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MathGroup Archive 1997

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Re: Re: Re: Just another bug in MMA 3.0

  • To: mathgroup at
  • Subject: [mg7587] Re: [mg7532] Re: [mg7491] Re: [mg7431] Just another bug in MMA 3.0
  • From: Allan Hayes <hay at>
  • Date: Thu, 19 Jun 1997 03:13:50 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

On 13 Jun 1997
Kai Koehler<koehler at>
in [mg7532] Re: [mg7491] Re: [mg7431] Just another bug in MMA 3.0
wrote as copied after **************

The original problem [mg7491] was that

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

The behaviour seems to be explained by the fact that with j set to  
1 (j =1) then 	
does not sum and so is returned unevaluated; similarly with j =  
2,3,4,5. Thus we end up adding together 5 copies of it.

The following two examples show up the processm (I change the 5 to  
3 to reduce the print out)

(1) If we put n = 2 then the summations are done


	Log[Log[2]]+2 Log[Log[3]]+2 Log[Log[4]]+Log[Log[5]]
(2) If we drop the outer Log then again the summations are done


Allan Hayes
hay at
voice:+44 (0)116 2714198
fax: +44 (0)116 2718642
12 Copse Close, Leicester, LE2 4FB, UK

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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