A-Level Maclaurin series: worked solution

Question

Given y=e^{π/4−arctan(eˣ)} has Maclaurin expansion y≈1−(1/2)x+(1/8)x². Deduce, with exact coefficients, the expansion of e^{π/4−arctan(eˣ)}/√(1−x) up to and including the x² term. [4]

The expression is a product of two things, so expand the second factor and multiply the two series together, keeping terms only as far as x². Step 1: expand 1/√(1-x) = (1-x)^(-1/2) by the binomial series. The series for (1+u)^n is 1 + nu + n(n-1)u²/2 + ... Here n = -1/2 and u = -x. nu = (-1/2)(-x) = x/2 n(n-1)u²/2 = [(-1/2)(-1/2 - 1)/2](-x)² = [(-1/2)(-3/2)/2]x² = (3/4)/2 x² = (3/8)x² So (1-x)^(-1/2) = 1 + (1/2)x + (3/8)x² + ... Step 2: multiply the two series. e^(π/4 - arctan(e^x))/√(1-x) = (1 - (1/2)x + (1/8)x²)(1 + (1/2)x + (3/8)x²) + ... Collect powers of x up to x²: constant term: 1 x 1 = 1 x term: 1 x (1/2) + (-1/2) x 1 = 1/2 - 1/2 = 0 x² term: 1 x (3/8) + (-1/2) x (1/2) + (1/8) x 1 = 3/8 - 1/4 + 1/8 = 3/8 - 2/8 + 1/8 = 2/8 = 1/4 Step 3: write the expansion. e^(π/4 - arctan(e^x))/√(1-x) = 1 + (1/4)x² + ... (The x term vanishes, which is why no x term appears in the answer.)