Difference between revisions of "1991 OIM Problems/Problem 3"
| Line 42: | Line 42: | ||
From property b: <math>f(19/3^7)=\frac{f(19/3^6)}{2}=5/2^7</math> and <math>f(20/3^7)=\frac{f(20/3^6)}{2}=5/2^7</math> | From property b: <math>f(19/3^7)=\frac{f(19/3^6)}{2}=5/2^7</math> and <math>f(20/3^7)=\frac{f(20/3^6)}{2}=5/2^7</math> | ||
| + | Since <math>\frac{19}{3^7} < \frac{18}{1991} < \frac{20}{3^7}</math> and <math>f(19/3^7)=f(20/3^7)</math>, | ||
| + | |||
| + | Then <math>f(18/1991)=f(19/3^7)=f(20/3^7)=5/2^7</math> | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
Revision as of 17:17, 13 December 2023
Problem
Let 
be an increasing function defined for every real number 
, 
, such that:
a. 
b. 
c. 
Find 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
From condition c: 
From property c: 
which means that 
From property b: 
From property c: 
Here is good to note the different between an increasing function an a strictly increasing function. In a strictly increasing function the function needs to be strictly increasing on all intervals. However, in an increasing function an interval that has the same constant value is allowed. In other words, it will not decrease in that interval.
Therefore, since in the interval 
the function can neither increase nor decrease and 
, then in the interval , 
From property b: 
and 
From property c: 
and 
From property b: 
and 
From property c: 
and 
From property b: 
and 
From property b: 
and 
From property b: 
and 
From property b: 
and 
Since 
and 
,
Then 
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
