Solution

For integer knots, the recurrence simplifies to

N_{i, k} (t) = \frac{t - i}{k - 1} N_{i, k - 1} (t) + \frac{i + k - t}{k - 1} N_{i + 1, k - 1} (t) .

Order-two splines are triangular:

N_{i, 2} (i) = N_{i, 2} (i + 2) = 0

and

N_{i, 2} (i + 1) = 1

. Applying the recurrence once more gives

t N_{0, 3} (t) N_{1, 3} (t) 1 1/2 0 2 1/2 1/2 30 1/2.

For example

N_{0, 3} (1) = \frac{1}{2} N_{0, 2} (1) + N_{1, 2} (1) = 1/2

and

N_{0, 3} (2) = N_{0, 2} (2) + \frac{1}{2} N_{1, 2} (2) = 1/2

N_{0, 4} (1) N_{0, 4} (2) N_{0, 4} (3) = \frac{1}{3} \frac{1}{2} + \frac{3}{3} 0 = \frac{1}{6}, = \frac{2}{3} \frac{1}{2} + \frac{2}{3} \frac{1}{2} = \frac{2}{3}, = \frac{3}{3} 0 + \frac{1}{3} \frac{1}{2} = \frac{1}{6} .

Therefore

(N_{0, 4} (1), N_{0, 4} (2), N_{0, 4} (3)) = (1/6, 2/3, 1/6) .

In particular its actual maximum is

2/3

in this standard normalization; dividing by that maximum would change the requested recurrence normalization.

 Ancestors (11)

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