Thread: Math Q&A Game View Single Post

 Quote by snipez90 Problem: Determine (with proof) for which $\alpha > 0$ and for which $x$ is the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by $$f_\alpha (x) =\begin{cases}\frac{1}{q^{\alpha}}&\text{if }x =\frac{p}{q}\neq 0, gcd(p,q) = 1, q > 0\\ 0 &\text{if }x = 0\,\, \mbox{or x is irrational}}\end{cases}$$ continuous. What about differentiability?
To me, there doesn't seem to be any $$\alpha > 0$$ that makes the function continuous (let alone differentiable). Choose any epsilon-neighborhood around a rational $$x = \frac p q$$, $$[a,b] = [x-\epsilon,x+\epsilon]$$ . Since we can make (b-a)2k arbitrarily large, we can make the difference between $$\lceil a 2^k \rceil$$ and $$\lfloor b 2^k \rfloor$$ arbitrarily large. Choose an odd integer m in the interval between them. $$y = \frac m {2^k}$$ is in the epsilon-neighborhood. That in turn means that we can make $$|f(x) - f(y)| = |\frac{1}{q^\alpha} - \frac{1}{2^{k\alpha}}| = \frac{2^{k\alpha}-q^{\alpha}}{q^{\alpha}2^{k\alpha}}$$ arbitrarily close to $$\frac{1}{q^\alpha}$$ for any epsilon-neighborhood, making f discontinuous at x (unless you allow $$\alpha = \infty$$, in which case f is the zero function and therefore both continuous and differentiable. $$C^\infty$$ in fact ;) )