Thanks! I thought I had copied too much traceback information, but we can
handle that later if necessary.
Your remark about unevaluated expressions looks like a very good place for
me to look, so I read the documentation and added a line of test code:
print('unevaluated part of exp', UnevaluatedExpr(exp))
Now the output looks like this:
after Eji -i*(i*j - 2*i*m + j**2 + 2*j*m + m**2)/(j*(2*j**2 + m**2))
after diff (-i*(i*j - 2*i*m + j**2 + 2*j*m + m**2)/(j*(2*j**2 +
m**2)))/factorial(0)
after subs (-1*(-2)*(m0**2 + 2*(-2)*m0 - 2*(-2)*m0 + (-2)**2 -
2*(-2))/((-2)**1*(m0**2 + 2*(-2)**2)**1))/factorial(0)
unevaluated part of exp (-1*(-2)*(m0**2 + 2*(-2)*m0 - 2*(-2)*m0 + (-2)**2 -
2*(-2))/((-2)**1*(m0**2 + 2*(-2)**2)**1))/factorial(0)
How can I avoid this?`
One more remark: I use this when I return a value that should not be
expanded, for example:
with evaluate(False):
exp3 = (m-m0)**(Rational(2,3))*exp3
but the evaluate(False) should not be in effect here, and if it is, I would
need to turn if off.
On Friday, April 22, 2022 at 8:49:26 PM UTC+2 [email protected] wrote:
> It's hard to say for sure without seeing the full traceback. The
> Mod(1*(2*0) + 1*(1*1), (1*1)*(1*2)) looks like it might be an
> unevaluated expression. If that's the case, those are known to cause
> issues because they can break the expectations of various functions.
>
> Aaron Meurer
>
> On Wed, Apr 20, 2022 at 2:23 AM Thomas Ligon <[email protected]> wrote:
> >
> > I am getting an exception where I don't see why it happens.
> >
> > The exception is:
> > invalid input: Mod(1*(2*0) + 1*(1*1), (1*1)*(1*2))
> >
> > The statement where it is being thrown is:
> > print(P3().subs(i, 0))
> > exp = factor(expand(P3().subs(i, 0)))
> >
> > P3() should always return the same thing:
> > def P3():
> > exp = 2*i**2 + 4*i*j + 8*i*m - j**2 +2*m*j + 9*m**2
> > return exp
> >
> > The print statement is just for debugging purposes. It shows me multiple
> times
> > -j**2 + 2*j*m + 9*m**2
> > then
> > 9*m**2 + j*(2*m) - j**2 + (8*0)*m + (4*0)*j + 2*0**2
> > invalid input: Mod(1*(2*0) + 1*(1*1), (1*1)*(1*2))
> >
> > So, I can't see why something that should work the same every time
> suddenly fails with an exception. The statement P3().subs(i, 0) should
> always return the same thing.
> >
> > A note on my coding style: P3 could easily be a global constant, but the
> function paradigm fits very well with the mathematics that I am
> reproducing. The most complex case I have runs recursively, which is very
> valuable. Also, I am using "exp" as a local variable that means
> "expression".
> >
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>
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