4 = (0.02) 3 13, = 0.25 because = 25. Simi-

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1 Theore. Let b and be integers greater than. If = (. a a 2 a i ) b,then for any t N, in base (b + t), the fraction has the digital representation = (. a a 2 a i ) b+t, where a i = a i + tk i with k i = (b i (od )). Proof. By the lea, a i = b (bi (od )) (bi (od )), and a i = b + t ((b + t)i (od )) ((b + t)i (od )). On the other hand, (b + t) i b i (od ) and(b + t) i b i (od ). Thus we get a i a i = t(b i (od )) = tk i, as claied. Earlier we discussed in bases 3,, and 7. As a second exaple, consider the 7 fraction. According to the theore, if we find the representation of in the bases 4 4 2, 3, 4, and 5, together with the corresponding keys, then we can easily get the digital representation of in any base. Recall that the key k 4 k l associated with in base b is defined by k i = (b i (od )), wherel is either the length of the fundaental period of or the length of its nontrivial fractional part. Thus 4 = (.) 2 2, 4 = (.) 4, 4 = (.2) 3 3, 4 = (.) 5. =.25 because = 25. Sii- Hence, = (.3) 4 6 since + 2 = 3, and 4 larly, using as key, one gets for instance 4 = (.2) 8, 4 = (.4) 6, and 4 = (.3) 3. In particular, in base 29, we have 4 = (.[52]) 29. References. J. Conway and R. Guy, The Book of Nubers, Copernicus, L. E. Dickson, History of the Theory of Nubers,Vol.I:Divisibility and priality, Chelsea, S. Guttan, On cyclic nubers, Aer. Math. Monthly 4 (934) A Waiting-Tie Surprise Richard Parris (rparris@exeter.edu), Phillips Exeter Acadey, Exeter, NH 3833 Let x, x 2, x 3,... be a sequence of nubers chosen randoly (and uniforly) fro the unit interval < x <. For each real nuber t, the first n for which VOL. 39, NO., JANUARY 28 THE COLLEGE MATHEMATICS JOURNAL 59

2 x + x 2 + +x n > t is a waiting-tie rando variable; let E(t) denote its expected value. In this note, we express E(t) as a su of eleentary functions of t, and show that it is asyptotic to a linear function. The data in Table was generated by doing a illion trials for each of four target values t. The frequency table shows how often each stopping index n was obtained, and the averages appear at the botto. Table. n t =.3 t =.5 t =.8 t = average Based on this data, the following result appears plausible: Theore. For t, E(t) = e t. Proof. For each positive integer n,letp n (t) denote the probability that x + x 2 + x 3 + +x n t < x + x 2 + x 3 + +x n. In other words, p n (t) is the probability that n is the waiting tie for target t. It is not difficult to see that p (t) = t. In general, the polynoial forula p n (t) = (n )! t n n! t n applies for all t. Once this forula for p n (t) is established, it is a routine exercise to show that E(t) = n p n (t) = e t. n= (n )! t n To establish the forula for p n (t), notice that is the volue of the (n )-diensional polytope defined by the inequalities x, x 2,..., x n, and x + x 2 + +x n t. It is also the volue of the n-diensional pris defined by the additional inequality x n. This pris includes all positive 6 c THE MATHEMATICAL ASSOCIATION OF AMERICA

3 solutions to the inequality x + x 2 + +x n t, whose probability is n! t n. Thus is the probability that (n )! t n n! t n = p n (t) x + x 2 + +x n t < x + x 2 + +x n. The case t = will be failiar to any. It appeared as a Putna proble in 958 (see []), and a discrete version of the proble was analyzed by Shultz [2]. It is clear that E(t) cannot be equal to e t when t is large. Although the geoetric approach used for t can be odified to cover additional values of t, itis ore efficient to assue that E(t) is continuous for t and to apply the ethods of calculus fro now on. Our recursive approach is to observe that, for t >, E(t) = + t E(u) du. () In other words, E(t) is ore than the siple average of all the expected waiting ties E(t x ) that could result fro choosing x ; we obtain the integral equation by replacing t x by u. When applied to (), the Fundaental Theore of Calculus gives E (t) = E(t) E(t ). (2) Notice that our result E(t) = e t for t < also follows fro (2) if we use the obvious values E(t) = fort < to extend the definition of E. We now outline an inductive proof that, for n andn t n, n E(t) = ( ) k et k k! (t k)k. (3) k= The upper liit of this su shows that only those ters for which t k is nonnegative are included. Assue first that < t 2, where we( know that E(t ) = e t.it is a straightforward application of (2) to show that d dt e t E(t) ) = e.frothisit follows that E(t) = e t (t )e t, because E is continuous at t = ande() = e. Thus (3) holds for n =. The induction step follows siilarly and is left to the reader. The jup discontinuity at t = forces E to be nondifferentiable at t = (thetwo one-sided derivatives are e fro the left and e fro the right), but forula (3) shows that E is differentiable everywhere else. If n >, the difference between the two forulas for E(n) is divisible by (t n) 2, forcing the two one-sided forulas for E (n) to agree. The recursive process akes use of the continuity of E when t is a positive integer. The values E(n) are interesting: E() = E(2) = E(3) = E(4) = E(5) = VOL. 39, NO., JANUARY 28 THE COLLEGE MATHEMATICS JOURNAL 6

