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Difference between revisions of "2020 CAMO Problems/Problem 1"
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~~Andy666
 
~~Andy666
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==Solution (2)==
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Let <math>P(x_0,y_0)</math> denote a substitution of <math>(x_0,y_0)</math> for <math>(x,y)</math> and <math>g</math> be the inverse of <math>f</math> when it exists.
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By <math>P(x,x)</math> we get <math>f(x)<1</math> so the domain <math>D</math> of <math>g</math>(x) must be in the <math>0<x<1</math> interval
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<math>P(g(x),g(y)); g(x)+g(y)=g(\frac{x+y}{1+xy})</math>(*) from here,
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Taking <math>\frac{\partial}{\partial x} and \frac{\partial}{\partial y};</math>
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<math>g^\prime(x)=\frac{1(1+xy)-y(x+y)}{(1+xy)^2}=\frac{1-y^2}{(1+xy)^2}, g^\prime(y)=\frac{1(1+xy)-x(x+y)}{(1+xy)^2}=\frac{1-x^2}{(1+xy)^2}</math>
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<math>\frac{g^\prime(x)}{g^\prime(y)}=\frac{1-y^2}{1-x^2}</math> so let <math>g^\prime(x)(1-x^2)=g^\prime(y)(1-y^2)=k</math> for some real constant <math>k>0</math>.
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<math>g(x)=\int{\frac{k}{1-x^2}}dx=kartanh(x)+c</math>
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by substitution into (*);  <math>kartanh(x)+kartanh(y)+2c=kartanh(\frac{x+y}{1+xy})+c</math> we know that <math>artanh(x)+artanh(y)=artanh(\frac{x+y}{1+xy})</math>  so <math>  \frac{c}{k}=0</math> so <math>c=0,k\neq0;g(x)=kartanh(x)</math>
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so <math> f(x)=tanh(\frac{x}{k})=\frac{{(e^\frac{1}{k})}^x-1}{{(e^\frac{1}{k})}^x+1}=\frac{a^x-1}{a^x-1}</math>
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where <math>a=e^{\frac{1}{k}}>1</math>
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-Shushninja
  
 
==See also==
 
==See also==

Latest revision as of 06:00, 13 September 2024

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

Solution (2)

Let $P(x_0,y_0)$ denote a substitution of $(x_0,y_0)$ for $(x,y)$ and $g$ be the inverse of $f$ when it exists.

By $P(x,x)$ we get $f(x)<1$ so the domain $D$ of $g$(x) must be in the $0<x<1$ interval

$P(g(x),g(y)); g(x)+g(y)=g(\frac{x+y}{1+xy})$(*) from here,

Taking $\frac{\partial}{\partial x} and \frac{\partial}{\partial y};$

$g^\prime(x)=\frac{1(1+xy)-y(x+y)}{(1+xy)^2}=\frac{1-y^2}{(1+xy)^2}, g^\prime(y)=\frac{1(1+xy)-x(x+y)}{(1+xy)^2}=\frac{1-x^2}{(1+xy)^2}$

$\frac{g^\prime(x)}{g^\prime(y)}=\frac{1-y^2}{1-x^2}$ so let $g^\prime(x)(1-x^2)=g^\prime(y)(1-y^2)=k$ for some real constant $k>0$.


$g(x)=\int{\frac{k}{1-x^2}}dx=kartanh(x)+c$


by substitution into (*); $kartanh(x)+kartanh(y)+2c=kartanh(\frac{x+y}{1+xy})+c$ we know that $artanh(x)+artanh(y)=artanh(\frac{x+y}{1+xy})$ so $\frac{c}{k}=0$ so $c=0,k\neq0;g(x)=kartanh(x)$ 

so $f(x)=tanh(\frac{x}{k})=\frac{{(e^\frac{1}{k})}^x-1}{{(e^\frac{1}{k})}^x+1}=\frac{a^x-1}{a^x-1}$

where $a=e^{\frac{1}{k}}>1$

-Shushninja

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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