text:
\section{\label{intro}Introduction}
The theory of cosmic inflation \cite{guth, linde, steinhardt} that has been favored by the latest observational data \cite{planck13, planck15, planck18, bkp}, is by now the most likely scenario for the early universe,  since it provides the explanation to flatness, horizon and monopole problems, among others, for the standard hot Bing Bang cosmology \cite{revlinde,liddle, riotto, lyth, mukhanov, baumann, nojirioo}. Inflation provides a detailed account of fluctuations that constitute the seeds for the large scale structure and the observed CMB anisotropies \cite{starobinsky1, mukhanov1, starobinsky3, hawking, starobinsky2, guth1, bardeen0, bardeen}, as well as predicts a nearly scale invariant power spectrum.\\
The inflation scenario can be realized by many models, starting from the simplest, the minimally coupled scalar field \cite{linde, steinhardt} and continuing with more elaborated models like non-minimally coupled scalar field \cite{futumase, unruh, barrow1, barrow2, bezrukov}, kinetic inflation \cite{picon}, vector inflation \cite{ford, koivisto, mukhanov2}, inflaton potential in supergravity \cite{kawasaki, davies, kallosh}, string theory inspired inflation \cite{soda1, soda2, maldacena, kallosh1, soda3, baumann2}, Dirac-Born-Infeld inflation model \cite{silver, silver1, chen, easson}, $\alpha$-attractor models originated in supergravity \cite{linde3, ferrara, sergei1, dimopoulos, akrami}. Apart from the DBI models of inflation, another class of ghost-free models has been recently considered, named "Galileon" models \cite{nicolis, deffayet}.
The main characteristic of these models is that the gravitational and scalar field equations remain as second-order differential equations.
The Galileon terms modify the kinetic term compared to the standard canonical scalar field, which in turn can relax the physical constraints on the potential. In the case of the Higgs-type potential, for instance, one of the effects of the higher derivative terms is the reduction of the self coupling of the Higgs boson, so that the spectra of primordial density perturbations are consistent with the present observational data \cite{kamada, ohashi}.
Galilean models of inflation have been considered in \cite{kamada, ohashi, yokoyama, mizuno, burrage, kobayashi1}. Some aspects of slow-roll inflation with non-minimal kinetic coupling have been analyzed in \cite{capozziello1, germani, granda3, stsujikawa, nanyang, gansu}.
For a sample papers devoted to the study of slow-roll inflation in the context of Gauss-Bonnet (GB) coupling see  \cite{Kanti:1998jd, soda4, Guo:2009uk, Guo:2010jr, Jiang:2013gza, Koh:2014bka, Kanti:2015pda, soda5, vandeBruck:2017voa, quiang, sergeioik, sergeioik1, chakraborty, tanmoy}.\\
\noindent This paper is dedicated to the study of the slow-roll inflation in the scalar-tensor model with non-minimal kinetic coupling to the Einstein tensor and coupling of the scalar field to the Gauss-Bonnet 4-dimensional invariant. The non-minimal couplings of the scalar field to curvature of the type considered in the present paper arise, among other couplings, in fundamental theories like supergravity and string theory after specific compactification to an effective four dimensional theory \cite{fradkin, gross, tseytlin, meissner, cartier}, where the scalar field is related to the size of the compact extra dimensions and the (exponential) potential is related to the curvature of the extra dimensional manifold.
Note however that for the potential  $V=V_0 e^{\lambda\phi}$, that  produces power-law $a\propto t^{2/\lambda^2}$, the constant $\lambda$ that appear from compactifications is usually of order $1$ or greater, which is insufficient to generate inflation. By considering non-minimal derivative and Gauss-Bonnet couplings, this problem can be avoided. This makes it appealing  to analyze the mechanism of slow-roll inflation in such  theories where the scalar field appears non-minimally coupled to curvature terms. This could provide a connection with fundamental  theories in a high curvature regime characteristic of inflation.\\
The  Gauss-Bonnet and non-minimal kinetic couplings have also been considered in the dark energy problem  \cite{nojiriodintsov, neupane, koivisto1, koivisto2, suchkov, lgranda}.
Particularly, couplings  with exponential form have been considered in \cite{grandaloaiza} to study the late time cosmological dynamics, where stable and saddle scaling solutions have been obtained and a critical points corresponding to de Sitter solution were found.  The de Sitter solutions correspond to the critical points {\bf C} and {\bf E} in \cite{grandaloaiza} with marginal stability, which could be considered as possible (saddle point)  inflationary solutions. So this model could describe an inflationary de Sitter solution that can evolve towards scaling solutions with saddle character or to stable attractor dominated by the scalar field, describing accelerated expansion \cite{grandaloaiza}. Thus, the present model can provide a connection between early time inflation and late time accelerated expansion (in \cite{granjimsan} special cases of quintessence and phantom solutions have been studied with exponential couplings). For unified description of early time inflation and late time accelerated expansion in the framework of scalar tensor theories see \cite{sergeinojiri}. \\
In the present work we consider models with exponential potential and exponential couplings. An important feature of the exponential potential (in the framework of minimally coupled scalar field model) is that under its dominance the universe expands following a power-law, which describes the asymptotic behavior of the background spacetime in different epochs. This is the case of the late time dark energy dominated universe, where the exponential potential can give rise to accelerated expansion  \cite{copeland1, copeland2}.
Applied to the study of the early universe, the exponential potential in the minimally coupled scalar field model, gives rise to power-law inflation \cite{abbott, lucchin, yokoyama1, liddle1, ratra} with constant slow-roll parameters. This implies that the exponential potential lacks a successful exit  from inflation, which added to the fact that the tensor-to-scalar ratio is larger than the limits set by Planck data, rules out the exponential potential in the standard canonical scalar field model. In the present paper we address the above shortcomings of the exponential potential, this time in the frame of scalar-tensor theories, taking into account non-minimal kinetic an GB couplings, which could  play relevant role in the high curvature regime typical for inflation. We find that the above couplings predict values for the scalar spectral index and the tensor-to-scalar ratio that fall in the region quoted by the latest observational data.
The paper is organized as follows. In the next section we introduce the model, the background field equations and define the slow-roll parameters. In section 3 we use quadratic action for the scalar and tensor perturbations to evaluate the primordial power spectra. In section 4 we analyze several models with exponential potential and exponential couplings. Some discussion is presented in section 5.
\section{The model and background equations}
We consider the following scalar-tensor model
\begin{equation}\label{eqm1}
S=\int d^4x\sqrt{-g}\left[\frac{1}{2}F(\phi)R-\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)+F_1(\phi)G_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi-F_2(\phi){\cal G}\right]
\end{equation}
where $G_{\mu\nu}$ is the Einstein's tensor, ${\cal G}$ is the GB 4-dimensional invariant given by
\begin{equation}\label{eqm2}
{\cal G}=R^2-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\lambda\rho}R^{\mu\nu\lambda\rho}
\end{equation}
\begin{equation}\label{eqm3}
F(\phi)=\frac{1}{\kappa^2}+f(\phi),
\end{equation}
and $\kappa^2=M_p^{-2}=8\pi G$.
One remarkable characteristic of this model is that it yields second-order field equations and can avoid Ostrogradski instabilities. In the spatially flat FRW background
\begin{equation}\label{eqm4}
ds^2=-dt^2+a(t)^2\left(dx^2+dy^2+dz^2\right),
\end{equation}
one can write the field equations as follows
\begin{equation}\label{eqm5}
3H^2F\left(1-\frac{3F_1\dot{\phi}^2}{F}-\frac{8H\dot{F_2}}{F}\right)=\frac{1}{2}\dot{\phi}^2+V-3H\dot{F}
\end{equation}
\begin{equation}\label{eqm6}
\begin{aligned}
2\dot{H}F\left(1-\frac{F_1\dot{\phi}^2}{F}-\frac{8H\dot{F_2}}{F}\right)&=-\dot{\phi}^2-\ddot{F}+H\dot{F}+8H^2\ddot{F_2}-8H^3\dot{F_2}\\
&-6H^2F_1\dot{\phi}^2+4HF_1\dot{\phi}\ddot{\phi}+2H\dot{F}_1\dot{\phi}^2
\end{aligned}
\end{equation}
\begin{equation}\label{eqm7}
\begin{aligned}
\ddot{\phi}+3H\dot{\phi}&+V'-3F'\left(2H^2+\dot{H}\right)+24H^2\left(H^2+\dot{H}\right)F_2'+18H^3F_1\dot{\phi}\\&+12H\dot{H}F_1\dot{\phi}+6H^2F_1\ddot{\phi}+3H^2F_1'\dot{\phi}^2=0
\end{aligned}
\end{equation}
where ($'$) denotes derivative with respect to the scalar field. Related to the different terms in the action (\ref{eqm1}) we define the following slow-roll parameters
\begin{equation}\label{eqm8}
\epsilon_0=-\frac{\dot{H}}{H^2},\;\;\; \epsilon_1=\frac{\dot{\epsilon}_0}{H\epsilon_0}
\end{equation}
\begin{equation}\label{eqm9}
\ell_0=\frac{\dot{F}}{HF},\;\;\; \ell_1=\frac{\dot{\ell}_0}{H\ell_0}
\end{equation}
\begin{equation}\label{eqm10}
k_0=\frac{3F_1\dot{\phi}^2}{F},\;\;\; k_1=\frac{\dot{k}_0}{Hk_0}
\end{equation}
\begin{equation}\label{eqm11}
\Delta_0=\frac{8H\dot{F_2}}{F},\;\;\; \Delta_1=\frac{\dot{\Delta}_0}{H\Delta_0}
\end{equation}
The slow-roll conditions in this model are satisfied if $\epsilon_0,\; \epsilon_1,\; \Delta_0,....<<1$.
From the cosmological equations (\ref{eqm5}) and (\ref{eqm6}) and using the parameters (\ref{eqm8})-(\ref{eqm11}) we can write the following expressions for $\dot{\phi}^2$ and $V$
\begin{equation}\label{eqm12}
\begin{aligned}
V=&H^2F\Big[3-\frac{5}{2}\Delta_0-2k_0-\epsilon_0+\frac{5}{2}\ell_0+\frac{1}{2}\ell_0\left(\ell_1-\epsilon_0+\ell_0\right)\\&-\frac{1}{2}\Delta_0\left(\Delta_1-\epsilon_0+\ell_0\right)-\frac{1}{3}k_0\left(k_1+\ell_0-\epsilon_0\right)\Big]
\end{aligned}
\end{equation}
\begin{equation}\label{eqm13}
\begin{aligned}
\dot{\phi}^2=&H^2F\Big[2\epsilon_0+\ell_0-\Delta_0-2k_0+\Delta_0\left(\Delta_1-\epsilon_0+\ell_0\right)-\\& \ell_0\left(\ell_1-\epsilon_0+\ell_0\right)+\frac{2}{3}k_0\left(k_1+\ell_0-\epsilon_0\right)\Big]
\end{aligned}
\end{equation}
where we used
\begin{equation}\label{eqm13a}
\ddot{F}=H^2F\ell_0\left(\ell_1-\epsilon_0+\ell_0\right),\;\;\; \ddot{F}_2=\frac{F\Delta_0}{8}\left(\Delta_1+\epsilon_0+\ell_0\right)
\end{equation}
It is also useful to define the variable $Y$ from Eq. (\ref{eqm13}) as
\begin{equation}\label{eqm13b}
Y=\frac{\dot{\phi}^2}{H^2F}
\end{equation}
where it follows that $Y = {\cal O}(\varepsilon )$.
Under the slow-roll conditions $\ddot{\phi}<<3H\dot{\phi}$ and $\ell_i, k_i, \Delta_i<<1$, it follows from the field equations (\ref{eqm5})-(\ref{eqm7}) that they can be reduced to
\begin{equation}\label{eqm14}
3H^2F\simeq V,
\end{equation}
\begin{equation}\label{eqm15}
2\dot{H}F\simeq -\dot{\phi}^2+H\dot{F}-6H^2F_1\dot{\phi}^2-8H^3\dot{F}_2,
\end{equation}
\begin{equation}\label{eqm16}
3H\dot{\phi}+V'-6H^2F'+18H^3F_1\dot{\phi}+24H^4F'_2\simeq 0,
\end{equation}
The scalar field equation (\ref{eqm16}) allows to determine the number of $e$-folds as
\begin{equation}\label{eqm17}
N=\int_{\phi_I}^{\phi_E}\frac{H}{\dot{\phi}}d\phi=\int_{\phi_I}^{\phi_E}\frac{H^2+6H^4F_1}{2H^2F'-8H^4F'_2-\frac{1}{3}V'}d\phi
\end{equation}
where $\phi_I$ and $\phi_E$ are the values of the scalar field at the beginning and end of inflation respectively.
\section{Second order action for the scalar and tensor perturbations}
{\bf Scalar Perturbations}.\\

