A simple proof of the algebraic version of a conjecture by Vogan

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The complex of words and Nakaoka stability

a rX iv:mat h /41261v2[mat h.KT]2M ar25The Complex of Words and Nakaoka Stability Moritz C.Kerz mokerz@students.uni-mainz.de Abstract We give a new simple proof of the exactness of the complex of injective words and use it to prove Nakaoka’s homology stability for symmetric groups.The methods are generalized to show acyclicity in low degrees for the complex of words in ”general position”.1.An elementary proof of Nakaoka stability We will be concerned with the complex C ∗(m ),where C n (m )is the free abelian group generated by elements (x 1,...,x n ),where the x i are pairwise diﬀerent natural numbers from 1to m .C 0(m )=Z .The diﬀerential is given by d (x 1,...,x n )=n j =1(−1)j +1(x 1,...,x j −1,x j +1,...,x n ).In discrete mathematics the following theorem is well established as the homol-ogy of C ∗(M )is given by the simplicial homology of the (shellable)poset of injective words.Shellability reduces the simplicial homology groups to those of a wedge of m-spheres.We refer to [1],[3],[7]for exact deﬁnitions and state-ments.Our proof of Theorem 1is new and rather straightforward compared to Farmer’s original elementary proof.Theorem 1(F.D.Farmer [3])The homology of C ∗(m )vanishes except in de-gree m .We have to introduce some notations which will be used throughout the paper.A chain c is called a term,if there exists an N ∈Z and x 1,...,x n ∈{1,...,m }with c =N (x 1,...,x n ).As C ∗(m )has a canonical basis all our sum decompostions correspond to parti-tions of the basis.We also speak about the appearance of numbers in a chain.For example 2appears in (2,3)+4(5,1)but 4does not.Although our complex C ∗(m )has no obvious cup product,we have a partially deﬁned product.If c ∈C n (m ),c ′∈C l (m )are elementary c =N (x 1,...x n ),c =M (x ′1,...,x ′l ),we deﬁne c c ′:=N M (x 1,...,x n ,x ′1,...x ′l )if the numbers x 1,...,x n ,x ′1,...x ′l are paiwise diﬀerent.This constuction extends bilinearly to arbitrary chains c ,c ′for which the numbers appearing in both of them are distinct.1There is a Leibniz rule for c∈C n(m)and c′∈C l(m)d(c c′)=d(c)c′+(−1)n c d(c′).Proof of Theorem1.The exactness in degree0is clear.For the rest we use induction on m.For the case m=2we have to check thatC2(2)−→C1(2)−→Zis exact.But ker(d:C1(2)→Z)is generated by elements of the form(x)−(x′)=d(x,x′)with x=x′.For the induction step we use a straightforward lemma:Lemma1If we have a number x∈{1,...,m}which does not appear in a cycle c∈C n(m),it is a boundary.Proof.According to the Leibniz rule c=c+(x)d(c)=d((x)c),since d(c)=0.Given an arbitrary cycle c of degree n<m we have to show that in order to apply the lemma we can eliminate a number from c by adding boundaries. Therefore we will push aﬁxed number x∈{1,...,m}to the right until it vanishes from the cycle.If x appears somewhere in the cycle at theﬁrst index,writec= j(x)c j+c′Where the c j are elementary and c′does not have x at theﬁrst index.To each c j choose a number x j∈{1,...,m},x j=x which does not appear in c j.Thenc−d( j(x j)(x)c j)=c′+ j(x j)c j−(x j)(x)dc j.Now x does not appear at theﬁrst index anymore.The next steps until the vanishing of x are similar.Suppose x does not appear at theﬁrst i>0indices of c.Now we can writec= j s j(x)c j+c′.Where the c j are elementary and diﬀerent,the s j have length i,x does not appear in s j and x does not appear at theﬁrst i+1entries of c′.One calculates: 0=d c= j (d s j)(x)c j+(−1)i s j c j+(−1)i+1s j(x)d c j +d c′.2If we forget those terms for which x does not appear at the i-th index,this equation implies d s j=0for all j.Since length(s j)=i<m−length((x)c j), there are by induction s′j with d s′j=s j and such that the following products make sense.z:= j s′j(x)c jIn the cyclec−d z=c′− j (−1)i+1s′j c j+(−1)i s′j(x)d c jx does not appear at theﬁrst i+1indices.Using the corresponding two hyperhomology spectral sequences for the natural action of the symmetric groupΣm on our complex C∗(m)(cf[2])one can now obtain a stability result due to Nakaoka.A similar proof was given by Maazen (thesis).Theorem2(Nakaoka[5])H m(Σn−1)=H m(Σn)for m<n/2.Proof.We use induction on n.It is well known for n=3H1(Σ2)=H1(Σ3)=Z/(2).For n≥4deﬁne C′l(n):=C l+1(n)for l≥0.ThenH m(Σn,Z)=H m(Σn,C′∗(n))when m<n−1because of Theorem1.The other spectral sequence of the bi-complex gives for E1∗,∗:···H2(Σn,C′0)H2(Σn,C′1)···H2(Σn,C′n−2)H2(Σn,C′n−1)H1(Σn,C′0)H1(Σn,C′1)···H1(Σn,C′n−2)H1(Σn,C′n−1)H0(Σn,C′0)H0(Σn,C′1)···H0(Σn,C′n−2)H0(Σn,C′n−1)Using Shapiro’s Lemma we get:···H2(Σn−1)H2(Σn−2)···H2(Σ1)0H1(Σn−1)H1(Σn−2)···H1(Σ1)0H0(Σn−1)H0(Σn−2)···H0(Σ1)ZThe horizontal arrows can be computed as0,1,0,1,···,since they are the sums of the signs in d′1,d′2,···.We have E2i,0=H i(Σn−1)for i≥0and by induction E2i,j=0for i<n−j−1Finally,parts of the spectral sequence degenerate;E∞i,0=H i(Σn−1)for n−i>2, such that H i(Σn)=H i(Σn−1)for n/2>i.2.Words in general positionLet n:={i|1≤i≤n}.We associate to every(nonempty)set X a complex F(X)∗,the so-called complex of words with alphabet X,as follows:(F(X))n=Z<{f:n→X}>d n(f)=nk=1(−1)n+1f◦δk.Hereδk:[n]→[n+1]are the coface mapsδk(j)= j if1≤j<kj+1if k≤j.It is immediate that the homology of F(X)vanishes.For if c∈F(X)n is a cycle,c=d((x0)c)for arbitrary x0∈X(see Lemma1).For given X certain subcomplexes of F(X)can be used in hyperhomology spec-tral sequences as above,if their homology vanishes to some extent.These sub-groups are determined by conditions which one could call”general position conditions”.For applications cf[8],[9].It could be asked,how to explain the fact that the vanishing of homology is not aﬀected by these conditions.In order to give a general result we translate the proof of Theorem1into our more complicated setting.Examples(i)Let X be aﬁnite set.The complex of injective words is(F inj(X))n:={f∈F(X)n|f injective}.According to Theorem1the homology of this subcomplex is zero except in degree m=card(X)whererank(H m(F(X)∗))=(−1)m(1−m−1i=0(−1)i m(m−1)···(m−i)).This equals the number ofﬁxed point free permutations of X.(ii)Let F be aﬁeld and V an F-vector space.The complex of vectors in general position is(F gen(V))n:={f∈F(V)n|f in general position}.If F is inﬁnite the complex is exact.A general vanishing result is contained in our main theorem.We introduce axioms for elements of a given(nonempty)set X to be in general position.4Deﬁnition1(General position)Let G l,m be relations in l+m variables from X;l>0,m≥0.The family(G l,m)l,m∈N is called a general position relation if the following properties are satisﬁed.Let x,y,z beﬁnite sequences of elements of X of length l,m,n:(i)G n,m is symmetric in theﬁrst n and last m arguments.(ii)If G l+m,n(x,y;z)then G l,m+n(x;y,z).If G l,m+n(x;y,z)then G l,n(x;z).(iii)If G l,m+n(x;y,z)and G m,n(y;z)then G l+m,n(x,y;z).Forﬁxed G we say x is in G-general position to y iﬀG l,m(x;y).Given x∈F(X)l and y∈F(X)m we say x is in G-general position to y if every term of x is in G-general postition to every term in y(for l=0we demand nothing).Deﬁnition2Given aﬁnite sequence b of elements of X the corresponding complex of chains in general position to b is(F G(X;b))n:={f∈F(X)n|f is in G-general position to b}.Before we can state the main theorem we have to introduce an invariant which determines an upper bound for the vanishing of the homology of F(X)∗.Deﬁnition3Given a general position relation G we deﬁne|G|to be the small-est natural number n≥0such that there is a sequence x of elements of X with length(x)=n such that there is no further element in X which is in general position to x.Examples(i)Example(i)is induced by saying x is in G inj-general position to y if theunderlying sets of x and y are disjoint and the entries of x are pairwise diﬀerent.We have|G inj|=card(X)and F G inj(X)∗=(F inj(X))∗(ii)Example(ii)is induced by saying(x1,...,x i)is in G vec-general position to(y1,...,y j)if we have for all a k∈F,k=1,...,i+j and only dim(V) many of them nonvanishinga1x1+···+a i x i+a i+1y1+···+a i+j y j=0implies a k=0for all k∈{1,...,i}.If card(F)=∞or dim(V)=∞then|G vec|=∞.Unfortunately the exact value of|G vec|is not known for allﬁnite dimen-sional vector spaces overﬁniteﬁelds.5Lemma2(a)For dim(V)≥card(F)we have|G vec|=dim(V)+1.(b)For dim(V)=2we have|G vec|=card(F)+1.Proof of(a).Let n=dim(V)+1and(e i)1≤i≤n−1be a basis of V.We show|G vec|≥nﬁrst.Otherwise we had a sequence(x i)1≤i≤n−1,x i∈V, such that there does not exist a vector x∈V in G vec-general position to(x i). This can only be true if(x i)is a basis of V.But if it is a basis the element x1+x2+···+x n−1would be in general position to it.Contradiction.Now the sequence f=(e1,...,e n−1,e1+e2+···+e n−1)is maximal in the sense of Deﬁnition3.To see this we have to show there is no vector x= a1e1+···+a n−1e n−1,a i∈F,in G vec-general position to f.If there is an index j such that a j=0,x is obviously not in general postion to f.But if no coeﬃcient in x vanishes two of them have to be equal since dim(V)≥card(F). In case a1=a2we can writex=a1(e1+···+e n−1)+(a3−a1)e3+···(a n−1−a1)e n−1.This shows f is in fact maximal.Proof of(b).For dim(V)=2we have x∈V in G vec-general position to y∈V iﬀx=0=y and[x]=[y]∈P(V).So|Gen vec|is simply the number of rational points in P(V)/F.Theorem3For a set X with general position condition G and aﬁnite se-quence a=(a1,...,a l)of elements of X the corresponding homology groups H m(F G(X;a))vanish for m≤(|G|−l−1)/2.In contrast to Theorem1one should notice that it is in general not possible to erase the factor1/2from the inequality of the theorem.Proof.Denote the degree by m.The exactness at m=0is trivial.We proceed by induction on m≥1.Let c be a cycle in F m(X;a1,...,a l).Case m=1:We can suppose c=(x)−(x′).Because1≤(|G|−l−1)/2we have l+2<|G| so that there exists y∈X in G-general position to(x,x′,a1,...,a l). According to Deﬁnition1(iii)(y,x,x′)is in G-general position to(a1,...,a l) and it is allowed to write d((y)c)=c.Induction step:Choose x∈X in G-general position to(a1,...,a l).The simplest case is c in G-general position to(x,a1,...,a l).Here we can apply a construction similar to Lemma1;we have c=d((x)c),since(x)c is in G-general position to(a1,...,a l).We reduce to this case by changing c by boundaries.To be more precise we introduce a number I(c)∈{0,...,m}which is m iﬀthe above applies,that is6c in G-general position to(x,a1,...,a l).By adding boundaries be will see that we can increase I(c).For d∈F G(X;a)n we deﬁne I(d)∈{0,...,n}as the the greatest natural number i≤n such thatπi(v)is in G-general position to(x,π′i(v),a1,...,a l) for any term v of d(πi denotes the projection to theﬁrst i entries andπ′i the projection to the last n−i entries).Reduction to I(c)>0:Suppose I(c)=0.Let c= j x j with x j elementary.Choose for every j y j∈X with y j in G-general position to(x,x j,a1,...,a l).This is possible since length(x,x j,a1,...,a l)=1+m+l<|G|.Clearly I(c−d( j(y j)x j))>0. Now suppose m>I(c)>0:Writec= j s j x j+x′such that exactly those terms v of c for which I(v)>I(c)are in x′,length(s j)=I(c)and all x j are elementary and pairwise diﬀerent.Lemma3d(s j)=0for all j.Proof.The terms v of d(c)such that I(v)<I(c)are exactly the terms ofd(s j)x j.In order to see this,notice that for a term v of d(c)we have I(v)=(v),a1,...,a l). I(c)−1iﬀthe I(c)-th entry of v is not in G-general position to(x,π′I(c)Now projecting the identity0=d(c)to the terms v with I(v)<I(c)we get 0= j s j x j.Lemma3is proven. According to our assumtion s j is in G-general position to(x,x j,a1,...,a l).By induction on m we know there exist s′j with d(s′j)=s j and s′j in G-general position to(x,x j,a1,...,a l),because2i+[(m−i)+l+1]≤2m+l≤|G|−1. NowI(c−d( j s′j x j))=I(x′+(−1)I(c)s′j d(x j))>I(c).Thisﬁnishes the reduction to I(c)=m and therefore the induction step for length(c)=m is accomplished.Theorem3could be useful in the generalization of Suslin’s GL-stability[8]toﬁniteﬁelds.A thorough treatment seems to indicate the following result:(i)Given m≥0and n≥m we have H m(GL n(F))=H m(GL n+1(F))for almostallﬁniteﬁelds F.(ii)For given m≥0H m(GL m−1(F))→H m(GL m(F))is surjective for almostallﬁniteﬁelds F.7In fact(i)has been proven–using other methods–by Quillen[6].He did even show,that the statement is true for allﬁelds with more then2elements.Similar results with weaker bounds are due to Maazen and van der Kallen[4].AcknowledgementThe above results were obtained in the Seminar M¨u ller-Stach2004at Mainz.I am indebted to Stefan M¨u ller-Stach,Volkmar Welker,W.van der Kallen, Philippe Elbaz-Vincent and Oliver Petras for their support.References[1]Bj¨o rner,Anders;Wachs,Michelle On lexicographically shellable posets. Trans.Amer.Math.Soc.277(1983),no.1,323–341.[2]Brown,Kenneth S.Cohomology of groups.Corrected reprint of the1982 original.Graduate Texts in Mathematics,87.Springer-Verlag,New York,1994 [3]Farmer,Frank D.Cellular homology for posets.Math.Japon.23(1978/79), no.6,607–613.[4]van der Kallen,Wilberd Homology stability for linear groups.Invent.Math. 60(1980),no.3,269–295.[5]Nakaoka,Minoru Decomposition theorem for homology groups of symmetric groups.Ann.of Math.(2)71196016–42.[6]Quillen,Daniel;unpublished.[7]Stanley,Richard binatorics and commutative algebra.Second edi-tion.Progress in Mathematics,41.Birkh¨a user Boston,Inc.,Boston,MA,1996.[8]Suslin,A.A.Homology of GL n,characteristic classes and Milnor K-theory. Algebraic K-theory,number theory,geometry and analysis(Bielefeld,1982), 357–375,Lecture Notes in Math.,1046,Springer,Berlin,1984.[9]Suslin,A.A.K3of aﬁeld,and the Bloch group.Proc.Steklov Inst.Math. 1991,no.4,217–239.8。

我们班的数学天才英语作文600字初中

我们班的数学天才英语作文600字初中全文共3篇示例，供读者参考篇1The Math Genius in Our ClassYou know that kid in every class who just gets math? Like, really gets it down to the core in a way the rest of us can't even fathom? Well, in my class that genius is Adam Smith. Math seems to flow through his veins more naturally than blood.From the first day of 6th grade when Mr. Patterson worked through some simple algebraic equations on the board, Adam's hand shot up before anyone could even processed what was going on. He explained the steps for solving the equations so clearly and effortlessly that even I, a kid who has always struggled with math, had a Light bulb moment."Oh, that's how you do it!" I thought as Adam walked us through algebraically isolating the variable. It all seemed so obvious when he laid it out systematically.That was just the start of Adam's math wizardry. As we moved into more complex concepts like graphing systems of equations, factoring polynomials, and applying the principles ofgeometry, he became increasingly indispensable. If anyone raised their hand with a question or got stuck on a problem set, we'd all turn our heads eagerly toward Adam waiting for him to share his insight.Despite being shy around school hallways, Adam transforms in math class. He leans forward in his seat, makes deliberate eye contact with the teacher, and articulates his thoughts with impressive coherence and precision. You can almost see the synapses firing in his brain as he works through the logicstep-by-step.Even Mr. Patterson regularly calls on Adam to explain certain problems or theorems to the rest of us when we look puzzled. I've lost count of how many times Adam has dropped a "Big Ah" moment on the class that allowed a breakthrough or clarified something we were all stuck on.What makes Adam even more remarkable is his humility and patience. Despite having a staggering mathematical ability, he never lorded it over anyone or acted arrogant. If someone didn't understand his explanation at first, Adam would re-word it orre-work the problem with a different approach until it clicked for that person.He spends plenty of free periods, study halls, and free time after school walking different classmates through the topics they struggled with. Almost everyone has taken advantage of Adam's tutoring at one point or another. His guidance has allowed so many of us to maintain solid grades in math rather than allowing the subject to become our downfall.While the rest of us dream about getting outdoors as soon as the weather turns nice each year, Adam looks most at peace and in his element when working through a challenge problem set or attempting to decipher a new mathematical proof. Books on advanced mathematics concepts are is his Preference for light reading over young adult fiction.Simply put, Adam is a savant when it comes to numbers, equations, geometry, and the wonderful world of mathematics. We're all just lucky he ended up in our class to shed is brilliant light on the subject and help raise the mathematical excellence of every student. Having a genius in our midst drove the rest of us to live up to our fullest potential as well.篇2My Class's Math GeniusThere's this kid in my class who is, simply put, a total math genius. His name is David and he has an incredible knack for numbers and problem-solving that is really mind-blowing. I'm not exaggerating when I say he could probably end up being one of the great mathematicians of our time if he keeps developing his skills.It's kind of crazy just how advanced David's math abilities are compared to the rest of us. I still have to use a calculator for most complicated calculations, but he can do enormously complex equations entirely in his head in just seconds. Once, when we were learning about algebraic expressions in class, the teacher put a really long and convoluted problem on the board to demonstrate. While the rest of us just stared at it blankly, David's hand shot up and he solved the whole thing, step-by-step, from memory. The teacher was stunned and so were we.David's parents told me he started showing an affinity for math basically from the moment he could talk. Apparently, as a toddler, he would constantly recite numbers and look for patterns and relationships between them. By the age of 4, he could already do basic addition, subtraction, multiplication and division. That's just not normal child development - the kid was clearly some kind of math prodigy from a very young age.In school, he breezes through math assignments and tests like they're nothing. He gets every question right, every time. Our math teacher once told me she plans to start giving David extra advanced coursework because the normal curriculum isn't enough of a challenge for his capabilities.Sometimes he even catches mistakes in the textbooks that the authors and editors all missed.You might think having such a talented student would be great, but it does create some difficulties too. Often the concepts we're supposed to be learning go completely over my head because David has already understood it in about 2 seconds. Then he gets bored waiting for the rest of us to catch up and starts distracting the class by joking around. The teacher has had to warn him about it multiple times.Still, David is a pretty humble, down-to-earth guy despite his incredible talents. He never brags or acts arrogant about his superior intellect. I've learned not to feel bad about not being as skilled as him because it's clear his brain is just wired differently for math aptitude. It would be like him feeling bad for not being as good at basketball as one of the players on the school team.For his future plans, David says he's torn between wanting to become a mathematics professor and researcher, or going into amore applied field like engineering or computer science. No matter what path he chooses, I'm confident he'll be hugely successful and make important contributions. David is truly one of the most gifted students in mathematics our school has ever seen. I'm just lucky I get to be in his class and witness his incredible abilities firsthand. Who knows, maybe I'm learning from the next Einstein or Ramanujan.篇3Our Class's Math GeniusYou know that kid in every class who just seems to get math effortlessly? The one who finishes tests in half the time while the rest of us are still struggling? Well, that person in our class is Tony. Tony is, without a doubt, a certified math genius.I've known Tony since we were little kids playing together at the neighborhood park. Even back then, you could tell he just thought differently about numbers and patterns. While the rest of us would count out objects one by one, Tony would immediately see the groupings and patterns. "That's two groups of four!" he would exclaim about a simple pile of rocks. His mind worked that way from the very start.In elementary school, you could already see the math genius emerging. He would race through worksheets of basic operations, making teachers do double-takes. By third grade, he was getting pulled out for an hour each day to work on higher level math meant for older kids. Can you imagine being 8 years old and doing algebra? That was Tony.When we got to middle school, that's when Tony's talents really went into overdrive. I'm in the highest level of math available, and even I can't keep up with that kid half the time. He breezes through complicated equations like a hot knife through butter. Just last week, our math teacher put an extremely tricky problem on the board involving exponential functions and logarithms. The rest of us sat there dumbfounded, but within 30 seconds, Tony's hand shot up with the correct solution fully worked out.Sometimes I think he gets bored in class because the material isn't challenging enough for his incredible mind. He'll start thinking about unsolved math problems or theorems that no one has cracked yet. Once Mrs. Robinson caught him scribbling some sort of crazy number theory conjecture during her lesson. When she asked him about it, he launched into thisintense explanation about prime numbers that went completely over our heads. Even she looked dazed by the end!While Tony's skills are clearly off the charts, he's also a really humble, down-to-earth guy which makes him even more likeable. You'd never know he was a genius unless you saw him do math. In a world where kid geniuses can sometimes be arrogant know-it-alls, Tony is the complete opposite. He never lords his abilities over others or makes us feel dumb in comparison. If someone is struggling, he's the first to try patiently explaining it in a new way. We all wish we had Tony's brain, but are just glad to be his friend.I have no idea what the future holds for our class's math whiz kid. Ivy League universities are definitely in his future, and perhaps he'll go on to win a Fields Medal or formulate the next big theory in mathematics. Wherever life takes him, I'm just happy to be able to witness Tony's incredible talents and be inspired by them. Having a bonafide genius in your midst makes you realize there's so much potential in this world still to be unlocked. Math may seem like a boring subject to some, but seeing it through Tony's eyes shows how exciting and full of possibilities it truly is.。

关于数学的英语作文80词

关于数学的英语作文80词Mathematics, often perceived as a cold and rigid discipline, is in reality a realm brimming with elegance and charm. It is the language through which the universe communicates its fundamental principles, offering a glimpse into the very fabric of existence. From the simple symmetryof a snowflake to the complex patterns woven by prime numbers, mathematics reveals an intricate tapestry of order and harmony.The study of mathematics begins with the most basic elements: numbers. Yet, these building blocks are not mere abstractions; they are the foundation upon which all else is built. The integers, rational and irrational numbers, form a hierarchy that mirrors the complexity of the world we inhabit. Each number carries its own unique properties, and theirinteractions give rise to the profound theories that defineour understanding of space, time, and matter.Geometry, another cornerstone of mathematics, explores shape, space, and relative position. It is the study of form and structure, revealing the beauty hidden in the straight lines, curves, and angles that surround us. Euclid's Elements, with its elegant postulates and proofs, laid the groundworkfor our understanding of geometric forms and spatial relationships.Algebra, on the other hand, deals with the manipulationof symbols to solve equations. It provides a framework for modeling real-world problems and offers tools for predicting future outcomes based on past trends. Algebraic structures underpin much of modern technology, from cryptography to data analysis.Calculus, perhaps one of the most powerful branches of mathematics, allows us to understand rates of change andaccumulation. It is the language of motion and growth, enabling us to describe phenomena ranging from the motion of planets to the spread of diseases.Beyond these traditional areas, mathematics continues to evolve. Topics such as topology, game theory, and chaos theory offer new lenses through which to view the world. They challenge our assumptions about what is knowable and reveal new pathways for exploration.The applications of mathematics are far-reaching and profound. In science, it forms the basis for understanding physical laws and natural phenomena. In engineering, it enables the design and construction of technological marvels. In economics, it informs decision-making processes and market analysis. In computer science, it underpins algorithms and data structures.However, the beauty of mathematics is not solely found in its applications. It lies in the pursuit of knowledge for itsown sake. The process of solving a mathematical problem, the eureka moments when patterns emerge from seeming chaos, the elegance of a proof that explains the unexplainable – these are the aspects that captivate the minds of mathematicians and enthusiasts alike.In conclusion, mathematics is a treasure trove of intellectual splendor. It is an art form that uses logic and reason to create a canvas of concepts and ideas. Its beauty is subtle yet profound, accessible to those who take the time to appreciate its intricacies. As the renowned mathematician Paul Erdős once said, "Mathematics is not just something I work on; it is part of my life, it is a pleasure, it is a passion." Truly, mathematics is a testament to the power and beauty of human intellect, a never-ending journey of discovery and wonder.。

