Difference between revisions of "1983 IMO Problems/Problem 1"
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+ | ==Problem== | ||
Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy: <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>. | Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy: <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>. | ||
+ | ==Solution== | ||
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>. Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>. | Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>. Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>. | ||
Revision as of 22:36, 31 January 2016
Problem
Find all functions defined on the set of positive reals which take positive real values and satisfy: for all ; and as .
Solution
Let and we have . Now, let and we have since we have .
Plug in and we have . If is the only solution to then we have . We prove that this is the only function by showing that there does not exist any other :
Suppose there did exist such an . Then, letting in the functional equation yields . Then, letting yields . Notice that since , one of is greater than . Let equal the one that is greater than . Then, we find similarly (since ) that . Putting into the equation, yields . Repeating this process we find that for all natural . But, since , as , we have that which contradicts the fact that as .
1983 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |