 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/16d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

\frac{1}{∣ x - y ∣} = \frac{1}{∣ x ∣} (1 - 2 \frac{x \cdot y}{∣ x ∣ ^{2}} + \frac{∣ y ∣ ^{2}}{∣ x ∣ ^{2}})^{- 1/2} .

(1 + s)^{- 1/2} = 1 - \frac{1}{2} s + \frac{3}{8} s^{2} + O (s^{3})

and retaining terms through second order in

∣ y ∣/∣ x ∣

gives

\frac{1}{∣ x - y ∣} = \frac{1}{∣ x ∣} [1 + \frac{x \cdot y}{∣ x ∣ ^{2}} + \frac{3 ( x \cdot y ) ^{2} - ∣ x ∣ ^{2} ∣ y ∣ ^{2}}{2∣ x ∣ ^{4}} + O (\frac{∣ y ∣ ^{3}}{∣ x ∣ ^{3}})] .

This is the beginning of the multipole expansion.

 New to topics? Read the docs here!