       Re: Product with p!=j

• To: mathgroup at smc.vnet.net
• Subject: [mg21712] Re: [mg21683] Product with p!=j
• From: BobHanlon at aol.com
• Date: Sun, 23 Jan 2000 01:52:31 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```In case you haven't tried simplifying the results; due to its symmetry, the
sum is zero except for the degenerate case of n=1. A much faster
implementation is then

q = 1;
q[_Integer] = 0;

Bob Hanlon
____________________________________________________

Break the product into two separate products.

q[n_] := Sum[
1/(Product[Subscript[k, p] - Subscript[k, j], {p, 1, j - 1}] *
Product[Subscript[k, p] - Subscript[k, j], {p, j + 1, n}]), {j, 1,
n}]

q

1/((Subscript[k, 2] - Subscript[k, 1])*(Subscript[k, 3] -
Subscript[k, 1])) +
1/((Subscript[k, 1] - Subscript[k, 2])*
(Subscript[k, 3] - Subscript[k, 2])) +
1/((Subscript[k, 1] - Subscript[k, 3])*
(Subscript[k, 2] - Subscript[k, 3]))

Bob Hanlon

In a message dated 1/21/2000 5:17:53 AM, guillerm at gugu.usal.es writes:

>I have the following problem:
>
>q[n_]:= Sum[1/Product[Subscript[k, p] - Subscript[k, j], {p, 1, n}], {j,
>1, n}], with p!=j
>
>Any suggestions?.
>

```

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