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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 C. Ullrich

Posts: 3,138
Registered: 12/13/04
Re: analysis question on periodic functions
Posted: Aug 18, 2013 9:59 AM
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On Sat, 17 Aug 2013 14:10:59 -0500, quasi <quasi@null.set> wrote:

>dullrich wrote:
>>David Bernier wrote:
>>>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?

>>>
>>>How about:
>>>
>>>(i) YES

>>
>>??? Are you sure about that?

>
>As I see it, the answer "yes" is correct.
>
>Let Z,R denote the usual sets (integers, reals).
>
>Assume the domain of f is R.
>
>The mix of
>
> f is odd
> f has period 1
>
>clinches it.
>
>Since f is odd, f(0) = 0.
>
>Since f has period 1, f(x) = 0 for all x in Z.
>
>Suppose x,y in R are such that x + y is in Z.
>
>Claim f(x) + f(y) = 0.
>
>If x in Z, then y in Z, so f(x) + f(y) = 0 + 0 = 0.
>
>Suppose x is not in Z.
>
>Let s = x - floor(x) and t = y - floor(y).
>
>Since x is not in Z, neither is y, hence s,t are positive,
>
>Since x + y is in Z, it follows that s + t = 1. Hence
>
> f(x) + f(y)
> = f(s) + f(t)
> = f(s) + f(1-s)
> = f(s) + f(-s)
> = 0
>
>as claimed.
>
>As a special case, f(1/2) = 0.
>
>Now suppose a is a positive rational.
>
>Write a = b/c as a fraction in lowest terms, with b,c in N.
>
>Now for j in N with j*a < (1/2)*c,
>
> f(j*a) + f((c-j)*a) = 0
>
>In the sum
>
> f(a) + f(2*a) + ... + f((c-1)*a)
>
>all terms except a possible middle term (which is present iff
>c is even) pair up and cancel. The middle term, if present,
>is also equal to 0.


Ah right, I missed the fact that the middle term must vanish.

>Noting that c*a = b which is in Z, it
>follows that
>
> f(a) + f(2*a) + ... + f(c*a) = 0.
>
>It's then easily seen that the partial sums
>
> S_1, S_2, S_3, ...
>
>repeat every c terms, that is, S_(k+c) = S_k.
>
>It follows that the partial sums are bounded, as claimed.
>
>quasi





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