Problem 23 · 2011 Math Kangaroo
Stretch
Algebra & Patterns
substitution
The sequence of functions \(f_{1}(x),\,f_{2}(x),\,\ldots\) satisfies \(f_{1}(x)=x\) and \(f_{n+1}(x)=\dfrac{1}{1-f_{n}(x)}\). Determine the value of \(f_{2011}(2011)\).
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Answer: A — 2011
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Hint 1 of 2
Compute f₂, f₃, f₄ and watch for a repeat.
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Hint 2 of 2
The map cycles with period 3, so reduce 2011 modulo 3.
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Approach: detect the period-3 cycle
- f₁(x)=x, f₂=1/(1−x), f₃=(x−1)/x, and f₄=x again — period 3.
- 2011 = 3·670 + 1, so f₂₀₁₁ = f₁, the identity.
- Therefore f₂₀₁₁(2011) = 2011.
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