Difference between revisions of "1983 IMO Problems/Problem 1"
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Suppose there did exist such an <math>a\ne1</math>. Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>. Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>. Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>. Let <math>b</math> equal the one that is greater than <math>1</math>. Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>. Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>. Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>. But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>. | Suppose there did exist such an <math>a\ne1</math>. Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>. Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>. Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>. Let <math>b</math> equal the one that is greater than <math>1</math>. Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>. Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>. Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>. But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>. | ||
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Revision as of 22:32, 31 January 2016
Find all functions defined on the set of positive reals which take positive real values and satisfy: for all ; and as .
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 |