Re: Puzzled by Sum

*To*: mathgroup at smc.vnet.net*Subject*: [mg124627] Re: Puzzled by Sum*From*: Dana DeLouis <dana01 at me.com>*Date*: Fri, 27 Jan 2012 06:10:28 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

> F[X_] := (Sum[A[i - j, j], {j, 1, i - 1}]) /. i -> X Hi. I believe the problem is that the Sum is evaluated First, and doesn't really have an idea what the variable i is. As Bob mentioned, using =91x in the Sum function works: a[i_,j_]:=i/j+j/i f[x_]:=(Sum[a[x-j,j],{j,1,x-1}]) f[2] 2 Perhaps an alternative to your 2 separate functions: g[n_]:= 2 n (HarmonicNumber[n]-1) Table[f[j],{j,10}] {0,2,5,26/3,77/6,87/5,223/10,962/35,4609/140,4861/126} Table[g[n],{n,10}] {0,2,5,26/3,77/6,87/5,223/10,962/35,4609/140,4861/126} %==%% True If the input is large, then it appears faster by not having to loop (By an apx. factor of 21) f[1000]//N//Timing {0.009008,12970.9} g[1000]//N//Timing {0.000427,12970.9} %% / % {21.096, 1.} = = = = = = = = = = = = = = HTH ;>) Dana DeLouis Mac, with Math 8 = = = = = = = = = = = = = = On Jan 25, 6:06 am, Themis Matsoukas <tmatsou... 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