[sage-devel] Re: Integral of piecewise functions
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 anxa(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 anxa(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 axb?
(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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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[sage-devel] Re: Integral of piecewise functions
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 anxa(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 anxa(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 axb?
(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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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[sage-devel] Re: Integral of piecewise functions
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 anxa(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 anxa(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 axb?
(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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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[sage-devel] Re: Integral of piecewise functions
This makes sense for compactly supported functions. How do you do that for something like f(x) = max(1,floor(x))? sage: f(x)=max(1,floor(x)) sage: f(10) 1 sage: type(f) class 'sage.calculus.calculus.CallableSymbolicExpression' huh, there are some problems with this function, first because it always return 1 and second because it's not built with Piecewise, which is the point of the discussion. Ronan --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: Integral of piecewise functions
On Dec 6, 2008, at 11:39 AM, Paul Butler wrote: Currently, taking the integral of a piecewise function in Sage gives you the definite integral. I've proposed on trac that the integral of piecewise functions be indefinite by default. This would be consistent with how integration works on other functions in Sage, as well as piecewise functions in Maple and Mathematica. The main concern is whether the integral of a piecewise function is even well-defined. It seems to me that at least for continuous piecewise functions, the indefinite integral is well-defined. The anti-derivative is well defined, and by the fundamental theorem of calculus, the indefinite integral of a continuous function is the anti-derivative. As for discontinuous piecewise functions, I'm finding it difficult to convince myself either way. The trac ticket is 4721 ( http://trac.sagemath.org/sage_trac/ticket/4721 ) This crops up regularly in solid mechanics, in particular in the bending of beams. One can represent the loading of a beam as derivatives and integrals of Heaviside functions and then integrate the expression to get the shear force, bending moment, and deflection of the beam. It's also one of the parts of Maple that's problematic. I wrote some code for my students a few years back to help them, but unfortunately, there are problems with some integrals of Heaviside functions. These can be discontinuous piecewise functions, at least the shear and moment graphs. Cheers, Tim. --- Tim Lahey PhD Candidate, Systems Design Engineering University of Waterloo http://www.linkedin.com/in/timlahey --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: Integral of piecewise functions
On Sat, Dec 6, 2008 at 12:37 PM, David Joyner [EMAIL PROTECTED] wrote:
On Sat, Dec 6, 2008 at 11:39 AM, Paul Butler [EMAIL PROTECTED] wrote:
Currently, taking the integral of a piecewise function in Sage gives
you the definite integral. I've proposed on trac that the integral of
piecewise functions be indefinite by default. This would be consistent
with how integration works on other functions in Sage, as well as
piecewise functions in Maple and Mathematica.
The main concern is whether the integral of a piecewise function is
even well-defined. It seems to me that at least for continuous
piecewise functions, the indefinite integral is well-defined. The
anti-derivative is well defined, and by the fundamental theorem of
calculus, the indefinite integral of a continuous function is the
anti-derivative. As for discontinuous piecewise functions, I'm finding
it difficult to convince myself either way.
First, the indefinite integral is the anti-derivative, by definition.
What the FTC says is that although the indefinite integral evalated
at b is not
well-defined, and the same antiderivative evalated at a is also not
well-defined,
their difference *is* well-defined. Moreover, this difference agrees with
the
definite integral defined by the Riemann sum between a and b.
Second, I think (but I am not sure), what you want when you say
indefinite intergal is not the indefinite integral but is a function who
derivavtive is the original function, defined as follows:
Actually, a function F which is simply the antiderivative of f is no use to
me, I need a function with the property that F(b) - F(a) is the Riemann sum
between a and b. (They are only guaranteed by the FTC to be the same
function if f is continuous.) Maybe there is a better term for this?
if the orginial function f(x) is
f1(x), a1x=a2,
f2(x), a2x=a3,
...
fn(x), anx=a{n+1}
(and 0 outside (a1,a{n+1}) then I guess you want to define the integral,
call it F, by
int_{a1}^x f1(t) dt, a1x=a2,
int_{a2}^x f2(t) dt, a2x=a3,
...
int_{an}^x fn(t) dt, anx=a{n+1}
Is this correct? This is not the antiderivative
but it does have the property that F'(x)=f(x).
I think you and I are defining antiderivative differently. I'm using the
definition that F is an antiderivative of f if F'(x) = f(x) for all x in the
domain of f(x). (Also stated here:
http://planetmath.org/encyclopedia/Antiderivative.html .)
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.)
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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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[sage-devel] Re: Integral of piecewise functions
If I had the expertise to implement it, I would do the following: The integration would return another Piecewise function in which the first interval is integrated normally, and the next ones would integrate the function in that interval and add the definite integral of the previous intervals. I think that makes some sense. Ronan Paixão Em Sáb, 2008-12-06 às 11:39 -0500, Paul Butler escreveu: Currently, taking the integral of a piecewise function in Sage gives you the definite integral. I've proposed on trac that the integral of piecewise functions be indefinite by default. This would be consistent with how integration works on other functions in Sage, as well as piecewise functions in Maple and Mathematica. The main concern is whether the integral of a piecewise function is even well-defined. It seems to me that at least for continuous piecewise functions, the indefinite integral is well-defined. The anti-derivative is well defined, and by the fundamental theorem of calculus, the indefinite integral of a continuous function is the anti-derivative. As for discontinuous piecewise functions, I'm finding it difficult to convince myself either way. The trac ticket is 4721 ( http://trac.sagemath.org/sage_trac/ticket/4721 ) -- Paul --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: Integral of piecewise functions
On Sat, Dec 6, 2008 at 3:56 PM, Ronan Paixão [EMAIL PROTECTED] wrote: If I had the expertise to implement it, I would do the following: The integration would return another Piecewise function in which the first interval is integrated normally, and the next ones would integrate the function in that interval and add the definite integral of the previous intervals. I think that makes some sense. This makes sense for compactly supported functions. How do you do that for something like f(x) = max(1,floor(x))? Ronan Paixão Em Sáb, 2008-12-06 às 11:39 -0500, Paul Butler escreveu: Currently, taking the integral of a piecewise function in Sage gives you the definite integral. I've proposed on trac that the integral of piecewise functions be indefinite by default. This would be consistent with how integration works on other functions in Sage, as well as piecewise functions in Maple and Mathematica. The main concern is whether the integral of a piecewise function is even well-defined. It seems to me that at least for continuous piecewise functions, the indefinite integral is well-defined. The anti-derivative is well defined, and by the fundamental theorem of calculus, the indefinite integral of a continuous function is the anti-derivative. As for discontinuous piecewise functions, I'm finding it difficult to convince myself either way. The trac ticket is 4721 ( http://trac.sagemath.org/sage_trac/ticket/4721 ) -- Paul --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: Integral of piecewise functions
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 axb?
(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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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[sage-devel] Re: Integral of piecewise functions
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 axb?
(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 , a1x=a2,
int f2(x) dx +C2, a2x=a3,
...
int fn(x) dx +Cn, anx=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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