MathematicalInduction.ppt
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argumentation
then 1 mile closer is still far. So 2 miles closer is far, and 3, and 4…. Therefore n miles is far
If 5 is a lot then 4 is a lot. If 4 is a lot, then 3 is a lot… If 1 is a lot then 0 is lot. If 0 is a lot then….
Example of Ad Hominem
Jessie: "I believe that file sharing of music should be stopped." Ben: “You say that, but I noticed that you downloaded an MP3 today." Jessie: "What about the arguments I gave to support my position?" Ben: "Those don't count. You don’t practice what you preach, so I can't believe what you say."
Ad Hominem
From Latin, "Ad Hominem" means "against the man" or "against the person." fallacies in which a claim or argument is rejected on the basis of some irrelevant fact about the person making the claim. Typically
chapter 1 recurrent problem 具体数学 第一章
Chapter 1 Recurrent Problems
1.1 The Tower of Hanoi Let‟s look first at a neat little puzzle called the Tower of Hanoi. We are given a tower of eight disks, initially stacked in decreasing size on one of three pegs:
Preface
• 具体数学(Concrete Mathematics)，从字面上它和传统 的“抽象数学”对立。名字其实是连续(CONtinuous)和离 散(disCRETE)的融合。不过名字不重要，内容主要是说在 计算机科学领域内遇到的问题和传统数学常常不太合拍， 传统数学的方法和理念往往不容易用来解决计算机问题。 具体数学主要致力如何解决计算机科学中的问题，为计算 机算法奠定一个数学基础，或者说给出一些可用的数学手 段／方法。具体数学最早是上个世纪70年代，由Donald E. Knuth在斯坦福开设的一门课，这本书可以说是这门课的 讲义之延伸。 书写的很有意思，第一章讲recurrence，用了几个经 典问题，写的还是挺吸引人的。最后那个约瑟夫环问题， 我看了后，的确是佩服的很。一般来说这个问题的程序的 写法是模拟求解，复杂度是O(n*n)，我自己推导过一个递 推的方法，可以把复杂度降到O(n)，而这本书里直接给出 了这个问题的解公式，强悍。
Chapter 1 Recurrent Problems
• The recurrence only gives indirect, local information. • A solution to the recurrence would make us much happier. With a closed form, we can understand what Tn really is. • So how do we solve a recurrence? 1. One way is to guess the correct solution, then to prove that our guess is correct.
数学归纳法ppt 通用
假设与递推
对所有的 n n N , n n 命题成 . 0
数 学归纳法适用于证明什么样的命 题 呢 ? 对于一些与无限多个正整 数 相关的命题, 如果不易用以前学习过 的方法证明, 用数学归纳法可能会收 到较好的效果 .
思考 如果要用数学归纳法证 明某命题 对于全体正整数都成立 ,应取 n ? 0 为何值 为什么 ?
当 n 5 时 ,共有 5 个点 ,记它为 P ,P ,P ,P ,P . 同前 , 1 2 3 4 5 在过点 P ,P ,P ,P 6 条直线的基础上 ,过 P ,P , 1 2 3 4的 1 2
P ,P 中任意一个点与点 P 作直线 , 共有 4 条 . 因此 , 3 4 5 过 5 个点共有 3 3 4 条直线 .
1 x
n 2 2 n n N , n 5 ,
n
1 nx x 1, n N .
在 高 考 中 , 这 类 问 题 也 是 经 常 出 现 ， 同 时 这 也 是 一 种 重 要 的 数 学 推 理 方 法 — — 数 学 归 纳 法 .
数学归纳法
一、提出问题
在数学 n n N 或不小于某个数 n0 的 任意正整数 n n N , n n0 , 都有某种不等 关系成立 .为表达这样的关系 , 就出现了与无 数多个正整数相关的不 等式 , 例如 : | sin n | n | sin | n N ,
k 1 k k 1
总结上述过程 , 我们用了两个步骤 : 第一步 , 证明n 1 时命题成立 , 从而奠定了命题成 立的一个起点 ; 第二步 , 先作归纳假设 ,然后 证明 "由前向后 "的递推关系 .由这两步保证 : 对于从起点向后的所有 正整数n N, 命题 都成立 .
人教版选修2-2第二章 3数学归纳法(上)-广东省肇庆市肇庆学院附属中学高二数学2020春(共25张
........ 10个圆片的时候: 210 1 问题3：n个圆片的时候是 ? 次
an 2n 1
二、数学归纳法类比 多米诺骨牌效应 探究一：多米诺骨牌都倒下的关键点是什么？
二、数学归纳法雏形：多米诺骨牌效应 （1）第一块骨牌倒下； （2）任意相邻的两块骨牌，前一块倒下一定导致后一块倒下.
