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MathGroup Archive 2004

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Re: Re: Derivative of Sum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47982] Re: [mg47976] Re: Derivative of Sum
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 4 May 2004 07:03:14 -0400 (EDT)
  • References: <c6q28u$p5l$1@smc.vnet.net> <c6updn$b9c$1@smc.vnet.net> <200405040508.BAA17814@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

There is no such thing, either in Mathematica or in any other symbolic 
algebra program I know (and that covers quite a few).
The reason why there isn't is that there are just no enough general 
algorithmic transformations one can do with sums or products of 
variable (symbolic) length that  work in sufficiently many cases for 
this to be useful. Most situations for which such general rules exist, 
like yours,  can be dealt with almost immediately by hand. In 
Mathematica
Sum[f[i],[i,a,b}] where a and/or b are symbolic is not really a sum but 
just an expression which only becomes one when both a and b are 
assigned numerical values  or when f is an actual function that 
evaluates to sum summable expression (and even then you will get an 
answer only if Mathematica "knows' the sum in question.)
You can try to define your own rules (like Brian Higgins did) that may 
work in the case that interests you although in almost every case you 
will be able to do the same by just forgetting about Sum and working 
only with the expression to be summed.
Of course for some purposes it will be enough to simply choose for your 
summation limit an integer larger than the indices you are intending to 
work with:

D[Sum[a[k]*b[k], {k, 1, 7}], a[5]]

b[5]

but I assume you already new that.

Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/



On 4 May 2004, at 14:08, Michal Kvasnicka wrote:

> 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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