
Re: analysis question on periodic functions
Posted:
Aug 17, 2013 12:32 PM


On Sat, 17 Aug 2013 11:03:01 0400, David Bernier <david250@videotron.ca> wrote:
>On 08/17/2013 06:21 AM, David Bernier wrote: >> Suppose we have an odd continuous function f: R > R >> with period 1 so that f(x+1) = f(x), and f(x) = f(x). >> >> Suppose f has mean zero over the unit interval, so >> int_{0, 1} f(x) dx = 0, but that >> int_{0, 1}  f(x)  dx > 0 (so it's not constantly zero). >> >> Given a real number a> 0, consider the series: >> >> f(a) + f(2a) + f(3a) + ... >> >> with partial sums >> >> S_k (a) = sum_{j = 1 ... k} f(ka). >> >> === >> >> >> (i) If a is rational, is it necessarily true that the S_k (a) are >> bounded in absolute value? >> >> >> if the answer to (i) were YES, then there's also (ii) below: >> if the answer were NO, (ii) might be forgotten. >> >> (ii) If the partial sums S_k (a) are bounded for some >> real number a>0, then does it follow that a is a rational >> number? >> >> >> > >How about: > >(i) YES
??? Are you sure about that?
> >(ii) NO. Let f(x):= sin(2pi*x) throughout below. > If a>0 , let z = exp(i*2pi*a). > > Then Im(z^k) = sin(2pi*a*k) = f(ka). > > z + z^2 + ... z^k = (z^(k+1)  z)/(z1). > > So, the partial sums S_k (a) = sin(2pi*a) + ... sin(2pi*k*a) will > be bounded except possibly if z = 1, so bounded except possibly if > a is a positive integer. If a is irrational, the S_k (a) are >bounded. (But the bound may change from a to another value of a).

