Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 4 / a

The right-hand side of the slow-roll curvature power spectrum is evaluated at cosmological horizon exit, $k=aH$. Usingrewrites it asAt horizon exit, $d\log k=d\log (aH)=(1-ϵ)dN$, so $d/d\log k=d/dN$ to first slow-roll order. With the first Hubble slow-roll parameter and second Hubble slow-roll parameter,The supplied relations $ϵ={ϵ}_V$ and ${\eta }_V=2ϵ-\eta /2$ imply $\eta =4{ϵ}_V-2{\eta }_V$. Therefore