On May 18th Derka Ladislav posted
"Simplifying x^a*y^a to (x*y)^a" to the list.
I just note that his new function expcontract
(repeated below from post on May 21st)
works on this summation.
exp_log_to_power(expr) := block([logconcoeffp: lambda([s],true)],
logcontract(expr))$
expcontract(expr) := block([logexpand: 'super],
scanmap(lambda([z], if atom(z) then z else
(exp_log_to_power(exp(logcontract(log(z)))))), expr)
)$
sum(binomial(k+1,2)^(n-k),k,1,n);
expcontract(%);
also, nothing new, but to get the
first 6 terms of the summation:
N:6$
sum(binomial(k+1,2)^(n-k),k,1,N);
and to evaluate for n=6
n:6$
sum(binomial(k+1,2)^(n-k),k,1,N);
414
L. Prevett
On Fri, May 24, 2013 at 10:04 AM, Przemek Klosowski <
przemek.klosowski at nist.gov> wrote:
> I am having trouble with finite sums. I started looking at some simple
> examples to try to get a better grasp--please let me know if there's a
> better way to handle sum() calculations. While 'sum(1,i,1,n)' returns 'n'
> as it should, sum(i,i,1,n) remains in open form. From describe(sum) I
> picked up that I need to do sum(i,i,1,n),simpsum to get the expected
>
> 2
> n + n
> (%o9) ------
> 2
>
>
> My first newbie question is what is that construct 'sum(),simpsum'? is
> simpsum a procedure evaluated on output of sum(), or some global flag?
> The help doesn't describe it, but just mention it in the examples.
>
> My real summation formula is sum(binomial(k+1,2)^(n-k),k,1,**n) (it has
> to do with a number of ways you can configure an electronic circiut with n
> components). I wasn't able to simplify this expression in any way; please
> let me know if you have any suggestions.
> ______________________________**_________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/**mailman/listinfo/maxima<http://www.math.utexas.edu/mailman/listinfo/maxima>;
>