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Re: simplifying inside sum, Mathematica 5.1

  • To: mathgroup at
  • Subject: [mg53728] Re: [mg53709] simplifying inside sum, Mathematica 5.1
  • From: DrBob <drbob at>
  • Date: Tue, 25 Jan 2005 05:03:29 -0500 (EST)
  • References: <>
  • Reply-to: drbob at
  • Sender: owner-wri-mathgroup at

The infinite sum can't be simplified (since the a[i] aren't known, after all), so of course

Sum[a[i]*x^i, {i, 0, Infinity}]

is unevaluated.

When you substitute x->0, yes, it COULD simplify to just one term--but 0^0 is Indeterminate, so it wouldn't serve any purpose.

This kind of thing works, however:

a[i_] = 1/i!;
Sum[a[i]*x^i, {i, 0, Infinity}]
% /. x -> 0


If we don't know the coefficients, term-by-term expansion of Taylor series is naturally doomed anywhere but 0.

In addition--and THIS is a Mathematica quirk--expansion at zero runs into the undeniably TRUE but inconvenient fact (in this case) that 0^0 is undefined. It's just a convention that 0^0 IN A TAYLOR SERIES means 1 -- but the built-in is Sum, not Taylor; it doesn't go by that convention except when the overall series simplifies through pattern-recognition. In a different problem, we wouldn't want Sum to accept 0^0 as 1.

A Taylor series defined only at zero isn't of much use anyway, I think.


On Mon, 24 Jan 2005 03:37:32 -0500 (EST), Richard Fateman <fateman at> wrote:

> Sum[a[i]*x^i,{i,0,Infinity}]
> %/. x->0
> leaves 0^i  inside the sum, unsimplified.
> So does the sum from 1 to Inf.
> The correct answers are presumably a[0] and 0.
> Is there a way to get Mathematica to do this?

DrBob at

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