Difference between revisions of "2002 USAMO Problems/Problem 4"

Latest revision as of 21:09, 18 September 2018

Problem

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that

$f(x^2 - y^2) = xf(x) - yf(y)$

for all pairs of real numbers $x$ and $y$ .

Solutions

Solution 1

We first prove that $f$ is odd.

Note that $f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$ , and for nonzero $y$ , $xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$ , or $yf(-y) = -yf(y)$ , which implies $f(-y) = -f(y)$ . Therefore $f$ is odd. Henceforth, we shall assume that all variables are non-negative.

If we let $y = 0$ , then we obtain $f(x^2) = xf(x)$ . Therefore the problem's condition becomes

$f(x^2 - y^2) + f(y^2) = f(x^2)$ .

But for any $a,b$ , we may set $x = \sqrt{a}$ , $y = \sqrt{b}$ to obtain

$f(a-b) + f(b) = f(a)$ .

(It is well known that the only continuous solutions to this functional equation are of the form $f(x) = kx$ , but there do exist other solutions to this which are not solutions to the equation of this problem.)

We may let $a = 2t$ , $b = t$ to obtain $2f(t) = f(2t)$ .

Letting $x = t+1$ and $y = t$ in the original condition yields

$\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \qquad \\ &=& (t+1)[f(t) + f(1) ] - tf(t) \\ &=& f(t) + (t+1)f(1) \qquad \qquad \end{matrix}$

But we know $f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1)$ , so we have $2f(t) + f(1) = f(t) + tf(1) + f(1)$ , or

$f(t) = tf(1)$ .

Hence all solutions to our equation are of the form $f(x) = kx$ . It is easy to see that real value of $k$ will suffice.

Solution 2

As in the first solution, we obtain the result that $f$ satisfies the condition

$f(a) + f(b) = f(a+b)$ .

We note that

$f(x) = f\left[ \left(\frac{x+1}{2}\right)^2 - \left( \frac{x-1}{2} \right)^2 \right] = \frac{x+1}{2} f \left( \frac{x+1}{2} \right) - \frac{x-1}{2} f \left( \frac{x-1}{2} \right)$ .

Since $f(2t) = 2f(t)$ , this is equal to

$\frac{(x+1)[f(x) +f(1)]}{4} - \frac{(x-1)[f(x) - f(1)]}{4} = \frac{xf(1) + f(x)}{2}$

It follows that $f$ must be of the form $f(x) = kx$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 77: / Line 77: @@
 It follows that <math>f </math> must be of the form <math>f(x) = kx </math>.
-===Solution 3===
-Let <math>y=0</math>, so that the functional equation becomes <math>f(x^2)=xf(x)</math>. For positive <math>x</math>, then, <math>f(x)=x^{\frac{1}{2}}f(x^{\frac{1}{2}})=x^{\frac{1}{2}}x^{\frac{1}{4}}f(x^{\frac{1}{4}})=x^{\frac{1}{2}}x^{\frac{1}{4}}x^{\frac{1}{8}}f(x^{\frac{1}{8}})=\cdots =x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}f(x^{\frac{1}{\infty}})</math>, which reduces to <math>xf(1)</math> for nonzero <math>x</math>. For <math>x=0</math>, we have <math>f(0)=0\cdot f(0)=0</math>. Thus, we have limited <math>f</math> to linear functions of the form <math>f(x)=kx</math> where <math>k</math> is a constant. We can verify that if <math>f(x)=kx</math>, then any value of <math>k</math> will work: <math>k(x^2-y^2)=x\cdot kx-y\cdot ky</math>, which is always true.
-=== Solution 4 ===
-We begin by defining <math>g(x)=\frac{f(x)}{x}</math>, so <math>f(x)=xg(x)</math>. Rewritting the given functional equation in terms of <math>g</math>, we find:
-<center>
-<math>
-(x^2-y^2)g(x^2-y^2)=x^2g(x)-y^2g(y)
-</math>
-</center>
-Setting <math>y=0</math> yields <math>x^2g(x^2)=x^2g(x)</math> or <math>g(x^2)=g(x)</math>. This can only be satisfied for all <math>x \in \mathbb{R}</math> if <math>g(x)</math> doesn't depend on <math>x</math>, i.e. <math>g(x)=k</math>. Back substituting, <math>f(x)=kx</math> is the only possible solution, and it can be easily confirmed that it satisfies the given condition for all real <math>k</math>.
 {{alternate solutions}}
@@ Line 97: / Line 86: @@
 [[Category:Olympiad Algebra Problems]]
+[[Category:Functional Equation Problems]]
 {{MAA Notice}}

2002 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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