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## 数学代写数值分析代写Numerical analysis代考|Forward and backward error

The first example shows that, in some cases, pencil and paper can still outperform a computer.

Use the Bisection Method to find the root of $f(x)=x^3-2 x^2+\frac{4}{3} x-\frac{8}{27}$ to within six correct significant digits.

Note that $f(0) f(1)=(-8 / 27)(1 / 27)<0$, so the Intermediate Value Theorem guarantees a solution in $[0,1]$. According to Example 1.2, 20 bisection steps should be sufficient for six correct places.

In fact, it is easy to check without a computer that $r=2 / 3=0.666666666 \ldots$ is a root:
$$f(2 / 3)=\frac{8}{27}-2\left(\frac{4}{9}\right)+\left(\frac{4}{3}\right)\left(\frac{2}{3}\right)-\frac{8}{27}=0 .$$

## 数学代写|数值分析代写Numerical analysis代考|The Wilkinson polynomial

A famous example with simple roots that are hard to determine numerically is discussed in Wilkinson [1994]. The Wilkinson polynomial is
$$W(x)=(x-1)(x-2) \cdots(x-20)$$
which, when multiplied out, is
\begin{aligned} W(x)= & x^{20}-210 x^{19}+20615 x^{18}-1256850 x^{17}+53327946 x^{16}-1672280820 x^{15} \ & +40171771630 x^{14}-756111184500 x^{13}+11310276995381 x^{12} \ & -135585182899530 x^{11}+1307535010540395 x^{10}-10142299865511450 x^9 \end{aligned}

\begin{aligned} & +63030812099294896 x^8-311333643161390640 x^7 \ & +1206647803780373360 x^6-3599979517947607200 x^5 \ & +8037811822645051776 x^4-12870931245150988800 x^3 \ & +13803759753640704000 x^2-8752948036761600000 x \ & +2432902008176640000 . \end{aligned}

## 数学代写数值分析代写Numerical analysis代考|Forward and backward error

$$f(2 / 3)=\frac{8}{27}-2\left(\frac{4}{9}\right)+\left(\frac{4}{3}\right)\left(\frac{2}{3}\right)-\frac{8}{27}=0 .$$

## 数学代写|数值分析代写Numerical analysis代考|The Wilkinson polynomial

Wilkinson [1994] 讨论了一个著名的例子，它的单根很难用数值确定。威尔金森多项式是
$$W(x)=(x-1)(x-2) \cdots(x-20)$$

\begin{aligned} & W(x)=x^{20}-210 x^{19}+20615 x^{18}-1256850 x^{17}+53327946 x^{16}-1672280820 x^{15}+40171771630 x^{14}-756111184500 x^{13}+11310276995381 x^{1 \digamma} \ & +63030812099294896 x^8-311333643161390640 x^7+1206647803780373360 x^6-3599979517947607200 x^5+8037811822645051776 x^4-128 \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写数值分析代写Numerical analysis代考|Bracketing a root

The function $f(x)$ has a root at $x=r$ if $f(r)=0$.
The first step to solving an equation is to verify that a root exists. One way to ensure this is to bracket the root: to find an interval $[a, b]$ on the real line for which one of the pair ${f(a), f(b)}$ is positive and the other is negative. This can be expressed as $f(a) f(b)<0$. If $f$ is a continuous function, then there will be a root: an $r$ between $a$ and $b$ for which $f(r)=0$. This fact is summarized in the following corollary of the Intermediate Value Theorem 0.4 :

Let $f$ be a continuous function on $[a, b]$, satisfying $f(a) f(b)<0$. Then $f$ has a root between $a$ and $b$, that is, there exists a number $r$ satisfying $a<r<b$ and $f(r)=0$.

In Figure $1.1, f(0) f(1)=(-1)(1)<0$. There is a root just to the left of 0.7 . How can we refine our first guess of the root’s location to more decimal places?

We’ll take a cue from the way our eye finds a solution when given a plot of a function. It is unlikely that we start at the left end of the interval and move to the right, stopping at the root. Perhaps a better model of what happens is that the eye first decides the general location, such as whether the root is toward the left or the right of the interval. It then follows that up by deciding more precisely just how far right or left the root lies and gradually improves its accuracy, just like looking up a name in the phone book. This general approach is made quite specific in the Bisection Method, shown in Figure 1.2.

## 数学代写|数值分析代写Numerical analysis代考|How accurate and how fast?

If $[a, b]$ is the starting interval, then after $n$ bisection steps, the interval $\left[a_n, b_n\right]$ has length $(b-a) / 2^n$. Choosing the midpoint $x_c=\left(a_n+b_n\right) / 2$ gives a best estimate of the solution $r$, which is within half the interval length of the true solution. Summarizing, after $n$ steps of the Bisection Method, we find that
Solution error $=\left|x_c-r\right|<\frac{b-a}{2^{n+1}}$
and
Function evaluations $=n+2$.
A good way to assess the efficiency of the Bisection Method is to ask how much accuracy can be bought per function evaluation. Each step, or each function evaluation, cuts the uncertainty in the root by a factor of two.

