On Sat, Dec 6, 2008 at 3:48 PM, Paul Butler <[EMAIL PROTECTED]> wrote:
>
>
> Either way, the property F'(x) = f(x) is not necessarily true for piecewise
> antiderivatives defined that way. Consider this function.
>
> f(x) = x, 0 <= x <= 1
> f(x) = 1, 1 < x
>
> If we use the definition you gave to find F = integral(f), F'(1) is
> undefined so it is not true that F'(x) == f(x) for all x.
>
> Instead, we use the definition that F=
>
> integrate(f1, t, a1, x), a1 < x <= a2
> integrate(f2, t, a2, x) + integrate(f1, t, a1, a2), a2 < x <= a3
> integrate(f3, t, a3, x) + integrate(f2, t, a2, a3) + integrate(f1, t, a1,
> a2), a3 < x <= a4
> ...
> integrate(fn, t, an, x) + integrate(f[n-1], t, a[n-1], an) + ... +
> integrate(f1, t, a1, a2), an < x
>
> (We also need a special case for when a1 = -infinity, which I didn't show.)
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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