 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/32a/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The result follows by induction. It is true for

n = 1

. If

F^{n} (x) = 2^{n} x (mod 1),

then

F^{n + 1} (x) = F (F^{n} (x)) = 2 (2^{n} x (mod 1)) (mod 1) = 2^{n + 1} x (mod 1) .

Therefore

F^{n} (x) = 2^{n} x (mod 1) .

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