Another Simple Proof of a Theorem of Chandler Davis
stoerwagner-mincut.[Stoer-Wagner,Prim,连通性,无向图,最小边割集]
A Simple Min-Cut AlgorithmMECHTHILD STOERTeleverkets Forskningsinstitutt,Kjeller,NorwayANDFRANK WAGNERFreie Universita¨t Berlin,Berlin-Dahlem,GermanyAbstract.We present an algorithm for finding the minimum cut of an undirected edge-weighted graph.It is simple in every respect.It has a short and compact description,is easy to implement,and has a surprisingly simple proof of correctness.Its runtime matches that of the fastest algorithm known.The runtime analysis is straightforward.In contrast to nearly all approaches so far,the algorithm uses no flow techniques.Roughly speaking,the algorithm consists of about͉V͉nearly identical phases each of which is a maximum adjacency search.Categories and Subject Descriptors:G.L.2[Discrete Mathematics]:Graph Theory—graph algorithms General Terms:AlgorithmsAdditional Key Words and Phrases:Min-Cut1.IntroductionGraph connectivity is one of the classical subjects in graph theory,and has many practical applications,for example,in chip and circuit design,reliability of communication networks,transportation planning,and cluster analysis.Finding the minimum cut of an undirected edge-weighted graph is a fundamental algorithmical problem.Precisely,it consists in finding a nontrivial partition of the graphs vertex set V into two parts such that the cut weight,the sum of the weights of the edges connecting the two parts,is minimum.A preliminary version of this paper appeared in Proceedings of the2nd Annual European Symposium on Algorithms.Lecture Notes in Computer Science,vol.855,1994,pp.141–147.This work was supported by the ESPRIT BRA Project ALCOM II.Authors’addresses:M.Stoer,Televerkets Forskningsinstitutt,Postboks83,2007Kjeller,Norway; e-mail:mechthild.stoer@nta.no.;F.Wagner,Institut fu¨r Informatik,Fachbereich Mathematik und Informatik,Freie Universita¨t Berlin,Takustraße9,Berlin-Dahlem,Germany;e-mail:wagner@inf.fu-berlin.de.Permission to make digital/hard copy of part or all of this work for personal or classroom use is granted without fee provided that the copies are not made or distributed for profit or commercial advantage,the copyright notice,the title of the publication,and its date appear,and notice is given that copying is by permission of the Association for Computing Machinery(ACM),Inc.To copy otherwise,to republish,to post on servers,or to redistribute to lists,requires prior specific permission and/or a fee.᭧1997ACM0004-5411/97/0700-0585$03.50Journal of the ACM,Vol.44,No.4,July1997,pp.585–591.586M.STOER AND F.WAGNER The usual approach to solve this problem is to use its close relationship to the maximum flow problem.The famous Max-Flow-Min-Cut-Theorem by Ford and Fulkerson[1956]showed the duality of the maximum flow and the so-called minimum s-t-cut.There,s and t are two vertices that are the source and the sink in the flow problem and have to be separated by the cut,that is,they have to lie in different parts of the partition.Until recently all cut algorithms were essentially flow algorithms using this duality.Finding a minimum cut without specified vertices to be separated can be done by finding minimum s-t-cuts for a fixed vertex s and all͉V͉Ϫ1possible choices of tʦVگ{s}and then selecting the lightest one.Recently Hao and Orlin[1992]showed how to use the maximum flow algorithm by Goldberg and Tarjan[1988]in order to solve the minimum cut problem in timeᏻ(͉VʈE͉log(͉V͉2/͉E͉),which is nearly as fast as the fastest maximum flow algorithms so far[Alon1990;Ahuja et al.1989;Cheriyan et al. 1990].Nagamochi and Ibaraki[1992a]published the first deterministic minimum cut algorithm that is not based on a flow algorithm,has the slightly better running time ofᏻ(͉VʈE͉ϩ͉V͉2log͉V͉),but is still rather complicated.In the unweighted case,they use a fast-search technique to decompose a graph’s edge set E into subsets E1,...,Esuch that the union of the first k E i’s is a k-edge-connected spanning subgraph of the given graph and has at most k͉V͉edges.They simulate this approach in the weighted case.Their work is one of a small number of papers treating questions of graph connectivity by non-flow-based methods [Nishizeki and Poljak1989;Nagamochi and Ibaraki1992a;Matula1992].Karger and Stein[1993]suggest a randomized algorithm that with high probability finds a minimum cut in timeᏻ(͉V͉2log͉V͉).In this context,we present in this paper a remarkably simple deterministic minimum cut algorithm with the fastest running time so far,established in Nagamochi and Ibaraki[1992b].We reduce the complexity of the algorithm of Nagamochi and Ibaraki by avoiding the unnecessary simulated decomposition of the edge set.This enables us to give a comparably straightforward proof of correctness avoiding,for example,the distinction between the unweighted, integer-,rational-,and real-weighted case.This algorithm was found independently by Frank[1994].Queyranne[1995]generalizes our simple approach to the minimization of submodular functions.The algorithm described in this paper was implemented by Kurt Mehlhorn from the Max-Planck-Institut,Saarbru¨cken and is part of the algorithms library LEDA[Mehlhorn and Na¨her1995].2.The AlgorithmThroughout the paper,we deal with an ordinary undirected graph G with vertex set V and edge set E.Every edge e has nonnegative real weight w(e).The simple key observation is that,if we know how to find two vertices s and t, and the weight of a minimum s-t-cut,we are nearly done:T HEOREM2.1.Let s and t be two vertices of a graph G.Let G/{s,t}be the graph obtained by merging s and t.Then a minimum cut of G can be obtained by taking the smaller of a minimum s-t-cut of G and a minimum cut of G/{s,t}.The theorem holds since either there is a minimum cut of G that separates s and t ,then a minimum s -t -cut of G is a minimum cut of G ;or there is none,then a minimum cut of G /{s ,t }does the job.So a procedure finding an arbitrary minimum s -t -cut can be used to construct a recursive algorithm to find a minimum cut of a graph.The following algorithm,known in the literature as maximum adjacency search or maximum cardinality search ,yields the desired s -t -cut.M INIMUM C UT P HASE (G ,w ,a )A 4{a }while A Vadd to A the most tightly connected vertexstore the cut-of-the-phase and shrink G by merging the two vertices added lastA subset A of the graphs vertices grows starting with an arbitrary single vertex until A is equal to V .In each step,the vertex outside of A most tightly connected with A is added.Formally,we add a vertexz ʦ͞A such that w ͑A ,z ͒ϭmax ͕w ͑A ,y ͉͒y ʦ͞A ͖,where w (A ,y )is the sum of the weights of all the edges between A and y .At the end of each such phase,the two vertices added last are merged ,that is,the two vertices are replaced by a new vertex,and any edges from the two vertices to a remaining vertex are replaced by an edge weighted by the sum of the weights of the previous two edges.Edges joining the merged nodes are removed.The cut of V that separates the vertex added last from the rest of the graph is called the cut-of-the-phase .The lightest of these cuts-of-the-phase is the result of the algorithm,the desired minimum cut:M INIMUM C UT (G ,w ,a )while ͉V ͉Ͼ1M INIMUM C UT P HASE (G ,w ,a )if the cut-of-the-phase is lighter than the current minimum cutthen store the cut-of-the-phase as the current minimum cutNotice that the starting vertex a stays the same throughout the whole algorithm.It can be selected arbitrarily in each phase instead.3.CorrectnessIn order to proof the correctness of our algorithms,we need to show the following somewhat surprising lemma.L EMMA 3.1.Each cut -of -the -phase is a minimum s -t -cut in the current graph ,where s and t are the two vertices added last in the phase .P ROOF .The run of a M INIMUM C UT P HASE orders the vertices of the current graph linearly,starting with a and ending with s and t ,according to their order of addition to A .Now we look at an arbitrary s -t -cut C of the current graph and show,that it is at least as heavy as the cut-of-the-phase.587A Simple Min-Cut Algorithm588M.STOER AND F.WAGNER We call a vertex v a active(with respect to C)when v and the vertex added just before v are in the two different parts of C.Let w(C)be the weight of C,A v the set of all vertices added before v(excluding v),C v the cut of A vഫ{v} induced by C,and w(C v)the weight of the induced cut.We show that for every active vertex vw͑A v,v͒Յw͑C v͒by induction on the set of active vertices:For the first active vertex,the inequality is satisfied with equality.Let the inequality be true for all active vertices added up to the active vertex v,and let u be the next active vertex that is added.Then we havew͑A u,u͒ϭw͑A v,u͒ϩw͑A uگA v,u͒ϭ:␣Now,w(A v,u)Յw(A v,v)as v was chosen as the vertex most tightly connected with A v.By induction w(A v,v)Յw(C v).All edges between A uگA v and u connect the different parts of C.Thus they contribute to w(C u)but not to w(C v).So␣Յw͑C v͒ϩw͑A uگA v,u͒Յw͑C u͒As t is always an active vertex with respect to C we can conclude that w(A t,t)Յw(C t)which says exactly that the cut-of-the-phase is at most as heavy as C.4.Running TimeAs the running time of the algorithm M INIMUM C UT is essentially equal to the added running time of the͉V͉Ϫ1runs of M INIMUM C UT P HASE,which is called on graphs with decreasing number of vertices and edges,it suffices to show that a single M INIMUM C UT P HASE needs at mostᏻ(͉E͉ϩ͉V͉log͉V͉)time yielding an overall running time ofᏻ(͉VʈE͉ϩ͉V͉2log͉V͉).The key to implementing a phase efficiently is to make it easy to select the next vertex to be added to the set A,the most tightly connected vertex.During execution of a phase,all vertices that are not in A reside in a priority queue based on a key field.The key of a vertex v is the sum of the weights of the edges connecting it to the current A,that is,w(A,v).Whenever a vertex v is added to A we have to perform an update of the queue.v has to be deleted from the queue,and the key of every vertex w not in A,connected to v has to be increased by the weight of the edge v w,if it exists.As this is done exactly once for every edge,overall we have to perform͉V͉E XTRACT M AX and͉E͉I NCREASE K EY ing Fibonacci heaps[Fredman and Tarjun1987],we can perform an E XTRACT M AX operation inᏻ(log͉V͉)amortized time and an I NCREASE K EY operation inᏻ(1)amortized time.Thus,the time we need for this key step that dominates the rest of the phase, isᏻ(͉E͉ϩ͉V͉log͉V͉).5.AnExample F IG .1.A graph G ϭ(V ,E )withedge-weights.F IG .2.The graph after the first M INIMUM C UT P HASE (G ,w ,a ),a ϭ2,and the induced ordering a ,b ,c ,d ,e ,f ,s ,t of the vertices.The first cut-of-the-phase corresponds