Lecture2

Lecture2 我们的身体——人体的科学(教案)一．教学目标1．知识性目标：大致了解身体结构；了解基本器官以及消化系统、呼吸系统的工作原理。

2．感受人体构造的“神奇”，培养探索性的好奇心。

3．介绍一些日常生活中自我保护和救助的小知识。

4．培养爱护身体的意识。

二．准备：1．课前在黑板上画人体结构图。

2．网球+网兜（模拟肺泡的结构）3．咀嚼、吞咽、消化、呼吸等的音效[注：六年级的教室里墙上有几个插座，事先不知，所以未带音箱去。

如果知道可以和mp3结合使用。

]三．教学设计Part I 引入：了解身体的意义我们今天要讲的话题是我们的身体。

可能有人觉得，这个话题太普通了，我们对自己的身体还不熟悉吗！是啊，我们每天每时每刻都在用自己的身体，可是我们真的很了解它吗？[跟着几个简单问题后，问：你有几颗牙齿？——资料：乳牙20颗，7-25换牙后（每换一颗牙需要半年）在乳牙后还会长出12 只恒磨牙，共32 颗，但有的人最后四颗磨牙终生不萌出或部份萌出。

全口牙齿总咬力男子为1408公斤，女子为936公斤；但如果失掉一颗牙齿，总咬力会减少200多公斤，缺二颗牙齿，总咬力减少50％-剩一半-牙齿需要配合才能最好地发挥作用，每一颗牙齿都要保护。

