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

m (→‎Solution: Reword final sentence nicely)
(→‎Solution: Making it more rigorous, although it does not change the core process of the solution)
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</cmath>
 
</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
+
Next, let us find the derivative of this function. First, with (2), we isolate <math>f(y)</math> one one side
  
 
<cmath>
 
<cmath>
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\begin{align*}
 
\begin{align*}
 
     \dfrac{\mathrm{d}f}{\mathrm{d}y}
 
     \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)) - 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] \\
 
     &= \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\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
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</cmath>
 
</cmath>
  
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The second derivative is as follows:
+
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 <math>f(1) \neq 0</math>. We know it is not zero because if it was, then (2) would become <math>f(y) = f(y) + 2</math>, implying that <math>0 = 2</math>, which is not true.
 +
 
 +
Therefore, the function <math>f'</math> must be a constant, and <math>f</math> 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:
  
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
     \dfrac{\mathrm{d}^2f}{\mathrm{d}y^2}
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     f(xy + f(x)) &= xf(y) + 2 \\
−
    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[\dfrac{\mathrm{d}f}{\mathrm{d}y}\right] \\
+
     f &= xf + 2 \\
−
    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f'(y + f(1))\right] \\
+
    f - xf &= 2 \\
−
     &= f''(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
+
    f(1-x) &= 2 \\
−
     f''(y) &= f''(y + f(1))\\
+
     f &= \dfrac{2}{1-x} \\
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</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 or a constant. We know it is not a constant because if it was, the problem could be reduced to a linear equation with two unknowns, <math>f</math> and <math>x</math>, making <math>f</math> depend on <math>x</math>, which is not a constant function. That means we can model <math>f(x)</math> like so:
+
where <math>f</math> is the constant from <math>f(x)</math>. As we see, <math>f</math> would depend on <math>x</math>, making it not a constant function. Thus, <math>f(x)</math> must be linear, meaning we can model it like so:
  
 
<cmath>
 
<cmath>

Revision as of 19:40, 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 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

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