A continuous antiderivative for the floor function
David W. Cantrell
square brackets
[x]
has jump discontinuities at all integers. As such, in giving an
antiderivative for it, we may choose a different constant of integration on
each interval (n, n+1). But certain choices are nicer than others. We
examine the possibilities and ask a question about terminology.
-----------------
A general antiderivative for [x] has the form
(1) x*[x] + c_n for n < x < n+1.
(Our n are always integer.) The constants c_n are arbitrary, of course. It
could be argued that we should stop here, that nothing further need be
said. But...
What about the derivative of (1) _at_ x = n? Well, we didn't even bother to
define (1) at integers. Even if we had, it would not have been
differentiable at integers... However, [x] is continuous from the right
everywhere. If we accordingly use an antiderivative of the form
(2) x*[x] + c_n for n <= x < n+1,
then it is right-differentiable everywhere. (A "right-antiderivative"?)
In a sense, (2) is nicer than (1): for _all_ real x, the right-derivative
of (2) is [x].
But we can still do better, in another sense. If we choose c_n
appropriately, we can get an antiderivative which is continuous on R. After
a little examination, it can be seen that such a continuous antiderivative
must be of the form
(3) F(x) = [x]*(x - ([x] + 1)/2) + C
for some arbitrary constant C. (By the way, at first glance, (3) might not
appear to be a special case of (2), but it is. Just note that, on any
interval [n, n+1), [x]*([x] + 1)/2 is indeed a _constant_.)
Continuity _per se_ is not the only reason that (3) is particularly nice.
Suppose that we wish to evaluate the definite integral of [x] wrt x from a
to b. That integral is now given simply by F(b) - F(a), as though we were
using the Fundamental Theorem.
Do such antiderivatives, which are continuous, even when the original
function is not, have a special name?
David Cantrell
Larry Hammick
David W. Cantrell wrote:
> Do such antiderivatives, which are continuous, even when the original
> function is not, have a special name?
version of Dieudonne's Treatise on Analysis VIII.7 :
"
Let f be a mapping of an interval I \subset R into a Banach space F. We
say that a continuous mapping g of I into F is a _primitive_ of f in I
if there exists a denumerable set D \subset I such that, for any \eta
\in I-D, g is differentiable at \eta and
g'(\eta) = f(\eta).
"
Note that he specifies that g is continous.
This is only my guess at an answer. Dieudonne's usage is not always
universal, and anyways, I had to look it up :)
LH
Gene Ward Smith
David W. Cantrell wrote:
> The floor (a.k.a. greatest integer) function, notated hereafter using
> square brackets
>
> [x]
>
> has jump discontinuities at all integers. As such, in giving an
> antiderivative for it, we may choose a different constant of integration on
> each interval (n, n+1). But certain choices are nicer than others. We
> examine the possibilities and ask a question about terminology.
Is there some reason why you don't simply use
F(x) = \int_0^x [u] du ?
Virgil
And how, as a practical matter, does one evaluate that expression for
values of x greater than 1? Darboux sums?
Proginoskes
The Riemann sum can be used; it exists, since the discontinuities of
[x] form a countable set. Lebesgue integration can also be done.
--- Christopher Heckman
Virgil
"Proginoskes" <CCHe...@gmail.com> wrote:
But an explicit formula, such as David suggested, would be considerably
more convenient for evaluating \int_a^b [u] du.
Particularly if one needed to do it many times with different limits of
integration.
Gene Ward Smith
Virgil wrote:
> But an explicit formula, such as David suggested, would be considerably
> more convenient for evaluating \int_a^b [u] du.
>
> Particularly if one needed to do it many times with different limits of
> integration.
If he'd said "here is an explicit formula for \int_0^x [u] du" then I
would have known what he was up to. *Is* that what he was trying to do?
Virgil
"Gene Ward Smith" <genewa...@gmail.com> wrote:
Other than allowing for other lower limits than 0 for the integral, yes.
His generalization is F(x) = [x]*(x - ([x] + 1)/2) + C
Taking C = 0, one gets F(x) = \int_0^x [u] du
The right sided derivative of F exists and equals [x] for all x and the
two sided derivative exists except when x is an integer, when the left
sides derivative is smaller by 1 that the right sided derivative.
Lim_{h -> 0+} (F(x+h)-f(x))/h = [x] for all x
Lim_{h -> 0} (F(x+h)-f(x))/h = [x] when x > [x]
Lim_{h -> 0-} (F(x+h)-f(x))/h = [x]-1 = [x-1] when x = [x]
David W. Cantrell
Thanks to everyone for replying and, in some cases, speaking for me. But I
need to answer Gene myself:
Yes, I did give an explicit formula for \int_0^x [u] du, in the sense that
just letting C = 0 in my (3) gives that formula. But that is not what I was
up to, ultimately. I was really just trying to lay the groundwork for a
simple question about terminology.
I realized, before making my original post, that one answer to my question
would be
\int_a^x f(u) du
but I was hoping for a nicer _name_ than that. Larry suggested "primitive",
but I think that most people consider that to be synonymous with
"antiderivative".
BTW, this thread was inspired by a question asked in the Mathematica
newgroup. Someone was needing to integrate {1/x}, where {t} denotes the
fractional part of t. Mathematica returns even Integrate[Floor[x], x]
unevaluated, and so it's not surprising that it does the same with
Integrate[FractionalPart[1/x], x]. But, by rewriting {t} in terms of
elementary functions first, Mathematica can then give an antiderivative
in closed form. Unfortunately, that antiderivative has jump
discontinuities. It's unfortunate because the person wanted to use his
antiderivative to evaluate _definite_ integrals of {1/x}, as if the
Fundamental Theorem were applicable. But I did supply him with a closed
form for an antiderivative which is continuous for x > 0:
Mod[1, x] + PolyGamma[Floor[1/x] + 1] + Log[x]
and so I hope that solves his problem. (BTW, PolyGamma[t] in Mathematica is
the digamma function.) Alas, I don't know how to get Mathematica to find
an antiderivative like the one I supplied.
David
michaeld
> The floor (a.k.a. greatest integer) function, notated hereafter using
> square brackets
>
> [x]
>
> has jump discontinuities at all integers. As such, in giving an
> antiderivative for it, we may choose a different constant of integration on
> each interval (n, n+1). But certain choices are nicer than others. We
> examine the possibilities and ask a question about terminology.
Just wondering... what's your definition of antiderivative? Needless to
say there is no differentiable F: R->R with F'(x) = [x] (for example,
because F' must be Darboux continuous). Are you defining it as a
differentiable function F: R-Z -> R with F'(x) = [x]? Or is there some
more general definition?
[...]
David W. Cantrell
> David W. Cantrell wrote:
> > The floor (a.k.a. greatest integer) function, notated hereafter using
> > square brackets
> >
> > [x]
> >
> > has jump discontinuities at all integers. As such, in giving an
> > antiderivative for it, we may choose a different constant of
> > integration on each interval (n, n+1). But certain choices are nicer
> > than others. We examine the possibilities and ask a question about
> > terminology.
>
> Just wondering... what's your definition of antiderivative?
Just the same as yours, I bet.
> Needless to say there is no differentiable F: R->R with F'(x) = [x]
Of course. (I must wonder if you read my post.) But I noted, as item (2),
that there is a right-differentiable F: R->R with F'(x) = [x].
> (for example, because F' must be Darboux continuous). Are you defining
> it as a differentiable function F: R-Z -> R with F'(x) = [x]?
See item (1).
There are different functions, all of which are antiderivatives (in the
usual sense) on R-Z. But some are nicer than others. For the nicest, see
item (3).
David
michaeld
David W. Cantrell wrote:
> "michaeld" <mich...@cantab.net> wrote:
> > David W. Cantrell wrote:
> > > The floor (a.k.a. greatest integer) function, notated hereafter using
> > > square brackets
> > >
> > > [x]
> > >
> > > has jump discontinuities at all integers. As such, in giving an
> > > antiderivative for it, we may choose a different constant of
> > > integration on each interval (n, n+1). But certain choices are nicer
> > > than others. We examine the possibilities and ask a question about
> > > terminology.
> >
> > Just wondering... what's your definition of antiderivative?
>
> Just the same as yours, I bet.
Probably not.
> > Needless to say there is no differentiable F: R->R with F'(x) = [x]
>
> Of course. (I must wonder if you read my post.)
I didn't say that you weren't aware of this (thus the "needless to
say") but my definition of an antiderivative of f is precisely: a
differentiable function F such that F' = f. Thus my puzzlement when you
talk about "antideriatives of [x]".
