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MathGroup Archive 1997

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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


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