By Daniel J. Velleman

**Read Online or Download American Mathematical Monthly, volume 116, number 1, january 2009 PDF**

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**Extra resources for American Mathematical Monthly, volume 116, number 1, january 2009**

**Example text**

It follows easily by induction that ti = i 2 , for 1 ≤ i ≤ k. In particular, the total weight of the stack is tk = k 2 ≤ w, as required. 4, we get that di = wi + ti 1 2 = 2(i − 1) + i2 1 2 = 3 2 − 2. 313. 5772156 is Euler’s gamma. Figure 13. A spinal stack with a shield. We next discuss a technique that can be used to convert loaded spinal stacks into standard stacks. This is of course done by constructing balancing sets that apply the required forces on the left-hand edges of the spine blocks.

2 f + 1, 2 f + 1, (r − 1) f (r + 1) f + 1 The forces in the first sequence can be regarded as acting on the r -slab contained in the (r + 1)-slab, which then, by the induction hypothesis, yield downward forces on the bottom row of r f + r − 1, 2r f + 2(r − 1), . . , 2r f + 2(r − 1), r f + r − 1 at positions − r −1 , − r −3 , . . , r −3 , r −1 . 2 2 2 2 36 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 The forces of the second sequence, together with the weights of the outermost blocks of the (r + 1)-rows, are passed straight down through the rigid structure of the r -slab to the bottom row.

1 and the small zero bias condition, then X In ,n → p 0. January 2009] THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM 55 Proof. For all n, 1 ≤ i ≤ n, and > 0, 2 2 2 σi,n = E(X i,n 1(|X i,n | < )) + E(X i,n 1(|X i,n | ≥ )) ≤ 2 + L n, . 1 we have lim sup m n ≤ 2 n→∞ and therefore lim m n = 0. 1. We are now ready to prove the forward direction of the Lindeberg-Feller CLT. 2. Let h ∈ Cc,0 that h, given by (23). Substituting Wn for w in (22), taking expectation, and using (11) we obtain E [h(Wn ) − Eh(Z)] = E f (Wn ) − Wn f (Wn ) = E f (Wn ) − f (Wn∗ ) (27) with Wn∗ given by (14).