> But I noted, as item (2),
> that there is a right-differentiable F: R->R with F'(x) = [x].
And not just one but infinitely many (even modulo the arbitrary
constant). F can have as many jump discontinuities as you like as long
as F is right continuous (note that the function 1_{x >= a} is
right-differentiable with right-derivative zero - and you can add on
any linear combination of these to F without changing its right
derivative).
> > (for example, because F' must be Darboux continuous). Are you defining
> > it as a differentiable function F: R-Z -> R with F'(x) = [x]?
>
> See item (1).
>
> There are different functions, all of which are antiderivatives (in the
> usual sense) on R-Z. But some are nicer than others. For the nicest, see
> item (3).
I have no disagreement with any of the mathematics in your post, but
your use of the term antiderivative seems puzzling to me. For example
if I'm allowed to let the equation F' = f be true except on a discrete
set then what's to stop me from having discontinuities outside Z?
Virgil
"michaeld" <mich...@cantab.net> wrote:
> David W. Cantrell wrote:
> > But I noted, as item (2),
> > that there is a right-differentiable F: R->R with F'(x) = [x].
>
> And not just one but infinitely many (even modulo the arbitrary
> constant).
Not if one wants continuity as well.
>
> I have no disagreement with any of the mathematics in your post, but
> your use of the term antiderivative seems puzzling to me. For example
> if I'm allowed to let the equation F' = f be true except on a discrete
> set then what's to stop me from having discontinuities outside Z?
And if one requires that one's antiderivative be continuous as well?
That certainly is a reasonable additional requirement for
"anitderivatives", even for functions which are not themselves
continuous but have only step discontinuities of the sort that [x] has.
michaeld
> In article <1148571799....@g10g2000cwb.googlegroups.com>,
> "michaeld" <mich...@cantab.net> wrote:
>
> > David W. Cantrell wrote:
>
> > > But I noted, as item (2),
> > > that there is a right-differentiable F: R->R with F'(x) = [x].
> >
> > And not just one but infinitely many (even modulo the arbitrary
> > constant).
>
> Not if one wants continuity as well.
Actually I don't think so. If my sketch construction is correct, it's
perfectly possible for F to be continuous and right-differentiable with
right-derivative zero, and yet not be a constant (in fact strictly
increasing). The idea is simple enough but it'd be a bit of a pain to
write out the construction here so I'll wait and see if anyone
supports/disputes this conclusion.
But anyway if that's correct then [x] would have infinitely many
continuous right-antiderivatives even modulo an overall constant.
David W. Cantrell
> David W. Cantrell wrote:
> > "michaeld" <mich...@cantab.net> wrote:
> > > David W. Cantrell wrote:
> > > > The floor (a.k.a. greatest integer) function, notated hereafter
> > > > using square brackets
> > > >
> > > > [x]
> > > >
> > > > has jump discontinuities at all integers. As such, in giving an
> > > > antiderivative for it, we may choose a different constant of
> > > > integration on each interval (n, n+1). But certain choices are
> > > > nicer than others. We examine the possibilities and ask a question
> > > > about terminology.
> > >
> > > Just wondering... what's your definition of antiderivative?
> >
> > Just the same as yours, I bet.
>
> Probably not.
Yes, they are the same. See below.
> > > Needless to say there is no differentiable F: R->R with F'(x) = [x]
> >
> > Of course. (I must wonder if you read my post.)
>
> I didn't say that you weren't aware of this (thus the "needless to
> say") but my definition of an antiderivative of f is precisely: a
> differentiable function F such that F' = f.
I agree.
> Thus my puzzlement when you talk about "antideriatives of [x]".
Sorry if I misspelled "antiderivatives". Other than that, I am puzzled at
your puzzlement.
> > But I noted, as item (2),
> > that there is a right-differentiable F: R->R with F'(x) = [x].
>
> And not just one but infinitely many (even modulo the arbitrary
> constant). F can have as many jump discontinuities as you like as long
> as F is right continuous (note that the function 1_{x >= a} is
> right-differentiable with right-derivative zero - and you can add on
> any linear combination of these to F without changing its right
> derivative).
Apparently, I was not adequately clear in my original post. Sorry.
