Sympy simplify sum of Kronecker delta

Question

How does one simplify sums of Kronecker delta expressions in sympy?

For example consider Sum(KroneckerDelta(i,j),(i,0,n-1)) or Sum(KroneckerDelta(i, j, (0, n - 1)), (i, 0, n - 1)):

from sympy import *
from sympy.concrete.delta import _simplify_delta
n = symbols('n')
j = tensor.Idx('j')
i = tensor.Idx('i')
_simplify_delta(simplify(Sum(KroneckerDelta(i,j),(i,0,n-1))))
_simplify_delta(simplify(Sum(KroneckerDelta(i,j,(0,n-1)),(i,0,n-1))))

Outputs Sum(KroneckerDelta(i, j), (i, 0, n - 1)) and Sum(KroneckerDelta(i, j, (0, n - 1)), (i, 0, n - 1))

If j is constrained to be between 0 and n-1 (how do I tell the sympy that?), then this should reduce to 1, which occurs at i==j. Moreover if the sum is more complicated, I would expect it to remove the sum and replace the sum variable i with the variable j.

Moreover, I would be interested in a resource for all sorts of simplifications in sympy for KroneckerDelta functions. I recently found out how to perform implicit matrix differentiation in sympy and KroneckerDelta functions appear everywhere.

Edit: I found a solution, kind of. Its not automated. I found more functions inside of sympy.concrete.delta using help(sympy.concrete.delta). If we copy the resulting expression and replace Sum with sympy.concrete.delta.deltasummation then the desired simplification happens. I am still curious if there is a delta simplification package that tries all these things automatically.

Answer 1

You can evaluate summations using Sum().doit() or summation:

In [1]: from sympy import *
   ...: from sympy.concrete.delta import _simplify_delta
   ...: 
   ...: n = symbols("n")
   ...: j = tensor.Idx("j")
   ...: i = tensor.Idx("i")

In [2]: s = Sum(KroneckerDelta(i, j), (i, 0, n - 1))

In [3]: s
Out[3]: 
n - 1     
 ___      
 ?        
  ?   ?   
  ?    i,j
 ?        
 ???      
i = 0     

In [4]: s.doit()
Out[4]: 
?1  for j ? 0 ? j ? n - 1
?                        
?0        otherwise      

In [5]: summation(KroneckerDelta(i, j), (i, 0, n - 1))
Out[5]: 
?1  for j ? 0 ? j ? n - 1
?                        
?0        otherwise 

There isn't currently a way to specify in assumptions that j<=n-1 although you can use j = symbols('j', nonnegative=True) to specify that j>=0. You can also manually replace those conditions with true though e.g.:

In [8]: s.doit().subs({j >= 0: True, j <= n-1: True})
Out[8]: 1

The second summation where you give bounds for the KroneckerDelta will compute automatically:

In [11]: s2 = Sum(KroneckerDelta(i, j, (0, n - 1)), (i, 0, n - 1))

In [12]: s2
Out[12]: 
n - 1     
 ___      
 ?        
  ?   ?   
  ?    i,j
 ?        
 ???      
i = 0     

In [13]: s2.doit()
Out[13]: 1