Difference between revisions of "2020 CAMO Problems/Problem 1"
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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> | 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> | 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 | + | 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*a^y-2}{2*a^x*a^y+2}</cmath> |
− | Then, <cmath>f(x+y)=\frac{a^x | + | Then, <cmath>f(x+y)=\frac{a^x*a^y-1}{a^x*a^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> | ||
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~~Andy666 | ~~Andy666 | ||
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+ | ==Solution (2)== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <math>P(g(x),g(y)); g(x)+g(y)=g(\frac{x+y}{1+xy})</math>(*) from here, | ||
+ | |||
+ | Taking <math>\frac{\partial}{\partial x} and \frac{\partial}{\partial y};</math> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | |||
+ | <math>g(x)=\int{\frac{k}{1-x^2}}dx=kartanh(x)+c</math> | ||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | where <math>a=e^{\frac{1}{k}}>1</math> | ||
+ | |||
+ | -Shushninja | ||
==See also== | ==See also== |
Latest revision as of 06:00, 13 September 2024
Contents
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
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
Solution (2)
Let denote a substitution of for and be the inverse of when it exists.
By we get so the domain of (x) must be in the interval
(*) from here,
Taking
so let for some real constant .
by substitution into (*); we know that so so
so
where
-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 |
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.