Re: [math-fun] Can a nonconstant continuous function on the reals
Here is my example of a continuous function which maps rationals to rationals, but is not a rational function:
F(x) = (-1)^floor(x) * frac(|x|) * (frac(|x|) - 1),
where frac(x) is the fractional part of x.
--well, that was too easy. I should have said "not a piecewise rational function" to make it harder. Or perhaps, as Dan Asimov suggested, demanding it be analytic, or Cinfinity. And one can similarly ask: is there an analytic or Cinfinity function that maps algebraic numbers to algebraic numbers, but is not an algebraic function? Here's one that comes close. Consider the "Klein j-invariant" https://en.wikipedia.org/wiki/J-invariant which is sort of the most fundamental "modular function." "In 1937 Theodor Schneider proved that if X is a quadratic irrational number in the upper half plane then j(X) is an algebraic integer. In addition he proved that if X is an algebraic number but not imaginary quadratic then j(X) is transcendental." Are all algebraic integers generated in this way? Here's another interesting real valued function F(X) of real X. Let [a0; a1,a2,a3,a4...] be the continued fraction expansion of X. For each a[j] which is (i) greater than 4, and (ii) both its neighbors a[j+1] and a[j-1] also are greater than 4, replace a[j] by a[j]+1. This editing process converts X into F(X). This function F(X) has the properties that (1) if X is rational, then F(X) is too. (2) F(X) is not a rational function. Nor is it a piecewise rational function (or anyhow if it were considered one then the number of pieces would be uncountably infinite and the breakpoints would be dense everywhere). (3) if X is quadratic irrational, which happens exactly when X's continued fraction is ultimately-periodic, then F(X) also is. But F(X) is not an algebraic function. (4) I cannot believe F(X) is analytic. (5) Is F(X) continuous? (I think yes. Certainly if X and Y are both rational and |X-Y|-->0, then |F(X)-F(Y)|-->0.) (6) If F(X) differentiable? Well, it certainly is possible to define a derivative F'(X) whenever X is a rational number. Indeed, I think the Kth derivative may be defined whenever X is a rational number, for any K=1,2,3,...? I'm not sure whether these are continuous functions of X; but if they all were then F(X) would be Cinfinity. There also are interesting variants of this construction, such as, a[j] is edited to become, not a[j]+1, but rather a[j]+2^(-j).
I'm curious how this applies to another sort of edit, the "rebasing" operations, for instance G(X) := replace 2^n in the binary expansion of X with 3^n. G maps rationals into rationals, the unit interval into itself, preserves ordering and so on.
(5) Is F(X) continuous? (I think yes. Certainly if X and Y are both rational and |X-Y|-->0, then |F(X)-F(Y)|-->0.)
Also true for G, I guess? However there are an awful lot of "holes" in the range -- every number that contains a 2 anywhere in its ternary expansion.
There also are interesting variants of this construction
I suppose we could hack up algebraic variants by editing representations of algebraic numbers in analogous ways, for instance twiddling the coefficients of the polynomial that X is the root of.
I'm not sure whether these are continuous functions of X
Heck, I'm not even sure I should continue to believe in the reals.
Let f: R —> R^3 be a C^oo one-to-one mapping of the reals into 3-space. (For convenience, assume WLOG that ||f'(t)|| is never 0.) Suppose further that a) The image f(R) is closed and bounded in R^3; and b) If for some sequence t_j in R we have lim f(t_j) = f(t) j—>oo for some t in R, then we also have convergence of the tangent vectors: lim f'(t_j) = f'(t) j—>oo ------------------------------------------------- Question: Does there exist such a curve as f ??? (Note: If not for b), an easy example would be for the image f(R) to be a figure-8.) —Dan
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I'll agree that the tangent vectors to any such curve can (probably) be extended to a nonzero vector field on some open set in R^3 containing it. But my understanding of the Lorenz attractor (which resembles two LP records that are mating), is sufficiently limited, and the Lorenz attractor sufficiently complicated, that I cannot say whether this provides an example or not. —Dan
On Feb 10, 2016, at 12:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system
On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I think the differential equations themselves are pretty much stated in the form of a tangent vector field. The tangent vector at (x,y,z) is (s(y-x), x(r-z)-y, xy-bz), with s, r, b being system parameters. On Wed, Feb 10, 2016 at 3:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll agree that the tangent vectors to any such curve can (probably) be extended to a nonzero vector field on some open set in R^3 containing it.
But my understanding of the Lorenz attractor (which resembles two LP records that are mating), is sufficiently limited, and the Lorenz attractor sufficiently complicated, that I cannot say whether this provides an example or not.
—Dan
On Feb 10, 2016, at 12:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system
On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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But that doesn't get us that the image f(R) is closed and bounded, does it? —Dan
On Feb 10, 2016, at 12:59 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I think the differential equations themselves are pretty much stated in the form of a tangent vector field. The tangent vector at (x,y,z) is (s(y-x), x(r-z)-y, xy-bz), with s, r, b being system parameters.
On Wed, Feb 10, 2016 at 3:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll agree that the tangent vectors to any such curve can (probably) be extended to a nonzero vector field on some open set in R^3 containing it.
But my understanding of the Lorenz attractor (which resembles two LP records that are mating), is sufficiently limited, and the Lorenz attractor sufficiently complicated, that I cannot say whether this provides an example or not.
—Dan
On Feb 10, 2016, at 12:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system
On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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For certain values of the system parameters, and for certain initial conditions, the forward orbit is closed and bounded. I thought that was all you were asking. So, I have a much simpler idea for a system that I think satisfies your conditions. First question: what if we only wanted R -> R? Wouldn't the principal value of arctangent satisfy your constraints? On Wed, Feb 10, 2016 at 7:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
But that doesn't get us that the image f(R) is closed and bounded, does it?