to the partition {1},{2,3,4,5,6,7,8}of V with weight w ϭ5.F IG .3.The graph after the second M INIMUM C UT P HASE (G ,w ,a ),and the induced ordering a ,b ,c ,d ,e ,s ,t of the vertices.The second cut-of-the-phase corresponds to the partition {8},{1,2,3,4,5,6,7}of V with weight w ϭ5.F IG .4.After the third M INIMUM C UT P HASE (G ,w ,a ).The third cut-of-the-phase corresponds to the partition {7,8},{1,2,3,4,5,6}of V with weight w ϭ7.589A Simple Min-Cut AlgorithmACKNOWLEDGMENT .The authors thank Dorothea Wagner for her helpful re-marks.REFERENCESA HUJA ,R.K.,O RLIN ,J.B.,AND T ARJAN ,R.E.1989.Improved time bounds for the maximum flow problem.SIAM put.18,939–954.A LON ,N.1990.Generating pseudo-random permutations and maximum flow algorithms.Inf.Proc.Lett.35,201–204.C HERIYAN ,J.,H AGERUP ,T.,AND M EHLHORN ,K.1990.Can a maximum flow be computed in o (nm )time?In Proceedings of the 17th International Colloquium on Automata,Languages and Programming .pp.235–248.F ORD ,L.R.,AND F ULKERSON ,D.R.1956.Maximal flow through a network.Can.J.Math.8,399–404.F RANK , A.1994.On the Edge-Connectivity Algorithm of Nagamochi and Ibaraki .Laboratoire Artemis,IMAG,Universite ´J.Fourier,Grenoble,Switzerland.F REDMAN ,M.L.,AND T ARJAN ,R.E.1987.Fibonacci heaps and their uses in improved network optimization algorithms.J.ACM 34,3(July),596–615.G OLDBERG ,A.V.,AND T ARJAN ,R.E.1988.A new approach to the maximum-flow problem.J.ACM 35,4(Oct.),921–940.H AO ,J.,AND O RLIN ,J.B.1992.A faster algorithm for finding the minimum cut in a graph.In Proceedings of the 3rd ACM-SIAM Symposium on Discrete Algorithms (Orlando,Fla.,Jan.27–29).ACM,New York,pp.165–174.K ARGER ,D.,AND S TEIN ,C.1993.An O˜(n 2)algorithm for minimum cuts.In Proceedings of the 25th ACM Symposium on the Theory of Computing (San Diego,Calif.,May 16–18).ACM,New York,pp.757–765.F IG .5.After the fourth and fifth M INIMUM C UT P HASE (G ,w ,a ),respectively.The fourth cut-of-the-phase corresponds to the partition {4,7,8},{1,2,3,5,6}.The fifth cut-of-the-phase corresponds to the partition {3,4,7,8},{1,2,5,6}with weight w ϭ4.F IG .6.After the sixth and seventh M INIMUM C UT P HASE (G ,w ,a ),respectively.The sixth cut-of-the-phase corresponds to the partition {1,5},{2,3,4,6,7,8}with weight w ϭ7.The last cut-of-the-phase corresponds to the partition {2},V گ{2};its weight is w ϭ9.The minimum cut of the graph G is the fifth cut-of-the-phase and the weight is w ϭ4.590M.STOER AND F.WAGNERM ATULA ,D.W.1993.A linear time 2ϩ⑀approximation algorithm for edge connectivity.In Proceedings of the 4th ACM–SIAM Symposium on Discrete Mathematics ACM,New York,pp.500–504.M EHLHORN ,K.,AND N ¨AHER ,S.1995.LEDA:a platform for combinatorial and geometric mun.ACM 38,96–102.N AGAMOCHI ,H.,AND I BARAKI ,T.1992a.Linear time algorithms for finding a sparse k -connected spanning subgraph of a k -connected graph.Algorithmica 7,583–596.N AGAMOCHI ,H.,AND I BARAKI ,puting edge-connectivity in multigraphs and capaci-tated graphs.SIAM J.Disc.Math.5,54–66.N ISHIZEKI ,T.,AND P OLJAK ,S.1989.Highly connected factors with a small number of edges.Preprint.Q UEYRANNE ,M.1995.A combinatorial algorithm for minimizing symmetric submodular functions.In Proceedings of the 6th ACM–SIAM Symposium on Discrete Mathematics ACM,New York,pp.98–101.RECEIVED APRIL 1995;REVISED FEBRUARY 1997;ACCEPTED JUNE 1997Journal of the ACM,Vol.44,No.4,July 1997.591A Simple Min-Cut Algorithm。
罗比塔法则
罗比塔法则罗比塔(L'Hospital)法则,是在一定条件下通过分子分母分别求导再求极限来确定未定式值的方法。
设(1)当x→a时,函数f(x)及F(x)都趋于零;(2)在点a的去心邻域内,f'(x)及F'(x)都存在且F'(x)≠0;(3)当x→a时lim f'(x)/F'(x)存在(或为无穷大),那么x→a时lim f(x)/F(x)=lim f'(x)/F'(x)。
再设(1)当x→∞时,函数f(x)及F(x)都趋于零;(2)当|x|>N时f'(x)及F'(x)都存在,且F'(x)≠0;(3)当x→∞时lim f'(x)/F'(x)存在(或为无穷大),那么x→∞时lim f(x)/F(x)=lim f'(x)/F'(x)。
利用罗比塔法则求未定式的极限是微分学中的重点之一,在解题中应注意:①在着手求极限以前,首先要检查是否满足0/0或∞/∞型,否则滥用罗比塔法则会出错。
当不存在时(不包括∞情形),就不能用罗比塔法则,这时称罗比塔法则失效,应从另外途径求极限。
比如利用泰勒公式求解。
②罗比塔法则可连续多次使用,直到求出极限为止。
③罗比塔法则是求未定式极限的有效工具,但是如果仅用罗比塔法则,往往计算会十分繁琐,因此一定要与其他方法相结合,比如及时将非零极限的乘积因子分离出来以简化计算、乘积因子用等价量替换等等.The Stolz-Cesàro theo rem is a similar result involving limits of sequences, but it uses finite difference operators rather than derivatives.In its simplest form, l'Hôpital's rule states that for functions ƒ and g:If or and exists,thenThe differentiation of the numerator and denominator often simplifies the quotient and/or converts it to a determinate form, allowing the limit to be evaluated more easily.Contents[hide]∙1General form∙2Requirement that limit exists∙3Examples∙4Other indeterminate forms∙5Other methods of evaluating limits∙6Logical circularity∙7Case where ƒ and g are differentiable at c∙8Geometric interpretation∙9Proof of l'Hôpital's ruleo9.1Zero over zeroo9.2Infinity over infinity∙10See also∙11Notes∙12External links[edit] General formThe general form of l'Hôpital's rule covers many more cases. Suppose that c and L are extended real numbers(i.e., real numbers, positive infinity, or negative infinity). Suppose that eitherorAnd suppose thatThenexists is essential. Differentiation of indeterminate forms can sometimes lead to limits that do not exist. If this happens, then l'Hôpital's rule does not apply. For example, if ƒ(x) = x + sin(x) and g(x) = x, thenwhich does not exist, whereasAlternatively, just observe that the limit is the definition of the derivative of the sine function at zero.∙This is a more elaborate example involving 0/0. Applying l'Hôpital's rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying the rule three times:∙This example involves 0/0. Suppose that b > 0. Then∙Here is another example involving 0/0:∙This example involves ∞/∞. Assume n is a positive integer. ThenRepeatedly apply l'Hôpital's rule until the exponent is zero to conclude that the limit is zero.∙Here is another example involving ∞/∞:∙One can also use l'Hôpital's rule to prove the following theorem. If is continuous at x, then∙Sometimes L'Hôpital's rule is invoked in a tricky way: suppose f(x) + f'(x) converges as . It follows:∙ :and so exists and[edit] Other indeterminate formsOther indeterminate forms, such as 1∞, 00, ∞0, 0·∞, and ∞−∞, can sometimes be evaluated using l'Hôpital's rule. For example, to evaluate a limit involving ∞−∞, convert the difference of two functions to a quotient:where l'Hôpital's rule was applied in going from (1) to (2) and then again in going from (3) to (4).l'Hôpital's rule can be used on indeterminate forms involving exponents by using logarithms to "move the exponent down". Here is an example involving the indeterminate form 00:It is valid to move the limit inside the exponential function because theexponential function is continuous. Now the exponent x has been "moved down". The limit is of the indeterminate form 0·(−∞), but as shown in an example above, l'Hôpital's rule may be used to determine thatThus[edit] Other methods of evaluating limitsAlthough l'Hôpital's rule is a powerful way of evaluating otherwise hard-to-evaluate limits, it is not always the easiest way. ConsiderThis limit may be evaluated using l'Hôpital's rule:It is valid to move the limit inside the cosine function because the cosine function is uniformly continuous.Another way to evaluate this limit is to use a substitution. Let y = 1/x. As |x| approaches infinity, y approaches zero. So,The final limit may be evaluated using l'Hôpital's rule or by noting that it is the definition of the derivative of the sine function at zero.Still another way to evaluate this limit is to use a Taylor series expansion:For |x| ≥ 1, the expression in parentheses is bounded, so the limit in the last line is zero.[edit] Logical circularityIn some cases it may constitute circular reasoning to use l'Hôpital's rule to evaluate a limit. ConsiderIf the purpose of evaluating this limit is to prove that if ƒ(x) = x n, thenand one uses l'Hôpital's rule and this same fact to evaluate the limit, then the argument uses the conclusion as an assumption (i.e., begging the question) and is therefore fallacious (even though the conclusion is true).[edit] Case where ƒ and g are differentiable at cThe proof of l'Hôpital's rule is simple in the case where ƒ and g are differentiable at the point c. It is not a proof of the general l'Hôpital's rule because it requires stronger hypotheses than does l'Hôpita l's rule, namely, the differentiability of ƒ and g at c.Suppose that ƒ and g are continuous and differentiable at c,ƒ(c) = g(c) = 0, and g′(c) ≠ 0. Then(remember that ƒ(c) = g(c) = 0). This follows from the limit rules for quotients and the definition of the derivative.This suggests the general case of l'Hôpital's rule, which does not require the functions ƒ and g to be differentiable at the point c and is proven below.[edit] Geometric interpretationConsider curve in the plane whose x-coordinate is given by g(t) and whose y-coordinate is given by ƒ(t) – i.e.Suppose ƒ(c) = g(c) = 0. The limit of the ratio ƒ(t)/g(t) as t→c is the slope of tangent to the curve at the point [0, 0]. The tangent to the curve at the point t is given by [g'(t),f'(t)]. L'Hôpital's rule thenstates that the slope of the tangent at 0 is the limit of the slopes of tangents at the points approaching 0.[edit] Proof of l'Hôpital's ruleA standard proof of l'Hôpital's rule uses Cauchy's mean value theorem. l'Hôpital's rule has many variations depending on whether c and L are finite or infinite, whether ƒ and g converge to zero or infinity, and whether the limits are one-sided or two-sided. All the variations follow from the two main variations below without, for the most part, requiring any new reasoning.[3][edit] Zero over zeroSuppose that c and L are finite and ƒ and g converge to zero.First, define (or redefine) ƒ(c) = 0 and g(c) = 0. This makes ƒ and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c). Sinceexists, there is an interval (c−δ, c + δ) such that for all x in the interval, with the possible exception of x = c, both and g'(x)exist and g'(x) is not zero.If x is in the interval (c, c + δ), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [c, x] (and a similar statement holds for x in the interval (c−δ, c)). The mean value theorem implies that g(x) is not zero (since otherwise there would be a y in the interval (c, x) with g'(y) = 0). Cauchy's mean value theorem now implies that there is a point ξx in (c, x) such thatIf x approaches c, then ξx approaches c(by the squeeze theorem). Sinceexists, it follows that[edit] Infinity over infinitySuppose that L is finite, c is positive infinity, and ƒ and g converge to positive infinity.For every ε > 0, there is an m such thatThe mean value theorem implies that if x > m, then g(x) ≠g(m) (since otherwise there would be a y in the interval (m, x) with g'(y) = 0). Cauchy's mean value theorem applied to the interval [m, x] now implies thatSince ƒ converges to positive infinity, if x is large enough, then ƒ(x) ≠ƒ(m). WriteNow,For x sufficiently large, this is less than ε and therefore*Note: Steps are missing.。