最擅长的学科数学英语作文

最擅长的学科数学英语作文Mathematics, a subject that has always fascinated me,is my forte in the academic arena. The beauty of mathematics lies in its simplicity, elegance, and the profound truths it unveils. From the basics of arithmetic to the complex theories of geometry and algebra, mathematics has always captivated my imagination and challenged my intellect.My journey with mathematics began at a tender age, when I was introduced to the basics of counting and addition. As I grew older, I delved deeper into the subject, discovering the wonders of fractions, decimals, and percentages. The logic and precision required in solving mathematical problems appealed to me, and I found myself enjoying the challenge of mastering new concepts.One of the aspects of mathematics that I particularly enjoy is its problem-solving aspect. The satisfaction of solving a complex mathematical puzzle, whether it's a geometry proof or an algebraic equation, is unparalleled. The process of breaking down a problem, analyzing it, and then finding a solution requires a great deal of thinkingand reasoning. This not only sharpens my mathematicalskills but also enhances my critical thinking abilities.Moreover, mathematics has a practical application in almost every field. It is the language of science, technology, engineering, and finance. Understanding mathematics enables me to comprehend and analyze real-world situations more effectively. Whether it's calculating the area of a room for painting or understanding the concept of interest rates in finance, mathematics plays a crucial role. In addition to its practical applications, mathematics also instills in me a sense of discipline and precision. It teaches me to approach problems with a methodical mindset,to break them down into smaller parts, and to solve them step by step. This discipline not only helps me in academics but also in my daily life, enabling me to tackle challenges with clarity and focus.In conclusion, mathematics is not just a subject for me; it is a passion. The joy of discovering new mathematical truths, the challenge of solving problems, and the satisfaction of mastering new concepts are what drive me to excel in this subject. Mathematics has not only shaped myacademic journey but also influenced my thinking and approach to life. As I continue to explore the depths ofthis fascinating subject, I am confident that it will continue to enrich my life and open new horizons of understanding and discovery.**数学：学术领域中的我的拿手好戏**数学，这个始终让我着迷的学科，是我在学术领域中的拿手好戏。

低维拓扑简介

考研论坛»数学»低维拓扑knight51发表于2005-7-28 08:34低维拓扑<P>下面说说低维拓扑的内容：低维拓扑是微分拓扑的一部分，主要研究3，4维流形与纽结理论。

又叫几何拓扑。

主要以代数拓扑与微分拓扑为工具。

它与微分几何和动力系统关系密切。

国外搞这个方向的也几乎都搞微分几何和动力系统。

我国这个方向北大最牛，美国是伯克利和普林斯顿最牛。

比起代数几何来，它比较好入门。

初学者只需要代数拓扑，微分拓扑，黎曼几何的知识就行了。

美国这方面比较牛，几乎每个搞基础数学研究的都会低维拓扑。

</P><DIV class=postcolor>纠正一下上面的错误，美国也不是每个搞基础数的都精通低维拓扑，而是懂一些低维拓扑的知识。

如果入门后还想更加深入了解它，那还需要读一些双曲几何和拓扑动力系统的书。

</DIV><DIV class=postcolor>下面介绍一下这方面的牛人：Bill Thurston studied at New College, Sarasota, Florida. He received his B.S. from there in 1967 and moved to the University of California at Berkeley to undertake research under Morris Hirsch's and Stephen Smale's supervision. He was awarded his doctorate in 1972 for a thesis entitled Foliations of 3-manifolds which are circle bundles. This work showed the existence of compact leaves in foliations of 3-dimensional manifolds.After completing his Ph.D., Thurston spent the academic year 1972-73 at the Institute for Advanced Study at Princeton. Then, in 1973, he was appointed an assistant professor of mathematics at Massachusetts Institute of Technology. In 1974 he was appointed professor of mathematics at Princeton University.Throughout this period Thurston worked on foliations. Lawson ([5]) sums up this work:-It is evident that Thurston's contributions to the field of foliations are of considerable depth. However, what sets them apart is their marvellous originality. This is also true of his subsequent work on Teichmüller space and the theory of 3-manifolds.In [8] Wall describes Thurston's contributions which led to him being awarded a Fields Medal in 1982. In fact the1982 Fields Medals were announced at a meeting of the General Assembly of the International Mathematical Union in Warsaw in early August 1982. They were not presented until the International Congress in Warsaw which could not be held in 1982 as scheduled and was delayed until the following year. Lectures on the work of Thurston which led to his receiving the Medal were made at the 1983 International Congress. Wall, giving that address, said:-Thurston has fantastic geometric insight and vision: his ideas have completely revolutionised the study of topology in 2 and 3 dimensions, and brought about a new and fruitful interplaybetween analysis, topology and geometry.Wall [8] goes on to describe Thurston's work in more detail:-The central new idea is that a very large class of closed 3-manifolds should carry a hyperbolic structure - be the quotient of hyperbolic space by a discrete group of isometries, or equivalently, carry a metric of constant negative curvature. Although this is a natural analogue of the situation for 2-manifolds, where such a result is given by Riemann's uniformisation theorem, it is much less plausible - even counter-intuitive - in the 3-dimensional situation.Kleinian groups, which are discrete isometry groups of hyperbolic 3-space, were first studied by Poincaré and a fundamental finiteness theorem was proved by Ahlfors. Thurston's work on Kleinian groups yielded many new results and established a well known conjecture. Sullivan describes this geometrical work in [6], giving the following summary:-Thurston's results are surprising and beautiful. The method is a new level of geometrical analysis - in the sense of powerful geometrical estimation on the one hand, and spatial visualisation and imagination on the other, which are truly remarkable.Thurston's work is summarised by Wall [8]:-Thurston's work has had an enormous influence on 3-dimensional topology. This area has a strong tradition of 'bare hands' techniques and relatively little interaction with other subjects. Direct arguments remain essential, but 3-dimensional topology has now firmly rejoined the main stream of mathematics.Thurston has received many honours in addition to the Fields Medal. He held a Alfred P Sloan Foundation Fellowship in 1974-75. In 1976 his work on foliations led to his being awarded the Oswald Veblen Geometry Prize of the American Mathematical Society. In 1979 he was awarded the Alan T Waterman Award, being the second mathematician to receive such an award (the first being Fefferman in 1976).</DIV>第2个牛人：Michael Freedman entered the University of California at Berkeley in 1968 and continued his studies at Princeton University in 1969. He was awarded a doctorate by Princeton in 1973 for his doctoral dissertation entitled Codimension-Two Surgery. His thesis supervisor was William Browder.After graduating Freedman was appointed a lecturer in the Department of Mathematics at the University of California at Berkeley. He held this post from 1973 until 1975 when he became a member of the Institute for Advanced Study at Princeton. In 1976 he was appointed as assistant professor in the Department of Mathematics at the University of California at San Diego.Freedman was promoted to associate professor at San Diego in 1979. He spent the year 1980/81 at the Institute for Advanced Study at Princeton returning to the University of California at San Diego where he was promoted to professor on 1982. He holds this post in addition to the Charles Lee Powell Chair of Mathematics which he was appointed to in 1985.Freedman was awarded a Fields Medal in 1986 for his work on the Poincaré conjecture. The Poincaré conjecture, one of the famous problems of 20th-century mathematics, asserts that a simply connected closed 3-dimensional manifold is a 3-dimensional sphere. The higher dimensional Poincaréconjecture claims that any closed n-manifold which is homotopy equivalent to the n-sphere must be the n-sphere. When n = 3 this is equivalent to the Poincaré conjecture. Smale proved the higher dimensional Poincaré conjecture in 1961 for n at least 5. Freedman proved the conjecture for n = 4 in 1982 but the original conjecture remains open.Milnor, describing Freedman's work which led to the award of a Fields Medal at the International Congress of Mathematicians in Berkeley in 1986, said:-Michael Freedman has not only proved the Poincaré hypothesis for 4-dimensional topological manifolds, thus characterising the sphere S4, but has also given us classification theorems, easy to state and to use but difficult to prove, for much more general 4-manifolds. The simple nature of his results in the topological case must be contrasted with the extreme complications which are now known to occur in the study of differentiable and piecewise linear 4-manifolds. ... Freedman's 1982 proof of the 4-dimensional Poincaré hypothesis was an extraordinary tour de force. His methods were so sharp as to actually provide a complete classification of all compact simply connected topological 4-manifolds, yielding many previously unknown examples of such manifolds, and many previously unknown homeomorphisms between known manifolds.Freedman has received many honours for his work. He was California Scientist of the Year in 1984 and, in the same year, he was made a MacArthur Foundation Fellow and also was elected to the National Academy of Sciences. In 1985 he was elected to the American Academy of Arts and Science. In addition to being awarded the Fields Medal in 1986, he also received the Veblen Prize from the American Mathematical Society in that year. The citation for the Veblen Prize reads (see [3]):-After the discovery in the early 60s of a proof for the Poincaré conjecture and other properties of simply connected manifolds of dimension greater than four, one of the biggest open problems, besides the three dimensional Poincaré conjecture, was the classification of closed simply connected four manifolds. In his paper, The topology of four-dimensional manifolds, published in the Journal of Differential Geometry (1982), Freedman solved this problem, and in particular, the four-dimensional Poincaré conjecture. The major innovation was the solution of the simply connected surgery problem by proving a homotopy theoretic condition suggested by Casson for embedding a 2-handle, i.e. a thickened disc in a four manifold with boundary.Besides these results about closed simply connected four manifolds, Freedman also proved:(a) Any four manifold properly equivalent to R4 is homeomorphic to R4; a related result holds for S3 R.(b) There is a nonsmoothable closed four manifold.© The four-dimensional Hauptvermutung is false; i.e. there are four manifolds with inequivalent combinatorial triangulations.Finally, we note that the results of the above mentioned paper, together with Donaldson's work, produced the startling example of an exotic smoothing of R4.In his reply Freedman thanked his teachers (who he said included his students) and also gave some fascinating views on mathematics [3]:-My primary interest in geometry is for the light it sheds on the topology of manifolds. Here it seems important to be open to the entire spectrum of geometry, from formal to concrete. By spectrum, I mean the variety of ways in which we can think about mathematical structures. At one extreme the intuition for problems arises almost entirely from mental pictures. At the other extreme the geometric burden is shifted to symbolic and algebraic thinking. Of course this extreme is only a middle ground from the viewpoint of algebra, which is prepared to go much further in the direction of formal operations and abandon geometric intuition altogether.In the same reply Freedman also talks about the influence mathematics can have on the world and the way that mathematicians should express their ideas:-In the nineteenth century there was a movement, of which Steiner was a principal exponent, to keep geometry pure and ward off the depredations of algebra. Today I think we feel that much of the power of mathematics comes from combining insights from seemingly distant branches of the discipline. Mathematics is not so much a collection of different subjects as a way of thinking. As such, it may be applied to any branch of knowledge. I want to applaud the efforts now being made by mathematicians to publish ideas on education, energy, economics, defence, and world peace. Experience inside mathematics shows that it isn't necessary to be an old hand in an area to make a contribution. Outside mathematics the situation is less clear, but I cannot help feeling that there, too, it is a mistake to leave important issues entirely to experts.In June 1987 Freedman was presented with the National Medal of Science at the White House by President Ronald Reagan. The following year he received the Humboldt Award and, in 1994, he received the Guggenheim Fellowship Award.<DIV class=postcolor>介绍第3个牛人：Simon Donaldson's secondary school education was at Sevenoaks School in Kent which he attended from 1970 to 1975. He then entered Pembroke College, Cambridge where he studied until 1980, receiving his B.A. in 1979. One of his tutors at Cambridge described him as a very good student but certainly not the top student in his year. Apparently he would always come to his tutorials carrying a violin case.In 1980 Donaldson began postgraduate work at Worcester College, Oxford, first under Nigel Hitchen's supervision and later under Atiyah's supervision. Atiyah writes in [2]:-In 1982, when he was a second-year graduate student, Simon Donaldson proved a result that stunned the mathematical world.This result was published by Donaldson in a paper Self-dual connections and the topology of smooth 4-manifolds which appeared in the Bulletin of the American Mathematical Society in 1983. Atiyah continues his description of Donaldson's work [2]:-Together with the important work of Michael Freedman ..., Donaldson's result implied that there are "exotic" 4-spaces, i.e. 4-dimensional differentiable manifolds which are topologically but not differentiably equivalent to the standard Euclidean 4-space R4. What makes this result so surprising is that n = 4 is the only value for which such exotic n-spaces exist. These exotic 4-spaces have the remarkable property that (unlike R4) they contain compact sets which cannot be contained inside any differentiably embedded 3-sphere !After being awarded his doctorate from Oxford in 1983, Donaldson was appointed a Junior Research Fellow at All Souls College, Oxford. He spent the academic year 1983-84 at the Institute for Advanced Study at Princeton, After returning to Oxford he was appointed Wallis Professor of Mathematics in 1985, a position he continues to hold.Donaldson has received many honours for his work. He received the Junior Whitehead Prize from the London Mathematical Society in 1985. In the following year he was elected a Fellow of the Royal Society and, also in 1986, he received a Fields Medal at the International Congress at Berkeley. In 1991 Donaldson received the Sir William Hopkins Prize from the Cambridge Philosophical Society. Then, the following year, he received the Royal Medal from the Royal Society. He also received the Crafoord Prize from the Royal Swedish Academy of Sciences in 1994:-... for his fundamental investigations in four-dimensional geometry through application of instantons, in particular his discovery of new differential invariants ...Atiyah describes the contribution which led to Donaldson's award of a Fields Medal in [2]. He sums up Donaldson's contribution:-When Donaldson produced his first few results on 4-manifolds, the ideas were so new and foreign to geometers and topologists that they merely gazed in bewildered admiration.Slowly the message has gotten across and now Donaldson's ideas are beginning to be used by others in a variety of ways. ... Donaldson has opened up an entirely new area; unexpected and mysterious phenomena about the geometry of 4-dimensions have been discovered. Moreover the methods are new and extremely subtle, using difficult nonlinear partial differential equations. On the other hand, this theory is firmly in the mainstream of mathematics, having intimate links with the past, incorporating ideas from theoretical physics, and tying in beautifully with algebraic geometry.The article [3] is very interesting and provides both a collection of reminiscences by Donaldson on how he came to make his major discoveries while a graduate student at Oxford and also a survey of areas which he has worked on in recent years. Donaldson writes in [3] that nearly all his work has all come under the headings:-(1) Differential geometry of holomorphic vector bundles.(2) Applications of gauge theory to 4-manifold topology.and he relates his contribution to that of many others in the field.Donaldson's work in summed up by R Stern in [6]:-In 1982 Simon Donaldson began a rich geometrical journey that is leading us to an exciting conclusion to this century. He has created an entirely new and exciting area of research through which much of mathematics passes and which continues to yield mysterious and unexpected phenomena about the topology and geometry of smooth 4-manifolds</DIV><DIV class=postcolor>下面continue介绍第4个牛人：Robion Kirby。

Communicated by Bruce Rothschild

Z Pk Z
2. ITERATED INTEGRALS
k 1Y
0 j =1
asj ?1 b
(5)
j =1 sj . Here, the j =1
x>t1 >t2 > >tn >y j =1
fj (tj ) dtj ;
j := fj (tj ) dtj
(6)
of 2] is used with a and b denoting the di erential 1-forms dt=t and dt=(1 ? t), respectively. Thus, for example, if f1 6= f2, we write 2 2 1 for 1 the integrand f1 (t1 )f1 (t2 )f2 (t3 )f1 (t4 ) dt1 dt2 dt3 dt4 : Furthermore, we shall agree that any iterated integral of an empty product of di erential 1-forms is equal to 1. This convention is mainly a notational convenience; nevertheless we shall nd it useful for stating results about iterated integrals more concisely and naturally than would be possible otherwise. Thus (6) reduces to 1 when n = 0 regardless of the values of x and y. Clearly the product of two iterated integrals of the form (6) consists of a sum of iterated integrals involving ? possible interlacings of the variall ables. Thus if we denote the set of all n+m permutations of the indices n f1; 2; : : :; n + mg satisfying ?1 (j ) < ?1 (k) for all 1 j < k n and n + 1 j < k n + m by Shu (n; m), then we have the self-evident formula