]其实这只是一个简单的例子，我们的身体中的奥秘还多着呢。

解答这些奥秘，有时候仅仅是为了满足我们的好奇心，但是另外有些时候是因为这些奥秘的答案实在太重要了，让人们不能忽视它。

比如，医生只有知道人体是怎么构造的，有哪些器官，这些器官在哪儿，都是怎么工作的，才能有效地诊断、治疗病人。

但是，在很长时间的一段时间里，医生连人体主要器官的位置都搞不清楚，因为当时解剖尸体是被禁止的（谁也不敢把人体“切开来”看看里面究竟是什么样）。

西方1800多年前有一位医学家叫盖仑，他认为解剖学对医生来说就像设计图纸对建筑师一样重要。

但是因为没法解剖人体，他只能解剖动物，比如猴子、猪、山羊、河马，然后从这些动物的结构上来猜测人体的结构。

Last Time:Defined knowledge, common knowledge, meet (of partitions), and reachability.Reminders:• E is common knowledge at ω if ()I K E ω∞∈.• “Reachability Lemma” :'()M ωω∈ if there is a chain of states 01,,...m 'ωωωωω== such that for each k ω there is a player i(k) s.t. ()()1()(i k k i k k h h )ωω+=:• Theorem: Event E is common knowledge at ωiff ()M E ω⊆.How does set of NE change with information structure?Suppose there is a finite number of payoff matrices 1,...,L u u for finite strategy sets 1,...,I S SState space Ω, common prior p, partitions , and a map i H λso that payoff functions in state ω are ()(.)u λω; the strategy spaces are maps from into . i H i SWhen the state space is finite, this is a finite game, and we know that NE is u.h.c. and generically l.h.c. in p. In particular, it will be l.h.c. at strict NE.The “coordinated attack” game8,810,11,100,0A B A B-- 0,010,11,108,8A B A B--a ub uΩ= 0,1,2,….In state 0: payoff functions are given by matrix ; bu In all other states payoff functions are given by . a upartitions of Ω1H : (0), (1,2), (3,4),… (2n-1,2n)... 2H (0,1),(2,3). ..(2n,2n+1)…Prior p : p(0)=2/3, p(k)= for k>0 and 1(1)/3k e e --(0,1)ε∈.Interpretation: coordinated attack/email:Player 1 observes Nature’s choice of payoff matrix, sends a message to player 2.Sending messages isn’t a strategic decision, it’s hard-coded.Suppose state is n=2k >0. Then 1 knows the payoffs, knows 2 knows them. Moreover 2 knows that 1knows that 2 knows, and so on up to strings of length k: . 1(0n I n K n -Î>)But there is no state at which n>0 is c.k. (to see this, use reachability…).When it is c.k. that payoff are given by , (A,A) is a NE. But.. auClaim: the only NE is “play B at every information set.”.Proof: player 1 plays B in state 0 (payoff matrix ) since it strictly dominates A. b uLet , and note that .(0|(0,1))q p =1/2q >Now consider player 2 at information set (0,1).Since player 1 plays B in state 0, and the lowest payoff 2 can get to B in state 1 is 0, player 2’s expected payoff to B at (0,1) is at least 8. qPlaying A gives at most 108(1)q q −+−, and since , playing B is better. 1/2q >Now look at player 1 at 1(1,2)h =. Let q'=p(1|1,2), and note that '1(1)q /2εεεε=>+−.Since 2 plays B in state 1, player 1's payoff to B is at least 8q';1’s payoff to A is at most -10q'+8(1-q) so 1 plays B Now iterate..Conclude that the unique NE is always B- there is no NE in which at some state the outcome is (A,A).But (A,A ) is a strict NE of the payoff matrix . a u And at large n, there is mutual knowledge of the payoffs to high order- 1 knows that 2 knows that …. n/2 times. So “mutual knowledge to large n” has different NE than c.k.Also, consider "expanded games" with state space . 0,1,....,...n Ω=∞For each small positive ε let the distribution p ε be as above: 1(0)2/3,()(1)/3n p p n ee e e -==- for 0 and n <<∞()0p ε∞=.Define distribution by *p *(0)2/3p =,. *()1/3p ∞=As 0ε→, probability mass moves to higher n, andthere is a sense in which is the limit of the *p p εas 0ε→.But if we do say that *p p ε→ we have a failure of lower hemi continuity at a strict NE.So maybe we don’t want to say *p p ε→, and we don’t want to use mutual knowledge to large n as a notion of almost common knowledge.So the questions:• When should we say that one information structure is close to another?• What should we mean by "almost common knowledge"?This last question is related because we would like to say that an information structure where a set of events E is common knowledge is close to another information structure where these events are almost common knowledge.Monderer-Samet: Player i r-believes E at ω if (|())i p E h r ω≥.()r i B E is the set of all ω where player i r- believesE; this is also denoted 1.()ri B ENow do an iterative definition in the style of c.k.: 11()()rr I i i B E B E =Ç (everyone r-believes E) 1(){|(()|())}n r n ri i I B E p B E h r w w -=³ ()()n r n rI i i B E B =ÇEE is common r belief at ω if ()rI B E w ¥ÎAs with c.k., common r-belief can be characterized in terms of public events:• An event is a common r-truism if everyone r -believes it when it occurs.