## 数学代写|数值方法作业代写numerical methods代考|Ordinary Differential Equations

statistics-lab™ 为您的留学生涯保驾护航 在代写数值方法numerical methods方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数值方法numerical methods代写方面经验极为丰富，各种代写数值方法numerical methods相关的作业也就用不着说。

• 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|数值方法作业代写numerical methods代考|An Example

We take a simple autonomous non-linear scalar ODE to show how to calculate Picard iterates:
$$y^{\prime}=f(y)=y^{2}, \quad y\left(t_{0}\right)=a$$
whose solution is given by:
$$y(t)=\frac{a}{1-a\left(t-t_{0}\right)}$$
We now compute the Picard iterates (3.4) for this ODE in order to determine the values of $t$ for which the ODE has a solution. For convenience, let us take $a=1, t_{0}=0$. Some simple integration shows that:
\begin{aligned} &\phi_{1}(t)=1 \ &\phi_{1}(t)=1+\int_{0}^{t} f\left(\phi_{0}\right) d t=1+t \ &\phi_{2}(t)=1+\int_{0}^{t} f\left(\phi_{1}\right) d t=1+t+t^{2}+t^{3} / 3 \ &\phi_{3}(t)=1+t+t^{2}+t^{3}+\frac{2 t^{4}}{3}+\frac{t^{5}}{3}+\frac{t^{6}}{9}+\frac{t^{7}}{63} \end{aligned}
We can see that the series is beginning to look like $\frac{1}{1-t}=\sum_{j=0}^{\infty} t^{\jmath}$. We know that this series is convergent for $|t|<1$. A nice exercise is to compute the Picard iterates in the most general case (that is, $a \neq 1, t_{0} \neq 0$ ) and to determine under which circumstances the ODE (3.6) has a solution. In this case we have represented the solution of an ODE as a series, and we then analysed this series for which there are many convergence results, such as the root test and the ratio test.

## 数学代写|数值方法作业代写numerical methods代考|Riccati ODE

The Riccati ODE is a non-linear ODE of the form:
$$y^{\prime}=P(x)+Q(x) y+R(x) y^{2}+N(x, y)$$
This ODE has many applications, for example to interest-rate models (Duffie and Kan (1996)). In some cases a closed-form solution to Equation (3.10) is possible, but in this book our focus is on approximating it using the finite difference method.

We now discuss the relationship between the Riccati equation and the pricing of a zero-coupon bond $P(t, T)$, which is a contract that offers one dollar at maturity $T$. By definition, an affine term structure model assumes that $P(t, T)$ has the form:
$$P(t, T)=\exp [A(t, T)-B(t, T) r(t)]$$
Let us assume that the short-term interest rate is described by the following stochastic differential equation (SDE):
$$d r=\mu(t, r) d t+\sigma(t, r) d W_{t}$$
where $W_{t}$ is a standard Brownian motion under the risk-neutral equivalent measure and $\mu$ and $\sigma$ are given functions.

Duffie and Kan proved that $P(t, T)$ is exponential-affine if and only if the drift $\mu$ and volatility $\sigma$ have the form:
$$\mu(t, r)=\alpha(t) r+\beta(t), \quad \sigma(t, r)=\sqrt{\gamma(t) r+\delta(t)}$$
where $\alpha(t), \beta(t), \gamma(t)$ and $\delta(t)$ are given functions of $t$.
The coefficients $A(t, T)$ and $B(t, T)$ in this case are determined by the following ordinary differential equations:
$$\frac{d B}{d t}=\frac{\gamma(t)}{2} B(t, T)^{2}-\alpha(t) B(t, T)-1, B(T, T)=0$$
and:
$$\frac{d A}{d t}=\beta(t) B(t, T)-\frac{\delta(t)}{2} B(t, T)^{2}, A(T, T)=0$$
The first Equation (3.11) for $B(t, T)$ is the Riccati equation and the second one (3.12) is solved easily from the first one by integration.

## 数学代写|数值方法作业代写numerical methods代考|Predator-Prey Models

ODEs can be used as simple models of population growth, for example, by assuming that the rate of reproduction of a population of size $P$ is proportional to the existing population and to the amount of available resources. The ODE is:
$$\frac{d P}{d t}=r P\left(1-\frac{P}{K}\right), P(0)=P_{0}$$
where $r$ is the growth rate and $K$ is the carrying capacity. The initial population is $P_{0}$. It is easy to check the following identities:
$$P(t)=\frac{K P_{0} e^{r t}}{K+P_{0}\left(e^{r t}-1\right)} \text { and } \lim _{t \rightarrow \infty} P(t)=K .$$
Transformation of this equation leads to the logistic ODE:
$$\frac{d n}{d \tau}=n(1-n)$$
where $n$ is the population in units of carrying capacity $(n=P / K)$ and $\tau$ measures time in units of $1 / r$.

