Re: Derivative of Sum

• To: mathgroup at smc.vnet.net
• Subject: [mg47976] Re: Derivative of Sum
• From: "Michal Kvasnicka" <michal.kvasnicka at _NO_ZpaMM-.quick.cz>
• Date: Tue, 4 May 2004 01:08:56 -0400 (EDT)
• References: <c6q28u\$p5l\$1@smc.vnet.net> <c6updn\$b9c\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```No, it not what I need. I am looking for general derivative of the general
sum.

Michal
"Brian Higgins" <bghiggins at ucdavis.edu> pí¹e v diskusním pøíspìvku
news:c6updn\$b9c\$1 at smc.vnet.net...
> Michal,
>
> I s this what you want:
>
> S[p_] := Sum[a[k]b[k], {k, 1, p - 1}] + a[p]b[p] + Sum[a[k]b[k], {k, p
> + 1, n} ]
>
> In[3]:=D[S[5],b[5]]
>
> Out[3]=a[5]
>
> Note that differentiating a series term-by-term may not always give
> you the incorrect answer.
>
> Cheers,
>
> Brian
>
>
> "Michal Kvasnicka" <michal.kvasnicka at _NO_ZpaMM-.quick.cz> wrote in message
news:<c6q28u\$p5l\$1 at smc.vnet.net>...
> > Is Mathematica 5 able to compute the folowing problem:
> > \!\(S = Sum[\(a\_k\) b\_k, {k, 1, n}]\)
> >
> > then should be
> >
> > \!\(\[PartialD]\_\(a\_i\)\ S = b\_i\) but the Mathematica gives 0.
> >
> > Thanks, Michal
>

```

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