PROBLEMS. Problems. x 1

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1 Irish Math. Soc. Bulletin Number 76, Winter 05, 9 96 ISSN PROBLEMS IAN SHORT We begin with three integrals. Problems Problem 76.. a 0 sinx dx b 0 x log x dx c cos x e /x + dx I learnt the next problem from a popular-mathematics lecture given by Vicy Neale of the University of Oxford. Problem 76.. For each point z on the unit circle, let l z denote the closed line segment from z to z. Consider the collection of those points in the closed unit disc that each lie at the intersection of two distinct line segments l z and l w. What shape is the complement in the unit disc of this collection of points? We finish with a problem proposed by Wenchang Chu of Universitá del Salento, Italy. Problem Evaluate n=0 x tan. n n Solutions Here are solutions to the problems from Bulletin Number 74. The first problem was solved by the North Kildare Mathematics Problem Club, and it is their solution that we present. The solution was also nown to Grahame Ersine of the Open University, who suggested the problem. Received on c 05 Irish Mathematical Society

2 9 I. SHORT Problem 74.. Given a positive integer A, let B be the number obtained by reversing the digits in the base n expansion of A. The integer A is called a reverse divisor in base n if it is a divisor of B that is not equal to B. For example, using decimal expansions, if we reverse the digits of the integer 5, then we obtain 5. Since 5 is not a divisor of 5, the integer 5 is not a reverse divisor in base 0. For which of the positive integers n between and 6, inclusive, is there a two-digit reverse divisor in base n? You may also wish to attempt the more difficult problem of classifying those positive integers n for which there is a two-digit reverse divisor in base n. Solution 74.. Given a positive integer n and two integers a and b with a, b < n, let [ab] n = an + b. The number [ab] n is defined to be a two-digit reverse divisor in base n if [ba] n is divisible by, but not equal to, [ab] n. By inspection, there are no two-digit reverse divisors in bases, or 3. For n > 3, we claim that there is a two-digit reverse divisor in base n if and only if n + is composite. To see this, suppose first that n + is composite. Choose positive integers a and with n + = a + +, and define b = n a. Then bn + a = an + b, so we have a two-digit reverse divisor, unless =. In the case =, we define a = and b = n to give b n+a = aa n+b. This time we certainly have a two-digit reverse divisor, because a = n / >. Conversely, suppose that n + is equal to a prime p, and suppose that [ab] n is a two-digit reverse divisor in base n. Let = bn + a/an + b. One calculates that pb a = b a +. Thus one of b a or + is divisible by p. But b a < p and < + < b + n + = p, so we have a contradiction. Hence, contrary to our earlier assumption, there does not exist a two-digit reverse divisor in base n. The second problem was solved by Henry Ricardo New Yor Math Circle, New Yor, USA, the North Kildare Mathematics Problem Club, and the proposer, Ángel Plaza Universidad de Las Palmas de Gran Canaria, Spain. All solutions were similar. The solution

3 PROBLEMS 93 we present is that of Henry Ricardo. Nesbitt s inequality features in the solution, which says that for positive real numbers a, b, and c, we have a b + c + b c + a + c a + b 3. This inequality can be proven by using two applications of the rearrangement inequality with the triples a, b, c and /a+b, /b+ c, /c + a. Problem 74.. Let f : R + R + be an increasing, convex function with f =, and let x, y, and z be positive real numbers. Prove that for any positive integer n, n n n x y z f + f + f 3. y + z z + x x + y Solution 74.. By Nesbitt s inequality, x y + z + y z + x + z x + y 3. Then, as the function gx = fx n is convex and increasing with g =, we can use Jensen s inequality to write x x y z y+z g + g + g 3g + y z+x + z x+y y + z z + x x + y 3 as required. 3g = 3, The third problem was solved by the North Kildare Mathematics Problem Club, Ángel Plaza, and the proposer Finbarr Holland University College Cor. Finbarr has pointed out that in fact the result in the problem can quicly be derived using Bernstein polynomials. Bernstein polynomials are useful for approximating continuous functions; they provide one way of proving the Weierstrass approximation theorem. To solve the problem, simply calculate the nth Bernstein polynomial of the function fx = x j and use the value x = /. Nonetheless, we present a complete, more elementary, solution here, which does not require Bernstein polynomials. The solution is essentially the same as that of the North Kildare Mathematics Problem Club and Ángel Plaza. In this problem, we use the standard notation fn gn as n,