In going from (1) to (2) and finally (3) I was never weakening or
dropping any requirements. Quite the opposite. I was progressively adding
requirements in stages. They _all_ had to be antiderivatives on R-Z. That
was the minimum requirement; it was never dropped. At stage (2), I added
the requirement that it be a "right-antiderivative" as well.
> > > (for example, because F' must be Darboux continuous). Are you
> > > defining it as a differentiable function F: R-Z -> R with F'(x) =
> > > [x]?
> >
> > See item (1).
> >
> > There are different functions, all of which are antiderivatives (in the
> > usual sense) on R-Z. But some are nicer than others. For the nicest,
> > see item (3).
>
> I have no disagreement with any of the mathematics in your post, but
> your use of the term antiderivative seems puzzling to me. For example
> if I'm allowed to let the equation F' = f be true except on a discrete
> set then what's to stop me from having discontinuities outside Z?
You're not allowed to do that if we wish to consider only, as I said,
"functions, all of which are antiderivatives (in the usual sense) on R-Z."
David
Virgil
"michaeld" <mich...@cantab.net> wrote:
> Virgil wrote:
> > In article <1148571799....@g10g2000cwb.googlegroups.com>,
> > "michaeld" <mich...@cantab.net> wrote:
> >
> > > David W. Cantrell wrote:
> >
> > > > But I noted, as item (2),
> > > > that there is a right-differentiable F: R->R with F'(x) = [x].
> > >
> > > And not just one but infinitely many (even modulo the arbitrary
> > > constant).
> >
> > Not if one wants continuity as well.
>
> Actually I don't think so. If my sketch construction is correct, it's
> perfectly possible for F to be continuous and right-differentiable with
> right-derivative zero, and yet not be a constant (in fact strictly
> increasing). The idea is simple enough but it'd be a bit of a pain to
> write out the construction here so I'll wait and see if anyone
> supports/disputes this conclusion.
Can you do it when the points of non-definition of the (two-sided)
derivative are all isolated points as they are for [x]?
michaeld
[...]
OK, it was just not clear to me from your original posting that you
were interested, specifically, in antiderivatives on R-Z (I think you
just talked about antiderivatives of [x] without mentioning the domain
explicitly). Now that that is established, all is clear. Thank you.
michaeld
Virgil wrote:
> In article <1148600633.5...@38g2000cwa.googlegroups.com>,
> "michaeld" <mich...@cantab.net> wrote:
>
> > Virgil wrote:
> > > In article <1148571799....@g10g2000cwb.googlegroups.com>,
> > > "michaeld" <mich...@cantab.net> wrote:
> > >
> > > > David W. Cantrell wrote:
> > >
> > > > > But I noted, as item (2),
> > > > > that there is a right-differentiable F: R->R with F'(x) = [x].
> > > >
> > > > And not just one but infinitely many (even modulo the arbitrary
> > > > constant).
> > >
> > > Not if one wants continuity as well.
> >
> > Actually I don't think so. If my sketch construction is correct, it's
> > perfectly possible for F to be continuous and right-differentiable with
> > right-derivative zero, and yet not be a constant (in fact strictly
> > increasing). The idea is simple enough but it'd be a bit of a pain to
> > write out the construction here so I'll wait and see if anyone
> > supports/disputes this conclusion.
>
> Can you do it when the points of non-definition of the (two-sided)
> derivative are all isolated points as they are for [x]?
No; of course if F is differentiable except at isolated points, and is
continuous and right differentiable everywhere with right derivative
zero, then F is constant.
It's still true though (modulo the correctness of my earlier claim)
that there are infinitely many right differentiable, continuous F with
right derivative [x] (even with F(0) fixed).
Zdislav V. Kovarik
I deleted the original inquiry, so: in Lebesgue sense (absolutely
continuous function whose derivative is the floor function almost
everywhere, and certainly at points of continuity of the integrand)
I had the patience to observe the pattern and came up with
x * floor(x) - floor(x) * (floor(x) + 1) / 2
(compare the left and right limits at integers).
Cheers, ZVK(Slavek).
David W. Cantrell
Yes, that's what I had given at the beginning of the thread. And I
found just yesterday that it's given at the Wolfram Functions site.
See the second result at
<http://functions.wolfram.com/IntegerFunctions/Floor/21/02/01/>.
Cheers,
David