Difference between revisions of "2023 USAMO Problems/Problem 2"

Revision as of 10:21, 7 April 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

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 first and second derivatives 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:

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

The second derivative is as follows:

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

For both of these derivatives, we see that the input to the function does not matter: it will return the same result regardless of input. Therefore, the functions $f'$ and $f''$ must be constants, and $f$ must be a linear equation. That means we can model $f(x)$ 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:

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

$\square$

~ cogsandsquigs

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Difference between revisions of "2023 USAMO Problems/Problem 2"

Revision as of 10:21, 7 April 2023

Problem 2

Solution

@@ Line 1: / Line 1: @@
-## Problem 2
+== Problem 2 ==
 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>
-## Solution
+== Solution ==
 First, let us plug in some special points; specifically, plugging in <math>x=0</math> and <math>x=1</math>, respectively:
-\begin{align}
+<cmath>
+\begin{align*}
      f(f(0)) &= 2 \\
      f(y + f(1)) &= f(y) + 2
-\end{align}
+\end{align*}
+</cmath>
 Next, let us find the first and second derivatives of this function. First, with (2), we isolate <math>f(y)</math> one one side
+<cmath>
 \begin{align*}
     f(y) = f(y + f(1)) - 2
 \end{align*}
+</cmath>
 and then take the derivative:
+<cmath>
 \begin{align*}
-     \dv{f}{y}
+     \dfrac{\mathrm{d}f}{\mathrm{d}y}
-     &= \dv{f}{y}\left[f(y + f(1)) - 2\right] \\
+     &=\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1)) - 2\right] \\
-     &= \dv{f}{y}\left[f(y + f(1))\right] - \dv{f}{y}\left[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\dv{f}{y}\left[y + f(1)\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*}
+</cmath>
 The second derivative is as follows:
+<cmath>
 \begin{align*}
-     \dv[2]{f}{y}
+     \dfrac{\mathrm{d}^2f}{\mathrm{d}y^2}
-     &= \dv{f}{y}\left[\dv{f}{y}\right] \\
+     &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[\dfrac{\mathrm{d}f}{\mathrm{d}y}\right] \\
-     &= \dv{f}{y}\left[f'(y + f(1))\right] \\
+     &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f'(y + f(1))\right] \\
-     &= f''(y + f(1))\cdot\dv{f}{y}\left[y + f(1)\right] \\
+     &= f''(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
      f''(y) &= f''(y + f(1))\\
 \end{align*}
+</cmath>
 For both of these derivatives, we see that the input to the function does not matter: it will return the same result regardless of input. Therefore, the functions <math>f'</math> and <math>f''</math> must be constants, and <math>f</math> must be a linear equation. That means we can model <math>f(x)</math> like so:
+<cmath>
 \begin{align}
      f(x) = ax + b
 \end{align}
+</cmath>
 Via (1), we get the following:
+<cmath>
 \begin{align*}
      f(f(0)) &= 2 \\
@@ Line 50: / Line 61: @@
      ab + b &= 2
 \end{align*}
+</cmath>
 And via (2),
+<cmath>
 \begin{align*}
      f(y + f(1)) &= f(y) + 2 \\
@@ Line 59: / Line 72: @@
      a^2 + ab &= 2 \\
 \end{align*}
+</cmath>
 Setting these equations equal to each other,
+<cmath>
 \begin{align*}
      ab + b &= a^2 + ab \\
      b &= a^2 \\
 \end{align*}
+</cmath>
 Therefore,
+<cmath>
 \begin{align*}
      ab + b &= 2 \\
      a^3 + a^2 &= 2 \\
 \end{align*}
+</cmath>
 There are three solutions to this equation: <math>a = 1</math>, <math>a = -1 + i</math>, and <math>a = -1 - i</math>. Knowing that <math>b = a^2</math>, the respective <math>b</math> values are <math>b = 1</math>, <math>b = -2i</math>, and <math>b = 2i</math>. Thus, <math>f(x)</math> could be the following:
+<cmath>
 \begin{align*}
      f(x) &= x + 1 \\
@@ Line 81: / Line 100: @@
      f(x) &= x(-1 - i) + 2i \\
 \end{align*}
+</cmath>
 <math>\square</math>
+~ cogsandsquigs