Re: Product/Sums
- To: mathgroup at smc.vnet.net
- Subject: [mg43022] Re: Product/Sums
- From: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>
- Date: Fri, 8 Aug 2003 00:26:36 -0400 (EDT)
- References: <bgsne4$np6$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Your first problem (the n=3 case) Sum[ k1=1 to q] Sum[k2=k1 to q] Sum[k3=k2
to q] (k1*k2*k3) can be written as follows:
Fold[Sum[#1, #2] &, k1 k2 k3, {{k3, k2, q}, {k2, k1, q}, {k1, 1, q}}]
which evaluates to
(1/48)*q^2*(1 + q)^2*(2 + q)*(3 + q)
In order to generalise this you first need to write it in the follwing way:
Fold[Sum[#1, #2] & , Product[k[i], {i, 3}], Table[{k[i], k[i - 1], q}, {i,
3, 1, -1}] /. k[0] -> 1]
which evaluates to the same result as above.
You can then replace the "3" by any specific other integer value n to obtain
the corresponding result. Here is a list of results I obtained for
n=1,...,5:
{(1/2)*q*(1 + q), (1/24)*q*(1 + q)*(2 + q)*(1 + 3*q), (1/48)*q^2*(1 +
q)^2*(2 + q)*
(3 + q), (q*(1 + q)*(2 + q)*(3 + q)*(4 + q)*(-2 + 5*q + 30*q^2 +
15*q^3))/5760,
(q^2*(1 + q)^2*(2 + q)*(3 + q)*(4 + q)*(5 + q)*(-2 + 7*q + 3*q^2))/11520}
--
Steve Luttrell
West Malvern, UK
"Blimbaum Jerry DLPC" <BlimbaumJE at ncsc.navy.mil> wrote in message
news:bgsne4$np6$1 at smc.vnet.net...
> Given the Sum product....where the number of sums is variable and
> nested...example....which I write logically rather then Mathematically,
> easier for me and I think for understanding....
>
> Given...
>
> Sum[ k1=1 to q] Sum[k2=k1 to q] Sum[k3=k2 to q] (k1*k2*k3)
>
> is there a way to write this as ...Product[Sum] ?
>
>
> I tried Product[a=1 to 3] Sum[k_subscript_a = k_subscript_(a-1) to q]
> but this didnt work...no nesting
>
> What I want is to be able to vary a = 1 to n.....
>
>
> any ideas?
>
>
> thanks....jerry blimbaum
>