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> | ||
+ | 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*a^y-2}{2*a^x*a^y+2}</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Then, we can solve this problem using mathematical induction | ||
+ | |||
+ | ~~Andy666 | ||
==See also== | ==See also== |
Latest revision as of 04:01, 28 November 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
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 |
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.