The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 12,067
Registered: 7/15/05
Re: analysis question on periodic functions
Posted: Aug 17, 2013 2:30 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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. 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.


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.