
Re: analysis question on periodic functions
Posted:
Aug 17, 2013 2:09 PM


On 08/17/2013 12:32 PM, dullrich@sprynet.com wrote: > 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?
No, I'm not sure about very much ...
> >> >> (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). >
 abc?

