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# 数学代写|离散数学代写DISCRETE MATHEMATICS代考|MATH215

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## 数学代写|离散数学代写Discrete Mathematics代考|INVERSE FUNCTION

Let $f: \mathrm{A} \rightarrow \mathrm{B}$ be a bijective function. Then the inverse of $f$, i.e. $f^{-1}$ be a function from $\mathrm{B}$ to $\mathrm{A}$. Since $f$ is a function from A to B, for every $x \in \mathrm{A}$, there exists unique $y \in \mathrm{B}$ such that $f(x)=y$.

Since $f^{-1}: \mathrm{B} \rightarrow$ A for every $y \in \mathrm{B}$ there exists unique $x \in$ A such that $f^{-1}(y)=x$, i.e. $f^{-1}(f(x))$ $=x$.

If $f: \mathrm{A} \rightarrow \mathrm{B}$ is bijective, then the function $f$ posses inverse mapping.
Proof: Suppose that $f: \mathrm{A} \rightarrow \mathrm{B}$ is not bijective and posses an inverse mapping, i.e. (i) $f$ is onto but not one-one. (ii) $f$ is one-one but not onto or (iii) $f$ is neither one-one nor onto.
Case ( $i$ ) Suppose that $f$ is onto but not one-one.
As $f$ is onto, so for every $y_1 \in \mathrm{B}$ there exists at least one $x_1 \in \mathrm{A}$ such that $f\left(x_1\right)=y_1$ and $\mathrm{R}(f)=\mathrm{B}$. Again as $f$ is not one-one we have $x_1 \neq x_2, x_1, x_2 \in$ A implies $y_1=f\left(x_1\right)=f\left(x_2\right)=y_2$.
Since $f^{-1}: \mathrm{B} \rightarrow \mathrm{A}$, so $\mathrm{D}\left(f^{-1}\right)=\mathrm{R}(f)=\mathrm{B}$, i.e. $\mathrm{D}\left(f^{-1}\right)=\mathrm{B}$. Also $\left(x_1, y_1\right),\left(x_2, y_2\right) \in f$ implies $\left(y_1, x_1\right),\left(y_2, x_2\right) \in f^{-1}$ with $x_1 \neq x_2$ as $y_1=y_2$. Hence $f^{-1}$ can not be a function.
Case (ii) Suppose that $f$ is one-one but not onto.
As $f$ is not onto, so for at least one $y_1 \in \mathrm{B}$ there exists no $x_1 \in \mathrm{A}$ such that $f\left(x_1\right)=y_1$ and $R(f)$ $\neq \mathrm{B}$. Since, $f^{-1}: \mathrm{B} \rightarrow \mathrm{A}$, so
$\mathrm{D}\left(f^{-1}\right)=\mathrm{R}(f) \neq \mathrm{B}$, i.e. $\mathrm{D}\left(f^{-1}\right) \neq \mathrm{B}$. Hence $f^{-1}$ can not be a function.
Case (iii) Similarly it can be proved that $f^{-1}$ can not be a function if $f$ is neither onto nor one-one.

Therefore, it is a contradiction. So our supposition is wrong. Hence $f: \mathrm{B} \rightarrow \mathrm{A}$ must be bijective to posses a inverse mapping.

## 数学代写|离散数学代写Discrete Mathematics代考|Theorem

Let $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. If both $f$ and $g$ are invertible, then $\left(g_0 f\right)^{-1}=f^{-1}$ ${ }_0 g^{-1}$.

Proof: Suppose that both $f$ and $g$ are invertible. This indicates that both $f$ and $g$ are bijective functions. So by theorem 4.5.1, $\left(g_0 f\right)$ is also bijective and hence invertible.

As $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ we have $\left(g_0 f\right): \mathrm{A} \rightarrow \mathrm{C}$ i.e. $\left(g_0 f\right)^{-1}: \mathrm{C} \rightarrow \mathrm{A}$. Also $f^{-1}: \mathrm{B} \rightarrow \mathrm{A}$ and $g^{-1}: \mathrm{C} \rightarrow \mathrm{B}$ we have $f^{-1}{ }_{\mathrm{o}} g^{-1}: \mathrm{C} \rightarrow \mathrm{A}$.

Hence first of all it is evident that both $\left(g_o f\right)^{-1}$ are $f^{-1}{ }0 g^{-1}$ are functions from the set $\mathrm{C}$ to the set $\mathrm{A}$ and $\left(g{\mathrm{o}} f\right)^{-1}(z)=x$ for $z \in \mathrm{C}$ and $x \in \mathrm{A}$.

Again $g^{-1}: \mathrm{C} \rightarrow \mathrm{B}$, so for every $z \in \mathrm{C}$ there exists unique $y \in \mathrm{B}$ such that $g^{-1}(z)=y$. Similarly $f^{-1}: \mathrm{B} \rightarrow \mathrm{A}$, so for every $y \in \mathrm{B}$ there exists $x \in \mathrm{A}$ such that $f^{-1}(y)=x$. Further i.e.
\begin{aligned} & f^{-1}{ }_0 g^{-1}(z)=f^{-1}\left(g^{-1}(z)\right)=f^{-1}(y)=x . \ & f^{-1}{ }_0 g^{-1}(z)=x=\left(g_0 f\right)^{-1}(z) . \text { Therefore }\left(g_0 f\right)^{-1}=f^{-1}{ }_0 g^{-1} . \end{aligned}

## 数学代写|离散数学代写Discrete Mathematics代考|INVERSE FUNCTION

$\mathrm{D}\left(f^{-1}\right)=\mathrm{R}(f) \neq \mathrm{B}$，即$\mathrm{D}\left(f^{-1}\right) \neq \mathrm{B}$。因此$f^{-1}$不能是函数。

## 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。