# 数学代写|最优化作业代写optimization theory代考|EECE506

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

Graeffe’s method is a global method, as it gives a simultaneous approximation of all roots.

Consider a monic polynomial $f$ of type (3.1.2). To $f$, the following adjoint function $\phi$ is associated
\begin{aligned} \phi(x) &=(-1)^n f(x) f(-x) \ &=\left(x^2-\alpha_1^2\right)\left(x^2-\alpha_2^2\right) \ldots\left(x^2-\alpha_n^2\right), \end{aligned}
where $\alpha_i$ are the searched roots, ordered by decreasing modulus. As $\phi(x)$ contains only even powers, a new function is defined
$$f_2(x)=\phi(\sqrt{x})=\left(x-\alpha_1^2\right)\left(x-\alpha_2^2\right) \ldots\left(x-\alpha_n^2\right)$$
which has the property that its roots are the squares of the roots of $f$. The operation can be repeated, and we obtain a sequence of polynomials $f_2, f_4, f_8 \ldots$ such that
$$f_m(x)=\left(x-\alpha_1^m\right)\left(x-\alpha_2^m\right) \ldots\left(x-\alpha_n^m\right)$$
where $m$ is an integer positive of 2 and $f_m$ has roots $\alpha_1^m, \alpha_2^m, \ldots, \alpha_n^m$. The aim of this sequence is to form an equation whose roots have very different orders of magnitude, that is, if the roots are real, the ratios $\left|\alpha_{i-1}^m / \alpha_i^m\right|$ can be made as small as desired when $m$ becomes large.
$f_m(x)$ can be developed
\begin{aligned} f_m(x) &=x^n-\left(\alpha_1^m+\ldots\right) x^{n-1}+\left(\alpha_1^m \alpha_2^m+\ldots\right) x^{n-2} \ &-\left(\alpha_1^m \alpha_2^m \alpha_3^m+\ldots\right) x^{n-3}+\cdots+(-1)^n\left(\alpha_1^m \alpha_2^m \ldots \alpha_n^m\right) \ &=x^n-A_1 x^{n-1}+\cdots+(-1)^i A_i x^{n-i}+\cdots+(-1)^n A_n \end{aligned}
the approximations result
$$\alpha_1^m \doteq A_1, \quad \alpha_2^m \doteq \frac{A_2}{A_1} \quad, \ldots \quad \alpha_n^m \doteq \frac{A_n}{A_{n-1}}$$
hence an approximation of the absolute values or of the moduli of the searched roots by taking the $m$ th root.

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

Bernoulli’s method to find a root $\alpha_k$ of the polynomial
$$P(x)=\sum_{i=0}^n a_i x^{n-i} \quad \text { with } a_0=1$$
first consists of building a sequence $\left{u_i\right}$ by associating to each monomial $x^{n-k}$ a term $u_{i-k}$ (thus $i \geq n$ ).

To understand the interest of the building of the sequence $u_i$, first consider the fact that the roots $\alpha_i$ of the polynomial $P(x)$ are supposed to be ordered according to their modulus $\left|\alpha_1\right|>\cdots>\left|\alpha_n\right|$.
Express that $\alpha_i$ is a root
\begin{aligned} &a_0 \alpha_1^n+a_1 \alpha_1^{n-1}+\cdots+a_{n-1} \alpha_1+a_n=0 \ &\vdots \ &a_0 \alpha_n^n+a_1 \alpha_n^{n-1}+\cdots+a_{n-1} \alpha_n+a_n=0 \end{aligned}

Each of the previous equations is then multiplied by an arbitrary coefficient $c_i$, we sum the rows, that is
$$c_1\left(a_0 \alpha_1^n+a_1 \alpha_1^{n-1}+\cdots+a_{n-1} \alpha_1+a_n\right)+\cdots+c_n\left(a_0 \alpha_n^n+a_1 \alpha_n^{n-1}+\cdots+a_{n-1} \alpha_n+a_n\right)=0$$
and then we order again with respect to the coefficients $a_i$, i.e.
$$a_0\left(c_1 \alpha_1^n+\cdots+c_n \alpha_n^n\right)+a_1\left(c_1 \alpha_1^{n-1}+\cdots+c_n \alpha_n^{n-1}\right)+\cdots+a_n\left(c_1+\cdots+c_n\right)=0 \text { (3.3.4) }$$
We set
$$u_i=c_1 \alpha_1^i+\cdots+c_n \alpha_n^i, \quad 0 \leq i \leq n$$
hence
$$a_0 u_n+a_1 u_{n-1}+\cdots+a_n u_0=0$$
To simplify the writing, we consider $a_0=1$, hence
$$u_n=-a_1 u_{n-1}-\cdots-a_n u_0=-\sum_{i=1}^n a_i u_{n-i}$$
By extension, the sequence is defined
$$u_i=-\sum_{j=1}^n a_j u_{i-j} \quad \text { for } i \geq n$$
Thus, it can be seen that to define this sequence, it suffices to arbitrarily choose the real numbers $u_i$ for $0 \leq i<n$. From the form of the solutions $u_i$ previously given
$$u_i=c_1 \alpha_1^i+\cdots+c_n \alpha_n^i, \quad 0 \leq i \leq n-1$$
a system of $n$ equations with $n$ unknowns $\alpha_j^i$ results (Vandermonde determinant).

