MIT基础数学讲义(计算机系)lecture19

合集下载

mit公开课《线性代数》

mit公开课《线性代数》线性代数是数学领域里一个基础且非常重要的课程，它主要探讨这样一种问题：有多少种不同的变量和它们之间的关系，以及如何找到最优的模型来描述它们的关系。

美国麻省理工学院（MIT）已经推出了一门免费的《线性代数》课程，这门课旨在让学生掌握线性代数的基本概念和方法，以便他们能够在解决实际问题的解决方案中使用线性代数的工具。

该课程主要涵盖了一系列理论方面的话题，包括矩阵代数、向量空间、线性变换、矩阵特征分解、行列式、谱分解和特征值分解。

在学习这些概念和理论之后，学生还会学习如何将这些概念应用到实际的应用中去，以帮助他们在实际的问题中运用线性代数的技能。

此外，该课程还涵盖了有关概率论和统计方面的话题，其中包括概率分布、分布函数、偏差、协方差和相关系数，以及概率分析中的各种方法，如极大似然估计和贝叶斯统计。

学生通过学习这些内容，可以从定量和定性数据中发现规律和模式，从而帮助他们解决各种问题。

本课程还包括有关凸优化的内容，其中包括对有关线性系统的数学模型的理解以及如何使用这些模型来解决实际问题的技术。

此外，学生还会学习如何使用数值方法来解决复杂的线性系统，如最小二乘法、拟牛顿法、拉格朗日方法等。

学习这些数学模型和数值方法，可以帮助学生了解如何实现更加优化的解决方案。

本课程还涉及优化技术在计算机科学中的应用，例如机器学习、信号处理、图像处理、大数据处理等。

学生们需要学习如何利用优化技术解决这些问题，从而向他们提供更有效的解决方案。

此外，本课程还将涉及有关可视化的概念，包括可视化技术如统计图表、折线图、散点图等，这些技术可以帮助学生们更好地理解数据和模型，也可以让他们更好地解决实际问题。

MIT的《线性代数》课程是一门非常实用和有价值的免费课程，可以帮助学生们更好地理解线性代数的概念、方法和技能，从而为他们在实际问题中运用线性代数提供有力的帮助。

通过学习线性代数的知识，学生们可以更好地解决复杂的实际问题，从而更好地应用数学在实践中。

MIT公开课-线性代数笔记

目录方程组的几何解释 (2)矩阵消元 (3)乘法和逆矩阵 (4)A的LU分解 (6)转置-置换-向量空间R (8)求解AX=0：主变量，特解 (9)求解AX=b：可解性和解的解构 (10)线性相关性、基、维数 (11)四个基本子空间 (12)矩阵空间、秩1矩阵和小世界图 (13)图和网络 (14)正交向量与子空间 (15)子空间投影 (18)投影矩阵与最小二乘 (20)正交矩阵和Gram-Schmidt正交化 (21)特征值与特征向量 (27)对角化和A的幂 (28)微分方程和exp(At)（待处理） (29)对称矩阵与正定性 (29)正定矩阵与最小值 (31)相似矩阵和若尔当型（未完成） (32)奇异值分解(SVD) (33)线性变换及对应矩阵 (34)基变换和图像压缩 (36)NOTATIONp：projection vectorP：projection matrixe：error vectorP：permutation matrixT：transport signC(A)：column spaceN(A)：null spaceU：upper triangularL：lower triangularE：elimination matrixQ：orthogonal matrix, which means the column vectors are orthogonalE：elementary/elimination matrix, which always appears in the elimination of matrix N：null space matrix, the “solution matrix” of AX=0R：reduced matrix, which always appears in the triangular matrix, “IF00”I：identity matrixS：eigenvector matrixΛ：eigenvalue matrixC：cofactor matrix关于LINER ALGEBA名垂青史的分析方法：由具象到抽象，由二维到高维。

MIT基础数学讲义(计算机系)lecture4

her zero or one edges joining a pair of vertices. In a multigraph, multiple edges are permitted between the same pair of vertices.
2 De nitions
A nuisance in rst learning graph theory is that there are so many de nitions. They all correspond to intuitive ideas, but can take a long time to absorb. Worse, the same thing often has several names and the same name sometimes means slightly di erent things to di erent people! It's a big mess, but muddle through.
2.2 Not-So-Simple Graphs
There are actually many variants on the de nition of a graph. The de nition in the preceding section really only describes simple graphs. There are many ways to complicate matters.
2.1 Simple Graphs
A graph is a pair of sets (V E ). The elements of V are called vertices. The elements of E are called edges. Each edge is a pair of distinct vertices. Graphs are also sometimes called networks. Vertices are also sometimes called nodes. Edges are sometimes called arcs. Graphs can be nicely represented with a diagram of dots and lines as shown in Figure 2 As noted in the de nition, each edge (u v ) 2 E is a pair of distinct vertices u v 2 V . Edge (u v ) is said to be incident to vertices u and v . Vertices u and v are said to be adjacent or neighbors. Phrases like, \an edge joins u and v " and \the edge between u and v " are comon. A computer network is can be modeled nicely as a graph. In this instance, the set of vertices V represents the set of computers in the network. There is an edge (u v) if there is a direct communication link between the computers corresponding to u and v .

MIT基础数学讲义(计算机系)lecture1

Massachusetts Institute of Technology 6.042J/18.062J: Mathematics for Computer Science Professor Tom Leighton
Lecture 1 5 Sep 97
What is a Proof? 1 What is a Proof?
Axiom 1 If a = b and b = c, then a = c.
This seems very reasonable! But sometimes the right choice of axiom is not clear.
Axiom 2 (Euclidean geometry) Given a line l and a point p not on l, there is exactly one line
many lines through p parallel to l.
No one of the three preceding axioms is better than the others all yield equally good proofs. Of course, a di erent choice of axioms makes di erent propositions true. Still, a set of axioms should not be chosen arbitrarily. In particular, there are two basic properties that one would want in any set of axioms it should be consistent and complete.

MIT基础数学讲义(计算机系)lecture15

every item of B, then A has k times as many items as B.
For example, suppose A is a set of students, B is a set of recitations, and f de nes the assignment
f f;1( x1x2 : : : xn]) = (x1 x2 x3 : : : xn)
(xn x1 x2 : : : xn;1)
(xn;1 xn x1 : : : xn;2)
: : :x1 appears (x2 x3 x4 : : :
in n di
x1)g
erent
places : : :
By the Division Rule, jAj = njBj. This gives:
de in
nition, however, f A to some element
;b12(bB) c,atnhebne
a set,
f ;1(b)
not just a single value. For exampe, if f
is actually the empty set. In the special
4
Lecture 15: Lecture Notes
1 The Division Rule
We will state the Division Rule twice, once informally and then again with more precise notation.
Theorem 1.1 (Division Rule) If B is a nite set and f : A 7! B maps precisely k items of A to

MIT基础数学讲义(计算机系)lecture16

(For example, let S be the set fA B C D Eg, with elements ordered alphabetically. Let R be the 7-combination with repetition fA B B B D E Eg. The stars-and-bars string corresponding to R is shown below.
n
+
r r
;
1!:
2
Lecture 16: Lecture Notes
In the example above, we found six ways to choose two elements from the set S = fA B Cg with
rseept eistit;i3o+n22;a1llo=we6d.. Sure enough, the theorem says that the number of 2-combinations of a 3-element
swthaircshainsd;nb+arrr;s1.
This .
is
the
number
of
ordinary
r-combinations
of
a
set
with
n
+
r
;
1
elements,
1.2 Triple-Scoop Ice Cream Cones
Baskin-Robbins is an ice cream store that has 31 di erent avors. How many di erent triple-scoop ice cream cones are possible at Baskin-Robbins? Two ice cream cones are considered the same if one can be obtained from the other by reordering the scoops. Of course, we are permitted to have two or even three scoops of the same avor.

MIT基础数学讲义(计算机系)lecture18

Yet some people felt that the probability of disaster was su ciently high that they staged vigorous protests and led suit in federal court to prevent the launch. How could there be such intense dispute about a probability question? Part of the problem is that probability is a tricky subject. In principle, probability boils down to a few simple rules. The di culty is that these rules often produce highly counterintuitive results \common sense" often leads to the wrong answer! We will see two examples of this today.
1 The Monty Hall Problem
In the 1970's, there was a game show called Let's Make a Deal, hosted by Monty Hall and his assistant Carol Merryl. At one stage of the game, a contestant is shown three doors. The contestant knows there is a prize behind one door and that there are goats behind the other two. The contestant picks a door. To build suspense, Carol always opens a di erent door, revealing a goat. The contestant can then stick with his original door or switch to the other unopened door. He wins the prize only if he now picks the correct door. Should the contestant \stick" with his original door, \switch" to the other door, or does it not matter? We will analyze the probability that the contestant wins with the \switch" strategy that is, the contestant chooses a random door initially and then always switches after Carol reveals a goat behind one door. The analysis can be broken down into four steps.

MIT基础数学讲义(计算机系)lecture8

1 Annuities
Would you prefer a million dollars today or $50,000 a year for the rest of your life? This is a question about the value of an annuity. An annuity is a nancial instrument that pays out a xed amount of money at the beginning of every year for some speci ed number of years. In particular, an n-year, m-payment annuity pays m dollars at the start of each year for n years. In some cases, n is nite, but not always. Examples include lottery payouts, student loans, and home mortgages. There are even Wall Street people who specialize in trading annuities. A key question is what an annuity is worth. For example, lotteries often pay out jackpots over many years. Intuitively, $50,000 a year for 20 years ought to be worth less than a million dollars right now. If you had all the cash right away, you could invest it and begin collecting interest. But what if the choice were between $50,000 a year for 20 years and a half million dollars today? Now it is not clear which option is better. In order to answer such questions, we need to know what a dollar paid out in the future is worth today. We will to assume that money can be invested at a xed annual interest rate p. These days a good estimate for p is around 8% we'll use this value for the rest of the lecture. Here is why the interest rate p matters. Ten dollars invested today at interest rate p will become (1 + p) = $10:80 in a year, (1 + p)2 $11:66 in two years, and so forth. Looked at another way, ten dollars paid out a year from now are only really worth 1=(1 + p) $9:26 today. The reason is that if we had the $9:26 today, we could invest it and would have $10.00 in a year anyway. Therefore, p determines the value of money paid out in the future.