—Dan
On Feb 10, 2016, at 12:59 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I think the differential equations themselves are pretty much stated in the form of a tangent vector field. The tangent vector at (x,y,z) is (s(y-x), x(r-z)-y, xy-bz), with s, r, b being system parameters.
On Wed, Feb 10, 2016 at 3:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll agree that the tangent vectors to any such curve can (probably) be extended to a nonzero vector field on some open set in R^3 containing it.
But my understanding of the Lorenz attractor (which resembles two LP records that are mating), is sufficiently limited, and the Lorenz attractor sufficiently complicated, that I cannot say whether this provides an example or not.
—Dan
On Feb 10, 2016, at 12:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system
On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Even if it were just the forward orbit, how do we conclude that it's closed? Arctangent(R) = (-pi/2, pi/2). So the image wouldn't be closed. Or did you mean something else? —Dan
On Feb 10, 2016, at 5:06 PM, Allan Wechsler <acwacw@gmail.com> wrote:
For certain values of the system parameters, and for certain initial conditions, the forward orbit is closed and bounded. I thought that was all you were asking.
So, I have a much simpler idea for a system that I think satisfies your conditions. First question: what if we only wanted R -> R? Wouldn't the principal value of arctangent satisfy your constraints?
On Wed, Feb 10, 2016 at 7:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
But that doesn't get us that the image f(R) is closed and bounded, does it?
—Dan
On Feb 10, 2016, at 12:59 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I think the differential equations themselves are pretty much stated in the form of a tangent vector field. The tangent vector at (x,y,z) is (s(y-x), x(r-z)-y, xy-bz), with s, r, b being system parameters.
On Wed, Feb 10, 2016 at 3:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll agree that the tangent vectors to any such curve can (probably) be extended to a nonzero vector field on some open set in R^3 containing it.
But my understanding of the Lorenz attractor (which resembles two LP records that are mating), is sufficiently limited, and the Lorenz attractor sufficiently complicated, that I cannot say whether this provides an example or not.
—Dan
On Feb 10, 2016, at 12:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
How about an orbit under the Lorenz system of differential equations? https://en.wikipedia.org/wiki/Lorenz_system
On Wed, Feb 10, 2016 at 2:57 PM, Dan Asimov <asimov@msri.org> wrote:
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Yes. Note that by composing with tanh, it suffices to find some f : (-1, 1) --> R^3 with these properties. We'll construct such an example with codomain R^2 (which embeds into R^3). Firstly, we take the 'bump function': b : (0, 1) --> R b(x) = exp(1/(x-1) - 1/x) which has the property that at both 0 and 1, it and all of its derivatives are 0. We now integrate it twice and normalise to give a C^infinity function: g : (0, 1) --> R which has g'(0) = 0, g'(1) = 1, and all further derivatives are zero at both endpoints. We can take the graph of this function and reflect in the normal to the point (1, g(1)) to obtain some curve in R^2 (a 'rounded corner') which has these properties: (a) The endpoints are (0, 0) and (c, c), where c = 1 + g(1). (b) The image of the curve is a subset of [0, c]^2, and only the endpoints lie on the boundary of this square. (c) The curve is C^infinity. Now, we can assemble rounded corners and line segments like so: /---v---\ |...|...| \---^---/ The resulting set is homeomorphic to the graph K_(3, 2). We now define, in the obvious way, a Eulerian path beginning at one of the degree-3 vertices and ending at the other, parametrised by arc-length. Now this is a C^infinity injection from (0, l) to R^2 with compact image, and we can compose with tanh and some boring linear polynomial to get a similar injection from R to R^2. Best wishes, Adam P. Goucher
Sent: Wednesday, February 10, 2016 at 7:57 PM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Does such an curve exist in 3-space?
Let
f: R —> R^3
be a C^oo one-to-one mapping of the reals into 3-space.
(For convenience, assume WLOG that ||f'(t)|| is never 0.)
Suppose further that
a) The image f(R) is closed and bounded in R^3;
and
b) If for some sequence t_j in R we have
lim f(t_j) = f(t) j—>oo
for some t in R, then we also have convergence of the tangent vectors:
lim f'(t_j) = f'(t) j—>oo
-------------------------------------------------
Question: Does there exist such a curve as f ???
(Note: If not for b), an easy example would be for the image f(R) to be a figure-8.)
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Wed, Feb 10, 2016 at 9:44 AM, Warren D Smith <warren.wds@gmail.com> wrote:
And one can similarly ask: is there an analytic or Cinfinity function that maps algebraic numbers to algebraic numbers, but is not an algebraic function?
Here's another interesting real valued function F(X) of real X. Let [a0; a1,a2,a3,a4...] be the continued fraction expansion of X. For each a[j] which is (i) greater than 4, and (ii) both its neighbors a[j+1] and a[j-1] also are greater than 4, replace a[j] by a[j]+1. This editing process converts X into F(X).
I don't think this is continuous. Let X be a rational number whose continued fraction expansion ends with [...2, 4, 4, 4]. f(X) has a continued fraction expansion that ends with [...2, 4, 5, 4]. But a number very slightly larger or smaller than X (depending on whether X has an even or odd number of convergents) will have a continued fraction expansion that has in this spot [...2, 4, 4, 3, 1, ...], and f maps this to itself.
(5) Is F(X) continuous? (I think yes. Certainly if X and Y are both rational and |X-Y|-->0, then |F(X)-F(Y)|-->0.)
General rule when looking for the flaw in an argument; look for the statement that something is "certainly" or "obviously" true. Andy Latto andy.latto@pobox.com
participants (7)
-
Adam P. Goucher -
Allan Wechsler -
Andy Latto -
Dan Asimov -
Dan Asimov -
Marc LeBrun -
Warren D Smith