Thread: Math Q&A Game
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laohu is offline
Jun17-10, 02:42 PM
P: 7
Quote Quote by snipez90 View Post

Determine (with proof) for which [itex]\alpha > 0[/itex] and for which [itex]x[/itex] is the function [itex]f:\mathbb{R} \rightarrow \mathbb{R}[/itex] defined by
[tex] f_\alpha (x) =\begin{cases}\frac{1}{q^{\alpha}}&\text{if }x =\frac{p}{q}\neq 0, gcd(p,q) = 1, q > 0\\ 0 &\text{if }x = 0\,\, \mbox{or x is irrational}}\end{cases} [/tex]
continuous. What about differentiability?
To me, there doesn't seem to be any [tex]\alpha > 0[/tex] that makes the function continuous (let alone differentiable). Choose any epsilon-neighborhood around a rational [tex]x = \frac p q[/tex], [tex][a,b] = [x-\epsilon,x+\epsilon][/tex] . Since we can make (b-a)2k arbitrarily large, we can make the difference between [tex]\lceil a 2^k \rceil[/tex] and [tex]\lfloor b 2^k \rfloor[/tex] arbitrarily large. Choose an odd integer m in the interval between them. [tex]y = \frac m {2^k}[/tex] is in the epsilon-neighborhood. That in turn means that we can make [tex]|f(x) - f(y)| = |\frac{1}{q^\alpha} - \frac{1}{2^{k\alpha}}| = \frac{2^{k\alpha}-q^{\alpha}}{q^{\alpha}2^{k\alpha}} [/tex] arbitrarily close to [tex]\frac{1}{q^\alpha}[/tex] for any epsilon-neighborhood, making f discontinuous at x (unless you allow [tex]\alpha = \infty[/tex], in which case f is the zero function and therefore both continuous and differentiable. [tex]C^\infty[/tex] in fact ;) )