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| | == 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> | | 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 1 ==
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| − | 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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| − | <cmath>
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| − | \begin{align}
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| − | f(f(0)) &= 2 \\
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| − | f(y + f(1)) &= f(y) + 2
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| − | \end{align}
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| − | </cmath>
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| − |
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| − | Next, let us find the derivative of this function. First, with (2), we isolate <math>f(y)</math> one one side
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(y) = f(y + f(1)) - 2
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | and then take the derivative:
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | \dfrac{\mathrm{d}f}{\mathrm{d}y}
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| − | &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1)) - 2\right] \\
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| − | &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1))\right] - \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[2\right] \\
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| − | &= f'(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
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| − | &= f'(y + f(1))\cdot(1)\\
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| − | f'(y) &= f'(y + f(1))\\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | 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.
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| − |
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| − | 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:
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(xy + f(x)) &= xf(y) + 2 \\
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| − | f &= xf + 2 \\
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| − | f - xf &= 2 \\
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| − | f(1-x) &= 2 \\
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| − | f &= \dfrac{2}{1-x} \\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | 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:
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(x) = ax + b
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | Via (1), we get the following:
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(f(0)) &= 2 \\
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| − | a(a(0) + b) + b &= 2 \\
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| − | ab + b &= 2
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | And via (2),
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(y + f(1)) &= f(y) + 2 \\
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| − | a(y + a(1) + b) + b &= ay + b + 2 \\
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| − | ay + a^2 + ab + b &= ay + b + 2 \\
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| − | a^2 + ab &= 2 \\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | Setting these equations equal to each other,
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | ab + b &= a^2 + ab \\
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| − | b &= a^2 \\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | Therefore,
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | ab + b &= 2 \\
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| − | a^3 + a^2 &= 2 \\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | 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:
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| − |
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| − | <cmath>
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| − | \begin{align*}
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| − | f(x) &= x + 1 \\
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| − | f(x) &= x(-1 + i) - 2i \\
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| − | f(x) &= x(-1 - i) + 2i \\
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| − | \end{align*}
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| − | </cmath>
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| − |
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| − | Because only the first function maps strictly to positive real numbers, the only solution that works is <math>f(x) = x + 1</math>. <math>\square</math>
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| − |
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| − | ~cogsandsquigs
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| − |
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| − | This solution is heavily flawed:
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| − |
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| − | 1. <math>f(0)</math> doesn't exist.
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| − |
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| − | 2. The function isn't necessarily continuous, let alone differentiable.
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| − |
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| − | 3. Even if it was differentiable, the equation
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| − | <cmath>
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| − | \begin{align*}
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| − | f'(y) &= f'(y + f(1))\\
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| − | \end{align*}
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| − | </cmath>
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| − | doesn't necessarily mean the derivative is constant either. It only implies the derivative is periodic with period <math>f(1)</math>
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| − |
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| − | spencerD.
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| | == Solution 2 == | | == Solution 2 == |
be the set of positive real numbers. Find all functions 
such that, for all 
,![\[f(xy + f(x)) = xf(y) + 2\]](http://latex.artofproblemsolving.com/c/6/2/c623d226331401de700d000f6c1cc868f268b65f.png)
, so we know that this function is linear for 
. Solving for the coefficients (in the same way as solution 1), we find that 
.
. Since 
, 
, so 
. It becomes clear then that 
as well, so 
