Difference between revisions of "2008 IMO Problems/Problem 4"
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== Problem == | == Problem == | ||
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Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that | Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that | ||
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<center> | <center> | ||
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<math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math> | <math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math> | ||
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</center> | </center> | ||
− | for all positive real | + | |
+ | for all positive real numbers <math>w,x,y,z,</math> satisfying <math>wx = yz.</math> | ||
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== Solution == | == Solution == | ||
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Considering <math>w=1</math> and <math>z=y=\sqrt{x}</math> which satisfy the constraint <math>wx=yz</math> we get the following equation: | Considering <math>w=1</math> and <math>z=y=\sqrt{x}</math> which satisfy the constraint <math>wx=yz</math> we get the following equation: | ||
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<cmath>\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2}) = (1+x^2)f(x)</cmath> | <cmath>\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2}) = (1+x^2)f(x)</cmath> | ||
− | At once considering <math>x=1</math> we get <math>(f(1))^2 = f(1)</math> and | + | |
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+ | At once considering <math>x=1</math> we get <math>(f(1))^2 = f(1)</math> and knowing that <math>f : \mathbb{R}^+ \rightarrow \mathbb{R}^+</math> the only possible solution is <math>f(1)=1</math> since <math>f(1)=0</math> is impossible. | ||
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So we get the quadratic equation: | So we get the quadratic equation: | ||
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<cmath> x(f(x))^2} - (1+x^2)f(x) + x = 0 </cmath> | <cmath> x(f(x))^2} - (1+x^2)f(x) + x = 0 </cmath> | ||
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Solving for <math>f(x)</math> as a function of <math>x</math> we get: | Solving for <math>f(x)</math> as a function of <math>x</math> we get: | ||
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<cmath> f(x) = \frac{1+x^2 \pm \sqrt{(1+x^2)^2-4x^2}}{2x} = \frac{1+x^2 \pm \sqrt{(1-x^2)^2}}{2x}</cmath> | <cmath> f(x) = \frac{1+x^2 \pm \sqrt{(1+x^2)^2-4x^2}}{2x} = \frac{1+x^2 \pm \sqrt{(1-x^2)^2}}{2x}</cmath> | ||
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At once we see that for one value of <math>x</math>, <math>f(x)</math> can only take one of 2 possible values: | At once we see that for one value of <math>x</math>, <math>f(x)</math> can only take one of 2 possible values: | ||
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<cmath> f(x) = x \vee f(x) = \frac{1}{x}</cmath>. | <cmath> f(x) = x \vee f(x) = \frac{1}{x}</cmath>. | ||
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Take into consideration that <math>f(2) = 2</math> but <math>f(3) = \frac{1}{3} </math> verifies the quadratic equation and thus so far we can't say that <math>f(x)=x \, \forall_{x \x \in \mathbb{R}^+}</math> or alternatively <math>f(x)=\frac{1}{x} \, \forall_{x \x \in \mathbb{R}^+\\}</math>. This is indeed the case but we haven't proved it yet. | Take into consideration that <math>f(2) = 2</math> but <math>f(3) = \frac{1}{3} </math> verifies the quadratic equation and thus so far we can't say that <math>f(x)=x \, \forall_{x \x \in \mathbb{R}^+}</math> or alternatively <math>f(x)=\frac{1}{x} \, \forall_{x \x \in \mathbb{R}^+\\}</math>. This is indeed the case but we haven't proved it yet. | ||
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To prove the previous assertion consider 2 values <math>a,b \in \mathbb{R}^+</math> such that <math>a\ne 1 \wedge b \ne 1 \wedge a \ne b</math> while having <math>f(a) = a \wedge f(b)=\frac{1}{b}</math> | To prove the previous assertion consider 2 values <math>a,b \in \mathbb{R}^+</math> such that <math>a\ne 1 \wedge b \ne 1 \wedge a \ne b</math> while having <math>f(a) = a \wedge f(b)=\frac{1}{b}</math> | ||
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Consider now the original functional equation with <math>w=a,\ x=b,\ y=z=\sqrt{ab}</math> which verifies the constraint. Substituting we have: | Consider now the original functional equation with <math>w=a,\ x=b,\ y=z=\sqrt{ab}</math> which verifies the constraint. Substituting we have: | ||
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<cmath> \frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}</cmath> | <cmath> \frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}</cmath> | ||
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Now either <math>f(ab)=ab</math> or <math>f(ab)=\frac{1}{ab}</math>. (notice that <math>ab \ne b \wedge ab\ne b</math> by hypothesis) | Now either <math>f(ab)=ab</math> or <math>f(ab)=\frac{1}{ab}</math>. (notice that <math>ab \ne b \wedge ab\ne b</math> by hypothesis) | ||
