Assignment 3, Math 3510 Due October 2 from the text by D.M. Burton, 7th edition Solutions Pages 103-104 Problems 8, 11c), 13 Pages 117-120 Problem 7, 18 Pages 137-138 Problem 3 Additional Problem not to be handed in (possibly a test problem). Write an essay about Thales and Pythagoras emphasizing their mathematics. Include comparisons of the two men’s mathematics as appropriate. Comment on their influence on later mathematics, again as appropriate. A handwritten page is sufficient. Problem 8, page 103 It is useful to have a list of the first few triangular numbers: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91. (The modern definition of a triangular number is tk = k(k + 1)/2 where k is any integer (possibly 0; t−k−1 = tk.) (1) V. Lebesgue proved in 1872 that: Every positive integer greater is the sum of a square number and two triangular numbers. Here we allow 0 = 02 as a square number. The triangular numbers should be distinct. We are to illustrate Lebesgue’s result by finding appropriate expressions for 9, 44, 81, 100 9 =02 +3+6 44 =12 + 15 + 28 = 42 + 0 + 28 81 = = 02 +3+78=02 + 15 + 66 = 02 + 36 + 45 = 42 + 10 + 55 = 62 + 0 + 45 100 =02 + 45 + 55 = 12 + 21 + 78 = 32 +0+91=32 + 36 + 55 = 42 + 6 + 78 =62 + 28 + 36 = 72 +6+45=72 + 36 + 15 = 82 + 0 + 36 Are there more? (2) Lebesgue further proved that every positive integer is the sum of two square numbers and a triangular number. This time the intention is that the square numbers be distinct. We also illustrate this result for 9, 44, 81, 100 9 =32 + 02 + 0 44 =02 + 42 + 28 = 22 + 52 + 15 = 32 + 52 + 10 = 42 + 52 + 3 81 =12 + 52 + 55 = 02 + 62 + 45 = 32 + 62 + 36 = 22 + 72 + 28 = 42 + 82 + 1 100 =02 + 32 + 91 = 32 + 52 + 66 = 32 + 62 + 55 = 02 + 82 + 36 = 22 + 92 + 15 =32 + 92 + 10 = 62 + 82 + 0 Are there more? Problem 11 (c), page 103 We follow the hint and use the formula 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)/6 derived in class and in the text (page 101). 1 2 + 2 3 + 3 4+ ... + n(n + 1) =1 (1+1)+2 (2+1)+3 (3+1)+ ... + n(n + 1) · · · · · · =12 +1+22 +2+32 +3+ ... + n2 + n =12 + 22 + 32 + ...n2 +1+2+3+ ... + n n(n + 1)(2n + 1) n(n + 1) = + 6 2 2 because the second sum is the nth triangular number. Simplify n(n + 1)(2n + 1) n(n + 1) n(n + 1) n(n + 1) n(n + 1)(n + 2) + = [2n +1+3]= [2(n +2)] = 6 2 6 6 3 Problem 13, page 104 We show (1+2+3+ ... + n)2 = 13 + 23 + 33 + ...n3 by applying two formulas from the book. One is the formula for the nth triangular number: 1 + 2 + 3+ ... + n = n(n + 1)/2 so that 2 n(n + 1) (1+2+3+ ... + n)2 = 2 and the other is for the sum of the first n cubes on page 99 so that 2 n(n + 1) 13 + 23 + 33 + ...n3 = 2 This verifies the equality. Problem 7, page 118 We consider the iteration process xn = 3xn−1 + 2yn−1 yn = 4xn−1 + 3yn−1 2 2 for n 2 which we initiate at (x1,y1)=(2, 3) which lies on the hyperbola y 2x = 1. Subsequent iterations≥ are − (x2,y2) =(12, 17) (x3,y3) =(70, 99) (x4,y4) =(408, 577) (x5,y5) =(2378, 3363) Consider 2 2 2 2 yn 2xn = (4xn−1 + 3yn−1) 2(3xn−1 + 2yn−1) − 2 − 2 2 2 2 2 = 16xn−1 + 24xn−1yn−1 + 9yn−1 18xn−1 24xn−1yn−1 8yn−1 = yn−1 2xn−1 − − − − 2 2 2 2 Therefore the iteration process preserves the value of yn 2xn and since y1 2x1 = 1, we have 2 2 − − yn 2xn = 1. The iteration process stays on the hyperbola. but of course xn and yn are getting larger − 2 and larger and therefore closer to the asymptote: y = √2x, indeed yn/xn = 2 + 1/xn approaches √2. p Problem 18 page 119: The harmonic mean h of two numbers a and b is defined by the relation h a a − = b h b − (a) Multiply both sides of the equation by (b h)b to clear the fractions − b(h a)= a(b h) or bh ba = ab ah − − − − Divide by abh: 1 1 1 1 = a − h h − b which establishes part (a). (b) Solve the equation in part (a) for 1/h: 1 1 1 1 1 b + a = + = h 2 a b 2 ab 3 by finding a common denominator. Invert to find 2ab h = a + b Notice that 1/h is the usual arithmetic mean of 1/a and 1/b and that the definition is symmetric in a and b: the harmonic mean of a and b is the harmonic mean of b and a (reversed order). Problem 3, Page 137 In the picture on page 137, accompanying the problem, join AF and EF with straight lines. The triangle AEF so formed is a right triangle because it is inscribed in a semi-circle. It follows that the right triangles ABF and FBE are similar; indeed both are similar to AFE. Consequently a/x = x/b so that x2 = ab. Therefore the square of sidelength x has the same area as the rectangle ab. x = √ab. For future reference, we note that, by choosing b = 1, we can construct the square root of any number a and so √a is constructible whenever a is. (The cube root of an arbitrary length is not constructible in general.).