11 Aug
2013
11 Aug
'13
3:43 p.m.
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