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Description
Steps To Reproduce
If we run
n=var('n')
a=0^n
b=a.simplify_full()
(a({n:0}),b({n:0}))
we get (1,0)
as a result
Expected Behavior
I would expect that simplify_full
applied at 0^n
would not convert it to 0
, since most (all?) mathematical formulas on non-negative integers are more compatible with 0^0=1
(as currently implemented, see the above-evaluation of a({n:0})
).
I currently think that simplifications should preserve the form 0^n
.
Actual Behavior
even with assumptions like
assume(n, 'integer')
assume(n>=0)
or even assume(n==0)
, the simplification of 0^n
yields to 0
.
Additional Information
This is a more general issue than #36838 where we (yet) replaced 0^n
by kronecker_delta(0, n)
.
Note that such basic replacement assumes n
to be a non-negative integer.
Environment
- **OS**: Ubuntu 20.04
- **Sage Version**: 9.0
- Jupyter Notebook
Checklist
- I have searched the existing issues for a bug report that matches the one I want to file, without success.
- I have read the documentation and troubleshoot guide