# 最优化代考

## 数学代写|最优化作业代写优化理论代考|格拉夫方法

Graeffe方法是一种全局方法，因为它给出了所有根的同时逼近

\begin{aligned} \phi(x) &=(-1)^n f(x) f(-x) \ &=\left(x^2-\alpha_1^2\right)\left(x^2-\alpha_2^2\right) \ldots\left(x^2-\alpha_n^2\right), \end{aligned}

$$f_2(x)=\phi(\sqrt{x})=\left(x-\alpha_1^2\right)\left(x-\alpha_2^2\right) \ldots\left(x-\alpha_n^2\right)$$
，它的根是$f$的根的平方。该操作可以重复，我们得到一个多项式序列$f_2, f_4, f_8 \ldots$，满足
$$f_m(x)=\left(x-\alpha_1^m\right)\left(x-\alpha_2^m\right) \ldots\left(x-\alpha_n^m\right)$$
，其中$m$是2的正整数，$f_m$有根$\alpha_1^m, \alpha_2^m, \ldots, \alpha_n^m$。这个序列的目的是形成一个方程，它的根具有非常不同的数量级，也就是说，如果根是实数，当$m$变大时，比值$\left|\alpha_{i-1}^m / \alpha_i^m\right|$可以取到所需的最小值。
$f_m(x)$可以发展
\begin{aligned} f_m(x) &=x^n-\left(\alpha_1^m+\ldots\right) x^{n-1}+\left(\alpha_1^m \alpha_2^m+\ldots\right) x^{n-2} \ &-\left(\alpha_1^m \alpha_2^m \alpha_3^m+\ldots\right) x^{n-3}+\cdots+(-1)^n\left(\alpha_1^m \alpha_2^m \ldots \alpha_n^m\right) \ &=x^n-A_1 x^{n-1}+\cdots+(-1)^i A_i x^{n-i}+\cdots+(-1)^n A_n \end{aligned}

$$\alpha_1^m \doteq A_1, \quad \alpha_2^m \doteq \frac{A_2}{A_1} \quad, \ldots \quad \alpha_n^m \doteq \frac{A_n}{A_{n-1}}$$

## 数学代写|最优化作业代写优化理论代考|伯努利方法

$$P(x)=\sum_{i=0}^n a_i x^{n-i} \quad \text { with } a_0=1$$

\begin{aligned} &a_0 \alpha_1^n+a_1 \alpha_1^{n-1}+\cdots+a_{n-1} \alpha_1+a_n=0 \ &\vdots \ &a_0 \alpha_n^n+a_1 \alpha_n^{n-1}+\cdots+a_{n-1} \alpha_n+a_n=0 \end{aligned} 前面的每个方程乘以一个任意系数 $c_i$，我们对行求和，结果是
$$c_1\left(a_0 \alpha_1^n+a_1 \alpha_1^{n-1}+\cdots+a_{n-1} \alpha_1+a_n\right)+\cdots+c_n\left(a_0 \alpha_n^n+a_1 \alpha_n^{n-1}+\cdots+a_{n-1} \alpha_n+a_n\right)=0$$

$$a_0\left(c_1 \alpha_1^n+\cdots+c_n \alpha_n^n\right)+a_1\left(c_1 \alpha_1^{n-1}+\cdots+c_n \alpha_n^{n-1}\right)+\cdots+a_n\left(c_1+\cdots+c_n\right)=0 \text { (3.3.4) }$$

$$u_i=c_1 \alpha_1^i+\cdots+c_n \alpha_n^i, \quad 0 \leq i \leq n$$

$$a_0 u_n+a_1 u_{n-1}+\cdots+a_n u_0=0$$

$$u_n=-a_1 u_{n-1}-\cdots-a_n u_0=-\sum_{i=1}^n a_i u_{n-i}$$通过扩展，该序列被定义为
$$u_i=-\sum_{j=1}^n a_j u_{i-j} \quad \text { for } i \geq n$$因此，可以看出，要定义这个序列，任意选择实数就足够了 $u_i$ 为 $0 \leq i<n$。从解的形式 $u_i$ 先前给出
$$u_i=c_1 \alpha_1^i+\cdots+c_n \alpha_n^i, \quad 0 \leq i \leq n-1$$

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