漫谈微分几何、多复变函数与代数几何(Differential geometry, functions

漫谈微分几何、多复变函数与代数几何（Differential geometry, functions of complex variable and algebraic geometry）Differential geometry and tensor analysis, developed with the development of differential geometry, are the basic tools for mastering general relativity. Because general relativity's success, to always obscure differential geometry has become one of the central discipline of mathematics.Since the invention of differential calculus, the birth of differential geometry was born. But the work of Euler, Clairaut and Monge really made differential geometry an independent discipline. In the work of geodesy, Euler has gradually obtained important research, and obtained the famous Euler formula for the calculation of normal curvature. The Clairaut curve of the curvature and torsion, Monge published "analysis is applied to the geometry of the loose leaf paper", the important properties of curves and surfaces are represented by differential equations, which makes the development of classical differential geometry to reach a peak. Gauss in the study of geodesic, through complicated calculation, in 1827 found two main curvature surfaces and its product in the periphery of the Euclidean shape of the space not only depends on its first fundamental form, the result is Gauss proudly called the wonderful theorem, created from the intrinsic geometry. The free surface of space from the periphery, the surface itself as a space to study. In 1854, Riemann made the hypothesis about geometric foundation, and extended the intrinsic geometry of Gauss in 2 dimensional curved surface, thus developing n-dimensional Riemann geometry, with the development of complex functions. A group of excellentmathematicians extended the research objects of differential geometry to complex manifolds and extended them to the complex analytic space theory including singularities. Each step of differential geometry faces not only the deepening of knowledge, but also the continuous expansion of the field of knowledge. Here, differential geometry and complex functions, Lie group theory, algebraic geometry, and PDE all interact profoundly with one another. Mathematics is constantly dividing and blending with each other.By shining the charming glory and the differential geometric function theory of several complex variables, unit circle and the upper half plane (the two conformal mapping establishment) defined on Poincare metric, complex function theory and the differential geometric relationships can be seen distinctly. Poincare metric is conformal invariant. The famous Schwarz theorem can be explained as follows: the Poincare metric on the unit circle does not increase under analytic mapping; if and only if the mapping is a fractional linear transformation, the Poincare metric does not change Poincare. Applying the hyperbolic geometry of Poincare metric, we can easily prove the famous Picard theorem. The proof of Picard theorem to modular function theory is hard to use, if using the differential geometric point of view, can also be in a very simple way to prove. Differential geometry permeates deep into the theory of complex functions. In the theory of multiple complex functions, the curvature of the real differential geometry and other series of calculations are followed by the analysis of the region definition metric of the complex affine space. In complex situations, all of the singular discrete distribution, and in more complex situations, because of the famous Hartogsdevelopment phenomenon, all isolated singularities are engulfed by a continuous region even in singularity formation is often destroyed, only the formation of real codimension 1 manifold can avoid the bad luck. But even this situation requires other restrictions to ensure safety". The singular properties of singularities in the theory of functions of complex functions make them destined to be manifolds. In 1922, Bergman introduced the famous Bergman kernel function, the more complex function or Weyl said its era, in addition to the famous Hartogs, Poincare, Levi of Cousin and several predecessors almost no substantive progress, injected a dynamic Bergman work will undoubtedly give this dead area. In many complex function domains in the Bergman metric metric in the one-dimensional case is the unit circle and Poincare on the upper half plane of the Poincare, which doomed the importance of the work of Bergman.The basic object of algebraic geometry is the properties of the common zeros (algebraic families) of any dimension, affine space, or algebraic equations of a projective space (defined equations),The definitions of algebraic clusters, the coefficients of equations, and the domains in which the points of an algebraic cluster are located are called base domains. An irreducible algebraic variety is a finite sub extension of its base domain. In our numerical domain, the linear space is the extension of the base field in the number field, and the dimension of the linear space is the number of the expansion. From this point of view, algebraic geometry can be viewed as a study of finite extension fields. The properties of algebraic clusters areclosely related to their base domains. The algebraic domain of complex affine space or complex projective space, the research process is not only a large number of concepts and differential geometry and complex function theory and applied to a large number of coincidence, the similar tools in the process of research. Every step of the complex manifold and the complex analytic space has the same influence on these subjects. Many masters in related fields, although they seem to study only one field, have consequences for other areas. For example: the Lerey study of algebraic topology that it has little effect on layer, in algebraic topology, but because of Serre, Weil and H? Cartan (E? Cartan, eldest son) introduction, has a profound impact on algebraic geometry and complex function theory. Chern studies the categories of Hermite spaces, but it also affects algebraic geometry, differential geometry and complex functions. Hironaka studies the singular point resolution in algebraic geometry, but the modification of complex manifold to complex analytic space and blow up affect the theory of complex analytic space. Yau proves that the Calabi conjecture not only affects algebraic geometry and differential geometry, but also affects classical general relativity. At the same time, we can see the important position of nonlinear ordinary differential equations and partial differential equations in differential geometry. Cartan study of symmetric Riemann space, the classification theorem is important, given 1, 2 and 3 dimensional space of a Homogeneous Bounded Domain complete classification, prove that they are all homogeneous symmetric domains at the same time, he guessed: This is also true in the n-dimensional equivalent relation. In 1959, Piatetski-Shapiro has two counterexample and find the domain theory of automorphic function study in symmetry, in the 4 and 5dimensional cases each find a homogeneous bounded domain, which is not a homogeneous symmetric domain, the domain he named Siegel domain, to commemorate the profound work on Siegel in 1943 of automorphic function. The results of Piatetski-Shapiro has profound impact on the theory of complex variable functions and automorphic function theory, and have a profound impact on the symmetry space theory and a series of topics. As we know, Cartan transforms the study of symmetric spaces into the study of Lie groups and Lie algebras, which is directly influenced by Klein and greatly develops the initial idea of Klein. Then it is Cartan developed the concept of Levi-Civita connection, the development of differential geometry in general contact theory, isomorphic mapping through tangent space at each point on the manifold, realize the dream of Klein and greatly promote the development of differential geometry. Cartan is the same, and concluded that the importance of the research in the holonomy manifold twists and turns, finally after his death in thirty years has proved to be correct. Here, we see the vast beauty of differential geometry.As we know, geodesic ties are associated with ODE (ordinary differential equations), minimal surfaces and high dimensional submanifolds are associated with PDE (partial differential equations). These equations are nonlinear equations, so they have high requirements for analysis. Complex PDE and complex analysis the relationship between Cauchy-Riemann equations coupling the famous function theory, in the complex case, the Cauchy- Riemann equations not only deepen the unprecedented contact and the qualitative super Cauchy-Riemann equations (the number of variables is greater than the number of equations) led to a strange phenomenon. This makes PDE and the theory ofmultiple complex functions closely integrated with differential geometry.Most of the scholars have been studying the differential geometry of the intrinsic geometry of the Gauss and Riemann extremely deep stun, by Cartan's method of moving frames is beautiful and concise dumping, by Chern's theory of characteristic classes of the broad and profound admiration, Yau deep exquisite geometric analysis skills to deter.When the young Chern faced the whole differentiation, he said he was like a mountain facing the shining golden light, but he couldn't reach the summit at one time. But then he was cast as a master in this field before Hopf and Weil.If the differential geometry Cartan development to gradually change the general relativistic geometric model, then the differential geometry of Chern et al not only affect the continuation of Cartan and to promote the development of fiber bundle in the form of gauge field theory. Differential geometry is still closely bound up with physics as in the age of Einstein and continues to acquire research topics from physicsWhy does the three-dimensional sphere not give flatness gauge, but can give conformal flatness gauge? Because 3D balls and other dimension as the ball to establish flat space isometric mapping, so it is impossible to establish a flatness gauge; and n-dimensional balls are usually single curvature space, thus can establish a conformal flat metric. In differential geometry, isometry means that the distance between the points on the manifold before and after the mapping remains the same. Whena manifold is equidistant from a flat space, the curvature of its Riemann cross section is always zero. Since the curvature of all spheres is positive constant, the n-dimensional sphere and other manifolds whose sectional curvature is nonzero can not be assigned to local flatness gauge.But there are locally conformally flat manifolds for this concept, two gauge G and G, if G=exp{is called G, P}? G between a and G transform is a conformal transformation. Weyl conformal curvature tensor remains unchanged under conformal transformation. It is a tensor field of (1,3) type on a manifold. When the Weyl conformal curvature tensor is zero, the curvature tensor of the manifold can be represented by the Ricci curvature tensor and the scalar curvature, so Penrose always emphasizes the curvature =Ricci+Weyl.The metric tensor g of an n-dimensional Riemann manifold is conformally equivalent to the flatness gauge locally, and is called conformally flat manifold. All Manifolds (constant curvature manifolds) whose curvature is constant are conformally flat, so they can be given conformal conformal metric. And all dimensions of the sphere (including thethree-dimensional sphere) are manifold of constant curvature, so they must be given conformal conformal metric. Conversely, conformally flat manifolds are not necessarily manifolds of constant curvature. But a wonderful result related to Einstein manifolds can make up for this regret: conformally conformally Einstein manifolds over 3 dimensions must be manifolds of constant curvature. That is to say, if we want conformally conformally flat manifolds to be manifolds of constant curvature, we must call Ric= lambda g, and this is thedefinition of Einstein manifolds. In the formula, Ric is the Ricci curvature tensor, G is the metric tensor, and lambda is constant. The scalar curvature S=m of Einstein manifolds is constant. Moreover, if S is nonzero, there is no nonzero parallel tangent vector field over it. Einstein introduction of the cosmological constant. So he missed the great achievements that the expansion of the universe, so Hubble is successful in the official career; but the vacuum gravitational field equation of cosmological term with had a Einstein manifold, which provides a new stage for mathematicians wit.For the 3 dimensional connected Einstein manifold, even if does not require the conformal flat, it is also the automatic constant curvature manifolds, other dimensions do not set up this wonderful nature, I only know that this is the tensor analysis summer learning, the feeling is a kind of enjoyment. The sectional curvature in the real manifold is different from the curvature of the Holomorphic cross section in the Kahler manifold, and thus produces different results. If the curvature of holomorphic section is constant, the Ricci curvature of the manifold must be constant, so it must be Einstein manifold, called Kahler- Einstein manifold, Kahler. Kahler manifolds are Kahler- Einstein manifolds, if and only if they are Riemann manifolds, Einstein manifolds. N dimensional complex vector space, complex projective space, complex torus and complex hyperbolic space are Kahler- and Einstein manifolds. The study of Kahler-Einstein manifolds becomes the intellectual enjoyment of geometer.Let's go back to an important result of isometric mapping.In this paper, we consider the isometric mapping between M and N and the mapping of the cut space between the two Riemann manifolds, take P at any point on M, and select two non tangent tangent vectors in its tangent space, and obtain its sectional curvature. In the mapping, the two tangent vectors on the P point and its tangent space are transformed into two other tangent vectors under the mapping, and the sectional curvature of the vector is also obtained. If the mapping is isometric mapping, then the curvature of the two cross sections is equal. Or, to be vague, isometric mapping does not change the curvature of the section.Conversely, if the arbitrary points are set, the curvature of the section does not change in nature, then the mapping is not isometric mapping The answer was No. Even in thethree-dimensional Euclidean space on the surface can not set up this property. In some cases, the limit of the geodesic line must be added, and the properties of the Jacobi field can be used to do so. This is the famous Cartan isometry theorem. This theorem is a wonderful application of the Jacobi field. Its wide range of promotion is made by Ambrose and Hicks, known as the Cartan-Ambrose-Hicks theorem.Differential geometry is full of infinite charm. We classify pseudo-Riemannian spaces by using Weyl conformal curvature tensor, which can be classified by Ricci curvature tensor, or classified into 9 types by Bianchi. And these things are all can be attributed to the study of differential geometry, this distant view Riemann and slightly closer to the Klein point of the perfect combination, it can be seen that the great wisdom Cartan, here you can see the profound influence of Einstein.From the Hermite symmetry space to the Kahler-Hodge manifold, differential geometry is not only closely linked with the Lie group, but also connected with algebra, geometry and topologyThink of the great 1895 Poicare wrote the great "position analysis" was founded combination topology unabashedly said differential geometry in high dimensional space is of little importance to this subject, he said: "the home has beautiful scenery, where Xuyuan for." (Chern) topology is the beauty of the home. Why do you have to work hard to compute the curvature of surfaces or even manifolds of high dimensions? But this versatile mathematician is wrong, but we can not say that the mathematical genius no major contribution to differential geometry? Can not. Let's see today's close relation between differential geometry and topology, we'll see. When is a closed form the proper form? The inverse of the Poicare lemma in the region of the homotopy point (the single connected region) tells us that it is automatically established. In the non simply connected region is de famous Rham theorem tells us how to set up, that is the integral differential form in all closed on zero.Even in the field of differential geometry ignored by Poicare, he is still in a casual way deeply affected by the subject, or rather is affecting the whole mathematics.The nature of any discipline that seeks to be generalized after its creation, as is differential geometry. From the curvature, Euclidean curvature of space straight to zero, geometry extended to normal curvature number (narrow Riemann space) andnegative constant space (Lobachevskii space), we know that the greatness of non Euclidean geometry is that it not only independent of the fifth postulate and other alternative to the new geometry. It can be the founder of triangle analysis on it. But the famous mathematician Milnor said that before differential geometry went into non Euclidean geometry, non Euclidean geometry was only the torso with no hands and no feet. The non Euclidean geometry is born only when the curvature is computed uniformly after the metric is defined. In his speech in 1854, Riemann wrote only one formula: that is, this formula unifies the positive curvature, negative curvature and zero curvature geometry. Most people think that the formula for "Riemann" is based on intuition. In fact, later people found the draft paper that he used to calculate the formula. Only then did he realize that talent should be diligent. Riemann has explored the curvature of manifolds of arbitrary curvature of any dimension, but the quantitative calculations go beyond the mathematical tools of that time, and he can only write the unified formula for manifolds of constant curvature. But we know,Even today, this result is still important, differential geometry "comparison theorem" a multitude of names are in constant curvature manifolds for comparison model.When Riemann had considered two differential forms the root of two, this is what we are familiar with the Riemann metric Riemannnian, derived from geometry, he specifically mentioned another case, is the root of four four differential forms (equivalent to four yuan product and four times square). This is the contact and the difference between the two. But he saidthat for this situation and the previous case, the study does not require substantially different methods. It also says that such studies are time consuming and that new insights cannot be added to space, and the results of calculations lack geometric meaning. So Riemann studied only what is now called Riemann metric. Why are future generations of Finsler interested in promoting the Riemann's not wanting to study? It may be that mathematicians are so good that they become a hobby. Cartan in Finsler geometry made efforts, but the effect was little, Chern on the geometric really high hopes also developed some achievements. But I still and general view on the international consensus, that is the Finsler geometry bleak. This is also the essential reason of Finsler geometry has been unable to enter the mainstream of differential geometry, it no beautiful properties really worth geometers to struggle, also do not have what big application value. Later K- exhibition space, Cartan space will not become mainstream, although they are the extension of Riemannnian geometry, but did not get what the big development.In fact, sometimes the promotion of things to get new content is not much, differential geometry is the same, not the object of study, the more ordinary the better, but should be appropriate to the special good. For example, in Riemann manifold, homogeneous Riemann manifold is more special, beautiful nature, homogeneous Riemann manifolds, symmetric Riemann manifold is more special, so nature more beautiful. This is from the analysis of manifold Lie group action angle.From the point of view of metric, the complex structure is given on the even dimensional Riemann manifold, and the complexmanifold is very elegant. Near complex manifolds are complex manifolds only when the near complex structure is integrable. The complex manifold must be orientable, because it is easy to find that its Jacobian must be nonnegative, whereas the real manifold does not have this property in general. To narrow the scope of the Kahler manifold has more good properties, all complex Submanifolds of Kahler manifolds are Kahler manifolds, and minimal submanifolds (Wirtinger theorem), the beautiful results captured the hearts of many differential geometry and algebraic geometry, because other more general manifolds do not set up this beautiful results. If the first Chern number of a three-dimensional Kahler manifold is zero, the Calabi-Yau manifold can be obtained, which is a very interesting manifold for theoretical physicists. The manifold of mirrors of Calabi-Yau manifolds is also a common subject of differential geometry in algebraic geometry. The popular Hodge structure is a subject of endless appeal.Differential geometry, an endless topic. Just as algebraic geometry requires double - rational equivalence as a luxury, differential geometry requires isometric transformations to be difficult. Taxonomy is an eternal subject of mathematics. In group theory, there are single group classification, multi complex function theory, regional classification, algebraic geometry in the classification of algebraic clusters, differential geometry is also classified.The hard question has led to a dash of young geometry and old scholars, and the prospect of differential geometry is very bright.。

机器证明-埃尔多斯-谢卡雷斯问题说明书

Mechanical Proving for ERDÖS-SZEKERES Problem1Meijing ShanInstitute of Information science and Technology,East China University of Political Science and Law, Shanghai, China. 201620Keywords: Erdös-Szekeres problem, Automated deduction, Mechanical provingAbstract：The Erdös-Szekeres problem was an open unsolved problem in computational geometry and related fields from 1935. Many results about it have been shown. The main concern of this paper is not only show how to prove this problem with automated deduction methods and tools but also contribute to the significance of automated theorem proving in mathematics using advanced computing technology. The present case is engaged in contributing to prove or disprove this conjecture and then solve this problem. The key advantage of our method is to utilize the mechanical proving instead of the traditional proof and this method could improve the arithmetic efficiency.IntroductionThe following famous problem has attracted more and more attention of many mathematicians [3, 6, 12, 16] due to its beauty and elementary character. Finding the exact value of N(n) turns out to be a very challenging problem. The problem is very easy to explain and understand.The Erdös-Szekeres Problem 1.1 [4, 15]. For any integer n ≥ 3, determine the smallest positive integer N(n) such that any set of at least N(n) points in generalposition in the plane contains n points that are the vertices of a convex n-gon.A set of points in the plane is said to be in the general position if it contains no three points on a line. This problem was also called Happy Ending Problem by Erdös, because two investigators Esther Klein and George Szekeres who first worked on the problem became engaged and subsequently married[8, 17].The interest of Erdös and Szekeres in this problem was initiated by Esther Klein(later Mrs. Szekeres), who observed that from 5 points of the plane of which no three lie on the same straight line it is always possible to select 4 points determining convex quadrilateral. There are three distinct types of five points in the plane, as shown in Figure 1.Figure 1. Three cases for 5 points.In any case of the Figure 1, one can find at least one convex quadrilateral determined by the points. Klein [4] suggested the following more general problem: Can we find for a given n a number N(n) such that from any set containing at least N points it is possible to select n points forming a convex polygon?As observed by Erdös and Szederes [4], there are two particular questions:(1) Does the number N corresponding to n exist?(2) If so, how is the least N(n) determined as a function of n?1T his work was financially supported by Humanity and Social Science Youth foundation of Ministry of Education of China (No. 14YJCZH020).They proved the existence of the number N(n) by two different methods. The first one is a combinatorial theorem of Ramsey. The foundation of the second one isbased on some geometrical and combinatorial considerations. And then they formulated the following conjecture.Conjecture 1.1 N (n ) = 2n −2 + 1 for all n ≥ 3.Despite its elementary characters and the efforts of many researchers, the ErdösSzekeres problem is solved for the value n = 3, 4 and 5 only. The case n = 3 istrivial, and n = 4 is due to Klein. The equality N (5) = 9 was proved by E. Makaiwhile the published proof by Kalbfleisch [11] and then Bonnice [2] and Lovasz [13]independently published the much simpler proofs. The bottle neck of this problem now is when n >5, how to prove or disprove the conjecture.About this problem, the best currently known bounds are(1)Where is a binomial coefficient. The lower bound was obtained by Erdös andSzekeres [4] and the upper bound is due to Töth and Valtr [18]. The lower bound issupposed to be sharp, according to conjecture 1.1.In this paper we use the automated deduction method and tools [1, 19-21] to establish a mechanical method for proving problem instead of the manual proof. We hope the method might give a rise to substantially promote this unsolved problem.The rest content of the paper is indicated by section headings as follows: Section 2 preliminaries, Section 3 main results, Section 4 conclusion and remarks.PreliminariesIn this Section, we present some algorithms that would help us develop our method in next section.Algorithm 2.1. Modified Cylindrical Algebraic Decomposition (CAD). Due to the problem statement, the proof should consider all kinds of points’ positions on the plane.The Cylindrical Algebraic Decomposition [5, 14] of R n adapted to a set ofpolynomials which is a partition of R d is cells (simple connected subsets of R d )such that each input polynomial has a constant sign on each cell. Basically, thealgorithm computes recursively at least one point in each cell (so that one can testthe cells that verify a fixed sign condition). The sample points Pi got by originalCAD are always more than one on each cell. We modify the procedure to have asample point on each cell of the final cell decomposition, by the rule that Pi has a constant sign on each cell. We elaborate the main idea under lying our method by showing how our main algorithm evolved from the original one.Here, we describe it as follows.Algorithm. MCADInput: A set F of polynomialsOutput: A F-sign-invariant CAD of R nStep 1. Projection. Compute the projection polynomials which using exclusivelyoperations, and receive some (n − 1) -variate polynomials.Step 2. Recur. Apply the algorithm recursively to compute a CAD of R n −1which Q(F )is sign-invariant.Step 3. Lifting. Lift the Q(F )-sign-invariant CAD of R n −1 up to a F-sign-invariant CAD of using the auxiliary polynomial Π(F ) of degree no largerthan d (F ) (d is the maximum degree of any polynomial in F).Step 4. Choice. Utilize the strategy that {Pi} has constant sign on each cell tochoose one sample point on each cell.Algorithm 2.2 (Graham Scan Algorithm) [9, 10, 22]We present one of the simplest algorithms used to find the convex hull fromsome points. Some basic definitions are provided in the field of Computational Geometry. This algorithm works in three phases:Input: A set S of pointsOutput: The convex hull of S.Step 1. Find an extreme point. The algorithm starts by picking a point in S known to be a vertex of the convex hull. This point is chosen to be with smallest y coordinate and guaranteed to be on the hull. If there are some points with the same smallest y coordinate, we will choose the point with largest x coordinate in them. In other words, we select the right most lowest point as the extreme point.Step 2. Sort the points. Having selected the base point which is called P0 , then the algorithm sorts the other points P in S by the increasing counter-clockwise (ccw) angle the line segment P0P makes with the x-axis. If there is a tie and two points have the same angle, discard the one that is closest to P0 .Step 3. Construct the convex hull. Build the hull by marching around the star shaped polygon, adding edges when we make a left turn, and back-tracking when we make a right turn. We end up with a star-shaped polygon, see Figure 3 (one in whichone special point, in this case the pivot, can “see” the whole polygon). Considering efficiency in Step 2, it is important to note that the comparison of sorting between two points P2 and P3 can be made without actually computing their angles. In fact, computing angles would use slow in accurate trigonometry functions, and doing these computations would be a bad mistake. Instead, one just observes that P2 would make a greater angle than P1 if (and only if) P2 lies on theleft side of the directed line segment P0P1 , see Figure 2.We make full use of this algorithm to judge whether the polygon received in every recursive step is a convex polygon or not. It is a decision method in our algorithm.To state the algorithm clearly, we will describe it in a style of pseudo-code.Algorithm. Graham Scan AlgorithmInput: A set S of points in the planeOutput: A list containing the vertex of the convex hullSelect the right most lowest point P0 in SFigure 3. Graham ScanFigure 2. Sort the points.Sort S angularly about P0 as a center.For ties, discard the closer points.Let S be the sorted array of points.Push S [1] = P0 and P1 onto a stack Ω.Let P1 = the top point on ΩLet P2 = the second top point on Ωwhile S [k ] P doif (S [k −1] is strictly left of the line S [k ] to S [k +1]), thenPush S [k −1] onto ΩelsePop the top point S [k ] off the stack Ω.fi;od;Main ResultsIn contrast with the traditional proof, this method presented following can show the convex polygons received in every step.The main idea of our algorithm: First, give randomly four points in general position in the plane, and then we use polynomials of points’ coordinates to represent the lines. We extend the 4-element set to some5-element sets. We do this by establishing the corresponding Modified Cylindrical Algebraic Decomposition (MCAD) and design an interactive program which allow the user to choose among the candidate one sample point in each cell. We use Graham Scan algorithm to determine whether or not there is a convex 5-gon (convexpentagon) in every set received. If some of the received 5-element sets have no convex 5-gon, then extend them to the 6-element sets by the strategy mentionedabove. Simultaneously, check whether or not each of the received 6-element sets has any convex hull at least with any convex 5-gon in the polygon. We implement theprogram repeatedly until can find a convex n-gon (n ≥ 5) in any set. We trace the processing of the extension and decision, and then draw a conclusion that N (5) = 9.To prove this approach, we write the following algorithm named “conv5”. Based on this algorithm can generate short and readable mechanical proving for the Erdös-Szekeres conjecture, including the case of n = 3, 4, 5. It consists of two main algorithms—Modified Cylindrical Algebraic Decomposition algorithm and Graham Scan algorithm and some sub-algorithms such as collinear, pol, sam, ponlist, min0,isleft, ord, conhull, point5, convex, G5,G6,Pmn.Algorithm Conv5Input: four points in the general position in the plane.Output: Any polygon with at least 9 points in general position in the planecontains a convex 5-gon. Step 1 [collinear]. Write the line polynomials with the given four points (basepoints).Step 2 [pol, sam, ponlist]. Illustrate how we utilize the CAD to find somesample points in the cell which built by the lines, and then with the base n points get n +1 -element sets.Step 3 [min0, isleft, conhull, point5, convex, G5]. Decide whether or not thereis a convex hull or a convex n-gon (n ≥ 5) in every set; if it is true, then stop; elsegoto Step 4.Step 4 [G6, Pmn]. Deal with the sets which have no convex n-gon (n ≥ 5).Recursively implement Step 2 and Step 3 process, until at least there is a convex 5-gon in any set. End Conv5.The key techniques of the algorithm are listed as follows:1. To reduce the complexity of the computation and increase the efficiency,when we check whether or not there is a convex 5-gon in the given points set,we utilize the strategy as follows:if there is a convex hull at least with 5 points in the points set thenpop this points setelif there is a convex 5-gon thenpop this points setelse “there is no convex 5-gon in this points set” go to next step2. Each convex n-gon (n ≥ 5) contains a convex 5-gon3. If there is no convex 4-gon, then there should be no convex 5-gon.Conclusion and RemarksBy the Maple procedure we have implemented the mechanical method for the conjecture in certain cases. Through observing the whole computational process, we obtain a certain answer that any set with at least 9 points in general position in the plane contains a convex 5-gon. This method can be generalized in an obvious way to arbitrary base points in the plane.For the mechanical method proposed here, on one hand it provided one of the promising direction for proving or disproving the Conjecture 1.1 (when n ≥ 6), even for handling with some unsolved problems in computational geometry. On the other hand, it gave one especially useful application of computer algebraic andautomated deduction. For further investigations, now we consider about the following problems:1. Does any set of at least 17 points in general position in the plane contains 6 points which are the vertices of a convex hexagon? Can we give the proof about N (6) existence and prove or disprovethe corresponding conclusion by mechanical proving? Now the best known conclusion about this is N (6) ≥ 27, if it exists.2. Erdös posed a similar problem on empty convex polygons. Whether or not we can give the automated proof to this problem?References[1] M. de Berg, M. van Kreveld, M. Overmars and O. Schwarzkopf, omputational GeometryAlgorithm and Applications, (2nd ed.), Spring-Verlag, Berlin, Heideberg, New York, 1997.[2] W. E. Bonnice, On convex polygons determined by a finie planar set, Amer. Math. Monthly.[3] F. R. K. Chung and R. L. Graham, Forced Convex n-gons in the Plane, Discr. Comput. Geom. 19(1998), 367-371.[4] P. Erdös and G. Szekeres, A combinatiorial problem in geometry, Comositio Mathematica 2(1935), 463-470.[5] E. Collins George, Quantifier elimination for the elementary theory of real closed fields by cylindrical algebraic decomposition Lecture Notes In Computer Sciencevol. 33, Springer-Verlag, Berlin, pp. 134-183.[6] Kráolyi Gyula, An Erdös-Szekeres type problem in the plane.[7] X. R. Hou and Z. B. Zeng, An efficient Algorithm for Finding Sample Points of Algebraic Decomposition of Plane, Computer Application, 1997, (in Chinese).[8] P. Hofiman, The Man Who loved Only Numbers Hyperion, New York, 1998.[9] /ah/alganim/version0/Graham.html.[10] http://cgm.cs.mcgill.ca/ beezer/cs507/main.html.[11] J. D. Kalbfleisch, J. G. Kalbfleisch and R. G. Stanton, A combinatorial problem on convexregions, Proc. Louisiana Conf. Combinatorics, Graph Theory and Computing, Louisianna StateUniv., Baton Touge, La, Congr. Numer. 1 (1970), 180-188.[12] D. Kleitman and L. Pachter, Finding convex sets among points in the plane, Discr. Comput.Geom. 19, (1998), 405-410.[13] L. Lovasz, Combinatorial Problem and Exercises North-Holland, msterdam, 1979.[14] Bhubaneswar Mishra, Algorithmic Algebra, Springer-Verlag, 2001.[15] W. Morris and V. Soltan, The Erdös-Szekeres Problem on Points in ConvexPosition- A, Survey Bulletin of the American Mathematical Society, vol. 37.[16] L. Graham Ronald and Yao Frances, Finding the Convex Hull of a SimplePolygon Report No. STAN-CS-81-887, 1998.[17] B. Schechter, My Brain is Open Simon Schuster, New York, 1998.[18] G. Tóth and P. Valtr, Note on the Erdös-Szekeres theorem, Discr. Comput. Geom. 19 (1998), 457-459.[19] W. T. Wu, Basic Principles of Mechanical Theorem Proving in Geometries Science Press,Beijing, 1984, (Part on elementary geometries, in Chinese).[20] L. Yang and B. C. Xia, Automated Deduction in Real Geometry, Geometric Computation, WorldScientific, 2004.[21] L. Yang and Z. Z. Jing and X. R. Hou, Nonlinear Algebraic Equation System and AutomatedTheorem Proving, Shanghai Scientific and Technological Education Publishing House, Shanghai,1996, (in Chinese).[22] P. D. Zhou, Computational Geometry Design and Analysis in Chinese, TsingHua UniversityPress, 2005.。