八年级数学理论英语阅读理解20题
八年级数学理论英语阅读理解20题1<背景文章>The Pythagorean theorem is one of the most fundamental theorems in mathematics. It has a long and interesting history. The theorem was named after the ancient Greek mathematician Pythagoras, who was the first to prove it, though there is evidence that the Babylonians knew about this relationship much earlier.The Pythagorean theorem states that in a right - angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. That is, for a right triangle with sides \(a\), \(b\), and hypotenuse \(c\), \(a^{2}+b^{2} = c^{2}\).There are many ways to prove the Pythagorean theorem. One of the most well - known proofs is by using the area of squares. We can construct squares on each side of the right triangle. The area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.In terms of applications, the Pythagorean theorem is widely used in various fields. For example, in architecture and engineering, it helps to calculate distances and lengths. If we know the lengths of two sides of aright - angled triangle in a building structure, we can easily find the length of the third side. In navigation, it can be used to calculate the shortest distance between two points.In modern mathematics, the Pythagorean theorem is the basis for many more advanced concepts. It is an important part of trigonometry, as it helps in understanding the relationships between the sides and angles of triangles. It also has implications in vector mathematics and complex number systems.1. <问题1>Who was the theorem named after?A. A Babylonian mathematician.B. An ancient Greek mathematician.C. A modern mathematician.D. An Egyptian mathematician.答案:B。
2023-2024学年上海市静安区高三上学期期末教学质量调研考试英语试题
2023-2024学年上海市静安区高三上学期期末教学质量调研考试英语试题Directions: After reading the passage below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Japan’s robot revolution in senior careJapan’s artificial intelligence expertise is transforming the elder care industry, with 1 (specialize) robotic care accomplishing more than just taking pressure off the critical shortage of caregivers. Senior care facilities across Japan are testing out such new robots 2 deliver a collection of social and physical health care and the government-backed initiative has been met with positive reviews by elderly residents.The rapidly graying population 3 (eye) by the government as a potential market for medical technology now. Disappointing government predictions show that by 2025, Japan's first baby boomers will have turned 75 and about 7 million people are likely to suffer from some form of dementia (痴呆). The nation won't be able to avoid a dementia crisis 4 an additional 380,000 senior care workers.The long-standing shortage of professional care workers has encouraged the Japanese government 5 (simplify) procedures for foreign caregivers to be trained and certified. The current Technical Intern Training Program between Vietnam, the Philippines, and Indonesia, under 6 Economic Partnership Agreement, was extended to include nursing care as well as agriculture, fishery, and construction sectors.7 the government made efforts to increase the numbers of senior care workers, the target number of foreign graduates has still fallen flat, with the national caregiver examination proving a major obstacle to pass. The success rate for foreign students was a merely 106 students last year, 8 has slightly improved to 216 students this year. Another depressing reality is that 19 to 38 percent of foreign nurses who pass the exam opt to leave the industry and return home, 9 (cite) tough work conditions and long hours. Given the challenges, this is 10 the government believes care robots will be able to step in.Directions: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.A. smoothingB. remainC. switchedD. likelihoodE. impactF. tipG. broadly H. headed I. booming J. positioning K. reliablySea-level rise predictionsA team of University of Idaho scientists is studying a fast-moving glacier in Alaska in hopes of developing better predictions on how quickly global sea levels will rise.Tim Bartholomaus, a professor in the Department of Geography and Geological Sciences, spent several weeks on Turner Glacier in Alaska’s southeastern 11 near Disenchantment Bay. The glacier is unique because, unlike other glaciers, it rises greatly every five to eight years.A surging glacier is defined, 12 , as one that starts flowing at least 10 times faster than normal. But the how and why of that glacial movement is poorly understood, although recent research suggests that global climate change increases the 13 of glacial surging.During Turner’s surges, the mass of ice and rock will increase its speed from roughly 3 feet a day to 65 feet per day.All of that is important because glaciers falling into the ocean are a major contributor to sea level rise, and current clima te change models don’t 14 account for these movements. For example, Greenland’s glaciers are one of the leading contributors to global sea-level rise. Since the early 2000s, Greenland 15 from not having any effect on world sea levels, to increasing sea level by about 1 millimeter per year. Half of that yearly increase is due to warmer average temperatures, which leads to more ice melting. The other half, however, is because glaciers in Greenland are, as a whole, moving faster and running into the ocean more frequently.Glacial movement has something to do with water running underneath the glacier. Glaciers are full of holes, and water runs through those holes. When the water pressure is high underneath a glacier, it starts to move, partly because it’s li fting the mass of ice and rock off the ground and partly because it’s 16 the underside of the glacier.But how exactly does that water move through the glacier, and how does the movement 17 the glacier’s speed? Those are the questions the scientists ho pe to answer.Bartholomaus, some graduate students and researchers from Boise State University, 18 onto the ice in August. They set up a base camp at the toe of the glacier and spent their days flying in on helicopters. They placed roughly 30 instruments, burying them deeply into the glacier and 19 them on rock outcroppings (露岩) alongside the glacier. This summer the team will return to get the instruments and replace batteries. Those instruments will 20 on and around the glacier until the glacier surge stops, providing researchers with before and after data.Investors probably expect that following the suggestions of stock analysts would make them better off than doing the exact opposite. _________, recent research by Nicola Gennaioli and his colleagues shows that the best way to gain excess return s would be to invest in the shares least favored by analysts. They compute that, during the last 35 years, investing in the 10 percent of U. S. stocks analysts were most _________ about would have yielded on average 3 percent a year._________, investing in the 10 percent of stocks analysts were most pessimistic about would have yielded a surprising 15 percent a year.Gennaioli and colleagues shed light on this _________ with the help of cognitive sciences and, in particular, using Kahneman and Tversky's concept of representativeness. Decision makers, according to this view, _________ the representative features of a group or a phenomenon. These are defined as the features that occur more frequently in that group than in a baseline reference group.After observing strong earnings growth—the explanation goes—analysts think that the firm may be the next Google. “Googles” are in fact more frequent among firms experiencing strong growth, which makes them _________. The problem is that “Googles” are very _________ in absolute terms. As a result, expectations become too optimistic, and future performance_________. A model of stock prices in which investor beliefs follow this logic can account both qualitatively and quantitatively for the beliefs of analysts and the dynamics (动态变化) of stock returns.In related work, the authors also show that the same model can _________ booms and busts in the volume of credit and interest rate spreads.These works are part of a research project aimed at taking insights from cognitive sciences and at__________them into economic models. Kahneman and Tversky's concept of “representativeness” lies at the heart of this effort. “In a classical example, we __________ to think of Irishmen as redheads because red hair is much more frequent among Irishmen than among the rest of the world,” Prof. Gennaioli says. “However, only 10 percent of Irishmen are redheads. In our work, we develop models of belief formation that show this logic and study the __________ of this important psychological force in different fields.”Representativeness helps describe __________ and behavior in different fields, not only in financial markets. One such field is the formation of stereotypes about social groups. In a recent experimental paper, Gennaioli and colleagues show that representativeness can explain self-confidence, and in particular the __________ of women to compete in traditionally male subjects, such as mathematics.A slight prevalence of __________ male math ability in the data is enough to make math ability un-representative for women, driving their under confidence in this particular subject.21.A.Consequently B.Furthermore C.Nevertheless D.Meanwhile22.A.curious B.controversial C.concerned D.optimistic23.A.In brief B.By contrast C.In addition D.Without doubt 24.A.engagement B.concentration C.puzzle D.definition25.A.memorize B.prioritize C.modernize D.fertilize26.A.representative B.argumentative C.executive D.sensitive27.A.harsh B.adaptable C.crucial D.rare28.A.cheers B.disappoints C.stabilizes D.improves29.A.account for B.count on C.suffer from D.hold up30.A.pouring B.admitting C.integrating D.tempting31.A.pretend B.afford C.offer D.tend32.A.effects B.delights C.intervals D.codes33.A.companions B.scales C.expectations D.findings34.A.necessity B.involvement C.perseverance D.reluctance35.A.equivalent B.exceptional C.mysterious D.distressing Montessori was born in Italy in 1870 with progressive parents, who frequently communicated with the country’s leading thinkers and scholars. This enlightened family environment provided Montessori with many advantages over other young girls of the time.Her mother’s support was vital for some impo rtant decisions, such as her enrolment in a technical school after her elementary education. Her parents’ support also proved to be essential for her decision to study medicine, a