Question

hello can someone help me answer this please

Exercise 2.5. Consider finding a zero of function F : D as the sum of a linear and nonlinear part: Rn → Rn that can be written F(x) BG(x) where B is a nonsingular matrix and G DCRR" is a general nonlinear function. At a point ak consider a general linear model Mk(x)Ak(x -), where the quantities ak E R" and Ak E Rn*n are to be determined. Define the linear model so that the conditions Mk(k) F(xk) and M(k)B hold. Derive the iteration obtained by defining k+ as the zero of M(x)

Answer 1

We are given a linear model

We need to find the vector  
ds
and the matrix  
ds
so that , where B is a given non singular matrix and F is a function from $\mathbb{R}^n$ to $\mathbb{R}^n$.

We use the given conditions:  

The second condition gives the derivative of the linear model at a point. Applying this condition tells us  .

To see this, first note that
Mk
is also a function from $\mathbb{R}^n$ to $\mathbb{R}^n$. So, we can write this function as an n-tuple of functions

Mk1 71, T2, Mkn(T1, r2 . ,in)

The matrix that denotes the derivative of this function is the n x n matrix whose i th row is the gradient of the i th component function.

${M_k}' (x) = \begin{bmatrix} \nabla M_k_1 (x)\\ \nabla M_k_2 (x)\\ \ldots \\ \nabla M_k_n (x) \end{bmatrix} = \begin{bmatrix} \partial M_k_1 / \partial x_1 & \partial M_k_1 /\partial x_2 & \ldots & \partial M_k_1 / \partial x_n \\ \partial M_k_2 / \partial x_1 & \partial M_k_2 /\partial x_2 & \ldots & \partial M_k_2 / \partial x_n \\ \ldots & \ldots & \ldots & \ldots \\ \partial M_k_n / \partial x_1 & \partial M_k_n /\partial x_2 & \ldots & \partial M_k_n / \partial x_n \end{bmatrix}$

Using the definition of the linear model  , we can write

Mk

$\begin{bmatrix} M_k_1 (x)\\ M_k_2 (x)\\ \cdots \\ M_k_n (x) \end{bmatrix} = \begin{bmatrix} a_k_1\\ a_k_2\\ \cdots \\ a_k_n \end{bmatrix} + A_k \begin{bmatrix} x_1 - x_k_1\\ x_2 - x_k_2\\ \cdots \\ x_n - x_k_n \end{bmatrix}$ where we wrote all the vectors using their components.

If we denote the i-j th component of A_k by $\left [ A_k \right ]_{i,j}$ , we can rewrite the above equation as

$\begin{bmatrix} M_k_1 (x)\\ M_k_2 (x)\\ \cdots \\ M_k_n (x) \end{bmatrix} = \begin{bmatrix} a_k_1\\ a_k_2\\ \cdots \\ a_k_n \end{bmatrix} + \begin{bmatrix} \sum_{j=1}^{n} \left [ A_k \right ]_{1,j} (x_j - x_k_j)\\ \sum_{j=1}^{n} \left [ A_k \right ]_{2,j} (x_j - x_k_j) \\ \cdots \\ \sum_{j=1}^{n} \left [ A_k \right ]_{n,j} (x_j - x_k_j)\end{bmatrix}$

So, we find

$M_k_i (x) = a_k_i + \sum_{j=1}^n \left [ A_k \right ]_{i,j} (x_j - x_k_j)$

The derivative of the above function with respect to the variable $x_t$ is the i-t th component of the derivative matrix of M_k ( from the definition of the derivative matrix given earlier).

This is just

$\left [ {M_k}' \right ]_{i,t}= \partial M_k_i (x) / \partial x_t = 0 + \partial / \partial x_t\sum_{i=1}^n \left [ A_k \right ]_{i,j} (x_j - x_k_j) = \left [ A_k \right ]_{i,t}$

But we were given that the derivative matrix of  M_k is B, so we find

$\left [ {M_k}' \right ]_{i,t} = \left [ B \right ]_{i,t} = \left [ A_k \right ]_{i,t}$ for all i, t in {1,2, ..., n}

Hence .

Applying the first condition, we find

$M_k(x_k) = F(x_k) \rightarrow a_k + B (x_k - x_k) = F(x_k)$

This gives us . As B is a matrix, B(0) = 0 which tells us that the constant vector

The linear model satisfying the given conditions is therefore

.

------------------------------------------------------------------------------------------------------------------------------------

Now, we can find the iteration step: $x_{k+1}$ is a zero of . Applying this condition,

Mk

We can now rearrange this equation to find

Since $F(x_k)$ is a known constant vector and B is given to be non singular, we can multiply both sides by the inverse of B:

$x_{k+1} - x_k = B^{-1} (-F(x_k)) \\ \rightarrow x_{k+1} = x_k + B^{-1} (-F(x_k))$

To conclude, the required iteration step is

.