By R. C. Baker

This publication launches the celebrated new sequence London Mathematical Society Monographs. the writer, famous for his paintings in the course of the mathematical neighborhood, right here offers an summary of the idea of nonlinear Diophantine approximation. He has targeting the real growth made within the final ten years through such participants as I. M. Vinogradov, H. Heilbronn, and W. M. Schmidt, discovering, for instance, that it truly is attainable to contemplate simultaneous approximation to integers via values of a suite of quadratic varieties, or a discrete analogue (small ideas of a approach of homogeneous congruences).

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71. A Fejer limit. Let f ∈ L1 [0, 1] and let g : [0, 1] → R be a continuous function. Prove that 1 1 f (x)g({nx})dx = lim n→∞ 0 0 f (x)dx 1 g(x)dx, 0 where {a} = a − a denotes the fractional part of a. 72. Let f , g : [0, 1] × [0, 1] → R be two continuous functions. Prove that 1 1 lim n→∞ 0 0 f ({nx} , {ny})g(x, y)dxdy = 1 0 1 0 f (x, y)dxdy · 1 1 g(x, y)dxdy, 0 0 where {a} = a − a denotes the fractional part of a. 73. Let f : [−1, 1] × [−1, 1] → R be a continuous function. Prove that 1 lim t→∞ −1 f (sin(xt), cos(xt)) dx = 1 π 2π 0 f (sin x, cos x)dx and 1 1 lim t→∞ −1 −1 f (sin(xt), cos(yt)) dxdy = 1 π2 2π 0 2π 0 f (sin x, cos y)dxdy.

13. It is worth mentioning the history of this problem. It appears that the case s = 1 is an old problem which Bromwich attributed to Joseph Wolstenholme (1829–1891) (see [20, Problem 19, p. 528]). The problem, as Bromwich records it, states that lim n→∞ 1 n n + 2 n n + ···+ n−1 n n = 1 , e−1 which we call the Wolstenholme limit. In 1982, I. J. Schoenberg proposed the problem of proving that limn→∞ Sn = 1/(e − 1) where Sn = (n + 1)−n ∑nk=1 kn (see [114]). However, it appears that this problem appeared previously in Knopp’s book [76, p.

1, 1 + 1n , . . , k − 2, k − 2 + 1 n−1 n , . . , k − 2 + n , k − 1. Since f is Riemann integrable we obtain that √ √ √ p n + p n + 1 + · · · + p kn √ = lim n→∞ npn k−1 0 √ p 1 + xdx = p · k(p+1)/p − 1 . 15), we get that xn + xn+1 + · · · + xkn p L−ε · k(p+1)/p − 1 ≤ lim · n→∞ L+ε p+1 nxn ≤ L+ε p · k(p+1)/p − 1 . · L−ε p+1 Since ε > 0 is arbitrary taken the result follows. 19. The problem, which is recorded as a lemma (see [91, Lemma, p. 434]), is used for introducing a new approach for determining asymptotic evaluations of products of the form f (n) = c1 c2 · · · cn .