解答
1
最適性条件より、最適な \(\{x_i\},\{\lambda_i\},\theta\) の下では、
$$
\begin{cases}
\begin{aligned}
\frac{\partial L}{\partial x_i} &= 0 & (i=1,\ldots,n+1)\\
\frac{\partial L}{\partial \lambda_i} &= 0 & (i=1,\ldots,n+1)\\
\frac{\partial L}{\partial\theta} &= 0
\end{aligned}
\end{cases}
$$
が成り立つので、
$$
\begin{aligned}
\frac{\partial L}{\partial x_i}
&= \frac{\partial J_D\left(x_1,\ldots,x_n\right)}{\partial x_i} - \left(\lambda_i - \lambda_{i+1}\frac{\partial F\left(x_i,\theta\right)}{\partial x_i}\right) = 0\\
\therefore \lambda_i &= \lambda_{i+1}\frac{\partial F\left(x_i,\theta\right)}{\partial x_i} + \frac{\partial J_D\left(x_1,\ldots,x_n\right)}{\partial x_i}
\end{aligned}
$$
となる。特に、\(i=n+1\) に関しては、
$$
\begin{aligned}
\frac{\partial L}{\partial x_{n+1}}
&= \frac{\partial J_D\left(x_1,\ldots,x_n\right)}{\partial x_{n+1}} - \lambda_{n+1} = -\lambda_{n+1} = 0\\
\therefore \lambda_{n+1} &= 0
\end{aligned}
$$
が成り立つ。
2
$$
\begin{aligned}
x_{i+1} &= F\left(x_i;\theta\right)\\
\frac{\partial x_{i+1}}{\partial\theta} &= \frac{\partial F\left(x_i;\theta\right)}{\partial\theta}\\
&= \frac{\partial F}{\partial x}\left(x_i;\theta\right)\frac{}{}
\end{aligned}
$$
a