field that was dominated by men.Soon after graduating, in 1896, Montessori began work as a voluntary assistant in a clinic at the University of Rome, where she cared for children with learning difficulties. The rooms were bare, with just a few pieces of furniture. One day, she found that the children were enthusiastically playing with breadcrumbs (面包屑) that had dropped on the floor. It then occurred to her that the origin of some intellectual disabilities could be related with poverty. With the right learning materials, these and other young minds could be nurtured, Montessori concluded.The observation would lead Montessori to develop a new method of education that focused on providing optimal stimulation during the sensitive periods of childhood.At its centre was the principle that all the learning materials should be child-sized and designed to appeal to all the senses. In addition, each child should also be allowed to move and act freely, and use their creativity and problem-solving skills. Teachers took the role of guides, supporting the children without press or control.Mont essori opened her first Children’s House in 1907. When the Fascists (法西斯主义者) first came into power in Italy in 1922, they initially embraced her movement. But they soon came to oppose the emphasis on the children’s freedom of expression. Montessori’s value s had always been about human respect, and the rights of children and women, but the Fascists wanted to use her work and her fame.Things reached a breaking point when the Fascist tried to influence the schools’ educational content, and in 1934 Montessori and her son decided to leave Italy. She didn’t return to her homeland until 1947, and she continued to write about and develop her method until her death in 1952, at the age of 81.36. The primary reason for Montessori to develop a new educational method was ______.A.her family’s supportive influence on her educationB.her experience as a voluntary assistant in a clinicC.her observation of children playing with breadcrumbs happilyD.her decision to study medicine, a field dominated by men37. What was a central principle of Montessori’s educational method as described in the passage?A.Providing standardized, one-size-fits-all learning materials.B.Encouraging strict discipline and control over children’s actions.C.Focusing on rote memorization and competition.D.Creating a free and children-centered learning environment.38. Montessori decided to leave Italy in 1934 because .A.she wanted to explore other countries and culturesB.she wanted to avoid the Fascist’s influence on her workC.she was offered a better job in a different countryD.she wanted to retire and enjoy a peaceful life in another country39. Which of the following words can best describe Montessori in this passage?A.Observant and innovative. B.Traditional and emotional.C.Progressive and dependent. D.Open-minded and indifferent. Reducing the workweek to four days could have a climate benefit. In addition to improving the well-being of workers, cutting working hours may reduce carbon emissions. But those benefits would depend on a number of factors, experts emphasize, including how people choose to spend nonworking time.Commuting and travelTransportation is the biggest contributor to greenhouse emissions. A November 2021 survey of2,000 employees and 500 business leaders in the United Kingdom found that if all organizations introduced a four-day week, the reduced trips to work would decrease travel overall by more than 691 million miles a week.But the climate benefits of less commuting could be eliminated, experts said, if people choose to spend their extra time off traveling, particularly if they do so by car or plane.Energy usageShorter working hours could lead to reductions in energy usage, experts said. According to a 2006 paper, if the United States adopted European work standards, the country would consume about 20 percent less energy.Energy could also be conserved if fewer resources are needed to heat and cool large office buildings, reducing demands on electricity. For example, if an entire workplace shuts down on the fifth day, that would help lower consumption — less so if the office stays open to accommodate employees taking different days off.Lifestyle changesIt’s possible that fewer working hours may lead some people to have a larger carbon footprint, bu t experts say research suggests that most people are likely to shift toward more sustainable lifestyles.One theory is that people who work more and have less free time tend to do things in more carbon-intensive ways, such as choosing faster modes of transportation or buying prepared foods. Convenience is often carbon-intensive and people tend to choose convenience when they're time-stressed. Meanwhile, some research suggests that those who work less are more likely to engage in traditionally low-carbon activities, such as spending time with family or sleeping.“When we talk about the four-day workweek and the environment, we focus on the tangible, but actually, in a way, the biggest potential benefit here is in the intangible,” experts said.40. What is identified as the leading cause of greenhouse emissions according to the passage?A.The well-being of employees.B.The conservation of energy.C.Commuting and travel.D.The European work standard.41. What can be inferred from the underlined sentence “the biggest potential benefit here is in the intangible” in the last paragraph?A.People will have big potential in achieving intangible benefits while working.B.People are more likely to engage in carbon-intensive activities due to time constraints.C.People may shift toward more sustainable lifestyles and lower carbon footprints.D.People may travel more frequently by car or plane during their extra time off.42. The passage is mainly written to .A.highlight the importance of shortening working time in the context of well-beingB.provide an overview of transportation emissions worldwideC.analyze the impact of reduced working hours on mode of businessD.illustrate factors affecting the climate benefits of a shorter workweekThe cultivation of plants by ants is more widespread than previously realized, and has evolved on at least 15 separate occasions.There are more than 200 species of ant in the Americas that farm fungi (真菌) for food, but this trait evolved just once sometime between 45 million and 65 million years ago. Biologists regard the cultivation of fungi by ants as true agriculture appearing earlier than human agriculture because it meets four criteria: the ants plant the fungus, care for it, harvest it and depend on it for food.By contrast, while thousands of ant species are known to have a wide variety of interdependent relationships with plants, none were regarded as true agriculture. But in 2016, Guillaume Chomicki and Susanne Renner at the University of Munich, Germany, discovered that an ant in Fungi cultivates several plants in a way that meets the four criteria for true agriculture.The ants collect the seeds of the plants and place them in cracks in the bark of trees. As the plants grow, they form hollow structures called domain that the ants nest in. The ants defecate (排便) at designated absorptive places in these domain, providing nutrients for the plant. In return, as well as shelter, the plant provides food in the form of fruit juice.This discovery prompted Chomicki and others to review the literature on ant-plant relationships to see if there are other examples of plant cultivation that have been overlooked. “They have never really been looked at in the framework of agriculture,” says Chomicki, who is now at the University of Sheffield in the UK. “It’s definitely widespread.”The team identified 37 examples of tree-living ants that cultivate plants that grow on trees, known as epiphytes (附生植物). By looking at the family trees of the ant species, the team was able to determine on how many occasions plant cultivation evolved and roughly when. Fifteen is a conservative estimate, says Campbell. All the systems evolved relatively recently, around 1million to 3 million years ago, she says.Whether the 37 examples of plant cultivation identified by the team count as true agriculture depends on the definitions used. Not all of the species get food from the plants, but they do rely on them for shelter, which is crucial for ants living in trees, says Campbell. So the team thinks the definition of true agriculture should include shelter as well as food.43. According to biologists, why is ant-fungus cultivation considered as a form of true agriculture?A.Because it occurred earlier than human agriculture.B.Because it fulfills the standards typical of agricultural practices.C.Because it redefines the four criteria for true human agriculture.D.Because it is less common than previously thought.44. What motivated Chomicki and others to review the literature on ant-plant relationships?A.They determined on new family trees of the ant species.B.They overlooked some tree-living ants that provided nutrients for the plants.C.They never studied the ant-plant relationships within the context of agriculture.D.They never identified any an t species that engaged in cultivation of fungi.45. Which of the following statements is supported by the team's findings according to the passage?A.Ants’ cultivation of plants is limited to a few specific species.B.The cultivation of fungi by ants is considered the earliest form of agriculture.C.True agriculture in ants involves only food-related interactions with plants.D.Ants have independently cultivated plants on at least 15 distinct occasions.46. What is the passage mainly about?A.The evolution of ants in the plant kingdom.B.The widespread occurrence of ant-plant cultivation.C.The discovery of a new ant species engaging in agriculture.D.The contrast between ant agriculture and human agriculture.What is the likelihood of you having someone who looks just like you? Would it be a good thing? And if you did have one, would you want to meet them?Consider how often your facial features are used to identify you. Your passport, ID card and driving license all feature your face. 47 You may need your face to unlock your smartphone and possibly even need it to exclude you from being present at a crime scene.The word “doppelgänger” refers to a person who looks the same as you, essentially sharing your features; those that you thought were unique to you and your identity. Not identical twins, as a doppelgänger has no relation to you. The idea originated in German folklore. 