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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: 3,755 Registered: 12/13/04
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?
>
>
>

(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)/(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?

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