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Re: Strange behavior of NSum on 4.0

• To: mathgroup at smc.vnet.net
• Subject: [mg28064] Re: Strange behavior of NSum on 4.0
• From: "Sebastien de Menten" <sdementen at hotmail.com>
• Date: Fri, 30 Mar 2001 04:12:18 -0500 (EST)
• References: <99uruh$5vv@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Hi,

NSum[] is very different of N[Sum[]].
NSum will give an approximation of the sum by an integral or by a fit (in
the help page : "NSum uses either the Euler­Maclaurin (Integrate) or Wynn
epsilon (Fit) method.")
So, it supposes that the arguments can be evaluated at fractional points and
not only integers. But trying to evaluate a[[41/2]] make no sense while
evaluating f[41/2] makes sense (in general).

Even if the Help says that N[Sum[]] calls NSum, I think they must be wrong.

I hope it helps.

Seb

"Ken Kikuchi" <kikuchi at neti.com> wrote in message
news:99uruh$5vv at smc.vnet.net...
> I encountered a strange behavior with NSum on Mathematica 4.0 on Mac.
> The following is a simplified example about sum of 1 to 25.
>
> In[1]:= a = Range[25]
> Out[1]=
> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
>   15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
>
> In[2]:= NSum[a[[i]], {i, 25}]
>
> I expected the result as 325 but received the following strange error
> messages.
>
> Part::pspec :
>     Part specification 41/2 is neither an integer nor a list of
> integers.
> ...
> NIntegrate::inum :
>     Integrand <<1>>[[41/2]] is not numerical at {1}={41/2}."
>
> However, the same code works fine up to 24.
> In[23]:= NSum[a[[i]], {i, 24}]
> Out[23]= 300.
>
> It works fine if not using Part.
> In[3]:= NSum[i, {i, 25}]
> Out[3]= 325.
>
> But Part works fine with N while manual noted -- N[Sum[...]] calls NSum.
> In[4]:= N[Sum[a[[i]], {i, 25}]]
> Out[4]= 325.
>
> And Sum works too.
> In[5]:= Sum[a[[i]], {i, 25}]
> Out[5]= 325
>
> My version is,
> In[6]:= $Version
> Out[6]= "4.0 for Power Macintosh (July 26, 1999)"
>
> I am not sure it's only my environmental problem or my incorrect usage.
> Could anyone help me?
>
> Many thanks,
> Ken
>
>