Constructing a newton sequence How may I construct the newton sequence for the following:
$(1) f(x_1,x_2) = x_1^4 + 2x_1^2x_2^2 + x_2^4$ with $x_0 = (1,1)$ and $x_0 = (1,0)$
$(2) f(t) = t^4 - 32t^2$ and $t_0 = 1$
To find $x_{k+1}$ do I need to use $\nabla^2f(x_k)(x_{k+1}-x_k) = - \nabla f(x_k)$
For finding the $x_{k+1}$ terms? If so, how would I go about solving that equation?
 A: For the first question, we are given an $f(x_1, x_2)$ and asked to find the minimum using Newton's method with starting points $X_0 = (x_1, x_2) = (1, 1)$ and $x_0 = (1, 0)$ as:
$$\tag 1 f(x_1, x_2) = x_1^4 + 2x_1^2x_2^2 + x_2^4$$
Newton's method can be extended for the minimization of multivariable functions.
The iteration is given by:
$$X_{n+1} = X_n - [J_n]^{-1} \nabla f_n$$
where:
$$ \nabla f(x_1,x_2) = \begin{bmatrix}
\dfrac{\partial f}{\partial x_1} \\ \dfrac{\partial f}{\partial x_2} 
\end{bmatrix} = \begin{bmatrix} 4 x_1^3 + 4 x_1 x_2^2 \\ 4 x_1^2 x_2 + 4 x_2^3 
\end{bmatrix}$$
$$J(x_1,x_2) = \begin{bmatrix}
\dfrac{\partial^2 f}{\partial x_1^2} & \dfrac{\partial^2 f}{\partial x_1 \partial x_2} \\ \dfrac{\partial^2 f}{\partial x_2 \partial x_1} & \dfrac{\partial^2 f}{\partial x_2^2}
\end{bmatrix} = \begin{bmatrix} 12 x_1^2 + 4 x_2^2 & 8 x_1 x_2 \\ 8 x_1 x_2 & 4 x_1^2 + 12 x_2^2
\end{bmatrix}$$
The iterates are:


*

*$X_0 = \begin{bmatrix}
 1.0 \\ 1.0
\end{bmatrix}$

*$X_1 = \begin{bmatrix}
 0.666667 \\ 0.666667
\end{bmatrix}$

*$X_2 = \begin{bmatrix}
 0.444445 \\ 0.444445
\end{bmatrix}$

*$X_3 = \begin{bmatrix}
 0.296297 \\ 0.296297
\end{bmatrix}$

*$\ldots$

*$X_n = \begin{bmatrix}
 0. \\ 0.
\end{bmatrix}$


A plot shows:

You try for $X_0 = (1, 0)$.
For the second problem, we can use Newton's method to find the minimum of a function. We need to derive a function $h(t)$ that has a root at the point where $f(t)$ achieves its minimum.
Write down the formula for Newton’s method applied to $h(t)$.
$$h(t) = f'(t) = 4t^3 - 64 t$$
Use Newton's Method to find the root (you should verify what you get as critical points with $h''(t)$, where you can also use Newton's Method):
$$t_{n+1} = t_n - \dfrac{h(t_n)}{h'(t_n)} = t_n - \dfrac{4t_n^3-64t_n}{12t_n^2-64}$$


*

*$t_0 = 1.0$

*$t_1 = -0.153846$

*$\ldots$

*$t_n = 0$


A plot shows:

We can see that using $t_0 = 1$, we found the local max. Had we chosen a better starting value, like $t_0 = 5$ or $t_0 = -5$, we would have landed on either of those two global minimum ($t = \pm 4$). I assume your problem likely meant local max for this part of the problem.
