> The function F(x) defined on the interval [0,1] as > the sum from n=1 to n=infinity of [nx]/n!, where [.] > denotes the greatest integer function, is strictly > increasing and therefore has a derivative almost > everywhere in [0,1]. The question is whether the > derivative can be computed anywhere. > > Clearly, F is discontinuous at every rational number, > but what can be said about its derivative at > irrational numbers? > > I posted this question a few years ago, but thought > in the interim someone may have some ideas or results > that might be helpful. > > The range of his function is strange, containining > only one rational number (F(0)=0). Noting that > F(1)=e, we are led to ask: Is 0 the only algebraic > number in the range of F?
Let x be a quadratic irrational, say x=1/sqrt(2). Given delta, you can get a lower bound Q on denominators q such that some fraction p/q is within delta of x: say | p/q - x | < delta implies q > Q(delta). This means that if a term [ny]/n! changes as y ranges through the interval [x-delta, x+delta], we must have n > Q(delta). That puts a bound on F(y)-F(x) for y in that interval. I won't spoil the fun of working out the details.