Bourbaki

Exercise 3, § 3

Functions of a Real Variable, VI, 297

Let ff be a continuous function with 2p+12p + 1 continuous derivatives on the interval [0,1][0, 1]. Establish the formula

∫01f(t) dt=1n(12f(0)+f(1n)+f(2n)+⋯+f(n−1n)+12f(1)) \int_0^1 f(t)\,dt = \frac{1}{n} \left( \frac{1}{2} f(0) + f\left(\frac{1}{n}\right) + f\left(\frac{2}{n}\right) + \cdots + f\left(\frac{n-1}{n}\right) + \frac{1}{2} f(1) \right)
−∑k=12pb2k(2k)!1n2k(f(2k−1)(1)−f(2k−1)(0))+Rp(n) - \sum_{k=1}^{2p} \frac{b_{2k}}{(2k)!} \frac{1}{n^{2k}} (f^{(2k-1)}(1) - f^{(2k-1)}(0)) + R_p(n)

with

∣Rp(n)∣≤4e2π(2π)2p+11n2p+1∫01∣f(2p+1)(t)∣ dt. |R_p(n)| \leq \frac{4e^{2\pi}}{(2\pi)^{2p+1}} \frac{1}{n^{2p+1}} \int_0^1 |f^{(2p+1)}(t)|\,dt.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.