 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-1/14b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Since

x = r cos θ

, the function on the left of the defining expansion is

e^{i x}

. Therefore

(\nabla^{2} + 1) e^{i r c o s θ} = (\partial_{x}^{2} + \partial_{y}^{2} + 1) e^{i x} = 0,

which is the two-dimensional Helmholtz equation.

\nabla^{2} = \frac{\partial ^{2}}{\partial r ^{2}} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r ^{2}} \frac{\partial ^{2}}{\partial θ ^{2}} .

Substitute the defining Fourier cosine series and equate the coefficient of each

cos (n θ)

. Its angular second derivative contributes

- n^{2}

J_{n}

satisfies

J_{n}^{''} + \frac{1}{r} J_{n}^{'} + (1 - \frac{n ^{2}}{r ^{2}}) J_{n} = 0,

r^{2} J_{n}^{''} + r J_{n}^{'} + (r^{2} - n^{2}) J_{n} = 0 .

Expand the exponential through cubic order and use

cos^{2} θ = \frac{1 + c o s 2 θ}{2}, cos^{3} θ = \frac{3 c o s θ + c o s 3 θ}{4} .

This gives

e^{i r c o s θ} = 1 - \frac{r ^{2}}{4} + i (r - \frac{r ^{3}}{8}) cos θ - \frac{r ^{2}}{4} cos 2 θ - \frac{i r ^{3}}{24} cos 3 θ + O (r^{4}) .

Comparing with

J_{0} + 2 \sum_{n \geq 1} i^{n} J_{n} cos (n θ)

yields

J_{0} (r) = 1 - \frac{r ^{2}}{4} + O (r^{4}),

J_{1} (r) = \frac{r}{2} - \frac{r ^{3}}{16} + O (r^{5}), J_{2} (r) = \frac{r ^{2}}{8} + O (r^{4}), J_{3} (r) = \frac{r ^{3}}{48} + O (r^{5}) .

 New to topics? Read the docs here!