4 94 I. SHORT where f and g are positive functions, to mean that fn gn as n. Problem Prove that for j = 0,,,..., j n j n j as n. =0 Solution Let f 0 x = + x n, and for j =,,... let f j x = x d f j x. d x We shall prove by induction that for j = 0,,,..., f j x = n j x j + x n j + p j x, n, where p j is a polynomial in x and n, and its degree as a polynomial in n is less than j. This equation holds when j = 0 p 0 is the zero polynomial, which we assume has degree in each variable. Suppose that it holds when j = m. Then one can chec that where f m x = n m x m + x n m + qx, n, qx, n = m n m x m + x n m + x x p m x, n. The degree of q as a polynomial in n is less than m, so the inductive proof is complete. We deduce that, for x, Now, so f j x n j x j + x n j, j = 0,,,.... f 0 x = + x n = =0 x, j = f j n j n j, j = 0,,,.... =0 Finally, we return to a problem of Niall Ryan University of Limeric from issue 7. Niall had a solution to this problem, but it was quite involved. The solution we include here, provided by Finbarr Holland, deals with the odd case only.

5 Problem 7.. For each integer n 0, let K m S n = m + n + + PROBLEMS 95 m n K m m n, where [ ] m! K m =. m m! Prove that 0, n odd, S n = K n log +, n even. Solution 7.. Let = Γ a n = + n Γ Γn + n n! = K n. We aim to prove that for n = 0,,,..., S n+ = a m m + n + + m n = 0. Clearly, it is enough to prove that T n = 0 for n = 0,,,..., where T n = j S j+ = 0. j To establish this, notice that j j z + j + n!γz + n = z + j + Γz + n +, and j j z j = n n n! n z j = n n!γz n+ Γz +. Next we appeal to a well-nown result about hypergeometric series, which can be found, for example, in Section 0.3 of An introduction to the theory of functions of a complex variable by E.T. Copson

6 96 I. SHORT OUP, 970. Recall that the hypergeometric series F a, b; c; z is defined for z < by the equation Γa + mγb + m z m Γc + m m! = ΓaΓb F a, b; c; z, Γc and it is defined for z by analytic continuation. In fact, by Abel s continuity theorem, the series above can also be used to define F a, b; c;. The result we need, from page 5 of Copson s text, is that ΓcΓc a b F a, b; c; = Γc aγc b if Re[c a b] > 0. Using this observation, we see that T n a Γm + n+ m Γn + + m + nγ + m Γ + m Γ + m Γ + m + nγ + m Γ n+ + m π m!γn + + m m!γ + m π π π = 0. Γ F, ; n + ; + nγ Γ n+ F, n+ ; ; Γn + Γ πγn + Γ n nγ n+ Γn + Γ n + 3 πγn + Γ n n πγn + sin n + π Γ n + 3 The well-nown functional equation ΓzΓ z = π/ sinπz is used in the second-last line. We invite readers to submit problems and solutions. Please submissions to imsproblems@gmail.com in any format we prefer Latex. Submissions for the summer Bulletin should arrive before the end of April, and submissions for the winter Bulletin should arrive by October. The solution to a problem is published two issues after the issue in which the problem first appeared. Please include solutions to any problems you submit, if you have them. Department of Mathematics and Statistics, The Open University, Milton Keynes MK7 6AA, United Kingdom

PROBLEMS IAN SHORT. ( 1) n x n

PROBLEMS IAN SHORT. ( 1) n x n Irish Math. Soc. Bulletin Number 8, Winter 27, 87 9 ISSN 79-5578 PROBLEMS IAN SHORT Problems The first problem this issue was contributed by Finbarr Holland of University College Cork. Problem 8.. Let