A sequence of real numbers {an}n is called a bs sequence if an=|an+1−an+2|, for all n≥0. Prove that a bs sequence is bounded if and only if the function f given by f(n,k)=anak(an−ak), for all n,k≥0 is the null function. Mihai Baluna - ISL 2004
Problem
Source: Romanian IMO TST 2005 - day 2, problem 3
Tags: function, induction, algebra proposed, algebra
01.04.2005 20:21
The condition simply says that the set {an|n∈N∗} has at most one non-null value. It's clear that if this condition hiolds, then the sequence is bounded, so the other implication is the interesting one. We assume the sequence is bounded. There are two cases: Case 1: the maximum M of {an} is reached. We can see that if an=M, then an+1 is either M or 0, and the same holds for an−1. Moreover, an+1=M⟺an−1=0, and we can construct the sequence in two possible ways, in both directions, according to whether an+1=M or 0. In both these situations, the sequence contains only two values, in both directions, 0 and M, so the conclusion is proven. Case 2: the maximum M is never reached. Let ε>0 be small enough (small enough that 2(M−ε)>M, for instance) so that we can find an=M−ε. It's easy to show that we can also find n s.t. an=M−ε and an+1=t, 0<t<M−ε. We thus have an+2=M−ε+t, and if an+3=M−ε, then an+4=2(M−ε)+t>M , which is a contradiction. This shows that, in fact, an+3=M−ε+2t,an+4=t, and we can repeat the procedure with ε replaced with ε−2t (which must necessarily be >0). We find that ε−2nt>0, ∀n≥0, which is absurd. Hope it's Ok.
01.04.2005 20:24
I hope my memory still works (I try to stay beyond Alzheimer's club of Pierre, but it becomes harder and harder), but I think this problem was already solved on the forum. I think Darij posted it, after a kind of TST of Germany. Probably I'm wrong, but it sounds too familiar to me and I do not have the ISL 2004.
01.04.2005 20:25
Yes, I believe that problem asked whether such a sequence could be bounded .
15.04.2005 04:28
yea it was with the conditions given that x0 and x1 and positive and not equal to each other. And the answer is no. Its in the ISL 2004, and came up in the British FST.
16.04.2005 13:05
Only the main idea: write all xi−s in terms of the first 2 distinct xi−s. By induction one can easily obtain that all the terms of the sequence are of the form mx1+nx2 with m,n∈N . There are finitely many (m,n)−s with m<Mx1,n<Mx2, with M being the bound so M is attained . Let xi+1 be the first term that achieves the maximum, so 0<xi<xi+1, xi+2=xi+1−xi, since it cannot be their sum (M is max) so xi+3=2xi+1−xi>xi+1=M, false. There are still some cases to be considered for the first few terms of the sequence and for the case 0 occurs in the sequence.
04.12.2005 17:36
ikap wrote: all the terms of the sequence are of the form mx1+nx2 with m,n∈N Why is that? (m,n∈N) I can only prove it with m,n∈Z and that doesn't help very much...