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Re: Puzzled by Sum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg124615] Re: Puzzled by Sum
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Thu, 26 Jan 2012 03:31:40 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201201251203.HAA05962@smc.vnet.net>

A[i_, j_] := i/j + j/i

Clear[F]

F[X_] := Sum[A[X - j, j], {j, 1, X - 1}]

F[2]

2

Clear[F]

F[X_] = Sum[A[X - j, j], {j, 1, X - 1}];

F[2]

2


Bob Hanlon

On Wed, Jan 25, 2012 at 7:03 AM, Themis Matsoukas <tmatsoukas at me.com> wrote:
> I am trying to define a function F[X] whose argument is the upper limit in a summation:
>
> A[i_, j_] := i/j + j/i
> F[X_] := (Sum[A[i - j, j], {j, 1, i - 1}]) /. i -> X
>
> but when I evaluate, say F[2], I get
>
> -2 DifferenceRoot[{\[FormalY],\[FormalN]}\[Function]{(\[FormalN]-2) \[FormalY](\[FormalN])+(3-2 \[FormalN]) \[FormalY](\[FormalN]+1)+(\[FormalN]-1) \[FormalY](\[FormalN]+2)==0,\[FormalY](0)==0,\[FormalY](1)==-(1/2)}][2]-1
>
> where the result should have been 2:
>
> ii = 2;
> Sum[A[ii - j, j], {j, 1, ii - 1}]
>
> 2
>
> If I use A[i,j]=i j or other simple functions, everything works as expected.
>
> Themis
>



-- 
Bob Hanlon



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