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MathGroup Archive 2004

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Re: Derivative of Sum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47935] Re: Derivative of Sum
  • From: bghiggins at ucdavis.edu (Brian Higgins)
  • Date: Fri, 30 Apr 2004 19:27:38 -0400 (EDT)
  • References: <c6q28u$p5l$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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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