2020 CAMO Problems/Problem 1: Difference between revisions
| Line 8: | Line 8: | ||
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+y-2}{2*a^x+y+2}</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> | Then, <cmath>f(x+y)=\frac{a^x+y -1}{a^x+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 09:27, 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 | ||