quasi wrote:
> On Sat, 26 Jan 2008 06:39:09 -0500, quasi <quasi@[EMAIL PROTECTED]
> wrote:
>
>> On Sat, 26 Jan 2008 03:07:45 -0800 (PST), Vladimir Bondarenko
>> <vb@[EMAIL PROTECTED]
> wrote:
>>
>>> ----------------------------------------------------------------
>>>
>>> An exact 1-D integration challenge - 48 -
>>> (go and give a kick to all stupid CASs!)
>>>
>>>
http://groups.google.com/group/sci.math.symbolic/browse_thread/thread/28416525f0de8f90/1d8cd83e96cb7f63?#1d8cd83e96cb7f63
>>>
>>> ----------------------------------------------------------------
>>>
>>> N[Integrate[
>>>
>>> Sqrt[Sqrt[2] + Sqrt[z] + Sqrt[2 + 2 Sqrt[2] Sqrt[z] + 2 z]],
>>>
>>> {z, 0, 1}]]
>>>
>>> ----------------------------------------------------------------
>>> VERSION OUTPUT RESOLUTION
>>> ----------------------------------------------------------------
>>>
>>> Mathematica 6.0 2.67602 <-------------------------------- BUG
>>>
>>> Mathematica 5.2 3.66533 <-------------------------------- BUG
>> Let f(z) = sqrt(sqrt(2) + sqrt(z) + sqrt(2 + 2*sqrt(2)*sqrt(z) + 2*z))
>>
>> Then it's completely obvious that f has domain [0,infinity) and that
>> is strictly increasing.
>>
>> Hence the integral of f on the interval [0,1] must be strictly between
>> f(0) and f(1).
>>
>> Approximately, we have
>>
>> f(0) = 1.681792831
>>
>> and
>>
>> f(1) = 2.242172940
>>
>> which makes it clear that Mathematica's outputs are ridiculous.
>>
>> Maple 9.5 gives a result of 2.062759840, which at least seems
>> reasonable. Whether it's correct or not, I'm not sure.
>>
>> quasi
>
> A CAS should never return a wrong answer. Returning no answer is
> better than a wrong one.
>
> For this example in particular, Mathematica really has no excuse for
> getting it wrong. It's not like the function has wild variations which
> might throw off a numerical integration. This is a positive,
> increasing function, easily evaluated. For a CAS, the numerical
> integration of such a function should be effortless.
>
> quasi
Finding a possible (numerical) error is only half the job
of a tester. I would expect him to locate it closer by
deforming the situation, easiest would be the bounds.
Then one could say something whether MMA is disturbed
by the vertical tangent in 0.
Changing z ---> t^2 should do it.
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