A-Level Maclaurin series: worked solution

Question

Using BC^2 approx 1 + sqrt3 theta + (3/2)theta^2 from part (b), show BC approx 1 + (sqrt3/2)theta + (3/8)theta^2.

Step 1: write BC as a bracket of the form (1 + u) raised to a power, so the binomial series can be used. BC = sqrt(BC²) = (1 + sqrt3 theta + (3/2)theta²)^(1/2) Put u = sqrt3 theta + (3/2)theta², so BC = (1 + u)^(1/2). Note u is of order theta, so u² is of order theta² and u³ is of order theta³ (which we may discard). Step 2: quote the binomial series for (1 + u)^n and set n = 1/2. (1 + u)^n = 1 + nu + n(n - 1)u²/2! + ... With n = 1/2: nu = (1/2)u n(n - 1)/2! = (1/2)(-1/2)/2 = -1/8 So (1 + u)^(1/2) = 1 + u/2 - u²/8 + ... Step 3: substitute u and keep only terms up to theta². u/2 = (1/2)(sqrt3 theta + (3/2)theta²) = (sqrt3/2)theta + (3/4)theta² u² = (sqrt3 theta + (3/2)theta²)² = 3theta² + 2(sqrt3 theta)((3/2)theta²) + ((3/2)theta²)² = 3theta² + 3sqrt3 theta³ + (9/4)theta⁴ The theta³ and theta⁴ terms are beyond the required order, so u² = 3theta² to this order. -u²/8 = -(3/8)theta² Step 4: add the pieces and collect the theta² terms. BC ≈ 1 + (sqrt3/2)theta + (3/4)theta² - (3/8)theta² Coefficient of theta²: 3/4 - 3/8 = 6/8 - 3/8 = 3/8 BC ≈ 1 + (sqrt3/2)theta + (3/8)theta² (as required)