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MathGroup Archive 2005

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Re: Re: Sum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55288] Re: [mg55215] Re: Sum
  • From: DrBob <drbob at bigfoot.com>
  • Date: Fri, 18 Mar 2005 05:34:14 -0500 (EST)
  • References: <d192qe$ng4$1@smc.vnet.net> <200503170828.DAA21649@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

>> x[k_]:= Exp[-k]
>> Sum[x, {k, 0, Infinity}]
>> - works fine

No, but this does:

x[k_] := Exp[-k]
Sum[x[k], {k, 0, Infinity}]
E/(-1 + E)

That's probably what you meant.

Bobby

On Thu, 17 Mar 2005 03:28:33 -0500 (EST), Ofek Shilon <ofek at simbionix.com> wrote:

> I believe the answer lies in an innocent looking line in the
> Mathematica book, under "Sum":
> "If a sum cannot be carried out explicitly by adding up a finite number
> of terms, Sum will attempt to find a symbolic result. In this case, f
> is first evaluated symbolically. "
> (f refers to the syntax Sum[f, {i, a}]).    i.e., when the Sum cannot
> be converted to Plus, f (your x) is evaluated BEFORE the iterator
> variable i (your k). Thus, x is converted to the form Exp[-k] first,
> and is not affected by the "change" in k - in an infinite sum, the
> iterator does not really iterates \inf times.
>
> an easy workaround is to define the k-dependence explicitly:
>
> x[k_]:= Exp[-k]
> Sum[x, {k, 0, Infinity}]
>
> - works fine  in Mathematica 5.1.0 on windows. no need to copy/paste
> long expressions, just add the index dependence as above.
>
>
>
>



-- 
DrBob at bigfoot.com


  • References:
    • Re: Sum
      • From: "Ofek Shilon" <ofek@simbionix.com>
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