Derivative of Indexed by index: does it make sense?

[Francesco Bonazzi]

Just a question:

We currently only have indexed derivatives by indexed objects:

>>> A[i].diff(A[j])
KroneckerDelta(i, j)

But, does it make sense to derive an indexed symbol by the index?

>>> A[i].diff(i)
A[1]

>>> A[i**2].diff(i)
A[2*i]

Currently the aforementioned operations return 0.

I thought we could add this kind of operation, but I'm not sure if it makes sense. Indices are supposed to be integers, but indices can also be any kind of expression.

[Björn Dahlgren]

On Thursday, 26 January 2017 10:32:46 UTC+1, Francesco Bonazzi wrote:

But, does it make sense to derive an indexed symbol by the index?

Do you have an applied example where this would help? If it simplifies users' code I think that would be an argument for it.

[Francesco Bonazzi]

This could solve the issue https://github.com/sympy/sympy/issues/11444

Expr.is_constant() checks whether the expression derivative by its free symbols is nonzero. Unfortunately A[i].diff(i) currently gives 0.

Free symbols of A[i] are A and i.

On Thursday, 26 January 2017 17:04:01 UTC+1, Björn Dahlgren wrote:

Do you have an applied example where this would help? If it simplifies users' code I think that would be an argument for it.

 Not really, indices are supposed to be integers, the derivative doesn't make any sense. On the other hand integer symbols are still derivable in SymPy.

Furthermore, nothing forbids to use indexed objects as very strange functions.

[Aaron Meurer]

I don't think it makes sense. The index is a discrete variable, but
derivatives only make sense for continuous variables.

Aaron Meurer

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[Björn Dahlgren]

On Thursday, 26 January 2017 20:40:10 UTC+1, Aaron Meurer wrote:

I don't think it makes sense. The index is a discrete variable, but
derivatives only make sense for continuous variables.

I agree with Aaron, I think making the Expr.is_constant method aware of Indexed would be the preferred path.

[Aaron Meurer]

Perhaps the derivative should raise an exception and is_constant should ignore the derivative check if the derivative raises an exception. 

Aaron Meurer 

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[Francesco Bonazzi]

On Monday, 30 January 2017 01:51:05 UTC+1, Aaron Meurer wrote:

Perhaps the derivative should raise an exception and is_constant should ignore the derivative check if the derivative raises an exception.

https://github.com/sympy/sympy/pull/12109

I made the derivative return S.NaN in case the symbol appears in the indices. It appears to work.