Hi Aaron,
Thanks for your help!
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
- Defining x1 as a symbol does nothing in this code, as you
immediately overwrite it with x.evalf(...).
OK. I'm still hazy on when Python allocates new storage for an object.
- S(-1) is unnecessary. You can just use -x.
OK.
- 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.
This assumption is correct: all of my expressions are polynomials in a,
and rational powers of those polynomials. So 'a' is the only variable
that appears.
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.
I only need equations that are valid in an infinitesimal neighborhood
around a=0.57
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]:
I can test at the endpoints of a small interval, say 0.56 < a < 0.58.
But the expressions are complicated enough that solving for the extrema
will be too expensive. Fortunately I do have a consistency test at the
end which should catch any mistakes.
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.
You are correct. Fortunately I only need results that are correct in an
open set around a specific value of a.
What I might do is to modify MyAbs so that it maintains a global
dictionary of all arguments it has been called with, adding new
arguments if they are not found in the dictionary. Later I can check
that the objects stored in the dictionary have no roots within the small
interval. Does that seem reasonable?
Cheers,
Bruce
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