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11) October 22, 2010 48 15:7 World Scientific Book - 9in x 6in Random Sequential Packing of Cubes then there exists a solution φ on [0, τ ] with |φ(t)| ≤ R hence a(h) > τ . In order to estimate the integrals we use the bound τ τ τ th−1 1 1 τh 0< C(t)2 dt ≤ dt ≤ th−1 dt ≤ 2 2 2 (log τ ) 0 (log τ ) h 0 0 (log t) τ and similarly for 0 C(t)dt. 10) and we have to use some specific methods. The method, we used for estimating a(h) uses monotonicity results for the coefficients of the equation. Assuming C is a constant, the differential equation is written as 1 1 d − = .

2 (i) G(s) converges for Re s > −a(h) and G has a pole at −a(h). (ii) the following relation holds H (s) G(s) = . 8) H(s) (iii) H satisfies 1 (1 − h) −(1+h)s e − 2 [e−(1+h)s − e−2s ] H. 9) H + 2H = s s (iv) H has only simple zeros. (v) H has no zero in the half plane Re z > −a(h). (vi) −a(h) is the only real zero of H. Proof. 6). 8) is rewritten as G(s) = (log H(s)) . 8) on the half plane Re s > −a(h). 5) by simple rewriting. e. 9) is actually removable as can be proved by a Taylor expansion. As a consequence, the solutions of the equation are naturally defined over C and thus H also.

Book October 22, 2010 15:7 World Scientific Book - 9in x 6in book On the minimum of gaps generated by 1-dimensional random packing Fig. 4 49 Lower and Upper bounds for a(h) Asymptotic analysis for a(h) In this section we use singular perturbation techniques to investigate the values of a(h) when h is small, and when h is close to 1. We use the Laplace transform and as before we write H (s) G(s) = . 12) +1 with ψ(u) = ue −e a function holomorphic over C. 1 we consider the Renewal Theory method that allows to find the first order of the expansion of a(h) when h → 1.