Why so offensive?
> $f$ does not depend on phi
So what?
The reason for such function --- i was trying to replicate things i was
doing in sage
(and in sage i was trying to imitate generic behaviour)
> if you want to report a bug, it is polite to find the simplest
demonstration
of the defect that you can.
Well, i didn't "report a bug", i started discussion
And there is "simple demo" in the first post
Maxima probably doesn't care about continuity, OK
But sage's wrapper does, i guess (since it was able to return correct
answer, and i can't imagine such coincidence)
So the thing that i was claiming is wrong,
is that sage returns different results for fundamentally same inputs in
different forms
Tell me if i am wrong
воскресенье, 10 апреля 2016 г., 7:01:38 UTC+3 пользователь rjf написал:
>
> I don't know what nonsense you are trying out, but
> f(r,phi):=signum(r^2-4) is a function that does not
> depend on phi.
>
> If you want to integrate functions that are discontinuous, there are two
> processes
> involved. One: find the continuous pieces and break up the problem.
> Two, integrate, as appropriate.
>
> So if your system is unable to find the continuous pieces, it is not a
> problem in integration, really.
>
> If I had a complicated expression S(x) that I thought was identically zero,
> but I couldn't prove it, I could make it into an integration problem...
>
> integrate(S(x),x,a,b)
>
> which would be zero for all a and b.
>
> So now should I report a failure to return a specific answer as an
> integration problem??
>
> Moral of the story: if you want to report a bug, it is polite to find the
> simplest demonstration
> of the defect that you can.
>
>
> Finally, this message demonstrates a major defect in the idea behind Sage.
> That is, people involved in the project might find a problem, but they have
> insufficient expertise to do anything substantive about it. If there are
> bugs in Maxima, go learn about Maxima. Maybe fix it.
>
> On Saturday, April 9, 2016 at 12:19:31 PM UTC-7, Dima Pasechnik wrote:
>>
>> maxima's definite integration is quite buggy, we see dozens bugs like
>> this a year, and report them upstream, with limited success...
>>
>>
> If you learned some Lisp, and maybe something about algorithms for
> integration
> you could fix them yourself.
>
>
>>
>> On Saturday, April 9, 2016 at 7:17:09 PM UTC+1, Sergey V Kozlukov wrote:
>>>
>>> (%i1) f(r,phi) := signum(r^2 - 4);
>>> 2
>>> (%o1) f(r, phi) := signum(r - 4)
>>> (%i2) integrate(integrate(r*f(r,phi), r, 0, 3), phi, 0, 2*%pi);
>>> (%o2) - 9 %pi
>>>
>>> That's strange
>>>
>>>
>>> суббота, 9 апреля 2016 г., 19:14:49 UTC+3 пользователь Dima Pasechnik
>>> написал:
>>>>
>>>> Try these computations directly in Maxima, and see whether it's still a
>>>> discrepancy there.
>>>>
>>>> On Saturday, April 9, 2016 at 1:31:44 PM UTC+1, Sergey V Kozlukov wrote:
>>>>>
>>>>> x, y, r, phi = var('x y r phi')
>>>>> f(x, y) = sign(x^2 + y^2 - 4)
>>>>> T(r, phi) = [r*cos(phi), r*sin(phi)]
>>>>> J = diff(T).det().simplify_full()
>>>>> T_f = f.substitute(x=T[0], y=T[1])
>>>>> int_f = integral(integral(T_f*abs(J), r, 0, 3), phi, 0, 2*pi).
>>>>> simplify_full()
>>>>> show(r"$\iint\limits_\Omega%s = %s$"%(latex(f(x)), latex(int_f)))
>>>>> Returns correct answer: $\pi$, while
>>>>> x, y, r, phi = var('x y r phi')
>>>>> f(x, y) = sign(x^2 + y^2 - 4)
>>>>> T(r, phi) = [r*cos(phi), r*sin(phi)]
>>>>> J = diff(T).det() #.simplify_full()
>>>>> T_f = f.substitute(x=T[0], y=T[1])
>>>>> int_f = integral(integral(T_f*abs(J), r, 0, 3), phi, 0, 2*pi).
>>>>> simplify_full()
>>>>> show(r"$\iint\limits_\Omega%s = %s$"%(latex(f(x)), latex(int_f)))
>>>>> Yields $-9\pi$
>>>>> The only thing simplify_full() changes here is it applies identity
>>>>> $\sin^2 + \cos^2 = 1$
>>>>>
>>>>> Code was executed in SMC worksheet
>>>>>
>>>>
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