Functional Equation Solution 3
Let $ \mathbb{Q}^+ $ denote the set of positive rational numbers. Let $ f : \mathbb{Q}^+ \to \mathbb{Q}^+ $ be a function such that
\[f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y\]for all $ x, $ $ y \in \mathbb{Q}^+ $.
Find all possible values of $ f \left( \frac{1}{3} \right) $. Enter all the possible values, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Setting $ y = x $ in the given functional equation, we get
\[f(x + 1) = f(x) + 1 + 2x.\quad (*)\]Then
\begin{align*}
f(x + 2) &= f(x + 1) + 1 + 2(x + 1) \\
&= f(x) + 1 + 2x + 1 + 2(x + 1) \\
&= f(x) + 4x + 4.\end{align*}Setting $ y = 2x, $ we get
\[f(x + 2) = f(x) + \frac{f(2x)}{f(x)} + 4x,\]so
\[f(x) + 4x + 4 = f(x) + \frac{f(2x)}{f(x)} + 4x.\]Hence, $ \frac{f(2x)}{f(x)} = 4, $ so $ f(2x) = 4f(x) $ for all $ x \in \mathbb{Q}^+ $.
In particular, $ f(2) = 4f(1) $. But from $ (*), $ $ f(2) = f(1) + 3 $. Solving, we find $ f(1) = 1 $ and $ f(2) = 4 $. Then
\[f(3) = f(2) + 1 + 2 \cdot 2 = 9.\]Setting $ x = 3 $ and $ y = 1, $ we get
\[f \left( 3 + \frac{1}{3} \right) = f(3) + \frac{f(1)}{f(3)} + 2 \cdot 1 = 9 + \frac{1}{9} + 2 = \frac{100}{9}.\]Then by repeated application of $ (*), $
\begin{align*}
f \left( 2 + \frac{1}{3} \right) &= f \left( 3 + \frac{1}{3} \right) - 1 - 2 \left( 2 + \frac{1}{3} \right) = \frac{49}{9}, \\
f \left( 1 + \frac{1}{3} \right) &= f \left( 2 + \frac{1}{3} \right) - 1 - 2 \left( 1 + \frac{1}{3} \right) = \frac{16}{9}, \\
f \left( \frac{1}{3} \right) &= f \left( 1 + \frac{1}{3} \right) - 1 - 2 \cdot \frac{1}{3} = \boxed{\frac{1}{9}}.\end{align*}More generally, we can prove that $ f(x) = x^2 $ for all $ x \in \mathbb{Q}^+ $.
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