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>