Difference between revisions of "1968 IMO Problems/Problem 5"
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(b) <math>f(x) = 1</math> when <math>2n\le x < 2n+1</math> for some integer <math>n</math>, and <math>f(x)=\frac{1}{2}</math> when <math>2n+1\le x < 2n+2</math> for some integer <math>n</math>. | (b) <math>f(x) = 1</math> when <math>2n\le x < 2n+1</math> for some integer <math>n</math>, and <math>f(x)=\frac{1}{2}</math> when <math>2n+1\le x < 2n+2</math> for some integer <math>n</math>. | ||
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| + | [[Category:Olympiad Algebra Problems]] | ||
| + | [[Category:Functional Equation Problems]] | ||
Revision as of 07:37, 19 July 2016
Problem 5
Let 
be a real-valued function defined for all real numbers 
such that, for some positive constant 
, the equation
![\[f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}\]](http://latex.artofproblemsolving.com/0/c/8/0c8b3bbc8a56354c4a9940dfe4076f5dad29e31a.png)
holds for all .
(a) Prove that the function is periodic (i.e., there exists a positive number 
such that 
for all ).
(b) For 
, give an example of a non-constant function with the required properties.
Solution
(a) Since
![\[f(x+a) \ge \frac{1}{2}\]](http://latex.artofproblemsolving.com/b/3/c/b3c0d8b1936802188854dee709b4b361158d36e2.png)
is true for any , and
![\[f(x+a)(1-f(x+a)) = \frac{1}{4} - (f(x)-(f(x))^2) = (\frac{1}{2}-f(x))^2\]](http://latex.artofproblemsolving.com/9/b/d/9bdd0bc38caa5ed7be000fb06296b00db4fd02da.png)
We have:
![\[f(x+2a) = \frac{1}{2} + \sqrt{(\frac{1}{2}-f(x))^2} = \frac{1}{2} + (f(x) - \frac{1}{2}) = f(x)\]](http://latex.artofproblemsolving.com/4/1/b/41b8de7a868b77d7f5d148e86e515659aa49cfb7.png)
Therefore is periodic, with 
as a period.
(b) 
when 
for some integer 
, and 
when 
for some integer .
