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

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David Bernier

Posts: 3,396
Registered: 12/13/04
Re: analysis question on periodic functions
Posted: Aug 17, 2013 2:09 PM
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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)/(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).

>


--
abc?



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