MIT机电工程与计算机科学系【本科生课程】6.042J.计算机科学数学

Mathematics for Computer ScienceAs taught in: Spring 2010The Fifteen Puzzle. See the Lecture 12 in-class problems for more information about this game. Image courtesy of Nick Matsakis.Instructors:Prof. Albert R. MeyerMIT Course Number:6.042J /18.062JLevel:UndergraduateCourse DescriptionThis subject offers an introduction to Discrete Mathematics oriented toward Computer Science and Engineering. Thesubject coverage divides roughly into thirds: 1. Fundamental concepts of mathematics: definitions proofs sets functions relations. 2. Discrete structures: graphs state machines modular arithmetic counting. 3. Discrete probability theory.On completion of 6.042 students will be able to explain and apply the basic methods of discrete noncontinuousmathematics in Computer Science. They will be able to use these methods in subsequent courses in the design andanalysis of algorithms computability theory software engineering and computer systems.SyllabusCourse Meeting TimesLectures: 3 sessions / week 1.5 hours / sessionContents Introduction Considerations for Taking the Subject This Term Problem Sets Online Tutor Problems Weekly Reading Comments Biweekly Mini-quizzes Final Exam Collaboration Grades LaTeX MacrosIntroductionThis subject offers an introduction to Discrete Mathematics oriented toward Computer Science and Engineering. Thesubject coverage divides roughly into thirds:1. Fundamental concepts of mathematics: definitions proofs sets functions relations.2. Discrete structures: graphs state machines modular arithmetic counting.3. Discrete probability theory.The prerequisite is 18.01 first term calculus in particular some familiarity with sequences and series limits anddifferentiation and integration of functions of one variable.The goals of the course are summarized in a statement of Course Objectives and Educational Outcomes.Considerations for Taking the Subject This TermThere are two main considerations for students in deciding to take 6.042J/18.062J this term—or at all.1. Team Problem Solving This term as in many previous terms the subject is being taught in Lecture/Team Problem Solving style. More than half the class meeting time will generally be devoted to problem solving in teams of 7 or 8 sitting around a table with a nearby whiteboard where a team can write their solutions. Each TA covers 3 tables acting as coach and providing feedback on students solutions. The Lecturer acts likewise circulating among all the tables. The coach will resist answering questions about the material always trying first to find a team member who can explain the answer to the rest of the team. Of course the coach will provide hints and explanations when the whole team is stuck. Problem solving sessions will generally be preceded by half hour presentations by the lecturer usually reviewing just the topics needed to understand the problems. These overviews are not intended as first-time introductions to the material nor as complete coverage of the assigned reading. The Good News is that the immediate active engagement in problem solving sessions is an effective and enjoyable way for most students to master the material. Team sessions also provide a supervised setting to acquire and practice technical communication skills. Student grades for problem solving sessions depend on degree of active prepared participation rather than problem solving success. Sessions are not aimed to test how well a student can solve the problems in class the goal is to have them understand how to solve them by the end of the session. Participation in team sessions counts for 20 of the final grade. In-class team problem solving works to solidify students understanding of material they have already seen. The Bad News is that this requires students to arrive prepared at the sessions: they need to have read though not carefully studied the assigned reading and done the OnlineTutor problems before class. There is no way to make up for not working with the team so it is necessary to keep up and be there—no focusing on something else for a month aiming to catch up afterward. If you cannot commit to keeping up you may prefer to take the subject some other term.2. This subject covers many mathematical topics that are essential in Computer Science but are not covered in the standard calculus curriculum. In addition the subject teaches students about careful mathematics: precisely stating assertions about well-defined mathematical objects and verifying these assertions using mathematically sound proofs. While some students have had earlier exposure to some of these topics in most cases they learn a lot more in 6.042J/18.062J.The subject is required of all Computer Science 6-3 majors and is in a required category for Math majors taking theComputer Science option 18C. But students with a firm understanding of sound proofs and who are familiar with manyof the covered topics should discuss substituting a more advanced Math subject for 6.042 with the Lecturer or theiradvisor.Problem SetsProblem Sets are normally due at the beginning of lecture on Fridays but a few may be due at alternate times becauseof holidays. Doing the problem sets is for most students crucial for mastering the course material. Solutions to theproblem sets will be posted immediately after the due date. Consequently late problem sets will not be accepted.Problem sets count for 25 of the final grade.Students are encouraged to collaborate on problem sets as on teams in class. The last page of each problem set has acollaboration statement to be completed and attached as the first page of a problem set submission:quotI worked alone and only with course materialsquotorquotI collaborated on this assignment with students in classgot help from people other than collaborators and course staffand referred to citations to sources other than the material from this termquot.No problem set will be given credit until it has a collaboration statement.Online Tutor ProblemsThere are weekly Online Tutor problems due before class on specified dates. These consist of straightforward questionsthat provide useful feedback about the assigned material. Tutor problems should take about 20 minutes after thereading has been completed. Some students prefer to try the tutor problems before doing the reading which is fine.Like team problem-solving in class online tutor problems are graded solely on participation: students receive full creditas long as they try the problem even if their answer is wrong. Tutor problems count for 5 of the final grade.Weekly Reading CommentsA comment on the weeks reading using the NB online annotation system is due on specified dates by9AM before class.A single comment citing some paragraph that specially catches your attention is all that is required. The commentshould indicate why this paragraph stood out for example because you found it especially difficult/confusing or surprising or mistaken pointing out typos amp suggesting corrections is appreciated or funny or boring or lacking Computer Science motivation or poorly written something youd like reviewed in class ments in response to other peoples comments are generally also OK to satisfy this requirement. Multiple commentsare welcomed.Note that global comments such as quotI understood everything in the reading found it all interesting and have noquestionsquot are not considered responsive. Even if you understood everything there must in the 15 to 30 pagesassigned each week have been something that stood out for you as suggested above. Similarly responses to othercomments such as quotI agreequot quotlolquot... are not sufficient.Reading comments count for 5 of the finalgrade.Collaboration and Outside SourcesWe encourage students to collaborate on homework as on in-class problems. Study groups can be a big help inmastering coursematerial besides being fun and a good way to make friends. However students must write upsolutions on their own neither copying solutions nor providing solutions to be copied. All collaborators must be citedand if a source beyond the course materials is used in a solution—for example an quotexpertquot consultant other than 6.042staff or another text—there must be a proper scholarly citation of the source.We discourage but do not forbid use of materials from prior terms other than those available on OCW. We repeathowever that use of material from any previous term requires a proper scholarly citation. As long as a student providesaccurate citation and collaboration statements a questionable submission will rarely be sanctioned—instead wellexplain why we judge the submission unsatisfactory and maybe deny credit for specific clearly copied solutions. Butomission of such a citation will be taken as a priori evidence of cheating with unpleasant consequences for everyone.Biweekly Mini-quizzesA 25-30 minute mini-quiz will generally be given every other week usually on Wednesdays. Mini-quizzes count for atotal of 20 of the final grade.Material to study for a mini-quiz is very well defined: a mini-quiz will cover only the material in problems from theprevious two weeks.Mini-quiz questions are often simply some parts of these online class and problem set problems.Students can prepare for a mini-quiz simply by reviewing the posted problem solutions for the previous two weeks.Final ExamThere will be a standard 3 hour final exam. This exam is worth 25 of the final course grade.GradesThe lowest mini-quiz score and problem set score and the lowest two team problem-solving scores will not count ingrade calculation. This effectively gives everyone 1 mini-quiz 1 problem set and 2 team problem-solving sessions theycan miss without excuse or penalty.Grades for the course will be based on the following weighting: ACTIVITIES PERCENTAGES ACTIVITIES PERCENTAGESProblem sets 25Final 25Class participation 20Mini-quizzes 20Weekly reading comments 5Online tutor problems 5LaTeX MacrosCourse handouts are formatted using LaTeX which is the preferred formatting system among Mathematicsprofessionals. Note that we do not think its worthwhile for students to use it for their class submissions.Calendar LEC TOPICS KEY DATESWeek 11 Good and bad proofs2 Proof by contradictionWeek 23 Well ordering principle4 Propositionallogic5 Sets and relations Problem set 1 dueWeek 36 Size of sets mapping lemma7 Predicates and quantifiers Mini-quiz 18 Set theory Russell paradox Problem set 2 dueWeek 49 Induction and strong induction10 Partial ordersLEC TOPICS KEY DATES11 Partial orders and scheduling Problem set 3 dueWeek 512 Digraphs13 State machines preserved invariants Mini-quiz 214 Derived variables termination Problem set 4 dueWeek 615 Stable matching16 Simple graphs degrees isomorphism17 Graph connectedness trees Problem set 5 dueWeek 718 Graph coloring bipartite matching19 Recursive data Mini-quiz 320 Planar graphs Problem set 6 dueWeek 821 GCD and integer linear combinations22 Modular arithmetic23 Inverses mod n RSA encryption Problem set 7 dueWeek 924 Harmonic sums Stirlings approximation25 Asymptotics Mini-quiz 426 Counting with bijections Problem set 8 dueWeek 1027 Pigeonhole and division rules28 Counting repetitions card magic29 Inclusion-exclusion counting practice Problem set 9 dueWeek 11 LEC TOPICS KEY DATES30 Binomial theorem combinatorial identities Mini-quiz 531 Generating functions: for counting Problem set 10 dueWeek 1232 Generating functions: for recurrences33 Introduction to probability34 Conditional probability independence Problem set 11 dueWeek 1335 Random variables36 Expectation Mini-quiz 637 Variance Problem set 12 dueWeek 1438Sampling and confidence39 Random processesReadingsThis section contains the complete course notes Mathematics for Computer Science.NB allows you to annotate PDFs for yourself or for sharing with others. We are piloting NB on this course and will use theresults of this pilot to decide whether this tool will be useful for other courses. CHAPTERS FILES NB PDFSComplete course notes PDF - 4.7MBChapter 1: What is a proof PDF NB PDFChapter 2: The well ordering principle PDF NB PDFChapter 3: Propositional formulas PDF NB PDFChapter 4: Mathematical data types PDF NB PDFChapter 5: First-order logic PDF NB PDFChapter 6: Induction PDF NB PDFC hapter 7: Partial orders PDF NB PDF CHAPTERS FILES NB PDFSChapter 8: Directed graphs PDF NB PDFChapter 9: State machines PDF NB PDFChapter 10: Simple graphs PDF NB PDFChapter 11: Recursive data types PDF NB PDFChapter 12: Planar graphs PDF NB PDFChapter 13: Communication networks PDF NB PDFChapter 14: Number theory PDF NB PDFChapter 15: Sums and asymptotics PDF NB PDFChapter 16: Counting PDF NB PDFChapter 17: Generating functions PDF NB PDFChapter 18: Introduction to probability PDF NB PDFChapter 19: Random processes PDF NB PDFChapter 20: Random variables PDF NB PDFChapter 21: Deviation from the mean PDF NB PDFLecture NotesThis section contains slides in-class problems and solutions. Students work in groups to solve the in-class problemssee the Syllabus for moreinformation.Powerpoint and LaTeX source files and LaTeX macros are available to instructors by request: email Prof. Albert Meyerat meyer at csail dot mit dot edu.NB allows you to a.。

MIT本科计算机教材

30 教材名称: COMPUTER SIMULATION OF LIOUIDS
作者: ALLEN
31 教材名称: CONTROL OF UNCERTAIN SYSTEMS
作者: DAHLEH
32 教材名称: CONTROL SYSTEM DESIGN
作者: FRIEDALND
作者: PERLMAN
62 教材名称: INTRO TO ALGORITHMS
作者: CORMEN
63 教材名称: INTRO.TO FORTRAN 90 F/ENGRS.+SCI.
作者: NYHOFF
64 教材名称: INTRO.TO MATLAB F/ENGRS+SCI.
作者: HOWE
76 教材名称: MICROSOFT ACCESS 2000 BIBLE
作者: PRAGUE
77 教材名称: MICROSYSTEM DESIGN
作者: SENTURIA
78 教材名称: MICRSFT,VISUAL BASIC:PROGRAMMER'S GDE
作者: MENEZES
59 教材名称: HOW COMPUTERS WORK,MILLENIUM ED.-W/CD
作者: WHITE
60 教材名称: HOW TO SET UP+MAINTAIN A WEB SITE-W/CD
作者: STEIN
61 教材名称: INTERCONNECTIONS:BRIDGES,ROUTERS…
作者: HERTZ
68 教材名称: Introduction to Random Signals & Applied Kalman Filtering 3/E (With Matlab Exercise & Solutions )

MIT基础数学讲义(计算机系)lecture2

Massachusetts Institute of Technology 6.042J/18.062J: Mathematics for Computer Science Professor Tom Leighton
Lecture 2 9 Sep 97
Lecture Notes 1 Proof by Induction
X
n i
=1
i
or
1
X
i
n
i
or
X
i
1
i
n
In this notation, the pattern of terms in the summation is made explicit. In two important special cases, the de nition of the summation 1 + 2 + 3 + : : : + n requires some care: If n = 1, then 1+2+3+ : : :+ n = 1. There is only one term in the summation the appearance of 2 and 3 and the presence of both 1 and n is misleading! If n 0, then 1 + 2 + 3 + : : : + n = 0. There are no terms at all in the summation, so the result is zero. The text of a proof by induction should consist of four parts. We'll discuss each part below and then write a proof for the preceding theorem following this outline. 1. State that the proof is by induction. This immediately conveys the general structure of the argument. 2. Specify the induction hypothesis, P (n). Sometimes, the choice of P (n) will come directly from the theorem statement. In this case P (n) will simply be the predicate 1+2+3+ : : : + n = n(n + 1)=2. Other times, the choice of P (n) is not obvious at all we will see an example of this at the end of the lecture. 3. The basis step: prove P (0). The \basis step" or \base case" is a proof of the predicate P (0). In this case, we must prove that 1 + 2 + 3 + : : : + n = n(n + 1)=2 when n = 0. This is trivial because both sides of the equation are zero. 4. The inductive step: prove that 8n 2 N P (n) ) P (n + 1). Begin the inductive step by writing, \For n 0, assume" P (n) \to prove" P (n + 1). (Substitute in the statements of the predicates P (n) and P (n + 1).) Then verify that for all n 0, P (n) indeed implies P (n + 1). A complete proof of the theorem that follows this four-part formula is given below. Proof. The proof is by induction. Let P (n) be the predicate 1 + 2 + 3 + : : : + n = n(n + 1)=2. In the base case, P (0) is true because both sides of the equation are zero. For n 0, assume 1 + 2 + 3 + : : : + n = n(n + 1)=2 to prove that 1 + 2 + 3 + : : : + n + (n + 1) = (n + 1)(n + 2)=2. We can now reason as follows. 1 + 2 + 3 + : : : + n + (n + 1) = ( + 1) + (n + 1) 2 2 n + n + 2n + 2 = 2 2 n + 3n + 2 = 2 (n + 1)(n + 2) = 2

MIT基础数学讲义(计算机系)lecture3

Massachusetts Institute of Technology 6.042J/18.062J: Mathematics for Computer Science Professor Tom Leighton
Lecture 3 11 Sep 97
Lecture Notes 1 Strong Induction
Claim 2.1
8n
8, it is possible to produce n cents of postage from 3 and 5 cents stamps.
Now let's preview the proof. A proof by strong induction will have the same four-part structure as an ordinary induction proof. The base case, P (8), will be easy because we can produce 8 cents of postage from a 3 cent stamp and a 5 cent stamp. In the inductive step we assume that we know how to produce 8 9 : : : n cents of postage, and we have to show how to produce n + 1 cents of postage. One way is rst to create n ; 2 cents of postage using strong induction and then to add a three cent stamp. We have to be careful there is a pitfall here. If n is 8 or 9, then we can not use the trick of creating n + 1 cents of postage from n ; 2 cents and a 3 cent stamp. In these cases, n ; 2 is less than 8. None of the assumptions that strong induction permits help us make less than 8 cents of postage. Fortunately, making n + 1 cents of postage when n is 8 or 9 is very easy we can do this directly. Proof. The proof is by strong induction. Let P (n) be the predicate that it is possible to produce n cents of postage from 3 and 5 cent stamps. In the basis step, P (8) is true because 8 cents of postage can be made from a 3 cent stamp and a 5 cent stamp.

