# Stokey, Lucas (1989) p 11 FOC

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

Your FOC is correct. They took the FOC a step further and concluded that $$\lambda_t=0$$ for $$t. I quote from page 11:
To obtain these conditions note that since $$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.