Difference between revisions of "1983 IMO Problems/Problem 1"
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− | 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>; | + | ==Problem== |
+ | Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy the conditions: | ||
+ | (i) <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; | ||
+ | (ii) <math>f(x)\to0</math> as <math>x\to \infty</math>. | ||
+ | ==Solution 1== | ||
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>. | ||
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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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>x=1</math> so <cmath>f(f(y))=yf(1).</cmath> If <math>f(a)=f(b),</math> then <math>af(1)=f(f(a))=f(f(b))=bf(1)\implies a=b</math> because <math>f(1)</math> goes to the real positive integers, not <math>0.</math> Hence, <math>f</math> is injective. Let <math>x=y</math> so | ||
+ | <cmath> f(xf(x))=xf(x)</cmath> so <math>xf(x)</math> is a fixed point of <math>f.</math> Then, let <math>y=1</math> so <math>f(xf(1))=f(x)\implies f(1)=1</math> as <math>x</math> can't be <math>0</math> so <math>1</math> is a fixed point of <math>f.</math> We claim <math>1</math> is the only fixed point of <math>f.</math> Suppose for the sake of contradiction that <math>a,b</math> be fixed points of <math>f</math> so <math>f(a)=a</math> and <math>f(b)=b.</math> Then, setting <math>x=a,y=b</math> in (i) gives <cmath>f(ab)=f(af(b))=bf(a)=ab</cmath> so <math>ab</math> is also a fixed point of <math>f.</math> Also, let <math>x=\frac{1}{a},y=a</math> so <cmath>1=f(1)=f(\frac{1}{a}\cdot a)=f(\frac{1}{a}\cdot f(a))=af(\frac{1}{a})\implies f(\frac{1}{a})=\frac{1}{a}</cmath> so <math>\frac{1}{a}</math> is a fixed point of <math>f.</math> If <math>f(a)=a</math> with <math>a>1,</math> then <math>f(a^n)</math> is a fixed point of <math>f</math>, contradicting (ii). If <math>f(a)=a</math> with <math>0<a<1,</math> then <math>f(\frac{1}{a^n})=\frac{1}{a^n}</math> so <math>\frac{1}{a^n}</math> is a fixed point, contradicting (ii). Hence, the only fixed point is <math>1</math> so <math>xf(x)=1</math> so <math>f(x)=\boxed{\frac{1}{x}}</math> and we can easily check that this solution works. | ||
+ | |||
+ | {{IMO box|year=1983|before=First question|num-a=2}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Revision as of 14:04, 27 December 2018
Problem
Find all functions defined on the set of positive reals which take positive real values and satisfy the conditions: (i) for all ; (ii) as .
Solution 1
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 .
Solution 2
Let so If then because goes to the real positive integers, not Hence, is injective. Let so so is a fixed point of Then, let so as can't be so is a fixed point of We claim is the only fixed point of Suppose for the sake of contradiction that be fixed points of so and Then, setting in (i) gives so is also a fixed point of Also, let so so is a fixed point of If with then is a fixed point of , contradicting (ii). If with then so is a fixed point, contradicting (ii). Hence, the only fixed point is so so and we can easily check that this solution works.
1983 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |