如果你也在 怎样代写离散数学Discrete Mathematics MA210这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。离散数学Discrete Mathematics是(理论)计算机科学、统计学、概率论和代数基础的重要组成部分。这些思想在微积分的不同部分反复出现。许多人认为离散数学是所有现代数学思想中最重要的组成部分。

离散数学Discrete Mathematics 揭秘解释了这一全景的想法在一步一步和可访问的方式。作者是一位著名的教师和说明者，他对将要阅读这本书的学生的水平、他们的背景和他们的优势有很强的认识，并且可以用学生可以自己学习的易于理解的片段来呈现材料。精心挑选的例子和同源练习将强化所呈现的观点。经常复习、评估和应用这些思想将有助于学生保留和内化所有重要的微积分概念。

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数学代写|离散数学作业代写discrete mathematics代考|GRAPH OF FUNCTION

Let $f$ be a function from A to B, i.e. for every $x \in \mathrm{A}$ there exists unique $y \in \mathrm{B}$ such that $y=f(x)$. Further note that using the functional notation, $f$ can be expressed as
$$
f={(x, f(x)): x \in \mathrm{A}} .
$$
This representation is known as the graph of the function $f$. Consider the functions $f_1: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f_1(x)=x+1$; and $f_2: R \rightarrow{-2,2}$ defined by $f_2(x)=\left{\begin{array}{cc}-2 & \text { if } x>0 \ 2 & \text { if } x<0\end{array}\right.$
The graphs of above functions are given below.

Now consider the relations $f_1:[-4,4] \rightarrow[-4,4]$ defined by $\left[f_1(x)\right]^2=16-x^2 ; x \in[-4,4]$ and $f_2: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\left[f_2(x)\right]^2=16 x ; x \in \mathrm{R}$. The graphs of above relations are given below.
These are nothing but a circle and parabola respectively. Where in figure $-1, y=f_1(x)$ and in figure $-2, y=f_2(x)$.

From the graph it is clear that for one value of $x$ in the domain set leads to two values in the range set. Hence these relations are not functions.

数学代写|离散数学作业代写discrete mathematics代考|COMPOSITION OF FUNCTIONS

Let $f$ be a function from the set $\mathrm{A}$ to the set $\mathrm{B}$ and $g$ be a function from the set $\mathrm{B}$ to the set $\mathrm{C}$. Then the composition of the functions $\mathrm{f}$ and $\mathrm{g}$ is given as $\left(g_0 f\right)$ or $g f$. This is a function from the set A to the set C. It may also be noted that domain of $g$ is equal to co-domain of $f$.

As $f$ is a function from the set A to the set B, then for every $x \in$ A there exists unique $y \in B$ such that $y=f(x)$. Similarly $g$ is a function from the set B to the set $\mathrm{C}$, then for every $y \in \mathrm{B}$ there exists unique $z \in \mathrm{C}$ such that $z=g(y)$. Again $\left(g_0 f\right)$ is a function from the set A to the set $\mathrm{C}$, so we get i.e. $\left(g_{\circ} f\right)(x)=z$ for all $x \in$ A.
i.e. $(g \circ f)(x)=g(y)$ $\left(g_{\circ} f\right)(x)=g(f(x))$
Consider two functions $f(x)=2 x+5$ and $g(x)=3 x$.
Therefore $\left(g_{\circ} f\right)(x)=g(f(x))$
i.e.
Similarly,
$=g(2 x+5)$
$=3(2 x+5)$
$\left(g_{\circ} f\right)(x)=6 x+15$
$\left(f_{\circ} g\right)(x)=f(g(x))$
$=f(3 x)$

i.e.
$$
=2(3 x)+5
$$
$$
\left(f_{\mathrm{o}} g\right)(x)=6 x+5
$$
4.5.1 Theorem
Let $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. Then $\left(g_0 f\right)$ is one-one if both $f$ and $g$ are one-one and $\left(g_o f\right)$ is onto if both $f$ and $g$ are onto.

Proof: Let $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. Since $f$ is a function from the set $\mathrm{A}$ to the set $\mathrm{B}$, then for every $x \in \mathrm{A}$ there exists unique $y \in \mathrm{B}$ such that $y=f(x)$. Similarly $g$ is a function from the set $\mathrm{B}$ to the set $\mathrm{C}$, then for every $y \in \mathrm{B}$ there exists unique $z \in \mathrm{C}$ such that $z=g(y)$.
Suppose the $f$ and $g$ are both one-one. Our claim is $\left(g_{\circ} f\right)$ is one-one. Since $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ we have $\left(g_{\circ} f\right): \mathrm{A} \rightarrow \mathrm{C}$.
Let $x_1, x_2 \in \mathrm{A}$ and $\left(g_{\circ} f\right)\left(x_1\right)=\left(g_{\circ} f\right)\left(x_2\right)$
This implies $\quad g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$
i.e.
$f\left(x_1\right)=f\left(x_2\right)$
i.e.
$x_1=x_2$
$[\because g$ is one-one $]$
$[\because f$ is one-one $]$
Therefore $g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$ implies $x_1=x_2$. So $\left(g_{\circ} f\right)$ is one-one.
Suppose that both $f$ and $g$ are onto. Since $g$ is onto, for every $z \in \mathrm{C}$ there is at least one $y \in \mathrm{B}$ such that $g(y)=z$. Again as $f$ is onto, for every $y \in \mathrm{B}$ there exists at least one $x \in$ A such that $f(x)=y$.

As a result for every $z \in \mathrm{C}$ there is at least one $x \in \mathrm{A}$ such that $\left(g_{\circ} f\right)(x)=z$. Therefore $\left(g{ }_{\circ} f\right)$ is onto.

离散数学代写

数学代写|离散数学作业代写discrete mathematics代考|GRAPH OF FUNCTION

设$f$为从a到B的函数，即对于每个$x \in \mathrm{A}$存在唯一的$y \in \mathrm{B}$，使得$y=f(x)$。进一步注意，使用函数表示法，$f$可以表示为
$$
f={(x, f(x)): x \in \mathrm{A}} .
$$
这种表示称为函数$f$的图形。考虑$f_1(x)=x+1$定义的函数$f_1: \mathrm{R} \rightarrow \mathrm{R}$;$f_2: R \rightarrow{-2,2}$由$f_2(x)=\left{\begin{array}{cc}-2 & \text { if } x>0 \ 2 & \text { if } x<0\end{array}\right.$定义
上述函数的曲线图如下所示。

现在考虑由$\left[f_1(x)\right]^2=16-x^2 ; x \in[-4,4]$定义的关系$f_1:[-4,4] \rightarrow[-4,4]$和由$\left[f_2(x)\right]^2=16 x ; x \in \mathrm{R}$定义的关系$f_2: \mathrm{R} \rightarrow \mathrm{R}$。上述关系的图表如下所示。
它们分别是圆和抛物线。在图$-1, y=f_1(x)$和图$-2, y=f_2(x)$中。

从图中可以清楚地看出，对于域集中的一个值$x$，会导致范围集中的两个值。因此，这些关系不是函数。

数学代写|离散数学作业代写discrete mathematics代考|COMPOSITION OF FUNCTIONS

设$f$是集合$\mathrm{A}$到集合$\mathrm{B}$的函数，$g$是集合$\mathrm{B}$到集合$\mathrm{C}$的函数。然后给出函数$\mathrm{f}$和$\mathrm{g}$的组成为$\left(g_0 f\right)$或$g f$。这是一个从集合a到集合c的函数，也可以注意到$g$的定义域等于$f$的上域。

因为$f$是从集合a到集合B的函数，那么对于每个$x \in$ a都存在唯一的$y \in B$，使得$y=f(x)$。类似地，$g$是从集合B到集合$\mathrm{C}$的函数，那么对于每个$y \in \mathrm{B}$存在唯一的$z \in \mathrm{C}$，使得$z=g(y)$。同样，$\left(g_0 f\right)$是从集合a到集合$\mathrm{C}$的函数，所以我们得到，对于所有的$x \in$ a都是$\left(g_{\circ} f\right)(x)=z$。
即$(g \circ f)(x)=g(y)$$\left(g_{\circ} f\right)(x)=g(f(x))$
考虑两个函数$f(x)=2 x+5$和$g(x)=3 x$。
因此$\left(g_{\circ} f\right)(x)=g(f(x))$
例如:
类似地，
$=g(2 x+5)$
$=3(2 x+5)$
$\left(g_{\circ} f\right)(x)=6 x+15$
$\left(f_{\circ} g\right)(x)=f(g(x))$
$=f(3 x)$

例如:
$$
=2(3 x)+5
$$
$$
\left(f_{\mathrm{o}} g\right)(x)=6 x+5
$$
4.5.1定理
设$f: \mathrm{A} \rightarrow \mathrm{B}$和$g: \mathrm{B} \rightarrow \mathrm{C}$为两个函数。如果$f$和$g$都是1:1，那么$\left(g_0 f\right)$就是1:1;如果$f$和$g$都是onto，那么$\left(g_o f\right)$就是onto。

证明:设$f: \mathrm{A} \rightarrow \mathrm{B}$和$g: \mathrm{B} \rightarrow \mathrm{C}$为两个函数。由于$f$是从集合$\mathrm{A}$到集合$\mathrm{B}$的函数，因此对于每个$x \in \mathrm{A}$都存在唯一的$y \in \mathrm{B}$，使得$y=f(x)$。类似地，$g$是一个从集合$\mathrm{B}$到集合$\mathrm{C}$的函数，那么对于每个$y \in \mathrm{B}$都存在唯一的$z \in \mathrm{C}$，使得$z=g(y)$。
假设$f$和$g$都是1 – 1。我们的主张是$\left(g_{\circ} f\right)$是1比1。因为$f: \mathrm{A} \rightarrow \mathrm{B}$和$g: \mathrm{B} \rightarrow \mathrm{C}$，我们有$\left(g_{\circ} f\right): \mathrm{A} \rightarrow \mathrm{C}$。
让$x_1, x_2 \in \mathrm{A}$和$\left(g_{\circ} f\right)\left(x_1\right)=\left(g_{\circ} f\right)\left(x_2\right)$
这意味着$\quad g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$
例如:
$f\left(x_1\right)=f\left(x_2\right)$
例如:
$x_1=x_2$
$[\because g$是1 – 1 $]$
$[\because f$是1 – 1 $]$
因此$g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$意味着$x_1=x_2$。所以$\left(g_{\circ} f\right)$是1 – 1。
假设$f$和$g$都在上。因为$g$是上的，所以对于每个$z \in \mathrm{C}$，至少有一个$y \in \mathrm{B}$使得$g(y)=z$。同样，由于$f$是对的，对于每个$y \in \mathrm{B}$，至少存在一个$x \in$ A，使得$f(x)=y$。

因此，对于每个$z \in \mathrm{C}$，至少有一个$x \in \mathrm{A}$使得$\left(g_{\circ} f\right)(x)=z$。因此$\left(g{ }_{\circ} f\right)$是on。

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