条件（1）的作用是什么？
基础自测2.用数学归纳法证明 3nn3(n3,nN)， 第一步应验证（ C ）
A.n 1
B.n 2
C.n 3
D.n 1
注意：有些问题中验证的初始值不一定是1.
基础自测3.在应用数学归纳法证明凸n边形的对角线有 第一步检证n等于（ C ）
1 2
n(n
3)
条时，
A.n 1
B.n 2
C.n 3
D.n 1
注意：当n<3时构不成多边形，且由
1 n(n 3) 2
可知 n 3
.
技能2：能弄清两端项的情况
例2 .用数学归纳法证明等式 1+2+3+…+(2n+1)=(n+1)(2n+1) 当n＝1时，左边所得项是 1_+_2_+__3___ ； 左边=6，右边= (1 1 )(2 1 ) 2 3 6 当n＝2时，左边所得项是 1_+_2_+__3_+_4_+_5__ ；左边=15，右边= (2 1 )(4 1 ) 3 5 1 5 当n＝k时，左边所得项是 1_+_2_+_3_+_…_+_(_2_k_+_1_)_=_左__边，； 右边= (k1)(2k1)
左边=右边， 当 n 1 时 ， 等 式 成 立 .
deduction, induction, contradiction 等证明观念
第一章
6
關於集合(set)
complementation: The complement of a set S consists of all elements not in S. S Universal set U: 代表所有可能的元素。
S {x : x U , x S}
空集合Φ或,empty set 或 null set：就是不 包含任何元素的集合。 S = S - = S, S=
第一章 8
關於集合(set)
有限集合 finite sets，元素個數為有限者； 其他的為 infinite sets 無限集合。 集合 S 的元素個數記成 |S|。 冪集合 powerset: 所有子集合所形成的集 合稱為冪集合。令 2S 為 S 的 powerset， 其元素個數為 2|S| 。 S = {a, b, c} 2S = {, {a}, {b}, {c}, {a,b}, {b, c}, {c, a}, {a, b, c} }.
自動機 (Automata)
Time:
1:10~2:00 Monday: homework practice, quiz 2:10~4:00 Wednesday: lecture
Textbook: (new!)
An Introduction to Formal Languages and Automata, 3rd Edition, Peter Linz
15% for 上課筆記
No lecture slides 上課要抄筆記，學期間會抽查三次算分數
第一章 2
1 Introduction to the theory of computation
Unit+4+Never+too+old+to+learn+词汇详解课件
be occupied in doing / with sth.
qualify /ˈkwɒlɪfaɪ/ vt. 使合格；使具备资格 vi. 取得资格（或学 历），合格（48）
牛津英语单元词汇详解 选择性 必修四
Unit 4 Never too old to learn
*auditory /ˈɔːdətri/ adj. 听的，听觉的（43） *verbal /ˈvɜːb(ə)l/ adj. 言语的，文字的，口头的（43） 【联想拓展】 verbal communication 语言通信 nonverbal 非言语的 verbally adv. 口头地，非书面地；用言辞地
He resolved not to tell her the truth.
refresh /rɪˈfreʃ/ vt.&vi. 更新；使恢复精力；使想起（45） 【联想拓展】 refresh one’s memory 提醒；提示；使想起 refreshment n. 茶点，点心（refreshments）；食物和饮料；恢复活 力，焕发精神 refreshing adj. 提神的；使清爽的；使人重新振作的 refreshed adj.感到神清气爽的，焕然一新的 refresher n. 可提神的人或物；补习课程；清凉饮料 【经典例句】 他冲了个凉水澡凉快凉快。
concrete /ˈkɒŋkriːt/ adj. 确定的，具体的；有形的，实在的 n. 混 凝土（45）
【联想拓展】 cement n.水泥 / sɪˈment / concrete evidence/proposals/proof 确凿的证据；具体的建议；确实 的证明 concrete jungle 水泥丛林 concretely adv. 具体地 concreteness n. 具体；具体性；确实 反义词：abstract 抽象的
qualify /ˈkwɒlɪfaɪ/ vt. 使合格；使具备资格 vi. 取得资格（或学 历），合格（48）
牛津英语单元词汇详解 选择性 必修四
Unit 4 Never too old to learn
*auditory /ˈɔːdətri/ adj. 听的，听觉的（43） *verbal /ˈvɜːb(ə)l/ adj. 言语的，文字的，口头的（43） 【联想拓展】 verbal communication 语言通信 nonverbal 非言语的 verbally adv. 口头地，非书面地；用言辞地
He resolved not to tell her the truth.
refresh /rɪˈfreʃ/ vt.&vi. 更新；使恢复精力；使想起（45） 【联想拓展】 refresh one’s memory 提醒；提示；使想起 refreshment n. 茶点，点心（refreshments）；食物和饮料；恢复活 力，焕发精神 refreshing adj. 提神的；使清爽的；使人重新振作的 refreshed adj.感到神清气爽的，焕然一新的 refresher n. 可提神的人或物；补习课程；清凉饮料 【经典例句】 他冲了个凉水澡凉快凉快。
concrete /ˈkɒŋkriːt/ adj. 确定的，具体的；有形的，实在的 n. 混 凝土（45）
【联想拓展】 cement n.水泥 / sɪˈment / concrete evidence/proposals/proof 确凿的证据；具体的建议；确实 的证明 concrete jungle 水泥丛林 concretely adv. 具体地 concreteness n. 具体；具体性；确实 反义词：abstract 抽象的
马尔科夫过程介绍PPT培训资料
40% in the suburbs. The decimals in x0 add up to 1 because they account for the entire population of the region.
Percentages are more convenient for our purpose here then
Thus the Markov chain is described by the first-order difference equation
When a Markov chain of vectors in Rn describes a system or a sequence of experiments, the entries in xk list, respectively, the probabilities that the system is in each of n possible states, or the probabilities that the outcome of the experiment that is one of n possible outcomes. For this reason, xk is often called a state vector.
n
nn
nn
nn
so cij ( aikbkj) ( aikbkj) ( aik)bkj
population totals.
A vector with nonnegative entries that add up to 1 is called a probability vector. A stochastic matrix is a square matrix whose columns are probability vectors. A Markov chain is a sequence of probability vectors x0, x1, x2, …, together with a stochastic matrix P, such that
Percentages are more convenient for our purpose here then
Thus the Markov chain is described by the first-order difference equation
When a Markov chain of vectors in Rn describes a system or a sequence of experiments, the entries in xk list, respectively, the probabilities that the system is in each of n possible states, or the probabilities that the outcome of the experiment that is one of n possible outcomes. For this reason, xk is often called a state vector.
n
nn
nn
nn
so cij ( aikbkj) ( aikbkj) ( aik)bkj
population totals.
A vector with nonnegative entries that add up to 1 is called a probability vector. A stochastic matrix is a square matrix whose columns are probability vectors. A Markov chain is a sequence of probability vectors x0, x1, x2, …, together with a stochastic matrix P, such that
离散数学英文版
• If you have a question, you can ask me face-to-face
– during the class,
Discrete Structures
CSCI-235 Dr. Kaitian Cao
Introduce Myself
• What is my name? • Where am I from? • What are my current research fields?
Prerequisite and Description
Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
Learning Outcomes
1. An ability to apply algorithmic concepts and constructs in problem analysis and design (Specific about algorithmic concepts and constructs)
• Prepare two notebooks for the assignments
– during the class,
Discrete Structures
CSCI-235 Dr. Kaitian Cao
Introduce Myself
• What is my name? • Where am I from? • What are my current research fields?
Prerequisite and Description
Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
Learning Outcomes
1. An ability to apply algorithmic concepts and constructs in problem analysis and design (Specific about algorithmic concepts and constructs)
• Prepare two notebooks for the assignments
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Mathematical Induction
CS 202 Epp, chapter 4 Aaron Bloomfield
1
How do you climb infinite stairs?
• Not a rhetorical question!
• First, you get to the base platform of the staircase
• Recall the inductive hypothesis:
k
2i 1 k 2
i 1
• Proof of inductive step:
k 1
2i 1 (k 1)2
i 1 k
2(k 1) 1 2i 1 k 2 2k 1 i 1 2k 1
• Three parts:
– Base case(s): show it is true for one element
• (get to the stair’s base platform)
– Inductive hypothesis: assume it is true for any given element
• Show that the sum of the first n odd
integers is n2
– Example: If n = 5, 1+3+5+7+9 = 25 = 52
– Formally, show:
n
n P(n) whereP(n) 2i 1 n2
i 1
• Base case: Show that P(1) is true
• Then, show that if it’s true for some value n, then it is true for n+1
– Show: P(n) P(n+1) – This is climbing the stairs – Let n=0. Since it’s true for P(0) (base case), it’s true for n=1 – Let n=1. Since it’s true for P(1) (previous bullet), it’s true for n=2 – Let n=2. Since it’s true for P(2) (previous bullet), it’s true for n=3 – Let n=3 … – And onwards to infinity
• Then repeat:
– From your current position, move one step up
2
Let’s use that as a proof method
• First, show P(x) is true for x=0
– This is the base of the stairs
• (assume you are on a stair) • Must be clearly labeled!!!
