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My Lagrangian is:

$L=\sum\limits_{t=0}^T \beta^tU(f(k_t)-k_{t+1})+\sum\limits_{t=0}^T\lambda_t(f(k_t)-k_{t+1}).$

My FOC for $[k_{t+1}]$ is:

$\beta^tU'(f(k_t)-k_{t_1}^*)(-1)-\lambda_t^*+\beta^{t+1}U'(f(k_{t+1}^*-k_{t+2})+\lambda^*_{t+1}f'(k_{t+1})^*=0$.

$\textbf{My Question:}$ I try to recover from their FOC and backtrack, but how did they get (5) as the Euler Equation on page 11?

Their EE is:

$\beta f'(k_t)U'[f(k_t-k_{t+1})]=U'[f(k_{t-1})-k_t]$ for $t=1,...,T$.

Reference:

Stokey, N., R. Lucas, and E. Prescott, Recursive Methods in Economic Dynamics, Harvard Univ. Press, 1989

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Your FOC is correct. They took the FOC a step further and concluded that $\lambda_t=0$ for $t<T$. I quote from page 11:

To obtain these conditions note that since $f(0)=0$ and $U'(0)=\infty$, it is clear that the inequality constraints in (4) do not bind and it is clear that $k_{T+1}=0$

The intuition is that the inequality constraints will only bind when you consume all capital today ($k_{t+1}=0$) or you invest all your capital and consume nothing today ($k_{t+1}=f(k_t))$. Both of these cases (save the former when $t=T$) imply that you will be consuming nothing for at least one period. Combining this with the MU being infinite at zero, obviously this cannot be optimal.

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