By Daniel J. Velleman

**Read Online or Download American Mathematical Monthly, volume 117, number 2, February 2010 PDF**

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**Extra info for American Mathematical Monthly, volume 117, number 2, February 2010**

**Example text**

Let Y be a random variable with mean zero and variance σ 2 such that |Y | ≤ κ. Then for any L > 0, E exp Y κL ≤1+ σ 2 e(L) σ 2 e(L) ≤ exp . 2. Suppose that X i = (X i, j )dj =1 satisfies X i ∞ ≤ κ, and let n Var(X i, j ). Then for any L > 0, bound for max1≤ j ≤d i=1 E Sn 2 ∞ ≤ κ L log(2d) + be an upper L e(L) . κ Now we return to our general random vectors X i ∈ Rd with mean zero and E X i 2∞ < ∞. They are split into two random vectors via truncation: X i = X i(a) + X i(b) with X i(a) := 1[ X i ∞ ≤κo ] X i and X i(b) := 1[ X i ∞ >κo ] X i for some constant κo > 0 to be specified later.

The family of circles {ξn } is recursively defined by letting ξ1 be the circle associated to curve MC and letting ξn+1 be the rescaled image of ξn . The family of circles {βn } is defined by letting βn be the largest circle between ξn and ξn+1 . From the two infinite families of circles, only a few that are close to ζ will be needed. So formally, let k be the natural number such that the horizontal coordinate of the center of ζ is between the horizontal coordinates of the centers ξk and ξk+1 . Roughly speaking, the region around ζ is the critical region that needs to be checked for intersection.

Note that this inscribed circle is concentric with the outer approximation constructed earlier. 4. If the inscribed circles associated to two curves overlap, then the von Koch curve self-intersects. Proof. If the inscribed circles overlap, then the domains that the two curves bound overlap. Thus, the snowflake domain overlaps, which is equivalent to the von Koch curve self-intersecting. We now have the machinery to prove the main result. 5. The set of c for which the (n, c)-von Koch curve is self-avoiding is not in general an interval.