Difference between revisions of "2016 APMO Problems/Problem 5"
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<cmath>(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))</cmath> | <cmath>(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))</cmath> | ||
− | This gives that <math>\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)</math>. Putting <math>x=\frac{1}{2}</math>, we get <math>\text{Im}(f) \in \left(\frac{1}{2},\infty\right)</math>. | + | This gives that <math>\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)</math>. Putting <math>x=\frac{1}{2}</math>, we get <math>\text{Im}(f) \in \left(\frac{1}{2},\infty\right)</math>. By induction, surjectivity is proved as <math>\lim_{m \to \infty}\frac{1}{2^m}=0</math> |
Revision as of 23:12, 12 July 2021
Problem
Find all functions such that
for all positive real numbers
.
Solution
We claim that is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let
be the assertion to the Functional Equation.
Claim 1: is injective.
Proof: Assume for some
. Now, from
and
we have:
Now comparing, we have as desired.
This gives us the power to compute . From
we get
and injectivity gives
.
Claim 2: is surjective.
Proof: gives
This gives that . Putting
, we get
. By induction, surjectivity is proved as