• An event is common r -belief at ω if it is implied by a common r-truism at ω.Now we have one version of "almost ck" : An event is almost ck if it is common r-belief for r near 1.MS show that if two player’s posteriors are common r-belief, they differ by at most 2(1-r): so Aumann's result is robust to almost ck, and holds in the limit.MS also that a strict NE of a game with knownpayoffs is still a NE when payoffs are "almost ck” - a form of lower hemi continuity.More formally:As before consider a family of games with fixed finite action spaces i A for each player i. a set of payoff matrices ,:l I u A R ->a state space W , that is now either finite or countably infinite, a prior p, a map such that :1,,,L l W®payoffs at ω are . ()(,)()w u a u a l w =Payoffs are common r-belief at ω if the event {|()}w l w l = is common r belief at ω.For each λ let λσ be a NE for common- knowledgepayoffs u .lDefine s * by *(())s l w w s =.This assigns each w a NE for the corresponding payoffs.In the email game, one such *s is . **(0)(,),()(,)s B B s n A A n ==0∀>If payoffs are c.k. at each ω, then s* is a NE of overall game G. (discuss)Theorem: Monder-Samet 1989Suppose that for each l , l s is a strict equilibrium for payoffs u λ.Then for any there is 0e >1r < and 1q < such that for all [,1]r r Î and [,1]q q Î,if there is probability q that payoffs are common r- belief, then there is a NE s of G with *(|()())1p s s ωωω=>ε−.Note that the conclusion of the theorem is false in the email game:there is no NE with an appreciable probability of playing A, even though (A,A) is a strict NE of the payoffs in every state but state 0.This is an indirect way of showing that the payoffs are never ACK in the email game.Now many payoff matrices don’t have strictequilibria, and this theorem doesn’t tell us anything about them.But can extend it to show that if for each state ω, *(s )ω is a Nash (but not necessarily strict Nash) equilibrium, then for any there is 0e >1r < and 1q < such that for all [,1]r r Î and [,1]q q Î, if payoffs are common r-belief with probability q, there is an “interim ε equilibria” of G where s * is played with probability 1ε−.Interim ε-equilibria:At each information set, the actions played are within epsilon of maxing expected payoff(((),())|())((',())|())i i i i i i i i E u s s h w E u s s h w w w w e-->=-Note that this implies the earlier result when *s specifies strict equilibria.Outline of proof:At states where some payoff function is common r-belief, specify that players follow s *. The key is that at these states, each player i r-believes that all other players r-believe the payoffs are common r-belief, so each expects the others to play according to s *.*ΩRegardless of play in the other states, playing this way is a best response, where k is a constant that depends on the set of possible payoff functions.4(1)k −rTo define play at states in */ΩΩconsider an artificial game where players are constrained to play s * in - and pick a NE of this game.*ΩThe overall strategy profile is an interim ε-equilibrium that plays like *s with probability q.To see the role of the infinite state space, consider the"truncated email game"player 2 does not respond after receiving n messages, so there are only 2n states.When 2n occurs: 2 knows it occurs.That is, . {}2(0,1),...(22,21,)(2)H n n =−−n n {}1(0),(1,2),...(21,2)H n =−.()2|(21,2)1p n n n ε−=−, so 2n is a "1-ε truism," and thus it is common 1-ε belief when it occurs.So there is an exact equilibrium where players playA in state 2n.More generally: on a finite state space, if the probability of an event is close to 1, then there is high probability that it is common r belief for r near 1.Not true on infinite state spaces…Lipman, “Finite order implications of the common prior assumption.”His point: there basically aren’t any!All of the "bite" of the CPA is in the tails.Set up: parameter Q that people "care about" States s S ∈,:f S →Θ specifies what the payoffs are at state s. Partitions of S, priors .i H i pPlayer i’s first order beliefs at s: the conditional distribution on Q given s.For B ⊆Θ,1()()i s B d =('|(')|())i i p s f s B h s ÎPlayer i’s second order beliefs: beliefs about Q and other players’ first order beliefs.