# 数学代写|最优化作业代写optimization theory代考|ISE520

## 数学代写|最优化作业代写optimization theory代考|Method of Successive Substitutions

Thẻ methood of successsivé substitutions can bé āppliéd tó any typé of function. Thé original equation
$$f(x)=0$$
is written under the modified form
$$x=F(x)$$
Supposing that $\alpha$ is a root of Equation (3.7.2), $\alpha$ is called a fixed point of $F$ and the method is also named as fixed point iteration method (Burden and Faires 2011; Fortin 2008).
We use the recurrence relation
$$x_{i+1}=F\left(x_i\right)$$
hoping that the sequence $x_i$ converges to the desired solution $\alpha$ of the equation. Cauchy’s criterion is applicable:
A sequence $u_i$ of $R^n$ converges $\Longleftrightarrow \forall \epsilon>0, \exists N(\epsilon)$ such that $\forall j, k>N(\epsilon), | u_j-$ $u_k |<\epsilon$

Take a given point $\alpha$ of $F$. For any vector $x_0$ taken in the neighborhood on $\mathcal{N}(\alpha)$, suppose that an inequality of the form
$$\left|x_{i+1}-\alpha\right| \leq C\left|x_i-\alpha\right|^p$$
is verified $\forall i$, with $0<C<1$ if $p=1$. In this case, the iteration method defined by $F$ is said to be a convergent method of order at least $p$ to determine $\alpha$. Any method of order $p$ to determine $\alpha$ is locally convergent (on $\mathcal{N}(\alpha)$ ). If $\mathcal{N}(\alpha)$ extends to $\mathcal{R}^n$, the method is globally convergent.
Let us reason on $\mathcal{R}, \alpha$ is such that $\alpha=F(\alpha)$. Suppose that we can write
$$|F(x)-F(\alpha)| \leq C|x-\alpha|$$
when $|x-\alpha| \leq\left|x_1-\alpha\right|$. It results
$$\left|x_2-\alpha\right|=\left|F\left(x_1\right)-\alpha\right|=\left|F\left(x_1\right)-F(\alpha)\right| \leq C\left|x_1-\alpha\right|$$
by iterating
$$\left|x_3-\alpha\right|=\left|F\left(x_2\right)-\alpha\right| \leq C\left|x_2-\alpha\right| \leq C^2\left|x_1-\alpha\right|$$
and in general
$$\left|x_i-\alpha\right| \leq C^{i-1}\left|x_1-\alpha\right|$$
hence
$$\lim _{i \rightarrow \infty} x_i=\alpha$$
provided that $0 \leq C<1$.

## 数学代写|最优化作业代写optimization theory代考|Newton’s Method

Newton’s method can be applied to functions of a real variable as well as to functions of a complex variable.

Consider a second degree Taylor polynomial for the function $f$ at point $\alpha$, sought solution, belonging to the neighborhood of a given point $x$
$$f(\alpha)=f(x)+(\alpha-x) f^{\prime}(x)+\frac{1}{2}(\alpha-x)^2 f^{\prime \prime}(x)+\frac{1}{3 !}(\alpha-x)^3 f^{(3)}(\xi), \quad \xi \text { between } x \text { and } \alpha$$
We know that $f(\alpha)=0$. If the series expansion is truncated at first order, it gives $$f(x)=(x-\hat{\alpha}) f^{\prime}(x) \quad \text { or } \hat{\alpha}=x-\frac{f(x)}{f^{\prime}(x)}$$
where $\hat{\alpha}$ is an estimation of the root $\alpha$. This requires that $f^{\prime}(\alpha) \neq 0$, i.e. $\alpha$ is not a multiple root of $f(x)=0$.
Newton’s iterative formula results
$$x_{k+1}=x_k-\frac{f\left(x_k\right)}{f^{\prime}\left(x_k\right)}$$
Thus, this formula amounts to drawing the tangent line to the curve $f(x)$ at the point of abscissa $x_k$ (Figure 3.6). Its intersection with the $x$-axis is equal to $x_{k+1}$.

