 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2007/p/3/d/solution

For nodes

0, 1

and

f (x) = e^{x}

h (x) = 1 + x + (2 e - 5) x^{2} + (3 - e) x^{3} .

Indeed its values at zero and one are

1, e

, and its derivatives there are also

1, e

∣ e^{x} - h (x) ∣ = \frac{e ^{ξ}}{4 !} x^{2} (1 - x)^{2} \leq \frac{e}{24} (\frac{1}{4})^{2} = \frac{e}{384},

since

0 \leq ξ \leq 1

and

x (1 - x) \leq 1/4

[0, 1]

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