University of South Carolina High School Math Contest/1993 Exam/Problem 24

Problem

If $f(x) = \frac{1 + x}{1 - 3x}, f_1(x) = f(f(x)), f_2(x) = f(f_1(x)),$ and in general $f_n(x) = f(f_{n-1}(x)),$ then $f_{1993}(3)=$

$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }1993 \qquad \mathrm{(C) \ }\frac 12 \qquad \mathrm{(D) \ }\frac 15 \qquad \mathrm{(E) \ } -2^{-1993}$

Solution

$f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}$. Then $f_1(3) = f(-\frac12) = \frac{1 - \frac12}{1 + 3\cdot\frac12} = \frac15$, $\displaystyle f_2(3) = f(\frac15) = \frac{1 + \frac15}{1 - 3\cdot\frac15} = 3$ and $f_3(3) = f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}$.

It follows immediately that the function cycles and $f_n(3) = -\frac12$ if $n = 3k$, $f_n(3) = \frac15$ if $n = 3k + 1$ and $f_n(3) = 3$ if $n = 3k + 2$. Since $1993 = 3\cdot 664 + 1$, $f_{1993}(3) = \frac 15 \Longrightarrow \mathrm{(D)}$.


Alternatively, $f_1(x) = f(f(x)) = \frac{1 + f(x)}{1 - 3f(x)} = \frac{1 + \frac{1 + x}{1 - 3x}}{1 - 3\frac{1 + x}{1 - 3x}} = \frac{(1 - 3x) + (1 + x)}{(1 - 3x) - 3(1 + x)} = \frac{2 - 2x}{-2 - 6x} = \frac{x - 1}{3x + 1}$. Thus $f_2(x) = f(f_1(x)) = \frac{1 + \frac{x - 1}{3x + 1}}{1 - 3\frac{x - 1}{3x + 1}} = \frac{(3x + 1) + (x - 1)}{(3x + 1) - 3(x - 1)} = \frac{4x}{4} = x$, so $f_{3k + 2}(x)  = f(f(f(f_{3k - 1})))(x) = f_{3k - 1}(x) = \ldots = f_2(x) = x$. Thus $f_{1993}(x) = f(f(f_{1991}(x))) = f(f(x))$ so $f_{1993}(3) = f(f(3)) = \frac15$.


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