Using the framework of the method of successive substitutions (Section 3.7) and setting
$$F(x)=x-\frac{f(x)}{f^{\prime}(x)}$$
Newton’s method can be considered as a fixed point method.

# 最优化代考

## 数学代写|最优化作业代写优化理论代考|连续替换法

.

Thẻ方法的successsivé替换可以bé āppliéd tó任何typé的函数。Thé原方程
$$f(x)=0$$

$$x=F(x)$$

$$x_{i+1}=F\left(x_i\right)$$
，希望序列$x_i$收敛于方程的期望解$\alpha$。柯西准则是适用的:
$R^n$的序列$u_i$收敛于$\Longleftrightarrow \forall \epsilon>0, \exists N(\epsilon)$，使$\forall j, k>N(\epsilon), | u_j-$$u_k |<\epsilon 取F中的一个给定点\alpha。对于\mathcal{N}(\alpha)附近的任何向量x_0，假设一个形式为$$ \left|x_{i+1}-\alpha\right| \leq C\left|x_i-\alpha\right|^p $$的不等式被验证为\forall i，如果p=1则验证为0<C<1。在本例中，F定义的迭代方法被认为是一个至少为p的收敛方法来确定\alpha。为p排序以确定\alpha的任何方法都是本地收敛的(在\mathcal{N}(\alpha)上)。如果\mathcal{N}(\alpha)扩展到\mathcal{R}^n，则该方法全局收敛。 让我们推论\mathcal{R}, \alpha是这样的\alpha=F(\alpha)。假设我们可以写$$ |F(x)-F(\alpha)| \leq C|x-\alpha| $$当|x-\alpha| \leq\left|x_1-\alpha\right|。它的结果是$$ \left|x_2-\alpha\right|=\left|F\left(x_1\right)-\alpha\right|=\left|F\left(x_1\right)-F(\alpha)\right| \leq C\left|x_1-\alpha\right| $$，通过迭代$$ \left|x_3-\alpha\right|=\left|F\left(x_2\right)-\alpha\right| \leq C\left|x_2-\alpha\right| \leq C^2\left|x_1-\alpha\right| $$和一般$$ \left|x_i-\alpha\right| \leq C^{i-1}\left|x_1-\alpha\right| $$因此$$ \lim _{i \rightarrow \infty} x_i=\alpha $$，如果0 \leq C<1 . ## 数学代写|最优化作业代写优化理论代考|牛顿方法 牛顿法既可以应用于实变量函数，也可以应用于复变量函数 考虑函数f在点\alpha处的二阶泰勒多项式，求解，属于一个给定点的邻域x$$ f(\alpha)=f(x)+(\alpha-x) f^{\prime}(x)+\frac{1}{2}(\alpha-x)^2 f^{\prime \prime}(x)+\frac{1}{3 !}(\alpha-x)^3 f^{(3)}(\xi), \quad \xi \text { between } x \text { and } \alpha $$我们知道f(\alpha)=0。如果级数展开被一阶截断，则给出$$ f(x)=(x-\hat{\alpha}) f^{\prime}(x) \quad \text { or } \hat{\alpha}=x-\frac{f(x)}{f^{\prime}(x)} $$，其中\hat{\alpha}是根节点\alpha的估计。这要求f^{\prime}(\alpha) \neq 0，即\alpha不是f(x)=0的多重根 牛顿迭代公式的结果$$ x_{k+1}=x_k-\frac{f\left(x_k\right)}{f^{\prime}\left(x_k\right)} $$因此，这个公式相当于在x_k的横坐标点绘制曲线f(x)的切线(图3.6)。它与x轴的交点等于x_{k+1}。 使用连续替换法的框架(第3.7节)并设置$$ F(x)=x-\frac{f(x)}{f^{\prime}(x)}$\$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: