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