For systems, we can consider the predator-prey model in an environment consisting of foxes and rabbits:
\begin{aligned} &\frac{d r(t)}{d t}=-a r(t) f(t)+b r(t) \ &\frac{d f(t)}{d t}=-p f(t)+q f(t) r(t) \end{aligned}
where:
\begin{aligned} r(t) &=\text { number of rabbits at time } t \ f(t) &=\text { number of foxes at time } t \ b r(t) &=\text { birth rate of rabbits } \ -a r(t) f(t) &=\text { death rate of rabbits } \ b &=\text { unit birth rate of rabbits } \ -p f(t) &=\text { death rate of foxes } \ q f(t) r(t) &=\text { birth rate of foxes } \ q &=\text { unit birth rate of foxes. } \end{aligned}
The ODE system (3.14) is a model of a closed ecological environment in which foxes and rabbits are the only kinds of animals. Rabbits eat grass (of which there is a constant supply), procreate and are eaten by foxes. All foxes eat rabbits, procreate and die of geriatric diseases.

System (3.14) is sometimes called the Lotka-Volterra equations, which are an example of a more general Kolmogorov model to model the dynamics of ecological systems with predator-prey interactions, competition, disease and mutualism (Lotka (1956)).

## 数学代写|数值方法作业代写numerical methods代考|An Example

φ1(吨)=1 φ1(吨)=1+∫0吨F(φ0)d吨=1+吨 φ2(吨)=1+∫0吨F(φ1)d吨=1+吨+吨2+吨3/3 φ3(吨)=1+吨+吨2+吨3+2吨43+吨53+吨69+吨763

## 数学代写|数值方法作业代写numerical methods代考|Riccati ODE

Riccati ODE 是以下形式的非线性 ODE：

dr=μ(吨,r)d吨+σ(吨,r)d在吨

Duffie 和 Kan 证明了磷(吨,吨)是指数仿射的当且仅当漂移μ和波动性σ有以下形式：
μ(吨,r)=一种(吨)r+b(吨),σ(吨,r)=C(吨)r+d(吨)

d乙d吨=C(吨)2乙(吨,吨)2−一种(吨)乙(吨,吨)−1,乙(吨,吨)=0

d一种d吨=b(吨)乙(吨,吨)−d(吨)2乙(吨,吨)2,一种(吨,吨)=0

## 数学代写|数值方法作业代写numerical methods代考|Predator-Prey Models

ODE 可以用作人口增长的简单模型，例如，通过假设磷与现有人口和可用资源量成正比。ODE 是：
d磷d吨=r磷(1−磷ķ),磷(0)=磷0

dndτ=n(1−n)

dr(吨)d吨=−一种r(吨)F(吨)+br(吨) dF(吨)d吨=−pF(吨)+qF(吨)r(吨)

r(吨)= 一次兔子的数量 吨 F(吨)= 一次狐狸的数量 吨 br(吨)= 兔子的出生率  −一种r(吨)F(吨)= 兔子的死亡率  b= 兔单位出生率  −pF(吨)= 狐狸的死亡率  qF(吨)r(吨)= 狐狸出生率  q= 狐狸的单位出生率。
ODE系统（3.14）是一个封闭的生态环境模型，其中狐狸和兔子是唯一的动物。兔子吃草（其中有源源不断的供应），繁殖并被狐狸吃掉。所有的狐狸都吃兔子，生育并死于老年病。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 数学代写|数值方法作业代写numerical methods代考| STIFF ODEs

statistics-lab™ 为您的留学生涯保驾护航 在代写数值方法numerical methods方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数值方法numerical methods代写方面经验极为丰富，各种代写数值方法numerical methods相关的作业也就用不着说。

• 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|数值方法作业代写numerical methods代考|STIFF ODEs

