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For $V=\lambda {\varphi }^n$ on $\varphi >0$, slow-roll approximation gives $ϵ\simeq n^2{M}_{\mathrm{Pl}}^2/(2{\varphi }^2)$ and $\eta \simeq 2n{M}_{\mathrm{Pl}}^2/{\varphi }^2$. The slow-roll e-fold count integrates to $N\simeq ({\varphi }_N^2-{\varphi }_e^2)/(2n{M}_{\mathrm{Pl}}^2)$. For $n=2,3,4$, the loose order-one conditions $ϵ\le 1$ and $|\eta |\le 1$ imply ${\varphi }^2\ge 2n{M}_{\mathrm{Pl}}^2$ in this approximation. Genuine slow roll requires a much stronger hierarchy. A prescribed endpoint ${\varphi }_e^2=2n{M}_{\mathrm{Pl}}^2$ gives ${\varphi }_N^2=2n(N+1)M_{\mathrm{Pl}}^2$ andThe actual end criterion for accelerated expansion is $ϵ=1$, whose leading potential estimate instead gives ${\varphi }_e^2\simeq n^2{M}_{\mathrm{Pl}}^2/2$. The prescribed endpoint is a convention, not an exact slow-roll or inflation boundary.

For the quadratic-potential slow-roll solution, $V^{\prime }/V=2/\varphi$, so $ϵ=\eta =2M_{\mathrm{pl}}^2/{\varphi }^2$. Inflation ends when $ϵ\simeq 1$, giving ${\varphi }_{\mathrm{end}}^2=2M_{\mathrm{pl}}^2$. The slow-roll e-fold count remaining before that endpoint isWith $N_{\ast }=50$, this gives ${\varphi }_{\ast }^2=202M_{\mathrm{pl}}^2$ and ${ϵ}_{\ast }={\eta }_{\ast }=1/101$. The quadratic-inflation amplitude calibration isMatching the supplied amplitude therefore givesTaking $M_{\mathrm{pl}}\simeq 2.4\times 10^{18}\mathrm{GeV}$, this is $m\simeq 1.7\times 10^{13}\mathrm{GeV}$. Retaining ${\varphi }_{\mathrm{end}}$ produces the small $+1$ correction; dropping it gives almost the same leading slow-roll estimate.

The scalar spectral index isIt agrees with the supplied central value and lies well inside the stated uncertainty interval. This conclusion concerns the scalar tilt; the tensor prediction is a separate test of the same model.

For the monomial inflation potential on $\varphi >0$, $V_{,\varphi }/V=n/\varphi$ and $V_{,\varphi \varphi }/V=n(n-1)/{\varphi }^2$. Thus, to leading slow-roll approximation,For $n=2,3,4$, $2n\ge n^2/2$. Therefore the loose order-one requirements $ϵ\le 1$ and $|\eta |\le 1$ giveThis is the interpretation under which the printed threshold follows. It is not a sufficient criterion for genuine slow roll, which requires much smaller parameters and hence ${\varphi }^2/M_{\mathrm{Pl}}^2\gg 2n$ for these fixed powers. Nor is its equality the actual end of acceleration for $n=2,3$: there $ϵ\simeq n/4<1$. For $n=4$ the leading estimate has $ϵ\simeq 1$. The potential-based order-one criterion $|{\eta }_V|<1$ also differs from the Hubble-flow criterion, especially at $n=4$. We use the endpoint the paper explicitly stipulates, rather than treating this boundary as a controlled slow-roll approximation.

Let ${\varphi }_e^2=2n{M}_{\mathrm{Pl}}^2$ by that prescription. The slow-roll e-fold count, using $\dot{\varphi }/H\simeq -n{M}_{\mathrm{Pl}}^2/\varphi$, isFor $N=60$, the monomial inflation with a prescribed endpoint result isApplying the leading scalar spectral index in Hubble slow-roll parameters and tensor-to-scalar ratio formulas givesThese are predictions at the chosen 60-e-fold initial point, normally identified with the pivot scale's horizon exit. A physical endpoint determined by $ϵ=1$ would instead have the leading estimate ${\varphi }_e^2\simeq n^2{M}_{\mathrm{Pl}}^2/2$; its change to the e-fold count and predictions must not be silently mixed with the stipulated $122n$ calculation.

The number of e-folds $N=\log a$ satisfiesWhen $\varphi$ is monotone, the chain rule gives $dt=d\varphi /\dot{\varphi }$, and henceUsing the two slow-roll equations, $3H\dot{\varphi }\simeq -V^{\prime }(\varphi )$ and $H^2\simeq 8\pi GV/3$, eliminates both $H$ and time:This is the differential form of the slow-roll e-fold count.

For a monotonic inflaton trajectory, the potential slow-roll parameter satisfiesThe remaining slow-roll e-fold count is the integral from the present field value to its end value along the direction of motion:

By definition,Using $dt=d\varphi /\dot{\varphi }$ and the result of part (a),Reversing the limits yields the slow-roll e-fold count

The slow-roll e-fold count before the end of inflation isAt leading order near the hilltop, $V/V^{\prime }\simeq -{\Lambda }^4/(m^2\varphi )$, and henceThis has $N({\varphi }_e)=0$ and gives