彭赛列闭合定理

Lorenz Halbeisen and Norbert Hungerb¨uhlerMSC:Primary51M04,Secondary51A20;key words:Poncelet’s Porism,Pascal’s TheoremJanuary2014]A SIMPLE PROOF OF PONCELET’S THEOREM1used Abel’s Theorem and the representation of elliptic curves by means of the Weier-strass ℘-function to establish the equivalence of Poncelet’s Theorem and the groupstructure on elliptic curves,see [11].Poncelet’s Theorem has a surprising mechani-cal interpretation for elliptic billiards in the language of dynamical systems:see [8]or [7]for an overview of this facet.A common approach to all four classical closingtheorems (the Poncelet porism,Steiner’s Theorem,the Zigzag Theorem,and Emch’sTheorem)has recently been established by Protasov in [18].King showed in [14],thatPoncelet’s porism is isomorphic to Tarski’s plank problem (a problem about geometricset-inclusion)and to Gelfand’s question (a number theoretic problem)via the con-struction of an invariant measure.However,according to Berger [1,p.203],all knownproofs of Poncelet’s Theorem are rather long and recondite.The aim of this paper is to give a simple proof of Poncelet’s Theorem in the realprojective plane.More precisely,we will show that Poncelet’s Theorem is a purelycombinatorial consequence of Pascal’s Theorem.Before we give several forms of thelatter,let us introduce some notation.For two points a and b ,let a −b denote the linethrough a and b ,and for two lines ℓ1and ℓ2,let ℓ1∧ℓ2denote the intersection pointof these lines in the projective plane.In abuse of notation,we often write a −b −c inorder to emphasize that the points a ,b ,c are collinear.In the sequel,points are oftenlabeled with numbers,and lines with encircled numbers like ③.In this terminology,Pascal’s Theorem and its equivalent forms read as follows.Pascal’s Theorem (cf.[16])314256Any six points 1,...,6lie on a conic if and only if (1−2)∧(4−5)(2−3)∧(5−6)(3−4)∧(6−1)are collinear.Carnot’s Theorem (cf.[3,no.396])514236Any six points 1,...,6lie on a conicif and only if[(1−2)∧(3−4)]−[(4−5)∧(6−1)](2−5)(3−6)are concurrent.2c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121Brianchon’s Theorem (cf.[2])①②③④⑤⑥Any six lines ①,...,⑥are tangent to a conic if and only if (①∧②)−(④∧⑤)(②∧③)−(⑤∧⑥)(③∧④)−(⑥∧①)are concurrent.Carnot’s Theorem ∗①②⑤④③⑥Any six lines ①,...,⑥are tangent to a conicif and only if[(①∧②)−(③∧④)]∧[(④∧⑤)−(⑥∧①)](②∧⑤)(③∧⑥)are collinear.As a matter of fact,we would like to mention that if the conic is not degenerate,then the collinear points in Pascal’s Theorem are always pairwise distinct (the sameapplies to the concurrent lines in Brianchon’s Theorem).Since the real projective plane is self-dual,Pascal’s Theorem and Brianchon’s The-orem are equivalent.Moreover Carnot’s Theorem and its dual Carnot’s Theorem ∗arejust reformulations of Pascal’s Theorem and Brianchon’s Theorem by exchanging thepoints 3and 5,and the lines ③and ⑤,respectively.Recall that if two adjacent points,say 1and 2,coincide,then the corresponding line 1−2becomes a tangent with 1ascontact point.Similarly,if two lines,say ①and ②,coincide,then ①∧②becomes thecontact point of the tangent ①.As a last remark,we would like to mention that a conicis in general determined by ﬁve points,by ﬁve tangents,or by a combination like threetangents and two contact points of these tangents.The paper is organized as follows.In Section 2,we prove Poncelet’s Theorem forthe special case of triangles and at the same time we develop the kind of combinato-rial arguments we shall use later.Section 3contains the crucial tool which allows toshow that Poncelet’s Theorem holds for an arbitrary number of edges.Finally,in Sec-tion 4,we use the developed combinatorial technics in order to prove some additionalsymmetry properties of Poncelet-polygons.2.PONCELET’S THEOREM FOR TRIANGLES In order to prove Poncelet’sTheorem for triangles,we will show that if the six vertices of two triangles lie ona conic K ,then the six sides of the triangles are tangents to some conic C .The crucial point in the proof of the following theorem (as well as in the proofsof the other theorems of this paper)is to ﬁnd the suitable numbering of points andedges,and to apply some form of Pascal’s Theorem in order to ﬁnd collinear points orconcurrent lines.Theorem 2.1.If two triangles are inscribed in a conic and the two triangles do nothave a common vertex,then the six sides of the triangles are tangent to a conic.January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 3a 1a 2a 3b 3b 2b 1Proof.Let K be a conic in which two triangles △a 1a 2a 3and △b 1b 2b 3are inscribedwhere the two triangles do not have a common vertex.First,we introduce the following three intersection points:I :=(a 1−a 2)∧(b 1−b 2),X :=(a 2−b 3)∧(b 2−a 3),I ′:=(a 3−a 1)∧(b 3−b 1).In order to visualize the intersection points I ,X ,and I ′,we break up the conic K anddraw it as two straight lines,one for each triangle as follows.I X I ′Ib 1a 1b 2a 2b 3a 3b 1a 1b 2a 2Now,we number the six points a 1,a 2,a 3,b 1,b 2,b 3on the conic K as shown by thefollowing ﬁgure.I X I ′Ib 14a 11b 25a 22b 33a 36b 14a 11b 25a 224c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121By Pascal’s Theorem we get that the three intersection points(1−2)∧(4−5),(2−3)∧(5−6),and (3−4)∧(6−1)are collinear,which is the same as saying that the points I −X −I ′are collinear.In the next step,we label the sides of the triangles as shown in the following ﬁgure.I X I ′Ib 1a 1b 2a 2b 3a 3b 1a 1b 2a 2②①⑥②⑤④③⑤By Carnot’s Theorem ∗we get that the six sides ①,...,⑥of the two triangles aretangents to a conic if and only if[(①∧②)−(③∧④)]∧[(④∧⑤)−(⑥∧①)],(②∧⑤),and(③∧⑥)are collinear.Now,this is the same as saying that the points X −I −I ′are collinear,which,as we have seen above,is equivalent to a 1,a 2,a 3,b 1,b 2,b 3lying on a conic.a 1a 2a 3b 3b 2b 1q.e.d.As an immediate consequence we get Poncelet’s Theorem for triangles.Corollary 2.2(Poncelet’s Theorem for triangles).Let K and C be nondegenerateconics.Suppose there is a triangle△a 1a 2a 3inscribed in K and circumscribed aboutC .Then for any point b 1of K for which two tangents to C exist,there is a triangle△b 1b 2b 3which is also inscribed in K and circumscribed about C .January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 5Proof.Let K and C be two conics with a triangle△a1a2a3inscribed in K and circumscribed about C.Let b1be an arbitrary point on K which is distinct froma1,a2,a3,and let b2and b3be distinct points on K such that b1−b2and b1−b3are two tangents to C.By construction,we get thatﬁve sides of the triangles△a1a2a3and△b1b2b3are tangents to C.On the other hand,by Theorem2.1,we know that allsix sides of these triangles are tangents to some conic C′.Now,since a conic is deter-mined byﬁve tangents,C′and C coincide,which implies that the triangle△b1b2b3is circumscribed about C.q.e.d.As a special case of Brianchon’s Theorem we get the following.Fact2.3.Let C be a conic and let the triangle△a1a2a3be circumscribed about C. Furthermore,let t1,t2,t3be the contact points of the three tangents a2−a3,a3−a1,a1−a2.Then the three lines a1−t1,a2−t2,and a3−t3meet in a point.bel the three sides of the triangles as follows:①=a2−a3=②,③=a3−a1=④,⑤=a1−a2=⑥.Then①∧②=t1,③∧④=t2,⑤∧⑥=t3,and by Brianchon’s Theorem we getthat a1−t1,a2−t2,a3−t3meet in a point.q.e.d.In general,for arbitrary n-gons tangent to C the analogous statement will be false. However,if n is even and if the n-gon is at the same time inscribed in a conic K,a similar phenomenon occurs(see Theorem4.2).3.THE GENERAL CASE Let K and C be nondegenerate conics in general posi-tion.We assume that there is an n-sided polygon a1,...,a n which is inscribed in K such that all its n sides a1−a2,a2−a3,...,a n−a1are tangent to C and none of its vertices belongs to C.Let us assume that n is minimal with this property(thus,in par-ticular,the points a1,...,a n are pairwise distinct).Further,let b1,...,b n be an(n−1)-sided polygonal chain on K where all n−1sides b1−b2,b2−b3,...,b n−1−b nare tangent to C and none of its vertices is one of a1,...,a n or belongs to C.We donot yet know that b n−b1is tangent to C too.If we break up the conic K and draw itas two straight lines,one for the polygon and one for the polygonal chain,we get the following situation.a n−1a n a1a2a n−1a n a1a2b n−1b n b1b2b n−1b n b1b2In order to prove Poncelet’s Theorem,we have to show that b n−b1is also tangent toC.This will follow easily from the following result.Lemma3.1.For n≥4,the three intersection pointsI:=(a1−a2)∧(b1−b2),X:=(a2−b n−1)∧(b2−a n−1),andI′:=(a n−1−a n)∧(b n−1−b n),6c THE MATHEMATICAL ASSOCIATION OF AMERICA[Monthly121I X I ′b 1b 2b n −1b na 1a 2a n −1a nare pairwise distinct and collinear,which is visualized above by the dashed line.Proof.Depending on the parity of n ,we have one of the following anchorings,fromwhich we will work step by step outwards.n even,with k =n2:I X I ′b k −1a k −1b k a kb k +1a k +1123456By Pascal’s Theorem we have thatI −X −I ′are collinear.For n ≥5,the lemma will follow from the following two claims.Claim 1.Let p and q be integers with 2≤p <q ≤n −1.Further,letI p −1:=(a p −1−a p )∧(b p −1−b p ),I p :=(a p −a p +1)∧(b p −b p +1),I q :=(a q −1−a q )∧(b q −1−b q ),I q +1:=(a q −a q +1)∧(b q −b q +1),and letX :=(a p −b q )∧(b p −a q ).If I p −X −I q are pairwise distinct and collinear,then Ip −1−X −Iq +1are alsopairwise distinct and collinear.This implication is visualized by the following ﬁgure.I p X I qb p −1b p b p +1b q −1b q b q +1a p −1a p a p +1a q −1a q a q +1⇓January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 7I p −1X I q +1b p −1b p b p +1b q −1b q b q +1a p −1a p a p +1a q −1a q a q +1Proof of Claim 1.εI p I qγαb p b qa p a q ④③②⑤⑥①(a)By Brianchon’s Theorem,the lines α,γ,εare pairwise distinct and concurrent.I p X I qβγαb p b qa p a q(b)The lines α,β,γmeet,by assumption,in X ,and they are pairwise distinct.By (a),we have that αand γare distinct,and by symmetry also βand γare distinct.Since a straight line meets a nondegenerate conic in at most two points,αand βarealso distinct.εI p −1I q +1βδb p −1b p b q −1b q b q +1a p −1a p a p +1a q a q +1⑥①②⑤④③(c)By Brianchon’s Theorem,the lines β,ε,δare pairwise distinct and concurrent.By (a)&(b)we get that α,β,and εmeet in X ,and by (c)we get that also α,β,andδmeet in X ,which implies that Ip −1−X −I q +1are collinear and pairwise distinct.If I p −1=I q +1,then the four lines a p −1−a p ,b p −1−b p ,a q −a q +1,b q −b q +1,whichare all tangent to C ,would be concurrent.But then these four lines are not pairwisedistinct,and since the eight points ap −1,ap ,aq ,a q +1,b p −1,b p ,bq ,bq +1are pairwisedistinct (recall that 1≤p −1<q +1≤n ),this contradicts our assumption that the8c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121conic K is nondegenerate.By similar arguments it follows that both I p −1and I q +1aredistinct from X .q.e.d.Claim 2.Let I p −1,I q +1,and X be as above,and letX ′:=(a p −1−b q +1)∧(b p −1−a q +1).If I p −1−X −I q +1are pairwise distinct and collinear,then Ip −1−X ′−Iq +1arepairwisedistinct and collinear too.This implication is visualized by the following ﬁgure.I p −1X I q +1b p −1b p b p +1b q −1b q b q +1a p −1a p a p +1a q −1a q a q +1⇓I p −1X ′I q +1b p −1b p b p +1b q −1b q b q +1a p −1a p a p +1a q −1a q a q +1Proof of Claim 2.I p −1X I q +1b p b qa p a q(a)By assumption,the points I p −1−X −I q +1are pairwise distinct and collinear.JI p −1X b p b qa p a q123456(b)By Pascal’s Theorem,the points I p −1−X −J are pairwise distinct andcollinear.January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 9J X ′I q +1b p b qa p a q126453(c)By Pascal’s Theorem,the points X ′−J −I q +1are pairwise distinct andcollinear.By (a)&(b)we get that I p −1−J −I q +1are collinear,and by (c)we get that X ′lies on J −Iq +1.Hence,Ip −1−X ′−Iq +1are collinear.By (a),(c)and a symmetricversion of (c),the three points I p −1,X ′,I q +1are pairwise distinct.q.e.d.By an iterative application of Claim 1&2,we ﬁnally get the situationI X I ′b 1b 2b n −1b na 1a 2a n −1a nin which I −X −I ′are pairwise distinct and collinear.q.e.d.With similar arguments as in the proof of Poncelet’s Theorem for triangles (Corol-lary 2.2),we can now prove the general case of Poncelet’s Theorem (Theorem 1.1):Proof of Poncelet’s Theorem.Let K and C be nondegenerate conics in general posi-tion.We assume that there is an n -sided polygon a 1,...,an which is inscribed in Ksuch that all its n sides a 1−a 2,a 2−a 3,...,a n −a 1are tangent to C and none of itsvertices belongs to C .Let us assume that n is minimal with this property.Further weassume that there is an (n −1)-sided polygonal chain b 1,...,bn whose n −1sidesare tangent to C and none of its vertices is one of a 1,...,an or belongs to C .We haveto show that b n −b 1is tangent to C .!!b n −1b n b 1b 2b n −1b n b 1b 2a n −1a n a 1a 2a n −1a n a 1a 2By Lemma 3.1we know that I −X −I ′are pairwise distinct and collinear,where I =(a 1−a 2)∧(b 1−b 2),I ′=(a n −1−a n )∧(b n −1−b n ),and X =(a 2−b n −1)∧(b 2−a n −1).In order to show that b n −b 1is tangent to C ,we have tointroduce two more intersection points:J :=(a n −1−a 1)∧(b n −1−b 1),X ′:=(a n −b 1)∧(b n −a 1).10c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121J I X I ′X ′Ib 1b 2b 1b 2b n −1b n a 1a 2a 1a 2a n −1a n We now apply Pascal’s Theorem twice as illustrated below.J I X Ib 1b 2b 1b 2b n −1b n a 1a 2a 1a 2a n −1a n 1122344556(a)By Pascal’s Theorem,the points I −X −J are pairwise distinct and collinear.J I ′X ′b 1b 2b 1b 2b n −1b n a 1a 2a 1a 2a n −1a n 126345(b)By Pascal’s Theorem,the points I ′−J −X ′are pairwise distinct and collinear.Since,by Lemma 3.1,I −X −I ′are pairwise distinct and collinear,by (a)we getthat I −X −J −I ′are collinear,and by (b)we ﬁnally get that I −X ′−I ′arecollinear.For the last step,we apply Carnot’s Theorem ∗.I ′X ′Ib 1b 2b n −1b n a 1a 2a n −1a n ②①⑥⑤④③Since I −X ′−I ′are collinear,by Carnot’s Theorem ∗we get that the six lines ①,...,⑥are tangent to some conic C ′.Now,since a conic is determined by ﬁve tangents,andthe ﬁve lines ①,②,③,⑤,⑥are tangent to C ,C ′and C coincide.This implies that ④istangent to C ,which is what we had to show.q.e.d.January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 114.SYMMETRIES IN PONCELET-POLYGONS In this section we present somesymmetries in 2n -sided polygons which are inscribed in some conic K and circum-scribed about another conic C .To keep the terminology short,we shall call such apolygon a 2n -Poncelet-polygon with respect to K &C .Theorem 4.1.Let K and C be nondegenerate conics in general position which neithermeet nor intersect and let a 1,...,a 2n be the vertices of a 2n -Poncelet-polygon withrespect to K &C .Further let t 1,...,t 2n be the contact points of the tangents a 1−a 2,...,a 2n −a 1.(a)All the n diagonals a 1−a n +1,a 2−a n +2,...,a n −a 2n meet in a point H 0.(b)All the n lines t 1−t n +1,t 2−t n +2,...,t n −t 2n meet in the same point H 0.Proof.(a)By the proof of Lemma 3.1,we get that the three points a 1,an +1,and(a 2−a n +2)∧(a n −a 2n )are collinear.a 1a n +1a 1a 2n a n +2a n +1a 1a 2a na n +1This is the same as saying that the three diagonals a 1−an +1,a 2−an +2,and an −a 2n meet in a point,say H 0.Now,by cyclic permutationwe get that all n diagonals meet in H 0.(b)By the proof of Lemma 3.1,we get that the three points t 1−H 0−t n +1arecollinear.t 1H 0t n +1a 2a 1a 2n a n +3a n +2a n +1a 1a 2a 3a n a n +1a n +2Thus,by cyclic permutation we get that all n lines t 1−tn +1,t 2−t n +2,...,tn −t 2npass through H 0,which implies that all n lines meet in H 0.q.e.d.In the last result,we show that the point H 0is independent of the particular 2n -Poncelet-polygon (compare with Poncelet’s results no.570&571in [17]).Theorem 4.2.Let K and C be nondegenerate conics in general position which nei-ther meet nor intersect and let a 1,...,a 2n and b 1,...,b 2n be the vertices of two2n -Poncelet-polygons with respect to K &C .Further let t 1,...,t 2n and t ′1,...,t′2nbe the contact points of the Poncelet-polygons.Then all 4n lines a 1−an +1,...,t 1−t n +1,...,b 1−b n +1,...,t ′1−t′n +1,...meet in a point H 0.Moreover,oppositesides of the Poncelet-polygons meet on a ﬁxed line h ,where h is the polar of H 0,bothwith respect to C and K .Proof.By Theorem 4.1we know that the 2n lines a 1−a n +1,...,t 1−tn +1,...meetin a point H 0.First,we show that the polar h of the pole H 0with respect to C is thesame as the polar h ′of H 0with respect to K ,and then we show that the point H 0isindependent of the choice of the 2n -Poncelet-polygon.12c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121a i +1a k +1a i +na k +n a i a k +n +1a k a i +n +1t kt k +n ti t i +nH 0PQFirst notice that in the ﬁgure above,H 0is on the polar p of P with respect to theconic C and that H 0is also on the polar p ′of P with respect to the conic K (see forexample Coxeter and Greitzer [6,Theorem 6.51]).Thus,P lies on the polar h of H 0with respect to C ,as well as on the polar h ′of H 0with respect ot K .Since the sameapplies to the point Q ,the polars h and h ′coincide,which shows that the pole H 0hasthe same polar with respect to both conics.The fact that H 0is independent of the choice of the 2n -Poncelet-polygon is just aconsequence of the following.Claim.Let H 0be as above and let h be the polar of H 0(with respect to K or C ).Choose an arbitrary point P on h .Let s 1&s 2be the two tangents from P to C andlet A &A ′and B &B ′be the intersection points of s 1and s 2with K .H 0PBA B ′A ′s 1s 2Then H 0=(A −B ′)∧(B −A ′).Proof of Claim.By a projective transformation,we may assume that h is the line atinﬁnity.Then,the pole H 0becomes the common center of both conics and the claimfollows by symmetry.q.e.d.January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 13Now,let a1,...,a2n and b1,...,b2n be the vertices of two2n-Poncelet-polygons with respect to K&C.Furthermore,let H0=(a1−a n+1)∧(a2−a n+2)and H′0= (b1−b n+1)∧(b2−b n+2),and let h and h′be their respective polars.Choose anypoint P which lies on both h and h′,and draw the two tangents from P to C which intersect K in the points A,A′,B,B′.If the conics K and C do not meet(what we assume),then these points are pairwise distinct and by the Claim we get H0= (A−B′)∧(B−A′)=H′.q.e.d.Notice,that for n=3,H0is the Brianchon point with respect to C of the Poncelet-hexagon,and h its Pascal line with respect to K.So,for n>3,the point H0is the generalized Brianchon point with respect to C of the2n-Poncelet-polygon,and h its generalized Pascal line with respect to K.,Bicentennial of the great Poncelet theorem(1813–2013):current advances,Bull.Amer.Math.Soc.(N.S.)51(3)(2014)373–445.9.L.Euler,Solutio facilis problematum quorundam geometricorum difﬁcillimorum,Novi Commentariiacademiae scientiarum imperialis Petropolitanae(1767)103–123.10.N.Fuss,De polygonis symmetrice irregularibus circulo simul inscriptis et circumscriptis,Nova ActaPetropol(1798)166–189.11.P.Grifﬁths and J.Harris,On Cayley’s explicit solution to Poncelet’s porism,L’EnseignementMath´e matique.Revue Internationale.IIe S´e rie24(1978)31–40.12. A.S.Hart,On the porism of the in-and-circumscribed triangle,The Quarterly journal of pure and appliedmathematics2(1858)143.13. C.G.J.Jacobi,Ueber die Anwendung der elliptischen Transcendenten auf ein bekanntes Problem derElementargeometrie:“Die Relation zwischen der Distanz der Mittelpuncte und den Radien zweier Kreise zuﬁnden,von denen der eine einem unregelm¨aßigen Polygon eingeschrieben,der andere demselben umgeschrieben ist”,J.Reine Angew.Math.3(1828)376–389.14.J.L.King,Three problems in search of a measure,The American Mathematical Monthly101(1994)609–628.15.H.Lebesgue,Les coniques,Gauthier-Villars,Paris,1942.16. B.Pascal,Essay pour les Coniques,(1640).17.J.-V.Poncelet,Trait´e des propri´e t´e s projectives desﬁgures,Bachelier,Paris,1822.18.V.Y.Protasov,Generalized closing theorems,Elemente der Mathematik66(2011)98–117.LORENZ HALBEISEN(lorenz.halbeisen@math.ethz.ch)received his PhD from the ETH Z¨u rich in1994. After research positions in France,Spain,and California,he became lecturer at Queen’s University Belfast and is now senior scientist at the ETH Z¨u rich.His interests range across set theory,ﬁnite and inﬁnite combinatorics, geometry,and number theory.Department of Mathematics,ETH Zentrum,R¨a mistrasse101,8092Z¨u rich,SwitzerlandNORBERT HUNGERB¨UHLER(norbert.hungerbuehler@math.ethz.ch)received his PhD at ETH Z¨u rich in 1994.After research positions in Germany and the US he became assistant professor at the University of Alabama at Birmingham and later professor at the University of Fribourg.Currently he is professor at ETH. His interests range across analysis,geometry,discrete mathematics and number theory.Department of Mathematics,ETH Zentrum,R¨a mistrasse101,8092Z¨u rich,Switzerland14c THE MATHEMATICAL ASSOCIATION OF AMERICA[Monthly121。