48So, let's get real. What are the chances of you having one in the first place? There's said to be a one in 135 chance of an exact match for you existing anywhere in the world, so the chances are pretty low, despite folk wisdom promising you otherwise. And the chances of meeting? The mathematical certainty of finding this particular person is supposedly less than one in a trillion.That said, these statistics may be a good thing. Historically, having a double wasn't always a positive. Back in 1999, an innocent American man, indistinguishable from the real criminal, was sent to prison for robbery, where he stayed for 19 years. 49 . In a different case, a woman in New York was accused of trying to poison her doppelgänger with deadly cheesecake so that she could steal her identity!50 The fascination with doppelgängers may be rooted in historical beliefs that facial resemblance meant they were from the same family or had a common ancestor. It leads to the hope that one day you will meet your lookalike, creating the thrill of a potentially strange meeting. However, as these encounters can be both interesting and disturbing, we understand that after such an experience, you might not want to meet your doppelgänger again.passage in no more than 60 words. Use your own words as far as possible.Competitive CheerleadingOver the years, cheerleading has taken two primary forms: game-time cheerleading and competitive cheerleading. Game-time cheerleaders’ main goal is to entertain the crowd and lead them with team cheers, which should not be considered a sport. However, competitive cheerleading is more than a form of entertainment. It is really a competitive sport.Competitive cheerleading includes lots of physical activity. The majority of the teams require a certain level of tumbling (翻腾运动) ability. It’s a very common thing for gymnasts, so it’s easy for them to go into competitive cheerleading. Usually these cheerleaders integrate lots of their gymnastics experience including their jumps, tumbling, and overall energy. They also perform lifts and throws.Competitive cheerleading is also an activity that is governed by rules under which a winner can be declared. It is awarded points for technique, creativity and sharpness. Usually the more difficult the action is, the better the score is. That’s why cheerleaders are trying to experience great difficulty in their performance. Besides, there is also a strict rule of time. The whole performance has to be completed in less than three minutes and fifteen seconds, during which the cheerleaders are required to stay within a certain area. Any performance beyond the limit of time is invalid.Another reason for the fact that competitive cheerleading is one of the hardest sports is that it has more reported injuries. According to some research, competitive cheerleading is the number one cause of serious sports injuries to women. Generally, these injuries affect all areas of the body, including wrists, shoulders, ankles, head, and neck.There can be no doubt that competitive cheerleading is a sport with professional skills. It should be noted that it is a team sport and even the smallest mistake made by one teammate can bring the score of the entire team down. So without working together to achieve the goal, first place is out of reach. ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ___________________________________________________________________________52. 如果不好好准备,周五的演讲可能会变得一塌糊涂。
Let x = (x1, x2,..., xn) N
On the Enumeration of Generalized ParkingFunctionsCatherine Huafei YanDepartment of MathematicsTexas A&M UniversityE-mail:cyan@AbstractLet x=(x1,x2,...,x n)∈n.Define a x-parking function to be a sequence(a1,a2,...,a n)of positive integers whose non-decreasing re-arrangement b1≤b2≤···≤b n satisfies b i≤x1+···+x i.Let P n(x)denote the number of x-parking functions.We discuss the enumer-ations of such generalized parking functions.In particular,We givethe explicit formulas and present two proofs,one by combinatorialargument,and one by recurrence,for the number of x-parking func-tions for x=(a,n−m−1b,...,b,c,m−10,...,0)and x=(a,n−m−1b,...,b,mc,...,c).1IntroductionThe notion of parking function was introduced by Konheim and Weiss [3]in their studying of an occupancy problem in computer science.Then the subject has been studied by numerous mathematicians.For example, Foata and Riordan[1],and Fran¸c on[2]constructed bijections from the set of parking functions to the set of acyclic functions on[n];Kreweras[4]in-vestigated the recurrence relations satisfied by the generating functions of parking functions and labeled trees;Stanley[8,9,10]and Pitman&Stan-ley[5]revealed the relations between parking functions and hyperplane arrangements,interval orders,non-crossing partitions,and polytopes.Re-cently it is also known that the parking functions has been of interests by statistician and probabilist from different approaches.A parking function of length n may be defined as a sequence(a1,a2,...,a n) of positive integers whose non-decreasing rearrangement b1≤b2≤···≤b n satisfies b i≤i.In particular,the number of parking functions of length n is(n+1)n−1,[1].1The notion of parking function can be generalized as follows[5].Let x=(x1,...,x n)∈n.Define a x-parking function to be a sequence (a1,...,a n)of positive integers whose non-decreasing rearrangement b1≤···≤b n satisfies b i≤x1+···+x i.Thus an ordinary parking function corresponds to the case x=(1,1,...,1).Let P n(x)denote the number of x-parking functions.Clearly that P n(x)=0if x1=0.In general,explicit formulas for the number of generalized parking func-tions are difficult to get.For x=(a,b,b,...,b),the number is known,(c.f. Theorem1).It can be proved by a simple combinatorial proof generalizing the proof of Pollak for the ordinary parking functions[1].A complete proof may be found,for example,in[7],Ex.5.49.Theorem1For x=(a,b,b,...,b),P n(x)=a(a+nb)n−1.In this paper we give the explicit formulas for the number of generalized parking functions P n(x),wherex=(a,n−m−1b,...,b,c,m−10,...,0),(1)andx=(a,n−m−1b,...,b,mc,...,c).(2)The formula for(1)was also obtained by Pitman and Stanley[5]by using results in the theory of empirical distributions and order statistics.In Sec. 2we give an easy combinatorial proof which can also be applied to(2). Another proof by recurrence relations is given in Sec.3.The combinatorial argument we used can be extended further to more complicated vectors x. As a corollary,we give the enumeration of x-parking functions forx=(a,n−m−1b,...,b,m−1c,...,c,d),andx=(a,n−m−1b,...,b,d,m−1c,...,c).2Combinatorial enumeration for x-parking functionsFirst we extend the notion of x-parking functions to include x=(x1,...,x n)∈n+.LetΠn(x)be the n-dimensional polytopeΠn(x):={y∈n:y i≥0and y1+···y i≤x1+···x i for all1≤i≤n}2The n-dimensional volumeV n(x):=Vol(Πn(x))is a homogeneous polynomial of degree n in the variables x1,x2,...,x n. Explicitly,Lemma1(Pitman&Stanley)For each n=1,2,···,1V n(x)=Any x -parking function α=(a 1,a 2,...,a n )satisfies the following con-ditions,1≤a i ≤a +(n −m −1)b +c,#{a i |a i ≤a +(k −1)b }≥k,for k =1,2,...,n −m.We decompose a x -parking function into two subsequences β,γby the following rule:Let t be the largest integer such that the condition#{a i |a i ≤a +(n −m −1+k )b }≥n −m +k (3)holds for all k =1,2,...,t .If no such t exists,we say that t =0.Let βbe the subsequence of all the terms a i which are less than or equal to a +(n −m −1+t )b ,and γbe the subsequence of the remaining terms.The x -parking function αis uniquely determined by the subsequences β,γand their positions in the original sequence α.From the condition (3),we note that βis a parking function of length n −m +t corresponding to x ′=(a,n −m +t −1 b,...,b ).There are f n −m +t such sequences.On the other hand,γis a sequence of length m −t in which every term lies between a +(n −m −1)b +(t +1)b +1and a +(n −m −1)b +c ,which contains c −(t +1)b integers.The terms of γmay take any m −t positions in α,therefore,P n (x )=m t =0n m −t c −(t +1)b m −t f n −m +t .Substitute into j =m −t ,we haveP n (x )=m j =0n j c −(m +1−j )b j f n −j =a m j =0n j c −(m +1−j )b j a +(n −j )b n −j −1.This finishes the proof.Theorem 3For x =(a,n −m −1 b,...,b,m c,...,c ),P n (x )=a m j =0 n j (m +1−j )(c −b )· (m +1)c −(m +1−j )b j −1 a +(n −j )b n −j −1.(4)4Proof.Again let f i=a(a+bi)i−1.Without loss of generality,we assume c>b.Any x-parking functionα=(a1,a2,...,a n)satisfies the following conditions,#{a i|a i≤a+(k−1)b}≥k,for k=1,2,...,n−m, #{a i|a i≤a+(n−m−1)b+ℓc}≥n−m+ℓ,forℓ=1,2,...,m.Similar to the preceding proof,we decompose a x-parking function into two subsequencesβ,γas follows:Let t be the largest integer such that the condition#{a i|a i≤a+(n−m−1+k)b}≥n−m+k(5) holds for all k=1,2,...,t.If no such t exists,we say that t=0.Let βbe the subsequence of all the terms a i which are less than or equal to a+(n−m−1+t)b,andγbe the subsequence of the remaining terms.The x-parking functionαis uniquely determined by the subsequences β,γand their positions inα.From the condition(5),we note thatβis aparking function of length n−m+t corresponding to x′=(a,n−m+t−1 b,...,b).There are f n−m+t such sequences.On the other hand,γis a sequence of length m−t satisfying the conditions1.Every term inγis greater than a+(n−m+t)b.2.#{a i∈γ|a i≤a+(n−m−1)b+ℓc}≥ℓ−t forℓ=t+1,t+2,...,m.Let1=(m−t1,...,1).The above conditions imply thatγ−(a+(n−m+t)b)1 is a parking function of length m−t corresponding to x=((t+1)(c−b),m−t−1c,...,c).There are(t+1)(c−b)[(m+1)c−(t+1)b]m−t−1such sequences. Therefore,P n(x)=mt=0 n m−t (t+1)(c−b) (m+1)c−(t+1)b m−t−1f n−m+t=amj=0 n j (m+1−j)(c−b) (m+1)c−(m+1−j)b j−1 a+(n−j)b n−j−1.In general,the formulas in Theorem2and3as summations of m+1 terms cannot be further simplified.5The method we used in Theorem 2and 3can be applied to more com-plicated vectors x .As a corollary,we list two more enumerations of P n (x )forx =(a,n −m −1 b,...,b,m −1 c,...,c,d )and x =(a,n −m −1 b,...,b,d,m −1 c,...,c ).Corollary 1Ifx =(a,n −m −1 b,...,b,m −1 c,...,c,d ),thenP n (x )=a m j =0 n j (m +1−j )(c −b ) a +(n −j ) n −j −1· (m +1)c −(m +1−j )bj −1+j (d −c ) mc −(m +1−j )b j −2 .Corollary 2Ifx =(a,n −m −1 b,...,b,d,m −1 c,...,c ),then P n (x )=a m j =0 n j d +(m −j )c −(m +1−j )· d +mc −(m +1−j )b j −1 a +(n −j )b n −j −1.The proof of Corollary 1and 2are straightforward computations using the proofs of Theorem 2and 3.3Proof by recurrence relationsBoth Theorem 2and 3can be proved by induction and using recurrence relations.In this section we present the inductive proofs.Inductive Proof of Theorem 2.Let g m (c )=P ((a,n −m −1 b,...,b,c,m −1 0,...,0))for fixed n ,a ,and b .From The-orem 1,g 0(c )=a (a +nb )n −1.(6)6Assume c>b.For a given x-parking functionα,consider the m th largest term b m.If b m≤a+(n−m)b,thenαis a parking function correspondingto x1=(a,n−mb,...,b,c−b,m−20,...,0).There are g m−1(c−b)many such parkingfunctions.If b m>a+(n−m)b,then the largest m terms ofαare ranging from a+(n−m)b+1to a+(n−m−1)b+c,and other terms form a parkingfunction of x2=(a,n−m−1b,...,b).There are n m (c−b)m a(a+(n−m)b)n−m−1many such parking functions.Thereforeg m(c)=g m−1(c−b)+ n m (c−b)m a(a+(n−m)b)n−m−1.(7)Theorem2follows from Equations(6)and(7)by iterating on m. Inductive Proof of Theorem3.Let h be the number of terms less than or equal to a+(n−m−1)b+c and let t equal h−(n−m).Forx=(a,n−m−1b,...,b,mc,...,c),any x-parking function can be formed by twosubsequencesβandγwhereβconsists of terms less than or equal to a+(n−m−1)b+c,andγconsists of all the remaining terms.Thenβis a parkingfunction of length n−m+t corresponding to x t=(a,n−m−1b,...,b,c,t−10,...,0),andγ−(a+(n−m−1)b+c)1is a parking function of length m−tcorresponding to x′t=(tc,m−t−1c,...,c).ThereforeP n((a,n−m−1b,...,b,mc,...,c))=mt=1 n m−t P m−t((tc,m−t−1 c,...,c))P n−m+t((a,n−m−1b,...,b,c,t−10,...,0)).(8)Substituting results in Theorem1and2into(8),one hasP n(x)=mt=1 n m−t (tc)(mc)m−t−1·tk=0 n−m−t t−k [c−(k+1)b]t−k[a+(n−m+k)b]n−m+k−1=mk=0(mc)m−k−1[a+(n−m+k)b]n−m+k−1c·T,(9)7whereT=nt=k n m−t n−m+tn−m+k t(mc)−t+k[c−(k+1)b]t−k= n m−k n t=k m−k t−k t(mc)−t+k[c−(k+1)b]t−k= n m−k [k(1+p)m−k+p(m−k)(1+p)m−k−1],(10) wherep=c−(k+1)b[8]R.Stanley,Hyperplane arrangements,interval orders,and trees,Proc.Nat.Acad.Sci.93(1996),2620–2625.[9]R.Stanley,Hyperplane arrangements and tree inversions,Mathemat-ical essays in honor of Gian-Carlo Rota(Cambridge,MA,1996),359–375,Progr.Math.,161,Birkh¨a user Boston,Boston,MA,1998. [10]R.Stanley,Parking functions and noncrossing partitions,Electronic J.Combinatorics4,R20(1997).[11]C.H.Yan,Generalized tree inversions and k-parking functions.J.Combin.Theory Ser.A79(1997),No.2,268–280.9。
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.。
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).。
ProofoftheFolkTheorem:的民间定理的证明
Proof of the Folk TheoremHere is a sketch of the proof of the folk theorem by Shoham,following Osborne and Rubinstein.First,some notation and definitions.Consider a game G =(N,(A i ),(u i )).Let v i is min max value for player i ,i.e.v i =min a −i ∈A −i max a i ∈A i u i (a −i ,a i )Notice that in this definition players are not allowed to randomize.If we use the standard definition of v i that allows randomizations,the Folk theorem would still hold,though proofs would become more involved.(Intuitively,randomization isn’t needed in the repeated setting,since we can simulate frequencies of play).We say that a payoffprofile (r i )is enforceable if r i ≥v i .We say that it is feasible if r i can be written as a ∈A αa u i (a ),where A is the joint action space of G ,for some αa ’s that are rational,non-negative,and a ∈A αa =1.(i.e.(r i )is a convex rational combination of all outcomes in G ).Now we are ready to prove two parts of the theorem.(The Folk Theorem:Part 1)For any game G and any payoffpair (r i ),if (r i )is a Nash equilibrium payoffprofile of the average reward infinitely repeated game of G then it is an enforceable payoffprofile of GProof:Suppose (r i )is not enforceable,i.e.r i <v i for some i .Then consider a deviation of player i to b i (s −i (h ))for any history h of the repeated game,where b i is any best-response action in the one-shot game and s −i (h )is the (repeated)equilibrium strategy of other players.By definition of b i ,player i would receive a payoffof at least v i in every stage game using this strategy.Thus,his payoffon average would also be at least v i >r i ,and hence (r i )couldn’t be a Nash equilibrium,completing the proof.(The Folk Theorem:Part 2)Suppose (r i )is a feasible enforceable payoffprofile of G .Then it is a payoffprofile of some equilibrium of the average reward infinitely repeated game of G .Proof:Suppose (r 1,r 2)is a feasible enforceable payoffprofile.Then we can write it as r i = a ∈A (βa γ)u i (a ),where βa and γare non-negative integers.(Recall that αa were required to be rational.So we can take γto be their common denominator).Since the combination was convex,we have γ= a ∈A βa .1We’re going to construct a strategy profile that will cycle through all outcomes a∈A of G with cycles of lengthγ,each cycle repeating action a exactlyβa times.Let(a t)be such a sequence of outcomes.Let’s define a strategy s i of player i to be a grim(trigger)version of playing(a t):if nobody deviates,then s i plays a t i in period t.However,if there was a period t in which some player j=i deviated,then s i will play(p−j)i,where(p−j)is a solution to the minimization problem in the definition of v j.First note,that if everybody plays according to s i,then,by construction,player i receives average payoffof r i(look at averages over periods of lengthγ).It is easy to see that this strategy profile is a Nash equilibrium:suppose everybody plays according to s i,and player j deviates at some point.Then,forever after,player j will receive his min max payoffv j≤r j,rendering the deviation unprofitable.。
Let Then
The Stickelberger IdealRobin Chapman7November1999I give a simple proof of the theorem of Iwasawa,that the index of an ideal defined via the Stickelberger element associated to the Galois group of Q(ζp n)/Q is the minus part of the class number of Q(ζp n).This is Theorem 6.19in[1].However Ifind Washington’s proof to be overcomplicated and inelegant.Instead of providing a global approach he gives a local proof at each prime.In effect that he proves the result three times,each case varying with the idiosyncrasy of the prime involved.In this note I give a direct global proof,essentially a simplification of Washington’s.I am indebted to Franz Lemmermeyer for pointing out some errors in an earlier version of this note.We need to define our notation.Let p n be a power of the odd prime p. Let G be a group isomorphic to(Z/p n Z)∗.Letσa be the element of G corresponding to the integer a coprime to p.(The notation comes from the standard identification of G with the Galois group of Q(ζp n)/Q,but we shall not need this.)Let R=Z[G]be the integral group ring of G.We define the Stickelberger element asθ=1p np na=1p aaσ−1a.Note thatθ∈Q[G]butθ/∈R.The Stickelberger ideal is defined as I= Rθ∩R.LetI ={α∈R:αθ∈R}so that I=I θ.Then I is an ideal of R which we wish to identify.To this end define a ring homomorphismφ:R→Z/p n Z byφ(σa)=a.Thenφis surjective so its kernel has index p n in R.Lemma1The ideal I is the kernel ofφ:R→Z/p n Z.Proof This is a reformulation of Lemma6.9in[1].1Letα=p nb=1p bx bσb∈R.Thenp nαθ=p na=1p ap nb=1p bax bσ−1aσb=p nc=1p cσcp na=1p aax ac.Ifαθ∈R then the coefficient ofσ1in p nαθis divisible by p n and sop na=1p aax a≡0(mod p n)or equivalentlyφ(α)=0.Conversely,ifφ(α)=0then the coefficient ofσ1 inαθis an integer.But the coefficient ofσc inαθis also the coefficient ofσ1inασ−1c θ.But asφis a homomorphism,φ(α)=0implies thatφ(ασ−1c)=0,and so the coefficient ofσc inαθis an integer.Henceαθ∈R.To summarize,αθ∈R if and only ifφ(α)=0,as required.Following Washington define J=σ−1,and letR−={α∈R:Jα=−α}.Suppose thatα=p na=1p ax aσa∈R.Thenα∈R−if and only if x p n−a=−x a for each a,equivalently,if and onlyifα=(1−J)(p n−1)/2a=1p ax aσa.Hence R−⊆(1−J)R,and the reverse inclusion is obvious,and so R−= (1−J)R.Define I−=I∩R−.Theorem6.19of[1]states:Theorem1(Iwasawa)We have|R−:I−|=h−(Q(ζp n)).Here h−(Q(ζp n))is the minus part of the class number of Q(ζp n),defined to be h(Q(ζp n))/h(Q(ζp n+ζ−1p n)).For the proof of theorem1we require a couple of lemmas.2Lemma2We have I ∩R−=(1−J)I and|R−:(1−J)I |=p n.Proof Certainly(1−J)I ⊆I ∩R−.Letα∈I ∩R−.Thenα=(1−J)βwhereβ∈R,since R−=(1−J)R.Then0=φ(α)=φ((1−J)β)=(1−φ(J))φ(β)=2φ(β).As2is a unit in Z/p n Z,thenφ(β)=0and soβ∈I .Thusα∈(1−J)I and so I ∩R−=(1−J)I .The set(1−J)I is thus the intersection of of R−and the kernel ofφ.But 1−J∈R−andφ(1−J)=2,which is a unit in Z/p n Z.Thusφ(R−)=Z/p n Z and so|R−:(1−J)I |=p n.Define a homomorphismψ:R→Z byψ(σa)=1for each a.LetN=σ∈Gσbe the norm element of R.Thenσa N=N for each a,and soαN=ψ(α)N for eachα∈R.Lemma3We have2I−⊆(1−J)I θand|(1−J)I θ:2I−|=2.Proof Letα∈I−.Thenα=βθfor someβ∈I .Also Jα=−αand so2α=(1−J)α=(1−J)βθ∈(1−J)I θ.Hence2I−⊆(1−J)I θ.Now letγ∈I .Then2γθ=(1+J)γθ+(1−J)γθ.But(1+J)θ=1p np na=1p a[a+(p n−a)]σa=N.Hence1+J∈I and N∈I.We thus have2γθ=γN+(1−J)γθ=gN+(1−J)γθwhere g=ψ(γ).If g is even,then(1−J)γθ=2γθ−gN∈2I and so (1−J)γθ∈2I−.If g is odd then(1−J)γθ−N=2γθ−(g+1)N∈2I and so(1−J)γθ/∈2R and a fortiori(1−J)γθ/∈2I−.3It follows that2I−={(1−J)γθ:γ∈I andψ(γ)is even}.As p n∈I andψ(p n)is odd,the set{(1−J)γθ:γ∈I andψ(γ)is odd}is nonempty,and is thus a coset of2I−disjoint from2I−.Thus|(1−J)I θ:2I−|=2which proves the lemma. Proof of Theorem1We calculate what we shall denote|(1−J)I : (1−J)I θ|.It is not clear whether this is an index,as it might not be the case that(1−J)I θ⊆(1−J)I .However if A and B are free abelian groups of rank r,each spanning a Q-vector space V of dimension r,then|A:A∩B| and|B:A∩B|are bothfinite and if we define|A:B|=|A:A∩B| |B:A∩B|then|A:B|has the same formal properties as the index.In particular if T: V→V is a non-singular linear transformation,then|A:T(A)|=|det(T)|. LetV=Q[G]−={α∈Q[G]:Jα=−α}.Then V is a Q-vector space of dimension r=|G|/2.Also R−⊆V and R−has rank r.As|R−:(1−J)I |=p n then(1−J)I has rank r too.Consider T:V→V given by T(α)=αθ.We can compute the determinant of T by extending T to a linear map onC[G]−={α∈C[G]:Jα=−α}.For a characterχ:G→C∗defineχ=1|G|σ∈Gχ(σ)σ−1.Then C[G]−has as a basis the set of χfor the odd charactersχ,those with χ(J)=−1.Butσ χ=χ(σ) χfor eachσandχ,and soχθ=B1,χ χ4whereB1,χ=1p np na=1p aaχ(σa).As these χform a basis of eigenvectors of T thendet(T)=χ(J)=−1B1,χ.Let h−=h−(Q(ζp n)).By an argument based on the analytic class number formula[1,Theorem4.17]h−=2p nχ(J)=−1(−B1,χ/2)and so|det(T)|=2r h−2p n=0.It follows that(1−J)I θhas rank r and|(1−J)I :(1−J)I θ|=2r h−/(2p n). But|R−:I−|=|R−:(1−J)I ||(1−J)I :(1−J)I θ||(1−J)I θ:2I−||I−:2I−|−1. We have seen that|R−:(1−J)I |=p n(Lemma2),|(1−J)I :(1−J)I θ|= 2r h−/(2p n)and|(1−J)I θ:2I−|=2(Lemma3).As I−has rank r then |I−:2I−|=2r.Putting all these pieces together we get|R−:I−|=h−,as required.Lemmermeyer has informed me that with suitable modifications this proof is also valid for p=2.If p=2and n≥2(to avoid trivialities)wefind that in Lemma2we|I ∩R−:(1−J)I |=2but that|R−:(1−J)I |=2n.In Lemma3wefind that(1−J)I θ=2I−.Finally in the proof of Theorem1 we need that the anlaytic class number formula gives us|det(T)|=2r h−/2n. References[1]Lawrence C.Washington,Introduction to Cyclotomic Fields,Springer-Verlag,1982,1997.5。
庞加莱关于数学发现的心理学的演讲
{经典演讲}庞加莱关于数学发现的心理学的演讲认知神经科学家,但演讲中的使用的类比以及描述基本上都是靠谱的。
其中最有意思的是他也提到了自己的几次顿悟的瞬间(其中有一次就是著名的踏上马车一瞬间想到解的那次)。
