Difference between revisions of "1983 IMO Problems/Problem 1"
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Suppose there did exist such an <math>a\ne1</math>. Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>. Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>. Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>. Let <math>b</math> equal the one that is greater than <math>1</math>. Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>. Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>. Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>. But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>. | Suppose there did exist such an <math>a\ne1</math>. Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>. Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>. Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>. Let <math>b</math> equal the one that is greater than <math>1</math>. Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>. Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>. Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>. But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>. | ||
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Revision as of 21:32, 31 January 2016
Find all functions defined on the set of positive reals which take positive real values and satisfy:
for all
; and
as
.
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 |