Information Processing Letters

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T his appendix derives various results for ordinary least squares estimation of themultiple linear regression model using matrix notation and matrix algebra (see Appendix D for a summary). The material presented here is much more ad-vanced than that in the text.E.1THE MODEL AND ORDINARY LEAST SQUARES ESTIMATIONThroughout this appendix,we use the t subscript to index observations and an n to denote the sample size. It is useful to write the multiple linear regression model with k parameters as follows:y t ϭ␤1ϩ␤2x t 2ϩ␤3x t 3ϩ… ϩ␤k x tk ϩu t ,t ϭ 1,2,…,n ,(E.1)where y t is the dependent variable for observation t ,and x tj ,j ϭ 2,3,…,k ,are the inde-pendent variables. Notice how our labeling convention here differs from the text:we call the intercept ␤1and let ␤2,…,␤k denote the slope parameters. This relabeling is not important,but it simplifies the matrix approach to multiple regression.For each t ,define a 1 ϫk vector,x t ϭ(1,x t 2,…,x tk ),and let ␤ϭ(␤1,␤2,…,␤k )Јbe the k ϫ1 vector of all parameters. Then,we can write (E.1) asy t ϭx t ␤ϩu t ,t ϭ 1,2,…,n .(E.2)[Some authors prefer to define x t as a column vector,in which case,x t is replaced with x t Јin (E.2). Mathematically,it makes more sense to define it as a row vector.] We can write (E.2) in full matrix notation by appropriately defining data vectors and matrices. Let y denote the n ϫ1 vector of observations on y :the t th element of y is y t .Let X be the n ϫk vector of observations on the explanatory variables. In other words,the t th row of X consists of the vector x t . Equivalently,the (t ,j )th element of X is simply x tj :755A p p e n d i x EThe Linear Regression Model inMatrix Formn X ϫ k ϵϭ .Finally,let u be the n ϫ 1 vector of unobservable disturbances. Then,we can write (E.2)for all n observations in matrix notation :y ϭX ␤ϩu .(E.3)Remember,because X is n ϫ k and ␤is k ϫ 1,X ␤is n ϫ 1.Estimation of ␤proceeds by minimizing the sum of squared residuals,as in Section3.2. Define the sum of squared residuals function for any possible k ϫ 1 parameter vec-tor b asSSR(b ) ϵ͚nt ϭ1(y t Ϫx t b )2.The k ϫ 1 vector of ordinary least squares estimates,␤ˆϭ(␤ˆ1,␤ˆ2,…,␤ˆk )؅,minimizes SSR(b ) over all possible k ϫ 1 vectors b . This is a problem in multivariable calculus.For ␤ˆto minimize the sum of squared residuals,it must solve the first order conditionѨSSR(␤ˆ)/Ѩb ϵ0.(E.4)Using the fact that the derivative of (y t Ϫx t b )2with respect to b is the 1ϫ k vector Ϫ2(y t Ϫx t b )x t ,(E.4) is equivalent to͚nt ϭ1xt Ј(y t Ϫx t ␤ˆ) ϵ0.(E.5)(We have divided by Ϫ2 and taken the transpose.) We can write this first order condi-tion as͚nt ϭ1(y t Ϫ␤ˆ1Ϫ␤ˆ2x t 2Ϫ… Ϫ␤ˆk x tk ) ϭ0͚nt ϭ1x t 2(y t Ϫ␤ˆ1Ϫ␤ˆ2x t 2Ϫ… Ϫ␤ˆk x tk ) ϭ0...͚nt ϭ1x tk (y t Ϫ␤ˆ1Ϫ␤ˆ2x t 2Ϫ… Ϫ␤ˆk x tk ) ϭ0,which,apart from the different labeling convention,is identical to the first order condi-tions in equation (3.13). We want to write these in matrix form to make them more use-ful. Using the formula for partitioned multiplication in Appendix D,we see that (E.5)is equivalent to΅1x 12x 13...x 1k1x 22x 23...x 2k...1x n 2x n 3...x nk ΄΅x 1x 2...x n ΄Appendix E The Linear Regression Model in Matrix Form756Appendix E The Linear Regression Model in Matrix FormXЈ(yϪX␤ˆ) ϭ0(E.6) or(XЈX)␤ˆϭXЈy.(E.7)It can be shown that (E.7) always has at least one solution. Multiple solutions do not help us,as we are looking for a unique set of OLS estimates given our data set. Assuming that the kϫ k symmetric matrix XЈX is nonsingular,we can premultiply both sides of (E.7) by (XЈX)Ϫ1to solve for the OLS estimator ␤ˆ:␤ˆϭ(XЈX)Ϫ1XЈy.(E.8)This is the critical formula for matrix analysis of the multiple linear regression model. The assumption that XЈX is invertible is equivalent to the assumption that rank(X) ϭk, which means that the columns of X must be linearly independent. This is the matrix ver-sion of MLR.4 in Chapter 3.Before we continue,(E.8) warrants a word of warning. It is tempting to simplify the formula for ␤ˆas follows:␤ˆϭ(XЈX)Ϫ1XЈyϭXϪ1(XЈ)Ϫ1XЈyϭXϪ1y.The flaw in this reasoning is that X is usually not a square matrix,and so it cannot be inverted. In other words,we cannot write (XЈX)Ϫ1ϭXϪ1(XЈ)Ϫ1unless nϭk,a case that virtually never arises in practice.The nϫ 1 vectors of OLS fitted values and residuals are given byyˆϭX␤ˆ,uˆϭyϪyˆϭyϪX␤ˆ.From (E.6) and the definition of uˆ,we can see that the first order condition for ␤ˆis the same asXЈuˆϭ0.(E.9) Because the first column of X consists entirely of ones,(E.9) implies that the OLS residuals always sum to zero when an intercept is included in the equation and that the sample covariance between each independent variable and the OLS residuals is zero. (We discussed both of these properties in Chapter 3.)The sum of squared residuals can be written asSSR ϭ͚n tϭ1uˆt2ϭuˆЈuˆϭ(yϪX␤ˆ)Ј(yϪX␤ˆ).(E.10)All of the algebraic properties from Chapter 3 can be derived using matrix algebra. For example,we can show that the total sum of squares is equal to the explained sum of squares plus the sum of squared residuals [see (3.27)]. The use of matrices does not pro-vide a simpler proof than summation notation,so we do not provide another derivation.757The matrix approach to multiple regression can be used as the basis for a geometri-cal interpretation of regression. This involves mathematical concepts that are even more advanced than those we covered in Appendix D. [See Goldberger (1991) or Greene (1997).]E.2FINITE SAMPLE PROPERTIES OF OLSDeriving the expected value and variance of the OLS estimator ␤ˆis facilitated by matrix algebra,but we must show some care in stating the assumptions.A S S U M P T I O N E.1(L I N E A R I N P A R A M E T E R S)The model can be written as in (E.3), where y is an observed nϫ 1 vector, X is an nϫ k observed matrix, and u is an nϫ 1 vector of unobserved errors or disturbances.A S S U M P T I O N E.2(Z E R O C O N D I T I O N A L M E A N)Conditional on the entire matrix X, each error ut has zero mean: E(ut͉X) ϭ0, tϭ1,2,…,n.In vector form,E(u͉X) ϭ0.(E.11) This assumption is implied by MLR.3 under the random sampling assumption,MLR.2.In time series applications,Assumption E.2 imposes strict exogeneity on the explana-tory variables,something discussed at length in Chapter 10. This rules out explanatory variables whose future values are correlated with ut; in particular,it eliminates laggeddependent variables. Under Assumption E.2,we can condition on the xtjwhen we com-pute the expected value of ␤ˆ.A S S U M P T I O N E.3(N O P E R F E C T C O L L I N E A R I T Y) The matrix X has rank k.This is a careful statement of the assumption that rules out linear dependencies among the explanatory variables. Under Assumption E.3,XЈX is nonsingular,and so ␤ˆis unique and can be written as in (E.8).T H E O R E M E.1(U N B I A S E D N E S S O F O L S)Under Assumptions E.1, E.2, and E.3, the OLS estimator ␤ˆis unbiased for ␤.P R O O F:Use Assumptions E.1 and E.3 and simple algebra to write␤ˆϭ(XЈX)Ϫ1XЈyϭ(XЈX)Ϫ1XЈ(X␤ϩu)ϭ(XЈX)Ϫ1(XЈX)␤ϩ(XЈX)Ϫ1XЈuϭ␤ϩ(XЈX)Ϫ1XЈu,(E.12)where we use the fact that (XЈX)Ϫ1(XЈX) ϭIk . Taking the expectation conditional on X givesAppendix E The Linear Regression Model in Matrix Form 758E(␤ˆ͉X)ϭ␤ϩ(XЈX)Ϫ1XЈE(u͉X)ϭ␤ϩ(XЈX)Ϫ1XЈ0ϭ␤,because E(u͉X) ϭ0under Assumption E.2. This argument clearly does not depend on the value of ␤, so we have shown that ␤ˆis unbiased.To obtain the simplest form of the variance-covariance matrix of ␤ˆ,we impose the assumptions of homoskedasticity and no serial correlation.A S S U M P T I O N E.4(H O M O S K E D A S T I C I T Y A N DN O S E R I A L C O R R E L A T I O N)(i) Var(ut͉X) ϭ␴2, t ϭ 1,2,…,n. (ii) Cov(u t,u s͉X) ϭ0, for all t s. In matrix form, we canwrite these two assumptions asVar(u͉X) ϭ␴2I n,(E.13)where Inis the nϫ n identity matrix.Part (i) of Assumption E.4 is the homoskedasticity assumption:the variance of utcan-not depend on any element of X,and the variance must be constant across observations, t. Part (ii) is the no serial correlation assumption:the errors cannot be correlated across observations. Under random sampling,and in any other cross-sectional sampling schemes with independent observations,part (ii) of Assumption E.4 automatically holds. For time series applications,part (ii) rules out correlation in the errors over time (both conditional on X and unconditionally).Because of (E.13),we often say that u has scalar variance-covariance matrix when Assumption E.4 holds. We can now derive the variance-covariance matrix of the OLS estimator.T H E O R E M E.2(V A R I A N C E-C O V A R I A N C EM A T R I X O F T H E O L S E S T I M A T O R)Under Assumptions E.1 through E.4,Var(␤ˆ͉X) ϭ␴2(XЈX)Ϫ1.(E.14)P R O O F:From the last formula in equation (E.12), we haveVar(␤ˆ͉X) ϭVar[(XЈX)Ϫ1XЈu͉X] ϭ(XЈX)Ϫ1XЈ[Var(u͉X)]X(XЈX)Ϫ1.Now, we use Assumption E.4 to getVar(␤ˆ͉X)ϭ(XЈX)Ϫ1XЈ(␴2I n)X(XЈX)Ϫ1ϭ␴2(XЈX)Ϫ1XЈX(XЈX)Ϫ1ϭ␴2(XЈX)Ϫ1.Appendix E The Linear Regression Model in Matrix Form759Formula (E.14) means that the variance of ␤ˆj (conditional on X ) is obtained by multi-plying ␴2by the j th diagonal element of (X ЈX )Ϫ1. For the slope coefficients,we gave an interpretable formula in equation (3.51). Equation (E.14) also tells us how to obtain the covariance between any two OLS estimates:multiply ␴2by the appropriate off diago-nal element of (X ЈX )Ϫ1. In Chapter 4,we showed how to avoid explicitly finding covariances for obtaining confidence intervals and hypotheses tests by appropriately rewriting the model.The Gauss-Markov Theorem,in its full generality,can be proven.T H E O R E M E .3 (G A U S S -M A R K O V T H E O R E M )Under Assumptions E.1 through E.4, ␤ˆis the best linear unbiased estimator.P R O O F :Any other linear estimator of ␤can be written as␤˜ ϭA Јy ,(E.15)where A is an n ϫ k matrix. In order for ␤˜to be unbiased conditional on X , A can consist of nonrandom numbers and functions of X . (For example, A cannot be a function of y .) To see what further restrictions on A are needed, write␤˜ϭA Ј(X ␤ϩu ) ϭ(A ЈX )␤ϩA Јu .(E.16)Then,E(␤˜͉X )ϭA ЈX ␤ϩE(A Јu ͉X )ϭA ЈX ␤ϩA ЈE(u ͉X ) since A is a function of XϭA ЈX ␤since E(u ͉X ) ϭ0.For ␤˜to be an unbiased estimator of ␤, it must be true that E(␤˜͉X ) ϭ␤for all k ϫ 1 vec-tors ␤, that is,A ЈX ␤ϭ␤for all k ϫ 1 vectors ␤.(E.17)Because A ЈX is a k ϫ k matrix, (E.17) holds if and only if A ЈX ϭI k . Equations (E.15) and (E.17) characterize the class of linear, unbiased estimators for ␤.Next, from (E.16), we haveVar(␤˜͉X ) ϭA Ј[Var(u ͉X )]A ϭ␴2A ЈA ,by Assumption E.4. Therefore,Var(␤˜͉X ) ϪVar(␤ˆ͉X )ϭ␴2[A ЈA Ϫ(X ЈX )Ϫ1]ϭ␴2[A ЈA ϪA ЈX (X ЈX )Ϫ1X ЈA ] because A ЈX ϭI kϭ␴2A Ј[I n ϪX (X ЈX )Ϫ1X Ј]Aϵ␴2A ЈMA ,where M ϵI n ϪX (X ЈX )Ϫ1X Ј. Because M is symmetric and idempotent, A ЈMA is positive semi-definite for any n ϫ k matrix A . This establishes that the OLS estimator ␤ˆis BLUE. How Appendix E The Linear Regression Model in Matrix Form 760Appendix E The Linear Regression Model in Matrix Formis this significant? Let c be any kϫ 1 vector and consider the linear combination cЈ␤ϭc1␤1ϩc2␤2ϩ… ϩc k␤k, which is a scalar. The unbiased estimators of cЈ␤are cЈ␤ˆand cЈ␤˜. ButVar(c␤˜͉X) ϪVar(cЈ␤ˆ͉X) ϭcЈ[Var(␤˜͉X) ϪVar(␤ˆ͉X)]cՆ0,because [Var(␤˜͉X) ϪVar(␤ˆ͉X)] is p.s.d. Therefore, when it is used for estimating any linear combination of ␤, OLS yields the smallest variance. In particular, Var(␤ˆj͉X) ՅVar(␤˜j͉X) for any other linear, unbiased estimator of ␤j.The unbiased estimator of the error variance ␴2can be written as␴ˆ2ϭuˆЈuˆ/(n Ϫk),where we have labeled the explanatory variables so that there are k total parameters, including the intercept.T H E O R E M E.4(U N B I A S E D N E S S O F␴ˆ2)Under Assumptions E.1 through E.4, ␴ˆ2is unbiased for ␴2: E(␴ˆ2͉X) ϭ␴2for all ␴2Ͼ0. P R O O F:Write uˆϭyϪX␤ˆϭyϪX(XЈX)Ϫ1XЈyϭM yϭM u, where MϭI nϪX(XЈX)Ϫ1XЈ,and the last equality follows because MXϭ0. Because M is symmetric and idempotent,uˆЈuˆϭuЈMЈM uϭuЈM u.Because uЈM u is a scalar, it equals its trace. Therefore,ϭE(uЈM u͉X)ϭE[tr(uЈM u)͉X] ϭE[tr(M uuЈ)͉X]ϭtr[E(M uuЈ|X)] ϭtr[M E(uuЈ|X)]ϭtr(M␴2I n) ϭ␴2tr(M) ϭ␴2(nϪ k).The last equality follows from tr(M) ϭtr(I) Ϫtr[X(XЈX)Ϫ1XЈ] ϭnϪtr[(XЈX)Ϫ1XЈX] ϭnϪn) ϭnϪk. Therefore,tr(IkE(␴ˆ2͉X) ϭE(uЈM u͉X)/(nϪ k) ϭ␴2.E.3STATISTICAL INFERENCEWhen we add the final classical linear model assumption,␤ˆhas a multivariate normal distribution,which leads to the t and F distributions for the standard test statistics cov-ered in Chapter 4.A S S U M P T I O N E.5(N O R M A L I T Y O F E R R O R S)are independent and identically distributed as Normal(0,␴2). Conditional on X, the utEquivalently, u given X is distributed as multivariate normal with mean zero and variance-covariance matrix ␴2I n: u~ Normal(0,␴2I n).761Appendix E The Linear Regression Model in Matrix Form Under Assumption E.5,each uis independent of the explanatory variables for all t. Inta time series setting,this is essentially the strict exogeneity assumption.T H E O R E M E.5(N O R M A L I T Y O F␤ˆ)Under the classical linear model Assumptions E.1 through E.5, ␤ˆconditional on X is dis-tributed as multivariate normal with mean ␤and variance-covariance matrix ␴2(XЈX)Ϫ1.Theorem E.5 is the basis for statistical inference involving ␤. In fact,along with the properties of the chi-square,t,and F distributions that we summarized in Appendix D, we can use Theorem E.5 to establish that t statistics have a t distribution under Assumptions E.1 through E.5 (under the null hypothesis) and likewise for F statistics. We illustrate with a proof for the t statistics.T H E O R E M E.6Under Assumptions E.1 through E.5,(␤ˆjϪ␤j)/se(␤ˆj) ~ t nϪk,j ϭ 1,2,…,k.P R O O F:The proof requires several steps; the following statements are initially conditional on X. First, by Theorem E.5, (␤ˆjϪ␤j)/sd(␤ˆ) ~ Normal(0,1), where sd(␤ˆj) ϭ␴͙ෆc jj, and c jj is the j th diagonal element of (XЈX)Ϫ1. Next, under Assumptions E.1 through E.5, conditional on X,(n Ϫ k)␴ˆ2/␴2~ ␹2nϪk.(E.18)This follows because (nϪk)␴ˆ2/␴2ϭ(u/␴)ЈM(u/␴), where M is the nϫn symmetric, idem-potent matrix defined in Theorem E.4. But u/␴~ Normal(0,I n) by Assumption E.5. It follows from Property 1 for the chi-square distribution in Appendix D that (u/␴)ЈM(u/␴) ~ ␹2nϪk (because M has rank nϪk).We also need to show that ␤ˆand ␴ˆ2are independent. But ␤ˆϭ␤ϩ(XЈX)Ϫ1XЈu, and ␴ˆ2ϭuЈM u/(nϪk). Now, [(XЈX)Ϫ1XЈ]Mϭ0because XЈMϭ0. It follows, from Property 5 of the multivariate normal distribution in Appendix D, that ␤ˆand M u are independent. Since ␴ˆ2is a function of M u, ␤ˆand ␴ˆ2are also independent.Finally, we can write(␤ˆjϪ␤j)/se(␤ˆj) ϭ[(␤ˆjϪ␤j)/sd(␤ˆj)]/(␴ˆ2/␴2)1/2,which is the ratio of a standard normal random variable and the square root of a ␹2nϪk/(nϪk) random variable. We just showed that these are independent, and so, by def-inition of a t random variable, (␤ˆjϪ␤j)/se(␤ˆj) has the t nϪk distribution. Because this distri-bution does not depend on X, it is the unconditional distribution of (␤ˆjϪ␤j)/se(␤ˆj) as well.From this theorem,we can plug in any hypothesized value for ␤j and use the t statistic for testing hypotheses,as usual.Under Assumptions E.1 through E.5,we can compute what is known as the Cramer-Rao lower bound for the variance-covariance matrix of unbiased estimators of ␤(again762conditional on X ) [see Greene (1997,Chapter 4)]. This can be shown to be ␴2(X ЈX )Ϫ1,which is exactly the variance-covariance matrix of the OLS estimator. This implies that ␤ˆis the minimum variance unbiased estimator of ␤(conditional on X ):Var(␤˜͉X ) ϪVar(␤ˆ͉X ) is positive semi-definite for any other unbiased estimator ␤˜; we no longer have to restrict our attention to estimators linear in y .It is easy to show that the OLS estimator is in fact the maximum likelihood estima-tor of ␤under Assumption E.5. For each t ,the distribution of y t given X is Normal(x t ␤,␴2). Because the y t are independent conditional on X ,the likelihood func-tion for the sample is obtained from the product of the densities:͟nt ϭ1(2␲␴2)Ϫ1/2exp[Ϫ(y t Ϫx t ␤)2/(2␴2)].Maximizing this function with respect to ␤and ␴2is the same as maximizing its nat-ural logarithm:͚nt ϭ1[Ϫ(1/2)log(2␲␴2) Ϫ(yt Ϫx t ␤)2/(2␴2)].For obtaining ␤ˆ,this is the same as minimizing͚nt ϭ1(y t Ϫx t ␤)2—the division by 2␴2does not affect the optimization—which is just the problem that OLS solves. The esti-mator of ␴2that we have used,SSR/(n Ϫk ),turns out not to be the MLE of ␴2; the MLE is SSR/n ,which is a biased estimator. Because the unbiased estimator of ␴2results in t and F statistics with exact t and F distributions under the null,it is always used instead of the MLE.SUMMARYThis appendix has provided a brief discussion of the linear regression model using matrix notation. This material is included for more advanced classes that use matrix algebra,but it is not needed to read the text. In effect,this appendix proves some of the results that we either stated without proof,proved only in special cases,or proved through a more cumbersome method of proof. Other topics—such as asymptotic prop-erties,instrumental variables estimation,and panel data models—can be given concise treatments using matrices. Advanced texts in econometrics,including Davidson and MacKinnon (1993),Greene (1997),and Wooldridge (1999),can be consulted for details.KEY TERMSAppendix E The Linear Regression Model in Matrix Form 763First Order Condition Matrix Notation Minimum Variance Unbiased Scalar Variance-Covariance MatrixVariance-Covariance Matrix of the OLS EstimatorPROBLEMSE.1Let x t be the 1ϫ k vector of explanatory variables for observation t . Show that the OLS estimator ␤ˆcan be written as␤ˆϭΘ͚n tϭ1xt Јx t ΙϪ1Θ͚nt ϭ1xt Јy t Ι.Dividing each summation by n shows that ␤ˆis a function of sample averages.E.2Let ␤ˆbe the k ϫ 1 vector of OLS estimates.(i)Show that for any k ϫ 1 vector b ,we can write the sum of squaredresiduals asSSR(b ) ϭu ˆЈu ˆϩ(␤ˆϪb )ЈX ЈX (␤ˆϪb ).[Hint :Write (y Ϫ X b )Ј(y ϪX b ) ϭ[u ˆϩX (␤ˆϪb )]Ј[u ˆϩX (␤ˆϪb )]and use the fact that X Јu ˆϭ0.](ii)Explain how the expression for SSR(b ) in part (i) proves that ␤ˆuniquely minimizes SSR(b ) over all possible values of b ,assuming Xhas rank k .E.3Let ␤ˆbe the OLS estimate from the regression of y on X . Let A be a k ϫ k non-singular matrix and define z t ϵx t A ,t ϭ 1,…,n . Therefore,z t is 1ϫ k and is a non-singular linear combination of x t . Let Z be the n ϫ k matrix with rows z t . Let ␤˜denote the OLS estimate from a regression ofy on Z .(i)Show that ␤˜ϭA Ϫ1␤ˆ.(ii)Let y ˆt be the fitted values from the original regression and let y ˜t be thefitted values from regressing y on Z . Show that y ˜t ϭy ˆt ,for all t ϭ1,2,…,n . How do the residuals from the two regressions compare?(iii)Show that the estimated variance matrix for ␤˜is ␴ˆ2A Ϫ1(X ЈX )Ϫ1A Ϫ1؅,where ␴ˆ2is the usual variance estimate from regressing y on X .(iv)Let the ␤ˆj be the OLS estimates from regressing y t on 1,x t 2,…,x tk ,andlet the ␤˜j be the OLS estimates from the regression of yt on 1,a 2x t 2,…,a k x tk ,where a j 0,j ϭ 2,…,k . Use the results from part (i)to find the relationship between the ␤˜j and the ␤ˆj .(v)Assuming the setup of part (iv),use part (iii) to show that se(␤˜j ) ϭse(␤ˆj )/͉a j ͉.(vi)Assuming the setup of part (iv),show that the absolute values of the tstatistics for ␤˜j and ␤ˆj are identical.Appendix E The Linear Regression Model in Matrix Form 764。

双语阅读：真相

双语阅读：真相·爱因斯坦?自闭症天才与花花公子可可英语2016-02-23 18:50:57阅读(2267)评论(0)声明：本文由入驻搜狐公众平台的作者撰写，除搜狐官方账号外，观点仅代表作者本人，不代表搜狐立场。

举报【新朋友】点击标题下面蓝字【可可英语】加关注【老朋友】点击手机右上角图标【转发】内容Everyone knows Albert Einstein as a wild-haired, violin-playing genius who revolutionized physics, and many have heard how he arrived at his groundbreaking theories via one ingenious thought experiment, or gedankenexperiment, after another. But did you know that he was also an eccentric who gleefully eschewed socks, dodged German military service and spurned social conventions? Or that he was an enthusiastic but third-rate sailor?在众人眼中阿尔伯特·爱因斯坦是一位伟大的物理学家，他梳着一头标志性的乱发，喜欢拉小提琴,开创了物理学的新纪元。

众所周知，他经过一次次巧妙的假想和试验，最终创立了相对论。

可是你了解这位天才的怪脾气吗？他从来不穿袜子、在德国逃避兵役、藐视社会习俗，他还是一个狂热的航海爱好者，不过他的航海技术实在让人不敢恭维。

Ever since solar eclipse observations in 1919 made him front-page news, we haven't been able to get enough of this guy. And why not? Einstein's influence extended beyond the scientific fields he revolutionized. His theories of relativity, which departed from the classical Newtonian view of the cosmos, came to symbolize a broader societal shift away from Enlightenment-influenced concepts of art, literature, morality and politics. More than that, thanks to his strong political and social views, often distilled into playful, philosophical and pithy quotes, he's been a mainstay of dorm-room posters and pop culture for decades. But with the revelations that accompanied the release of his private papers 30 years after his death, do we finally have too much of Einstein? Do they remind us to never meet our heroes, or merely that all geniuses are, finally, human?1919年，一场日食观测活动让他成为头条新闻,可是我们对这位伟大科学家的了解还不够多。

organizedby

Manifolds with nonnegative sectional curvatureorganized byKristopher Tapp and Wolfgang ZillerWorkshop SummaryIn the past few years,the study of Riemannian manifolds with nonnegative and positive curvature has been reinvigorated by breakthroughs and by new connections to other topics, including Ricciﬂow and Alexandrov Geometry.Our workshop brought together experts and newcomers to theﬁeld,including5graduate students and researchers representing a diverse range of sub-specialties.Our goal was to discuss future directions for theﬁeld and to initiate progress solving signiﬁcant open problems.We had on average two talks every morning.These talks were primarily surveys em-phasizing open problems and possible future directions for continued progress.The talks were roughly divided between the following three sub-topics,which we identiﬁed as key to continued progress in theﬁeld:(1)Riemannian submersions and group actions in nonnegative curvature(2)Alexandrov Geometry and Collapse(3)RicciﬂowOn Monday afternoon,all participants gathered to list and discuss open problems related to nonnegative curvature.The lively discussion lasted almost3hours,and resulted in a preliminary list of about30open problems.Many of these problems prompted interesting discussions.Participants continued to add problems to this list during subsequent days of the workshop.The list will continue to evolve,and has the potential to become a useful resource for future researchers in theﬁeld.On Tuesday afternoon,we divided the workshop participants into three groups,cor-responding to the three sub-topics enumerated above.This subdivision remained roughly constant through the remainder of the week,with a few participants choosing toﬂoat be-tween the groups.The groups learned of each other’s activities informally each evening during happy hour,and more formally through group reports on Friday afternoon.Theﬁrst group explored Riemannian submersions and group actions.This group was the largest,and its members decided to further subdivide.They began by brainstorming pos-sible problems to attack in smaller subgroups.Before splitting up,they scheduled mini-talks to explain some recent unpublished work.These mini-talks helped participants(especially newcomers to theﬁeld)decide which subgroup they felt best equipped to join.One sub-group formed to begin classifying the Riemannian submersions from a compact Lie group with a bi-invariant metric.This subgroup quickly discovered an interesting non-homogeneous example,which contradicts the naive conjecture that all such submersions are bi-quotient submersions.This subgroup then spent most of the remaining time considering the case of totally geodesicﬁbers.This collaboration will likely lead to a paper in the coming months.A second subgroup formed to bound the dimension of a torus acting freely on a manifold12with nonnegative curvature,and to consider related problems.The third(largest)subgroup investigated cohomogeneity-one manifolds with nonnegative curvature.They discussed this topic from several angles.They considered cohomogeneity one manifolds with a totally ge-odesic principle orbit,and came to believe that the classiﬁcation of such spaces is within reach.A collaboration on this problem will continue and probably lead to the complete solution in the near future.They also considered obstructions to metrics of nonnegative curvature and smoothness conditions for cohomogeneity one actions.Finally,some of the members of this subgroup considered topological aspect of cohomogeneity-one manifolds,in-cluding topological invariants of known and candidate examples and the problem ofﬁnding cohomogeneity-one manifolds which are topologically interesting,and for which the prob-lem of constructing new metrics with nonnegative curvature or obstructions should thus be investigated.The second group explored Alexandrov geometry and collapse.This group began by generating a list of about20interesting open problems.They then chose three of these problems to explore in more depth.Theﬁrst of these problems was to extend(the dual version of)Wilking’s connectivity lemma to Alexandrov spaces.The group mapped out a proposal involving Morse functions to solve this problem.This work will hopefully lead to a collaborative solution in the near future.The second problem was to discover topological properties of an Alexandrov space which sits at the top of aﬁnite tower ofﬁber bundles. The third problem was the conjecture that all manifolds with almost nonnegative curvature are rationally elliptic.The group discussed a rough strategy for how a proof by induction on dimension might go.One important step in such a proof would be to show that any Alexandrov space which collapses to a point also admits nontrivial collapse.In exploring this issue,the group constructed an essentially complete proof that the torus does not collapse to an interval.The third group studied Ricciﬂow.Recent progress in the applications of the Ricci ﬂow to manifolds with positive curvature operator,positive isotropic curvature and manifolds with1/4pinching were discussed in the morning survey talks.The Ricciﬂow subsection gave a simple proof of Tachibana’s theorem that an Einstein metric with positive curvature operator is a space form.Two of the participants generalized recent work by Boehm and Wilking on even dimensional manifolds with small Weyl tensor to the odd dimensional case and gave a simple proof of the algebraic part needed in the proof of the weakly1/4pinching theorem.Furthermore,existence and stability of singularity models and the nonexistence of noncompact3dimensional shrinkers was discussed.They also discussed the problem of ruling out noncompact gradient shrinking solitons with positive curvature operator,or more generally classify the gradient shrinking solitons with certain positivity of the curvature.The participants were almost unanimous in feeling that the workshop was successful. One participant stated that“all conferences should be structured this way”.Of course one should add that this is only possible with a narrowly focused research area.The afternoon group-work varied between brainstorming ideas for solving very diﬃcult open problems and solving easier problems.Work at either extreme of this spectrum was felt to be productive and meaningful.Often the groups continued working past the5:00beginning of happy hour (even the most beer-loving of the groups)which demonstrates the energy that the group members felt.We expect that new collaborations will develop as a result of the workshop. Further,participants are returning home with new ideas that could shape the long term development of theﬁeld in less tangible ways.3 We are very thankful for the generous support of the AIM.We appreciate the guidance and hard work of the AIM staﬀin helping us conduct a successful workshop.。

