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

(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...")
 
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==Solution==
 
==Solution==
 
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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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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>
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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>
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Then, <cmath>f(x+y)=\frac{a^(x+y)-1}{a^(x+y)+1}</cmath>
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The fraction is also satisfies for $f(x+y)
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Then, we can solve this problem using mathematical induction
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 +
~~Andy666
  
 
==See also==
 
==See also==

Revision as of 09:26, 3 October 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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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

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

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

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