• Support PF! Buy your school textbooks, materials and every day products Here!

Little(?) problem

  • Thread starter penguin007
  • Start date
  • #1
77
0
Hi everyone,

I’m studying the correction of an exercise and I there are some points I can’t see:
In this exercise, we introduce a irrational x and function f so that f(kx)=kx-[kx] with k an integer included in the interval [0,N].

It is said that f has N+1 values (Ok) that are included in the intervals [i/N,(i+1)/N] for i integers included in [0,N-1]… I can’t see this last point. If anyone could help me understand this point…

Thanks in advance!
 

Answers and Replies

  • #2
[itex]y-[y]\in [0,1][/itex]
[itex]\cup\{[i/N,i+1/N]:i\in \mathbb{Z}\cap [0,N-1]\}=[0,1][/itex].
Is it saying anything more than that?
 
  • #3
77
0
Nothing else, they just say that for each k integer included in [0,N], x an irrational, there exists an i integer included in [0,N-1] so that f(kx) is included in the interval [i/N,(i+1)/N] (and then, for the rest of the exercise, since there are N+1 values and N intervals, there are two values included in the same [i/N,(i+1)/N]).

But I don't know why f(kx) is included in [i/N,(i+1)/N]...
 
  • #4
It's not saying f(kx) is included in any particular [i/N,(i+1)/N], only that it's somewhere in [0,1] and since the sets [i/N,(i+1)/N] cover the whole of [0,1] it's got to be in one of them.
 
  • #5
77
0
Ok Martin I got it. (In fact, the important in this exercise is that there exists two values k1 and k2 so that abs(f(k1x)-f(k2x))<=1/N.)

Thanks very much!!
 
  • #6
Yes I guessed that. Glad to be of help.
 

Related Threads for: Little(?) problem

  • Last Post
Replies
2
Views
955
  • Last Post
Replies
3
Views
2K
Replies
6
Views
2K
Replies
3
Views
2K
Replies
3
Views
966
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
840
  • Last Post
Replies
1
Views
2K
Top