秩1修正矩阵特征值问题的推广及其应用（英文）

秩1修正矩阵特征值问题的推广及其应用（英文）吕海玲;明清河【摘要】本文给出了秩1修正矩阵特征值问题推广的新证明,证明过程主要应用了一个行列恒等式.在此基础上,把秩1修正矩阵的特征值问题推广到块特征值问题.最后给出一个应用说明结论的重要性.%We prove a spectral perturbation theorem for an extension eigenvalues of rank-one update matrix of special structure.The main idea behind our proof is from the simple relation between the determinants for a matrix and this result.Furthermore,we extent this theorem to the block eigenvalues problem.At last one application of the result is given to illustrate the usefulness of the theorem.【期刊名称】《枣庄学院学报》【年(卷),期】2011(000)005【总页数】4页(P29-32)【关键词】秩1更新;行列式;谱【作者】吕海玲;明清河【作者单位】枣庄学院信息科学与工程学院,山东枣庄277160;枣庄学院数学与统计学院,山东枣庄277160【正文语种】中文【中图分类】O151.211 IntroductionIn this paper we prove a spectral perturbation theorem for an extension eigenvalues of rank－ one update matrix of special structure，which shows how to modify r eigenvalues of a matrix of order n，(r≤n)，ia a rank－k updated matrix，without changing any of the n－rremaining eigenvalues．This theorem plays a relevan t role in the study of the nonnegative inverse eigenvalue problem(NIEP)．The main idea behind our proof is from the simple relation between the determinants of a matrix and this result，using a well known determinant identity．Furthermore，we extent this theorem to the block eigenvalues problem．By using this extension，we give a Application on eigenvalues problem of matrix perturbation of special structure．Because we apply a classic determinant equality to our spectral analysis，we are able to find explicit expression of the characteristic polynomial of the rank－r update matrix．All eigenvalues of the matrix are immediately available．Lemma1 If A is an invertible n×n matrix，and u and v are two n－dimensional column vectors，thenProof．We may assume A=I，the n × n identity matrix，sincethen(1)follows fromin the general case．In this special case，the result comes from theequalityso(2)becomesRemark1 If A is an invertible n×n matrix，B is a n×r matrix，C is a r×n matrix，thenIn the next section we present the main result．2 Main resultLet A be an n×n matrix．The eigenvalues of A are all the complex zeros of the characteristic polynomial pA(λ)=det(λI－ A)of A．Letσ(A)= {λ1，λ2，…，λn }be the set of the eigenvalues of A，counting algebraic multiplicity，that is spectrum of A．Theorem 1［1］ Let u and v are two n － dimensional column vectors such that u is an eigenvector of A associated with eigenvalue λ1． Then，the eigenvalues of A + uvT are {λ1+vT u，λ2，…，λn}，counting algebraic multiplicity．The following result is an extension of the theorem 1．This extension shows how to change r eigenvalues λ1，λ2，…，λr，r≤ n，of a matrix A of order n，via a rank － k updated matrix，without changing any of the n －rremaining eigenvaluesλr+1，λr+2，…，λn．Theorem 2 Let A be an n × n matrix with eigenvalues λ1，λ2，…，λn．LetX =[x1 x2 …xr ]be such that rank(X)=r and AX=Xdiag [λ1，λ2，…，λr]，r≤n．Let C be a r × n matrix．Then the matrix A+XC has eigenvalues γ1，γ2，…，γr，λr+1，λr+2，…，λn．where γ1，γ2，…，γr are eigenvalues of the matrix K+CX with K=diag [λ1，λ2，…，λr]．Proof Letλ ∉ σ(A)be any comple x number．Then，by applying remark 1 to the equalityW e haveThe condition AX=Xdiag [λ1，λ2，…，λr]implies thatso(7)becomesSince the above equality is true for allλ ∉ σ(A)，the theorem is p roved．Rem ark2．1 By Theorem 2．1，the characteristic polynomial of A+XC isRemark2．2 Since A and AT have the same eigenvalues counting algebraic multiplicity，the conclusion of Theorem 2．1 also holds for A+XC，whereX= [x1 x2 … xr ]be such that rank(X)=r and AX=Xdiag [λ1，λ2，…，λr]．Furthermore，we extent this theorem to the block eigenvalues problem Definition 1［4］．A matrix X of order n is a block eigenvalue of a matrix A of order mn，if there exists a block vector V of full rank，such that AV=VX，X is a block eigenvector of A．The matrix A is partitioned into m ×m blocks of or der n，and the block vector V．Definition 2［4］．A set of block eigenvalues of a block matrix is acomplete set if the set of all the eigenvalues of these block eigenvalues is the set of the matrix．Let us suppose now that we have computed mn scalar eigenvalues of a partitioned matrix A．We can construct a complete set of block eigenvalues by taking m matrix of order n in Jordan form where the diagonal elements are those scalar eigenvalues．Furthermore，if the scalar eigenvalues of A are distinct，these m matrix are diagonal matrix as is shown in the following construction:where theλi，i=1，…，mn，are the eigenvaluesof A．The proof that the matrix Xj，j=1，…，m，are a complete set of block eigenvalues of A is in ［1，p．74］．Theorem 2．If the scalar eigenvalues of A are distinct，let V and C be the block vectors such that V is a block eigenvector of A associated with block eigenvalues X1，Then，the eigenvalues of A + VCT are μ1，…，μn，λn+1，…，λ2n，…λ(m－1)n+1，…，λmn where μ1，…，μn are eigenvalues of the matrix K+CT V with K=diag［λ1，…，λn］．Proof．The same to theorem 1．3 Application of the theoremA direct consequence of Theorem 2．1 is the following．One Application of the result is given to illustrate the eigenvalues problem with the perturbation matrix．Proposition 3．1Let A，B，C，D ∈ Cn×n，D=A+B，where B is the perturbation of A．If B=XC，where X= ［x1，x2，…，xn］，xi is aneigenvector of A dissociate with eigenvalue xi，i=1，2，…，n．So thatthen，the eigenvalues of A+B are the eigenvalues of the matrix diag ［λ1，λ2，…，λn］+CX．References［1］Jiu D，Zhou A H．Eigenvalues of rank －one updated matrix with some applications［J］．Applied Mathematics Letters，2007，20:1223－1226．［2］Ricrdo L S，Oscar R．Applications of a Brauer theorem in the nonnegative inverse eigenvalue problem［J］．Linear Algebra and its Applications，2006，416:844 －856．［3］Bapat R B，Raghavan E S．Nonnegative Matrices and Applications，Cambridge University press，1997．［4］Dennis J E，Traub J F and Weber R．P．On the matrix polynomial，lambda－ matrix and block eigenvalue problem，Tech．Rep．71 － 109，Computer Science Department，Cornell Univ，Ithaca，NY and Carnegie －Mellon Univ．，Pitsburgh，PA，(1971)．。

数学专业英语课后部分习题答案

数学专业英语课后部分习题答案数学、方程与比例（1）数学来源于人类的社会实践，包括工农业的劳动，商业、军事和科学技术研究等活动。

Mathematics comes from man’s social practice, for example, industrial and agricultural production, commercial activities, military operations and scientific and technological researches. （2）如果没有运用数学，任何一个科学技术分支都不可能正常地发展。

No modern scientific and technological branches could be regularly developed without the application of mathematics.（3）符号在数学中起着非常重要的作用，它常用于表示概念和命题。

Notations are a special and powerful tool of mathematics and are used to express conceptions and propositions very often.（4）17 世纪之前，人们局限于初等数学，即几何、三角和代数，那时只考虑常数。

Before 17th century, man confined himself to the elementary mathematics, i. e. , geometry, trigonometry and algebra, in which only the constants were considered.（5）方程与算数的等式不同在于它含有可以参加运算的未知量。

Equation is different from arithmetic identity in that it contains unknown quantity which can join operations.（6）方程又称为条件等式，因为其中的未知量通常只允许取某些特定的值。

阿西莫夫科学指南英文

阿西莫夫科学指南英文English:Isaac Asimov's "Guide to Science" is a nonfiction book that aims to introduce readers to the world of science. The book is divided into three parts: the physical sciences, the biological sciences, and the social sciences. In each of these sections, Asimov provides an overview of the major areas of study, as well as some of the key discoveries that have shaped our understanding of these fields. One of the strengths of the book is Asimov's ability to explain complex scientific concepts in a way that is accessible to readers of all ages. Additionally, the book is filled with interesting anecdotes and historical tidbits that help to contextualize the scientific information. Overall, "Guide to Science" is a fantastic resource for anyone looking to learn more about the sciences and their impact on our world.中文翻译:《阿西莫夫的科学指南》是一本非虚构性的书，旨在向读者介绍科学世界。