\noindent The details of the first and second order perturbations fro the model (\ref{eqm1}) are given in \cite{grajim}. The second order action for the scalar perturbations is given by the following expression
\begin{equation}\label{slr1}
\delta S_s^{2}=\int dt d^3xa^3\left[{\cal G}_s\dot{\xi}^2-\frac{{\cal F}_s}{a^2}\left(\nabla\xi\right)^2\right]
\end{equation}
where
\begin{equation}\label{slr2}
{\cal G}_s=\frac{\Sigma}{\Theta^2}{\cal G}_T^2+3{\cal G}_T
\end{equation}
\begin{equation}\label{slr3}
{\cal F}_s=\frac{1}{a}\frac{d}{dt}\left(\frac{a}{\Theta}{\cal G}_T^2\right)-{\cal F}_T
\end{equation}
with
\begin{equation}\label{slr4}
{\cal G}_T=F-F_1\dot{\phi}^2-8H\dot{F}_2.
\end{equation}
\begin{equation}\label{slr5}
{\cal F}_T=F+F_1\dot{\phi}^2-8\ddot{F}_2
\end{equation}
\begin{equation}\label{slr6}
\Theta=FH+\frac{1}{2}\dot{F}-3HF_1\dot{\phi}^2-12H^2\dot{F}_2
\end{equation}
\begin{equation}\label{slr7}
\Sigma=-3FH^2-3H\dot{F}+\frac{1}{2}\dot{\phi}^2+18H^2F_1\dot{\phi}^2+48H^3\dot {F}_2
\end{equation}
And the sound speed of scalar perturbations is given by
\begin{equation}\label{slr8}
c_S^2=\frac{{\cal F}_S}{{\cal G}_S}
\end{equation}
The conditions for avoidance of ghost and Laplacian instabilities as seen from the action (\ref{slr1}) are
$$ {\cal F}>0,\;\;\;\; {\cal G}>0 $$
We can rewrite ${\cal G}_T$, ${\cal F}_T$, $\Theta$ and $\Sigma$ in terms of the slow-roll parameters (\ref{eqm8})-(\ref{eqm11}) and using Eqs. (\ref{eqm13}) and (\ref{eqm13a}), as follows
\begin{equation}\label{slr8a}
{\cal G}_T=F\left(1-\frac{1}{3}k_0-\Delta_0\right)
\end{equation}
\begin{equation}\label{slr8b}
{\cal F}_T=F\left(1+\frac{1}{3}k_0-\Delta_0\left(\Delta_1+\epsilon_0+\ell_0\right)\right)
\end{equation}
\begin{equation}\label{slr8c}
\Theta=FH\left(1+\frac{1}{2}\ell_0-k_0-\frac{3}{2}\Delta_0\right)
\end{equation}
\begin{equation}\label{slr8d}
\begin{aligned}
\Sigma=&-FH^2\Big[3-\epsilon_0+\frac{5}{2}\ell_0-5k_0-\frac{11}{2}\Delta_0+\frac{1}{2}\ell_0\left(\ell_1-\epsilon_0+\ell_0\right)\\&-\frac{1}{3}k_0\left(k_1-\epsilon_0+\ell_0\right)-\frac{1}{2}\Delta_0\left(\Delta_1-\epsilon_0+\ell_0\right)\Big]
\end{aligned}
\end{equation}
The expressions for ${\cal G}_S$ and $c_S^2$ in terms of the slow roll parameters can be written as
\begin{equation}\label{slowGS}
{{\cal G}_S} = \frac{{F\left( {\frac{1}{2}Y + {k_0} + \frac{3}{4}{W^2}(1 - {\Delta _0} - \frac{1}{3}{k_0})} \right)}}{{{{\left( {1 + \frac{1}{2}W} \right)}^2}}}
\end{equation}
\begin{equation}\label{slowCS}
c_S^2 = 1 + \frac{{{W^2}\left( {\frac{1}{2}{\Delta _0}({\Delta _1} + {\varepsilon _0} + {l_0} - 1) - \frac{1}{3}{k_0}} \right) + W\left( {\frac{2}{3}{k_0}\left( {2 - {k_1} - {l_0}} \right) + 2{\Delta _0}{\varepsilon _0}} \right) - \frac{4}{3}{k_0}{\varepsilon _0}}}{{Y + 2{k_0} + \frac{3}{2}{W^2}(1 - {\Delta _0} - \frac{1}{3}{k_0})}}
\end{equation}
where

\begin{equation}\label{slowW}
W = \frac{\ell_0-\Delta_0-\frac{4}{3}k_0}{1-\Delta_0-\frac{1}{3}k_0}
\end{equation}
Notice that in general ${\cal G}_S=F {\cal O}(\varepsilon )$ and $c_S^2 = 1 +{\cal O}(\varepsilon )$. Also in absence of the kinetic coupling it follows that $c_S^2 = 1 +{\cal O}(\varepsilon^2 )$. Keeping first order terms in slow-roll parameters, the expressions for ${\cal G}_S$ y $c_S^2$ reduce to
\begin{equation}\label{aproxGS}
{\cal G}_S = F\left( {{\varepsilon _0} + \frac{1}{2}{l_0} - \frac{1}{2}{\Delta _0}} \right)
\end{equation}
\begin{equation}\label{aproxCS}
c_S^2 = 1 + \frac{{\frac{4}{3}{k_0}\left( {{l_0} - {\Delta _0} - \frac{4}{3}{k_0}} \right) - \frac{4}{3}{k_0}{\varepsilon _0}}}{{2{\varepsilon _0} + {l_0} - {\Delta _0}}}
\end{equation}
\noindent After the appropriate change of variables  to normalize the action (\ref{eqm1}), we find the equation of motion, working in the Fourier representation, as (\cite{kobayashi1}) (see (F.7) of \cite{grajim})
\begin{equation}\label{slr12}
\tilde{U}_{\vec{k}}''+\left(k^2-\frac{\tilde{z''}}{\tilde{z}}\right)\tilde{U}_{\vec{k}}=0
\end{equation}
where
\begin{equation}\label{slr9}
d\tau_s=\frac{c_S}{a}dt,\;\;\, \tilde{z}=\sqrt{2}a\left({\cal F}_S{\cal G}_S\right)^{1/4},\;\;\; \tilde{U}=\xi\tilde{z}
\end{equation}
From (\ref{slr9}), and keeping up to first-order terms in slow-roll variables in (\ref{slowGS}) and (\ref{slowCS}), we find the following expression for $\tilde{z}''/\tilde{z}$
\begin{equation}\label{slr14}
\frac{\tilde{z}''}{\tilde{z}}=\frac{a^2H^2}{c_S^2}\Big[2-\epsilon_0+\frac{3}{2}\ell_0+\frac{3}{2}\frac{2\epsilon_0\epsilon_1+\ell_0\ell_1-\Delta_0\Delta_1}{2\epsilon_0+\ell_0-\Delta_0}\Big].
\end{equation}
Taking into account the slow-roll parameters we can rewrite the Eq. (\ref{slr12}) in the form
\begin{equation}\label{slr20}
\tilde{U}_k''+k^2\tilde{U}_{k}+\frac{1}{\tau_s^2}\left(\mu_s^2-\frac{1}{4}\right)\tilde{U}_{k}=0
\end{equation}
where
\begin{equation}\label{slr21}
\mu_s^2=\frac{9}{4}\left[1+\frac{4}{3}\epsilon_0+\frac{2}{3}\ell_0+\frac{2}{3}\frac{2\epsilon_0\epsilon_1+\ell_0\ell_1-\Delta_0\Delta_1}{2\epsilon_0+\ell_0-\Delta_0} \right],
\end{equation}
After the integration of (\ref{slr20}) using the slow-roll formalism (see \cite{grajim} for details) we find, at super horizon scales ($c_S k<<aH$), the following asymptotic solution
\begin{equation}\label{slr24}
\tilde{U}_k=\frac{1}{\sqrt{2}}e^{i\frac{\pi}{2}(\mu_s-\frac{1}{2})}2^{\mu_s-\frac{3}{2}}\frac{\Gamma(\mu_s)}{\Gamma(3/2)}\sqrt{-\tau_s}(-k\tau_s)^{-\mu_s}.
\end{equation}
On the other hand, from the relationship
\begin{equation}\label{slr25}
\frac{\tilde{z}'}{\tilde{z}}=-\frac{1}{(1-\epsilon_0)\tau_s}\Big[1+\frac{1}{2}\ell_0+\frac{1}{2}\frac{2\epsilon_0\epsilon_1+\ell_0\ell_1-\Delta_0\Delta_1}{2\epsilon_0+\ell_0-\Delta_0}\Big]=-\frac{1}{\tau_s}\left(\mu_s-\frac{1}{2}\right),
\end{equation}
and after integrating in the slow-roll approximation we find
\begin{equation}\label{slr26}
\tilde{z}\propto \tau_s^{\frac{1}{2}-\mu_s},
\end{equation}
which gives in the super horizon regime, from from (\ref{slr9}), the following $k$-dependence for the amplitude of the scalar perturbations
\begin{equation}\label{slr27}
\xi_k=\frac{\tilde{U}_k}{\tilde{z}}\propto k^{-\mu_s}
\end{equation}
Then, from the power spectra for the scalar perturbations
\begin{equation}\label{slr28}
P_{\xi}=\frac{k^3}{2\pi^2}|\xi_k|^2
\end{equation}
we find the spectral index, in first order in slow-roll parameters
\begin{equation}\label{slr29}
n_s-1=\frac{d\ln P_{\xi}}{d\ln k}=3-2\mu_s=-2\epsilon_0-\ell_0-\frac{2\epsilon_0\epsilon_1+\ell_0\ell_1-\Delta_0\Delta_1}{2\epsilon_0+\ell_0-\Delta_0}
\end{equation}

\noindent {\bf Tensor perturbations}.\\

\noindent The second order action for the tensor perturbations is given by (\cite{grajim})
\begin{equation}\label{slrt1}
\delta S_2=\frac{1}{8}\int d^3xdt{\cal G}_T a^2\left[\left(\dot{h}_{ij}\right)^2-\frac{c_T^2}{a^2}\left(\nabla h_{ij}\right)^2\right]
\end{equation}
where ${\cal G}_T$ and ${\cal F}_T$ are defined in (\ref{slr4}) and (\ref{slr5}) (in terms of the slow-roll variables (\ref{eqm8})-(\ref{eqm11})). The velocity of tensor perturbations is given by
\begin{equation}\label{slrt2}
c_T^2=\frac{{\cal F}_T}{{\cal G}_T}=\frac{3+k_0-3\Delta_0\left(\Delta_1+\epsilon_0+\ell_0\right)}{3-k_0-3\Delta_0}.
\end{equation}
Following the same lines as for the scalar perturbations and introducing the following variables
\begin{equation}\label{slrt3}
d\tau_T=\frac{c_T}{a}dt,\;\;\; z_T=\frac{a}{2}\left({\cal F}_T{\cal G}_T\right)^{1/4},\;\;\; v_{ij}=z_T h_{ij},
\end{equation}
that lead to the equation
\begin{equation}\label{slrt6}
v''_{(k)ij}+\left(k^2-\frac{z''_T}{z_T}\right)v_{(k)ij}=0,
\end{equation}
The deduction of the power spectrum for primordial tensor perturbations follows the same pattern as for the scalar perturbations. At super horizon scales ($c_Tk<<aH$) the tensor modes (\ref{slrt1}) have the same functional form for the asymptotic behavior as the scalar modes (\ref{slr24}), and therefore we can write power spectrum for tensor perturbations as
\begin{equation}\label{slrt11}
P_T=\frac{k^3}{2\pi^2}|h^{(k)}_{ij}|^2
\end{equation}
where, in first order in slow-roll parameters, the tensor spectral index has the following form \cite{grajim}
\begin{equation}\label{slrt11}
n_T=3-2\mu_T=-2\epsilon_0-\ell_0
\end{equation}
where
\begin{equation}\label{slr11a}
\mu_T=\frac{3}{2}+\epsilon_0+\frac{1}{2}\ell_0.
\end{equation}
An important quantity is the relative contribution to the power spectra of tensor and scalar perturbations, defined as the tensor/scalar ratio $r$
\begin{equation}\label{slrt12}
r=\frac{P_T(k)}{P_{\xi}(k)}.
\end{equation}
For the scalar perturbations, using (\ref{slr28}), we can write the power spectra as
\begin{equation}\label{slrt13}
P_{\xi}=A_S\frac{H^2}{(2\pi)^2}\frac{{\cal G}_S^{1/2}}{{\cal F}_S^{3/2}}
\end{equation}
where
$$A_S=\frac{1}{2}2^{2\mu_s-3}\Big|\frac{\Gamma(\mu_s)}{\Gamma(3/2)}\Big|^2$$
and all magnitudes are evaluated at the moment of horizon exit when $c_s k=aH$ ($k\tau_s=-1$). For $\tilde{z}$ we used (\ref{slr9}) with $a=c_S k/H$. In analogous way we can write the power spectra for tensor perturbations as
\begin{equation}\label{slrt14}
P_T= 16A_T\frac{H^2}{(2\pi)^2}\frac{{\cal G}_T^{1/2}}{{\cal F}_T^{3/2}}
\end{equation}
where
$$A_T=\frac{1}{2}2^{2\mu_T-3}\Big|\frac{\Gamma(\mu_T)}{\Gamma(3/2)}\Big|^2.$$
Noticing that $A_T/A_S\simeq 1$ when evaluated at the limit $\epsilon_0,\ell_0,\Delta_0,...<<1$, as follows from (\ref{slr21}) and (\ref{slr11a}), we can write the tensor/scalar ratio as follows
\begin{equation}\label{slrt15}
r=16\frac{{\cal G}_T^{1/2}{\cal F}_S^{3/2}}{{\cal G}_S^{1/2}{\cal F}_T^{3/2}}=16\frac{c_S^3{\cal G}_S}{c_T^3{\cal G}_T}
\end{equation}
taking into account the expressions for ${\cal G}_T, {\cal F}_T, {\cal G}_S,{\cal F}_S$ given in (\ref{slr8a})-(\ref{slowW}), up to first order, and using the condition $\epsilon_0,\ell_0,k_0,\Delta_0<<1$, then  we can see that $c_T\simeq c_S\simeq 1$ (in fact in the limit $\ell_0\rightarrow 0$, $c_S=1$ independently of the values of $\epsilon_0$ and $\Delta_0$) and we can make the approximation
\begin{equation}\label{slrt16}
r=8\left(\frac{2\epsilon_0+\ell_0-\Delta_0}{1-\frac{1}{3}k_0-\Delta_0}\right)\simeq 8\left(2\epsilon_0+\ell_0-\Delta_0\right)
\end{equation}
which is a modified consistency relation due to the non-minimal and GB couplings. In the limit $\ell_0,\Delta_0\rightarrow 0$ it gives the standard consistency relation for the single canonical scalar field inflation
\begin{equation}\label{slrt17}
r= -8n_T,
\end{equation}
with $n_T=-2\epsilon_0$. Note that if the model contains only non-minimal coupling $F(\phi)$, then $r\simeq 8(2\epsilon_0+\ell_0)$, and from (\ref{slrt11}) it follows that the standard consistency relation (\ref{slrt17}) remains valid in presence of non-minimal coupling.
In the general case from (\ref{slrt16}) we find the deviation from the standard consistency relation in the form
\begin{equation}\label{slrt18}
r=-8n_T+\delta r,\;\;\; \delta r=-8\Delta_0,
\end{equation}
with $n_T$ given by (\ref{slrt11}). Here for standard consistency relation we mean the relation (\ref{slrt17}) independently of the content of $n_T$. This expression can also be written as
\begin{equation}\label{consistency}
r=-8n_T\left(1+\frac{\Delta_0}{n_T}\right)=-8n_T\left(1-\frac{\Delta_0}{2\epsilon_0+\ell_0}\right)=-8\gamma n_T
\end{equation}
where $\gamma=1-\Delta_0/(2\epsilon_0+\ell_0)$  characterizes the deviation from the standard consistency relation. Note that this deviation in first-order approximation is independent of $k_0$.
Thus, the consistency relation still valid in the case of non-minimal coupling ($\Delta_0=0$), and a deviation from the standard consistency relation can reveal the effect of interactions beyond the simple canonical or non-minimally coupled scalar field.

