Difference between revisions of "2002 USAMO Problems/Problem 4"
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===Solution 3=== | ===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. | 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(xg(x))-y(yg(y)) \implies (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}} | {{alternate solutions}} |
Revision as of 01:44, 14 January 2015
Contents
Problem
Let be the set of real numbers. Determine all functions
such that
for all pairs of real numbers and
.
Solutions
Solution 1
We first prove that is odd.
Note that , and for nonzero
,
, or
, which implies
. Therefore
is odd. Henceforth, we shall assume that all variables are non-negative.
If we let , then we obtain
. Therefore the problem's condition becomes
.
But for any , we may set
,
to obtain
.
(It is well known that the only continuous solutions to this functional equation are of the form , but there do exist other solutions to this which are not solutions to the equation of this problem.)
We may let ,
to obtain
.
Letting and
in the original condition yields
But we know , so we have
, or
.
Hence all solutions to our equation are of the form . It is easy to see that real value of
will suffice.
Solution 2
As in the first solution, we obtain the result that satisfies the condition
.
We note that
.
Since , this is equal to
It follows that must be of the form
.
Solution 3
Let , so that the functional equation becomes
. For positive
, then,
, which reduces to
for nonzero
. For
, we have
. Thus, we have limited
to linear functions of the form
where
is a constant. We can verify that if
, then any value of
will work:
, which is always true.
Solution 4
We begin by defining , so
. Rewritting the given functional equation in terms of
, we find:
<\center>
Setting
yields
or
. This can only be satisfied for all
if
doesn't depend on
, i.e.
. Back substituting,
is the only possible solution, and it can be easily confirmed that it satisfies the given condition for all real
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.