()21()(){'|(('),('))}|()i i j i s B p s f s s B h d d =Îs and so on.The main point can be seen in his exampleTwo possible values of an unknown parameter r .1q q = o 2qStart with a model w/o common prior, relate it to a model with common prior.Starting model has only two states 12{,}S s s =. Each player has the trivial partition- ie no info beyond the prior.1122()()2/3p s p s ==.example: Player 1 owns an asset whose value is 1 at 1θ and 2 at 2θ; ()i i f s θ=.At each state, 1's expected value of the asset 4/3, 2's is 5/3, so it’s common knowledge that there are gains from trade.Lipman shows we can match the players’ beliefs, beliefs about beliefs, etc. to arbitrarily high order in a common prior model.Fix an integer N. construct the Nth model as followsState space'S ={1,...2}N S ´Common prior is that all states equally likely.The value of θ at (s,k) is determined by the s- component.Now we specify the partitions of each player in such a way that the beliefs, beliefs about beliefs, look like the simple model w/o common prior.1's partition: events112{(,1),(,2),(,1)}...s s s 112{(,21),(,2),(,)}s k s k s k -for k up to ; the “left-over” 12N -2s states go into 122{(,21),...(,2)}N N s s -+.At every event but the last one, 1 thinks the probability of is 2/3.1qThe partition for player 2 is similar but reversed: 221{(,21),(,2),(,)}s k s k s k - for k up to . 12N -And at all info sets but one, player 2 thinks the prob. of is 1/3.1qNow we look at beliefs at the state 1(,1)s .We matched the first-order beliefs (beliefs about θ) by construction)Now look at player 1's second-order beliefs.1 thinks there are 3 possible states 1(,1)s , 1(,2)s , 2(,1)s .At 1(,1)s , player 2 knows {1(,1)s ,2(,1)s ,(,}. 22)s At 1(,2)s , 2 knows . 122{(,2),(,3),(,4)}s s s At 2(,1)s , 2 knows {1(,2)s , 2(,1)s ,(,}. 22)sThe support of 1's second-order beliefs at 1(,1)s is the set of 2's beliefs at these info sets.And at each of them 2's beliefs are (1/3 1θ, 2/3 2θ). Same argument works up to N:The point is that the N-state models are "like" the original one in that beliefs at some states are the same as beliefs in the original model to high but finite order.(Beliefs at other states are very different- namely atθ or 2 is sure the states where 1 is sure that state is2θ.)it’s1Conclusion: if we assume that beliefs at a given state are generated by updating from a common prior, this doesn’t pin down their finite order behavior. So the main force of the CPA is on the entire infinite hierarchy of beliefs.Lipman goes on from this to make a point that is correct but potentially misleading: he says that "almost all" priors are close to a common. I think its misleading because here he uses the product topology on the set of hierarchies of beliefs- a.k.a topology of pointwise convergence.And two types that are close in this product topology can have very different behavior in a NE- so in a sense NE is not continuous in this topology.The email game is a counterexample. “Product Belief Convergence”:A sequence of types converges to if thesequence converges pointwise. That is, if for each k,, in t *i t ,,i i k n k *δδ→.Now consider the expanded version of the email game, where we added the state ∞.Let be the hierarchy of beliefs of player 1 when he has sent n messages, and let be the hierarchy atthe point ∞, where it is common knowledge that the payoff matrix is .in t ,*i t a uClaim: the sequence converges pointwise to . in t ,*i t Proof: At , i’s zero-order beliefs assignprobability 1 to , his first-order beliefs assignprobability 1 to ( and j knows it is ) and so onup to level n-1. Hence as n goes to infinity, thehierarchy of beliefs converges pointwise to common knowledge of .in t a u a u a u a uIn other words, if the number of levels of mutual knowledge go to infinity, then beliefs converge to common knowledge in the product topology. But we know that mutual knowledge to high order is not the same as almost common knowledge, and types that are close in the product topology can play very differently in Nash equilibrium.Put differently, the product topology on countably infinite sequences is insensitive to the tail of the sequence, but we know that the tail of the belief hierarchy can matter.Next : B-D JET 93 "Hierarchies of belief and Common Knowledge”.Here the hierarchies of belief are motivated by Harsanyi's idea of modelling incomplete information as imperfect information.Harsanyi introduced the idea of a player's "type" which summarizes the player's beliefs, beliefs about beliefs etc- that is, the infinite belief hierarchy we were working with in Lipman's paper.In Lipman we were taking the state space Ω as given.Harsanyi argued that given any element of the hierarchy of beliefs could be summarized by a single datum called the "type" of the player, so that there was no loss of generality in working with types instead of working explicitly with the hierarchies.I think that the first proof is due to Mertens and Zamir. B-D prove essentially the same result, but they do it in a much clearer and shorter paper.The paper is much more accessible than MZ but