\section{Inflation Driven by Exponential Potential and Exponential Couplings}
The exponential potential leads to scaling solutions important to describe different epochs of cosmological evolution, including solutions with accelerated expansion. In the standard minimally coupled scalar field it leads to inflationary solutions with constant slow-roll parameters, that lead to eternal inflation, which added to the strong signal of gravitational waves ($r>0.1$), makes the model inviable. As stated in the introduction, the exponential potential and couplings appear in a number of compactifications from higher dimensional fundamental theories  such as supergravity and string theory, where the scalar field encodes the size of the extra dimensions.  Although these couplings are inspired by higher-dimensional gravitational theories, in the present study we are not trying to  match any specific model coming directly from higher dimensional compactifications. The viability of the present model is probed by the fact that it leads to graceful exit from inflation, and after estimating the power spectra of scalar and tensor perturbations it gives the main inflationary observables $n_s$ and $r$ in the region quoted by the latest observational data. \\

\noindent {\bf Kinetic coupling.}\\

\noindent Let us start with the model (\ref{eqm1}) with the explicit form of the couplings given by\\
\begin{equation}\label{expo1}
F\left( \phi  \right) = \frac{1}{{{\kappa ^2}}},\,\,\,\,\,\,\,\,\,V\left( \phi  \right) = V_0 e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,\,{F_1}\left( \phi  \right) =f_{k} e^{-\eta\kappa\phi} ,\,\,\,\,\,\,\,\,{F_2}\left( \phi  \right) = 0.
\end{equation}
from (\ref{eqm8})-(\ref{eqm11}), using (\ref{eqm14})-(\ref{eqm16}) we find the slow-roll parameters
\begin{equation}\nonumber
\epsilon_0=\frac{\lambda^2}{2\left(2\alpha e^{-(\lambda+\eta)\phi}+1\right)},\;\;\; \epsilon_1=\frac{2\alpha\lambda(\lambda+\eta)e^{(\lambda+\eta)\phi}}{\left(e^{(\lambda+\eta)\phi}+2\alpha\right)^2},
\end{equation}
\begin{equation}\label{epsilons}
k_0=\frac{\alpha\lambda^2 e^{(\lambda+\eta)\phi}}{\left(e^{(\lambda+\eta)\phi}+2\alpha\right)^2},\;\;\; k_1=-\frac{\lambda(\lambda+\eta)e^{(\lambda+\eta)\phi}\left(e^{(\lambda+\eta)\phi}-2\alpha\right)}{\left(e^{(\lambda+\eta)\phi}+2\alpha\right)^2}
\end{equation}
where we have set $\kappa=1$ and $\alpha=V_0f_k$. In standard slow-roll inflation ($f_k=0,\; \eta=0$) the condition $\lambda^2<<1$ is required, while in the presence of kinetic coupling this condition can be avoided due to the $\phi$ dependence in the slow-roll parameters. This $\phi$-dependence of the slow-roll parameters also allows the graceful exit from inflation.
Using the condition $\epsilon_0(\phi_E)=1$ we find the expression for the scalar field at the end of inflation as
\begin{equation}\label{fielde}
\phi_E=\frac{1}{\lambda+\eta}\ln\left[\frac{4\alpha}{\lambda^2-2}\right].
\end{equation}
With $f_k$ being positive, this field is well defined whenever $\lambda>\sqrt{2}$. It is clear from this expression that the larger $\eta$, the smaller $\phi_E$ can be. It also follows that  $\phi_E$ varies very slowly with the increment of $\alpha$ because of the logarithm dependence. Assuming for instance $\lambda=2, \eta=5, \alpha=10^3$, give $\phi_E\simeq 1.08 M_p$, and $\lambda=2, \eta=5, \alpha=10^2$ give $\phi_E\simeq 0.76 M_p$.\\
The Eq. (\ref{eqm17}) gives the number of e-foldings as
\begin{equation}\label{efolds}
N=\frac{1}{2\lambda(\lambda+\eta)}\left(2-\lambda^2+2\ln\left[\frac{4\alpha}{\lambda^2-2}\right]\right)-\left(\frac{\phi_I}{\lambda}-\frac{2\alpha e^{-(\lambda+\eta)\phi_I}}{\lambda(\lambda+\eta)}\right)
\end{equation}
where $\phi_I$ is the scalar field $N$ $e$-folds before the end of inflation. Solving this equation gives the explicit form of $\phi_I$
\begin{equation}\label{fieldi}
\phi_I=\frac{1}{2(\lambda+\eta)}\left[2\ln\left(\frac{4\alpha}{\lambda^2-2}\right)-2\lambda N\left(\lambda+\eta\right)-\lambda^2+2W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\right].
\end{equation}
For the scalar spectral index, we see from (\ref{slr29}) that up to first order in slow-roll parameters $n_s$ does not depend on $k_0$ and $k_1$. So, if the model contains only non-minimal kinetic coupling the scalar spectral index becomes $n_s=1-2\epsilon_0-\epsilon_1$, and for the same reason from (\ref{slrt16}) follows that $r=16\epsilon_0$. However, both $\epsilon_0$ and $\epsilon_1$ depend on all the parameters of the model. The analytical expression for $n_s$ is given by
\begin{equation}\label{ns1}
n_s=1-\frac{\lambda^2}{2\alpha e^{-(\lambda+\eta)\phi_I}+1}-\frac{2\alpha\lambda(\lambda+\eta)e^{(\lambda+\eta)\phi_I}}{\left(e^{(\lambda+\eta)\phi_I}+2\alpha\right)^2},
\end{equation}
where $\phi_I$ is given by (\ref{fieldi}). And for the tensor-to-scalar-ratio it is found
\begin{equation}\label{r1}
r=\frac{8\lambda^2}{1+2\alpha e^{-(\lambda+\eta)\phi_I}}
\end{equation}
Notice that the kinetic coupling constant $f_k$ and $V_0$ appear only in the combination (reestablishing $\kappa$) $\alpha=\kappa^2V_0 f_k=V_0 f_k/M_p^2$. Taking into account the dimensionality of the kinetic coupling one can set $f_k=1/M^2$ and then, $\alpha=V_0/(M^2M_p^2)$.
By replacing $\phi_I$ from (\ref{fieldi}) into (\ref{ns1}) and (\ref{r1}) we find the exact analytical expressions for the scalar spectral index and the tensor-to-scalar ratio in terms of the model parameters and the number of $e$-foldings in the slow-roll approximation:
\begin{equation}\label{nsexplicit}
n_s=1-\frac{\lambda^2+\lambda(2\lambda+\eta)W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]}{\Big(1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\Big)^2}
\end{equation}
\begin{equation}\label{rexplicit}
r=\frac{8\lambda^2}{1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]}
\end{equation}
And for the slow-roll parameters $N$ $e$-folds before the end of inflation, we find the following analytical expressions
\begin{equation}\label{epsilk0}
\epsilon_0=\frac{\lambda^2}{2\Big(1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\Big)}
\end{equation}
\begin{equation}\label{epsilk1}
\epsilon_1=\frac{\lambda(\lambda+\eta)W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]}{\Big(1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\Big)^2}
\end{equation}
\begin{equation}\label{deltak0}
\Delta_0=\frac{\lambda^2W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]}{2\Big(1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\Big)^2}
\end{equation}
\begin{equation}\label{deltak1}
\Delta_1=\frac{\lambda(\lambda+\eta)\Big(W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]-1\Big)}{\Big(1+W\left[\frac{1}{2}\left(\lambda^2-2\right)e^{\frac{\lambda^2}{2}+\lambda N(\lambda+\eta)-1}\right]\Big)^2}
\end{equation}
An interesting result from these equations is that the two observables $n_s$ and $r$ and the slow-roll parameters (at the horizon crossing) do not depend on $\alpha$. So, the behavior of $n_s$ and $r$ is controlled exclusively by the dimensionless constants $\lambda$ and $\eta$ and by the number of $e$-foldings $N$. However $\alpha$ is important to define the scalar field at the beginning and the end of inflation as follows from (\ref{fielde}) and (\ref{fieldi}). Having fixed $\alpha=V_0 f_k/M_p^2=V_0/(M^2M_p^2)$ by the initial conditions on the scalar field, we still have freedom to fix $V_0$ by using the COBE-WMAP normalization \cite{cobe,wmap}, which sets the scale of $M$.
The restrictions imposed by the COBE-WMAP normalization and the tensor-to-scalar ratio allows to set the set the scales of Hubble parameter and the energy involved in the inflation. From  (\ref{slrt13})
\begin{equation}\label{powersd}
P_{\xi}=A_S\frac{H^2}{(2\pi)^2}\frac{{\cal G}_S^{1/2}}{{\cal F}_S^{3/2}}\sim \frac{H^2}{2(2\pi)^2}\frac{1}{{\cal F}_S}\sim  \frac{H^2}{8\pi^2}\frac{1}{\epsilon_0}
\end{equation}
where we used the limit $(\epsilon_0,\epsilon_1,...)\rightarrow 0$ that gives $A_S\rightarrow 1/2$ and $c_S^2\rightarrow 1$. Taking for instance the case $N=60$, $\lambda=2$, $\eta=1.5$ we find $\epsilon_0\sim 0.0048$ and $r\sim 0.077$. Taking into account the COBE-WMAP normalization  we find
\begin{equation}\label{cobeh}
P_{\xi}\simeq 2.5\times 10^{-9}\sim \frac{H^2}{8\pi^2}\frac{1}{0.0048}\;\; \Rightarrow H\sim 3\times 10^{-5}M_p\sim 7\times 10^{13} Gev.
\end{equation}
And using the tensor-to-scalar ratio under the same approximations done for $P_S$
\begin{equation}\label{energyscale}
P_T=rP_S\sim 2\frac{H^2}{\pi^2M_p^2}\sim \frac{2V}{3\pi^2M_p^4}\sim (r) 2.5\times 10^{-9}\sim \;\;\;  \Rightarrow V^{1/4}\sim 7\times 10^{-3}M_p\sim 10^{16}Gev.
\end{equation}
Given $V\sim 3\times 10^{-9}M_p^4$ and $\alpha=10^3$, the mass $M$ takes the value $M\sim 10^{-6}M_p$.
In Fig1 we show the behavior of $n_s$ and $r$ assuming $N=60$ for some numerical values of the constants.
\begin{figure}
\centering
\includegraphics[scale=0.7]{Exponential-ns-r1.pdf}
\caption{$n_s$ vs $r$ for $N=60$, and $\eta$ varying in the interval $1/2<\eta<10$. The red line corresponds to $\lambda=2$ and the blue line to $\lambda=1.42$.}
\end{figure}
In Fig 2 we illustrate the behavior of the slow-roll parameters that show the successful exit from inflation.\\
\begin{figure}
\centering
\includegraphics[scale=0.7]{slow-roll-par.pdf}
\caption{The evolution of slow-roll parameters in the interval $\phi_I<\phi<\phi_E$, for $N=60$, $\alpha=10^3$, $\lambda=2$ and $\eta=1/2$.  This behavior allows the exit form inflation. The values of the slow-roll parameters 60 $e$-folds before the end on inflation are:  $\epsilon_0\simeq 0.0067$, $\epsilon_1\simeq 0.016$, $k_0\simeq 0.0067$, $k_1\simeq 0.016$.}
\end{figure}
\noindent A special case takes place when $\eta=-\lambda$. As seen from (\ref{epsilons}) the slow-roll parameters become constant and the model leads to eternal inflation. In this case $N$ and $\phi_I$ are not well defined as follows from (\ref{efolds}) and (\ref{fieldi}) but $n_s$ and $r$ can be found from (\ref{ns1}), (\ref{r1}) and become constants given by
\begin{equation}\label{nsreternal}
n_s=1-\frac{\lambda^2}{2\alpha+1},\;\;\; r=\frac{8\lambda^2}{2\alpha+1}
\end{equation}
which gives the relationship
\begin{equation}\label{nsvsr}
n_s=1-\frac{1}{8}r,
\end{equation}
which imply that $n_s$ and $r$ can not simultaneously satisfy the observational restrictions, and therefore this case is discarded. \\

