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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

## 电子工程代写|并行计算代写Parallel Computing代考|Floating Point Numbers

The IEEE standard makes computers work with floating point numbers. The point will be floating around within the number representation. It is conceptionally close to our second variant from the introductory thought experiment: However, we write down the format with an explicit exponent, and we let the exponent determine where the decimal point is. The point does not float around in the bit representation, but it floats around in the real representation. Lets derive this format step by step.

We start our introduction into machine data representations with a simple observation: we need normalisation. It is obvious that $a \cdot b c d \cdot 10^{4}=a b . c d \cdot 10^{3}$. Whenever we write down a number, we have some degree of freedom. Lets exploit this degree of freedom-well, lets actually remove it-and always move the point to the left such that the leading digit is 1 . This makes the notation unique, as we work in a binary world. So let $a b .\left.c d \cdot 2^{4}\right|{2}$ be a number that is not normalised. The same number $a .\left.b c d \cdot 2^{5}\right|{2}=1 .\left.b c d \cdot 2^{5}\right|_{2}$ is normalised.

Definition $4.2$ (Normalised number representation) If a number is written down as
$$\hat{\imath}=(-1)^{\hat{s}}\left(1+\sum_{i=1}^{S-1} \hat{x}_{i} 2^{-i}\right) \cdot 2 \cdots,$$
it is in its normalised representation.
Normalisation is nice to realise (in hardware or software): It requires bit shifts which is something computers are really good at:

In the upper representations, there are four zeroes to the left of the leftmost digit $1 .$ If we assume that the one and only bit left of the decimal point should be a one, then we have to move all the digits four positions to the left. This is called a bit shift. Before the shift, we had an exponent of $\left.4\right|{10}=\left.100\right|{2}$ which we have to reduce in return.

I emphasise that the discussion of the exponent above is “wrong”. We’ll discuss that in a minute. However, the principle idea holds: Hardware shifts around the bits, a process we call normalisation.

## 电子工程代写|并行计算代写Parallel Computing代考|The IEEE Format

Once we write down any (non-integer) number as
$$\hat{x}=(-1)^{\hat{s}}\left(1+\sum_{i=1}^{S-1} \hat{x}{i} 2^{-i}\right) \cdot 2^{E}$$ the bit sequence is a unique representation of the number $x$. It consists of $S$ significant bits of which one bit is sacrificed for the sign bit $\hat{s}$. The $\hat{x}{i}$ are the bits after the decimal point. The significand ${ }^{1}$ is supplemented by $E$ exponent bits for the exponent. Other names for significand are fraction or mantissa.

Problems result from the fact that we have to squeeze this representation into a finite number of bits. Lets assume that we have 24 bits for the significand plus sign.

One bit is “lost” for the sign. This leaves us with 23 bits. We get one back, as we do not have to store the leading 1 bit of the normalised representation. We know it has to be there as we have defined normalisation that way. There is no point in storing it. As a consequence, we know exactly how many bits we have available to store the significand. For a 32-bit floating point number, we use 23 bits for all the significant bits right from the decimal point. This is C’s $\mathrm{f}$ loat. For a 64-bit number, a $C$ double, we use 52 bits (Table $4.1)$.

Definition $4.3$ (Truncation as rounding) If we take a normalised floating point number of any bit count and squeeze it into the (IEEE) floating point number, we might throw bits away. We truncate the representation such that it fits into our predefined number of bits. Effectively, this is rounding or chopping off. The other way round, we might have to add bits as we move the significand to the left. In this case, we add either $1 \mathrm{~s}$ or $0 \mathrm{~s}$. Both of them might be the wrong thing to add (compared to the exact math), so we effectively add garbage.

# 并行计算代考

## 电子工程代写|并行计算代写Parallel Computing代考|Floating Point Numbers

IEEE 标准使计算机能够处理浮点数。该点将在数字表示中浮动。它在概念上接近于介绍性思想实验的第二个变体：但是，我们用明确的指数写下格式，并让指数确 定小数点的位置。该点在位表示中不浮动，但在实际表示中浮动。让我们一步一步推导出这种格式。

$$\hat{\imath}=(-1)^{\hat{s}}\left(1+\sum_{i=1}^{S-1} \hat{x}_{i} 2^{-i}\right) \cdot 2 \cdots$$

## 电子工程代写|并行计算代写Parallel Computing代考|The IEEE Format

$$\hat{x}=(-1)^{\hat{s}}\left(1+\sum_{i=1}^{S-1} \hat{x} i 2^{-i}\right) \cdot 2^{E}$$

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## 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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