By Abel Flint

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**Additional resources for A system of geometry and trigonometry: Together with a treatise on surveying : teaching various ways of taking the survey of a field, also to protract ... using them : compiled from various authors**

**Sample text**

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.