### 数学代写|数值分析代写numerical analysis代考|MATHS7104

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数值分析代写numerical analysis代考|THE REPRESENTATION OF INTEGERS

In everyday life we use numbers based on the decimal system. Thus the number 257 , for example, is expressible as
\begin{aligned} 257 & =2 \cdot 100+5 \cdot 10+7 \cdot 1 \ & =2 \cdot 10^2+5 \cdot 10^1+7 \cdot 100^0 \end{aligned}
We call 10 the base of this system. Any integer is expressible as a polynomial in the base 10 with integral coefficients between 0 and 9 . We use the notation
\begin{aligned} N & =\left(a_n a_{n-1} \cdots a_0\right){10} \ & =a_n 10^n+a{n-1} 10^{n-1}+\cdots+a_0 10^0 \end{aligned}
to denote any positive integer in the base 10 . There is no intrinsic reason to use 10 as a base. Other civilizations have used other bases such as 12, 20 , or 60 . Modern computers read pulses sent by electrical components. The state of an electrical impulse is either on or off. It is therefore convenient to represent numbers in computers in the binary system. Here the base is 2 , and the integer coefficients may take the values 0 or 1 .

A nonnegative integer $N$ will be represented in the binary system as
\begin{aligned} N & =\left(a_n a_{n-1} \cdots a_1 a_0\right)2 \ & =a_n 2^n+a{n-1} 2^{n-1}+\cdots+a_1 2^1+a_0 2^0 \end{aligned}
where the coefficients $a_k$ are either 0 or 1 . Note that $N$ is again represented as a polynomial, but now in the base 2 . Many computers used in scientific work operate internally in the binary system. Users of computers, however, prefer to work in the more familiar decimal system. It is therefore necessary to have some means of converting from decimal to binary when information is submitted to the computer, and from binary to decimal for output purposes.

Conversion of a binary number to decimal form may be accomplished directly from the definition (1.2). As examples we have
$$\begin{gathered} (11)_2=1 \cdot 2^1+1 \cdot 2^0=3 \ (1101)_2=1 \cdot 2^3+1 \cdot 2^2+0 \cdot 2^1+1 \cdot 2^0=13 \end{gathered}$$
The conversion of integers from a base $\beta$ to the base 10 can also be accomplished by the following algorithm, which is derived in Chap. 2.

## 数学代写|数值分析代写numerical analysis代考|THE REPRESENTATION OF FRACTIONS

If $x$ is a positive real number, then its integral part $x_I$ is the largest integer less than or equal to $x$, while
$$x_F=x-x_I$$
is its fractional part. The fractional part can always be written as a decimal fraction:
$$x_F=\sum_{k=1}^{\infty} b_k 10^{-k}$$
where each $b_k$ is a nonnegative integer less than 10. If $b_k=0$ for all $k$ greater than a certain integer, then the fraction is said to terminate. Thus
$$\frac{1}{4}=0.25=2 \cdot 10^{-1}+5 \cdot 10^{-2}$$
is a terminating decimal fraction, while
$$\frac{1}{3}=0.333 \cdots=3 \cdot 10^{-1}+3 \cdot 10^{-2}+3 \cdot 10^{-3}+\cdots$$
is not.
If the integral part of $x$ is given as a decimal integer by
$$x_I=\left(a_n a_{n-1} \cdots a_0\right)_{10}$$

while the fractional part is given by (1.4), it is customary to write the two representations one after the other, separated by a point, the “decimal point”:
$$x=\left(a_n a_{n-1} \cdots a_0 \cdot b_1 b_2 b_3 \cdots\right){10}$$ Completely analogously, one can write the fractional part of $x$ as a binary fraction: $$x_F=\sum{k=1}^{\infty} b_k 2^{-k}$$
where each $b_k$ is a nonnegative integer less than 2, i.e., either zero or one. If the integral part of $x$ is given by the binary integer
$$x_I=\left(a_n a_{n-1} \cdots a_0\right)2$$ then we write $$x=\left(a_n a{n-1} \cdots a_0 \cdot b_1 b_2 b_3 \cdots\right)_2$$
using a “binary point.”

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|THE REPRESENTATION OF INTEGERS

$$257=2 \cdot 100+5 \cdot 10+7 \cdot 1 \quad=2 \cdot 10^2+5 \cdot 10^1+7 \cdot 100^0$$

$\$ \begin{aligned } N \& =|left(a_n a_{n-1} Icdots a_1 a_0uright) 2\left|\&=a _n 2^{\wedge} n+a{n-1} 2^{\wedge}{n-1}+\right| c d o t s+a _1 2^{\wedge} 1+a _02^{\wedge} 0 lend{aligned} \ \

$$(11)_2=1 \cdot 2^1+1 \cdot 2^0=3(1101)_2=1 \cdot 2^3+1 \cdot 2^2+0 \cdot 2^1+1 \cdot 2^0=13$$

## 数学代写|数值分析代写numerical analysis代考|THE REPRESENTATION OF FRACTIONS

$$x_F=x-x_I$$

$$x_F=\sum_{k=1}^{\infty} b_k 10^{-k}$$

$$\frac{1}{4}=0.25=2 \cdot 10^{-1}+5 \cdot 10^{-2}$$

$$\frac{1}{3}=0.333 \cdots=3 \cdot 10^{-1}+3 \cdot 10^{-2}+3 \cdot 10^{-3}+\cdots$$

$$x_I=\left(a_n a_{n-1} \cdots a_0\right){10}$$ 而小数部分由 (1.4) 给出，习惯上将两种表示形式依次书写，用一个点分隔，即“小数点”: $$x=\left(a_n a{n-1} \cdots a_0 \cdot b_1 b_2 b_3 \cdots\right) 10$$

$$x_F=\sum k=1^{\infty} b_k 2^{-k}$$

$$x_I=\left(a_n a_{n-1} \cdots a_0\right) 2$$

$$x=\left(a_n a n-1 \cdots a_0 \cdot b_1 b_2 b_3 \cdots\right)_2$$

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## MATLAB代写

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