6
$\begingroup$

I have a summation function, for which I have defined derivation via

Custom`sum /: D[Custom`sum[idx_, a_, b_, exprs_], y_] := Custom`sum[idx, a, b D[expr, y]];

where idx is the index with bounds a and b. This does not seem to work, though when the sum is inside another function. For example, I get

D[Sqrt[Custom`sum[idx, 1, 2, func[idx, y]]], y]

Mathematica graphics

where via chain rule I would expect

1/Sqrt[Custom`sum[idx, 1, 2, func[idx, y]]]*1/2*Custom`sum[idx, 1, 2, Derivative[0,1][func][idx, y]]]

To give an even simpler example, look what happens with Plus

D[f[g[x,1]+g[x,2]],x] (* (f^′)[g[x,1]+g[x,2]] ((g^(1,0))[x,1]+(g^(1,0))[x,2]) *)

How can I achieve the same with

D[f[sum[g[x,i]]],x] (* (f^′)[sum[g[x,i]]] (sum^′)[g[x,i]] (g^(1,0))[x,i] *)

Is there a way to get the chain rule to work for these cases?

Improve this question
$\endgroup$
2
  • $\begingroup$ I fixed the typos in your question because I believe it is a good one. I currently see no direct way to do this, but defining UpValues to sum with D will not work. If you have a wrapping function like your Sqrt, then the intermediate expressions only consist of Derivative and there is no D anymore that matches AFAIK. But I might of course be wrong. $\endgroup$ Commented Jun 16, 2018 at 0:59
  • 1
    $\begingroup$ You might also want to look at this question as it is very similar to yours, but I'm not sure your problem can be solved with the information there. So don't claim it is a duplicate, but it might be worth reading. $\endgroup$ Commented Jun 16, 2018 at 1:04

2 Answers 2

Reset to default
7
$\begingroup$

There is a system option that provides a list of excluded functions for differentiation. If you add Custom`sum to the list you should be able to use your definition as expected.

exf = "ExcludedFunctions" /. ("DifferentiationOptions" /. SystemOptions[]); SetSystemOptions[ "DifferentiationOptions" -> "ExcludedFunctions" -> Append[exf, Custom`sum]]; Custom`sum /: D[Custom`sum[idx_, a_, b_, exprs_], y_] := Custom`sum[idx, a, b, D[exprs, y]] D[Custom`sum[idx, 1, 2, func[idx, y]], y] // InputForm (* Custom`sum[idx, 1, 2, Derivative[0, 1][func][idx, y]] *) D[Sqrt @ Custom`sum[idx, 1, 2, func[idx, y]], y] // InputForm (* Custom`sum[idx, 1, 2, Derivative[0, 1][func][idx, y]] / (2*Sqrt[Custom`sum[idx, 1, 2, func[idx, y]]]) *)
Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much :) $\endgroup$ Commented Jun 16, 2018 at 11:46
  • $\begingroup$ Thanks! It seems, though, as if the other definitions for Custom'sum are messing with this method, so that I get zero again for the Sqrt case. Therefore I'll probably stick with the NonConstant solution I posted below. $\endgroup$ Commented Jun 16, 2018 at 13:18
1
$\begingroup$

The question halirutan linked to was very helpful. By setting 

SetOptions[D, NonConstants -> {Custom`sum}];

a D[Custom'sum[idx,a,b,func[idx,y],y,NonConstants->{Custom'sum}] appears during the evaluation of the derivative, and the UpValue, slightly modified to include the NonConstants->{Custom'sum}, works again:

SetOptions[D, NonConstants -> {Custom`sum}]; Custom`sum /: D[Custom`sum[idx_, a_, b_, exprs_], y_, c___] := Custom`sum[idx, a, b , D[exprs, y, c]]; D[Sqrt @ Custom`sum[idx, 1, 2, func1[idx, x]], x] // InputForm (* Custom`sum[idx, 1, 2, Derivative[0, 1][func1][idx, x]]/(2*Sqrt[Cutom`sum[idx, 1, 2, func1[idx, x]]]) *)
Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.