\noindent {\bf Gauss-Bonnet coupling.}\\
\begin{equation}\label{expogb}
F\left( \phi  \right) = \frac{1}{{{\kappa ^2}}},\,\,\,\,\,\,\,\,\,V\left( \phi  \right) = V_0 e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,\,{F_1}\left( \phi  \right) =0 ,\,\,\,\,\,\,\,\,{F_2}\left( \phi  \right) = f_{g} e^{-\eta\kappa\phi}.
\end{equation}
which from (\ref{eqm8}) gives the following slow-roll parameters
\begin{equation}\nonumber
\epsilon_0=\frac{1}{6}\lambda\left(8\eta\beta e^{-(\lambda+\eta)\phi}+3\lambda\right),\;\;\; \epsilon_1=-\frac{8}{3}\beta\eta\left(\lambda+\eta\right)e^{-(\lambda+\eta)\phi}
\end{equation}
\begin{equation}\label{epsilong}
\Delta_0=-\frac{8}{9}\beta\eta\left(3\lambda e^{(\lambda+\eta)\phi}+8\beta\eta\right)e^{-2(\lambda+\eta)\phi},\;\;\; \Delta_1=-\frac{1}{3}\left(\lambda+\eta\right)\left(3\lambda e^{(\lambda+\eta)\phi}+16\beta\eta\right)e^{-(\lambda+\eta)\phi}.
\end{equation}
where $\kappa=1$ and $\beta=V_0f_g$. The scalar field at the end of inflation, from the condition $\epsilon_0=1$, takes the form
\begin{equation}\label{fieldeg}
\phi_E=\frac{1}{\lambda+\eta}\ln\left[\frac{8\beta\eta\lambda}{3(2-\lambda^2)}\right]
\end{equation}
And the number of $e$-foldings from (\ref{eqm17}) is given by
\begin{equation}\label{efoldsg}
N=\frac{1}{\lambda(\lambda+\eta)}\left(\ln\left[\frac{16\beta\eta}{2-\lambda^2}\right]-\ln\left[8\beta\eta+3\lambda e^{(\lambda+\eta)\phi}\right]\right).
\end{equation}
Solving this equation with respect to the scalar field we find the scalar field $N$ $e$-folds before the end of inflation as
\begin{equation}\label{fieldig}
\phi_I=\frac{1}{\lambda+\eta}\ln\left[\frac{16\beta\eta e^{-\lambda N(\lambda+\eta)}}{3\lambda(2-\lambda^2)}-\frac{8\beta\eta}{3\lambda}\right]
\end{equation}
Using (\ref{epsilong})  in (\ref{slr29}) gives the expression for scalar spectral index as follows
\begin{equation}\label{nsg}
n_s=1-\lambda^2+\frac{8}{3}\beta\eta\left(\lambda+2\eta\right)e^{-(\eta+\lambda)\phi_I}.
\end{equation}
And replacing  (\ref{epsilong}) into (\ref{slrt16}) gives the tensor-to-scalar ratio as
\begin{equation}\label{rg}
r=\frac{8}{9}\left(8\beta\eta+3\lambda e^{(\eta+\lambda)\phi_I}\right)^2 e^{-2(\eta+\lambda)\phi_I}.
\end{equation}
taking into account the above expression for $\phi_I$, it is found
\begin{equation}\label{nsgi}
n_s=1-\lambda^2-\frac{\lambda(2\eta+\lambda)}{1-\frac{2e^{-\lambda N(\lambda+\eta)}}{2-\lambda^2}},
\end{equation}
and
\begin{equation}\label{rgi}
r=\frac{32\lambda^2}{\left(2+(\lambda^2-2)e^{\lambda N(\lambda+\eta)}\right)^2}
\end{equation}
Notice that neither $n_s$ nor $r$ depend on $\beta$, which appears only in the expressions for $\phi_I$ and $\phi_E$. In order to appreciate the order of the parameters involved in inflation, having in mind that at the end of inflation the slow-roll parameters should be of order $1$, we can evaluate the slow-roll parameters at the end of inflation by replacing $\phi_E$ into Eqs. (\ref{epsilong}), giving
\begin{equation}\label{epsilonge}
\epsilon_0=1,\;\;\; \epsilon_1=\frac{(\lambda+\eta)(\lambda^2-2)}{\lambda},\;\;\; \Delta_0=2-\frac{4}{\lambda^2}, \;\;\; \Delta_1=\frac{(\lambda+\eta)(\lambda^2-4)}{\lambda}.
\end{equation}
Replacing $\Phi_I$ into (\ref{epsilong}) gives the slow-roll parameters $N$ $e$-foldings before the end of inflation as

\begin{equation}\nonumber
\epsilon_0=\frac{\lambda^2}{2+(\lambda^2-2)e^{N\lambda(\lambda+\eta)}},\;\;\;  \epsilon_1=\frac{\lambda(\lambda+\eta)}{1+\frac{2e^{-N\lambda(\lambda+\eta)}}{\lambda^2-1}}
\end{equation}
\begin{equation}\label{epsilongi}
\Delta_0=\frac{2\lambda^2(\lambda^2-2)e^{N\lambda(\lambda+\eta)}}{\left(2+(\lambda^2-2)e^{N\lambda(\lambda+\eta)}\right)^2},\;\; \Delta_1=\frac{\lambda(\lambda+\eta)\left((\lambda^2-2)e^{N\lambda(\lambda+\eta)}-2\right)}{2+(\lambda^2-2)e^{N\lambda(\lambda+\eta)}}
\end{equation}

\noindent According to (\ref{epsilonge}), in order to keep $\Delta_0\sim 1$, $\lambda$ should be close to $2$, and from the expressions for $\epsilon_1, \Delta_1$ follows that $\eta\sim -1$. All these approximations are valid under the condition that $\epsilon=1$ at the end of inflation. On the other hand, the exponential in the expressions for $n_s$ and $r$ makes a big difference between $n_s$ and $r$ provided $N\sim 60$ for the above approximations for $\eta$ and $\lambda$. In fact it provides a wrong value for $n_s$ and $r\sim 0$.
One can also consider the region of parameters where the exponent $e^{-\lambda N(\lambda+\eta)}$ is of order $1$. In this case, numerical analysis shows that if one assumes, for instance the values $\lambda=-0.001$ and $\eta=1$, then $n_s$ and $r$ fall in the appropriate region according to the latest observational data. For $N$ varying in the interval $[50,60]$, $n_s$ and $r$ take values $0.961\le n_s\le 0.967$ and $0.002\le r\le 0.003$.
But in this same interval, the final field (\ref{fieldeg}) which depends on $\eta, \; \lambda,\; \beta$ takes the value $\phi_E\simeq 0.065 M_p$ (assuming $\lambda=-0.001$, $\eta=1$, $\beta=-8\times 10^2$). And the initial field (\ref{fieldig}), which depends  additionally on $N$, varies in the interval $11.6M_p\le \phi_I \le 11.8 M_p$. As we can see the difference between the initial and final fields is almost two orders of magnitude. Besides this, according to (\ref{epsilonge}) when the scalar field reaches the final value, the slow-roll parameters $\epsilon_1, |\Delta_0|, \Delta_1>>1$ ($\epsilon_0=1$) indicating that the slow-roll regime is broken long before the field reaches the value $\phi_E\simeq 0.065 M_p$.
Numerical analysis shows that at $\phi\simeq 7.6 M_p$ the slow-roll parameters  $\epsilon_1, |\Delta_0|, \Delta_1\sim1$ while $\epsilon<<1$. But given these values of the parameters, for the scalar field it takes $N\approx 1$ to evolve from $\phi_I=11.8 M_p$ to $7.6 M_p$, making the slow-roll mechanism impracticable under the condition $\epsilon_0=1$ at the end of inflation.
It is also possible to assume that the inflation ends when any of the main slow-roll parameters becomes of order $1$, which in our case would be $\Delta_0$, and have viable inflation (see \cite{sergeioik}). If the condition to end the inflation is imposed on the GB slow-roll parameter $\Delta_0$ (notice that $\epsilon_0$ and $\Delta_0$ enter with the same hierarchy in the expression for the potential (\ref{eqm12})), then the following results can be obtained. First, from the condition $\Delta_0=-1$ the scalar field at the end of inflation takes the value
\begin{equation}\label{fieldgde}
\phi_E=\frac{1}{\lambda+\eta}\ln\left[\frac{4}{3}\left(\eta\lambda \beta+\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)\right].
\end{equation}
From (\ref{eqm17}) we find the scalar field $N$ $e$-folds before the end of inflation as
\begin{equation}\label{fieldgid}
\phi_I=\frac{1}{\lambda+\eta}\ln\left[\frac{1}{3\lambda}\left(4e^{-\lambda N(\lambda+\eta)}-8\beta\eta\right)\left(\eta\beta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)\right].
\end{equation}
This expression for $\phi_I$ leads to the following $n_s$ and $r$ according to (\ref{nsg}) and (\ref{rg}) respectively
\begin{equation}\label{nsgd}
n_s=1-\lambda^2+\frac{2\beta\eta\lambda(2\eta+\lambda) e^{\lambda N(\lambda+\eta)}}{\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}+\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)},
\end{equation}
\begin{equation}\label{rgd}
r=\frac{8\lambda^2\left(\beta\eta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)^2}{\left(\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}+\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)\right)^2}
\end{equation}
The slow-roll parameters can be also explicitly written in terms of the model parameters when evaluated at the end and at the beginning of inflation.
By replacing (\ref{fieldgde}) into (\ref{epsilong}) we find the following expressions at the end of inflation
\begin{equation}\nonumber
\epsilon_0=\frac{\lambda\left(\beta\eta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)}{2\left(\left(\beta\eta\lambda+\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)\right)},\;\; \epsilon_1=-\frac{2\beta\eta(\eta+\lambda)}{\beta\eta\lambda+\sqrt{\beta^2\eta^2(\lambda^2+4)}}
\end{equation}
\begin{equation}\label{epsilongde}
\Delta_0=-1, \;\; \Delta_1=-\frac{(\eta+\lambda)\left(\beta\eta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)}{\beta\eta\lambda+\sqrt{\beta^2\eta^2(\lambda^2+4)}}
\end{equation}
And replacing (\ref{fieldgid}) into  (\ref{epsilong}) gives the slow-roll parameters $N$ $e$-foldings before the end of inflation as
\begin{equation}\label{epsgd0}
\epsilon_0=\frac{\lambda^2\left(\beta\eta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)}{2\left(\left(\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)\right)},
\end{equation}
\begin{equation}\label{epsgd1}
\epsilon_1=-\frac{2\beta\eta\lambda(\eta+\lambda) e^{\lambda N(\lambda+\eta)}}{\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}+\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)},
\end{equation}
\begin{equation}\label{delta0}
\Delta_0=-\frac{2\beta\eta\lambda^2\left(\beta\eta(\lambda^2+2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)e^{N\Lambda(\lambda+\eta)}}{\left(\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)^2},
\end{equation}
\begin{equation}\label{delta1}
\Delta_1=-\frac{\lambda(\lambda+\eta)\left(\beta\eta(\lambda^2+2+2e^{N\lambda(\lambda+\eta)})+\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}\right)}{\lambda\sqrt{\beta^2\eta^2(\lambda^2+4)}+\beta\eta(2-2e^{\lambda N(\lambda+\eta)}+\lambda^2)}.
\end{equation}
Assuming for instance, $N=50$, $\lambda=-0.005$, $\eta=1$, $f_2=-1$, we find
$$n_s=0.965,\;\; r=0.004,\;\; \phi_I=5.06,\;\; \phi_E=0.988.$$
The slow-roll parameters $50$ $e$-foldings before the end of inflation take the values
\begin{equation}\label{sample}
(\epsilon_0, \; \epsilon_1,\; \Delta_0,\; \Delta_1)\big|_{\phi_I}=(0.000056, 0.017, -0.00039, 0.039),
\end{equation}
and at the end of inflation
\begin{equation}\label{sample1}
(\epsilon_0, \; \epsilon_1,\; \Delta_0,\; \Delta_1)\big|_{\phi_E}=(0.0025, 0.99, -1, 1.99).
\end{equation}
In Fig.3 we show the $n_s-r$ trajectory for $50\le N\le 60$.
\begin{figure}
\centering
\includegraphics[scale=0.7]{ns-r-Delta.pdf}
\caption{The scalar spectral index $n_s$ and tensor/scalar ratio $r$, for  $\lambda=-0.001,\; \eta=1,\; \beta=-1$ (blue) and  $\lambda=-0.004,\; \eta=1,\; \beta=-1$ (red), for $N$ varying between $50\le N\le 60$. Both curves fall in the region constrained by the latest observations.}
\end{figure}
\noindent The above results show that the scalar potential in the frame of scalar-tensor models is not the only magnitude that drives the inflation. The  effect of interactions terms amounts to the effect of an effective potential since the scalar field also rolls down the coupling functions.
Concerning the restrictions imposed by the COBE-WMAP normalization, we find from  (\ref{slrt13})
\begin{equation}\label{powersd}
P_{\xi}=A_S\frac{H^2}{(2\pi)^2}\frac{{\cal G}_S^{1/2}}{{\cal F}_S^{3/2}}\sim \frac{H^2}{2(2\pi)^2}\frac{1}{{\cal F}_S}\sim  \frac{H^2}{(2\pi)^2}\frac{1}{2\epsilon_0-\Delta_0}
\end{equation}
Taking into account the values for the sample (\ref{sample}), where $\Delta_0$ is larger than $\epsilon$, and using the COBE normalization for the power spectrum $P_{\xi}$, we can write
\begin{equation}\label{hubblescale}
P_{\xi}\simeq 2.5\times 10^{-9}\sim \frac{H^2}{(2\pi)^2}\frac{1}{4\times 10^{-4}}\;\;\;  \Rightarrow H\sim 6.3\times 10^{-6}M_p\sim 10^{13}Gev.
\end{equation}
And from the tensor-to-scalar ratio it is found
\begin{equation}\label{energyscale}
P_T=rP_S\sim 2\frac{H^2}{\pi^2M_p^2}\sim \frac{2V}{3\pi^2M_p^4}\sim (r) 2.5\times 10^{-9} \;\;\;  \Rightarrow V^{1/4}\sim 3\times 10^{-3}M_p\sim 7\times 10^{15}Gev,
\end{equation}
where we used the value for $r$ given in the sample (\ref{sample1}). This imply, given that $\beta=\frac{f_gV_0}{M_p^4}=-1$, that the GB coupling constant $|f|\sim 10^{12}$. \\
A special case takes place when $\eta=-\lambda$ in (\ref{expogb}). The slow-roll parameters become constants given by
\begin{equation}\label{epsilconst}
\epsilon_0=\frac{1}{6}\left(3-8\beta\right)\lambda^2,\;\; \epsilon_1=0,\;\; \Delta_0=\frac{8}{9}\left(3-8\beta\right)\beta\lambda^2
\end{equation}
The spectral index and tensor-to-scalar ratio are given by
\begin{equation}\label{nsr}
n_s=1-\left(1-\frac{8}{3}\beta\right)\lambda^2,\;\;\; r=8\left(1-\frac{8}{3}\beta\right)^2\lambda^2
\end{equation}
Solving these equations with respect to $\beta$ and $\lambda$ gives
\begin{equation}\label{betalambda}
\beta=\frac{3(8n_s+r-8)}{64(n_s-1)},\;\;\; \lambda=\pm\frac{2\sqrt{2}(1-n_s)}{\sqrt{r}}.
\end{equation}
Thus, given the values of the observables $n_s$ and $r$, we can find the model parameters. Taking for instance $n_s=0.968$ and $r=10^{-2}$, give $\beta\simeq 0.36$ and $\lambda\simeq \pm 0.9$. According to (\ref{hubblescale}) and (\ref{energyscale})
$$P_{\xi}\sim  \frac{H^2}{(2\pi)^2}\frac{1}{2\epsilon_0-\Delta_0}\sim  \frac{H^2}{(2\pi)^2}\frac{1}{1.3\times 10^{-3}}\sim \simeq 2.5\times 10^{-9}\Rightarrow H\sim 10^{-5}M_p$$
and
$$P_T=rP_S\sim \frac{2V}{3\pi^2M_p^4}\sim (10^{-2}) 2.5\times 10^{-9}\;\; \Rightarrow V^{1/4}\sim 4.4\times 10^{-3}M_p$$\\
The issue with this case is the constancy of the slow-roll parameters that leads to eternal inflation, unless an alternative mechanism to trigger the graceful exit from inflation is provided.\\

