MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Nasty bug in Integrate (version 5.0)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg45982] Re: [mg45958] Re: Nasty bug in Integrate (version 5.0)
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 31 Jan 2004 05:21:02 -0500 (EST)
  • References: <200401281019.FAA18530@smc.vnet.net> <bvapk6$aat$1@smc.vnet.net> <200401300917.EAA05062@smc.vnet.net> <68A351B5-5333-11D8-94B3-00039311C1CC@mimuw.edu.pl> <opr2mc5zdhamtwdy@smtp.cox-internet.com>
  • Sender: owner-wri-mathgroup at wolfram.com

On 31 Jan 2004, at 03:57, Dr Bob wrote:

>>> if you are going to use it only for problems that are trivial to do 
>>> by hand what is the point of using it at all?
>
> Why use it only for the ones that are too big to check? That way, we 
> don't know whether we're wrong or not. We wouldn't know about this 
> bug, for instance.

 From reading the Mathgroup I get the impression that more people are 
"debugging" Mathematica than using it. This may not bad thing for those 
who are actually using it, since it presumably helps to improve the 
program, but I am surprised that so many find this sort of thing worth 
their while.

Andrzej



>
> Bobby
>
> On Fri, 30 Jan 2004 15:48:47 +0100, Andrzej Kozlowski 
> <akoz at mimuw.edu.pl> wrote:
>
>> But then if you are going to use it only for problems that are 
>> trivial to do by hand what is the point of using it at all? Computers 
>> are supposed to be good at computing, we are supposed to be good at 
>> thinking.
>>
>> Andrzej
>>
>>
>> On 30 Jan 2004, at 10:17, Bobby R. Treat wrote:
>>
>>> The only reason we know the answer is wrong is BECAUSE it's trivial 
>>> by
>>> hand. It makes no sense to trust Mathematica for problems that are 
>>> too hard to
>>> check.
>>>
>>> Bobby
>>>
>>> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote in message 
>>> news:<bvapk6$aat$1 at smc.vnet.net>...
>>>> Mathematica seems to get lost in the large number of checks that it
>>>> attempts to perform to find the exact conditions  (in the complex
>>>> plane) for the convergence of this integral. Even if a and b are
>>>> assumed to be real there is a problem that 0 may lie in between 
>>>> them.
>>>> Only in the case when you specify that the limits are both positive 
>>>> or
>>>> both negative you get the correct answers because Mathematica can
>>>> quickly deal with this problem.
>>>> If you really want to see how much checking Mathematica attempts to 
>>>> do
>>>> just evaluate:
>>>>
>>>> Trace[Integrate[1/x + x^c, {x, a, b}], TraceInternal->True]
>>>>
>>>> and be prepared to wait and obtain a huge output.
>>>>
>>>> Anyway, this is clearly a  bug. However, it is in general a good 
>>>> idea
>>>> not to ask a computer program to do what is trivial to do by hand,
>>>> which in this case means evaluating a definite rather than an
>>>> indefinite integral. Note that
>>>>
>>>>
>>>> Integrate[1/x + x^c, x]
>>>>
>>>> x^(c + 1)/(c + 1) + Log[x]
>>>>
>>>> and the answer is obtained instantly. If this is what you really 
>>>> were
>>>> after (you can substitute in the limits for example by using:
>>>>
>>>>
>>>> Subtract @@ (Integrate[1/x + x^c, x] /. x -> #1 & ) /@ {b, a}
>>>>
>>>>
>>>> -(a^(c + 1)/(c + 1)) - Log[a] + Log[b] + b^(c + 1)/(c + 1)
>>>>
>>>>
>>>> Of course this now makes sense only under certain conditions on a,b 
>>>> and
>>>> c which you now have to determine yourself.
>>>>
>>>>
>>>> On 28 Jan 2004, at 11:19, Math User wrote:
>>>>
>>>>> Hi,
>>>>>
>>>>> Sorry if this has been discussed before. Could anyone explain why
>>>>> Mathematica 5.0 is ignoring the term 1/x in the first and second
>>>>> answers?
>>>>>
>>>>> Thanks!
>>>>>
>>>>>
>>>>> In[1]:= Integrate[1/x + x^c, {x, a, b}]
>>>>>
>>>>> Out[1]= (-a^(1 + c) + b^(1 + c))/(1 + c)
>>>>>
>>>>> In[2]:= Integrate[1/x + x^c, {x, a, b}, Assumptions -> {Element[a,
>>>>> Reals], Element[b, Reals]}]
>>>>>
>>>>> Out[2]= (-a^(1 + c) + b^(1 + c))/(1 + c)
>>>>>
>>>>> In[3]:= Integrate[1/x + x^c, {x, a, b}, Assumptions -> {a > 0, b > 
>>>>> 0}]
>>>>>
>>>>> Out[3]= (-a^(1 + c) + b^(1 + c))/(1 + c) + Log[b/a]
>>>>>
>>>>>
>>>
>>>
>>
>
>
>
>


  • Prev by Date: Re: Simplifying a second order eq. system
  • Next by Date: Re: one liner for a function?
  • Previous by thread: Re: Re: Nasty bug in Integrate (version 5.0)
  • Next by thread: Re: Nasty bug in Integrate (version 5.0)