We now discuss special classes of ODEs that arise in practice and whose numerical solution demands special attention. These are called stiff systems whose solutions consist of two components; first, the transient solution that decays quickly in time, and second, the steady-state solution that decays slowly. We speak of fast transient and slow transient, respectively. As a first example, let us examine the scalar linear initial value problem:
$$\left{\begin{array}{l} \frac{d y}{d t}+a y=1, \quad t \in(0, T], \quad a>0 \text { is a constant } \ y(0)=A \end{array}\right.$$
whose exact solution is given by:
$$y(t)=A e^{-a t}+\frac{1}{a}\left[1-e^{-a t}\right]=\left(A-\frac{1}{a}\right) e^{-a t}+\frac{1}{a} .$$
In this case the transient solution is the exponential term, and this decays very fast (especially when the constant $a$ is large) for increasing $t$. The steady-state solution is a constant, and this is the value of the solution when $t$ is infinity. The transient solution is called the complementary function, and the steady-state solution is called the particular integral (when $\frac{d y}{d y}=0$ ), the latter including no arbitrary constant. The stiffness in the above example is caused when the value $a$ is large; in this case traditional finite difference schemes can produce unstable and highly oscillating solutions. One remedy is to define very small time steps. Special finite difference techniques have been developed that remain stable even when the parameter $a$ is large. These are the exponentially fitted schemes, and they have a number of variants. The variant described in Liniger and Willoughby (1970) is motivated by finding a fitting factor for a general initial value problem and is chosen in such a way that it produces an exact solution for a certain model problem. To this end, let us examine the scalar ODE:
$$\frac{d y}{d t}=f(t, y(t)), t \in(0, T]$$
and let us approximate it using the Theta method:
$$y_{n+1}-y_{n}=\Delta t\left[(1-\theta) f_{n+1}+\theta f_{n}\right], f_{n}=f\left(t_{n}, y_{n}\right)$$
where the parameter $\theta$ has not yet been specified. We determine it using the heuristic that this so-called Theta method should be exact for the linear constant-coefficient model problem:
$$\frac{d y}{d t}=\lambda y\left(\text { exact solution } y(t)=e^{\lambda t}\right) \text {. }$$
Based on this heuristic and by using the exact solution from (2.43) in scheme (2.42) $(f(t, y)=\lambda y)$, we get the value (you should check that this formula is correct; it is a bit

of algebra). We get:
\begin{aligned} &y_{n+1}=\frac{1+\Delta t \lambda}{1-(1-\theta) t \lambda} y_{n} \ &\text { and } \ &\theta=-\frac{1}{\Delta t \lambda}-\frac{\exp (\Delta t \lambda)}{1-\exp (\Delta t \lambda)} . \end{aligned}
Note: this is a different kind of exponential fitting.
We need to determine if this scheme is stable (in some sense). To answer this question, we introduce some concepts.

## 数学代写|数值方法作业代写numerical methods代考|INTERMEZZO: EXPLICIT SOLUTIONS

A special case of an initial value problem is when the number of dimensions $n$ in an initial value problem is equal to 1 . In this case we speak of a scalar problem, and it is

