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Difficulty: HardFunction Evaluation, Composition, and Properties

Let the function ff be defined by f(x)=x−1x+1f(x) = \frac{x - 1}{x + 1} for all real numbers x≠−1x \neq -1. Let f1(x)=f(x)f^1(x) = f(x), and let fn(x)=f(fn−1(x))f^n(x) = f(f^{n-1}(x)) for all integers n≥2n \geq 2. What is the value of 30⋅f2026(3)30 \cdot f^{2026}(3)?

Answer: -10

Answer

The value of the expression is -10.
Evaluating successive iterations of f(3)f(3) reveals a repeating sequence: f1(3)=1/2f^1(3) = 1/2, f2(3)=−1/3f^2(3) = -1/3, f3(3)=−2f^3(3) = -2, f4(3)=3f^4(3) = 3, and then f5(3)=1/2f^5(3) = 1/2. This indicates a cycle of period 4. Dividing the power 2026 by the period 4 gives a remainder of 2, meaning f2026(3)f^{2026}(3) is equal to f2(3)=−1/3f^2(3) = -1/3. Multiplying this by 30 yields the final answer of -10.

Step-by-Step Solution

1
Calculate the first few compositions of the function evaluated at the given input x=3x = 3.
f1(3)=1/2f^1(3) = 1/2, f2(3)=−1/3f^2(3) = -1/3, f3(3)=−2f^3(3) = -2, and f4(3)=3f^4(3) = 3.
To look for a repeating pattern or periodic behavior in the iterated function composition.
2
Identify the period of the repeating cycle.
The cycle has a length of 4, repeating the values [1/2,−1/3,−2,3][1/2, -1/3, -2, 3].
Since f4(3)=3f^4(3) = 3, evaluating further iterations will yield the same sequence of values.
3
Use modular arithmetic to find the value of the 2026th composition.
2026≡2(mod4)2026 \equiv 2 \pmod 4, meaning f2026(3)=f2(3)=−1/3f^{2026}(3) = f^2(3) = -1/3.
Since the cycle repeats every 4 iterations, dividing 2026 by 4 yields a remainder of 2, indicating the second value in the sequence.
4
Multiply the computed composition value by the given coefficient.
30⋅(−1/3)=−1030 \cdot (-1/3) = -10.
To find the final value of the expression 30⋅f2026(3)30 \cdot f^{2026}(3).

Key Concept

Evaluating repeated function compositions using periodicity and modular arithmetic.
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Topics

Function Evaluation, Composition, and Properties

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