Difference between revisions of "2020 CAMO Problems/Problem 1"

Revision as of 03:01, 28 November 2023

Problem 1

Let $f:\mathbb R_{>0}\to\mathbb R_{>0}$ (meaning $f$ takes positive real numbers to positive real numbers) be a nonconstant function such that for any positive real numbers $x$ and $y$ , $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y).$ Prove that there is a constant $a>1$ such that $f(x)=\frac{a^x-1}{a^x+1}$ for all positive real numbers $x$ .

Solution

Because $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y)$ , we can find that $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ It's obvious that if there exists two real numbers $x$ and $y$ , which satisfies $f(x)=\frac{a^x-1}{a^x+1}$ and $f(y)=\frac{a^y-1}{a^y+1}$

Then, for $f(x+y)$ , $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ , $f(x+y)=\frac{2*a^x*a^y-2}{2*a^x*a^y+2}$

Then, $f(x+y)=\frac{a^x*a^y-1}{a^x*a^y+1}$

The fraction is also satisfies for $f(x+y)$

Then, we can solve this problem using mathematical induction

~~Andy666

@@ Line 3: / Line 3: @@
 ==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>

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Difference between revisions of "2020 CAMO Problems/Problem 1"

Revision as of 03:01, 28 November 2023

Problem 1

Solution

See also