On Tue, Mar 30, 2021 at 9:39 AM 'B A' via sympy <sympy@googlegroups.com> wrote:
>
> Here is one solution that seems to work. To simplify Z I use Z.replace(Abs,
> MyAbs) with
>
> def MyAbs(x):
> x1=symbols('x1',real=True,positive=True)
> x1 = x.evalf(subs={a:0.573})
> if x1 < 0.0:
> return S(-1)*x
> else:
> return x
> Is this a reasonable way to go, or are there gotchas that I should be aware
> of?
A few minor points.
- Defining x1 as a symbol does nothing in this code, as you
immediately overwrite it with x.evalf(...).
- S(-1) is unnecessary. You can just use -x.
- Your code assumes that the expression is a function in the variable
a (and only a). But perhaps that assumption is fine for your use-case.
The main gotcha however is that the code is wrong if the abs is
positive for that specific value of a but not for the general domain
of a. A more robust way would be to use the maximum() and minimum()
functions to get the maximum and minimum values of the abs argument on
the interval [0, 1]:
>>> maximum(2*a**2 - 1, a, Interval(0, 1))
1
>>> minimum(2*a**2 - 1, a, Interval(0, 1))
-1
This tells you that abs(2*a**2 - 1) cannot be simplified like this for
a in [0, 1], because it is both positive and negative on that
interval. To simplify it, you would need to factor it or write it as a
piecewise.
Aaron Meurer
> On Sunday, March 28, 2021 at 11:14:57 AM UTC+2 B A wrote:
>>
>> I am a sympy beginner, and fairly new to python, so I suspect that my
>> question has a simple answer, but have not been able to figure it out
>> myself.
>>
>> I have sympy expressions containing the built-in Abs function. The arguments
>> of Abs() are polynomials in a=symbol('a', real=True,positive=True) . Here
>> are a three examples:
>>
>> Abs(a**2 + 2)
>> sqrt(2)*Abs(2*a**2 - 1)/3
>> sqrt(2)*(a + 1)*Abs(a - 1)/3
>>
>> Here's the point: I know that 'a' is approximately 0.6. So in cases where
>> the argument of Abs has an unambiguous sign (within floating-point
>> precision) I would like to simplify the absolute value. For example in the
>> three cases above I would like:
>>
>> Abs(a**2 + 2) -> a**2 + 2
>> sqrt(2)*Abs(2*a**2 - 1)/3 -> sqrt(2)*(1- 2*a**2)/3
>> sqrt(2)*(a + 1)*Abs(a - 1)/3 -> sqrt(2)*(a + 1)*(1-a)/3
>>
>> where the right arrow '->' indicates replacement or simplification. Note
>> the change of sign in the second and third examples, because (for example) I
>> know that (a-1) is negative.
>>
>> Is there a (simple) way to implement this?
>>
>> Thank you!
>> Bruce
>
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