\noindent {\bf Kinetic and Gauss-Bonnet couplings I.}\\

\noindent The following model includes both, the non-minimal kinetic and Gauss-Bonnet couplings.\\
\begin{equation}\label{expokgb}
F\left( \phi  \right) = \frac{1}{{{\kappa ^2}}},\,\,\,\,\,\,\,\,\,V\left( \phi  \right) = V_0 e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,\,{F_1}\left( \phi  \right) =f_k e^{\lambda\kappa\phi},\,\,\,\,\,\,\,\,{F_2}\left( \phi  \right) = f_{g} e^{-\lambda\kappa\phi}.
\end{equation}
The slow-roll parameters in terms of the scalar field take the form
\begin{equation}\nonumber
\epsilon_0=\frac{(8\beta e^{-2\lambda\phi}+3)\lambda^2}{12\alpha+6},\;\;\; \epsilon_1=-\frac{16\beta\lambda^2 e^{-2\lambda\phi}}{6\alpha+3}
\end{equation}
\begin{equation}\nonumber
\Delta_0=-\frac{8\beta e^{-4\lambda\phi}(8\beta +3e^{2\lambda\phi})\lambda^2}{9(2\alpha+1)},\;\;\; \Delta_1=-\frac{2(16\beta e^{-2\lambda\phi}+3)\lambda^2}{6\alpha+3},
\end{equation}
\begin{equation}\label{epsilonkgb}
k_0=\frac{\beta e^{-4\lambda\phi}(8\beta +3e^{2\lambda\phi})^2\lambda^2}{9(2\alpha+1)^2},\;\;\; k_1=-\frac{32\beta\lambda^2 e^{-2\lambda\phi}}{6\alpha+3}
\end{equation}
Where $\alpha$ and $\beta$ are defined as before, i.e.  $\alpha=f_k V_0$ and $\beta=f_g V_0$. By solving the condition to end the inflation, $\epsilon_0=1$ we find
\begin{equation}\label{phiendkgb}
\phi_E=\frac{1}{2\lambda}\ln\left[\frac{8\beta\lambda^2}{3(4\alpha-\lambda^2+2)}\right].
\end{equation}
From (\ref{eqm17}) we find
\begin{equation}\label{efoldskgb}
N=\frac{3}{\lambda^2(3-8\beta)}\left(\lambda\phi-\alpha e^{-2\lambda\phi}\right)\Big|_{\phi_I}^{\phi_E},
\end{equation}
which gives the scalar field $N$ $e$-folds before the end of inflation as
\begin{equation}\label{phiinikgb}
\begin{aligned}
\phi_I=&\frac{1}{2\lambda}\left(\ln\left[\frac{8\beta\lambda^2}{3(4\alpha-\lambda^2+2)}\right]+W\left[\frac{3\alpha(4\alpha-\lambda^2+2)}{4\beta\lambda^2} e^{\frac{36\alpha^2+8N\beta(3-8\beta)\lambda^4-9\alpha(\lambda^2-2)}{12\beta\lambda^2}}\right]\right)\\ &
+\frac{9\alpha\left(\lambda^2-4\alpha-2\right)+8N\beta\lambda^4(8\beta-3)}{24\beta\lambda^3}
\end{aligned}
\end{equation}
writing the scalar spectral index in terms of the scalar field from (\ref{slr29}) and using  (\ref{epsilonkgb}) we find
\begin{equation}\label{nskgb}
n_s=1+\frac{\left(8\beta e^{-2\lambda\phi}-1\right)\lambda^2}{2\alpha+1}
\end{equation}
and for the tensor-to-scalar ratio from (\ref{slrt16}) and (\ref{epsilonkgb}) we find
\begin{equation}\label{rkgb}
r=\frac{8\lambda^2\left(3 e^{2\lambda\phi}+8\beta\right)^2 e^{-4\lambda\phi}}{9(2\alpha+1)}.
\end{equation}
At the horizon crossing, $N$ $e$-folds before the end of inflation, we find the following expressions for $n_s$ and $r$
\begin{equation}\label{nskgbhc}
n_s=1-\frac{\lambda^2}{2\alpha+1}+\frac{4\beta\lambda^2}{\alpha(2\alpha+1)}W\left[\frac{3\alpha(4\alpha-\lambda^2+2)}{4\beta\lambda^2} e^{\frac{36\alpha^2+8N\beta(3-8\beta)\lambda^4-9\alpha(\lambda^2-2)}{12\beta\lambda^2}}\right],
\end{equation}
and
\begin{equation}\label{rkgbhc}
r=\frac{8\lambda^2}{9\alpha^2(2\alpha+1)}\left(3\alpha+4\beta W\left[\frac{3\alpha(4\alpha-\lambda^2+2)}{4\beta\lambda^2} e^{\frac{36\alpha^2+8N\beta(3-8\beta)\lambda^4-9\alpha(\lambda^2-2)}{12\beta\lambda^2}}\right]\right)^2
\end{equation}
Notice that setting $\beta=0$ we obtain the previous results (\ref{nsreternal}) for $n_s$ and $r$. Evaluating the slow-roll parameters at the end of inflation (under the condition $\epsilon=1)$ we find
\begin{equation}\nonumber
\epsilon_0=1,\epsilon_1=\frac{6(\lambda^2-4\alpha-2)}{6\alpha+3},\;\; \Delta_0=\frac{2(\lambda^2-4\alpha-2)}{\lambda^2}
\end{equation}
\begin{equation}\label{epsilonkgbhc}
\Delta_1=\frac{2(\lambda^2-8\alpha-4)}{2\alpha+1},\;\; k_0=\frac{4\alpha}{\lambda^2},\;\; k_1=\frac{4(\lambda^2-4\alpha-2)}{2\alpha+1}
\end{equation}
Looking at the expressions (\ref{nskgbhc}) and (\ref{rkgbhc}) it can be seen that to obtain the observable values for $n_s$ and $r$, $\lambda$ should be small or of the order $1$ and $\alpha$ should be large. But, according to (\ref{epsilonkgbhc}), this will make $\Delta_0>>1$ and $k_0>>1$, meaning that they become of the order $1$ long before the end of inflation, spoiling the slow-roll approximation. Therefore it is not possible to satisfy the condition that all slow-roll parameters maintain in the region of $\pm 1$ at the end of inflation, for appropriate values of $\lambda$ and $\alpha$. Better results are obtained with the following model.\\

