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Re: Sum
*To*: mathgroup at smc.vnet.net
*Subject*: [mg55215] Re: Sum
*From*: "Ofek Shilon" <ofek at simbionix.com>
*Date*: Thu, 17 Mar 2005 03:28:33 -0500 (EST)
*References*: <d192qe$ng4$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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.
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