Difference between revisions of "2020 CAMO Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
+ | |||
+ | |||
Because <math>f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y)</math>, we can find that <cmath>f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}</cmath> | Because <math>f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y)</math>, we can find that <cmath>f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}</cmath> | ||
It's obvious that if there exists two real numbers <math>x</math> and <math>y</math>, which satisfies <cmath>f(x)=\frac{a^x-1}{a^x+1}</cmath> and <cmath>f(y)=\frac{a^y-1}{a^y+1}</cmath> | It's obvious that if there exists two real numbers <math>x</math> and <math>y</math>, which satisfies <cmath>f(x)=\frac{a^x-1}{a^x+1}</cmath> and <cmath>f(y)=\frac{a^y-1}{a^y+1}</cmath> | ||
− | Then, for <math>f(x+y)</math>, <cmath>f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}</cmath>, <cmath>f(x+y)=\frac{2*a^x | + | Then, for <math>f(x+y)</math>, <cmath>f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}</cmath>, <cmath>f(x+y)=\frac{2*a^x*a^y-2}{2*a^x*a^y+2}</cmath> |
− | Then, <cmath>f(x+y)=\frac{a^x | + | Then, <cmath>f(x+y)=\frac{a^x*a^y-1}{a^x*a^y+1}</cmath> |
The fraction is also satisfies for <math>f(x+y)</math> | The fraction is also satisfies for <math>f(x+y)</math> |
Revision as of 09:29, 3 October 2023
Problem 1
Let (meaning takes positive real numbers to positive real numbers) be a nonconstant function such that for any positive real numbers and , Prove that there is a constant such that for all positive real numbers .
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Because , we can find that It's obvious that if there exists two real numbers and , which satisfies and
Then, for , ,
Then,
The fraction is also satisfies for
Then, we can solve this problem using mathematical induction
~~Andy666
See also
2020 CAMO (Problems • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All CAMO Problems and Solutions |
2020 CJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All CJMO Problems and Solutions |
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.