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An elementary treatise on curve tracing by Percival Frost

By Percival Frost

This obtainable remedy covers orders of small amounts, varieties of parabolic curves at an enormous distance, sorts of curves in the community of the starting place, and types of branches whose tangents on the beginning are the coordinate axes. extra themes contain asymptotes, analytical triangle, singular issues, extra. 1960 variation.

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45 Proof: With a view to induction, assume that for some constants (n-independent) d(a), g(a) and for n ∈ [1, N − 1] we have qn < dng . 1) mean that An > f (a)/n for an n-independent f . Then we have a bound for the next qN : f qN > d(N − 1)g + d (N − 2)g . N Since for g < 1, 0 < x ≤ 1/2 we have (1 − x)g > 1 − gx − gx2, and the constants d, g can be arranged so that qN > dN g and the induction goes through. 1) The Gauss fraction for R(a) exhibits (at least) geometric/linear convergence. 2) So does the original Ramanujan form R1(a, b) when a/b or b/a is (significantly) greater than unity .

52 • An observation that led to the results below is that we have implicitly used, for positive reals a = b and perforce for the Jacobian parameter q := min(a, b) ∈ [0, 1), max(a, b) the fact that θ2(q) 0≤ < 1. θ3(q) • If, however, one plots complex q with this ratio of absolute value less than one, a complicated fractal structure emerges, as shown in the Figures below—this leads to the theory of modular forms [BB]. 1 are suspect for complex q. 53 • Numerically, the identities appear to fail when |θ2(q)/θ3(q)| exceeds unity as graphed in white for |q| < 1: 54 • Such fractal behaviour is ubiquitous.

2). 1: R(a) := R1(a, a) converges iff a ∈ I. That is, the fraction diverges if and only if a is pure imaginary. Moreover, for a ∈ C\I the fraction converges to a holomorphic function of a in the appropriate open half-plane. 2: R1(a, b) converges for all real pairs; that is whenever Im(a) = Im(b) = 0. ) converge to distinct limits. (ii) There are Re(a), Re(b) > 0 such that R1(a, b) diverges. Define • H := {z ∈ C : √ 2 z 1+z < 1}, 2z • K := {z ∈ C : 1+z 2 < 1}. 4: If a/b ∈ K then both R1(a, b) and R1(b, a) converge.

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