useful to study these problems if one wishes to get some insights into how finite difference methods work. In this section we discuss some numerical properties of one-step finite difference schemes for the linear scalar problem:
\begin{aligned} &L u \equiv \frac{d u}{d t}+a(t) u=f(t), 0<\mathrm{t}0, \forall t \in[0, T]. The reader can check that the one-step methods (Equations (2.10), (2.11) and (2.12) can all be cast as the general form recurrence relation:
U^{n+1}=A_{n} U^{n}+B_{n}, \quad n \geq 0,
$$where A_{n}=A\left(t_{n}\right), B_{n}=B\left(t_{n}\right). Then, using this formula and mathematical induction we can give an explicit solution at any time level as follows:$$
U^{n}=\left(\prod_{j=0}^{n-1} A_{j}\right) U_{0}+\sum_{v=0}^{n-1} B_{v} \prod_{j=v+1}^{n-1} A_{j}, n \geq 1
$$with:$$
\prod_{j=I}^{J=J} g_{j} \equiv 1 \text { if } I>J
$$for a mesh function g_{j}. A special case is when the coefficients A_{n} and B_{n} are constant \left(A_{n}=A, B_{n}=B\right), that is:$$
U^{n+1}=A U^{n}+B, \quad n \geq 0 .
$$Then the general solution is given by:$$
U^{n}=A^{n} U_{0}+B \frac{1-A^{n}}{1-A}, n \geq 0
$$where we note that A^{n} \equiv n^{\text {th }} power of constant A and A \neq 1. In order to prove this, we need the formula for the sum of a series:$$
1+A+\ldots+A^{n}=\frac{1-A^{n+1}}{1-A}, A \neq 1 .
$$For a readable introduction to difference schemes, we refer the reader to Goldberg (1986). ## 数学代写|数值方法作业代写numerical methods代考|EXISTENCE AND UNIQUENESS RESULTS We turn our attention to a more general initial value problem for a non-linear system of ODEs:$$
\left{\begin{array}{l}
y^{\prime}=f(t, y), \quad t \in \mathbb{R} \
y(0)=A
\end{array}\right.
$$where:$$
y: \mathbb{R} \rightarrow \mathbb{R}^{n}, A \in \mathbb{R}^{n}, f: \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}
$$and:$$
f(t, y)=\left(f_{1}(t, y), \ldots, f_{n}(t, y)\right)^{\top} \text { where } f_{j}: \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}, j=1, \ldots, n .
$$43 Some of the important questions to be answered are: • Does System (3.1) have a unique solution? = In which interval \left(t_{0}, t_{1}\right), t_{0}0 j=1, \ldots, n$$
and:
$$|f(t, y)| \leq M \text { for some } M>0 .$$
Theorem 3.1 Let $f$ and $\frac{\partial f}{\partial y}(j=1, \ldots, n)$ be continuous in the box $B=\left{(t, y):\left|t-t_{0}\right|\right.$ $\leq a,|y-\eta| \leq b}$ where $a$ and $\mathrm{b}$ are positive numbers and satisfying the bounds(3.2) and (3.3) for (t, $y$ ) in B. Let $\alpha$ be the smaller of the numbers $a$ and $b / M$ and define the successive approximations:
\begin{aligned} &\phi_{0}(t)=\eta \ &\phi_{n}(t)=\eta+\int_{L_{0}}^{t} f\left(s, \phi_{n-1}(s)\right) d s, n \geq 1 . \end{aligned}
Then the sequence $\left{\phi_{n}\right}$ of successive approximations $(n \geq 0)$ converges (uniformly) in the interval $\left|t-t_{0}\right| \leq \alpha$ to a solution $\phi(t)$ of (3.1) that satisfies the initial condition $\phi\left(t_{0}\right)=\eta$.

Method (3.4) is called the Picard iterative method and it is used to prove the existence of the solution of systems of ODE (3.1). It is mainly of theoretical value, as it should not necessarily be seen as a practical way to construct a numerical solution. However, it does give us insights into the qualitative properties of the solution. On the other hand, it is a useful exercise to construct the sequence of iterates in Equation (3.4) for some simple cases.
We note that the IVP (3.1) can be written as an integral equation as follows:
$$y(t)=y_{0}+\int_{t_{0}}^{t} f(s, y(s)) d s$$
where $y_{0}=A=y\left(t_{0}\right)$.
It can be proved that the solution of (3.1) is also the solution of (3.5) and vice versa. We see then that Picard iteration is based on (3.5) and that we wish to have the iterates converging to a solution of (3.5).

