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$f(x) $ is an injective function . The definition of $f(x)$ is like following:

$$ f:[0, \infty[\to \Bbb R-\{0\}, f\left(x + \frac{1}{f(y)}\right) = \frac{f(x)f(y)}{f(x) + f(y)} $$

If $f(0) = 1$ then what is the value of $ f(2012)$?

Can you help me to solve this problem ?

Hashir Omer
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Way to infinity
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2 Answers2

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You can check that $$ f\left(0 + \frac{1}{f(0)}\right) = \frac{f(0)f(0)}{f(0) + f(0)} \Rightarrow f(1) = \frac{1}{2} $$ Now assume that $f(n-1) = 1/n$, then $$ f(n) = f\left(0 + \frac{1}{f(n-1)}\right) = \frac{f(0)f(n-1)}{f(0) + f(n-1)} = \frac{1}{n+1} $$ Hence, $$ f(2012) = \frac{1}{2013} $$

Prahlad Vaidyanathan
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$f\left(x+ \frac{1}{f(y)} \right) = \frac{f(x)f(y)}{f(x)+f(y)}$ and $f(0)=1$

i.e., for $x=-\frac{1}{f(y)}$ we have $\frac{f(x).\frac{-1}{x}}{f(x)-\frac{1}{x}}=1$

$\frac{-f(x)}{xf(x)-1}=1$ i.e.,$-f(x)=xf(x)-1$ i.e., $f(x)(x+1)=1$ i.e., $f(x)=\frac{1}{x+1}$

This would give $f(x)$ for any $x$ in the domain.

in particular $f(2012)=\frac{1}{2013}$