
Re: analysis question on periodic functions
Posted:
Aug 17, 2013 11:03 AM


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
(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?