庞加莱认为下意识里面会对问题的各个元素(条件)进行组合,然后根据人对于知识的某种美感上的偏好筛选出来,那些足够"美"的东西就会浮上意识层面,于是产生顿悟。
这也是我看了一些认知科学的书之后得到的说法。
但此外庞加莱同时也认为下意识进行的探索是相当多的,他认为也许远远大于意识层面进行的探索(组合)。
而我倾向于认为下意识层面能进行的逻辑推理是有限远的,一般一到两步就了不得了。
下意识里面更多的进行的是某种模糊的模式匹配,或者说模糊联想。
这就是为什么对问题有一个全局感性认识那么重要的原因,这样的认识足够模糊足够全局,有助于提取出重要的相关知识来。
此外,一个总体的认识往往包含了问题的最重要(往往也是最本质的)要素,将这些要素同时装进工作记忆有着非常重要的意义——使它们有机会组合在一起,衍生出新的知识。
否则就是陷在在问题的某个局部(某几个局部条件)下,得到不相干的知识。
另外他也提到了对问题整体理解的另一个好处:当你对解的大致过程有了一个整体认识之后,即便缺乏某个局部的细节,也可以在这个整体视图的指导下将其推导出来(填充出brain is it that can compose the propositions and systems of mathematics?How do the mental processes of the geometer or algebraist compare with those of the musician,the poet,the painter,the chess player?In mathematical creation which are the key elements?Intuition?An exquisite sense of space and time?The precision of a calculating machine?A powerful memory? Formidable skill in following complex logical sequences?A supreme capacity for concentration?The essay below,delivered in the firstyears of this century as a lecture before the Psychological Society in Paris,is the most celebrated of the attempts to describe what goes on in the mathematician's brain.Its author,Henri Poincaré,cousin of Raymond, the politician,was peculiarly fitted to undertake the task.One of the foremost mathematicians of all time,unrivaled as an analyst and mathematical physicist,Poincaréwas known also as a brilliantly lucid expositor of the philosophy of science.These writings are of the first importance as professional treatises for scientists and are at the same time accessible,in large part,to the understanding of the thoughtful layman.Poincaréon Mathematical CreationThe genesis of mathematical creation is a problem which should intensely interest the psychologist.It is the activity in which the human mind seems to take least from the outside world,in which it acts or seems to act only of itself and on itself,so that in studying the procedure of geometric thought we may hope to reach what is most essential in man's mind...A first fact should surprise us,or rather would surprise us if we were not so used to it. How does it happen there are people who donot understand mathematics?If mathematics invokes only the rules of logic,such as are accepted by all normal minds;if its evidence is based on principles common to all men, and that none could deny without being mad, how does it come about that so many persons are here refractory?That not every one can invent is nowise mysterious.That not every one can retain a demonstration once learned may also pass. But that not every one can understand mathematical reasoning when explained appears very surprising when we think of it.And yet those who can follow this reasoning only with difficulty are in the majority;that is undeniable,and will surely not be gainsaid by the experience of secondary-school teachers.And further:how is error possible in mathematics?A sane mind should not be guilty of a logical fallacy,and yet there are very fine minds who do not trip in brief reasoning such as occurs in the ordinary doings of life,and who are incapable of following or repeating without error the mathematical demonstrations which arelonger,but which after all are only an accumulation of brief reasonings wholly analogous to those they make so easily. Need we add that mathematicians themselves are not infallible?...As for myself,I must confess,I am absolutely incapable even of adding without mistakes...My memory is not bad,but it would be insufficient to make me a good chess-player.Why then does it not fail me in a difficult piece of mathematical reasoning where most chess-players would lose themselves?Evidently because it is guidedby the general march of the reasoning.A mathematical demonstration is not a simple juxtaposition of syllogisms,it is syllogisms placed in a certain order,and the order in which these elements are placed is much more important than the elements themselves.If I have the feeling,the intuition, so to speak,of this order,so as to perceive at a glance the reasoning as a whole,I need no longer fear lest I forget one of the elements, for each of them will take its allotted place in the array,and that without any effort of memory on my part.We know that this feeling,this intuition of mathematical order,that makes us divine hidden harmonies and relations,cannot be possessed by every one.Some will not have either this delicate feeling so difficult to define, or a strength of memory and attention beyond the ordinary,and then they will be absolutely incapable of understanding higher mathematics.Such are the majority.Others will have this feeling only in a slight degree, but they will be gifted with an uncommon memory and a great power of attention.They will learn by heart the details one after another;they can understand mathematics and sometimes make applications,but they cannot create.Others,finally,will possess ina less or greater degree the special intuition referred to,and then not only can they understand mathematics even if their memory is nothing extraordinary,but they may become creators and try to invent with more or less success according as this intuition is more or less developed in them.In fact,what is mathematical creation?It does not consist in making new combinations with mathematical entities already known. Anyone could do that,but the combinations so made would be infinite in number and most of them absolutely without interest.Tocreate consists precisely in not making useless combinations and in making those which are useful and which are only a small minority.Invention is discernment,choice.It is time to penetrate deeper and to see what goes on in the very soul of the mathematician.For this,I believe,I can do best by recalling memories of my own.But I shall limit myself to telling how I wrote my first memoir on Fuchsian functions.I beg the reader's pardon;I am about to use some technical expressions,but they need not frighten him,for he is not obliged tounderstand them.I shall say,for example, that I have found the demonstration of such a theorem under such circumstances.This theorem will have a barbarous name, unfamiliar to many,but that is unimportant; what is of interest for the psychologist is not the theorem but the circumstances.For fifteen days I strove to prove that there could not be any functions like those I have since called Fuchsian functions.I was then very ignorant;every day I seated myself at my work table,stayed an hour or two,tried a great number of combinations and reachedno results.One evening,contrary to my custom,I drank black coffee and could not sleep.Ideas rose in crowds;I felt them collide until pairs interlocked,so to speak,making a stable combination.By the next morning I had established the existence of a class of Fuchsian functions,those which come from the hypergeometric series;I had only to write out the results,which took but a few hours.Then I wanted to represent these functions by the quotient of two series;this idea was perfectly conscious and deliberate, the analogy with elliptic functions guided me.I asked myself what properties these series must have if they existed,and I succeeded without difficulty in forming the series I have called theta-Fuchsian.Just at this time I left Caen,where I was then living,to go on a geologic excursion under the auspices of the school of mines. The changes of travel made me forget my mathematical work.Having reached Coutances,we entered an omnibus to go some place or other.At the moment when I put my foot on the step the idea came to me, without anything in my former thoughtsseeming to have paved the way for it,that the transformations I had used to define the Fuchsian functions were identical with those of non-Euclidean geometry.I did not verify the idea;I should not have had time,as,upon taking my seat in the omnibus,I went on with a conversation already commenced,but I felt a perfect certainty.On my return to Caen,for conscience's sake I verified the result at my leisure.Then I turned my attention to the study of some arithmetical questions apparently without much success and without asuspicion of any connection with my preceding researches.Disgusted with my failure,I went to spend a few days at the seaside,and thought of something else.One morning,walking on the bluff,the idea came to me,with just the same characteristics of brevity,suddenness and immediate certainty that the arithmetic transformations of indeterminate ternary quadratic forms were identical with those of non-Euclidean geometry.Returned to Caen,I meditated on this result and deduced the consequences.Theexample of quadratic forms showed me that there were Fuchsian groups other than those corresponding to the hypergeometric series;I saw that I could apply to them the theory of theta-Fuchsian series and that consequently there existed Fuchsian functions other than those from the hypergeometric series,the ones I then knew.Naturally I set myself to form all these functions.I made a systematic attack upon them and carried all the outworks, one after another.There was one,however, that still held out,whose fall would involve that of the whole place.But all my efforts only served at first the better to show me the difficulty,which indeed was something.All this work was perfectly conscious.Thereupon I left for Mont-Valérien,where I was to go through my military service;so I was very differently occupied.One day,going along the street,the solution of the difficulty which had stopped me suddenly appeared to me.I did not try to go deep into it immediately, and only after my service did I again take up the question.I had all the elements and had only to arrange them and put them together. So I wrote out my final memoir at a single stroke and without difficulty.I shall limit myself to this single example;it is useless to multiply them...Most striking at first is this appearance of sudden illumination,a manifest sign of long, unconscious prior work.The role of this unconscious work in mathematical invention appears to me incontestable,and traces of it would be found in other cases where it is less evident.Often when one works at a hard question,nothing good is accomplished at the first attack.Then one takes a rest,longer or shorter,and sits down anew to the work. During the first half-hour,as before,nothing is found,and then all of a sudden the decisiveidea presents itself to the mind...There is another remark to be made about the conditions of this unconscious work; it is possible,and of a certainty it is only fruitful,if it is on the one hand preceded and on the other hand followed by a period of conscious work.These sudden inspirations (and the examples already cited prove this) never happen except after some days of voluntary effort which