Statistics-lab™可以为您提供tesu.edu MAT270 discrete mathematics离散数学课程的代写代考和辅导服务！
MAT270 discrete mathematics课程简介
Discrete Mathematics is designed to meet the needs not only of students majoring in computer science but of a wider audience, especially students in mathematics and science. The course provides tools for formal reasoning. Topics include counting rules, propositional and first-order logic, set theory, functions (with an emphasis on recursive functions), partial order and equivalence relations, Boolean algebra, and switching circuits. Graphs and trees are also introduced. With an emphasis on communication skills, students are required to interpret, describe, discuss, and justify conclusions based on logical reasoning. While the particular focus of the course is on reasoning related to computer programs, no knowledge of programming is required.
Advisory: It is advisable to have knowledge in a course equivalent to MAT-121: College Algebra with a grade of C or better to succeed in this course. Students are responsible for making sure that they have the necessary knowledge.
PREREQUISITES
Discrete Mathematics is a fundamental course that provides a solid foundation for computer science, mathematics, and science students. The course covers a wide range of topics, including counting rules, logic, set theory, functions, relations, Boolean algebra, and graph theory. The emphasis on formal reasoning and communication skills makes this course valuable to all students.
To succeed in this course, it is advisable to have a good understanding of college algebra, as it forms the basis of many concepts covered in the course. Students should ensure that they have the necessary prerequisite knowledge before enrolling in the course.
The course provides tools for formal reasoning, which are essential for students pursuing computer science or related fields. These tools enable students to analyze and design algorithms, write correct programs, and prove the correctness of programs.
Overall, discrete mathematics is a challenging but rewarding course that prepares students for further study in computer science, mathematics, and science. The skills and knowledge gained in this course are valuable not only in academia but also in industry and other fields.
MAT270 discrete mathematics HELP（EXAM HELP， ONLINE TUTOR）
- A stack of cans (see Figure 1) refers to an arrangement of $n$ (equal-sized) cans in rows such that (i) the cans in each row form a single continuous block, and (ii) any can, except the ones in the bottom row, touches exactly two cans from the row beneath it. If the bottom row contains $k$ cans, this is a $(n, k)$ stack of cans. Figure 1 shows a $(28,12)$ stack of cans.
Your article should give a formula (as a function of $k$ ) for the number $s_k$ of stack of cans with exactly $k$ cans in the bottom row (and an arbitrary total number of cans). Your article should at least contain:
- A generating function for $\left(s_k\right)_{k \geq 0}$,
- A closed-form formula for $s_k$ for any $k \geq 0$,
- A relation between $s_k$ and another well-known sequence of numbers.
If you can find a bijective proof for $s_k$, you could also provide it. This is however not compulsory and should not replace any of the items above.
A stack of cans is an arrangement of $n$ cans in rows such that each row contains a single continuous block, and except for the bottom row, each can in the stack touches exactly two cans from the row below it. We call a stack with $k$ cans in the bottom row a $(n,k)$ stack. In this article, we will derive a formula for the number of $(n,k)$ stacks, denoted by $s_k$.
Let $s_n$ be the number of $(n,k)$ stacks. We can define the generating function $S(x) = \sum_{n \geq 0} s_n x^n$. Consider the first can in the bottom row of a $(n,k)$ stack. We can place this can in the leftmost or rightmost position of the bottom row, so there are $k$ choices for the position of this can. The cans above this can can be partitioned into two stacks, one to the left and one to the right. Let the sizes of these stacks be $a$ and $b$ respectively, so $a + b + 1 = n – k$. The stack to the left must have at least one can, so $a \geq 1$. The generating function $S(x)$ can then be expressed as:
S(x) = x^k \sum_{a=1}^{n-k-1} \sum_{b=0}^{n-k-1-a} s_a x^a s_b x^b.S(x)=xka=1∑n−k−1b=0∑n−k−1−asaxasbxb.
The factor of $x^k$ comes from the choice of position for the first can in the bottom row, and the limits of the summations ensure that the sizes of the left and right stacks are nonnegative and sum to $n-k-1$.