4 Moreover, the eerging pattern is not confined to integer values of t, as the exaple E(4.85) = = 2(4.85) shows. The ain purpose of this note is to establish the following asyptotic result: Theore 2. The function E defined by equation (3) satisfies li (E(t) 2t) = 2 3. Equation (3) does not see to be of uch help in establishing the asyptotic behavior of E(t), but the integral equation () and the derived equation (2) do pay dividends. Notice, in particular, that the derivatives of E also satisfy (2). This suggests that we look for an integral equation, siilar to (), that applies to the. We are thus led to consider functions f that have the average-value property,which eans that f is continuous for t and f (t) = f (u) du holds for t 2. Unless t f is constant, it is clear that f ust attain values above and below f (t) on the interval (t, t). It is plausible that the continuous averaging process dissipates this variability, forcing f (t) to approach a liit as t.(sinceweexpect f = E to approach a liit, this is exactly what we want to happen.) Furtherore, this liit (if it exists) is deterined by the values of f on any unit interval, and it is therefore reasonable to try to express the liit as a weighted average of these values. The recursive nature of f suggests that the weighting function should increase linearly, starting with at the lower liit of the integral. In the trivial case where f is constant, it is easily seen that the forula 2(u t + ) f (u) du produces the correct value. Furtherore, for all t functions of interest, the value produced by this integration forula does not depend on the choice of interval: Lea. Assue that the continuous function f has the average-value property. Then the function F(t) = (u t + ) f (u) du is constant for t 2. t Proof. As above, the Fundaental Theore of Calculus yields f (t) = f (t) f (t ) for t 2. Notice that F(t) = uf(u) du (t ) f (t). A short calculation now shows that F (t) = t. It is shown next that the coon value of these integrals is the desired liit. Lea 2. If f has the average-value property, then li f (t) = 2(u a) f (u) du. Proof. Let L = 2(u a) f (u) du,andletg(t) = f (t) L. It is routine to verify that g also has the average-value property, and that 2(u a)g(u) du =, so there is no loss of generality in assuing that L =. Furtherore, nothing is lost by assuing that f (t) is not constant. In this case, Lea iplies that f has both positive and negative values on every interval of length. We now show that the axia and inia of f on the intervals I n =[n, n] approach as n, and fro this the lea follows. Since the two arguents are essentially the sae, it suffices to give only one. Let M n = f (a n ) be the axiu value of f on I n. It follows fro (2) and the differentiability of f that f (a n ) = f (a n ),andsom n M n. The non-increasing 62 c THE MATHEMATICAL ASSOCIATION OF AMERICA

5 sequence {M n } is bounded below by, thus it ust have a liit M. Suppose that M >. For sufficiently large t, M f (t) < 2M. Fro it follows that f (t) = f (t) = t t f (u) du and t (t u) f (u) du < 2M which contradicts the definition of M. (u t + ) f (u) du =, t (t u) du = M, Corollary. Let F be continuous for t, and let k be constant. If for all t, then F(t) = k + t F(u) du li (F(t) 2kt) = 4k 3 + 2uF(u) du. Proof. We first apply Lea 2 to the function f (t) = F (t), which is easily seen to have the average-value property. A routine integration by parts leads us to li F (t) = 2(u )F (u) du = 2k. Another short calculation shows that g(t) = F(t) 2kt also has the average-value property. It therefore follows fro Lea 2 that li (F(t) 2kt) = 2(u )(F(u) 2ku) du = 2(u )(F (u) + F(u ) 2ku) du = 4k 3 + 2wF(w) dw. Finally, we return to the objective function E expressed in equation (3). Because E satisfies (), the conclusion of Theore 2 follows fro the corollary and the easily verified calculation ue u du = 2 3. References. L. E. Bush, The Willia Lowell Putna Copetition, Aer. Math. Monthly 68 (96) Harris S. Shultz, An Expected Value Proble, Two-Year College Math. J. (979) VOL. 39, NO., JANUARY 28 THE COLLEGE MATHEMATICS JOURNAL 63