it is still a bit technical; also, it involves some hard but important concepts. (Add hindsight disclaimer…)Review of math definitions:A sequence of probability distributions converges weakly to p ifn p n fdp fdp ®òò for every bounded continuous function f. This defines the topology of weak convergence.In the case of distributions on a finite space, this is the same as the usual idea of convergence in norm.A metric space X is complete if every Cauchy sequence in X converges to a point of X.A space X is separable if it has a countable dense subset.A homeomorphism is a map f between two spaces that is 1-1, and onto ( an isomorphism ) and such that f and f-inverse are continuous.The Borel sigma algebra on a topological space S is the sigma-algebra generated by the open sets. (note that this depends on the topology.)Now for Brandenburger-DekelTwo individuals (extension to more is easy)Common underlying space of uncertainty S ( this is called in Lipman)ΘAssume S is a complete separable metric space. (“Polish”)For any metric space, let ()Z D be all probability measures on Borel field of Z, endowed with the topology of weak convergence. ( the “weak topology.”)000111()()()n n n X S X X X X X X --=D =´D =´DSo n X is the space of n-th order beliefs; a point in n X specifies (n-1)st order beliefs and beliefs about the opponent’s (n-1)st order beliefs.A type for player i is a== 0012(,,,...)()n i i i i n t X d d d =¥=δD0T .Now there is the possibility of further iteration: what about i's belief about j's type? Do we need to add more levels of i's beliefs about j, or is i's belief about j's type already pinned down by i's type ?Harsanyi’s insight is that we don't need to iterate further; this is what B-D prove formally.Coherency: a type is coherent if for every n>=2, 21marg n X n n d d --=.So the n and (n-1)st order beliefs agree on the lower orders. We impose this because it’s not clear how to interpret incoherent hierarchies..Let 1T be the set of all coherent typesProposition (Brandenburger-Dekel) : There is a homeomorphism between 1T and . 0()S T D ´.The basis of the proposition is the following Lemma: Suppose n Z are a collection of Polish spaces and let021201...1{(,,...):(...)1, and marg .n n n Z Z n n D Z Z n d d d d d --´´-=ÎD ´"³=Then there is a homeomorphism0:(nn )f D Z ¥=®D ´This is basically the same as Kolmogorov'sextension theorem- the theorem that says that there is a unique product measure on a countable product space that corresponds to specified marginaldistributions and the assumption that each component is independent.To apply the lemma, let 00Z X =, and 1()n n Z X -=D .Then 0...n n Z Z X ´´= and 00n Z S T ¥´=´.If S is complete separable metric than so is .()S DD is the set of coherent types; we have shown it is homeomorphic to the set of beliefs over state and opponent’s type.In words: coherency implies that i's type determines i's belief over j's type.But what about i's belief about j's belief about i's type? This needn’t be determined by i’s type if i thinks that j might not be coherent. So B-D impose “common knowledge of coherency.”Define T T ´ to be the subset of 11T T ´ where coherency is common knowledge.Proposition (Brandenburger-Dekel) : There is a homeomorphism between T and . ()S T D ´Loosely speaking, this says (a) the “universal type space is big enough” and (b) common knowledge of coherency implies that the information structure is common knowledge in an informal sense: each of i’s types can calculate j’s beliefs about i’s first-order beliefs, j’s beliefs about i’s beliefs about j’s beliefs, etc.Caveats:1) In the continuity part of the homeomorphism the argument uses the product topology on types. The drawbacks of the product topology make the homeomorphism part less important, but theisomorphism part of the theorem is independent of the topology on T.2) The space that is identified as“universal” depends on the sigma-algebra used on . Does this matter?(S T D ´)S T ×Loose ideas and conjectures…• There can’t be an isomorphism between a setX and the power set 2X , so something aboutmeasures as opposed to possibilities is being used.• The “right topology” on types looks more like the topology of uniform convergence than the product topology. (this claim isn’t meant to be obvious. the “right topology” hasn’t yet been found, and there may not be one. But Morris’ “Typical Types” suggests that something like this might be true.)•The topology of uniform convergence generates the same Borel sigma-algebra as the product topology, so maybe B-D worked with the right set of types after all.。

【托福听力备考】TPO3听力文本——Lecture 2对于很多学生来说，托福TPO材料是备考托福听力最好的材料。

相信众多备考托福的同学也一直在练习这套材料，那么在以下内容中我们就为大家带来托福TPO听力练习的文本，希望能为大家的备考带来帮助。

Lecture 2 Film historyNarrator：Listen to part of a lecture in a film history class.Professor：Okay, we’ve been discussing films in the 1920s and 30s, and how back then film categories, as we know them today, had not yet been established. We said that by today’s standards, many of the films of the 20s and 30s would be considered hybrids, that is, a mixture of styles that wouldn’t exactly fit into any of today’s categories. And in that context, today we are going to talk about a film-maker who began making very unique films in the late 1920s. He was French, and his name was Jean Painlevé.Jean Painlevé was born in 1902. He made his first film in 1928. Now in a way, Painlevé’s films conform to norms of the 20s and 30s, that is, they don’t fit very neatly into the categories we use to classify films today. That said, even by the standards of the 20s and 30s, Painlevé’s films were a unique hybrid of styles. He had a special way of fusing, or some people might say, confusing, science and fiction. His films begin with facts, but then they become more and more fictional. They gradually add more and more fictional elements. In fact, Painlevé was known for saying that science is fiction.Painlevé was a pioneer in underwater film-making, and a lot of his short films focused on the aquatic animal world. He liked to show small underwater creatures, displaying what seemed like familiar human characteristics – what we think of as unique to humans. He might take a clip of a mollusk going up and down in the water and set it to music. You know, to make it look as if the mollusk were dancing to the music like a human being – that sort of thing. But then he suddenly changed the image or narration to remind us how different the animals are, how unlike humans.He confused his audience in the way he portrayed the animals he filmed, mixing up our notions of the categories human and animal. The films make us a little uncomfortable at times because we are uncertain about what we are seeing. It gives him films an uncanny feature: the familiar made unfamiliar, the normal made suspicious. He liked twists, he liked the unusual. In fact, one of his favorite sea animals was the seahorse because with seahorses, it’s the male that carries the eggs, and he thought that was great. His first and most celebrated underwater film is about the seahorse.Susan, you have a question?Student 1：But underwater film-making wasn’t that unusual, was it? I mean, weren’t there other people making movies underwater?Professor：Well, actually, it was pretty rare at that time. I mean, we are talking the early 1930s here.Student 1：But what about Jacques Cousteau? Was he like an innovator, you know, with underwater photography too?Professor： Ah, Jacques Cousteau. Well, Painlevé and Cousteau did both film underwater, and they were both innovators, so you are right in that sense. But that’s pretty much where the similarities end.First of all, Painlevé was about 20 years ahead of Cousteau. And Cousteau’s adventures were high-tech, with lots of fancy equipment, whereas Painlevé kind of patchedequipment together as he needed it. Cousteau usually filmed large animals, usually in the open sea, whereas Painlevé generally filmed smaller animals, and he liked to film in shallow water.Uh, what else? Oh well, the main difference was that Cousteau simply investigated and presented the facts – he didn’t mix in fiction. He was a strict documentarist. He set the standard really for the nature documentary. Painlevé, on the other hand, as we said before, mixed in elements of fiction. And his films are much more artistic, incorporating music as an important element.John, you have a question?Student 2：Well, maybe I shouldn’t be asking this, but if Painlevé’s films are so special, so good, why haven’t we ever heard of them? I mean, everyone’s heard of Jacques Cousteau.Professor： Well, that’s a fair question. Uh, the short answer is that Painlev é’s style just never caught on with the general public. I mean, it probably goes back at least in part to what we mentioned earlier, that people didn’t know what to make of his films – they were confused by them, whereas Cousteau’s documentaries were very straightforward, met people’s expectations more than Painlevé’s films did. But you true film history buffs know about him. And Painlevé is still highly respected in many circles.。

Lecture 2 心理偏誤與展望理論認知心理學家透過實驗研究提出相當多的心理偏誤，且諸多學者有不同的分類方式，本講次參考Shefrin (2007)依序介紹認知偏誤(biases)、捷思或經驗法則(heuristics)與框架效應(framing effect)，最後說明展望理論。

I. 心理偏誤1.認知偏誤(Biases)(1)過度樂觀: 高估有利出象的機率，低估不利出象的機率。

在診斷性問題中列出了18個事件，請評估與同性別的他人相比，該事件發生在你身上的機率是比平均水準低(1~6)、等於平均水準(7)或是高於平均水準(8~15)。

結果發現: Typically the average rating for the unfavorable events is below 7, while the average rating for the favorable events is above 7.(2)過度自信：人們所自己的所知(the limits of their knowledge)和自己的能力(their own ability)容易有過度自信的傾向。

在診斷性問題中列出了10個困難的題目，受實驗者必須對每個題目回答best guess, 以及90%信心水準下的low guess和high guess (即認為正確答案有90%的機率落於所設定的上下界之間)。

結果發現: Typically the most frequent number of hits, and the average number of hits, is about 4.(3)確認性偏誤：人們容易忽視與自己觀點不一致的訊息，而接受尋找與自己觀點一致的訊息。

