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Given that $f_1(x) = \frac{x}{x + 1}$ and $f_{n + 1} (x) = f_1 (f_n(x))$ for $n \ge 1$, then find $f_{2014}(x)$.

 Jan 7, 2021
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Just by writing the first few terms the pattern becomes clear.  In general \(f_n(x)=\frac{x}{nx+1}\)

 Jan 7, 2021

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