By Marek Jarnicki, Peter Pflug

This e-book covers the development, research, and thought of continuing nowhere differentiable services, comprehensively and accessibly. After illuminating the importance of the topic via an summary of its historical past, the reader is brought to the delicate toolkit of principles and methods used to review the categorical non-stop nowhere differentiable capabilities of Weierstrass, Takagi–van der Waerden, Bolzano, and others. sleek instruments of practical research, degree idea, and Fourier research are utilized to check the time-honored nature of constant nowhere differentiable services, in addition to linear constructions in the (nonlinear) area of constant nowhere differentiable features. To around out the presentation, complicated innovations from numerous components of arithmetic are introduced jointly to provide a state of the art research of Riemann’s non-stop, and purportedly nowhere differentiable, function.

For the reader’s profit, claims requiring elaboration, and open difficulties, are basically indicated. An appendix comfortably offers history fabric from research and quantity concept, and complete indices of symbols, difficulties, and figures increase the book’s software as a reference paintings. scholars and researchers of study will price this particular booklet as a self-contained consultant to the topic and its methods.

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1(6)) will give αC (b) ≤ (3+2ε)π 2 cos(πε) , ε := 1 b−1 . Note that if b ∈ 2N2 + 1, then (d) is not better than (c). 6034. 9669. b + κ(b) b Thus for b ≥ 213, the estimate (e) is better than (d). 1373. In particular, for b ∈ 2N + 1, b ≥ 243, the estimate (e) is better than (c). Exact values of αC (b) and αS (b) are not known ? 5 Weierstrass’s Method The following sections will present diﬀerent attempts to get the nowhere diﬀerentiability of Wp,a,b,θ for certain conﬁgurations of the parameters p, a, b, θ.

Nite or inﬁnite) right-sided derivative (Wθ )+ (0) does not exist. H. Hardy was the ﬁrst to notice that in general, Wp,a,b,θ ∈ / ND∞ (R). 1 (Cf. [Har16]). If ab ≥ 1 and a(b + 1) < 2, then S a,b (0) = +∞. Notice that a(b + 1) < 2, provided that ab = 1. Proof . Put f := S a,b . It suﬃces to show that f+ (0) = +∞. Take an h ∈ R, 0 < h ≤ 14 . Let N = N (h) ∈ N be such that bN −1 h ≤ 14 < bN h. Then f (h) 1 = h h N −1 an sin(2πbn h) + n=0 1 h ∞ an sin(2πbn h) =: f1 (h) + f2 (h). n=N We have N −1 f1 (h) ≥ 4 (ab)n = n=0 4N, if ab = 1 N −1 4 (ab) ab−1 , if ab > 1 , |f2 (h)| ≤ 1 aN .

Observe that 2 qb ∞ ∞ an cos(2πbn t + θ) = n=m ∞ an cos(πbn−m p + θ) = n=m 1 −m , 2 pb an (−1)p cos θ n=m = (−1)p cos θ am . 1(f), we obtain Δf (t, u) = 2 cos θ 2(ab)m (−1)q + M (m, p, q) , q−p 1−a 3π 3π where |M (m, p, q)| ≤ ab−1 . The condition ab > 1 + 2 cos θ (1 − a) implies that sgn Δf (t, u) = q sgn(−1) . Now ﬁx x ∈ R and δ > 0. Let m ∈ N be such that 2b−m < δ and let r := 2xbm . If r < 2xbm , then we put ti,m = 12 pi,m b−m , pi,m := r + 1 − i, ui,m = 12 qi,m b−m , i = 1, 2, q1,m := p1,m + 1, q2,m := p2,m + 3.