2020 CAMO Problems/Problem 1: Difference between revisions
Created page with "==Problem 1== Let <math>f:\mathbb R_{>0}\to\mathbb R_{>0}</math> (meaning <math>f</math> takes positive real numbers to positive real numbers) be a nonconstant function such t..." |
No edit summary |
||
| 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> | |||
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+y)-2}{2*a^(x+y)+2}</cmath> | |||
Then, <cmath>f(x+y)=\frac{a^(x+y)-1}{a^(x+y)+1}</cmath> | |||
The fraction is also satisfies for $f(x+y) | |||
Then, we can solve this problem using mathematical induction | |||
~~Andy666 | |||
==See also== | ==See also== | ||
Revision as of 09:26, 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 
, ![\[f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y).\]](http://latex.artofproblemsolving.com/f/b/f/fbfc1781f9e0e0468cfe295fe4242ac8843e1cd2.png)
Prove that there is a constant 
such that ![\[f(x)=\frac{a^x-1}{a^x+1}\]](http://latex.artofproblemsolving.com/f/5/3/f53c551cb8a20af7ac1680b0189a9195cae3b7fe.png)
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 ![\[f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}\]](http://latex.artofproblemsolving.com/c/c/1/cc11c69980791dce2103382a323199dff5ecde09.png)
It's obvious that if there exists two real numbers and , which satisfies and ![\[f(y)=\frac{a^y-1}{a^y+1}\]](http://latex.artofproblemsolving.com/f/d/3/fd3de3e55e294e80aa44dca9f7ab616ac963ec0f.png)
Then, for 
, , ![\[f(x+y)=\frac{2*a^(x+y)-2}{2*a^(x+y)+2}\]](http://latex.artofproblemsolving.com/5/0/7/5077c6e95a498bb086896d491f460a87c3c668cf.png)
Then, ![\[f(x+y)=\frac{a^(x+y)-1}{a^(x+y)+1}\]](http://latex.artofproblemsolving.com/0/b/1/0b1a93c099e888632bd41446cc13165317b06180.png)
The fraction is also satisfies for $f(x+y)
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 | ||