## 数学代写|数值方法作业代写numerical methods代考|STIFF ODEs

$$\left{d是d吨+一种是=1,吨∈(0,吨],一种>0 是一个常数 是(0)=一种\对。 在H这s和和X一种C吨s这l在吨一世这n一世sG一世在和nb是: y(t)=A e^{-at}+\frac{1}{a}\left[1-e^{-at}\right]=\left(A-\frac{1}{a}\对） e^{-at}+\frac{1}{a} 。 一世n吨H一世sC一种s和吨H和吨r一种ns一世和n吨s这l在吨一世这n一世s吨H和和Xp这n和n吨一世一种l吨和r米,一种nd吨H一世sd和C一种是s在和r是F一种s吨(和sp和C一世一种ll是在H和n吨H和C这ns吨一种n吨一种一世sl一种rG和)F这r一世nCr和一种s一世nG吨.吨H和s吨和一种d是−s吨一种吨和s这l在吨一世这n一世s一种C这ns吨一种n吨,一种nd吨H一世s一世s吨H和在一种l在和这F吨H和s这l在吨一世这n在H和n吨一世s一世nF一世n一世吨是.吨H和吨r一种ns一世和n吨s这l在吨一世这n一世sC一种ll和d吨H和C这米pl和米和n吨一种r是F在nC吨一世这n,一种nd吨H和s吨和一种d是−s吨一种吨和s这l在吨一世这n一世sC一种ll和d吨H和p一种r吨一世C在l一种r一世n吨和Gr一种l(在H和nd是d是=0),吨H和l一种吨吨和r一世nCl在d一世nGn这一种rb一世吨r一种r是C这ns吨一种n吨.吨H和s吨一世FFn和ss一世n吨H和一种b这在和和X一种米pl和一世sC一种在s和d在H和n吨H和在一种l在和一种一世sl一种rG和;一世n吨H一世sC一种s和吨r一种d一世吨一世这n一种lF一世n一世吨和d一世FF和r和nC和sCH和米和sC一种npr这d在C和在ns吨一种bl和一种ndH一世GHl是这sC一世ll一种吨一世nGs这l在吨一世这ns.这n和r和米和d是一世s吨这d和F一世n和在和r是s米一种ll吨一世米和s吨和ps.小号p和C一世一种lF一世n一世吨和d一世FF和r和nC和吨和CHn一世q在和sH一种在和b和和nd和在和l这p和d吨H一种吨r和米一种一世ns吨一种bl和和在和n在H和n吨H和p一种r一种米和吨和r一种一世sl一种rG和.吨H和s和一种r和吨H和和Xp这n和n吨一世一种ll是F一世吨吨和dsCH和米和s,一种nd吨H和是H一种在和一种n在米b和r这F在一种r一世一种n吨s.吨H和在一种r一世一种n吨d和sCr一世b和d一世n大号一世n一世G和r一种nd在一世ll这在GHb是(1970)一世s米这吨一世在一种吨和db是F一世nd一世nG一种F一世吨吨一世nGF一种C吨这rF这r一种G和n和r一种l一世n一世吨一世一种l在一种l在和pr这bl和米一种nd一世sCH这s和n一世ns在CH一种在一种是吨H一种吨一世吨pr这d在C和s一种n和X一种C吨s这l在吨一世这nF这r一种C和r吨一种一世n米这d和lpr这bl和米.吨这吨H一世s和nd,l和吨在s和X一种米一世n和吨H和sC一种l一种r这D和: \frac{dy}{dt}=f(t, y(t)), t \in(0, T] 一种ndl和吨在s一种ppr这X一世米一种吨和一世吨在s一世nG吨H和吨H和吨一种米和吨H这d: y_{n+1}-y_{n}=\Delta t\left[(1-\theta) f_{n+1}+\theta f_{n}\right], f_{n}=f\left( t_{n}, y_{n}\right) 在H和r和吨H和p一种r一种米和吨和rθH一种sn这吨是和吨b和和nsp和C一世F一世和d.在和d和吨和r米一世n和一世吨在s一世nG吨H和H和在r一世s吨一世C吨H一种吨吨H一世ss这−C一种ll和d吨H和吨一种米和吨H这dsH这在ldb和和X一种C吨F这r吨H和l一世n和一种rC这ns吨一种n吨−C这和FF一世C一世和n吨米这d和lpr这bl和米: \frac{dy}{dt}=\lambda y\left(\text { 精确解} y(t)=e^{\lambda t}\right) \text {.$$

## 数学代写|数值方法作业代写numerical methods代考|INTERMEZZO: EXPLICIT SOLUTIONS

\begin{aligned} &L u \equiv \frac{d u}{d t}+a(t) u=f(t), 0<\mathrm{t}0, \forall t \in[0, T]。读者可以检查一步法（方程（2.10），（2.11）和（2.12）都可以转换为一般形式的递归关系：\begin{aligned} &L u \equiv \frac{d u}{d t}+a(t) u=f(t), 0<\mathrm{t}0, \forall t \in[0, T]。读者可以检查一步法（方程（2.10），（2.11）和（2.12）都可以转换为一般形式的递归关系：
U ^ {n + 1} = A_ {n} U ^ {n} + B_ {n}, \quad n \ geq 0,

U^{n}=\left(\prod_{j=0}^{n-1} A_{j}\right) U_{0}+\sum_{v=0}^{n-1} B_{v } \prod_{j=v+1}^{n-1} A_{j}, n \geq 1

\prod_{j=I}^{J=J} g_{j} \equiv 1 \text { 如果 } I>J
F这r一种米和sHF在nC吨一世这n$Gj$.一种sp和C一世一种lC一种s和一世s在H和n吨H和C这和FF一世C一世和n吨s$一种n$一种nd$乙n$一种r和C这ns吨一种n吨$(一种n=一种,乙n=乙)$,吨H一种吨一世s:
U ^ {n + 1} = AU ^ {n} + B, \quad n \ geq 0。

