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I searched already a lot about indexed Variables, and it tends to be the most applicable way to use tensor notation. But I am having a hard time to solve for undefined indexed variables:

Here my stylized problem:

F := Table[f[i], {i, 1, n}];

The Derivative gives the correct answer if applied for a specified i=1

D[Sum[f[i], {i, 1, 100}], f[1]]

=1

If I want to solve more generally let say $ i \in [1, 100] $

D[Sum[f[i], {i, 1, 100}], f[i]]

=0

Obviously does not work, how can I solve such a problem generally?

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The variable i is a dummy one. The evaluated expression:

Sum[f[i], {i, 1, 10}]

f[1] + f[2] + f[3] + f[4] + f[5] + f[6] + f[7] + f[8] + f[9] + f[10]

contains no explicit variable f[i], hence, the result is 0.

Try to first Inactivate the sum, and only then to calculate the derivative:

expr1 = D[Inactivate[Sum[f[i], {i, 1, 100}], Sum], f[i]]

The result is

enter image description here

Now let us activate it and select a certain value of i:

Activate[expr1] /. i -> 20

(*  1   *)

Now, one can combine all that into a function, like this:

dif[f_, j_] := (D[Inactivate[Sum[f[i], {i, 1, 100}], Sum], f[i]] // 
     Activate) /. i -> j;

Let us check it up:

dif[g, 30]

(*  1  *)

Have fun!

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  • $\begingroup$ Hey, thanks for the fast help! I see that this workaround works good for that specific case. In my field, however, I am exposed quite a lot to unspecified indexed variables. Is there maybe a more direct way i.e to put a placeholder on the specific dummy variable place. Something like: D[Sum[f[i], {i, 1, 100}], f[0<i<100]] Or: D[Sum[f[i], {i, 1, 100}], f[i_PositiveInteger]] ? $\endgroup$
    – oyy
    Nov 26 '20 at 11:32
  • $\begingroup$ This way seems to be (1) quite direct and (2) rather flexible. Another thing that from your example I cannot imagine other terms that you need to treat. You had in mind probably - a more short way? Not that I know. The only approach that seems me reasonable is to help Mma a bit. For example, as a first step calculate the derivative of a single term on the screen prior of doing the summation (*step1*) expr2=D[expr1,f[i]]. As a second one, make a summation on the screen.(*step2*) Sum[expr2,{i,1,100}]] $\endgroup$ Nov 26 '20 at 12:23
  • $\begingroup$ (A corrected comment. Discard the previous one) This way seems to be (1) quite direct and (2) rather flexible. Another thing that from your example I cannot imagine other terms that you need to treat. You had in mind probably - a more short way? Not that I know. The only approach that seems me reasonable is to help Mma a bit. For example, calculate the derivative of a single term on the screen without doing the summation expr2=D[expr1,f[i]]. Then see, if the summation is still necessary, and do it, if yes. $\endgroup$ Nov 26 '20 at 12:30
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    $\begingroup$ Exactly this I tried to propose. Instead of differentiating Sum[expr[f[i]],{i,1,100}]], you can differentiate the term of the sum: D[expr[f[i]], f[i]]. This yields the desired result in your case. In your example expr[f[i]]=f[i] but in general, it may be any function of f[i]. If you have a double summation or another more complex expression, you may need to later still calculate a sum. But do it after you have already calculated the derivative. $\endgroup$ Nov 26 '20 at 13:43
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    $\begingroup$ No this is impossible. The reason is simple. In your last code you have Solve[f[x[1],x[2] ]==0,x[i] ]. That is, the left-hand part of the equation, f[x[1],x[2] ] depends on x[1] and x[2] . Mma, however, will look for its dependence on x[i] in the explicit form. And it will not find such. Hence, the result will be "No solution". If instead you try Solve[f[x[i],x[j] ]==0,x[i] ] it will work. $\endgroup$ Nov 26 '20 at 15:03

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