证明哥德巴赫猜想的原文(英文)并附译后汉语

此文发表在：Advances in Theoretical and Applied Mathematics (A TAM), ISSN 0793-4554, V ol. 7, №4, 2012, pp.417-424Proving Goldbach’s Conjecture by Two Number Axes’ Positive Half Lines which Reverse from Each Other’s DirectionsZhang TianshuNanhai west oil corporation,China offshore Petroleum,Zhanjiang city, Guangdong province, P.R.ChinaEmail: tianshu_zhang507@;AbstractWe know that every positive even number 2n(n≥3) can express in a sum which 3 plus an odd number 2k+1(k≥1) makes. And then, for any odd point 2k+1 (k≥1)at the number axis, if 2k+1 is an odd prime point, of course even number 3+(2k+1) is equal to the sum which odd prime number 2k+1 plus odd prime number 3makes; If 2k+1 is an odd composite point, then let 3<B<2k+1, where B is an odd prime point, and enable line segment B(2k+1) to equal line segment 3C. If C is an odd prime point, then even number 3+(2k+1) is equal to the sum which odd prime number B plus odd prime number C makes. So the proof for Goldbach’s Conjecture is converted to prove there be certainly such an odd prime point B at the number axis’s a line segment which take odd point 3 and odd point 2k+1 as ends, so as to prove the conjecture by such a method indirectly.KeywordsNumber theory, Goldbach’s Conjecture, Even number, Odd prime number, Mathematical induction, Two number axes’ positive half lines which reverse from each other’s direction, OD, PL, CL, and RPL.Basic ConceptsGoldbach’s conjecture states that every ev en number 2N is a sum of two prime numbers, and every odd number 2N+3 is a sum of three prime numbers, where N≥2.We shall prove the Goldbach’s conjecture thereinafter by odd points at two number axes’ positive half lines wh ich reverse from each other’s directions and which begin with odd point 3.First we must understand listed below basic concepts before the proof of the conjecture, in order to apply them in the proof.Axiom．Each and every even number 2n (n≥3) can express in a sum which 3 plus each odd number 2k+1 (k≥1) makes.Definition 1．A line segment which takes two odd points as two ends at the number axis’s positive half line which begins with odd point 3 is called an odd distance. “OD” is abbreviated from “odd distance”.The OD between odd point N and odd point N+2t is written as OD N(N+2t), where N≥3, and t≥1.A integer which the length of OD between two consecutive odd points expresses is 2.A length of OD between odd point 3 and each odd point is unique.Definition 2．An OD between odd point 3 and each odd prime point at the number axis’s positive half line which begins with odd point 3, otherwise called a prime length. “PL” is abbreviated from “prime length”, and “PLS”denotes the plural of PL.An integer which each length of from small to large PL expresses be successively 2, 4, 8, 10, 14, 16, 20, 26. . .Definition 3．An OD between odd point 3 and each odd composite point at the number axis’s positive half line which begins with odd point 3, otherwise called a composite length. “CL” is abbreviated from “composite length”.An integer which each length of from small to large CL expresses be successively 6, 12, 18, 22, 24, 30, 32. . .We know that positive integers and positive integers’ points at the number axis’s positive half line are one-to-one correspondence, namely each integer’s point at the number axis’s positive half line represen ts only a positive integer. The value of a positive integer expresses the length of the line segment between point 0 and the positive integer’s point here. When the line segment is longer, it can express in a sum of some shorter line segments; correspondingly the positive integer can also express in a sum of some smaller integers.Since each and every line segment between two consecutive integer’s points and the line segment between point 0 and point 1 have an identical length, hence when use the length as a unit to measure a line segment between two integer’s points or between point 0 and any integer’s point, the line segment has some such unit length, then the integer which the line segment expresses is exactly some.Since the proof for the conjecture relate merely to positive integers which are not less than 3, hence we take only the number axis’s positive half line which begins with odd point 3. However we stipulate that an integer which each integer’s point represents expresses yet the length of the line segment between the integer’s point and point 0. For example, an odd prime value which the right end’s point of any PL represents expresses yet the length of the line segment between the odd prime point and point 0 really.We can prove next three theorems easier according to above-mentioned some relations among line segments, integers’ points and integers.Theorem 1．If the OD which takes odd point F and odd prime point P S as two ends is equal to a PL, then even number3+F can express in a sum of two odd prime numbers, where F>P S.Proof．Odd prime point P S represents odd prime number P S, it expresses the length of the line segment from odd prime point P S to point 0.Though lack the line segment from odd point 3 to point 0 at the number axis’s positive half line which begins with odd point 3, but odd prime point P S represents yet odd prime P S according to above-mentioned stipulation;Let OD P S F=PL 3P b, odd prime point P b represents odd prime number P b, it expresses the length of the line segment from odd prime point P b to point 0. Since PL 3P b lack the segment from odd point 3 to point 0, therefore the integer which the length of PL 3P b expresses is even number P b-3, namely the integer which the length of OD P S F expresses is even number P b-3. Consequently there is F=P S+(P b-3), i.e. 3+F=odd prime P S + odd prime P b .Theorem 2．If even number 3+F can express in a sun of two odd prime numbers, then the OD which takes odd point 3 and odd point F as ends can express in a sun of two PLS, where F is an odd number which is more than 3.Proof．Suppose the two odd prime numbers are P b and P d, then there be 3+F= P b+P d.It is obvious that there be OD 3F=PL 3P b + OD P b F at the number axis’s positive half line which begins with odd point 3.Odd prime point P b represents odd prime number P b according to above-mentioned stipulation, then the length of line segment P b(3+F) is precisely P d, nevertheless P d expresses also the length of the line segment from odd prime point P d to point 0. Thereupon cut down 3 unit lengths of line segment P b(3+F), we obtain OD P b F; again cut down 3 unit lengths of the line segment from odd prime point P d to point 0, we obtain PL 3P d, then there be OD P b F=PL 3P d.Consequently there be OD 3F=PL 3P b + PL 3P d.Theorem 3．If the OD between odd point F and odd point 3can express in a sum of two PLS, then even number 3+F can express in a sum of two odd prime numbers, where F is an odd number which is more than 3. Proof．Suppose one of the two PLS is PL 3P S, then there be F＞P S,and the OD between odd point F and odd prime point P S is another PL. Consequently even number 3+F can express in a sum of two odd prime numbers according to theorem 1.The ProofFirst let us give ordinal number K to from small to large each and every odd number 2k+1, where k≥1,then from small to large each and every even number which is not less than 6is equal to 3+(2k+1).We shall prove this conjecture by the mathematical induction thereinafter.1．When k=1, 2, 3 and 4, we getting even number be orderly 3+(2*1+1)=6=3+3, 3+(2*2+1)=8=3+5, 3+(2*3+1)=10=3+7 and 3+(2*4+1)=12=5+7. This shows that each of them can express in a sum of two odd prime numbers.2．Suppose k=m, the even number which3 plus №m odd number makes, i.e. 3+(2m+1) can express in a sum of two odd prime numbers, where m≥4.3．Prove that when k=m+1, the even number which 3 plus №(m+1) odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers.Proof．In case 2m+3 is an odd prime number, naturally even number 3+（2m+3）is the sum of odd prime number 3 plus odd prime number 2m+3 makes.When 2m+3 is an odd composite number, suppose that the greatest odd prime number which is less than 2m+3 is P m, then the OD between odd prime point P m and odd composite point 2m+3 is either a PL or a CL. When the OD between odd prime point P m and odd composite point 2m+3 is a PL, the even number 3+（2m+3）can express in a sum of two odd prime numbers according to theorem 1.If the OD between odd prime point P m and odd composite point 2m+3 is a CL, then we need to prove that OD 3（2m+3）can express in a sum of two PLS, on purpose to use the theorem 3.When OD P m(2m+3) is a CL, from small to large odd composite number 2m+3 be successively 95, 119, 125, 145. . .First let us adopt two number axes’ positive half lines which reverse from each other’s directions and which begin with odd point 3.At first, enable end point 3 of either half line to coincide with odd point 2m+1 of another half line. Please, see first illustration:3 5 7 2m-3 2m+12m+1 2m-3 7 5 3First IllustrationSuch a coincident line segment can shorten or elongate, namely end point 3 of either half line can coincide with any odd point of another half line.This proof will perform at some such coincident line segments. And for certain of odd points at such a coincident line segment, we use usually names which mark at the rightward direction’s half line.We call PLS which belong both in the leftward direction’s half line and in a coincident line se gment “reverse PLS”. “RPLS” is abbreviated from “reverse PLS”, and “RPL” denotes the singular of RPLS.The RPLS whereby odd point 2k+1 at the rightward direction’s half line acts as the common right endmost point are written as RPLS2k+1,and RPL2k+1 denotes the singular, where k>1.This is known that each and every OD at a line segment which takes oddpoint 2m+1 and odd point 3 as two ends can express in a sum of a PL and a RPL according to preceding theorem 2 and the supposition of №2 step of the mathematical induction.We consider a PL and the RPL2k+1 wherewith to express together the length of OD 3(2k+1)as a pair of PLS, where k >1. One of the pair’s PLS is a PL which takes odd point 3 as the left endmost point, and another is a RPL2k+1 which takes odd point 2k+1 as the right endmost point. We consider the RPL2k+1 and another RPL2k+1 which equals the PL as twin RPLS2k+1.For a pair of PLS, the PL is either unequal or equal to the RPL2k+1. If the PL is unequal to the RPL2k+1, then longer one is more than a half of OD 3(2k+1), yet another is less than the half. If the PL is equal to the RPL2k+1, then either is equal to the half. A pair of PLS has a common end’s point.Since each of RPLS2k-1 is equal to a RPL2k+1,and their both left endmost points are consecutive odd points, and their both right endmost points are consecutive odd points too. So seriatim leftwards move RPLS2k+1to become RPLS2k-y, then part left endmost points of RPLS2k+1 plus RPLS2k-y coincide monogamously with part odd prime points at OD 3(2k+1), where y=1, 3, 5, ...Thus let us begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, where y= 1, 3, 5, 7 ... ỹ ...Suppose that y increases orderly to odd number ỹ,and∑ part left endmostpoints of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously, then there are altogether(ỹ+3)/2 odd points at OD (2m-ỹ)(2m+1), and let μ=(ỹ+3)/2.Let us separate seriatim OD 3(2m-y+2) (y=1, 3, 5,…) from each coincident line segment of two such half lines, and arrange them from top to bottom orderly. After that, put an odd prime number which each odd prime point at the rightward direction’s half line expresses to on the odd prime point, and put another odd prime number which each left endmost point of RPLS2m-y+2 at the leftward direction’s half line expresses to beneath the odd prime point.For example, when 2m+3=95, 2m+1=93, 2m-1=91 and 2m-ỹ =89，μ=3. For the distributer of odd prime points which coincide monogamously with left endmost points of RPLS95, RPLS93, RPLS91, RPLS89 and RPLS87, please see second illustration:OD 3(95)19 31 37 61 67 7979 67 61 37 31 19OD 3(93)7 13 17 23 29 37 43 53 59 67 73 79 83 8989 83 79 73 67 59 53 43 37 29 23 17 13 7 OD 3(91) 5 11 23 41 47 53 71 83 8989 83 71 53 47 41 23 11 5 OD 3(89)13 19 31 61 73 7979 73 61 31 19 13OD 3(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 8383 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7Second IllustrationTwo left endmost points of twin RPLS2m-y+2 at OD 3(2m-y+2) coincide monogamously with two odd prime points, they assume always bilateral symmetry whereby the centric point of OD 3(2m-y+2)acts as symmetriccentric. If the centric point is an odd prime point, then it is both the left endmost point of RPL2m-y+2 and the odd prime point which coincides with the left endmost point, e.g. centric point 47 of OD3(91) in above-cited that example.We consider each odd prime point which coincides with a left endmost point of RPLS2m-ỹ alone as a characteristic odd prime point, at OD 3(2m+1). Thus it can seen, there is at least one characteristic odd prime point at OD 3(2m-ỹ) according to aforesaid the way of making things, e.g. odd prime points 19, 31 and 61 at OD 3(89) in above-cited that example.Whereas there is not any such characteristic odd prime point in odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹ+2.In other words, every characteristic odd prime point is not any left endmost point of RPLS2m+1plus RPLS2m-1 ... plus RPLS2k-ỹ+2.Moreover left endmost points of RPLS2m-y are №1 odd points on the lefts of left endmost points of RPLS2m-y+2monogamously, where y is an odd number≥1.Consequently, №1 odd point on the left of each and every characteristic odd prime point isn’t a ny left endmost point of RPLS2m-1plus RPLS2m-3…plus RPLS2m-ỹ. — {1}Since each RPL2m-y+2 is equal to a PL at OD 3(2m-y+2).In addition, odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹare all odd prime points at OD 3(2m+1).Hence considering length, at OD 3(2m+1) RPLS whose left endmost points coincide monogamously with all odd prime points are all RPLS at OD 3(2m+1), irrespective of the frequency of RPLS on identical length. Evidently the longest RPL at OD 3(2m+1)is equal to PL 3P m.When OD P m(2m+3) is a CL, let us review aforesaid the way of making thing once again, namely begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, and∑ part left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously.Which one of left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincides first with odd prime point 3? Naturally it can only be the left endmost point of the longest RPL P m whereby odd prime point P m acts as the right endmost point.Besides all coincidences for odd prime points at OD 3(2m+1) begin with left endmost points of RPLS2m+1, whereas left endmost points of RPLS2m-ỹare final one series in the event that all odd prime points at OD 3(2m+1) are coincided just right by left endmost points of RPLS.Therefore odd point 2m-ỹ as the common right endmost point of RPLS2m-ỹcannot lie on the right of odd prime point P m, then odd point 2m-ỹ-2 can only lie on the left of odd prime point P m . This shows that every RPL2m-ỹ-2 at OD 3(2m-ỹ-2) is shorter than PL 3 P m.In addition, №1odd point on the left of a left endmost point of each and every RPL2k-ỹ is a left endmost point of RPLS2k-ỹ-2.Therefore each and every RPL2m-ỹ-2can extend contrary into at least one RPL2m-y+2, where y is a positive odd number ≤ỹ+2.That is to say, every left endmost point of RPLS2m-ỹ-2 is surely at least one left endmost point of RPLS2k-y+2.Since left endmost points of RPLS2m-ỹ-2 lie monogamously at №1 odd point on the left of left endmost points of RPLS2m-ỹ including characteristic odd prime points.Consequently, №1 odd point on the left of each and every characteristic odd prime point is surely a left endmost point of RPLS2m-y+2, where 1≤y≤ỹ+2.—— {2}So we draw inevitably such a conclusion that №1 odd point on the left of each and every characteristic odd prime point can only be a left endmost point of RPLS2m+1 under these qualifications which satisfy both above-reached conclusion {1}and above-reached conclusion {2}.Such being the case, let us rightwards move a RPL2m+1 whose left endmost point lies at №1odd point on the left of any characteristic odd prime point to adjacent odd points, then the RPL2m+1 is moved into a RPL2m+3. Evidently the left endmost point of the RPL2m+3is the characteristic odd prime point, and its right endmost point is odd point 2m+3.So OD 3(2m+3) can express in a sum of two PLS, and the commonendmost point of the two PLS is exactly the characteristic odd prime point. Thus far we have proven that even if OD P m(2m+3) is a CL, likewise OD 3(2m+3) can also express in a sum of two PLS.Consequently even number which 3 plus №(m+1)odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers according to aforementioned theorem 3.Proceed from a proven conclusion to prove a larger even number for each once, then via infinite many an once, namely let k to equal each and every natural number, we reach exactly a conclusion that every even number 3+(2k+1) can express in a sum of two odd prime numbers, where k≥1.To wit every even number 2N can express in a sum of two odd prime numbers, where N>2.In addition let N =2, get 2N=4=even prime number 2+even prime number 2. Consequently every even number 2N can express in a sum of two prime numbers, where N≥2.Since every odd number 2N+3 can express in a sum which a prime number plus the even number makes, consequently every odd number 2N+3 can express in a sum of three prime numbers, where N≥2.To sum up, we have proven that two propositions of the Goldbach’s conjecture are tenable, thus Goldbach’s conjecture holds water.附，翻译成汉语：利用互为反向数轴的正射线证明哥德巴赫猜想张天树Tianshu_zhang507@摘要我们知道,依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.于是,对于数轴上的任意一个奇数点2k+1（k≧1）,如果2k+1是一个奇素数点,当然,偶数3+(2k+1）可等于奇素数2k+1 与奇素数 3 之和;如果2k+1是一个奇合数点,那么,取3<B<2k+1,这里,B是一个奇素数点,且使线段B(2k+1)等于线段3C. 如果C是一个奇素数点,那么, 偶数3+（2k+1）等于奇素数B与奇素数C之和.于是对哥德巴赫猜想的证明就变换成了去证明：在数轴的以奇数点3和2k+1为端点的线段上，总是存在着这样的奇素数点B，以此方法来间接地证明哥德巴赫猜想.关键词数论、哥德巴赫猜想、数学归纳法、偶数、奇素数、互为反方向的数轴的正射线、奇距、素长、合长和反向素长.基本慨念哥德巴赫猜想表为:每一个偶数2N都是两个素数之和,每一个奇数2N+3都是三个素数之和,这里N≧2.本文将用互为反方向的两条数轴的从奇数点3开始的正方向射线上的奇数点来证明这个猜想.在证明这个猜想之前，我们先要熟知下述的基本慨念，以便在证明的过程中应用它们.公理.依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.定义1. 在数轴的从奇数点3开始的正方向射线上，以任意两个奇数点为端点的线段,称为这两个奇数点之间的距离，简称奇距．我们用符号“OD”表示奇距.奇数点N与奇数点N+2t之间的奇距,写成OD N(N+2t)，这里N≧3,t≧1. 相邻两个奇数点之间的奇距是2，奇数点3与各个奇数点之间的奇距长度,都是唯一的.定义2.在数轴的从奇数点3开始的正方向射线上，以任意一个奇素数点和奇数点3为端点的奇距，也被称为这个奇素数点的素长，并用符号“PL”表示一条素长，和符号“PLS”表示至少两条素长，即素长的复数.从小到大的素长依次是: 2、4、8、10、14、16、20、26. . . . . .定义3.在数轴的从奇数点3开始的正方向射线上，以任意一个奇合数点和奇数点3为端点的奇距，又被称为这个奇合数点的合长，并用符号“C L”表示一条合长.从小到大的合长依次是: 6、12、18、22、24、30、32. . . . . .我们知道，在数轴的正方向射线上的整数点与正整数是一一对应的.也就是说，在数轴的正方向射线上，每一个整数点只代表一个整数，这个整数的值在这里表示这个整数点距0点的线段长度. 当这条线段较长时，它可表为若干条线段之和. 因此，这个整数也可表为若干个整数之和. 且因为0点与1点、以及相邻两个整数点之间的线段长度都是相等的,当我们把这些相等线段每条的长度作为1个长度单位去度量任意一个整数点距0点的线段或任意两个整数点之间的线段长度时,该线段有多少个这样的长度单位，它就代表多大的整数.因为本文的证明仅仅用到不小于3的整数，所以，我们只取数轴的从奇数点3开始的正方向射线，但我们规定：在这射线上的每个整数点代表的整数值仍然是指这个整数点距0点的线段长度. 例如，在这射线上任意一条素长的右端点所代表的奇素数值，实际上是指这个奇素数点距0点的线段长度.根据以上所述，我们很容易地证明以下三条定理:定理1.如果以奇数点F和奇素数点P S 为端点的奇距等于一条素长，那么，偶数3+F可表为两个奇素数之和，这里F＞P S.证明：奇素数点P S代表奇素数P S，P S的值表示奇素数点P S到0点的长度.在数轴的从奇数点3开始的正方向射线上虽然缺少从奇数点3到0点的一段，但是按照前面的规定，奇素数点P S仍然代表奇素数P S；令OD P S F=PL 3P b，奇素数点P b代表奇素数P b，P b的值表示奇素数点P b 到0点的长度. 但是，PL 3P b缺少从奇数点3到0点的一段，因此，PL 3P b 的长度表示的整数是偶数P b-3，即OD P S F的长度表示的整数是偶数P b-3. 所以, F=P S+(P b-3)，即3+F=奇素数P S+奇素数P b.定理2. 如果偶数3+F可表为两个奇素数之和，那么，以奇数点F和奇数点3为端点的奇距可表为两条素长之和，这里F是一个大于3的奇数.证明：假设这两个奇素数为P b和P d，则有3+F= P b+P d. 显然，在数轴的从奇数点3开始的正方向射线上，OD 3F=PL 3P b + OD P b F. 根据前面的规定，奇素数点P b代表奇素数P b，那么，线段P b(3+F)的长度就是P d，该长度是指从奇素数点P d到0点的线段长度.在线段P b(3+F)上去掉3个单位长度，得到OD P b F；在从奇素数点P d到0点的线段上去掉3个单位长度，得到PL 3P d，于是，OD P b F=PL 3P d.所以, OD 3F=PL 3P b + PL 3P d.定理3.如果奇数点F与奇数点3之间的奇距可表为两条素长之和，那么，偶数3+F可表为两个奇素数之和，这里F是一个大于3的奇数.证明：假设这两条素长中的一条是PL 3P S，那么，F＞P S，且奇数点F与奇素数点P S 之间的奇距是另一条素长. 根据定理1，偶数3+F可表为两个奇素数之和.证明首先,让我们以自然数k给每一个依次增大的奇数2k+1编上序号，这里k≧1,那么，每一个不小于6的依次增大的偶数等于3+(2k+1). 在下文中, 我们将运用数学归纳法来证明这个猜想.1.当k= 1、2、3和4时，我们依次得到的偶数: 3+（2×1+1）=6=3+3，3+（2×2+1）=8=3+5，3+（2×3+1）=10=3+7或5+5，3+（2×4+1）=12=5+7都可表为两个奇素数之和.2.假设当k=m时，3加上第m个奇数所得的偶数、即3+（2m+1）可表为两个奇素数之和，这里m≧4.3.证明:当k=m+1时，3加上第m+1个奇数所得的偶数、即3+（2m+3）也可表为两个奇素数之和.证明如果2m+3是一个奇素数，当然，偶数3+（2m+3）可表为奇素数3与奇素数2m+3之和.当2m+3是一个奇合数时，假设小于2m+3的最大奇素数为P m，那么，奇合数点2m+3与奇素数点P m之间的奇距或是一条素长，或是一条合长.当奇合数点2m+3与奇素数点P m之间的奇距是一条素长时，根据定理1，偶数3+（2m+3）可表为两个奇素数之和.如果奇合数点2m+3与奇素数点P m之间的奇距是一条合长，我们则需要证明OD 3（2m+3）可表为两条素长之和，以便应用定理3.当OD P m(2m+3)是一条合长时，2m+3从小到大的值依次是95、119、125、145. . . . . .首先，让我们采用两条互为相反方向的数轴的从奇数点3开始的正方向射线，最初,让一条射线上的奇数点3与另一条射线上的奇数点2m+1重合. 请参看第一图：3 5 7 2m-3 2m+12m+1 2m-3 7 5 3第一图这两条射线互相重合的线段能够缩短或伸长，也就是说，一条射线的端点3能够重合另一条射线的任意一个奇数点.本文的整个证明将在这样互相重合、且可伸缩的线段上施行，并且，对这线段上互相重合后的一个奇数点，我们主要使用它在向右方向射线上的名称.我们把既属于向左方向射线，又在两条射线互相重合的线段上的素长称为“反向素长”. 一条反向素长用符号“RPL”表示，它的复数表为“RPLS”. 以向右方向射线上的奇数点2k+1作为共同右端点的至少两条反向素长被写作RPLS2k+1,以及RPL2k+1表示其中的一条，这里k＞1.根据前列的定理2和数学归纳法中第二步的假设，我们知道，在以奇数点2m+1和奇数点3为端点的线段上的各条奇距，都能够表为一条素长与一条反向素长之和. 我们把表示OD 3(2k+1)为两条素长之和的两条素长看作是一对素长，这里k＞1. 在这对素长中，一条是以奇数点3为左端点的素长，另一条则是以奇数点2k+1为右端点的反向素长. 我们把这对素长中的反向素长和在长度上等于这对素长中素长的另一条以奇数点2k+1为右端点的反向素长看作是孪生的反向素长.对于共表OD 3(2k+1)长度的一对素长，如果它们不等，那么，较长的一条比OD 3（2k+1）的一半长，而另一条则比这一半短. 如果这两条素长相等，那么，每一条都等于OD 3（2k+1）的一半.共表OD 3（2k+1）长度的一对素长，它们有一个共同的端点.因为在长度上每一条RPL2k-1 等于一条RPL2k+1. 并且，它们的左端点是相邻的奇数点，右端点也是相邻的奇数点. 于是,逐点向左移动RPLS2k+1使之成为RPLS2k-y,那么，RPLS2k+1和RPLS2k-y 的部分左端点就一对一地重合OD 3（2k+1）上的奇素数点，这儿y=1, 3, 5, …...因此,让我们在数轴的从奇数点3开始的正方向射线上，从奇数点2m+1开始，向左依次取每一个奇数点2m-y+2作为反向素长的共同右端点，这里y=1、3、5、7、...ỹ、...假设y依次增大到奇数ỹ，且RPLS2m+1、RPLS2m-1 ... RPLS2m-ỹ+2和RPLS2m-ỹ的部份左端点一对一地恰好重合完OD 3（2m+1）上的全部奇素数点，那么，在OD（2m- ỹ）（2m+1）上共有（ỹ+3）/2个奇数点，并且令μ =（ỹ+3）/2.然后，我们按照从长到短的次序，从两条射线的重合段中依次分离出OD 3（2m-y+2），并把它们从上到下依次排列起来，还把向右方向射线上表示奇素数点的奇素数放在这个奇素数点的上方，把重合这奇素数点的表示反向素长左端点的奇素数放在它的下方.例如:当2m+3=95时，2m+1=93, 2m-1=91和2m-ỹ=89，μ=3. 对于OD 3(95)、OD 3(93)、OD 3(91) 、OD 3(89) 和OD 3(87)上反向素长左端点重合奇素数点的分布情况，如下图：OD 5(95)19 31 37 61 67 7979 67 61 37 31 19OD 5(93) 7 13 17 23 29 37 43 53 59 67 73 79 83 8989 83 79 73 67 59 53 43 37 29 23 17 13 7OD 5(91)5 11 23 41 47 53 71 83 8989 83 71 53 47 41 23 11 5OD 5(89) 13 19 31 61 73 7979 73 61 31 19 13OD 5(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 8383 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7第二图在OD 3（2m-y+2）上，孪生反向素长的两个左端点一对一重合的两个奇素数点，以该奇距的中点为对称中心左右对称. 如果该奇距的中点是一个奇素数点，那么,它既是反向素长的左端点，又是这个左端点重合的奇素数点，例如，在上面引用例子中，OD 3（91）的中点47.在OD 3（2m+1）上，我们仅仅把唯一由RPLS2m-ỹ的左端点重合的奇素数点，即在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的奇素数点中没有的奇素数点看作是特有的奇素数点. 那么，根据上述的作法，在RPLS2m-ỹ的左端点一一重合的奇素数点中，有至少一个特有的奇素数点.例如，上面所举那个例子中，在OD 3(89)上的奇素数点19、31和61.因为在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的奇素数点中，没有一个特有的奇素数点. 换言之，每一个特有的奇素数点都不是RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点.又RPLS2m-y+2的左端点左边的第一个奇数点是RPLS2m-y的左端点，这里y是≧1的正奇数.所以,在每一个特有奇素数点左边的第一个奇数点不是RPLS2m-1、RPLS2m-3、...和RPLS2m-ỹ的左端点.--- {1}因为在OD 3(2m-y+2)上，每一条反向素长等于一条素长. 加之，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.因此，仅以长度而言，在OD 3(2m+1)上，其左端点一对一地重合完全部奇素数点的反向素长是OD 3(2m+1)上全部的反向素长，而不与具有同一长度的反向素长的条数有关.显然，在OD 3(2m+1)上的全部反向素长中，最长的反向素长等于PL 3P m.当奇合数点2m+3与奇素数点P m之间的奇距是一条合长时，让我们回顾前述的作法，即从2m+1开始，向左依次取每一个奇数点作为反向素长的共同右端点，并且，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一对一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.究竟哪一条反向素长的左端点首先重合奇素数点3呢？显然，它仅仅能够是以奇素数点P m为右端点的最长的反向素长的左端点.又从RPLS2m+1的左端点开始，RPLS2m-ỹ的左端点是重合完OD 3（2m+1）上全部奇素数点的最后一组.所以，作为RPLS2m-ỹ的共同右端点的奇数点2m-ỹ不能位于奇素数点P m的右边，于是，奇数点2m-ỹ-2位于奇素数点P m的左边. 由此可知，在OD 3（2m- ỹ-2）上的每一条反向素长都比PL 3P m短.又因为在RPLS2m-ỹ的左端点左边的第一个奇数点全部是RPLS2m-ỹ-2的左端点.所以，每一条RPL2m-ỹ-2都能反向延长成为至少一条RPL2m-y+2. 那就是说，RPLS2m-ỹ-2的每个左端点都一定是RPLS2m-y+2的至少一条的左端点，这里y是≦ỹ+2的正奇数.因为RPLS2m-ỹ-2的左端点一对一地全部在RPLS2m-ỹ左端点（包括特有奇素数点）左边的第一个奇数点上，于是，在特有奇素数点左边的第一个奇数点一定是RPLS2m-y+2的左端点，这里y是≦ỹ+2的正奇数. ---{2}所以,在既要满足结论{1}、又要满足结论{2}的情况下，我们只能够得出: 在特有奇素数点左边的第一个奇数点是RPLS2m+1的左端点.既然如此，让我们向右移动任意一条这样的PLS2m+1到相邻的奇数点，那么,这条PLS2m+1被移动成了一条PLS2m+3. 显然，这条PLS2m+3的左端点重合这个特有奇素数点，而它的右端点是奇数点2m+3.这样，OD 3（2m+3）可以被表为两条素长之和，这两条素长的共同端点就是这个特有奇素数点.到此为止，我们已经证明了: 当OD P m（2m+3）是一条合长时，OD 3（2m+3）也能表为两条素长之和.那么，根据定理3，3加上第m+1个奇数所得的偶数,即3+(2m+3)可以表示成两个奇素数之和.每次都从已证得的结论出发，就可以推出:当k等于每一个自然数时，每一个偶数3+(2k+1)都可以表示成两个奇素数之和. 即当N>2时，每一个偶数2N都可以表示成两个奇素数之和.又当N=2时,有2N=4=偶素数2+偶素数2.所以,每一个偶数2N都可以表示成两个素数之和，这里N≧2.因为每一个奇数2N+3等于偶数2N加上奇素数3，因此，每一个奇数2N+3都可以表示成三个素数之和，这里N≧2.综上所述，我们已经证明了哥德巴赫猜想的两个命题都是站得住脚的，因此，哥德巴赫猜想成立.。

广东省广州市天河区2024届高三下学期综合测试(二) 英语含答案

2024届天河区普通高中毕业班综合测试（二）英语（答案在最后）本试卷共8页，满分120分。

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不按以上要求作答的答案无效。

4.考生必须保证答题卡的整洁，考试结束后，将试卷和答题卡一并交回。

第二部分阅读（共两节，满分50分）第一节（共15小题；每小题2.5分，满分37.5分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项。

ASome of the greatest discoveries in science have been completely accidental.Then again,maybe science had a little help...Ice CandyOne of America’s favorite summertime treats came about thanks to a particularly cold night in the San Francisco Bay area in1905.That’s when11-year-old Frank Epperson forgot a glass containing powdered soda mix,water and a wooden stick out at the doorway overnight.The next morning,Frank discovered the mixture had frozen around the stick.He popped it out of the cup,licked it…and summers were never the same!Safety GlassIn1903,French chemistÉdouard Bénédictus accidentally dropped a glass bottle.To his surprise,the glass cracked but wasn’t broken into pieces.Out of curiosity,he checked it carefully.As it turned out,the bottle hadn’t been cleaned properly and was coated with a thin film of a kind of liquid plastic.A happy accident led to the discovery of safety glass.YInMn BlueIn2009,Mas Subramanian,a professor of materials science at Oregon State University,was testing out new materials for electronics in his lab with graduate students.A mixture of metal s went into the furnace.What came out was a striking blue mixture.The first new blue since cobalt(深蓝)was discovered in1802,is called“YInMn Blue.”PenicillinIn1928,Scottish researcher Alexander Fleming returned to his lab after a two-week vacation only to discover that mold(霉菌)from an open window had made an uncovered dish of bacteria dirty.Strangely enough,the mold stopped the bacteria from growing,giving the world its first antibiotic.21.Which discovery has a relatively short history?A.Ice candy.B.Safety glass.C.YInMn Blue.D.Penicillin.22.What do the four discoveries have in common?A Science helped a lot. b experiments laid a basis.C.They were made by scientists.D.They were made by chance.23.Who are the intended readers of the text?A.Science students.B.Curiosity seekers.C.Science researchers.D.The general public.BOccasionally,doctors become patients too.While I wouldn’t wish ill-health on anyone,it can be an inspiring lesson for medical professionals to suddenly be on the other side.This happened to me a few months ago when I had a kidney stone,which had decided to make its unwelcome presence known in a rather romantic fashion just as I was going out for dinner.The pain came from absolutely nowhere but within minutes I was incapacitated.I was quite taken ab ack by how astonishingly painful it was.As a doctor,I initially refused to believe that anything was seriously wrong,because I have witnessed countless individuals come to the emergency room convinced they are dying only for nothing more than trapped wind and then creep(蹑手蹑脚)out as they burp(打嗝)loudly and the pain disappears.However,the unbearable pain continued to exist,forcing me to acknowledge that this was indeed more than just a stubborn burp,I was particularly pleased to read afterwards in a medical textbook that renal colic,as the pain is known,is the “most painful event a person can endure,often described as being worse than childbirth.”It’s no wonder chronic(慢性的)pain drives people mad.While doctors talk about pain and its management,it’s hard to put into words how exhausting it really is until you’ve experienced it for yourself.During my time in hospital,I interacted with numerous doctors,each exhibiting professionalism and kindness but one stuck out in my mind.He was actually the most junior of them all,but something about his manner was incredibly calming and comforting.Whenever he came to my bed to speak to me,he knelt down so he was at my eye level.This simple act rid me of the stress that other healthcare professionals unintentionally projected.Just kneeling down made all the difference.Experiencing the role reversal of doctor-turned-patient provided me with profound insights into the realities of pain and effective patient care.24.What does the underlined phrase taken ab ack in paragraph2mean?A.Shocked.B.Impressed.C.Annoyed.D.Moved.25.What did the author initially think of those patients in the emergency room?A.Pitiful.B.Understandable.C.Embarrassing.D.Ridiculous.26.What impressed the author most during his stay in hospital?A.The intense pain caused by the kidney stone.B.A small gesture bringing ease and relief.C.Stress caused by the healthcare workers.D.Doctors with professionalism and kindness.27.What lesson did the author learn from this experience?A.Experience must be bought.B.Actions speak louder than words.C.Put yourself in someone else’s shoes.D.Health is not valued till sickness comes.CUS forests could become a“substantial carbon source”by2070,suggesting that forests could worsen global warming instead of easing it,a new Agriculture Department report says.US forests currently absorb11percent of US carbon emissions(释放),equal to the combined emissions from40 coal power plants.The report predicts that the ability of forests to absorb carbon will start declining after2025and that forests could emit up to100million metric tons of carbon a year as their emissions from decaying(腐烂)trees go beyond their carbon absorption.The prediction suggests that this will require the US to cut emissions more rapidly to reach net zero,according to Lynn Riley,a senior manager of climate science.“Eleven percent of our domestic emissions.That is a really significant portion,”Riley said.“As we work to decarbonize...forests are one of the greatest tools available.If we were to lose that,it means the US will contribute that much more”in emissions.The report also assesses and predicts the extent of renewable resources provided by the nation’s forests and undeveloped landscapes,including farmlands, wetlands and grasslands.According to the report,the loss of carbon absorption is driven in part by natural disasters which are increasing in frequency and strength as global temperatures rise.Development in forested areas,which the report predicts will continue to increase,is having the same effect as people increasingly move to the so-called wildland urban areas. Aging forests also contribute as older,mature trees absorb less carbon than younger trees of the same species,and the US forests are rapidly aging.More aggressive forest management can help by cutting down a small portion of aging forests to make ways for younger trees that absorb more carbon.A thorough study of each forest should be done before removing older trees, Riley said.28.Why could US forests become a“substantial carbon source”?A.They fail to absorb enough carbon.B.They begin to emit carbon increasingly.C.They may emit more carbon than absorb.D.Mature trees outnumber young trees.29.Which of the following may Lynn Riley agree with? forests have lost their role in carbon absorption.B.It is urgent for the US to reduce carbon emissions.C.The US has contributed to11%carbon emissions.D.The loss of forests contributes to less carbon emissions.30.What leads to the loss of carbon absorption?A.Improper tree species.B.Removal of aging trees.C.Decreasing urban development.D.Frequent severe natural disasters.31.Which could be a suitable title for the text?A.Aging Forests—a Major Emitter of Carbon.B.Growing Forests—a Solution to Carbon Emissions.C.Reducing Forests—a Threat to Global Warming.D.Decaying Forests—a Consequence of Global Warming.DArtists everywhere are getting“understandably nervous”about recent advances in artificial st month,a winner of an art prize at the Colorado State Fair“sparked a violent protest”when he posted the news and explained that he’d created his image using an AI program.Critics quickly accused39-year-old Lance Allen of cheating.To be fair,Allen had won in the digital art category and made no secret of how the image had been produced. But the rules of art making are clearly changing.Allen’s creative process,to be clear,“was not a push-button operation,”said Jason Blain in Forbes.He claims to have spent80hours on his entry,first on fine-tuning his text prompts(提示),then by touching up the final image using Photoshop and similar tools,then arranging to print the image on canvas.He made the finished product using AI much as a photographer creates an image using a camera.But Allen,a tabletop game developer,is awed by AI’s capabilities and urges artists and illustrators to welcome the technology rather than fight it.“Art is dead,”he says.“AI won.Humans lost.”A more inspiring lesson to take from his victory,though,is that image generators are likely to“expand the appreciation for and creation of art”by opening the field to people,like him,who could never draw anything as detailed as his award-winning image.“If anything,we will have more artists,”and as the technology progresses,“we might see the emergence of art styles that none have seen before.”You can’t blame traditional artists if they’re unhappy.Image generators work their magic,after all,by analyzing the aesthetics(美学)of millions of pre-existing images.One of the most complicated image generators“makes crystal clear just how destructive this technology will be,”said Loz Eliot in New Atlas.Given a specific prompt,it can produce an image of just about anything you can imagine and even follow the style of a favorite artist’s work.Its arrival marks“an incredible popularization of visual creativity”while aiming“a knife to the heart of anyone who’s spent decades improving their artistic techniques hoping to make a living from them.”32.Why are artists getting nervous about AI recently?A.A winner of an art prize used AI.nce Allen cheated in the art competition.C.The digital art will soon dominate.D.There will be great changes in art creation.33.What does the author intend to tell us in paragraph2?A.It was no easy work for Allen even with Al.B.Allen worked as a photographer creating an image.C.AI played a key role in Allen’s art creation.D.Although with AI,Allen’s creation counted a lot.34.What lesson can we draw from Allen’s winning?A.Human has been beaten by AI.B.AI will make art more popular.C.Greater artists and new art styles will appear.D.AI enables amateurs to win art competitions.35.Why does Loz Eliot say the new technology will be destructive?A.It works by analyzing images created by human.B.It can produce images beyond people’s imagination.C.It makes artists’long-time effort meaningless.D.It makes it impossible for artists to make a living.第二节（共5小题；每小题2.5分，满分12.5分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

英语单词精解系列[高中译林模块6单元1]七十一

英语单词精解系列[高中译林模块6单元1]七十一burst in释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 闯入；突然出现；打断短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ burst-in:煲机burst t in:闯进burst bleeding in operation:术中大出血Plants Burst out in profusion:草木怒生Burst Sword in Bow Stance:弓步崩剑burst fire in single firing:单打连water burst in face:工作面涌水burst in anger:英语翻译deep breathing释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 深呼吸的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ deep-breathing:深呼吸的deep-breathing exercise:深呼吸操练Deep Abdominal Breathing:至腹部呼吸Deep Slow Breathing:深缓地呼吸coughing g and deep-breathing exercises:咳嗽呼吸练习coughing and deep-breathing exercises:咳嗽呼吸练习deep breathing exercises:气功疗法Deep Breathing Yoga:呼吸冥想瑜伽Do deep breathing:做一下深呼吸guarantee音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[gær(ə)n’tiː] 美[,ɡærən’ti]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 过去式guaranteed 过去分词guaranteed 现在分词guaranteeing ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 保证；担保；保证人；保证书；抵押品vt. 保证；担保短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Average guarantee:海损担保书;海损保证书guarantee agreement:担保协议;保证协议guarantee something:打保票bid guarantee:投标保证书;承包担保;投标保函term guarantee:中长期担保banking guarantee:银行保证书guarantee letter:保函;保证函investment guarantee:投资保证guarantee work:保固工程;保修工程;保修工作例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.V-T If one thing guarantees another, the first is certain to cause the second thing to happen. 确保2.N-COUNT Something that is a guarantee of something else makes it certain that it will happen or that it is true. 保证3.V-T If you guarantee something, you promise that it will definitely happen, or that you will do or provide it for someone. 保证…必定发生; 保证…做某事或提供某物4.N-COUNT Guarantee is also a noun. 保证5.N-COUNT A guarantee is a written promise by a company to replace or repair a product free of charge if it has any faults within a particular time. 保修单[also ’under’ N]6.V-T If a company guarantees its product or work, they provide a guarantee for it. 担保7.N-COUNT A guarantee is money or something valuable that you give to someone to show that you will do what you have promised. 担保物make fun of释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 取笑短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ make e fun of:取笑;取笑某人;拿……开玩笑make a fun of sb:对某人开玩笑;嘲笑别人joke make fun of:戏谑;开玩笑;耍笑M make fun of:取笑nto make fun of:寻开心make fun of sth:取笑make fun of sb:取笑某人;拿某人开玩笑;同某人开玩笑;某人make fun of banter:拿人开玩笑make fun of tease:调笑positive音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[ˈpɒzətɪv] 美[’pɑzətɪv]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 比较级more positive 最高级most positive ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 积极的；[数] 正的，[医][化学] 阳性的；确定的，肯定的；实际的，真实的；绝对的n. 正数；[摄] 正片短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Positive Psychology:积极心理学;正面心理学positive feedback:正反馈;正回馈;正回输positive number:正数;叫做正数;负数;阳性容器数positive ion:阳离子;离子positive reinforcement:正强化;正增强;正面强化Positive Discipline:正面管教;正向管教;正面教育;积极惩罚positive electron:阳电子;positive logic:正电子;翻译Positive interdependence:正逻辑;翻译例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ If you are positive about things, you are hopeful and confident, and think of the good aspects of a situation rather than the bad ones. (心态) 积极的2.positively ADV 积极地[ADV after v]3.ADJ A positive fact, situation, or experience is pleasant and helpful to you in some way. 积极的(事实、情形、经历)4.N-SING The positive in a situation is the good and pleasant aspects of it. 积极面[’the’ N]5.ADJ If you make a positive decision or take positive action, you do something definite in order to deal with a task or problem. 积极的(决定、行动等)6.ADJ A positive response to something indicates agreement, approval, or encouragement. 积极的(回应)7.positively ADV 积极地[ADV after v]8.ADJ If you are positive about something, you are completely sure about it. 肯定的[v-link ADJ]9.ADJ Positive evidence gives definite proof of the truth or identity of something. 确凿的(证据) [ADJ n]10.positively ADV 确凿地[ADV with v]11.ADJ If a medical or scientific test is positive , it shows that something has happened or is present. (化验结果) 阳性的12.HIV positive →see HIV13.ADJ A positive number is greater than zero. 正的(数字) [ADJ n]14.ADJ If something has a positive electrical charge, it has the same charge as a proton and the opposite charge to a neutron. 正的(电荷) [技术]variety音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[və’raɪətɪ] 美[və’raɪəti]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 多样；种类；杂耍；变化，多样化短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ variety show:综艺节目;杂耍;综合节目algebraic variety:代数簇;代数族;代数多样体;代数曲体Shimura variety:志村簇Product variety:产品选择;shoes variety:积簇;产品种类;产品品种group variety:鞋子种类;鞋子鞋材颜色;鞋子品种;闉嫔瓙绉岖被local variety:群簇;群曲体Variety identification:地方品种;本地种;本地品种;地域变种regular variety:品种鉴别;例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-UNCOUNT If something has variety , it consists of things that are different from each other. 多样化2.N-SING A variety of things is a number of different kinds or examples of the same thing. (同一事物的) 一些不同种类3.N-COUNT A variety of something is a type of it. 品种foolishness释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 愚蠢；可笑短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ sheer foolishness:简直胡闹pretend foolishness:装憨儿foolishness n:愚蠢oppose foolishness:反对愚昧Extreme foolishness:愚蠢My Foolishness:我傻气Tender Foolishness:温柔傻事While Foolishness Has None:而愚蠢则没有The foolishness examination questions:傻气的试题textbook音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’teks(t)bʊk] 美[’tɛkstbʊk]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 教科书，课本短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Foothold textbook:立足课本Textbook contents:教材内容textbook compilation:教材编写;教科书编制textbook analysis:教材分析English textbook:英语教材;初中英语课本;英语读本development textbook:发展教科书Persian Textbook:波斯教科书Planning textbook:规划文本Knsh Textbook:康轩文教教科书例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-COUNT A textbook is a book containing facts about a particular subject that is used by people studying that subject. 教科书2.ADJ If you say that something is a textbook case or example, you are emphasizing that it provides a clear example of a type of situation or event. 典型的[ADJ n] [强调]initial音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[ɪ’nɪʃəl] 美[ɪ’nɪʃəl]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 过去式initialed或-tialled 过去分词initialed或-tialled 现在分词initialing或initialling ] 释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 最初的；字首的vt. 用姓名的首字母签名n. 词首大写字母短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ initial margin:初始保证金;INITIAL SAMPLE:原始保证金;首期保证金;基本按金initial offerings:原办;初版;首次样品;初办initial point:原始股initial set:始点;起点;起始点;initial stage:初始点initial speed:初凝;始集;初始化设置;初始设置initial current:初级阶段;开始后不久的一段时期;原始期;初期Initial results:初速;例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ You use initial to describe something that happens at the beginning of a process. 最初的[ADJ n]2.N-COUNT Initials are the capital letters that begin each word of a name. For example, if your full name is Michael Dennis Stocks, your initials are M.D.S. (姓名的) 大写首字母3.V-T If someone initials an official document, they write their initials on it, to show that they have seen it or that they accept or agree with it. 签姓名的首字母于attain音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[ə’teɪn] 美[ə’ten]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ vt. 达到，实现；获得；到达vi. 达到；获得；到达n. 成就短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ achieve attain:努力取得attain liberation:得脱attain empire:实现帝国attain arhatship:得阿罗汉attain desire:实现欲望attain wealth:获得财富Attain Extreme:达到极端attain 3:达到例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.V-T If you attain something, you gain it or achieve it, often after a lot of effort. (常指经过努力) 获得[正式]bathtub音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’bɑːθtʌb] 美[bæθtʌb]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 浴缸短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Massage Bathtub:按摩浴缸;家用小浴缸产品;按摩缸;按摩冲浪浴缸BATHTUB SERIES:浴缸系列;普通缸系列bathtub drainer:浴缸下水Bathtub Faucet:浴缸龙头Leggera Bathtub:新式浴缸Infinity Bathtub:花瓣浴缸Bathtub Effect:浴缸效应;很多人都说他们学英语的时候最大的障碍就是单词记不住Hydromassage bathtub:喷水按摩浴缸bathtub theory:浴缸原理例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-COUNT A bathtub is a long, usually rectangular container that you fill with water and sit in to wash your body. 浴缸[美国英语]style音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[staɪl] 美[staɪl]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 过去式styled 过去分词styled 现在分词styling ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 风格；时尚；类型；字体vt. 设计；称呼；使合潮流vi. 设计式样；用刻刀作装饰画n. (Style)人名；(英)斯泰尔短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Style :作风Text Style:字体样式;字布局;文字样式;文字布局Wandaba Style:妄想科学美少女;科学妄想美少女;妄想科学EXCELLENT STYLE:漂亮的款式;标致的技俩;标致的款式;美丽的款式style sheet:样式表;样式单;风格样式表;式样表in style:时髦的;流行的;别具风格的Mediterranean Style:地中海风格;地中海作风;地中海建筑风格;地中海风情Management Style:管理风格;治理作风;管理模式;classic style:管理作风例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-COUNT The style of something is the general way in which it is done or presented, which often shows the attitudes of the people involved. 方式2.N-UNCOUNT If people or places have style , they are fashionable and elegant. 风度; 格调3.N-VAR The style of a product is its design. 款式4.N-COUNT In the arts, a particular style is characteristic of a particular period or group of people. (某个时期或群体的艺术) 风格5.V-T If something such as a piece of clothing, a vehicle, or someone’s hair is styled in a particular way, it is designed or shaped in that way. 设计式样[usu passive]6.→ see also self-styledcrowded音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’kraʊdɪd] 美[’kraʊdɪd]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 比较级more crowded 最高级most crowded ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 拥挤的；塞满的v. 拥挤（crowd的过去分词）短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Crowded trade:拥挤交易crowded item:重叠项crowded traffic:拥堵的交通;拥挤的交通less crowded:不那么拥挤;松快;那么拥挤;不拥挤Crowded Tired:广场午后crowded d:拥挤的;拥挤;人多的;挤满的Crowded Planets:拥挤的行星Teeth crowded:牙列拥挤extremely crowded:人山人海例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ If a place is crowded , it is full of people. 拥挤的2.ADJ If a place is crowded , a lot of people live there. 人口密集的3.ADJ If your schedule, your life, or your mind is crowded , it is full of events, activities, or thoughts. 排满的technical音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’teknɪk(ə)l] 美[’tɛknɪkl]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 比较级more technical 最高级most technical ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 工艺的，科技的；技术上的；专门的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Technical information:技术信息;技术资料;Technical Engineer:技术情报;故障代码Technical Director:技术工程师;技能工程师;技术支持工程师;应用工程师Technical intelligence:技术总监;技术指导;经理;技术主管Technical Parameters:科技情报;技术情报;技术信息Technical Sergeant:技术参数;操作参数;技术指标;主要技术参数TECHNICAL CHARACTERISTICS:技术军士;上士;专业军士;技能军士Technical skills:主要技术特性;技术特点;技术特征;技术特性Technical Management:技术技能;技术能力;专业技术能力;专业技能例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ Technical means involving the sorts of machines, processes, and materials that are used in industry, transportation, and communications. 技术的; 工艺的2.technically ADV 技术上地; 工艺上地[ADV adj]3.ADJ You use technical to describe the practical skills and methods used to do an activity such as an art, a craft, or a sport. 技巧的4.technically ADV 技巧上地[ADV adj]5.ADJ Technical language involves using special words to describe the details of a specialized activity. 专业的(语言)6.→ see also technicallysaying音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’seɪɪŋ] 美[’seɪŋ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 谚语，格言，警句；（尤指政治或宗教名人）格言集；言论，话v. 说（say 的现在分词形式）短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Saying goes:常言说;俗话说得好Saying hello:打招呼;问好;打个招呼;第一次问候Saying Sorry:道歉;说抱歉;对自己说抱歉;唱片名Just saying:不过是说说而已;说说而已;织呀嘛织毛衣The saying:俗语说;说法部份;有俗语说;谚语说allegorical saying:歇后语Western Saying:西方谚语Popular saying:俗话例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-COUNT A saying is a sentence that people often say and that gives advice or information about human life and experience. 谚语; 格言stand-up释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 站立的；直立的；说单口相声的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Stand Put-up:完成外径Stand-up comedy:单口喜剧;脱口秀;单口相声stand-up comedian:喜剧演员;栋笃笑;搞笑艺人;脱口秀喜剧演员Stand-up Scooter:动力滑板车stand-up edian:栋笃笑stand-up meal:站着吃的便餐stand-up shot:站拍例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ A stand-up comic or comedian stands alone in front of an audience and tells jokes. 单人喜剧表演的[ADJ n]2.N-UNCOUNT Stand-up is stand-up comedy. 单人表演的喜剧3.N-COUNT A stand-up is a stand-up comedian. 单人喜剧演员amuse音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[ə’mjuːz] 美[ə’mjuz]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 过去式amused 过去分词amused 现在分词amusing ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ vt. 娱乐；消遣；使发笑；使愉快短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ amuse bouche:餐前开胃小吃;开胃小点;餐前小食;开胃小吃Amuse Yourself:自娱自乐amuse powerfully:十分逗乐amuse hugely:很有趣amuse mildly:稍微觉得好笑amuse irritatingly:烦躁地消遣amuse by:以culture amuse:文化娱乐Amuse itself:自娱自乐例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.V-T If something amuses you, it makes you want to laugh or smile. 使发笑2.V-T If you amuse yourself , you do something in order to pass the time and not become bored. 使消遣3.→ see also amused , amusingtake on释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 承担；呈现；具有；流行；接纳；雇用；穿上短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ take pity on:怜悯;怜惜;同情;可怜take out on:对;发泄;向别人发泄自己的情绪;对…take sth on:接受;承担;答应;决定做某事take revenge on:向…报仇;向某人复仇;向;向…take it on board:纳入考虑范围take pity on us:怜悯take hold on:握住;抓住take pity on people:悯民take out on someone:是例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.PHRASAL VERB If you take on a job or responsibility, especially a difficult one, you accept it. 承担2.PHRASAL VERB If something takes on a new appearance or quality, it develops that appearance or quality. 呈现[no passive]3.PHRASAL VERB If a vehicle such as a bus or ship takes on passengers, goods, or fuel, it stops in order to allow them to get on or to be loaded on. (交通工具) 停下装载(乘客、货物、燃料)4.PHRASAL VERB If you take someone on , you employ them to do a job. 聘用5.PHRASAL VERB If you take someone on , you fight them or compete against them, especially when they are bigger or more powerful than you are. 与(尤指强于自己的人) 较量[no passive]6.PHRASAL VERB If you take something on or upon yourself , you decide to do it without asking anyone for permission or approval. 擅自决定做[no passive]little-known释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 鲜为人知的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ little-known fact:鲜为人知的事实a little-known painter:一个藉藉无名画家little-known local artists:鲜为人知的当地艺术家a little-known museum:一个不太出名的博物馆one little-known fact:鲜为人知的事实;一个鲜为人知的事实He is a little-known painter:他是一个没有什么名气的画家Monroe little-known those things:梦露鲜为人知的那些事cosy音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’kəʊzɪ] 美[ˈkozi]附加_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ [ 复数cosies 比较级cosier 最高级cosiest ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 舒适的；惬意的n. 保温罩短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cosy Cups:毛绒马克杯系列Hotel Cosy:柯西酒店;温馨酒店COSY SA:比利时进口商;地区Simple Cosy:朴质闲适d cosy:一维copyCosy Zone:舒适地带Cosy Gum:金考拉COSY cafe:东田私享咖啡;东田私享;三里屯店;丽都水岸店例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ A house or room that is cosy is comfortable and warm. (指房子或房间) 温暖舒适的2.ADJ If you are cosy , you are comfortable and warm. (指人) 温暖舒适的[v-link ADJ]3.ADJ You use cosy to describe activities that are pleasant and friendly, and involve people who know each other well. 亲切的4.N-COUNT A cosy or a tea cosy is a soft knitted or fabric cover which you put over a teapot in order to keep the tea hot. 茶壶套howl音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[haʊl] 美[haʊl]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ vi. 咆哮；怒吼；狂吠vt. 狂喊着说；对…吼叫n. 嗥叫；怒号；嚎哭短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Coyote howl:郊狼的嚎叫First Howl:狼人豪尔Furious Howl:狂怒之嚎Intimidating Howl:威吓嚎叫howl blatantly:粗暴地吼叫Frustrating Howl:沮丧嚎叫Daily Howl:每日号叫;每日嚎叫Lupine Howl:表演者例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.V-I If an animal such as a wolf or a dog howls , it makes a long, loud, crying sound. (狼、狗等动物) 嗥叫2.N-COUNT Howl is also a noun. 嗥叫3.V-I If a person howls , they make a long, loud cry expressing pain, anger, or unhappiness. (人因痛苦、愤怒、不愉快等) 嚎叫4.N-COUNT Howl is also a noun. 嚎叫5.V-I When the wind howls , it blows hard and makes a loud noise. 呼啸6.V-T If you howl something, you say it in a very loud voice. 高声嚷[非正式]7.V-I If you howl with laughter, you laugh very loudly. 放声大笑8.N-COUNT Howl is also a noun. 大笑armchair音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[ɑːm’tʃeə; ’ɑːm-] 美[’ɑrmtʃɛr]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 扶手椅，单人沙发adj. 不切实际的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Tulip Armchair:郁金香扶手椅;郁金香椅;餐椅;休闲椅armchair strategist:空想战略家;纸上谈兵的战略家Armchair Apocrypha:吹牛话野史;唱片名Millennium Armchair:干禧扶手椅Armchair Thriller:心慌慌Plastic Armchair:塑料扶手椅Shell Armchair:鸟巢休闲躺椅Armchair Detectives:安乐椅神探Leather Armchair:真皮扶手椅例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-COUNT An armchair is a big comfortable chair that has a support on each side for your arms. 扶手椅2.ADJ An armchair critic, fan, or traveller knows about a particular subject from reading or hearing about it rather than from practical experience. 足不出户的[ADJ n]empty-handed释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 空手的；徒手的adv. 空手地；一无所有地短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ return empty-handed:空手而归fled empty-handed:一无所得地逃走empty- handed:空手start empty-handed:自食其力;白手起家empty-handed unarmed:徒手的go empty-handed:空手而去empty-handed adj:空手的empty-handed training:徒手训练Empty-handed Drawing Tool:徒手绘图工具例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ If you come away from somewhere empty-handed , you have failed to get what you wanted. 一无所获的[ADJ after v]anger音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’æŋgə] 美[’æŋɡɚ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 怒，愤怒；忿怒vt. 使发怒，激怒；恼火vi. 发怒；恼火n. (Anger)人名；(罗)安杰尔；(法)安热；(德、捷、瑞典)安格尔短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ show anger:显示愤怒;显出怒容in anger:含怒;气愤地;生气地;愤怒Trait Anger:特质愤怒;怒特质;愤怒特质ANGER BIRDS:愤怒的小鸟;愤怒鸟;愤怒的小鸟纸模型indignant anger:出离愤怒;钓鱼of anger:论愤怒Anger Blind:盲目愤怒righteous anger:愤慨;义怒anger r:愤怒;怒;气愤;生气例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-UNCOUNT Anger is the strong emotion that you feel when you think that someone has behaved in an unfair, cruel, or unacceptable way. 愤怒2.V-T If something angers you, it makes you feel angry. 使气愤polish音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’pɒlɪʃ] 美[’pɑlɪʃ]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n. 磨光，擦亮；擦亮剂；优雅，精良vi. 擦亮，变光滑vt. 磨光，使发亮；使完美; 改进v. 磨光；修改；润色adj. 波兰的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Polish Requiem:波兰安魂曲shoe polish:鞋油;鞋擦亮剂;皮鞋油;寻求与皮鞋抛光业Polish notation:波兰表示法;波兰式Polish Zloty:波兰兹罗提;兹罗提dull polish:粗磨光;磨砂;翻译;详细翻译Car polish:汽车抛光腊;汽车打光腊;上光用品polish up:改善;擦亮;提高Polish Remover:去光水;洗甲水;相关推荐boot polish:鞋油;擦鞋膏;擦例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-MASS Polish is a substance that you put on the surface of an object in order to clean it, protect it, and make it shine. 抛光剂; 亮漆2.V-T If you polish something, you put polish on it or rub it with a cloth to make it shine. (用抛光剂等) 擦亮3.N-SING Polish is also a noun. 擦亮; 磨光4.polished ADJ 擦亮的; 磨光的5.N-UNCOUNT If you say that a performance or piece of work has polish , you mean that it is of a very high standard. 完美; 上乘境界[表赞许]6.V-T If you polish your technique, performance, or skill at doing something, you work on improving it. 使完美;改进7.PHRASAL VERB Polish up means the same as . 使完美; 改进8.→ see also polishedmourn音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[mɔːn] 美[mɔrn]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ v. 哀悼；忧伤；服丧短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ mourn sb:哀悼某人to mourn:悼念;哀痛mourn n:哀悼;哀痛;哀伤;悲悼Mourn Prostitutes:悼妓Sometime Mourn:有时哀恸mourn text:哀思的mourn verb:哀痛And mourn:并哀悼nationally mourn:举国哀悼例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.V-T/V-I If you mourn someone who has died or mourn for them, you are very sad that they have died and show your sorrow in the way that you behave. 悼念2.V-T/V-I If you mourn something or mourn for it, you regret that you no longer have it and show your regret in the way that you behave. 为(失去) …而痛惜visual音标_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 英[’vɪʒjʊəl; -zj-] 美[’vɪʒʊəl]释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ adj. 视觉的，视力的；栩栩如生的短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Visual Arts:视觉艺术;可视艺术;视觉艺术科;视艺Visual Effects:视觉效果;最佳视觉效果;然后找到;视觉特效visual literacy:图像素养;视觉素养;视觉认识能力;视觉文化visual editor:可视编辑器;可视化编辑器;安装;编辑器Visual cliff:视觉悬崖;visual anthropology:视觉壁;视觉悬涯visual impairment:视觉人类学;影视人类学;人类学visual prosthesis:视力障碍;视觉障碍;视觉缺陷;视力损害Visual Design:视觉假体;人工视觉假体例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.ADJ Visual means relating to sight, or to things that you can see. 视觉的2.visually ADV 视觉地3.N-COUNT A visual is something such as a picture, diagram, or piece of film that is used to show or explain something. 视觉资料(展示或解释用的图画、图表或电影片段等)toilet paper释义_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 卫生纸，厕纸短语_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ toilet t paper:卫生纸;手纸;擦屁屁的厕纸;厕纸toilet-paper:卫生纸toilet t paper fixture:卫生纸盒Toilet Martin Paper Parr:厕所Toilet Tissue Paper:卫生纸TOILET TISSUE PAPER ROLL:卷筒卫生纸架Toilet seat paper:马桶垫纸toilet t paper holder:卫生纸架toilet t paper with reel:卷筒式卫生纸例句_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _1.N-UNCOUNT Toilet paper is thin soft paper that people use to clean themselves after they have got rid of urine or faeces from their body. 手纸; 卫生纸。