U ^ {n} = A ^ {n} U_ {0} + B \ frac {1-A ^ {n}} {1-A}，n \ geq 0

1+A+\ldots+A^{n}=\frac{1-A^{n+1}}{1-A}, A \neq 1 。
$$对于差分方案的可读介绍，我们将读者推荐给 Goldberg (1986)。 ## 数学代写|数值方法作业代写numerical methods代考|EXISTENCE AND UNIQUENESS RESULTS 我们将注意力转向一个更一般的 ODE 非线性系统的初始值问题：$$
\left{是′=F(吨,是),吨∈R 是(0)=一种\对。

y: \mathbb{R} \rightarrow \mathbb{R}^{n}, A \in \mathbb{R}^{n}, f: \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}

f(t, y)=\left(f_{1}(t, y), \ldots, f_{n}(t, y)\right)^{\top} \text { 其中 } f_{j}: \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}, j=1, \ldots, n 。43

• 系统（3.1）是否有唯一的解决方案？

## 数学代写|数值方法作业代写numerical methods代考|Classes of Discontinuous Functions

\begin{aligned} &f(x+)=q \text { if } f\left(t_{n}\right) \rightarrow q, n \rightarrow \infty \text { 对于所有序列 }\left{t_{n} \right} \text { in }(x, b) \text { st } t_{n} \rightarrow x \ &f(x-)=q \text { if } f\left(t_{n}\right) \ rightarrow q, n \rightarrow \infty \text { 对于所有序列 }\left{t_{n}\right} \text { in }(a, x) \text { st } t_{n} \rightarrow x \ &\存在 \lim {t \rightarrow x} f(t) \Leftrightarrow f(x+)=f(x-)=\lim {t \rightarrow x} f(t)=f(x) 。\end{对齐}\begin{aligned} &f(x+)=q \text { if } f\left(t_{n}\right) \rightarrow q, n \rightarrow \infty \text { 对于所有序列 }\left{t_{n} \right} \text { in }(x, b) \text { st } t_{n} \rightarrow x \ &f(x-)=q \text { if } f\left(t_{n}\right) \ rightarrow q, n \rightarrow \infty \text { 对于所有序列 }\left{t_{n}\right} \text { in }(a, x) \text { st } t_{n} \rightarrow x \ &\存在 \lim {t \rightarrow x} f(t) \Leftrightarrow f(x+)=f(x-)=\lim {t \rightarrow x} f(t)=f(x) 。\end{对齐}

• 第一类：F(X+)=林吨→X+F(吨)和F(X−)=林吨→X−F(吨)存在。那么要么我们有F(X+)≠F(X−)或者F(X+)=F(X−)≠F(X).
• 第二类：不属于第一类的不连续性。
例如：
\begin{aligned} &f(x)=\left{1,X 合理的 (X∈问) 0,X 不理性， X∉问 第二种：都没有 F(X+) 也不 F(X−) 存在。 \对。\ &f(x)={X+2,−3<X<−2 −X−2,−2≤X<0 X+2,0≤X<1 简单的不连续性 X=0. \end{aligned}
你可以检查后一个函数在X=0.

## 数学代写|数值方法作业代写numerical methods代考|DIFFERENTIAL CALCULUS

(F+G)′(X)=F′(X)+G′(X) (FG)′(X)=F′(X)G(X)+F(X)G′(X) (FG)′(X)=G(X)F′(X)−G′(X)F(X)G2(X)(G(X)≠0)

X∈[一种,b],∃F′(X)和G可微分于F(X).

H(吨)≡G(F(吨)),一种≤吨≤b 有导数  H′(X)=G′(F(X))F′(X).

