### 数学代写|微积分代写Calculus代写|MATH1111

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• 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

## 数学代写|微积分代写Calculus代写|Nonlinear difference equations

In Section 1.4 we discussed the difference equation
$$x_{n+1}=\alpha x_n,$$
$n=0,1,2, \ldots$, as a model for either growth or decay and we saw that its solution is given by
$$x_n=\alpha^n x_0,$$
$n=0,1,2, \ldots$. Now
$$\lim {n \rightarrow \infty} \alpha^n= \begin{cases}0, & \text { if } 0<\alpha<1, \ 1, & \text { if } \alpha=1, \ \infty, & \text { if } \alpha>1,\end{cases}$$ from which it follows that if $\left{x_n\right}$ is a solution of (1.5.1) with $x_0>0$, then $$\lim {n \rightarrow \infty} x_n=x_0 \lim {n \rightarrow \infty} \alpha^n= \begin{cases}0, & \text { if } 0<\alpha<1, \ x_0, & \text { if } \alpha=1, \ \infty, & \text { if } \alpha>1 .\end{cases}$$ These limiting values are consistent with our radioactive decay cxample since, in that case, $0<\alpha<1$ and we would expect the amount of a radioactive element to decline toward 0 over time. The case $0<\alpha<1$ also may make sense for a population model if the population is declining and heading toward extinction. However, the unbounded growth indefinitely into the future implied by the case $\alpha>1$ is very unlikely for a population model: eventually ecological or even sociological problems come to the forefront, such as when the population begins to overreach the resources available to it, and the rate of growth of the population changes. Even for bacteria growing in a Petri dish, diminishing food and space eventually cause a change in the rate of growth. Hence the equation $$x{n+1}=\alpha x_n$$
for $n=0,1,2, \ldots$ and $\alpha>1$, called the uninhibited, or natural, growth model, although often accurate as a model of population growth over short periods of time, is usually too simplistic for predictions over long time spans.

## 数学代写|微积分代写Calculus代写|The inhibited growth model

Suppose we wish to model the growth of a certain population which, without ecological constraints, would grow at a rate of $100 \beta \%$ per unit of time. That is, if $x_n$ represents the size of the population after $n$ units of time and there are no constraints on the size of the population, then
$$x_{n+1}-x_n=\beta x_n$$
for $n=0,1,2, \ldots$. However, suppose that, because of the limitation of resources, the population will begin to decline if it ever has more than $M$ individuals. We call $M$ the carrying capacity of the available resources, the maximum population which is sustainable over time. Then it would be reasonable to modify our model by forcing the amount of increase over a unit of time to decrease as the size of the population approaches $M$ and to become negative if the size of the population ever exceeds $M$. One way to accomplish this is to multiply the term $\beta x_n$ in (1.5.5) by
$$\frac{M-x_n}{M},$$
a ratio which is close to 1 when $x_n$ is small, close to 0 when $x_n$ is close to $M$, and negative when $x_n$ exceeds $M$. This leads us to the difference equation
$$x_{n+1}-x_n=\beta x_n\left(\frac{M-x_n}{M}\right),$$
$n=0,1,2, \ldots$, or, equivalently,
$$x_{n+1}=x_n+\frac{\beta}{M} x_n\left(M-x_n\right),$$
$n=0,1,2, \ldots$, which we call the inhibited growth model, also known as the discrete logistic equation. This is an example of a nonlinear difference equation because if we multiply out the right-hand side of the equation we have a quadratic term, namely, $\frac{\beta}{M} x_n^2$. Such equations are, in general, far more difficult to solve than the linear difference equations we considered in Section 1.4; in fact, many nonlinear difference equations are not solvable in terms of the elementary functions of calculus. Hence we will not consider any methods for solving such equations, relying instead on computing specific solutions by iterating the equation using a calculator or, preferably, a computer.

## 数学代写|微积分代写Calculus代写|Nonlinear difference equations

$$x_{n+1}=\alpha x_n,$$
$n=0,1,2, \ldots$ ，作为增长或衰退的模型，我们看到它的解决方案由下式给出
$$x_n=\alpha^n x_0,$$
$n=0,1,2, \ldots$ 现在
$$\lim n \rightarrow \infty \alpha^n={0, \quad \text { if } 0<\alpha<1,1, \quad \text { if } \alpha=1, \infty, \quad \text { if } \alpha>1,$$

$\lim n \rightarrow \infty x_n=x_0 \lim n \rightarrow \infty \alpha^n=\left{0, \quad\right.$ if $0<\alpha<1, x_0, \quad$ if $\alpha=1, \infty, \quad$ if $\alpha>1$.

$$x n+1=\alpha x_n$$

## 数学代写|微积分代写Calculus代写|The inhibited growth model

$$x_{n+1}-x_n=\beta x_n$$

$$\frac{M-x_n}{M},$$

$$x_{n+1}-x_n=\beta x_n\left(\frac{M-x_n}{M}\right),$$
$n=0,1,2, \ldots$ ，或者，等价地，
$$x_{n+1}=x_n+\frac{\beta}{M} x_n\left(M-x_n\right),$$
$n=0,1,2, \ldots$ ，我们称之为抑制增长模型，也称为离散逻辑方程。这是一个非线性差分方程的例子， 因为如果我们将方程的右边相乘，就会得到一个二次项，即 $\frac{\beta}{M} x_n^2$.一般来说，这些方程比我们在 $1.4$ 节 中考虑的线性差分方程更难求解；事实上，许多非线性差分方程无法用微积分的初等函数求解。因此，我 们不会考虑任何求解此类方程的方法，而是依赖于使用计算器或最好是计算机迭代方程来计算特定解。

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## MATLAB代写

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