Difference between revisions of "1983 IMO Problems/Problem 1"
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+ | ==Problem== | ||
Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy: <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>. | Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy: <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>. | ||
+ | ==Solution== | ||
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>. Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>. | Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>. Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>. | ||
Revision as of 21:36, 31 January 2016
Problem
Find all functions defined on the set of positive reals which take positive real values and satisfy:
for all
; and
as
.
Solution
Let and we have
. Now, let
and we have
since
we have
.
Plug in and we have
. If
is the only solution to
then we have
. We prove that this is the only function by showing that there does not exist any other
:
Suppose there did exist such an . Then, letting
in the functional equation yields
. Then, letting
yields
. Notice that since
, one of
is greater than
. Let
equal the one that is greater than
. Then, we find similarly (since
) that
. Putting
into the equation, yields
. Repeating this process we find that
for all natural
. But, since
, as
, we have that
which contradicts the fact that
as
.
1983 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |