A-Level Maclaurin series: worked solution

Question

(a) Expand 1/sqrt(1 + 5x) as a series as far as the term in x^2, and give the values of x for which it is valid. [3] (b) By putting x = 1/20, obtain a fractional approximation for sqrt(5). [2] (c) Explain, without further working, whether x = -4/25 would give a closer approximation to sqrt(5) than the value used in (b). [1]

(a) [3 marks] Write the expression as a power: 1/sqrt(1 + 5x) = (1 + 5x)^(-1/2). Use the binomial series (1 + u)^n = 1 + nu + [n(n - 1)/2!]u² + ... with u = 5x and n = -1/2: (1 + 5x)^(-1/2) = 1 + (-1/2)(5x) + [(-1/2)(-3/2)/2!](5x)² + ... The coefficient of the squared term is (3/4)/2 = 3/8, and (5x)² = 25x², so = 1 - (5/2)x + (3/8)(25x²) + ... = 1 - (5/2)x + (75/8)x² + ... The binomial series with n not a positive integer converges only when |u| < 1, so here |5x| < 1, that is |x| < 1/5: valid for -1/5 < x < 1/5. (b) [2 marks] x = 1/20 = 0.05 lies inside the valid interval. The exact left-hand side at x = 1/20 is 1/sqrt(1 + 5/20) = 1/sqrt(5/4) = 2/sqrt(5). The series at x = 1/20 gives 1 - (5/2)(1/20) + (75/8)(1/20)² = 1 - 1/8 + 75/3200 = 1 - 1/8 + 3/128 = (128 - 16 + 3)/128 = 115/128. So 2/sqrt(5) is approximately 115/128. To get sqrt(5) itself, note sqrt(5) = 5/sqrt(5) = (5/2) × (2/sqrt(5)). Multiplying both sides of the approximation by 5/2: sqrt(5) is approximately (5/2)(115/128) = 575/256. (c) [1 mark] No. Both x = 1/20 = 0.05 and x = -4/25 = -0.16 lie inside the valid interval -1/5 < x < 1/5, but a series truncated after the x² term is more accurate the smaller |x| is, because the discarded terms carry higher powers of x. Since |1/20| = 0.05 is smaller than |-4/25| = 0.16, x = -4/25 would give a worse approximation, not a closer one.