Re: Just another bug in MMA 3.0

*To*: mathgroup at smc.vnet.net*Subject*: [mg7568] Re: [mg7520] Just another bug in MMA 3.0*From*: BobHanlon at aol.com*Date*: Sun, 15 Jun 1997 16:32:50 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

The iterator for the outer summation must appear first, Consequently, the specified Sum should be written as Sum[Log[Log[k+j]], {j, 1, 5}, {k, 1, n}]; vice Sum[Log[Log[k+j]], {k, 1, n}, {j, 1, 5}]; A double summation should work the same whether written with one or two Sum functions. The problem appears to arise due to the fact that the inner sum's upper limit (n) is symbolic. For example, if the inner sum's upper limit is specified Table[Sum[Sum[Log[k+j], {k, 1, n}], {j, 1, 5}], {n, 10}] == Table[Sum[Log[k+j], {j, 1, 5}, {k, 1, n}], {n, 10}] True Re-writing the summations, Sum[Log[Product[k+j, {k, 1, n}]], {j, 1, 5}] \!\(Log[Gamma[2 + n]] + Log[1\/2\ Gamma[3 + n]] + Log[1\/6\ Gamma[4 + n]] + Log[1\/24\ Gamma[5 + n]] + Log[1\/120\ Gamma[6 + n]]\) Which is the sum Sum[Log[Gamma[j+n+1]/j!], {j, 1, 5}]; Again, verifying several values: (Table[Sum[Log[k+j], {j, 1, 5}, {k, 1, n}], {n, 1, 10}] // Simplify) == (Table[Sum[Log[Gamma[j+n+1]/j!], {j, 1, 5}], {n, 1, 10}] // Simplify) True The last form can also be written as Sum[Log[Pochhammer[j+1, n]], {j, 1, 5}]; Sum[Log[Pochhammer[j+1, n]], {j, 1, 5}] == Sum[Log[Gamma[j+n+1]/j!], {j, 1, 5}] // FullSimplify True Bob Hanlon