David, thanks.
What do you mean by an+ and an- in the notation?
>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.
-- 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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