– Show that if it true for the next highest element
• (show you can move to the next stair)
4
Induction example
7
What did we show
• Base case: P(1) • If P(k) was true, then P(k+1) is true
– i.e., P(k) → P(k+1)
• We know it’s true for P(1) • Because of P(k) → P(k+1), if it’s true for P(1), then it’s true for P(2) • Because of P(k) → P(k+1), if it’s true for P(2), then it’s true for P(3) • Because of P(k) → P(k+1), if it’s true for P(3), then it’s true for P(4) • Because of P(k) → P(k+1), if it’s true for P(4), then it’s true for P(5) • And onwards to infinity
• Thus, we have shown it to be true for all non-negative numbers
3
What is induction?
• A method of proof • It does not generate answers:
it only can prove them
• Thus, it is true for all possible values of n
• In other words, we showed that:
P(1) kP(k) P(k 1) n P(n)
8
The idea behind inductive proofs
• Show the base case • Show the inductive hypothesis • Manipulate the inductive step so that you
i 1
– Note: we don’t yet know if this is true or not!
• Inductive step: show true for k+1
– We want to show that:
k 1
2i 1 (k 1)2 6 i 1
Induction example, continued
1
P(1) 2(i) 1 12 i 1
1 1
5
Induction example, continued
• Inductive hypothesis: assume true for k
– Thus, we assume that P(k) is true, or that
k
2i 1 k 2
CS 202 Epp, chapter 4 Aaron Bloomfield
1
How do you climb infinite stairs?
• Not a rhetorical question!
• First, you get to the base platform of the staircase
• Recall the inductive hypothesis:
k
2i 1 k 2
i 1
• Proof of inductive step:
k 1
2i 1 (k 1)2
i 1 k
2(k 1) 1 2i 1 k 2 2k 1 i 1 2k 1
• Three parts:
– Base case(s): show it is true for one element
• (get to the stair’s base platform)
– Inductive hypothesis: assume it is true for any given element
• Show that the sum of the first n odd
integers is n2
– Example: If n = 5, 1+3+5+7+9 = 25 = 52
– Formally, show:
n
n P(n) whereP(n) 2i 1 n2
i 1
• Base case: Show that P(1) is true
• Then, show that if it’s true for some value n, then it is true for n+1
– Show: P(n) P(n+1) – This is climbing the stairs – Let n=0. Since it’s true for P(0) (base case), it’s true for n=1 – Let n=1. Since it’s true for P(1) (previous bullet), it’s true for n=2 – Let n=2. Since it’s true for P(2) (previous bullet), it’s true for n=3 – Let n=3 … – And onwards to infinity
• Then repeat:
– From your current position, move one step up
2
Let’s use that as a proof method
• First, show P(x) is true for x=0
– This is the base of the stairs
• (assume you are on a stair) • Must be clearly labeled!!!
– Show that if it true for the next highest element
• (show you can move to the next stair)
4
Induction example
7
What did we show
• Base case: P(1) • If P(k) was true, then P(k+1) is true
– i.e., P(k) → P(k+1)
• We know it’s true for P(1) • Because of P(k) → P(k+1), if it’s true for P(1), then it’s true for P(2) • Because of P(k) → P(k+1), if it’s true for P(2), then it’s true for P(3) • Because of P(k) → P(k+1), if it’s true for P(3), then it’s true for P(4) • Because of P(k) → P(k+1), if it’s true for P(4), then it’s true for P(5) • And onwards to infinity
• Thus, we have shown it to be true for all non-negative numbers
3
What is induction?
• A method of proof • It does not generate answers:
it only can prove them
• Thus, it is true for all possible values of n
• In other words, we showed that:
P(1) kP(k) P(k 1) n P(n)
8
The idea behind inductive proofs
• Show the base case • Show the inductive hypothesis • Manipulate the inductive step so that you
i 1
– Note: we don’t yet know if this is true or not!
• Inductive step: show true for k+1
– We want to show that:
k 1
2i 1 (k 1)2 6 i 1
Induction example, continued
1
P(1) 2(i) 1 12 i 1
1 1
5
Induction example, continued
• Inductive hypothesis: assume true for k
– Thus, we assume that P(k) is true, or that
k
2i 1 k 2