Hi Everyone,
I've been poking around in the sympy source, and I've noticed that the `
simplify` command does not deal with expressions like the following:
>>> from sympy import *
>>> from sympy.abc import x
>>> simplify(abs(cosh(x)))
Abs(cosh(x))
A simple glance at the graph of cosh(x) reveals that Abs(cosh(x)) ==
cosh(x). So, in case it's not obvious the above expression should return:
cosh(x)
I'd like to take a stab at implementing this but I need some direction. I
hope this isn't duplicating something I have stupidly missed (the
inequality solvers don't seem to identify cases where absolute value
brackets are unnecessary. There are three key cases (as far as I can see):
1. The argument within the abs brackets is non-negative over all
combinations of all independent variables. Therefore, the absolute value
brackets redundant / unnecessary and may be removed.
2. The argument within the abs brackets is non-positive over all
combinations of all independent variables. Therefore, the absolute value
brackets may be removed and the expression may be multiplied by -1 for the
same effect.
3. The argument within the abs brackets contains both positive and
negative values depending on the values of the independent variables.
Ok, so to implement the above rules in the general sense it makes sense to
me to perform the following steps given the expression
Abs(f(x_1,x_2,...x_n)):
1. Determine the number of real roots of f(x).
1. If there are one or more real roots then for each root:
1. Determine whether the gradient of f(x) is zero:
1. If non-zero slope at root, then argument expression obtains
opposite signed value at some point, so return Abs(f(x))
2. Slope of f(x) at root is 0.
3. Determine if the root is an inflection point (need to figure
out exactly how to test for this over multiple variables in the
expression)
1. If at inflection point then the expression will still become
opposite signed on either side of the root, so return Abs(f(x))
2. Any and all roots coincide with extrema values of f(x).
Therefore f(x) may be represented without absolute value brackets.
2. If f(x) >= 0 remove the absolute value brackets and return the
argument expression.
3. If f(x) < 0 remove the absolute value brackets, multiply the
expression by -1 and return it.
There are a few things which I'm not sure how it will work out:
- Imaginary numbers. Does anyone know if I will need to write special
code for this, or should the above procedure work out anyway?
- The case where a symbol in the expression has been defined with the
positive flag:
>>> y = Symbol('y')
>>> simplify(abs(sinh(y)))
Abs(sinh(y))
>>> y = Symbol('y',positive=True)
>>> simplify(abs(sinh(y)))
sinh(y)
- Are there sneaky ways of determining in a precise manner whether a
function which cannot be reduced (e.g. cosh(x)+cos(x)) has real roots, even
if finding those roots would only be possible numerically? Is there another
sympy module which can help with this?
- What about variables which produce no real roots over a given range?
Is there a way to handle those? E.g.
>>> y = Symbol('y',range=[0,pi])
>>> simplify(abs(sin(y)))
sin(y)
- Redundant absolute value brackets are removed somewhere. Can anyone
tell me where exactly in the code this happens in the simplify function? I
can't seem to find it:
>>> from sympy.abc import x
>>> simplify(abs(abs(x)+1))
Abs(x) + 1
>>> simplify(abs(x+1))
Abs(x + 1)
So, right now I have forked the sympy repo (see here
<https://github.com/NauticalMile64/sympy>) and set up my own little
function in sympy.symplify called abssimp.py (just copied combsimp.py and
started from there), and added an appropriate if-absolute check in the main
simplify function. Is this the right way to go about adding such a feature?
Would the code that I write here also be used in solve or something?
Any guidance / advice would be appreciated.
Thanks!
Nolan Dyck
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