\noindent {\bf Kinetic and Gauss-Bonnet couplings II.}\\
\begin{equation}\label{expokgb1}
F\left( \phi  \right) = \frac{1}{{{\kappa ^2}}},\,\,\,\,\,\,\,\,\,V\left( \phi  \right) = V_0 e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,\,{F_1}\left( \phi  \right) =f_k e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,{F_2}\left( \phi  \right) = f_{g} e^{\lambda\kappa\phi}.
\end{equation}
where the slow-roll parameters in terms of the scalar field take the form
\begin{equation}\nonumber
\epsilon_0=\frac{(3-8\beta)\lambda^2 e^{2\lambda\phi}}{6(e^{2\lambda\phi}+2\alpha)},\;\;\; \epsilon_1=\frac{4\alpha(3-8\beta)\lambda^2 e^{2\lambda\phi}}{3(e^{2\lambda\phi}+2\alpha)^2}
\end{equation}
\begin{equation}\nonumber
\Delta_0=\frac{8\beta(3-8\beta)\lambda^2 e^{2\lambda\phi}}{9(e^{2\lambda\phi}+2\alpha)},\;\;\; \Delta_1=\frac{4\alpha(3-8\beta)\lambda^2 e^{2\lambda\phi}}{3(e^{2\lambda\phi}+2\alpha)^2},
\end{equation}
\begin{equation}\label{epsilonkgb1}
k_0=\frac{\alpha(3-8\beta)^2\lambda^2 e^{2\lambda\phi}}{9(e^{2\lambda\phi}+2\alpha)^2},\;\;\; k_1=-\frac{2(3-8\beta)\lambda^2 e^{2\lambda\phi}(e^{2\lambda\phi}-2\alpha)}{3(e^{2\lambda\phi}+2\alpha)^2}
\end{equation}
The end of inflation takes place for the scalar field $\phi_E$ given by
\begin{equation}\label{phiendkgb1}
\phi_E=\frac{1}{2\lambda}\ln\left[\frac{12\alpha}{\lambda^2(3-8\beta)-6}\right].
\end{equation}
The number of $e$-folds, from  (\ref{eqm17}) is given by
\begin{equation}\label{efoldskgb1}
N=\frac{3}{\lambda^2(3-8\beta)}\left(\lambda\phi-\alpha e^{-2\lambda\phi}\right)\Big|_{\phi_I}^{\phi_E}.
\end{equation}
Replacing $\phi_E$ and solving with respect to $\phi_I$ we find
\begin{equation}\label{phiinikgb1}
\begin{aligned}
\phi_I=&\frac{1}{12\lambda}\Big(6\ln\left[\frac{12\alpha}{\lambda^2(3-8\beta)-6}\right]+6W\left[\frac{1}{6}\left((3-8\beta)\lambda^2-6\right) e^{\frac{1}{6}(3-8\beta)(1+4N)\lambda^2-1}\right]\\ &+\lambda^2\left(32\beta N-12N+8\beta-3\right)+6\Big).
\end{aligned}
\end{equation}
The scalar spectral index in terms of the scalar field is given by the following expression (from (\ref{slr29}) and (\ref{epsilonkgb1}))
\begin{equation}\label{nsphikgb}
n_s=\frac{12\alpha^2+6\alpha((8\beta-3)\lambda^2+2)e^{2\lambda\phi}+((8\beta-3)\lambda^2+3))e^{4\lambda\phi}}{3(e^{2\lambda\phi}+2\alpha)^2},
\end{equation}
and for the tensor/scalar ratio (using (\ref{slrt16}) and (\ref{epsilonkgb1})) it is found
\begin{equation}\label{rphikgb}
r=\frac{8\lambda^2(3-8\beta)^2 e^{2\lambda\phi}}{9(e^{2\lambda\phi}+2\alpha)}.
\end{equation}
The observed values of $n_s$ and $r$ are found through the evaluation of the above expressions $N$ $e$-foldings before the end of inflation, leading to
\begin{equation}\label{nskgb1}
n_s=1-\frac{(3-8\beta)\lambda^2\Big(1+3W\left[\frac{1}{6}\left((3-8\beta)\lambda^2-6\right) e^{\frac{1}{6}(3-8\beta)(1+4N)\lambda^2-1}\right]\Big)}{3\Big(1+W\left[\frac{1}{6}\left((3-8\beta)\lambda^2-6\right) e^{\frac{1}{6}(3-8\beta)(1+4N)\lambda^2-1}\right]\Big)^2}
\end{equation}
and
\begin{equation}\label{rkgb1}
r=\frac{8\lambda^2(3-8\beta)^2}{9\Big(1+W\left[\frac{1}{6}\left((3-8\beta)\lambda^2-6\right) e^{\frac{1}{6}(3-8\beta)(1+4N)\lambda^2-1}\right]\Big)}
\end{equation}
Notice that in fact the dependence of $n_s$ and $r$ on $\alpha$ disappears when evaluated at the horizon crossing. Replacing (\ref{phiendkgb1}) into
(\ref{epsilonkgb1}) we find the expressions for the slow-roll parameters at the end of inflation
\begin{equation}\label{epsikgbend}
\begin{aligned}
\epsilon_0=1,\;\; \epsilon_1=&4+\frac{24}{(8\beta-3)\lambda^2},\;\; \Delta_0=\frac{16\beta}{3},\;\; \Delta_1=4+\frac{24}{(8\beta-3)\lambda^2},\\
&  k_0=1-\frac{2}{\lambda^2}-\frac{8\beta}{3},\;\; k_1=4+\frac{48}{(8\beta-3)\lambda^2}.
\end{aligned}
\end{equation}
Analyzing these results, it can be seen that it is possible to find values $\epsilon_1, \Delta_0, ...\sim 1$ (guaranteeing graceful exit from inflation), assuming $|f_2|<<1$ and $\lambda\sim 1$, which at the same time give adequate values for the observables $n_s$ and $r$. In Fig. 4 we show the evolution of $n_s$ and $r$ for the number of $e$-foldings in the interval $50\le N\le 60$.
\begin{figure}
\centering
\includegraphics[scale=0.7]{ns-r-kgb.pdf}
\caption{The scalar spectral index $n_s$ and tensor/scalar ratio $r$, for  $\lambda=1.42,\; \beta=-0.01$ (blue) and  $\lambda=1.42,\; \beta=-0.001$ (red), for $N$ varying between $50\le N\le 60$. $n_s$ and $r$ do not depend on $\alpha$, which is used to set the values of $\phi_I$ and $\phi_E$.}
\end{figure}
taking for instance $N=60,\; \lambda=1.42,\; \alpha=10^3,\; \beta=-0.001$, the slow-roll parameters at the beginning of inflation take the values
$$
\epsilon_0\simeq 0.0043,\; \epsilon_1\simeq 0.017,\; \Delta_0\simeq -0.000023,\; \Delta_1\simeq 0.017,\; k_0\simeq 0.0043,\; k_1\simeq 0.017.
$$
\noindent Following the same lines as in the previous cases, we can evaluate the size of the Hubble parameter and the energy involved during inflation, obtaining that $H\sim 3\times 10^{-5} M_p$ and $V^{1/4}\sim 7\times 10^{-3} M_p$ (taking into account the above slow-roll parameters).
The evolution of the slow-roll parameters for this case is shown in Fig. 5, where $\phi_I\simeq 0.76 M_p$ and $\phi_E\simeq 4.2 M_p$.
 \begin{figure}
\centering
\includegraphics[scale=0.7]{slow-roll1.pdf}
\caption{The variation of the slow-roll parameters between $\phi_I\simeq 0.76 M_p$ and $\phi_E\simeq 4.2 M_p$ obtained for $N=60,\; \lambda=1.42,\; \alpha=10^3,\; \beta=-0.001$.}
\end{figure}
Observing Fig. 5 we can see that the slow-roll dynamics can also be consistent if one imposes the condition to end the inflation on the slow-roll parameters $\epsilon_1=\Delta_1=1$. This leads (from (\ref{epsilonkgb1})) to the following scalar field at the end of inflation
\begin{equation}\label{phiendkgb2}
\phi_E=\frac{1}{2\lambda}\ln\left[\frac{2}{3}\left(\alpha((3-8\beta)\lambda^2-3)+\sqrt{\alpha^2\lambda^2(8\beta-3)(6+(8\beta-3)\lambda^2)}\right)\right]
\end{equation}
Then, from (\ref{efoldskgb1}) and replacing $\phi_E$ given by (\ref{phiendkgb2}) it is found
\begin{equation}\label{phiikgb2}
\phi_I=\frac{1}{3}(8\beta-3)(N-f_N)\lambda+\frac{1}{2\lambda}W\left[2\alpha e^{-\frac{2}{3}(8\beta-3)(N-f_N)\lambda^2}\right],
\end{equation}
which leads, from (\ref{nsphikgb}) and (\ref{rphikgb}), to the following expressions for $n_s$ and $r$
\begin{equation}\label{nskbg2}
n_s=1-\frac{(3-8\beta)\lambda^2\Big(1+3W\Big[2\alpha e^{-\frac{2}{3}(8\beta-3)(N-f_N)\lambda^2}\Big]\Big)}{3\Big(1+3W\Big[2\alpha e^{-\frac{2}{3}(8\beta-3)(N-f_N)\lambda^2}\Big]\Big)^2}
\end{equation}
and
\begin{equation}\label{rkgb2}
r=\frac{8\lambda^2}{9\alpha^2(2\alpha+1)}\left(3\alpha+4\beta W\left[2\alpha e^{-\frac{2}{3}(8\beta-3)(N-f_N)\lambda^2}\right]\right)^2
\end{equation}
Fig. 6 shows the behavior of  $n_s$ and $r$ for $50\le N\le 60$.
\begin{figure}
\centering
\includegraphics[scale=0.7]{ns-r-kgb2.pdf}
\caption{The scalar spectral index $n_s$ and tensor/scalar ratio $r$, for  $\lambda=1.42,\; \alpha=10^3,\; \beta=-0.1$ (blue) and  $\lambda=1.42,\; \alpha=10^3, \; \beta=-0.05$ (red), for $N$ in the interval $50\le N\le 60$. The tensor/scalar ratio is an order of magnitude smaller than the case depicted in Fig.4.}
\end{figure}
The variation of the slow-roll parameters between the beginning and the end of inflation is shown in fig 7.
\begin{figure}
\centering
\includegraphics[scale=0.7]{slow-roll2.pdf}
\caption{The variation of slow-roll parameters between $\phi_I\simeq 0.67 M_p$ and $\phi_E\simeq 3 M_p$ obtained for $60$ $e$-foldings, with $\lambda=1.42,\; \alpha=10^3,\; \beta=-0.1$. The growth towards values of the order of $\pm 1$ is more homogeneous than the one depicted in Fig.5}
\end{figure}
The values of the slow-roll parameters $60$ $e$-folds before the end of inflation ($\lambda=1.42,\; \alpha=10^3,\; \beta=-0.1$) are
$$
\epsilon_0\simeq 0.0042,\; \epsilon_1\simeq 0.017,\; \Delta_0\simeq -0.0023,\; \Delta_1\simeq 0.017,\; k_0\simeq 0.005,\; k_1\simeq 0.017.
$$
\noindent The Hubble and energy scales involved in the process of inflation, for the present case, are $H\sim 3\times 10^{-5} M_p$ and $V^{1/4}\sim 4\times 10^{-3} M_p$ .\\

\noindent {\bf Kinetic and Gauss-Bonnet couplings III.}\\
\begin{equation}\label{expokgb1}
F\left( \phi  \right) = \frac{1}{{{\kappa ^2}}},\,\,\,\,\,\,\,\,\,V\left( \phi  \right) = V_0 e^{-\lambda\kappa\phi},\,\,\,\,\,\,\,\,\,{F_1}\left( \phi  \right) =f_k e^{\lambda\kappa\phi},\,\,\,\,\,\,\,\,{F_2}\left( \phi  \right) = f_{g} e^{\lambda\kappa\phi}.
\end{equation}
This model leads to exact power-law inflation with the constant slow-roll parameters given by
\begin{equation}\nonumber
\epsilon_0=\frac{(3-8\beta)\lambda^2}{6(\alpha+1)},\;\; \epsilon_1=0,\;\; \Delta_0=\frac{8\beta(3-8\beta)\lambda^2}{9(2\alpha+1)}
\end{equation}
\begin{equation}\label{epsilonpwl}
\Delta_1=0,\;\; k_0=\frac{\alpha(3-8\beta)^2\lambda^2}{9(2\alpha+1)^2},\;\; k_1=0,
\end{equation}
which predict the scalar spectral index and tensor/scalar ratio given by the following expressions
\begin{equation}\label{nsr-kgb}
n_s=1-\frac{(3-8\beta)\lambda^2}{3(2\alpha+1)},\;\;\; r=\frac{8(3-8\beta)^2\lambda^2}{9(2\alpha+1)}.
\end{equation}
These equations can be solved with respect to $\alpha$ and $\beta$, resulting in
\begin{equation}\label{couplings}
 \alpha=\frac{16n_s-8n_s^2+\lambda^2 r-8}{16(n_s-1)^2},\;\;\; \beta=\frac{3(8n_s+r-8)}{64(n_s-1)}
 \end{equation}
Thus, for a given $\lambda$ we can always find adequate values for $\alpha$ and $\beta$ that satisfy the observed values of $n_s$ and $r$. Taking for instance, $\lambda=2$, $ns=0.968$ and $r=0.01$, the reconstructed couplings acquire the values $\alpha=1.94$ and $\beta=0.36$, and the slow-roll parameters take the values $\epsilon_0\simeq 0.016,\; \Delta_0\simeq 0.0031,\; k_0\simeq 0.00005,\; \epsilon_1=\Delta_1=k_1=0 $. Taking $\lambda=0.5$ gives $\alpha\simeq 304,\; \beta\simeq 0.3$ and the slow-roll parameters  $\epsilon_0\simeq 4\times 10^{-5},\; \Delta_0\simeq 6.4\times 10^{-5},\; k_0\simeq 7.8\times 10^{-6},\; \epsilon_1=\Delta_1=k_1=0 $.

