 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-66/1/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Applying the preceding derivative identity repeatedly yields the Bézier derivative control polygon for every order:

P^{(r)} (t) = \frac{n !}{( n - r )!} j = 0 \sum n - r Δ^{r} P_{j} B_{j}^{n - r} (t), 0 \leq r \leq n,

where the forward finite difference is

Δ P_{j} = P_{j + 1} - P_{j}

and

Δ^{r} P_{j} = \sum_{k = 0}^{r} (- 1)^{r - k} (k r) P_{j + k} .

The induction step multiplies by the remaining degree

n - r

and takes one more finite difference. Derivatives above order

n

vanish. In particular,

P^{(r)} (0) = \frac{n !}{( n - r )!} Δ^{r} P_{0}, P^{(r)} (1) = \frac{n !}{( n - r )!} Δ^{r} P_{n - r} .

If the actual parameter interval has length

h

and

t

is its normalized coordinate, derivatives in the original parameter acquire the additional factor

h^{- r}

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