Hi,
I made a simplication function for divsum.
It works when second argument of divsum is 1.
I hope that someone will enjoy it.
Thanks,
-
Yuji
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On 2011/07/02, at 10:54, ICHIKAWA, Yuji wrote:
> Fateman-san,
>
> Thank you for your comments.
> I recognized that there is no reliable way to attach a condition for matching that involves 2 parameters in the match.
>
> By the way, you can see a^b even if a, b are numbers, when you use a returned value of "factor".
> It works well in the first case, though you need to change divsum to noun in order to see the result.
> I think that the case of a*b is same.
>
> Thanks,
> -
> Yuji
>
> On 2011/07/02, at 9:18, Richard Fateman wrote:
>
>> On 7/1/2011 5:12 PM, ICHIKAWA, Yuji wrote:
>>> Hi,
>>>
>>> I am trying to make simplification rules for divisor function divsum.
>>>
>>> 1. divsum(p^n) = (p^(n+1)-1)/(p-1) when p is prime.
>>> defrule was great to implement this rule though I was happier if there is a feature of "prime".
>>>
>>> (example)
>>> matchdeclare(pp, primep, nn, lambda([x], (integerp(x) or featurep(x, integer)) and x> 0))$
>>> defrule(sigma_simp1, divsum(pp^nn, 1), (pp^(nn + 1) - 1)/(pp - 1))$
>>>
>>> declare(n, integer)$
>>> assume(n> 0)$
>>> sigma_simp1(divsum(3^n));
>>> => (3^(n+1)-1)/2
>>>
>>> 2. divsum(a*b) = divsum(a)*divsum(b) if a and b are relatively prime.
>>>
>>> I lost my way for this simplification.
>>> Are there any manners to let Maxima check if two variables are relatively prime?
>>>
>>> Thanks,
>>> -
>>> Yuji
>>> _______________________________________________
>>> Maxima mailing list
>>> Maxima at math.utexas.edu
>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>> a. gcd(R,S) will be 1 if R and S are relatively prime.
>> b. You will probably never see a*b in Maxima if a, b are numbers. They will be multiplied together.
>> c. You will probably never see a^b in Maxima if a, b are numbers, either.
>> d. There is no reliable way to attach a condition for matching that involves 2 parameters in the match.
>>
>> RJF
>>
>