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Difference between revisions of "2023 USAMO Problems/Problem 2"
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Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath>
 
Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath>
  
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== Solution ==
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== Solution 1 ==
 
First, let us plug in some special points; specifically, plugging in <math>x=0</math> and <math>x=1</math>, respectively:
 
First, let us plug in some special points; specifically, plugging in <math>x=0</math> and <math>x=1</math>, respectively:
  
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spencerD.
 
spencerD.
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== Solution 2 ==
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Make the following substitutions to the equation:
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1. <math>(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2</math>
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2. <math>(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4</math>
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3. <math>(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf(1 + \frac{f(1)}{x}) + 2</math>
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It then follows from (2) and (3) that <math>f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}</math>, so we know that this function is linear for <math>x > 1</math>. Solving for the coefficients (in the same way as solution 1), we find that <math>f(x) = x + 1 \forall x > 1</math>.
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Now, we can let <math>x > 1</math> and <math>y \le 1</math>. Since <math>f(x) = x + 1</math>, <math>xy + f(x) > x > 1</math>, so <math>f(xy + f(x)) = xy + x + 2 = xf(y) + 2</math>. It becomes clear then that <math>f(y) = y + 1</math> as well, so <math>f(x) = x + 1</math> is the only solution to the functional equation.
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~jkmmm3

Revision as of 12:50, 17 May 2023

Problem 2

Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$,\[f(xy + f(x)) = xf(y) + 2\]

Solution 1

First, let us plug in some special points; specifically, plugging in $x=0$ and $x=1$, respectively:

\begin{align} f(f(0)) &= 2 \\ f(y + f(1)) &= f(y) + 2 \end{align}

Next, let us find the derivative of this function. First, with (2), we isolate $f(y)$ one one side

\begin{align*} f(y) = f(y + f(1)) - 2 \end{align*}

and then take the derivative:

\begin{align*} \dfrac{\mathrm{d}f}{\mathrm{d}y} &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1)) - 2\right] \\ &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1))\right] - \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[2\right] \\ &= f'(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\ &= f'(y + f(1))\cdot(1)\\ f'(y) &= f'(y + f(1))\\ \end{align*}

With the derivative, we see that the input to the function does not matter: it will return the same result regardless of input, assuming that $f(1) \neq 0$. We know it is not zero because if it was, then (2) would become $f(y) = f(y) + 2$, implying that $0 = 2$, which is not true.

Therefore, the function $f'$ must be a constant, and $f$ must be a linear equation or a constant. We know it is not a constant because if it was, the problem could be reduced to the following:

\begin{align*} f(xy + f(x)) &= xf(y) + 2 \\ f &= xf + 2 \\ f - xf &= 2 \\ f(1-x) &= 2 \\ f &= \dfrac{2}{1-x} \\ \end{align*}

where $f$ is the constant from $f(x)$. As we see, $f$ would depend on $x$, making it not a constant function. Thus, $f(x)$ must be linear, meaning we can model it like so:

\begin{align*} f(x) = ax + b \end{align*}

Via (1), we get the following:

\begin{align*} f(f(0)) &= 2 \\ a(a(0) + b) + b &= 2 \\ ab + b &= 2 \end{align*}

And via (2),

\begin{align*} f(y + f(1)) &= f(y) + 2 \\ a(y + a(1) + b) + b &= ay + b + 2 \\ ay + a^2 + ab + b &= ay + b + 2 \\ a^2 + ab &= 2 \\ \end{align*}

Setting these equations equal to each other,

\begin{align*} ab + b &= a^2 + ab \\ b &= a^2 \\ \end{align*}

Therefore,

\begin{align*} ab + b &= 2 \\ a^3 + a^2 &= 2 \\ \end{align*}

There are three solutions to this equation: $a = 1$, $a = -1 + i$, and $a = -1 - i$. Knowing that $b = a^2$, the respective $b$ values are $b = 1$, $b = -2i$, and $b = 2i$. Thus, $f(x)$ could be the following:

\begin{align*} f(x) &= x + 1 \\ f(x) &= x(-1 + i) - 2i \\ f(x) &= x(-1 - i) + 2i \\ \end{align*}

Because only the first function maps strictly to positive real numbers, the only solution that works is $f(x) = x + 1$. $\square$

~cogsandsquigs

This solution is heavily flawed:

1. $f(0)$ doesn't exist.

2. The function isn't necessarily continuous, let alone differentiable.

3. Even if it was differentiable, the equation \begin{align*} f'(y) &= f'(y + f(1))\\ \end{align*} doesn't necessarily mean the derivative is constant either. It only implies the derivative is periodic with period $f(1)$

spencerD.

Solution 2

Make the following substitutions to the equation:

1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$

2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$

3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf(1 + \frac{f(1)}{x}) + 2$

It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$, so we know that this function is linear for $x > 1$. Solving for the coefficients (in the same way as solution 1), we find that $f(x) = x + 1 \forall x > 1$.

Now, we can let $x > 1$ and $y \le 1$. Since $f(x) = x + 1$, $xy + f(x) > x > 1$, so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$. It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.

~jkmmm3

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