Exercise 2, § 3 Functions of a Real Variable, VI, 296 Establish the formula 12f(x)−f(x+1)+f(x+2)−⋯−f(x+2n−1)+12f(x+2n) \frac{1}{2} f(x) - f(x+1) + f(x+2) - \cdots - f(x+2n-1) + \frac{1}{2} f(x+2n) 21f(x)−f(x+1)+f(x+2)−⋯−f(x+2n−1)+21f(x+2n) =∑k=1pb2k(2k)!(22k−1)(f(2k−1)(x+2n)−f(2k−1)(x))+Vp(x,n) = \sum_{k=1}^{p} \frac{b_{2k}}{(2k)!} \left( 2^{2k} - 1 \right) \left( f^{(2k-1)}(x+2n) - f^{(2k-1)}(x) \right) + \mathrm{V}_p(x, n) =k=1∑p(2k)!b2k(22k−1)(f(2k−1)(x+2n)−f(2k−1)(x))+Vp(x,n) with ∣Vp(x,n)∣⩽4e2π(22p+1+1)(2π)2p+1∫xx+2n∣f(2p+1)(t)∣dt \left| \mathrm{V}_p(x, n) \right| \leqslant \frac{4 e^{2\pi} \left( 2^{2p+1} + 1 \right)}{(2\pi)^{2p+1}} \int_x^{x+2n} \left| f^{(2p+1)}(t) \right| dt ∣Vp(x,n)∣⩽(2π)2p+14e2π(22p+1+1)∫xx+2nf(2p+1)(t)dt (apply the Euler-Maclaurin formula to g(x)=f(2x)g(x) = f(2x)g(x)=f(2x)). Solution No solution has been written yet.