 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/4/c/ii/solution

On the panel

[2 j / n, 2 (j + 1) / n]

, the maximum fourth derivative is at the left endpoint because it is decreasing. Thus

E_{n} \leq \frac{1}{90 n ^{5}} \sum_{j = 0}^{n /2 - 1} (e^{2 j / n} - \frac{2 e j}{n}) .

The geometric series and the sum of consecutive integers give

\sum_{j = 0}^{n /2 - 1} e^{2 j / n} = \frac{e - 1}{e ^{2/ n} - 1}, \frac{2 e}{n} \sum_{j = 0}^{n /2 - 1} j = \frac{e n}{4} - \frac{e}{2} .

Substitution proves

E_{n} \leq B (n) := \frac{1}{90 n ^{5}} (\frac{e - 1}{e ^{2/ n} - 1} - \frac{e n}{4} + \frac{e}{2}) .

For

n > 2

, at least one panel's maximum is smaller than the global maximum

1

, so this bound is strictly tighter than the global bound. For

n = 2

, the two bounds coincide.

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