 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/32c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

V_{1} = \partial_{x}

, one has

ξ = 1

η = 0

, and hence

D_{x} ξ = 0

. The recursion from part a gives

η_{j} = 0

for every

j

For

V_{2} = x \partial_{x}

, one has

D_{x} ξ = 1

. Starting from

η_{0} = 0

, induction gives

η_{j} = - j u_{j} .

For

V_{3} = x^{2} \partial_{x}

, one has

D_{x} ξ = 2 x

η_{1} = - 2 x u_{1}

and

η_{2} = - 2 u_{1} - 4 x u_{2}

, and induction gives

η_{j} = - 2 j x u_{j} - j (j - 1) u_{j - 1} .

pr^{(n)} V_{1} pr^{(n)} V_{2} pr^{(n)} V_{3} = \partial_{x}, = x \partial_{x} - j = 1 \sum n j u_{j} \partial_{u_{j}}, = x^{2} \partial_{x} - j = 1 \sum n (2 j x u_{j} + j (j - 1) u_{j - 1}) \partial_{u_{j}} .

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