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
>>
>> 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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