Difference between revisions of "1991 IMO Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
− | The argument above would not work for <math>a=1</math>, since <math>{\textstyle\sum \frac{1}{k^a}}</math> only converges for <math>a>1</math>. But Osmun Nal argues [https://www.youtube.com/watch?v=3v3CMQS_5Cc in this video] that <math>x_k=k\sqrt{2}-\lfloor k\sqrt{2}\rfloor</math> satisfies the stronger inequality <math>|x_i-x_j|\cdot |i-j|\geq 1</math> for all distinct <math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously. | + | The argument above would not work for <math>a=1</math>, since <math>{\textstyle\sum \frac{1}{k^a}}</math> only converges for <math>a>1</math>. But Osmun Nal argues [https://www.youtube.com/watch?v=3v3CMQS_5Cc in this video] that <math>x_k=C\cdot\left(k\sqrt{2}-\lfloor k\sqrt{2}\rfloor\right), C=2\sqrt2+1</math> satisfies the stronger inequality <math>|x_i-x_j|\cdot |i-j|\geq 1</math> for all distinct <math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously. |
== Solution 3 == | == Solution 3 == |
Latest revision as of 03:20, 24 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 |