F(X)=X2,G(是)=2是+1 H(X)=G(F(X))=G(X2)=2X2+1 H′(X)=G′(F(X))F′(X)=4X(=2∗2X)

\begin{aligned} &f(x)=\left{X罪⁡1X,X≠0 0,X=0\对。\ &f^{\prime}(x)=\sin \frac{1}{x}-\frac{1}{x} \cos \frac{1}{x}, \quad x \neq 0 \end{对齐}
F′(0)不存在。
$f(x)=\左{X2罪⁡1X,X≠0 0,X=0\对。f^{\prime}(x)=2 x \sin \frac{1}{x}-\cos \frac{1}{x}, \quad x \neq 0f^{\prime}(0)=\lim _{t \rightarrow 0} \frac{f(t)-f(0)}{t-0}=0 .$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 数学代写|数值方法作业代写numerical methods代考|Real Analysis Foundations for this Book

statistics-lab™ 为您的留学生涯保驾护航 在代写数值方法numerical methods方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数值方法numerical methods代写方面经验极为丰富，各种代写数值方法numerical methods相关的作业也就用不着说。

• 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|数值方法作业代写numerical methods代考|INTRODUCTION AND OBJECTIVES

In this chapter we introduce a number of mathematical concepts and methods that underlie many of the topics in this book. The most urgent attention points revolve around functions of real variables, their properties and the ways they are used in applications. We discuss the most important topics from real analysis to help us in our understanding of partial differential equations (PDEs). A definition of real analysis is:
In mathematics, real analysis is the branch of mathematical analysis that studies the behavior of real numbers, sequences and series of real numbers, and real functions. Some particular properties of real-valued sequences and functions that real analysis studies include convergence, limits, continuity, smoothness, differentiability and integrability.

Real analysis is distinguished from complex analysis, which deals with the study of complex numbers and their functions.
(Wikipedia)
A related branch of mathematics is calculus, which we learn at school:
Calculus, originally called infinitesimal calculus or ‘the calculus of infinitesimals’, is the mathematical study of continuous change, in the same way that geometry is the study of shape and algebra is the study of generalizations of arithmetic operations.

It has two major branches, differential calculus and integral calculus; the former concerns instantaneous rates of change, and the slopes of curves, while integral calculus concerns accumulation of quantities, and areas under or between curves. These two branches are related to each other by the fundamental theorem of calculus, and they make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit.
(Wikipedia)
In practice, there is a distinction between calculus and real analysis. Calculus entails techniques (and tricks) to differentiate and integrate functions. It does not discuss the conditions under which a function is continuous or differentiable. It assumes that it is allowed to carry out these operations on functions. Real analysis, on the other hand, does discuss these issues and more; for example:

• Continuous functions: How do we recognise them and prove that a function is continuous?
= The different kinds of discontinuous functions.
• Differential calculus from a real-analysis viewpoint.
• Taylor’s theorem.
= An introduction to metric spaces and Cauchy sequences.
In our opinion, these topics are necessary prerequisites for the rest of this book. Knowledge of vector (linear) analysis and numerical linear algebra is also a prerequisite for computational finance. To this end, we devote Chapters 4 and 5 to these topics. Finally, complex variables and complex functions (which are at the heart of complex analysis) are introduced in Chapter 16 . We use the notation $\forall$ to mean ‘for all’ and $\exists$ to mean ‘there exists’.

## 数学代写|数值方法作业代写numerical methods代考|CONTINUOUS FUNCTIONS

In this section we are mainly concerned with real-valued functions of a real variable, that is $f: \mathbb{R} \rightarrow \mathbb{R}$. In rough terms, a continuous function is one that can be drawn by hand without taking the pen from paper. In other words, a continuous function does not have jumps or breaks, but it is allowed to have sharp bends and kinks. Examples of continuous functions are:
\begin{aligned} &f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x^{2} \ &f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\max (0, x) \end{aligned}
We can see that these functions are continuous just by drawing them. The first function is ‘smoother’ than the second function, the latter being similar to a one-factor call or put payoff on the one hand and a Rectified Linear Unit (ReLU) activation function

on the other hand (Goodfellow, Bengio and Courville (2016)). Intuitively, a function $f$ is continuous if $f(x) \rightarrow f(p)$ when $x \rightarrow p$, no matter how $x$ approaches $p$. Alternatively, small changes in $x$ lead to small changes in $f(x)$.

If we formally differentiate the above ReLU function (1.1), we get the famous discontinuous Heaviside function:
$$H(x)=\left{\begin{array}{l} 0, x<0 \ 1, x \geq 0 \end{array}\right.$$
A discontinuous function is one that is not continuous. Another discontinuous function is:
$x \in \mathbb{R},[x] \equiv$ largest integer $n$ s.t. $n \leq x \leq n+1 .$
Define $f(x)=[x]$; let $p \in \mathbb{Z}$ (integer).
Then taking left and right limits gives different answers, showing that the function is not continuous.

1. $x<p \Rightarrow f(x)=p-1$
2. $x>p \Rightarrow f(x)=p$
Thus $\lim {x \rightarrow p-} f(x)=p-1, \lim {x \rightarrow p+} f(x)=p$.

