[math-fun] Describe f(a/b)=a/(b+1)?
Dumb questions I¹m not sure even how to ask: could someone help describe or visualize the following function f? For rational x = a/b (in lowest terms) let f(x) = a/(b+1) So for instance we can say things like x/2 <= f(x) < x. Next, to extend f to real x, we observe that in sequences of rationals x_n ³converging on² x the denominators (almost always) grow, so that f(x_n) ³approaches² x. So let¹s define f as the identity on irrational values. What else can we say about f? Is it continuous? What about its derivative? Fourier transform? It¹s sure ³bumpy², but is it ³fractally²?
On Sat, Aug 10, 2013 at 11:10 AM, Marc LeBrun <mlb@well.com> wrote:
Dumb questions I¹m not sure even how to ask: could someone help describe or visualize the following function f?
For rational x = a/b (in lowest terms) let f(x) = a/(b+1)
So for instance we can say things like x/2 <= f(x) < x.
Next, to extend f to real x, we observe that in sequences of rationals x_n ³converging on² x the denominators (almost always) grow, so that f(x_n) ³approaches² x. So let¹s define f as the identity on irrational values.
What else can we say about f? Is it continuous?
It's continuous at every irrational number, discontinuous at every rational value. This follows pretty directly from the definition of continuity and the epsilon-delta definition of a limit.
What about its derivative?
A typical irrational is about 1/n away from a place where you have changed the value by 1/n^2, so it is differentiable (with derivative 0) almost everywhere. I think there's a set of measure 0, of irrationals well approximated by rationals where the function is continuous but not differentiable. Fourier transform? If you define Fourier transform using the Riemann integral, the Fourier transform does not exist, since the function is not Riemann-integrable. If you define it using Lebesgue integration, then it has the same Fourier transform as the identity function, since it only differs at countably many points, which doesn't effect the value of the integral. It¹s sure ³bumpy², but is it ³fractally²? Not sure what you mean. The Hausdorff dimension of the graph is 1, if that's what you're asking. Andy
I expect Andy meant "with derivative 1". But suppose we attempt to differentiate at irrational x. (f(x+h)-f(x))/h, when x+h is rational, will be ((x+h)(q/(q+1))-x)/h, where q is the denominator of x+h in lowest terms. This will be -x/((q+1)h) + q/(q+1). Now, the second term clearly approaches 1, because as h gets small it excludes more and more values of q. But I don't see why the first term approaches 0. --ms On 2013-08-10 13:55, Andy Latto wrote: > On Sat, Aug 10, 2013 at 11:10 AM, Marc LeBrun <mlb@well.com> wrote: >> Dumb questions I¹m not sure even how to ask: could someone help describe or >> visualize the following function f? >> >> For rational x = a/b (in lowest terms) let f(x) = a/(b+1) >> >> So for instance we can say things like x/2 <= f(x) < x. >> >> Next, to extend f to real x, we observe that in sequences of rationals x_n >> ³converging on² x the denominators (almost always) grow, so that f(x_n) >> ³approaches² x. So let¹s define f as the identity on irrational values. >> >> What else can we say about f? Is it continuous? > It's continuous at every irrational number, discontinuous at every > rational value. This follows pretty directly from the definition of > continuity and the epsilon-delta definition of a limit. > >> What about its >> derivative? > A typical irrational is about 1/n away from a place where you have > changed the value by 1/n^2, so it is differentiable (with derivative > 0) almost everywhere. I think there's a set of measure 0, of > irrationals well approximated by rationals where the function is > continuous but not differentiable. > > Fourier transform? > > If you define Fourier transform using the Riemann integral, the > Fourier transform does not exist, since the function is not > Riemann-integrable. If you define it using Lebesgue integration, then > it has the same Fourier transform as the identity function, since it > only differs at countably many points, which doesn't effect the value > of the integral. > > It¹s sure ³bumpy², but is it ³fractally²? > > Not sure what you mean. The Hausdorff dimension of the graph is 1, if > that's what you're asking. > > Andy > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Of course, integral Hausdorff dimension = n doesn't have to mean something is a normal n-manifold. E.g., The 2D Cantor set obtained by removing middle p's (with 0 < p < 1) . . . meaning removing the plus-sign-shaped 5p^2 fraction of a square, iteratively, results in a set with Hausdorff dimension d satisfying 1 = 4 ((1-p)/2)^d, which if my arithmetic is correct implies that d = 1 for p = 1/2. Yet this middle-1/2 2D Cantor set is quite fractally. --Dan Andy wrote: ----- [Marc wrote]: << It¹s sure ³bumpy², but is it ³fractally²? Not sure what you mean. The Hausdorff dimension of the graph is 1, if that's what you're asking. -----
(Did you receive my first e-mail about this?) P.S. I was wrong (in that first e-mail) and Andy's right: Your function is continuous at each irrational number. Sorry about that. --Dan On 2013-08-10, at 8:10 AM, Marc LeBrun wrote:
Dumb questions I’m not sure even how to ask: could someone help describe or visualize the following function f?
For rational x = a/b (in lowest terms) let f(x) = a/(b+1)
So for instance we can say things like x/2 <= f(x) < x.
Next, to extend f to real x, we observe that in sequences of rationals x_n “converging on” x the denominators (almost always) grow, so that f(x_n) “approaches” x. So let’s define f as the identity on irrational values.
What else can we say about f? Is it continuous? What about its derivative? Fourier transform? It’s sure “bumpy”, but is it “fractally”?
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participants (4)
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Andy Latto -
Dan Asimov -
Marc LeBrun -
Mike Speciner