Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 5 / 2 / a

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Solution

The integral form of the Taylor theorem about the left endpoint is

f (x) = q_{k - 1} (x) + \frac{1}{( k - 1 )!} \int_{a}^{b} (x - t)_{+}^{k - 1} f^{(k)} (t) d t,

where

q_{k - 1} \in P_{k - 1}

. Apply the order-

k

divided difference at

x_{i}, \dots, x_{i + k}

. The polynomial term vanishes, and linearity permits interchange with the integral:

f [x_{i}, \dots, x_{i + k}] = \frac{1}{( k - 1 )!} \int_{a}^{b} [x_{i}, \dots, x_{i + k}] (\cdot - t)_{+}^{k - 1} f^{(k)} (t) d t .

By the definition

M_{i} (t) = k [x_{i}, \dots, x_{i + k}] (\cdot - t)_{+}^{k - 1}

,

f [x_{i}, \dots, x_{i + k}] = \frac{1}{k !} \int_{a}^{b} M_{i} (t) f^{(k)} (t) d t .

This is the Peano kernel theorem for the divided-difference functional.

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