 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-310/4/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

d lo g k / d N = 1 - ϵ ≃ 1

n_{s} - 1 = \frac{d l o g Δ _{R}^{2}}{d l o g k} = - 2 ϵ - η .

The supplied relations imply

ϵ = \frac{M _{Pl}^{2}}{2} (\frac{V ^{'}}{V})^{2}, η = 4 ϵ - 2 M_{Pl}^{2} \frac{V ^{''}}{V} .

n_{s} - 1 = - 3 M_{Pl}^{2} (\frac{V ^{'}}{V})^{2} + 2 M_{Pl}^{2} \frac{V ^{''}}{V} .

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