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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