 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-62/1/b/solution

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

Write

j^{\underline{m}} = j (j - 1) \dots (j - m + 1)

for the falling factorial. At the sampling points,

f_{nm} (j / n) = n^{- m} j^{\underline{m}}

, which vanishes for

j < m

. For

j \geq m

, cancellation of the factorials gives

j^{\underline{m}} (j n) = \frac{n !}{( n - m )!} (j - m n - m) = n^{\underline{m}} (j - m n - m) .

Insert this into the Bernstein polynomial and set

ℓ = j - m

B_{n} (f_{nm}, x) = \frac{n ^{\underline{m}}}{n ^{m}} x^{m} ℓ = 0 \sum n - m (ℓ n - m) x^{ℓ} (1 - x)^{n - m - ℓ} = \frac{n ^{\underline{m}}}{n ^{m}} x^{m} = f_{nm} (1) x^{m} .

The binomial theorem evaluates the remaining sum. For

m = 0

, both sides equal one. This proves the Bernstein falling-factorial identity $B_{n} (f_{nm}, x) = f_{nm} (1) x^{m}$ .

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