An Average Maximum Distance: Problem 89-8Author(s): John G. WatsonSource: SIAM Review, Vol. 32, No. 2 (Jun., 1990), pp. 304-305Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2030532 .

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304 PROBLEMS AND SOLUTIONS

symmetry argument it follows that E(p) = 2P(p) + 2P(-p) for -1 < p < 1. Now as- sume that E(p) = c, a constant. Then P(p) = 2C - P(-p) for -1 < p < 1, while P(p) is bounded and analytic in p for Re (p) I -1 + e (e> 0) and while I4c - P(- p) is bounded and analytic in p for Re (p) ' 1 - c. Hence the analytic continuation of P(p) is just a constant, by Liouville's theorem; its value is zero, since limp,. P(p) = 0 (by dominated convergence). Thus c = E(p) = 0 and, since f > 0, we have f(x, y) = 0 almost everywhere. D

To obtain the statement of the original problem, let g denote the indicator function of A (g(x, y) = 1 if (x, y) E A and g(x, y) = 0 otherwise) and let f(x, y) = g(x, y) + g(-x, y) + g(x, -y) + g(-x, -y). Then E(p) = 2I(p) + 2I(-p) is constant when I(p) is, and, if f(x, y) = 0 almost everywhere, then g(x, y) = 0 almost everywhere and X(A) = 0.

Also solved by the proposers.

An Average Maximum Distance

Problem 89-8*, by PETER SENN (ETH-Zentrum, Zurich, Switzerland). Consider a cube with edges of length 2a whose center is located on a given plane.

We now randomize the orientation of the cube relative to the plane keeping its center fixed. Let the resulting distances of the eight vertices of the cube from the plane be d1, d2, ,d8 and let

D= max(d1, d2, ,d8).

It follows easily that a _ D-aV'3. Prove or disprove that the expected value for D is 3a/2. This value agrees with random walk simulations.

The problem has arisen in the design for a Monte Carlo algorithm for computing areas of irregularly shaped surfaces.

Solution by JOHN G. WATSON (Applied Mathematics, Menlo Park, California). The following calculation gives a formula for the average maximum distance of

an n-dimensional cube. The formula confirms the three-dimensional conjecture. For this calculation, we take the cube to be fixed, and average over all possible orientations of a plane passing through its center.

Consider the n-dimensional cube centered at the origin and with edge length 2a. We position the cube so that its vertices are generated by the n-dimensional vector a(?+, +1, ... + 1) using each of the 2n possible choices of signs. Generally, the distance between any point and a plane through the origin with unit normal u is equal to the magnitude of the dot product of u and the position vector from the origin to the point. In particular, the distance between a vertex and such a plane is given by

d= a ?U1 U2? *?Un,

where the us are the components of the unit vector u, i.e., they are direction cosines. The maximum distance is equal to

D(u) = a(l ul I + I U2 I + ***+ I Un I)

since a vertex can be chosen to reinforce the signs of the components of u. We note that a ' D a a'/n.

The expected value of D is obtained by averaging over all planar orientations u. The realizations of u cover the n-dimensional unit sphere uniformly, and the density distribution of u is given by dA/Sn, where dA is an element of area and Sn is the total

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PROBLEMS AND SOLUTIONS 305

surface area of the n-dimensional unit sphere. Hence the expected value of D is

(D) = D(u) dA

anr Sn J k k u I dA,

where the second equality follows from the first by symmetry and k is any conveniently chosen unit vector.

The plane through the center of the unit sphere with normal k splits the plane into two equal hemispheres. Now consider the projection of either hemisphere onto the plane. The integral appearing in the second expression for (D) is equal to twice thg area of this projection AP. The projection is the (n - 1)-dimensional unit sphere and it follows that AP = V,_1, where

m/2

m((m/2) + 1)

is the volume of the m-dimensional unit sphere. Since Sn = nV", the expected value of D simplifies to

(D) 2anV,._ Sn

2a Vni Vn

2a F((n + 2)/2) ]F ((n+ 1)/2)

In particular, (D) = 4a/ir for n = 2 and (D) = 3a/2 for n = 3.

Also solved by M. X. GOEMANS (Massachusetts Institute of Technology), J. A. GRZESIK (TRW Inc., Redondo Beach, CA), D. A. KURTZE (Clarkson University), T. S. LEWIS (Eastman Kodak Company, Rochester, NY), 0. P. LOSSERS (Eindhoven University of Technology, Eindhoven, the Netherlands), I. MAREELS (University of Newcastle, Newcastle, Australia), F. MATHIS (Baylor University), D. ULLMAN (George Washington University), and J. R. VARMA (Indian Institute of Manage- ment, Ahmedabed, India).

Photon Transport in Plant Canopies

Problem 89-9, by BARRY D. GANAPOL (University of Arizona). In the study of photon transport in plant canopies, with all leaves oriented at the

same polar inclination (AL = cos 6L) and distributed uniformly in the azimuth, the intercept function [1, 2] giving the azimuthally averaged projected area normal to a beam (of inclination , = cos 0) is given by { IA/1LI for IcotOcotOLI > 1,

/I/IL[2,D(/)/1r - 1] + (2/ir)V7 VlA- sin 41(p) otherwise,

where '= cos [-cot 0 cot OL].

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