A-Level Maclaurin series: worked solution

Question

You may quote standard expansions from MF27. With a>0, expand a/(a-x)-1 as a series in ascending powers of x as far as the x² term, giving coefficients in terms of a, and state (in terms of a) the values of x for which the expansion holds.

The standard expansions in MF27 are written for (1 + u)ⁿ, so the first job is to force the expression into that shape. Step 1: take the factor a out of the bracket in the denominator, then cancel it. a/(a − x) = a / (a(1 − x/a)) = 1/(1 − x/a) = (1 − x/a)⁻¹ Step 2: quote the standard expansion (1 + u)ⁿ = 1 + nu + n(n − 1)u²/2! + ... with n = −1 and u = −x/a: (1 − x/a)⁻¹ = 1 + (−1)(−x/a) + [(−1)(−2)/2!](−x/a)² + ... Simplify each term. The first bracket gives +x/a. In the second, (−1)(−2)/2 = 1 and (−x/a)² = x²/a² (the square removes the minus sign): (1 − x/a)⁻¹ = 1 + x/a + x²/a² + ... Step 3: subtract the 1 required by the question. a/(a − x) − 1 = (1 + x/a + x²/a² + ...) − 1 = x/a + x²/a² + ... Step 4: state the validity. The expansion of (1 + u)ⁿ for non-positive-integer n is valid only when |u| < 1, and here u = −x/a: |−x/a| < 1 |x|/a < 1 (a > 0, so |a| = a) |x| < a, that is −a < x < a.