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If <math>f(ab)=ab</math> then we have <math> ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1</math> and since <math>b>0</math> the only solution is <math>b=1</math>. | If <math>f(ab)=ab</math> then we have <math> ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1</math> and since <math>b>0</math> the only solution is <math>b=1</math>. | ||
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If <math>f(ab)=\frac{1}{ab}</math> then we have <math> \frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1</math> and since <math>a>0</math> the only solution is <math>a=1</math>. | If <math>f(ab)=\frac{1}{ab}</math> then we have <math> \frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1</math> and since <math>a>0</math> the only solution is <math>a=1</math>. | ||
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So the only solutions are <math>a=1</math> or <math>b=1</math> in which case both alternatives imply <math>f(1)=1</math>. Thus we conclude that solutions to the functional equation are a subset of <math>\left\{f(x)=x \ \forall_{x \x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+} \right\}</math>. | So the only solutions are <math>a=1</math> or <math>b=1</math> in which case both alternatives imply <math>f(1)=1</math>. Thus we conclude that solutions to the functional equation are a subset of <math>\left\{f(x)=x \ \forall_{x \x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+} \right\}</math>. | ||
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Finally plug each of these 2 functions into the functional equation and verify that they indeed are solutions. | Finally plug each of these 2 functions into the functional equation and verify that they indeed are solutions. | ||
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This is trivial since <math>f(x)=x</math> is an obvious solution and for <math>f(x)=\frac{1}{x}</math> we have: | This is trivial since <math>f(x)=x</math> is an obvious solution and for <math>f(x)=\frac{1}{x}</math> we have: | ||
− | <cmath> \frac{ \frac{1}{w^2} + \frac{1}{x^2} }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2} }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2} </cmath> provided that <math>(wx)^2 = (yz)^2</math> which | + | |
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+ | <cmath> \frac{ \frac{1}{w^2} + \frac{1}{x^2} }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2} }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2} </cmath> provided that <math>(wx)^2 = (yz)^2</math> which verifies the original constraint. | ||
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So the functional equation has 2 solutions: | So the functional equation has 2 solutions: | ||
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<cmath>f(x) = x\ \forall_{x \x \in \mathbb{R}^+}\ \vee\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+}</cmath> | <cmath>f(x) = x\ \forall_{x \x \in \mathbb{R}^+}\ \vee\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+}</cmath> |
Revision as of 06:46, 25 August 2008
Problem
Find all functions (so is a function from the positive real numbers) such that
for all positive real numbers satisfying
Solution
Considering and which satisfy the constraint we get the following equation:
\[\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2}) = (1+x^2)f(x)\] (Error compiling LaTeX. Unknown error_msg)
At once considering we get and knowing that the only possible solution is since is impossible.
So we get the quadratic equation:
\[x(f(x))^2} - (1+x^2)f(x) + x = 0\] (Error compiling LaTeX. Unknown error_msg)
Solving for as a function of we get:
At once we see that for one value of , can only take one of 2 possible values:
.
Take into consideration that but verifies the quadratic equation and thus so far we can't say that $f(x)=x \, \forall_{x \x \in \mathbb{R}^+}$ (Error compiling LaTeX. Unknown error_msg) or alternatively $f(x)=\frac{1}{x} \, \forall_{x \x \in \mathbb{R}^+\\}$ (Error compiling LaTeX. Unknown error_msg). This is indeed the case but we haven't proved it yet.
To prove the previous assertion consider 2 values such that while having
Consider now the original functional equation with which verifies the constraint. Substituting we have:
Now either or . (notice that by hypothesis)
If then we have and since the only solution is .
If then we have and since the only solution is .
So the only solutions are or in which case both alternatives imply . Thus we conclude that solutions to the functional equation are a subset of $\left\{f(x)=x \ \forall_{x \x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+} \right\}$ (Error compiling LaTeX. Unknown error_msg).
Finally plug each of these 2 functions into the functional equation and verify that they indeed are solutions.
This is trivial since is an obvious solution and for we have:
provided that which verifies the original constraint.
So the functional equation has 2 solutions:
\[f(x) = x\ \forall_{x \x \in \mathbb{R}^+}\ \vee\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+}\] (Error compiling LaTeX. Unknown error_msg)