診斷性問題：桌上有四張卡(p. A1-5)，每張卡的一面是字母，另一面是數字，受實驗者被詢問檢測以下假說“Any card having a vowel on one side has an even number on the other side.”在四張卡中受實驗者至少要掀開幾張卡才能決定此假說是否為真?結果發現: Most people turn over the card with the a,and some turn over the cardwith the 2 as well.(4)控制性幻覺：人們傾向於高估本身對結果或出象的控制力。

【托福听力备考】TPO6听力文本——Lecture 2众所周知，托福TPO材料是备考托福听力最好的材料。

相信众多备考托福的同学也一直在练习这套材料，那么在以下内容中我们就为大家带来托福TPO听力练习的文本，希望能为大家的备考带来帮助。

TPO 6 Lecture 2 BiologyNarrator：Listen to part of a lecture in a biology class.Professor：Ok, I have an interesting plant species to discuss with you today.Uh…it’s a species of a very rare tree that grows in Australia, Eidotheahardeniana, but it’s better known as the Nightcap Oak.Now, it was discovered only very recently, just a few years ago. Um… itremained hidden for so long because it’s so rare. There are only about 200 ofthem in existence. They grow in a rain forest, in a mountain rage…range in thenorth part of New South Wales which is uh… a state in Australia. So just 200individual trees in all.Now another interesting thing about the Nightcap Oak is that it is…itrepresents…uh…a very old type…uh…kind of tree that grew a hundred million yearsago. Um, we found fossils that old that bear remarkable resemblance to the tree.So, it’s a primitive tree. A…a living fossil you might say. It’s relic fromearlier times and it has survived all these years without much change. Andit…it’s probably a kind of tree from which other trees that grow in Australiatoday evolved.Just to give you an idea of what we are talking about. Here’s a picture ofthe leaves of the tree and its flowers. I don’t know how well you can see theflowers. They’re those little clusters sitting at the base of the leaves.Okay, what have we tried to find out about the tree since we’ve discoveredit? Hmm…or how…why is…is it so rare? It’s one of the first questions. Um…how isit…um…how does it reproduce? It’s another question. Um, maybe those two questions are actually related. Jim?Student：Hmm …I don’t know. But I can imagine that…for instance, seed dispersal might be a factor. I mean if the…er…you know, if the seeds cannot really disperse in the wild area, then, you know, the tree may not colonize new areas. It can’t spread from the area where it’s growing.Professor：Right. That’s…that’s actually a very good answer. Uh, of course, you might think there might not be many areas where the tree could spread into,er…because…um…well, it’s very specialized in terms of the habitat. But, that’snot really the case here. Um…the suitable habitat, that is, the actualrainforest is much larger than the few hectares where the Nightcap Oakgrows.Now this tree is a flowering tree as I showed you. Um…um…it produces a fruit,much like a plum. On the inci…inside there’s a seed with a hard shell. It…itappears that the shell has to crack open or break down somewhat to allow the seed to soak up water. You know, if the Nightcap Oak remains…if their seedsremain locked inside their shell, they will not germinate. Actually, theseeds…er…they don’t retain the power to germinate for very long, maybe two years. So there’s actually quite a short window of opportunity for the seed togerminate. So the shell somehow has to be broken down before this…um…germinationability expires. And…and then there’s a kind of rat that likes to feed on the seeds as well. So, given all these limitations, not many seeds that the tree produces will actually germinate. So this is a possible explanation for why the tree does not spread. It doesn’t necessarily explain how it became so rare, but it explains why it doesn’t increase.OK, so it seems to be the case that the species, this Nightcap Oak is notvery good at spreading. However, it seems, though we can’t be sure, that it’svery good at persisting as a population. Um…we…there’s some indications to suggest that the population of the Nightcap Oak has not declined over the last.er…you know, many hundreds of years. So it’s stayed quite stable. It’s not aremnant of some huge population that is dwindled in the last few hundred years for some reason. It’s not necessarily a species in retreat. Ok, so it cannotspread very well, but it’s good at maintaining itself. It’s rare, but it’s notdisappearing.Ok, the next thing we might want to ask about a plant like that is whatchances does it have to survive into the future. Let’s look at that.。