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Topic: analysis question on periodic functions
Replies: 7   Last Post: Aug 19, 2013 2:25 PM

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 3,555 Registered: 12/13/04
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?
>>
>>
>>

>
>
>(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)/(z-1).
>
> 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).

Date Subject Author
8/17/13 David Bernier
8/17/13 David Bernier
8/17/13 David C. Ullrich
8/17/13 David Bernier
8/17/13 quasi
8/18/13 David C. Ullrich
8/18/13 quasi
8/19/13 quasi