A solution is correct within $p$ decimal places if the error is less than $0.5 \times 10^{-p}$.
Use the Bisection Method to find a root of $f(x)=\cos x-x$ in the interval $[0,1]$ to within six correct places.

First we decide how many steps of bisection are required. According to (1.1), the error after $n$ steps is $(b-a) / 2^{n+1}=1 / 2^{n+1}$. From the definition of $p$ decimal places, we require that
\begin{aligned} \frac{1}{2^{n+1}} & <0.5 \times 10^{-6} \ n & >\frac{6}{\log _{10} 2} \approx \frac{6}{0.301}=19.9 \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:LU分解代写, Numerical analysis, 多项式插值方法代写, 数值分析, 数值积分代写, 数学代写, 最小二乘法代写

## avatest™帮您通过考试

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## 数学代写数值分析代写Numerical analysis代考|Machine representation

So far, we have described a floating point representation in the abstract. Here are a few more details about how this representation is implemented on a computer. Again, in this section we will discuss the double precision format; the other formats are very similar.

Each double precision floating point number is assigned an 8-byte word, or 64 bits, to store its three parts. Each such word has the form
$$s e_1 e_2 \ldots e_{11} b_1 b_2 \ldots b_{52}$$
where the sign is stored, followed by 11 bits representing the exponent and the 52 bits following the decimal point, representing the mantissa. The sign bit $s$ is 0 for a positive number and 1 for a negative number. The 11 bits representing the exponent come from the positive binary integer resulting from adding $2^{10}-1=1023$ to the exponent, at least for exponents between -1022 and 1023 . This covers values of $e_1 \ldots e_{11}$ from 1 to 2046 , leaving 0 and 2047 for special purposes, which we will return to later.

The number 1023 is called the exponent bias of the double precision format. It is used to convert both positive and negative exponents to positive binary numbers for storage in the exponent bits. For single and long-double precision, the exponent bias values are 127 and 16383 , respectively.

MATLAB’s format hex consists simply of expressing the 64 bits of the machine number $(0.10)$ as 16 successive hexadecimal, or base 16 , numbers. Thus, the first 3 hex numerals represent the sign and exponent combined, while the last 13 contain the mantissa.

## 数学代写|数值分析代写Numerical analysis代考|Addition of floating point numbers

Machine addition consists of lining up the decimal points of the two numbers to be added, adding them, and then storing the result again as a floating point number. The addition itself can be done in higher precision (with more than 52 bits) since it takes place in a register dedicated just to that purpose. Following the addition, the result must be rounded back to 52 bits beyond the binary point for storage as a machine number.
For example, adding 1 to $2^{-53}$ would appear as follows:
\begin{aligned} & 1.00 \ldots 0 \times 2^0+1.00 \ldots 0 \times 2^{-53} \ = & 1.0000000000000000000000000000000000000000000000000000 \times 2^0 \ • & 0.0000000000000000000000000000000000000000000000000000 \times 2^0 \ = & 1.00000000000000000000000000000000000000000000000000001 \times 2^0 \end{aligned}
This is saved as $1 . \times 2^0=1$, according to the rounding rule. Therefore, $1+2^{-53}$ is equal to 1 in double precision IEEE arithmetic. Note that $2^{-53}$ is the largest floating point number with this property; anything larger added to 1 would result in a sum greater than 1 under computer arithmetic.

The fact that $\epsilon_{\text {mach }}=2^{-52}$ does not mean that numbers smaller than $\epsilon_{\text {mach }}$ are negligible in the IEEE model. As long as they are representable in the model, computations with numbers of this size are just as accurate, assuming that they are not added or subtracted to numbers of unit size.

## 数学代写数值分析代写Numerical analysis代考|Machine representation

$$s e_1 e_2 \ldots e_{11} b_1 b_2 \ldots b_{52}$$

MATLAB 的 format hex 只是表示机器号的 64 位 $(0.10)$ 作为 16 个连续的十六进制数或以 16 为基数的数字。 因此，前 3 个十六进制数字代表符号和指数的组合，而后 13 个包含尾数。

## 数学代写|数值分析代写Numerical analysis代考|Addition of floating point numbers

$\$ \|begin { aligned } \& 1.00 \backslash Idots 0 \backslash times 2 \wedge 0+1.00 \backslash Idots 0 \backslash times 2 \wedge{-53} \backslash=\& 1.0000000000000000000000000000000000000000000000000000000 |times 2 \wedge 0 1 • \& 0.0000000000000000000000000000000000000000000000000000000 \backslash times 2 \wedge 0 \backslash =\& 1.000000000000000000000000000000000000000000000000000000001 \backslash times保存为 \

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。