On Tue, Dec 9, 2008 at 9:30 PM, Paul Butler <[EMAIL PROTECTED]> wrote:
>
> David, thanks.
>
> What do you mean by an+ and an- in the notation?
In general, by c+ I mean the limit as you approach c from the right,
and c- is the limit as you approach c from the left.
>
> From what I can tell, that does look like the same algorithm I plan to
> use. I have some working code now, I just want to review it and make
> sure I didn't miss any edge cases before re-posting the patch. The
> code posted as a patch now uses the same algorithm, but doesn't
> properly handle unbounded functions.
Great. I discussed the algorithm with a few colleagues and I am pretty sure it
will work even if (when?) piecewise functions with infinitely many parts are
implemented in Sage. It's cool that once you fix one of the constants
of integration
they are all determined, thanks to the FTC, which is consistent with how the
inverse of the derivative map should behave.
Looking forward to your patch.
>
> -- Paul
>
> On Tue, Dec 9, 2008 at 8:32 AM, David Joyner <[EMAIL PROTECTED]> wrote:
>>
>> Paul:
>> It was not at all clear to me when you first started working on the
>> antiderivatives of piecewise defined functions that it was possible or
>> made sense.
>> Now I have changed my mind to an extent. If you think of the
>> antiderivative as the
>> (multi-valued) inverse of the differentiation map, as I usually do, then I'm
>> not sure what to do. But if you just want a function whose derivavtive is the
>> original and which satisfies the FTC, then I think I have an algorithm which
>> might work.
>>
>> Let f(x) be the piecewise defined function which is fn(x) for an<x<a(n+1),
>> for n in ZZ (and let us ignore endpots for now). Here a_n is an
>> infinite sequence
>> a_{-infty}=-infty and a_{infty}=+infty and each f_n(x) is Riemann integrable
>> with antiderivative Fn(x)+Cn. The question is can we piece the Cns's
>> together so that
>> F(x) = Fn(x)+Cn for an<x<a(n+1) satisfies the FTC? These constants must
>> satisfy
>>
>> Cn = F_{n-1}(an-)-Fn(an+)+C_{n-1},
>>
>> If you fix C0 = C to be an arbitrary constant (you could take this to
>> be 0 for example),
>> this is a 1-step recursionrelation which can be solved.
>>
>> I think this should do the job.
>>
>> Thanks for raising this issue!
>>
>> ++++++++++++++++++++++++++++++++++++
>>
>>
>> On Sat, Dec 6, 2008 at 5:06 PM, Paul Butler <[EMAIL PROTECTED]> wrote:
>>> When a1 = -infinity, I would make F1 = integrate(f, x, a2, x) instead of
>>> integrate(f, x, a1, x). Then I would not calculate the definite integral of
>>> the first interval, which would align my constants so that F(a2) = 0. When I
>>> get a chance, I'll add this to my code.
>>>
>>> Functions like floor with an infinite number of pieces are beyond the scope
>>> of the Piecewise class, because piecewise functions in Sage can only (at the
>>> moment) have finitely many pieces. I can't give an algorithm for integration
>>> of all piecewise functions with multiple pieces off the top of my head, but
>>> I'll give it some thought.
>>>
>>> -- Paul
>>>
>>> On Sat, Dec 6, 2008 at 4:13 PM, David Joyner <[EMAIL PROTECTED]> wrote:
>>>>
>>>> Okay, this helps me understand what you mean.
>>>>
>>>> Still, the case a1=-ifinty is precisely the special case which I don't
>>>> understand.
>>>>
>>>> For example, take a function such as f(x) = max(1,floor(x)), x real.
>>>> How do you define an antiderivative F(x) so that
>>>> F(b)-F(a) = area under the y=f(x) for a<x<b?
>>>> (And mayeb you can do it for that special function,
>>>> and let us ignore points of discontinuity for the sake of
>>>> discussion.) In other words, I am asking for the algorithmic procedure
>>>> you would use to create an "area function" of a piecewise-defined
>>>> function on the reals.
>>>>
>>>>
>>>> >
>>>> > With this definition, F(b) - F(a) can be used to find the Riemann sum
>>>> > between a and b. Also, F'(x) = f(x) seems to hold, except at points
>>>> > where
>>>> > f(x) goes from defined to undefined or vice-versa.
>>>> >
>>>> >> The antiderivative is only well-defined up to an additive constant.
>>>> >> IMHO, the piecewise defined function of antiderivavtives
>>>> >>
>>>> >> int f1(x) dx +C1 , a1<x<=a2,
>>>> >> int f2(x) dx +C2, a2<x<=a3,
>>>> >> ...
>>>> >> int fn(x) dx +Cn, an<x<=a{n+1}
>>>> >>
>>>> >> does not make sense.
>>>> >
>>>> > I agree that it doesn't make sense where C1 .. Cn are arbitrary
>>>> > constants.
>>>> >
>>>> > -- Paul
>>>> >
>>>> > >
>>>> >
>>>>
>>>>
>>>
>>>
>>> >
>>>
>>
>> >
>>
>
> >
>
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