The Number of Functions

Algebra Level 4

Find the number of functions f f from the positive reals to the positive reals such that f ( x + f ( y ) ) = y x y + 1 f(x+ f(y)) = \frac{y}{xy+1} for all x , y > 0. x, y > 0.

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Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Oct 12, 2015

Let x = f ( x ) x = f(x) , f ( f ( x ) + f ( y ) ) = y f ( x ) y + 1 f(f(x) + f(y)) = \frac{y}{f(x)y+1}

Switching x , y x, y gets us f ( f ( x ) + f ( y ) ) = x f ( y ) x + 1 f(f(x) + f(y)) = \frac{x}{f(y)x + 1}

This implies that x f ( y ) x + 1 = y f ( x ) y + 1 f ( x ) − 1 x = f ( y ) − 1 y = c f ( x ) = 1 x + c \begin{aligned} \frac{x}{f(y)x + 1} & = \frac{y}{f(x)y+1} \\ f(x) - \frac{1}{x} & = f(y) - \frac{1}{y} = c \\ f(x) & = \frac{1}{x} + c \end{aligned}

Plugging this back into our function and letting x = y = 1 x = y = 1 , we find that c = 0 c = 0 or − 3 2 -\frac{3}{2} . However, f ( x ) = x − 3 2 f(x) = x - \frac{3}{2} yields negative values, therefore f ( x ) = 1 x f(x) = \frac{1}{x} is the only valid function and checking it to the original function validates that it is the correct function.

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Awesome problem. However, I'm not sure if it's just me, but when I substitute different values of x x and y y , I get different different values for c c . I don't have time to play with it right now, but I'll leave this page up to mess with it later to see if I made a mistake.

James Wilson - 3 years, 5 months ago

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But c = 0 c=0 seems to always come out. So, I think maybe c = 0 c=0 is really the only possibility.

James Wilson - 3 years, 5 months ago

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I had some extra time, so I took the liberty of determining the value of c c . 1 x + 1 y + c + c ≡ y x y + 1 ⇒ y x y + 1 + c y + c ≡ y x y + 1 \frac{1}{x+\frac{1}{y}+c}+c\equiv\frac{y}{xy+1}\Rightarrow\frac{y}{xy+1+cy}+c\equiv\frac{y}{xy+1} ⇒ ( x y + 1 ) ( y + c ( x y + 1 + c y ) ) ≡ y ( x y + 1 + c y ) \Rightarrow(xy+1)(y+c(xy+1+cy))\equiv y(xy+1+cy) ⇒ x y 2 + c x 2 y 2 + c x y + c 2 x y 2 + y + c x y + c + c 2 y ≡ x y 2 + y + c y 2 \Rightarrow xy^2+cx^2y^2+cxy+c^2xy^2+y+cxy+c+c^2y\equiv xy^2+y+cy^2 ⇒ c x 2 y 2 + c 2 x y 2 + 2 c x y − c y 2 + c 2 y + c ≡ 0 \Rightarrow cx^2y^2+c^2xy^2+2cxy-cy^2+c^2y+c\equiv 0 By equating the coefficients on both sides, we can conclude c = 0 c=0 .

James Wilson - 3 years, 5 months ago

How can you assume first x = f ( x ) x=f(x) then u prove f ( x ) = 1 / x f(x)=1/x .This is not a god way I think

Kushal Bose - 4 years, 2 months ago

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I substituted f ( x ) f(x) into x x of the functional equation. I'm not saying that x = f ( x ) x = f(x) .

Alan Yan - 4 years, 2 months ago

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