keyword which basically makes certain assumptions like this for you without you having to declare your variables as real. But be careful with the result -- you will not want to use it when those assumptions are not warranted.
>>> log(exp(x)).expand(force=True)
x
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