We can simplify this expression using the fact that $s_0 = 1$ (there is only one way to stack zero cans), so:
\begin{aligned} S(x) &= x^k \sum_{a=1}^{n-k-1} \sum_{b=0}^{n-k-1-a} s_a x^a s_b x^b \\ &= x^k \sum_{a=0}^{n-k-1} \sum_{b=0}^{n-k-1-a} s_a x^a s_b x^b – s_0 x^k \\ &= x^k \left(\sum_{a=0}^{n-k} s_a x^a\right)^2 – x^k. \end{aligned}S(x)=xka=1∑n−k−1b=0∑n−k−1−asaxasbxb=xka=0∑n−k−1b=0∑n−k−1−asaxasbxb−s0xk=xk(a=0∑n−ksaxa)2−xk.
Thus, we have the formula:
S(x) = \frac{x^k}{1 – \left(\sum_{a=0}^{n-k} s_a x^a\right)^2}.S(x)=1−(∑a=0n−ksaxa)2xk.
We can use partial fractions to expand the generating function $S(x)$ into a sum of simpler terms. Let $\alpha_i$ be the distinct roots of the polynomial $\sum_{a=0}^{n-k} s_a x^a$, and let $\beta_i$ be the multiplicities of these roots. Then:
S(x) = \frac{x^k}{\prod_{i=1}^m (1 – \alpha_i x)^{\beta_i}} = \sum_{i=1}^m \sum_{j=1}^{\beta_i} \frac{c_{i,j}}{(1 – \alpha_i x)^j},S(x)=∏i=1m(1−αix)βixk=i=1∑mj=1∑βi(1−αix)jci,j,
where $c_{i,j}$ are constants to be determined. Equating the coefficients of $x^n$ on both sides
- Recall the inclusion / exclusion formula, let $A_1, A_2, \ldots, A_n$ be events, and
$$
S_k:=\sum_{1 \leq i_1<i_2<\cdots<i_k \leq n} \mathbb{P}\left(A_{i_1} \wedge A_{i_2} \wedge \ldots \wedge A_{i_k}\right) .
$$
The inclusion/exclusion formula gives
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right)=\sum_{j=1}^n(-1)^{j-1} S_j
$$
In this problem we will develop and use an extension of it.
- Show that for all odd $k$ in ${1, \ldots, n}$,
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right) \leq \sum_{j=1}^k(-1)^{j-1} S_j
$$
and describe how the union bound is a special case of this. - Show that for all even $k$ in ${1, \ldots, n}$,
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right) \geq \sum_{j=1}^k(-1)^{j-1} S_j
$$
We begin by observing that for any $k\in{1,\ldots,n}$, we have \begin{align*} \mathbb{P}\left(\bigvee_{i=1}^n A_i\right) &= \sum_{i=1}^n\mathbb{P}(A_i) – \sum_{1\le i<j\le n}\mathbb{P}(A_i\wedge A_j) + \sum_{1\le i<j<k\le n}\mathbb{P}(A_i\wedge A_j\wedge A_k) – \cdots\ &\qquad\cdots + (-1)^{n-1}\mathbb{P}(A_1\wedge A_2\wedge\cdots\wedge A_n). \end{align*} We denote the $k$th term of this expansion by $T_k$.
Now suppose that $k$ is odd. We want to show that $\mathbb{P}(\bigvee_{i=1}^n A_i)\leq \sum_{j=1}^k(-1)^{j-1}S_j$. We have \begin{align*} \sum_{j=1}^k(-1)^{j-1}S_j &= T_1 – T_2 + T_3 – \cdots + (-1)^{k-1}T_k\ &= \mathbb{P}(A_1) – \sum_{1\le i<j\le n}\mathbb{P}(A_i\wedge A_j) + \cdots + (-1)^{k-1}\sum_{1\le i_1<i_2<\cdots<i_k\le n}\mathbb{P}(A_{i_1}\wedge A_{i_2}\wedge\cdots\wedge A_{i_k})\ &\ge \mathbb{P}(A_1) – \sum_{1\le i<j\le n}\mathbb{P}(A_i\wedge A_j) + \cdots + (-1)^{k-1}\mathbb{P}\left(\bigvee_{i=1}^n A_i\right)\ &= \sum_{j=1}^{k-1}(-1)^{j-1}S_j + (-1)^{k-1}\mathbb{P}\left(\bigvee_{i=1}^n A_i\right)\ &\ge (-1)^{k-1}\mathbb{P}\left(\bigvee_{i=1}^n A_i\right), \end{align*} where we have used the fact that $\mathbb{P}(\bigvee_{i=1}^n A_i)\geq 0$ and that $(-1)^{j-1}S_j\geq 0$ for all $j\in{1,\ldots,k}$.
Textbooks
• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.
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