MIT基础数学讲义(计算机系)lecture 25

1.1 Markov's Inequality
We can obtain a weak bound on the probability that at least one of the events A1 A2 : : : AN occurs using Markov's Theorem:
2 Ex(T ) 1
1.2 An Alternate Proof
We can obtain Fact 1 in another way. Recall Boole's Inequality, which says that for any events A1 A2 : : : AN , we have: Pr(A1 A2 : : : AN ) Pr(A1 ) + Pr(A2) + : : : + Pr(AN )
Massachusetts Institute of Technology 6.042J/18.062J: Mathematics for Computer Science Professor Tom Leighton
Lecture 25 4 Dec 97
Lecture Notes
This lecture is devoted to one rather general probability question. Let A1 A2 : : : AN be mutually independent events over the same sample space. Let the random variable T be the number of these events that occur. What is the probability that at least k events occur? That is, what is Pr(T k)? This question comes up often. For example, suppose we want to know the probability that at least k heads come up in N tosses of a coin. Here Ai is the event that that the coin is heads on the i-th toss, T is the total number of heads, and Pr(T k) is the probability that at least k heads come up. As a second example, suppose that we want the probability of a student answering at least k questions correctly on an exam with N questions. In this case, Ai is the event that the student answers the i-th question correctly, T is the total number of questions answered correctly, and Pr(T k) is the probability that the student answers at least k questions correctly. There is an important di erence between these two examples. The rst example is a special case in that all events Ai have equal probability that is, the coin is as likely to come up heads on one ip as on another. In particular, suppose that the coin comes up heads with probability p. Then all events Ai have probability p, and T has the now-familiar binomial distribution: ! n pk (1 ; p)n;k = f (k) Pr(T = k) = k np We studied the binomial distribution extensively two weeks ago. Therefore, we already have an answer to the question posed above in the special case where all events Ai have equal probability. In the second example, however, some exam questions might be more di cult than others. If question 1 is easier than question 2, then the probability of event A1 is greater than the probability of event A2. This lecture focuses on questions of this more general type in which the events Ai may have di erent probabilities.

MIT基础数学讲义(计算机系)lecture5

De nition A tree is a connected n-node graph with exactly n ; 1 edges.
The vertices in a tree can be classi ed into two categories. Vertices of degree at most one are called leaves, and vertices of degree greater than one are called internal nodes. Trees are usually drawn as in Figure 1 with the leaves on the bottom. Keep this convention in mind otherwise, phrases like \all the vertices below..." will be confusing. (The English mathematician Littlewood once remarked that he found such directional terms particularly bothersome, since he habitually read mathematics reclined on his back!) Trees arise in many problems. For example, the le structure in a computer system can be naturally represented by a tree. In this case, each internal node corresponds to a directory, and each leaf corresponds to a le. If one directory contains another, there there is an edge between the associated internal nodes. If a directory contains a le, then there there is an edge between the internal node and a leaf. There are several ways to describe trees that are equivalent to the preceding de nition.

MIT基础数学讲义(计算机系)lecture13

pi j m ) pi j p1 p2 ) pi j 1
pn + 1
2
Lecture 13: Lecture Notes
The rst implication follows by substituting the de nition of m. The second follows because pi divides the product p1p2 pn, and so must divide 1 in order to divide the sum. But no prime divides 1, so we have a contradiction. Therefore, there are an in nite number of primes. Proving that the set of primes is in nite is relatively easy, but the next example shows that determining whether a set is nite or in nite can be tricky.
De nition An in nite set S is said to be countably in nite i there exists a bijection f : N 7! S .
A set is countable if it is nite or countably in nite. Many familiar sets are countable: N , the even numbers, primes, integers modulo a constant k, etc. The formal de nition of a countable set is equivalent to the notion of a listable set, since listing the elements of a set gives a bijection with N and vice-versa. If we can list the elements of S as s0 s1 s2 : : : , then we can construct a bijection f : N 7! S de ned by f (i) = si. In the reverse direction, if we have such a bijection, then f (0) f (1) f (2) : : : is a list of all the elements of S . Sometimes we can list the elements of a set S , but the corresponding bijection f : N 7! S is very hard to compute. For example, in principle we could list the elements in any subset of N . However, most of these subsets are bizarre and di cult to describe writing a program to compute f (i) might be di cult or even impossible. The good news is that to prove that S is countable, we must only prove that a bijection f : N 7! S exists we do not have to write f explicitly or to give an algorithm to compute f (i).

MIT多变量微积分讲义目录文档

讲义下面的注记是Auroux 教授写给习题课教师的内容摘要.讲座主题讲义I. 向量和矩阵0 向量第1周内容摘要1 点积2 行列式与叉积3 矩阵与逆矩阵4 二次系统与平面方程第2周内容摘要5 直线和曲线的参数方程速度和加速度6开普勒第二定律第3周内容摘要7 复习II. 偏导数8 等位线,偏导数与切平面近似9 最大值最小值问题,最小二乘法第4周内容摘要10 二阶导数判定,边界和无穷远11 微分,链式法则第5周内容摘要12 梯度,方向导数与切平面13 拉格朗日乘数法14 非独立变量第6周内容摘要15 偏微分方程,复习III. 二重积分与平面上的线积分16 二重积分第7周内容摘要17 极坐标下的二重积分及其应用18 变量替换19 向量场与平面上的线积分第8周内容摘要20 线积分与路径无关问题,保守场21 梯度场与势函数22 格林公式第9周内容摘要23 通量,格林公式的法式24 单连通区域,复习第10周内容摘要IV. 三重积分与三维空间的面积分25 直角坐标和柱坐标下的三重积分第10周内容摘要讲座主题讲义26 球坐标与表面积27 三维空间中的向量场,面积分与通量第11周内容摘要28 散度公式29 散度公式的应用及证明第12周内容摘要30 空间线积分,旋度,恰当性与势31 斯托克斯公式第13周内容摘要32 斯托克斯公式 (续),复习拓扑研究33麦克斯韦方程第14周内容摘要34 期终复习35 期终复习 (续)。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

