Difference between revisions of "2002 USAMO Problems/Problem 4"
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<center> | <center> | ||
<math> | <math> | ||
− | + | f(x^2 - y^2) = xf(x) - yf(y) | |
</math> | </math> | ||
</center> | </center> | ||
− | for all pairs of real numbers <math> | + | for all pairs of real numbers <math>x </math> and <math>y </math>. |
== Solutions == | == Solutions == | ||
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It follows that <math>f </math> must be of the form <math>f(x) = kx </math>. | 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}{2}+\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. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 14:14, 24 April 2010
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.