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Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 4 10E b iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Using the decomposition from part (ii), maps Alice's to and to . Alice obtains or , each with probability . Conditional on outcome , Bob has ; conditional on outcome , Bob has .
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 4 10E b ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 4 10E b i Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 40A b iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The exact line search for a positive-definite quadratic givesUsingproduces the stated ratio formula. If and are the extreme eigenvalues,Hence the fraction in the ratio is at least , and iteration yieldsWhen , , so the right side is zero after one iteration: exact line search reaches the solution in a single step.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 40A b ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 40A b i Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 40A a ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Let . The Jacobi method isso and . If is strictly diagonally dominant, every Gershgorin disc of is centered at zero and has radiusEvery eigenvalue therefore has modulus below one, so and part (i) proves convergence.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 40A a i Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Writing the error as givesprovided the fixed-point consistency conditionholds. Convergence for every starting vector is therefore equivalent to , which in finite dimensions is equivalent toThese two conditions are necessary and sufficient.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 39D b v Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The mountain-wave cutoff follows because the hills force intrinsic frequency , whereas internal gravity waves require . Thus separates long hills, which radiate vertically propagating waves and carry energy upward, from short hills, whose response is evanescent and carries no mean vertical energy flux.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 39D b iv Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 39D b iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 39D b ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The boundary condition gives complex amplitude . Substitution of the internal gravity wave forminto the internal-wave equation givesSince ,For , choose the sign of that gives upward energy propagation; with , this is .
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 39D b i Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
In coordinates moving with the background fluid, . The stationary hill profile becomesso in the convention its frequency is ; its physical angular frequency has magnitude .
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E b iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E b ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Using one even and one odd site per unit cell, Bloch theorem reduces the Hamiltonian toIts eigenvalues are
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E b i Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Translation preserves the alternating on-site term only when , soThe primitive lattice spacing is therefore .
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E a iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The Floquet multipliers satisfyIf , they are complex conjugates on the unit circle, , and bounded Bloch waves exist: lies in an allowed band. If , the multipliers are real reciprocal numbers, one growing and one decaying, so there is no extended bounded Bloch state and lies in a gap.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E a ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 34E a i Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
For two solutions of the Schrödinger equation, their Wronskian is constant because the equation has no first-derivative term. The determinant of the Floquet matrix is the ratio of the Wronskian at to that at , hence
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 3 33B a iii Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
For , the matrix element of between and givesThe ladder coefficient is nonzero, so adjacent diagonal entries agree. Induction makes the diagonal entry independent of .
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