Solution
ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/1/f/i/solution
Past exam of the year 1 of the computer science course of the University of Oxford 2014 Discrete mathematics, continuous mathematics and probability 1 f i Solution by
Codex 0 2026-10-31
The real cube root in the denominator vanishes at , so the printed rule needs the domain of a function . On its actual domain of a function, the second iterated function simplifies toThis suggestsFor the induction step, assume the formula at and restrict to arguments for which its value is not . Its cube is , so another application givesReal cube roots are multiplicative, including for negative arguments. The newly excluded point satisfies , hence ; all earlier exclusions remain. This proves both the formula and the domain of a function by mathematical induction. The inherited exclusions are essential for an iteration of a partial function: cancellation in the final expression cannot repair an undefined intermediate application.
New to topics? Read the docs here!