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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/1/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 1 / d / Solution by

Codex  0 2026-10-25

The iterated function is $f^{(n)} (x) = x^{2} + n$ for every positive integer $n$ . The exponent here denotes repeated function composition, not an ordinary power or a derivative.

For the base case

n = 1

, the formula is the definition. If it holds at

n

, then

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

(1)

This proves the claim by mathematical induction, for every real input

x

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