By Whitehead J., Zhou Y., Patterson S.
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Extra resources for [Article] Easy-to-implement Bayesian methods for dose-escalation studies in healthy volunteers
Is a sequence of strictly positive numbers. We shall denote by C(M) the set of all indefinitely 23 Weierstrass-Stone theorem and generalisations - a brief survey differentiable complex-valued functions f, each defined on some open interval I C R (I depending on f), and satisfying the following estimates for its successive derivatives: for every compact subset K c I, 3C > 0, and 3c >0 3 Vx E I< and m = 0,1,2,. .. t C(M) is a quasi-analytic class if the following is true: i f f E C(M) and 3a E I such that f ( " ) ( a ) = 0 V m = O , l , 2 ,...
Suppose Cb(E) C CV,(E) and that A is a subalgebra of Cb(E) which is separating on E , contains the constant function 1 and is self-adjoint in the complex case. Then A is dense in CVm(E). 3. Let y 2 0 upper-semicontinuous on R be 3 3C y(t) Then y E Proof. 5 > 0,3c > 0 satisfying: Ce'l*l V t E R. rl. ) is rapidly decreasing at infinity, since trne-'ltl any rn E R. Let t , x , y E R, z = z + i y E C, and define e , --f 0 aa t --f 00, for E C(R;C) by e L ( t ) = eiZt for Chapter I 20 t E R. We then note provided (yI < c.
In general, having constructed the cubes of K,-1, we divide each cube which is now present but not in UfZ;Ki into 2" cubes of side from A are at least &; let K , be the set of all these cubes whose distances $ (if any). The following facts concerning this subdivision of E - A will be needed. 2. Proof. < -(s 2 1). For it lies in a cube C' of the previous subdivision not belonging to K,-l and whose distance from A is therefore < = -. 3. cubes of K,+1. Proof. 1ZJ;; This is true, because the distance d(C,A ) 2 p , the distance from any point of C' to A < g, and the diameter of any cube of IC,+l is &,which means that any 47 Strong approximation in finitedimensional spaces cube C’ of K,+2 is separated from any cube of K , by definitely more than 3 cubes of K,+l; this number of intervening cubes of I<,+, has to be a whole number of cubes and hence at least four.