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2002 USAMO Problems/Problem 4: Difference between revisions

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<center>
<center>
<math>
<math>
\displaystyle f(x^2 - y^2) = xf(x) - yf(y)
f(x^2 - y^2) = xf(x) - yf(y)
</math>
</math>
</center>
</center>
for all pairs of real numbers <math> \displaystyle x </math> and <math> \displaystyle y </math>.
for all pairs of real numbers <math>x </math> and <math>y </math>.


== Solutiona ==
== Solutions ==


=== Solution 1 ===
=== Solution 1 ===
We first prove that <math> \displaystyle f </math> is [[odd function | odd]].
We first prove that <math>f </math> is [[odd function | odd]].


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


If we let <math> \displaystyle y = 0 </math>, then we obtain <math> \displaystyle f(x^2) = xf(x) </math>.  Therefore the problem's condition becomes
If we let <math>y = 0 </math>, then we obtain <math>f(x^2) = xf(x) </math>.  Therefore the problem's condition becomes
<center>
<center>
<math>
<math>
\displaystyle f(x^2 - y^2) + f(y^2) = f(x^2)
f(x^2 - y^2) + f(y^2) = f(x^2)
</math>.
</math>.
</center>
</center>


But for any <math> \displaystyle a,b </math>, we may set <math> x = \sqrt{a}</math>, <math> y = \sqrt{b} </math> to obtain
But for any <math>a,b </math>, we may set <math> x = \sqrt{a}</math>, <math> y = \sqrt{b} </math> to obtain
<center>
<center>
<math>
<math>
\displaystyle f(a-b) + f(b) = f(a)
f(a-b) + f(b) = f(a)
</math>.
</math>.
</center>
</center>
(It is well known that the only [[continuous]] solutions to this functional equation are of the form <math> \displaystyle f(x) = kx </math>, but there do exist other solutions to this which are not solutions to the equation of this problem.)
(It is well known that the only [[continuous]] solutions to this functional equation are of the form <math>f(x) = kx </math>, but there do exist other solutions to this which are not solutions to the equation of this problem.)


We may let <math> \displaystyle a = 2t </math>, <math> \displaystyle b = t </math> to obtain <math> \displaystyle 2f(t) = f(2t) </math>.
We may let <math>a = 2t </math>, <math>b = t </math> to obtain <math>2f(t) = f(2t) </math>.


Letting <math> \displaystyle x = t+1 </math> and <math> \displaystyle y = t </math> in the original condition yields
Letting <math>x = t+1 </math> and <math>y = t </math> in the original condition yields
<center>
<center>
<math>
<math>
\displaystyle
 
\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \qquad \\
\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \qquad \\
&=& (t+1)[f(t) + f(1) ] - tf(t) \\
&=& (t+1)[f(t) + f(1) ] - tf(t) \\
Line 44: Line 44:
</center>
</center>


But we know <math> \displaystyle f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1) </math>, so we have <math> \displaystyle 2f(t) + f(1) = f(t) + tf(1) + f(1) </math>, or
But we know <math>f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1) </math>, so we have <math>2f(t) + f(1) = f(t) + tf(1) + f(1) </math>, or
<center>
<center>
<math>
<math>
\displaystyle f(t) = tf(1)
f(t) = tf(1)
</math>.
</math>.
</center>
</center>


Hence all solutions to our equation are of the form <math> \displaystyle f(x) = kx</math>.  It is easy to see that real value of <math> \displaystyle k </math> will suffice.
Hence all solutions to our equation are of the form <math>f(x) = kx</math>.  It is easy to see that real value of <math>k </math> will suffice.


=== Solution 2 ===
=== Solution 2 ===


As in the first solution, we obtain the result that <math> \displaystyle f </math> satisfies the condition
As in the first solution, we obtain the result that <math>f </math> satisfies the condition
<center>
<center>
<math>
<math>
\displaystyle f(a) + f(b) = f(a+b)
f(a) + f(b) = f(a+b)
</math>.
</math>.
</center>
</center>
Line 69: Line 69:
</center>
</center>


Since <math> \displaystyle f(2t) = 2f(t) </math>, this is equal to
Since <math>f(2t) = 2f(t) </math>, this is equal to
<center>
<center>
<math>
<math>
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</math>
</math>
</center>
</center>
It follows that <math> \displaystyle f </math> must be of the form <math> \displaystyle f(x) = kx </math>.
It follows that <math>f </math> must be of the form <math>f(x) = kx </math>.
 




{{alternate solutions}}
{{alternate solutions}}


== Resources ==
== See also ==
 
{{USAMO newbox|year=2002|num-b=3|num-a=5}}
* [[2002 USAMO Problems]]
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=337857#p337857 Discussion on AoPS/MathLinks]




[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Algebra Problems]]
[[Category:Functional Equation Problems]]
{{MAA Notice}}

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.

See also

2002 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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