How does sympy simplify ln((exp(x)+1)/exp(x)) to log(1+exp(-x))?

Question

If I use

simplify()

function in sympy,

log((exp(x)+1)/exp(x))

does simplify to

log(1+exp(-x))

, however, as I read the doc, the simplify function is "can be unnecessarily slow", I tried other simplification methods, but non of them works, so I'm wondering how do I simplify

ln((exp(x)+1)/exp(x))

to the form like this

log(1+exp(-x))

without calling simplify().

Answer 1

You can more directly just use sympy.polys.polytools.cancel(), which is available as a method on your expression with .cancel().

>>> from sympy.abc import x
>>> from sympy import *
>>> my_expr = log((exp(x)+1)/exp(x))
>>> my_expr.cancel()
log(1 + exp(-x))

This is what is doing the work of simplifying your expression inside simplify().

A very naive benchmark:

>>> import timeit
>>> %timeit my_expr.simplify()
100 loops, best of 3: 7.78 ms per loop
>>> %timeit my_expr.cancel()
1000 loops, best of 3: 972 µs per loop

How does sympy simplify ln((exp(x)+1)/exp(x)) to log(1+exp(-x))?

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