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Re: Re: Just another bug in MMA 3.0

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  • Subject: [mg7532] Re: [mg7491] Re: [mg7431] Just another bug in MMA 3.0
  • From: koehler at (Kai Koehler)
  • Date: Fri, 13 Jun 1997 03:08:46 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

In article <5nfpv3$5qp at>, Paulo Mouat <mouat at> wrote:

> Kai Koehler wrote:

> > Sum[Sum[Log[Log[k+j]],{k,1,n}],{j,1,5}]
> > 
> > gives
> > 
> > 5*Sum[Log[Log[k + j]], {k, 1, n}].

> If you want to do a multiple sum, the input should read
> Sum[Log[Log[k+j]],{k,1,n},{j,1,5}]
> What you have typed is a simple sum over k with a function that has an
> unknown j.  The j on the outer Sum is a dummy variable, with no
> relation to the one in Log[k+j].
> This is not a bug.  Mathematica simply interpreted what you did type,
> which is not quite what you intended to do.

If this where true,


should give 5 n j as output. Instead you get 15 n (correctly, IMHO).
Also, in StandardForm, the difference between




is just the insertion of a factor (e.g. 1) between the two sums:

\!\(?\+\(j = 1\)\%5 1 \(?\+\(k = 1\)\%n Log[Log[k + j]]\)\)

gives a different output then

\!\(?\+\(j = 1\)\%5\(?\+\(k = 1\)\%n Log[Log[k + j]]\)\)


\!\(?\+\(j = 1\)\%5\((\ ?\+\(k = 1\)\%n Log[Log[k + j]])\)\)

gives a different output then

\!\(?\+\(j = 1\)\%5\((\ ?\+\(k = 1\)\%n Log[Log[k + j]] + 1)\)\)

What you wrote is actually what Michael Trott from Wolfram support mailed
me 2 days ago: It's not a bug, it's a feature. This really worries me: If
they do not even recognize a bug when you show it to them, then what can
you expect them to do?

Kai Koehler

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