2008 IMO Problems/Problem 4: Difference between revisions
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== Problem == | |||
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 | |||
<center> | |||
<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> | |||
</center> | |||
for all positive real numbes <math>w,x,y,z,</math> satisfying <math>wx = yz.</math> | |||
== Solution == | |||
Considering <math>x=t,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>x=t,w=1</math> and <math>z=y=\sqrt{x}</math> which satisfy the constraint <math>wx=yz</math> we get the following equation: | ||
Revision as of 10:33, 23 August 2008
Problem
Find all functions 
(so 
is a function from the positive real numbers) such that
for all positive real numbes 
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 kowing 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:
![\[f(x) = \frac{1+x^2 \pm \sqrt{(1+x^2)^2-4x^2}}{2x} = \frac{1+x^2 \pm \sqrt{(1-x^2)^2}}{2x}\]](http://latex.artofproblemsolving.com/3/0/9/309c9fe04878910bac3f6e1ad2a2cdd3fb47e568.png)
At once we see that for one value of , can only take one of 2 possible values:
![\[f(x) = x \vee f(x) = \frac{1}{x}\]](http://latex.artofproblemsolving.com/4/6/4/464f2b5c6e124f95eb1561934c0800d9766fc22b.png)
.
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:
![\[\frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}\]](http://latex.artofproblemsolving.com/a/9/7/a97206cb25ca26c403442693dc098627111b97e6.png)
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:
![\[\frac{ \frac{1}{w^2} + \frac{1}{x^2} }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2} }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2}\]](http://latex.artofproblemsolving.com/d/9/b/d9b58f215cc8a4448e03de6886bb52bd5ba9d5bb.png)
provided that 
which is 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)
or $f(x) = x^{\pm 1}\ \forall_{x \x \in \mathbb{R}^+}$ (Error compiling LaTeX. Unknown error_msg) if you prefer.