\section{Discussion}
We have analyzed the slow-roll dynamics for the scalar-tensor model with non-minimal kinetic and GB couplings, where the potential and the functional form of the couplings are given by exponential functions of the scalar field. These type of couplings appear in a number of compactifications from higher dimensional fundamental theories  such as supergravity and string theory, where the scalar field encodes the size of the extra dimensions.
In the frame of the standard canonical scalar field, the exponential potential leads to important scaling solutions that describe different epochs of cosmological evolution, including solutions with late time accelerated expansion. It also leads to early time inflationary solutions, though it lacks successful exit  from inflation and leads to tensor-to-scalar ratio larger than the current observational limits.
With the Introduction of additional interactions like the non-minimal kinetic coupling and GB coupling (GB), we address the above shortcomings of the exponential potential and show that the tensor-to-scalar ratio can be lowered to values that are consistent with latest observational constraints \cite{planck15, planck18} and that the model leads to a graceful exit from inflation.\\
First we considered a model with potential $V_0 e^{-\kappa\lambda\phi}$ and kinetic coupling $f_k e^{-\kappa\eta\phi}$ and have found that the observable magnitudes $n_s$and $r$ do not depend on $\alpha=V_0f_k$, and depend only on the number of $e$-foldings and the exponential powers $\lambda$ and $\eta$. The constants $\alpha$ and $\eta$ can be used to set the values of the scalar field at the end and beginning of inflation, obtaining that $\phi_E\lesssim M_p$. A typical behavior of $n_s$ and $r$ in this case is shown in Fig.1. In the particular case $\eta=-\lambda$, the slow-roll parameters become constant, but the obtained relationship between $n_s$ and $r$ (\ref{nsvsr}) makes it imposible to simultaneously satisfy the observational restrictions, making the model non viable for $\eta=-\lambda$.
In the second case we considered the GB coupling given by $F_2=f_g e^{-\kappa\eta\phi}$, and it was found that, similar to the previous case, neither $n_s$ nor $r$ depend on $\beta=V_0 f_g$, but this parameter can be used to set  $\phi_E$ and $\phi_I$. Considering the region of parameters where $e^{-\lambda N(\lambda+\eta)}\sim 1$ it was found that, for $50\le N\le 60$, $n_s$ and $r$ can take values in the intervals $0.961\le n_s\le 0.967$ and $0.002\le r\le 0.003$, and the scalar field at the end of inflation can be as small as $\phi_E\sim 0.07 M_p$. However, in this case some of the slow-roll parameters become larger than $1$ long before $\epsilon_0\sim 1$, breaking the slow-roll conditions. To fix this problem we have chosen to break the slow-roll conditions when $\Delta_0=-1$, which gives excellent results as sown in Fig. 3 and is consistent with the slow-roll formalism according to the values obtained in (\ref{sample}) and (\ref{sample1}). Considering the case $\eta=-\lambda$ it was found that it leads to constant slow-roll inflation, but contrary to the case of kinetic coupling, it is always possible to find adequate values for the scalar spectral index and the tensor-to-scalar ratio.
In the model with non-minimal kinetic coupling $F_1=f_k e^{\kappa\lambda\phi}$ and GB coupling $F_2=f_g e^{-\kappa\lambda\phi}$ it was found that in order to obtain viable values of $n_s$ (\ref{nskgbhc}) and $r$ (\ref{rkgbhc}), the conditions $\lambda\lesssim 1$ and $\alpha>>1$  should be satisfied, but this imply according to (\ref{epsilonkgbhc}), that some slow-roll parameters reach values $\sim 1$ long before the end of inflation, spoiling the slow-roll approximation. Better result is obtained with the model  $F_1=f_k e^{-\kappa\lambda\phi}$ and $F_2=f_g e^{\kappa\lambda\phi}$, where there is appropriate slow-roll approximation for values of $\lambda$ and $\beta$ that lead to $n_s$ and $r$ in the range quoted by observations, as seen in Figs. 4 and 5.
But more appropriate behavior of the slow-roll parameters was found if the condition to end the inflation is assumed as $\epsilon_1=\Delta_1=1$. In the proposed numerical example, the scalar-to-tensor ratio decreases to values $r\sim 0.008$ as shown in Fig. 6, and the growth of the slow-roll parameters toward values of the order of $\pm 1$ at the end of inflation is more homogeneous than in the previous case, as seen in Fig. 7. Finally, the model with $V=V_0 e^{-\kappa\lambda\phi}$, $F_1=f_k e^{\kappa\lambda\phi}$ and $F_2=f_g e^{\kappa\lambda\phi}$ was analyzed. This model leads to inflation with constant slow-roll parameters and, as follows from (\ref{nsr-kgb}) and (\ref{couplings}), it is always possible to find adequate values of $\lambda$, $\alpha$ and $\beta$ that give the observational values of $n_s$ and $r$.
In all models considered in the present paper it was possible to find exact analytical expressions for the scalar spectral index and the tensor-to-scalar ratio, which facilitated the analysis. In all considered numerical examples the Hubble and energy scales involved in the inflationary process were of the order of $H\sim 10^{-5} M_p$ and $V^{1/4}\sim 10^{-3}$. \\
The slow-roll analysis for the exponential potential, in the frame of the scalar-tensor theories with non-minimal kinetic and GB couplings, allows to find the scalar spectral index and tensor-to-scalar ratio in the range set by the latest observational data, and lead to successful exit from inflation.
The advance in the future observations will allow to establish more accurate restrictions on the inflationary models with non-minimal couplings of the type considered in the present model and reaffirm or rule out its viability.

\section*{Acknowledgments}
\noindent This work was supported by Universidad del Valle under project CI 71187.
 DFJ acknowledges support from COLCIENCIAS, Colombia.

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Tyson Foods and Auburn University partner to build first stand-alone solar-powered poultry house
Off-the-grid broiler poultry house will be key to informing sustainable energy practices at scale
Auburn University's National Poultry Technology Center (NPTC), a leader in poultry housing and associated technologies for more than a decade, and Tyson Foods Inc. (NYSE: TSN) today announced the opening of the largest stand-alone solar powered poultry house to be operated completely off the grid. The 54-foot by 500-foot poultry house is located in Cullman County, Alabama, and capable of housing 36,000 broilers.
The poultry house will be one of two identical houses on Tim and Selena Butts' farm where 5.50-pound broilers will be grown. One house will be the control house while the other will be operated exclusively by solar power, also known as Stand-Alone Solar for Poultry (SASP).
"Auburn University's NPTC will work closely with Tyson Foods and Southern Solar Systems to provide leadership in the application of solar power technology to broiler production houses," said Paul Patterson, dean of Auburn's College of Agriculture. "The research will provide important, new information on how solar power technology can improve environmental sustainability and profits for farmers."
Introducing the largest solar-powered poultry house to be built completely off-the-grid
The house's power will derive from three components: the photovoltaic (PV) panel or solar cell, a battery set and a generator. On-site researchers will compare its energy use regularly with the normal operation of the twin house located next door over a 12-month cycle.
The data and insights gleaned from this project will be an important next step in identifying sustainable practices and new forms of energy for the poultry industry at large.
"Ultimately, this project will allow us to identify how solar houses might improve farmer profitability and bring increased efficiency to the poultry industry," said Chip Miller, vice president of poultry live operations for Tyson Foods. "Through our partnership with Auburn University's NPTC, we are creating a model for the future of the industry—one that is more sustainable and brings critical value and insights, previously unavailable, to poultry farmers."
"The combination of solar and batteries along with the other technologies are converting power to usable alternating current (AC) that's identical to grid power," said Dennis Brothers, extension specialist with NPTC. "Electricity drives all functions in poultry houses and is the largest variable cost for poultry farmers. We believe this new system may reduce costs for farmers while increasing efficiency."
Building a competitive system
The rising cost of electricity coupled with the unpredictability of long-term grid power has created an opportunity for Tyson Foods to explore solutions to help alleviate the effect of climbing prices.
"Looking ahead, we are eager to evaluate the efficacy of the solar house and its impact on farmer profitability," concluded Miller. "We expect this pilot to be the first of many, as we continue to leverage the power of collaboration to drive progress in the poultry industry."
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Outdoor Industrial Trash Bags For Pet Waste Stations - 10 Gallon Bags!
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Trash Can Liners for Receptacles from Pet Waste Eliminator are a necessary part of your Dog Waste Disposal Station! Fits our 10 Gallon Steel Trash Receptacles (Item E10, sold separately).


text:
// Stub class generated by rmic, do not edit.
// Contents subject to change without notice.

public final class CheckBoomerangs_Stub
    extends java.rmi.server.RemoteStub
    implements Boomerang, java.rmi.Remote
{
    private static final long serialVersionUID = 2;

    private static java.lang.reflect.Method $method_getObject_0;
    private static java.lang.reflect.Method $method_setChildPassed_1;
    private static java.lang.reflect.Method $method_setObject_2;

    static {
	try {
	    $method_getObject_0 = Boomerang.class.getMethod("getObject", new java.lang.Class[] {});
	    $method_setChildPassed_1 = Boomerang.class.getMethod("setChildPassed", new java.lang.Class[] {boolean.class});
	    $method_setObject_2 = Boomerang.class.getMethod("setObject", new java.lang.Class[] {java.lang.Object.class});
	} catch (java.lang.NoSuchMethodException e) {
	    throw new java.lang.NoSuchMethodError(
		"stub class initialization failed");
	}
    }

    // constructors
    public CheckBoomerangs_Stub(java.rmi.server.RemoteRef ref) {
	super(ref);
    }

    // methods from remote interfaces

    // implementation of getObject()
    public java.lang.Object getObject()
	throws java.rmi.RemoteException
    {
	try {
	    Object $result = ref.invoke(this, $method_getObject_0, null, -8225975311538388041L);
	    return ((java.lang.Object) $result);
	} catch (java.lang.RuntimeException e) {
	    throw e;
	} catch (java.rmi.RemoteException e) {
	    throw e;
	} catch (java.lang.Exception e) {
	    throw new java.rmi.UnexpectedException("undeclared checked exception", e);
	}
    }

    // implementation of setChildPassed(boolean)
    public void setChildPassed(boolean $param_boolean_1)
	throws java.rmi.RemoteException
    {
	try {
	    ref.invoke(this, $method_setChildPassed_1, new java.lang.Object[] {new java.lang.Boolean($param_boolean_1)}, 4082040846063646613L);
	} catch (java.lang.RuntimeException e) {
	    throw e;
	} catch (java.rmi.RemoteException e) {
	    throw e;
	} catch (java.lang.Exception e) {
	    throw new java.rmi.UnexpectedException("undeclared checked exception", e);
	}
    }

    // implementation of setObject(Object)
    public void setObject(java.lang.Object $param_Object_1)
	throws java.rmi.RemoteException
    {
	try {
	    ref.invoke(this, $method_setObject_2, new java.lang.Object[] {$param_Object_1}, 456464208915235554L);
	} catch (java.lang.RuntimeException e) {
	    throw e;
	} catch (java.rmi.RemoteException e) {
	    throw e;
	} catch (java.lang.Exception e) {
	    throw new java.rmi.UnexpectedException("undeclared checked exception", e);
	}
    }
}

redpajama_set_name: RedPajamaGithub


Here's a LIVE set I rocked recently at the Hip Hop Barbershop – just over an hour of me blending and scratching old school classics, with a couple of more recent songs, remixes & mashups thrown in the mix! Video above, audio & tracklist below.
Shout to DJ Pavlos & Porters Barbers for getting me down!
21. Busta Rhymes – Woo Hah!!


Offered for sale is my 1993 Ford F-350 Crew Cab Longbed 4x4 with 125305 miles. This is one of the cleanest and most rust free Ford trucks of this vintage I have ever seen. When is the last time you saw one stock and unmodified? The pictures tell the entire story of this truck. I purchased it as government surplus where it was fleet maintained and only used seasonally (not used on salty winter roads!). I was the first person to title the truck and it has not seen salty winter roads since I've owned it either. This is not your typical beat on work truck, it is clear whoever drove it looked after it and took care of it. Accordingly as you can see while it is certainly not perfect, this is an extremely nice straight and clean older ford without any typical Ford rot these trucks are known for.
F-350 has the 351 V8 and automatic transmission. Runs/drives/stops great and very reliably. I made a 400+ mile trip towing a boat earlier this year and I take it on my 130 mile commute once a month to keep it active. In the past year I have put in new front rotors, new rear brake line, new starter, new battery, new wiper motor. Starts every time even after sitting for a little while. Both gas tanks and switch work great. The only issue I know of currently is that occasionally you need to push up on the column shift while turning the key to get it to start. Never bothered me and doesn't happen a lot so I never did anything to fix.
Passed MA inspection without issue, check engine light is on but its pre OBD so I covered with electrical tape and forgot about it, just trying to be transparent as possible. I replaced one tire last year because it developed a bump in the sidewall, tire shop advised other tires did not need to be replaced at that time, have probably put less than 2500 miles on it since then. I would feel confident driving this truck anywhere.
The pictures tell the story of the condition better than I can hope to. The truck is unbelievably straight and rust free considering it's age. I did not cover everything in tire shine and undercoat the whole underside black like most trucks you see on eBay so you can get an accurate idea of the actual condition of this truck. The spray painted areas were done by myself last year purely as a preventative measure as the paint had worn through to primer. I sanded a small area of surface rust on the hood and a small area of surface rust on the rear passenger side door but otherwise there was zero rust under the painted areas. Note the dent in drivers door jamb area and underneath drivers side headlight. I was planning on repainting the whole truck. I believe the paint on it is otherwise original and is showing age as you can see.
All of the lights on the truck work including trailer connection. Has a trailer receiver and tows awesome thanks to the long wheelbase. Four wheel drive engages as it should and works well. I've never done more with 4 wheel drive engaged than move it around in my driveway in the snow but that was enough to make it clear this truck will not be stopped in 4x4. Heat and blower motor work great, a/c does not blow cold, may need a charge. Radio with Aux Connection works great.
All in all this is a super cool and unique old truck that is very usable as is or ready for a paint job if you want to take it to the next level. If you're still reading this you know you cannot find these trucks with low original miles without a ton of rust that have not been beat on, especially in the northeast! I think this truck will make it's next owner very happy, it kills me to sell it but since I purchased this I replaced my commuter vehicle with a newer truck and don't use this as much as I would like.
Terms of Sale: I'm selling the F-350 as-is, where is with no warranty expressed or implied and no returns. I require a $500 deposit via Paypal within 24 hours and the balance by either cash, certified bank check or bank wire. Truck is sold on a Massachusetts title in my name. Under 15 bidders please contact me before bidding or your bid will be canceled. Shipping is the responsibility of the buyer, truck needs to be picked up within two weeks of auction end. Your bid signifies your understanding of and agreement to these terms. Thanks.
1992 FORD F350 CREW CAB XLT LONG BED 4X4 7.5L V8 GAS ONLY 41,338 ORIGINAL MILES!


