Download Limits, Series, and Fractional Part Integrals: Problems in by Ovidiu Furdui PDF

By Ovidiu Furdui

This publication gains not easy difficulties of classical research that invite the reader to discover a bunch of ideas and instruments used for fixing difficulties of contemporary subject matters in actual research. This quantity deals an strange selection of difficulties — a lot of them unique — focusing on 3 subject matters of mathematical research: limits, sequence, and fractional half integrals.

The paintings is split into 3 elements, each one containing a bankruptcy facing a specific challenge sort in addition to a really brief portion of tricks to pick difficulties. the 1st bankruptcy collects difficulties on limits of unique sequences and Riemann integrals; the second one chapter focuses on the calculation of fractional half integrals with a unique part referred to as ‘Quickies’ which incorporates difficulties that experience had unforeseen succinct recommendations. the ultimate bankruptcy deals the reader an collection of issues of a style in the direction of the computational elements of endless sequence and distinct items, lots of that are new to the literature. every one bankruptcy features a component of tough difficulties that are influenced via different difficulties within the e-book. those ‘Open difficulties’ could be thought of study initiatives for college kids who're learning complex calculus, and which are meant to stimulate creativity and the invention of recent and unique equipment for proving identified effects and developing new ones.

This stimulating number of difficulties is meant for undergraduate scholars with a robust history in research; graduate scholars in arithmetic, physics, and engineering; researchers; and a person who works on themes on the crossroad among natural and utilized arithmetic. furthermore, the extent of difficulties is acceptable for college kids enthusiastic about the Putnam pageant and different excessive point mathematical contests.

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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 .

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