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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/6/c/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 6 / c / ii / Solution by

Codex  0 2026-10-31

Define the one-step local truncation error as the defect obtained by inserting the exact solution into an update:

δ_{n + 1} = y (x_{n} + h) - y (x_{n}) - h f (x_{n}, y (x_{n})) .

For a smooth solution, Taylor's theorem and

y^{'} = f (x, y)

give

δ_{n + 1} = \frac{h ^{2}}{2} y^{''} (x_{n}) + O (h^{3}) = O (h^{2}) .

Thus $m = 2$ for the one-step defect convention. The leading derivative can also be written

y^{''} = f_{x} + f_{y} f

by the chain rule.

Some conventions instead call

τ_{n + 1} = δ_{n + 1} / h

the truncation error. That normalized local truncation error is

O (h)

, so $m = 1$ under the normalized convention. Both describe the same Euler one-step defect; neither should be confused with an unqualified global error claim.

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