1985 IMO Problems/Problem 6
Problem
For every real number , construct the sequence by setting for each . Prove that there exists exactly one value of for which for every .
Solution 1
By recursive substitution, one can write , where is a polynomial with non-negative coefficients and zero constant term. Thus, , is strictly increasing in , and . We can therefore define the inverse of on . It follows that , , is strictly increasing in , and .
Denote by and . By the monotonicity of we have for each . Note that:
(a) ; (b) .
Thus, holds if and only if , or . We need to show that is a singleton. We have:
(c) if , then , which implies that , and . It follows that ; and (d) if , then , which implies that , and . It follows that ; and
Thus, . Therefore, the two sequences and converge, and their limits and satisfy . Hence, is non-empty, which demonstrates the existence of .
Now, suppose that . We have for each , so that for each . However, , so that , which implies that . Therefore, , proving unicity.
This solution was posted and copyrighted by DAFR. The original thread for this problem can be found here: [1]
Solution 2
For each let be the interval of real numbers such that if , we will have . (That is, the values of which make "work"). We must prove these intervals intersect at one point.
First, I prove they intersect. Notice that as increases, and do too. So if it is easy to see will give and will give (the "extremal" values can take are given by the extremal values can take).
But if we see that and so and similarly gives which gives a too big. So basically when only barely works for , it won't work for . So is a subset of .
So we have an infinite sequence of open intervals, each contained inside the previous one. Therefore their intersection must contain at least one number. This guarantees the existence of .
But, we must prove that can't take two different values. Suppose it could. Suppose both work. Let be the sequence generated by and the sequence generated by , and let (wlog ). Then we see for since since . So there exists a constant such that for . But since , this is impossible. So we're done.
This solution was posted and copyrighted by JuanOrtiz. The original thread for this problem can be found here: [2]
Solution 3
Consider, now, the following observations:
(a) If, at any point, , then from here we know that and therefore the sequence will be monotonically decreasing after one point.
(b) If some does satisfy for all , then for all , and therefore by Squeeze's theorem .
(c) If for some , then and so which then gives .
(d) If , then for each , . If then for each (fixed), . Thus, we can denote a mapping that maps to , which is continuous and monotonically increasing, with so is bijective.
Let's first show uniqueness. Suppose that and are both such sequences. We have and suppose that . Then for all , so inductively, with . However, gives , contradiction. Hence, the that satisfies this must be unique.
Next, let's show existence. We have seen from above that, implies , and that if for some then is monotonically increasing after some point. Suppose no such exists. Let \[ A=\{x_1: \exists n: x_{n+1}<x_n\}\qquad B=\{x_1: \exists n: x_n>1\} \]then if , for all and similarly , for all . Notice also an wasy fact that , so is bounded. Define, now, . As we assumed , this implies that and . It remains to ask whether or .
If , then for this , and so . Let be such that . By above, there's exactly one with , and notice that by the monotonicity of . This means that there exists , contradicting the definition of glb. A similar contradiction (but opposite direction) can also be established for the case .
Hence is neither in or , means that should satisfy the problem condition. Q.E.D.
This solution was posted and copyrighted by Anzoteh. The original thread for this problem can be found here: [3]
See Also
1985 IMO (Problems) • Resources | ||
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