### 统计代写|最优控制作业代写optimal control代考|EE501T

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等楖率论
• Advanced Mathematical Statistics 高等数理统计学
• (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

## 统计代写|最优控制作业代写optimal control代考|History of Optimal Control Theory

Optimal control theory is an extension of the calculus of variations (see Appendix B), so we discuss the history of the latter first.

The creation of the calculus of variations occurred almost immediately after the formalization of calculus by Newton and Leibniz in the seventeenth century. An important problem in calculus is to find an argument of a function at which the function takes on its maximum or minimum. The extension of this problem posed in the calculus of variations is to find a function which maximizes or minimizes the value of an integral or functional of that function. As might be expected, the extremum problem in the calculus of variations is much harder than the extremum problem in differential calculus. Euler and Lagrange are generally considered to be the founders of the calculus of variations. Newton, Legendre, and the Bernoulli brothers also contributed much to the early development of the field.

A celebrated problem first solved using the calculus of variations was the path of least time or the Brachistochrone problem. The problem is illustrated in Fig. 1.1. It involves finding the shape of a curve $\Gamma$ connecting the two points A and B in the vertical plane with the property that a bead sliding along the curve under the influence of gravity will move from A to B in the shortest possible time. The problem was posed by Johann Bernoulli in 1696, and it played an important part in the development of calculus of variations. It was solved by Johann Bernoulli, Jakob Bernoulli, Newton, Leibnitz, and L’Hôpital. In Sect. B.4, we provide a solution to the Brachistochrone problem by using what is known as the Euler-Lagrange equation, stated in Sect. B.2, and show that the shape of the solution curve is represented by a cycloid.

## 统计代写|最优控制作业代写optimal control代考|Notation and Concepts Used

In order to make the book readable, we will adopt the following notation which will hold throughout the book. In addition, we will define some important concepts that are required, including those of concave, convex and affine functions, and saddle points.

We use the symbol “=” to mean “is equal to” or “is defined to be equal to” or “is identically equal to” depending on the context. The symbol “:=” means “is defined to be equal to,” the symbol ” $\equiv$ ” means “is identically equal to,” and the symbol ” $\approx$ ” means “is approximately equal to.” The double arrow ” $\Rightarrow$ ” means “implies,” ” $\nabla$ ” means “for all,” and ” $\in$ ” means “is a member of.” The symbol $\square$ indicates the end of a proof.

Let $y$ be an $n$-component column vector and $z$ be an $m$-component row vector, i.e.,
$$y=\left[\begin{array}{c} y_{1} \ y_{2} \ \vdots \ y_{n} \end{array}\right]=\left(y_{1}, \ldots, y_{n}\right)^{T} \text { and } z=\left(z_{1}, \ldots, z_{m}\right)$$
where the superscript ${ }^{T}$ on a vector (or, a matrix) denotes the transpose of the vector (or, the matrix). At times, when convenient and not confusing, we will use the superscript ${ }^{\prime}$ for the transpose operation. If $y$ and $z$ are functions of time $t$, a scalar, then the time derivatives $\dot{y}:=d y / d t$ and $\dot{z},=d z / d t$ are defined as
$$\dot{y}=\frac{d y}{d t}=\left(\dot{y}{1}, \cdots, \dot{y}{n}\right)^{T} \text { and } \dot{z}=\frac{d z}{d t}=\left(\dot{z}{1}, \ldots, \dot{z}{m}\right)$$
where $\dot{y}{i}$ and $\dot{z}{j}$ denote the time derivatives $d y_{i} / d t$ and $d z_{j} / d t$, respectively. Whēn $n=m$, we cann défine the iñner product
$$z y=\Sigma_{i=1}^{n} z_{i} y_{i} .$$

## 统计代写|最优控制作业代写optimal control代考|Notation and Concepts Used

$$y=\left[\begin{array}{ll} y_{1} y_{2} & \vdots y_{n} \end{array}\right]=\left(y_{1}, \ldots, y_{n}\right)^{T} \text { and } z=\left(z_{1}, \ldots, z_{m}\right)$$

$$\dot{y}=\frac{d y}{d t}=(\dot{y} 1, \cdots, \dot{y} n)^{T} \text { and } \dot{z}=\frac{d z}{d t}=(\dot{z} 1, \ldots, \dot{z} m)$$

$$z y=\Sigma_{i=1}^{n} z_{i} y_{i} .$$

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## MATLAB代写

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