2
Lecture 19: Lecture Notes
How do we compute Pr(A j B )? Since we are given that the person lives in Cambridge, all outcomes outside of event B are irrelevant these irrelevant outcomes are diagonally shaded in the gure. Intuitively, Pr(A j B ) should be the fraction of Cambridge residents that are also MIT students. That is, the answer should be the probability that the person is in set A \ B (horizontally shaded) divided by the probability that the person is in set B . This gives:
the person is a Cambridge resident?
This is a conditional probability question. It can be concisely expressed in a special notation. In general, Pr(A j B ) denotes the probability of event A, given event B . In this example, Pr(A j B ) is the probability that the person is an MIT student, given that he or she is a Cambridge resident.
\ Pr(B j B ) = Pr(B B )B ) Pr( Pr(B ) = Pr(B ) = 1
The Product Rule can be generalized to many events.
Rule 1.2 (Product Rule, general case) Let A1 A2 : : : An be events.
Pr(A \ B ) = Pr(B ) Pr(A j B ) (We are now using the term \Product Rule" for two separate ideas. One is the de nition of Pr(A j B ) given above, and the other is the formula for the cardinality of a product of sets. In this lecture, the term always refers to the rule above.) As an example, what is Pr(B j B )? That is, what is the probability of event B , given that event B happens? Intuitively, this ought to be 1! The Product Rule gives exactly this result:
Lecture 19: Lecture Notes
3
2.1 A Two-out-of-Three Series
The MIT computer science department's famed D-league hockey team, The Halting Problem, is playing a 2-out-of-3 series. That is, they play games until one team wins a total of two games. 1 The probability that The Halting Problem wins the rst game is 2 . For subsequent games, the probability of winning depends on the outcome of the preceding game the team is energized by victory and demoralized by defeat. Speci cally, if The Halting Problem wins a game, then they 2 have a 3 chance of winning the next game. On the other hand, if the team loses, then they have 1 only a 3 chance of winning the following game. What is the probability that The Halting Problem wins the 2-out-of-3 series, given that they win the rst game? This problem involves two types of conditioning. First, we are told that the probability of the team winning a game is 2=3, given that they won the preceding game. Second, we are asked the odds of The Halting Problem winning the series, given that they win the rst game.
\ Pr(A j B ) = Pr(A B )B ) Pr(
Rearranging terms gives the following rule, which we will regard as the de nition of Pr(A j B ).
Rule 1.1 (Product Rule, special case) Let A and B be events.
2/3 WW 1/3 WLW 1/3 1/18
W
1/2
L 1/3
W L W
W
2/3 2/3
WLL LWW1/9 1/9源自L1/2W L
1/3
L 1/3
2/3
LWL
1/18 1/3
LL
1st game outcome
2nd game outcome
event A: event B: outcome 3rd game outcome win the win the outcome series? 1st game? probability
2 Conditional Probability Examples
This section contains ve examples of conditional probability problems. In each case, the solution requires only the Product Rule and the four-step method that we applied to the Monty Hall problem.
Pr(A1 \ A2 \ : : : \ An ) = Pr(A1) Pr(A2 j A1) Pr(A3 j A1 \ A2 ) : : : Pr(An j A1 \ : : : \ An;1) This generalization of the Product Rule can be proven from the special case by induction on n, the number of sets. We will not give a proof here, however.
Step 1: Find the Sample Space
The sample space for the hockey series is worked out with a tree diagram in Figure 2. Each internal node has two children, one corresponding to a win for The Halting Problem (labeled W ) and one corresponding to a loss (labeled L). The sample space consists of six outcomes, since there are six leaves in the tree diagram.
Massachusetts Institute of Technology 6.042J/18.062J: Mathematics for Computer Science Professor Tom Leighton
Lecture 19 6 Nov 97
Lecture Notes
This lecture concerns the probability of one event, given that some other event happens. This is called a conditional probability. Trying to solve conditional probability problems by intuition can be very di cult. On the other hand, we can chew through these problems with the same four-step method that we used for the Monty Hall problem: 1. 2. 3. 4. Find the sample space. De ne events of interest. Compute outcome probabilities. Compute event probabilities.