By Daniel J. Velleman
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Extra resources for American Mathematical Monthly, volume 116, number 1, january 2009
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.
Now, by (28), η(|Wn∗ − Wn |) → p 0, (29) and by (27) and the triangle inequality |Eh(Wn ) − Eh(Z)| = E( f (Wn ) − f (Wn∗ )) ≤ E f (Wn ) − f (Wn∗ ) ≤ Eη(|Wn − Wn∗ |). Therefore lim Eh(Wn ) = Eh(Z) n→∞ by (29) and the bounded convergence theorem. 1 finishes the proof. 56 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 5. PARTIAL NECESSITY. 3, showing in what sense the Lindeberg condition, in its equivalent zero bias form, is necessary. We begin with Slutsky’s lemma, see , which states that Un →d U and Vn → p 0 implies Un + Vn →d U.
We let yi be the x-coordinate of the left edge of Bi , for 1 ≤ i ≤ k − 1. Note that xi+1 − 1 < yi ≤ xi − 1, where xi is the x-coordinate of the left edge of Bi . 32 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 Shield block Bi applies a downward force of wi+1 on Bi+1 . , at the left edge of Bi+1 . Block Bi also applies a downward force of u i+1 on Bi+1 at z i+1 , where yi ≤ z i+1 ≤ yi+1 + 1. Similarly, block Bi−1 applies a downward force of u i on Bi at z i . Finally a downward external force of vi is applied on the left edge of Bi .