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The Newton Method

Q:- Minimize $f(x_1,x_2)=2x_1-2x_2+2x_1^2+2x_1x_2+x_2^2 $ starting from the point

$\begin{equation}X_1=\begin{bmatrix}0\\0\end{bmatrix}\end{equation}$

Solution:- $X_{1+i} = X_i-[J_i]^{-1}\nabla f_i$

and

$\begin{equation} \nabla f=\begin{bmatrix} \frac{\partial f}{\partial x_1}\\ \frac{\partial f}{\partial x_2}\end{bmatrix}=\begin{bmatrix} 2+4x_1+2x_2\\-2+2x_1+2x_2\end{bmatrix}\end{equation}$

Hence $\begin{equation} [J_1] =\begin{bmatrix} \frac {\partial^2 f}{\partial x_1^2}&\frac {\partial^2 f}{\partial x_1\partial x_2}\\\frac {\partial^2 f}{\partial x_1\partial x_2}&\frac{\partial^2 f}{\partial x_2^2} \end{bmatrix} _{X_1}=\begin{bmatrix} 4&2\\2&2 \end{bmatrix} \end{equation}$

$\begin{aligned} &=[J_1]^{-1}=1/4\begin{bmatrix} 2&-2\\-2&4\end{bmatrix} &=\begin{bmatrix} \frac 12&\frac{-1}2\\\frac{-1}2&1\end{bmatrix} \end{aligned}$

$\begin{equation} g_1=\nabla f_1= \begin{bmatrix} \frac{\partial f}{\partial x_1}\\\frac{\partial f}{\partial x_2}\end{bmatrix}_{X_1}= \begin{bmatrix} 2+4x_1+2x_2 \\ -2+2x_1+2x_2\end{bmatrix}_{(0,0)}=\begin{bmatrix}2\\-2 \end{bmatrix} \end{equation}$

Thus,

$\begin{aligned} &=X_2=X_1-[J_1]^{-1}g_1 \\ &=\begin{bmatrix}0\\0\end{bmatrix}-\begin{bmatrix}\frac12&\frac{-1}2\\\frac{-1}2&1\end{bmatrix}\begin{bmatrix}2\\-2\end{bmatrix} \\ &=\begin{bmatrix}0\\0\end{bmatrix}-\begin{bmatrix}2\\-3\end{bmatrix} \\ &=\begin{bmatrix}-2\\3\end{bmatrix}\\ \end{aligned}$

To check if $X_2$ is the optimum point ,we evaluate,

$\begin{aligned} &=\nabla f_2 \\ &=\begin{bmatrix}\frac{\partial f}{\partial x_1}\\ \frac{\partial f}{\partial x_2}\end{bmatrix}_{X_2} \\ &=\begin{bmatrix}2+4\times-2+2\times3\\ -2+2\times-2+2\times3\end{bmatrix} \\ &=\begin{bmatrix}0\\0\end{bmatrix}\\ \end{aligned}$

Thus,$X_2$ is the optimum point.

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