Many people never give eye care.
If eye problems seem to run in your family, visit an eyecare professional. Diseases need to be brought to the attention of your optometrist. Knowing about an illness in the early stages is always a benefit.
Always wear sunglasses to keep your eyes protected from the ever present threat of the sun. Get a quality pair with UV lenses and take them everywhere with you. The sun's radiation can harm your eyes and the skin around them. Don't take unnecessary chances when it comes to your eyes.
Sunglasses are not created equal, with some doing nothing to offer eye protection. Make sure that they block all UVA rays as well as all UVB rays. Regardless of whether you want to be stylish or not, sunglasses have a purpose, which is to protect your vision.
You can actually prevent certain issues with your eyes. Studies show that food that contain Omega-3 fatty acids, zinc, along with other eye maladies. Salmon, nuts, beans, beans, tuna and leafy, and oranges are just a few of the foods that have these helpful nutrients.
Your daily diet can have much to do with taking proper care of your eyes. Studies have shown that eating foods that are abundant in Vitamins E and C, zinc, and omega-3 fatty acids can help to prevent macular degeneration, cataracts, and other maladies. Foods that are rich in these nutrients are leafy green vegetables, nuts, beans, oranges, tuna and salmon.
You should always be aware of any eye diseases that your family tree. Some conditions concerning eyes are hereditary; knowing what your risk is can help with treatment.Ask your family members so you will know.
Be sure you know whether or not eye disease runs in your family. A lot of the time these things are hereditary and can be helped if you go to a doctor. Speak with older family members to know about any history with eye problems.
Omega-3 fatty acids are great for excellent eye care. It is a good idea to make use of foods that contain these acids. Some of these foods are halibut, dark, salmon, and salmon. Get a serving per day you can.
If you blink often, it might not necessarily be your eyes. Dry eyes or a nervous tic can cause frequent blinking. If this happens, try to relax. If this does not work, consult a professional.
Help your eyes out by wearing a pair of good sunglasses. These glasses can serve as a shield for the sun's rays. Get sunglasses that block out 100% of the UVA and UVB rays. If you enjoy driving, polarized lenses are important. They help to reduce glare. No matter if your contacts come equipped with UV protection or not, sunglasses should be worn.
Your eyes need to be checked on a regularly basis by a doctor. This is why you should always have regular checkups are important. Some conditions can be remedied if they are easily treated early enough.
If you frequently use a computer, keep in mind this can harm your eyes. Blink your eyes often if you are suffering from dry eyes. Cut down your screen glare too. It is easy to find anti-glare screens. Finally, position the monitor so that your eyes are even with the top edge of the screen. Your view should be looking down towards the screen.
If you blink a lot you may have an eye problem.If you're eyes aren't dry, it just may be the result of a nervous tic you are experiencing. If you are sure it is not just a tic, consult with your ophthalmologist.
Always have some saline solution at home. It's easy to protect your eyes if you wear goggles. However, most people don't wear goggles for cleaning up in the home. If any cleaning chemicals or soap gets into the eyes, you will need to wash eyes with the saline.
You should have eye exams regularly to maintain great eye health. As you age your eyes should get looked at more often. As you begin to age, your risk of cataracts and glaucoma increases. Keep informed at all times about the health of your eyes.
Use eye drops sparingly if you need them at all. They can relieve your eyes, but overuse leads to other issues. If eye drops aren't adequately soothing your eyes, see an eyecare professional to determine if other treatments might help more.
Know your family eye health history. A lot of conditions are hereditary. Because of this, it is important information to have about your family history when it concerns eye care health. This will allow you to follow a plan that will diminish your chances of facing problems because of early detection.
Replace the makeup you use every two months. The reason is that bacteria builds up when you keep the eyes leading to damage. After a few months, you could be rubbing bacteria onto your eyes every time you use your makeup. This risks all kinds of harm to your eyes and their surrounding skin.
As you age, it's normal for eyes to become dry more frequently. Omega-3 foods are an important part of your diet. Keep in mind that cold or hot air can make the issue worse. Keep your car vents pointed in the other direction from your eyes, and do not position yourself near fans or vents at work.
Stop smoking today if you want good long-term health for your eyes. Smoking increases your likelihood of developing cataracts, optic nerve damage and other health issues. If you've unsuccessfully tried giving up the habit in the past, quit again.
Stop smoking today if you want to keep your eyes healthy over the long term. Smoking makes it much more likely that you will have either optic nerve damage, cataracts or macular degeneration. If you have been unsuccessful when you tried to quit smoking, try again. You are more likely to succeed your second time at quitting than your first time.
Walking around every now and better your blood flowing.
If your eyelids are puffy, you can use cucumber slices to help prevent water retention. Cut a cold cucumber into quarter-inch slices, then place on the eyelids for 10 minutes or so as you relax with the eyes closed. Soaking green tea bags in water is great for reducing eye puffiness along with inflammation.
Cucumber slices can help with puffy swollen eyes to prevent water retention.You can reduce inflammation with green tea bags; just allow them to sit in cool water.
Take lots of breaks if you work using a computer. Your eyes need to rest so that can re-energize and remain healthy. Go out for a drive or eat lunch with your friends to get the break that you require.
It helps relieve any pain or irritation caused by redness you may have. It can also gives a burst of energy to help wake you up.
If your eyes get irritated and dry while you sleep, try an eye ointment. Ointments are great for lubrication and last longer as they aren't water based. Only use prior to bedtime due to potential blurriness.
If your eyelids are inflamed from oil, oils or debris, then have and eye scrub handy. This can help loosen particles and bacterial infection. Use this scrub when you have irritated eyes or as a preventative measure.
Care for your contact lenses properly. Studies have proven that a huge number of wearers do not care for their lenses the right way. Of course this situation can cause an infection as well as vision loss and eye irritation. Always use the proper solution. Your mouth contains bacteria that can result in infections. Plus, rest your eyes from wearing contacts and wear your glasses once a week.
Now that you completely understand all about how to care for your eyes, you can get to work. You should make use of this information each day to make sure your eyes are serving you for a lifetime. By altering your daily habits, you can enjoy clear vision for years to come.
Utilize a scrub formulated for the eyes if you have trapped particles. This product can get rid of dirt or soothe dry skin quickly and easily. Use this scrub when you feel irritated skin around your eyes, or regularly, to prevent irritation.


text:
10 Resources to Help You Become an Architect
4/29/19 by Salo Aburto
The road to becoming an architect is less complicated than you might think, and NCARB is here to help. We have the guidance and resources you'll need to advance from student to practicing architect.
Before you can call yourself an architect, you have to earn a license by completing your state's specific requirements. Each of the 55 U.S. licensing boards sets requirements for licensure in their jurisdiction, but they all include three core components: education, experience, and examination.
Learn more about the basics of becoming an architect by familiarizing yourself with the "three E's."
To become an architect, you'll need to meet your jurisdiction's education requirement. Most licensing boards require that architects hold a professional degree from a program accredited by the National Architectural Accrediting Board (NAAB).
NAAB-accredited Programs: You can search the NAAB's list of accredited programs, which include Bachelor of Architecture, Master of Architecture, and Doctor of Architecture programs.
Integrated Path to Architectural Licensure (IPAL): IPAL enables students to gain real-world experience and take the national licensing exam while in school. Students who graduate from IPAL options can be eligible for licensure soon after graduation.
NCARB Record: Your NCARB Record is a detailed, verified account of your education, experience, and examination history. Over half of students start their Record in school, so they can earn full credit for internships and eligible experience.
No matter where you intend to practice, you'll have to earn and document a certain amount of professional experience to ensure you're ready to protect the public. Our Architectural Experience Program® (AXP™) provides a framework to guide you through recording that experience.
AXP Guidelines: Essential reading for all candidates completing the experience program, this document guides you through the program's rules and requirements.
Your Guide to the AXP: This quick overview will fill you in on the basics of the experience program, including some FAQs.
AXP Supervisors: Need help navigating your candidate/supervisor relationship? Visit the Supervisors and Mentors section of our website for more tools and resources.
Another key part of becoming an architect is completing the Architect Registration Examination® (ARE®). With exam divisions designed to reflect the current profession, the ARE ensures that you're ready to practice architecture independently.
ARE 5.0 Guidelines: Learn everything you need to know about starting the ARE, scheduling an exam, going to the test center, and other important examination policies.
ARE 5.0 Handbook: The ARE 5.0 Handbook walks you through each division, providing specifics on objectives, sample questions, suggested references, and more.
Your Guide to the ARE: Explore the basics of preparing for the exam, arriving at the test center, and more with this quick overview.
ARE 5.0 Community: Through the community, you'll have access to exclusive prep videos, advice from NCARB experts and fellow test-takers, and more.
Stay up-to-date on the latest news by signing up for our newsletters filled with news, tools, and tips to support you along the path to licensure.
Explore the 2020 Annual Report
A Guide to Architecture Organizations
Carly De La Hoz: An Insight Into the Re-Think Tank Experience


This page notes details of Noble Arms - Caliburn (Spell Card/Equip) : decks, tips, effect and rulings. Learn and enjoy playing Yu-Gi-Oh! Duel Links!
You can only control 1 "Noble Arms - Caliburn". Equip only to a Warrior-Type monster. It gains 500 ATK. Once per turn: You can gain 500 LP. If this face-up card on the field is destroyed and sent to the Graveyard: You can target 1 Warrior-Type "Noble Knight" monster you control; equip this card to that target. You can only use this effect of "Noble Arms - Caliburn" once per turn.
Retrieves itself from the graveyard.
Heals 500 LP every turn.
Use these cards to search out Noble Arms cards from your deck. Noble Knight Bedwyr sends Noble Arms to the graveyard when he is summoned, you can then equip that Noble Arms card from your graveyard to a Noble Knight monster with Bedwyr's effect.
Noble Knights usually gain their effects only when they are equipped with a Noble Arms card.
Use this card when activating monster effects that need a "Noble Arms" as a tribute to be able to use the LP gain twice (once before activating the monster effect and then again after re-equipping).
Background of this card reminds me of the Lost woods from the Zelda games.


The Love is Grand™ Bouquet brings together everyones favorite blooms to celebrate the love you share with your special someone. Brilliant red roses are surrounded by hot pink spray roses, pink carnations and pink mini carnations to create a fantastic flower arrangement. Presented in a classic clear glass vase, this mixed flower bouquet will send your sweetest wishes to your special someone.Approximately 17H x 13W.


Language Advocacy Day is February 14-15 in Washington, D.C., and here at Wayside we have been thinking about the many ways we can all advocate for the teaching and learning of languages.
Organized by the Joint National Committee for Languages (JNCL), Language Advocacy Day brings together professionals from all over the United States concerned about the state of language learning in this country.
If you can't make it to Washington, join us on Twitter using #speakupforlanguages and tell us why you think world languages are important and how you speak up for language education in your community. We will be sharing some of our thoughts next week as well, so be sure to follow @WaysidePublish.
Need ideas on how to be a language advocate? The JNCL website offers great resources on contacting your local representatives and connecting with other language professionals.


text:
Hey Space Geeks: Help This Guy Build a Real Starship Enterprise [VIDEO]
By Sam Laird 2012-05-18 21:22:59 UTC
There are those who look up at the stars and decide to become astronomers, while others want to be astronauts. Then there are those who want to build the Starship Enterprise.
That's right, a guy known only as BTE Dan (stands for Build the Enterprise) believes a replica of the iconic Star Trek spacecraft can be built to scale within the next 20 years. He's even started a website — BuildTheEnterprise.org — to rally support for the cause.
"This Gen1 Enterprise can go on missions to key points of interest in our solar system, like Mars and Venus," Dan writes. "It will be the biggest ship of any kind ever built by humans, and it will be larger than the tallest building in the world.
"It's possible to build the Enterprise, and it would be a monumental achievement for us humans who inhabit the planet earth."
TBE Dan's site covers pretty much all the bases, with detailed specs including size (3,150 feet), maximum capacity (1,000 people) and hull design (triple walls at all points). The site also presents a white paper on how the Enterprise could be used to help build bases on Mars and the moon, a process that would involve "laser diggers," a modified gravity wheel and a nuclear reactor.
SEE ALSO: How One Actor Used an iPhone to Land a Role in the 'Star Trek' Sequel [VIDEO]
The Enterprise would be built entirely in space, Dan says, and "will cost no more than $1 trillion spent over twenty years." A lengthy chart shows where that trillion bucks would go. He argues the United States should set aside 0.27% of its GDP each year to fund the project.
It would be easy to dismiss Dan as some Trekkie living in his mother's basement. But you shouldn't. According to his site bio, the man has spent the past 30 years working as a systems engineer and electrical engineer for a Fortune 500 tech company.
So how can you help in his quest to build a real Starship Enterprise?
"It's by simply doing this," Dan writes. "Tell someone about the ideas on this website. This matters because the only hope for building the 1st generation USS Enterprise will ultimately depend on an enthusiastic core of people who are encouraging the undertaking.
"Enthusiasm for an idea can become contagious."
Do you think it's possible to build an actual Starship Enterprise within 20 years? Let us know in the comments.
Topics: Star Trek, Startups, Tech


text:
Puntís d'Inard (francès Pointis-Inard) és un municipi del departament francès de l'Alta Garona, a la regió d'Occitània.

Referències

Municipis del cantó de Sent Gaudenç

redpajama_set_name: RedPajamaWikipedia