Difference between revisions of "1991 IMO Problems/Problem 6"
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Then we have <math>\left| i - j \right| \ge n^\ell</math> | Then we have <math>\left| i - j \right| \ge n^\ell</math> | ||
− | Without loss of generality, assume that <math>i_\ell > j_\ell</math>, then we have <math>i_\ell - j_\ell \ge 1, j_{\ell + 1} - i_{\ell + 1}, j_{\ell + 2} - i_{\ell + 2}, \cdots, j_m - i_m \le | + | Without loss of generality, assume that <math>i_\ell > j_\ell</math>, then we have <math>i_\ell - j_\ell \ge 1, j_{\ell + 1} - i_{\ell + 1}, j_{\ell + 2} - i_{\ell + 2}, \cdots, j_m - i_m \le n - 1</math> |
<math>\left| a_i - a_j \right| = n \left( \left( i_\ell - j_\ell \right) \left( n + 1 \right)^{-\ell} - \left( j_{\ell+1} - i_{\ell+1} \right) \left( n + 1 \right)^{-\ell-1} - \left( j_{\ell+2} - i_{\ell+2} \right) \left( n + 1 \right)^{-\ell-2} - \cdots - \left( j_m - i_m \right) \left( n + 1 \right)^{-m} \right) \\~\\ \ge n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( \left( n + 1 \right)^{-\ell-1} + \left( n + 1 \right)^{-\ell-2} + \cdots + \left( n + 1 \right)^{-m} \right) \\~\\ > n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( n + 1 \right)^{-\ell-1} \dfrac{1}{1 - \left( n + 1 \right)^{-1}} = \left( n + 1 \right)^{-\ell}</math> | <math>\left| a_i - a_j \right| = n \left( \left( i_\ell - j_\ell \right) \left( n + 1 \right)^{-\ell} - \left( j_{\ell+1} - i_{\ell+1} \right) \left( n + 1 \right)^{-\ell-1} - \left( j_{\ell+2} - i_{\ell+2} \right) \left( n + 1 \right)^{-\ell-2} - \cdots - \left( j_m - i_m \right) \left( n + 1 \right)^{-m} \right) \\~\\ \ge n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( \left( n + 1 \right)^{-\ell-1} + \left( n + 1 \right)^{-\ell-2} + \cdots + \left( n + 1 \right)^{-m} \right) \\~\\ > n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( n + 1 \right)^{-\ell-1} \dfrac{1}{1 - \left( n + 1 \right)^{-1}} = \left( n + 1 \right)^{-\ell}</math> |
Revision as of 06:09, 23 January 2024
Problem
An infinite sequence of real numbers is said to be bounded if there is a constant
such that
for every
.
Given any real number , construct a bounded infinite sequence
such that
for every pair of distinct nonnegative integers
.
Solution 1
Since , the series
is convergent; let
be the sum of this convergent series. Let
be the interval
(or any bounded subset of measure
).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose
such that
holds for all distinct
. The only new cases are when one number (WLOG
) is equal to
, so we must guarantee that
for all
.
Let be the interval
, of length
. The points that are valid choices for
are precisely the points of
, so we must show that this set is nonempty. The total length
is at most the sum of the lengths
. This is
.
Therefore the total measure of is
, so
has positive measure and thus is nonempty. Choosing any
and continuing by induction constructs the desired sequence.
Solution 2
The argument above would not work for , since
only converges for
. But Osmun Nal argues in this video that
satisfies the stronger inequality
for all distinct
; in other words, this sequence simultaneously solves the problem for all
simultaneously.
Solution 3
Let , then we have
Thus
For any non-negative integer , represent it as base-
positional notation:
, where
We directly construct
We allow leading zeros for representing due to the leading zeros do not effect the value of
Then
Thus , which is bounded
For any two distinct non-negative integers , represent them as base-
positional notation:
where
Let be the minimum number such that
, that is,
Then we have
Without loss of generality, assume that , then we have
Thus we have
~Joseph Tsai, mgtsai@gmail.com
See Also
1991 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |