Specimen project question M256 This is a specimen project question. Each group project and each individual project will consist of two questions with similar length and style. However, the topics will be different. In the group and individual projects, both questions will have analytical components, but only one will have an implementation (Maple) component. Below we show one sample question with three sub-questions and Maple components. Let G be a positive integer. This problem is concerned with the series SG = ∞∑ j=0 ( 1)j (1 + j) h, with h = 1 + (1 +G) 1. (a) (i) Write a procedure that takes as its arguments the parameter G and a positive number δ. The procedure should sum the series SG using a do loop, stopping when (2 + n) h < ∣∣∣(SG)n∣∣∣× δ, where (SG)n is the partial sum up to j = n (so the left-hand side is the first term neglected). It should return the approximate value of the series and the index at which the loop was broken as its results. (ii) Experiment with your procedure using different values of G and δ. Does it provide an accurate and efficient method for computing SG Justify your answer. (b) It is sometimes possible to improve the rate at which a series converges by rearranging the terms. Some preliminary results are needed to achieve this. (i) Use repeated differentiation to obtain the full Maclaurin series for (1 + x) p 1, for |x| < 1 and p = 0, 1, . . . Hence deduce that ∞∑ n=p ( n p ) 1 2n+1 = 1, n = 0, 1, . . . where ( n p ) = n! p!(n p)! . (ii) Use induction to show that if aj = aj aj+1, 2aj = ( aj), etc. then naj = n∑ p=0 ( 1)p ( n p ) aj+p. (iii) Use the above results to prove that ∞∑ n=0 na0 2n+1 = ∞∑ p=0 ( 1)pap. In your answer you may use the fact that ∞∑ n=0 n∑ p=0 ≡ ∞∑ p=0 ∞∑ n=p . You are not asked to prove this, but you may like to derive it (it’s similar to reversing the order of a double integral). (c) The analysis in part (b) suggests an alternative means for summing the series SG. Let aj = ( 1)j(1 + j) h and consider the following table. p 0 1 2 3 0 a0 1 a1 a0 j 2 a2 a1 2a0 3 a3 a2 2a1 3a0 ... ... ... ... ... . . . 1 Write a second Maple procedure with arguments and return values as in part (a). The procedure should work as follows. Start with S = 0 and then for each j: Compute the entries in row j, using the fact that n+1aj = n(aj aj+1). Break the loop if | jaj|/2j+1 < δ × |S|. Otherwise add jaj/2j+1 to S. Repeat the experiments you carried out in part (a) with the new procedure, and comment on the results. 2