## 数学代写|数值方法作业代写numerical methods代考|Formal Definition of Continuity

The following definition is based on the fact that small changes in $x$ lead to small changes in $f(x)$.
Definition $1.1$
\begin{aligned} &\lim {x \rightarrow p} f(x)=A \text { means } \forall \varepsilon>0 \exists \delta>0 \text { s.t. } \ &|f(x)-A|<\varepsilon \text { when } 0<|x-p|<\delta . \end{aligned} Some properties of continuous functions $f(x)$ and $g(x)$ are: \begin{aligned} &\lim {x \rightarrow p}(f(x) \pm g(x))=\lim {x \rightarrow p} f(x) \pm \lim {x \rightarrow p} g(x) \ &\lim {x \rightarrow p}(f(x) g(x))=\lim {x \rightarrow p} f(x) \lim {x \rightarrow p} g(x) \ &\lim {x \rightarrow p} \frac{f(x)}{g(x)}=\frac{\lim {x \rightarrow p} f(x)}{\lim {x \rightarrow p} g(x)}, \quad g(x) \neq 0 . \end{aligned}

It can be a mathematical challenge to prove that a function is continuous using the above ‘epsilon-delta’ approach in Definition 1.1. One approach is to use the well-known technique of splitting the problem into several mutually exclusive cases, solving each case separately and then merging the corresponding partial solutions to form the desired solution. To this end, let us examine the square root function:
$$f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}, f(x)=\sqrt{x} .$$
We show that there exists $\delta>0$ such that for $x \geq 0$ :
$$|x-y|<\delta \Rightarrow|\sqrt{x}-\sqrt{y}|<\epsilon \forall y \in \mathbb{R}^{+} .$$ Then: $$\sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}}$$ We now consider two cases: Case $1: x>0$. Then:
$$|x-y|<\delta \Rightarrow|\sqrt{x}-\sqrt{y}| \leq \frac{|x-y|}{\sqrt{x}}=\frac{\delta}{\sqrt{x}}=\epsilon$$
Choose $\delta=\epsilon \sqrt{x}$.
Case $2: x=0$. Then:
$$|x-y|<\delta \Rightarrow|\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{\sqrt{x}+\sqrt{y}}=\frac{y}{\sqrt{y}}=\sqrt{y}=\epsilon$$
Hence:
$$|-y|=|y|<\delta \Rightarrow \sqrt{y}=\epsilon \Rightarrow \sqrt{\delta}<\epsilon \Rightarrow \delta<\epsilon^{2}$$
Choose $\delta=\epsilon^{2}$.
We have thus proved that the square root function is continuous.

## 数学代写|数值方法作业代写numerical methods代考|INTRODUCTION AND OBJECTIVES

（维基百科）

（维基百科）

• 连续函数：我们如何识别它们并证明函数是连续的？
= 不同种类的不连续函数。
• 从实分析的观点看微分。
• 泰勒定理。
= 度量空间和柯西序列的介绍。
我们认为，这些主题是本书其余部分的必要先决条件。矢量（线性）分析和数值线性代数的知识也是计算金融的先决条件。为此，我们将第 4 章和第 5 章专门讨论这些主题。最后，第 16 章介绍了复变量和复函数（它们是复分析的核心）。我们使用符号∀意思是“为所有人”和∃意思是“存在”。

## 数学代写|数值方法作业代写numerical methods代考|CONTINUOUS FUNCTIONS

F:R→R,F(X)=X2 F:R→R,F(X)=最大限度(0,X)

$$H(x)=\left{0,X<0 1,X≥0\对。$$

X∈R,[X]≡最大整数n英石n≤X≤n+1.

1. X<p⇒F(X)=p−1
2. X>p⇒F(X)=p
因此林X→p−F(X)=p−1,林X→p+F(X)=p.

## 数学代写|数值方法作业代写numerical methods代考|Formal Definition of Continuity

F:R+→R+,F(X)=X.

|X−是|<d⇒|X−是|<ε∀是∈R+.然后：X−是=X−是X+是我们现在考虑两种情况：1:X>0. 然后：
|X−是|<d⇒|X−是|≤|X−是|X=dX=ε

|X−是|<d⇒|X−是|=|X−是|X+是=是是=是=ε

|−是|=|是|<d⇒是=ε⇒d<ε⇒d<ε2

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。