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I'm having the following problem: I define

series = Exp[Sum[Subscript[J, n]/n t^n, {n, 1, \[Infinity]}]]

and it's all fine. Also when I ask Mathematica

D[series, t]

The answer is correct:

 E^Sum[(t^n*Subscript[J, n])/
     n, {n, 1, Infinity}]*
Sum[t^(-1 + n)*Subscript[J, 
     n], {n, 1, Infinity}]

but when I evaluate at t=0 I have problems

 D[series, t]/.{t->0}

of course it encounters 0^0 indeterminate form because it cannot evaluate the infinite sum before.

How can I find a clever solution to that (without redefine Power)?

Improve this question
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  • $\begingroup$ Whether the derivative is defined for t == 0 depends on whether Limit[D[series, t], t -> 0] exists. $\endgroup$
    – Bob Hanlon
    Commented Mar 3, 2017 at 16:46
  • $\begingroup$ I cannot see what is your proposal. Of course the series is well drfined for t=0. This is because first you write t^0=1 and then you set t=0, but 1(0)=1. Here the problem is that Mathematica cannot evaluate the infinite sum (if you substitute the upper limit by 10 for example, the evaluation at t=0 yields the correct result). $\endgroup$
    – MaPo
    Commented Mar 3, 2017 at 16:58
  • $\begingroup$ I am saying that in general the derivative may or may not be defined at t == 0 so the only way to express it without knowing more is to represent it as a Limit $\endgroup$
    – Bob Hanlon
    Commented Mar 3, 2017 at 17:04
  • $\begingroup$ I tried, but it cannot evaluate the limit: the output is just the input. $\endgroup$
    – MaPo
    Commented Mar 3, 2017 at 17:56
  • $\begingroup$ Without defining Subscript[J, n] there is no way to determine the Limit and know if the derivative is defined at t == 0. The unevaluated Limit accurately reflects what is known. $\endgroup$
    – Bob Hanlon
    Commented Mar 3, 2017 at 18:44

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