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MathGroup Archive 2000

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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] = 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[3]

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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