has appeared absolutely fruitless and whence nothing good seems to have come,where the way taken seems totally astray.These efforts then havenot been as sterile as one thinks;they have set agoing the unconscious machine and without them it would not have moved and would have produced nothing...Such are the realities;now for the thoughts they force upon us.The unconscious,or,as we say,the subliminal self plays an important role in mathematical creation;this follows from what we have said. But usually the subliminal self is considered as purely automatic.Now we have seen that mathematical work is not simply mechanical, that it could not be done by a machine,however perfect.It is not merely a question of applying rules,of making the most combinations possible according to certain fixed laws.The combinations so obtained would be exceedingly numerous,useless and cumbersome.The true work of the inventor consists in choosing among these combinations so as to eliminate the useless ones or rather to avoid the trouble of making them,and the rules which must guide this choice are extremely fine and delicate.It is almost impossible to state them precisely; they are felt rather than formulated.Under these conditions,how imagine a sieve capable of applying them mechanically?A first hypothesis now presents itself;the subliminal self is in no way inferior to the conscious self;it is not purely automatic;it is capable of discernment;it has tact,delicacy; it knows how to choose,to divine.What do I say?It knows better how to divine than the conscious self,since it succeeds where that has failed.In a word,is not the subliminal self superior to the conscious self?You recognize the full importance of this question...Is this affirmative answer forced upon us by the facts I have just given?I confess that,for my part,I should hate to accept it.Re-examine the facts then and see if they are not compatible with another explanation.It is certain that the combinations which present themselves to the mind in a sort of sudden illumination,after an unconscious working somewhat prolonged,are generally useful and fertile combinations,which seem the result of a first impression.Does it follow that the subliminal self,having divined by a delicate intuition that these combinations would be useful,has formed only these,or has it rather formed many others which werelacking in interest and have remained unconscious?In this second way of looking at it,all the combinations would be formed in consequence of the automatism of the subliminal self,but only the interesting ones would break into the domain of consciousness.And this is still very mysterious.What is the cause that,among the thousand products of our unconscious activity,some are called to pass the threshold, while others remain below?Is it a simple chance which confers this privilege?Evidently not;among all the stimuli of our senses,for example,only the most intense fix our attention,unless it has been drawn to them by other causes.More generally the privileged unconscious phenomena,those susceptible of becoming conscious,are those which,directly or indirectly,affect most profoundly our emotional sensibility.It may be surprising to see emotional sensibility invokedàpropos of mathematical demonstrations which,it would seem,can interest only the intellect.This would be to forget the feeling of mathematical beauty,ofthe harmony of numbers and forms,of geometric elegance.This is a true esthetic feeling that all real mathematicians know, and surely it belongs to emotional sensibility.Now,what are the mathematic entities to which we attribute this character of beauty and elegance,and which are capable of developing in us a sort of esthetic emotion? They are those whose elements are harmoniously disposed so that the mind without effort can embrace their totality while realizing the details.This harmony is at once a satisfaction of our esthetic needs and anaid to the mind,sustaining and guiding.And at the same time,in putting under our eyes a well-ordered whole,it makes us foresee a mathematical law...Thus it is this special esthetic sensibility which plays the role of the delicate sieve of which I spoke,and that sufficiently explains why the one lacking it will never be a real creator.Yet all the difficulties have not disappeared.The conscious self is narrowly limited,and as for the subliminal self we know not its limitations,and this is why we are not too reluctant in supposing that it has beenable in a short time to make more different combinations than the whole life of a conscious being could encompass.Yet these limitations exist.Is it likely that it is able to form all the possible combinations,whose number would frighten the imagination? Nevertheless that would seem necessary, because if it produces only a small part of these combinations,and if it makes them at random,there would be small chance that the good,the one we should choose,would be found among them.Perhaps we ought to seek theexplanation in that preliminary period of conscious work which always precedes all fruitful unconscious labor.Permit me a rough comparison.Figure the future elements of our combinations as something like the hooked atoms of Epicurus.During the complete repose of the mind,these atoms are motionless,they are,so to speak,hooked to the wall...On the other hand,during a period of apparent rest and unconscious work,certain of them are detached from the wall and put in motion.They flash in every direction throughthe space(I was about to say the room) where they are enclosed,as would,for example,a swarm of gnats or,if you prefer a more learned comparison,like the molecules of gas in the kinematic theory of gases.Then their mutual impacts may produce new combinations.What is the role of the preliminary conscious work?It is evidently to mobilize certain of these atoms,to unhook them from the wall and put them in swing.We think we have done no good,because we have moved these elements a thousand different ways inseeking to assemble them,and have found no satisfactory aggregate.But,after this shaking up imposed upon them by our will, these atoms do not return to their primitive rest.They freely continue their dance.Now,our will did not choose them at random;it pursued a perfectly determined aim.The mobilized atoms are therefore not any atoms whatsoever;they are those from which we might reasonably expect the desired solution.Then the mobilized atoms undergo impacts which make them enter into combinations among themselves or withother atoms at rest which they struck against in their course.Again I beg pardon,my comparison is very rough,but I scarcely know how otherwise to make my thought understood.However it may be,the only combinations that have a chance of forming are those where at least one of the elements is one of those atoms freely chosen by our will.Now,it is evidently among these that is found what I called the good combination. Perhaps this is a way of lessening the paradoxical in the original hypothesis...I shall make a last remark:when above I made certain personal observations,I spoke of a night of excitement when I worked in spite of myself.Such cases are frequent,and it is not necessary that the abnormal cerebral activity be caused by a physical excitant as in that I mentioned.It seems,in such cases, that one is present at his own unconscious work,made partially perceptible to theover-excited consciousness,yet without having changed its nature.Then we vaguely comprehend what distinguishes the two mechanisms or,if you wish,the working methods of the two egos.And thepsychologic observations I have been able thus to make seem to me to confirm in their general outlines the views I have given.Surely they have need of[confirmation], for they are and remain in spite of all very hypothetical:the interest of the questions is so great that I do not repent of having submitted them to the reader.。
数学专业英语课后部分习题答案
2.1 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。
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, then the OD between odd prime number which is less than 2m+3 is Pmand odd composite point 2m+3 is either a PL or a CL. prime point PmWhen the OD between odd prime point Pand odd composite point 2m+3 isma 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 Pand odd composite point 2m+3 is amCL, 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 3Pm.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 ofthe longest RPL Pm whereby odd prime point Pmacts as the right endmostpoint.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 Pm, then odd point 2m-ỹ-2 canonly lie on the left of odd prime point Pm. This shows that every RPL2m-ỹ-2 atOD 3(2m-ỹ-2) is shorter than PL 3 Pm.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.(2m+3) is a CL, likewise OD Thus far we have proven that even if OD Pm3(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.。
数学专业英语 第2章课后答案
2.12.比:ratio 比例:proportion 利率:interest rate 速率:speed 除:divide 除法:division 商:quotient 同类量:like quantity 项:term 线段:line segment 角:angle 长度:length 宽:width高度:height 维数:dimension 单位:unit 分数:fraction 百分数:percentage3.(1)一条线段和一个角的比没有意义,他们不是相同类型的量.(2)比较式通过说明一个量是另一个量的多少倍做出的,并且这两个量必须依据相同的单位.(5)为了解一个方程,我们必须移项,直到未知项独自处在方程的一边,这样就可以使它等于另一边的某量.4.(1)Measuring the length of a desk, is actually comparing the length of the desk to that of a ruler.(3)Ratio is different from the measurement, it has no units. The ratio of the length and the width of the same book does not vary when the measurement unit changes.(5)60 percent of students in a school are female students, which mean that 60 students out of every 100 students are female students.2.22.初等几何:elementary geometry 三角学:trigonometry 余弦定理:Law of cosines 勾股定理/毕达哥拉斯定理:Gou-Gu theorem/Pythagoras theorem 角:angle 锐角:acute angle 直角:right angle 同终边的角:conterminal angles 仰角:angle of elevation 俯角:angle of depression 全等:congruence 夹角:included angle 三角形:triangle 三角函数:trigonometric function直角边:leg 斜边:hypotenuse 对边:opposite side 临边:adjacent side 始边:initial side 解三角形:solve a triangle 互相依赖:mutually dependent 表示成:be denoted as 定义为:be defined as3.(1)Trigonometric function of the acute angle shows the mutually dependent relations between each sides and acute angle of the right triangle.(3)If two sides and the included angle of an oblique triangle areknown, then the unknown sides and angles can be found by using the law of cosines.(5)Knowing the length of two sides and the measure of the included angle can determine the shape and size of the triangle. In other words, the two triangles made by these data are congruent.4.(1)如果一个角的顶点在一个笛卡尔坐标系的原点并且它的始边沿着x轴正方向,这个角被称为处于标准位置.(3)仰角和俯角是以一条以水平线为参考位置来测量的,如果正被观测的物体在观测者的上方,那么由水平线和视线所形成的角叫做仰角.如果正被观测的物体在观测者的下方,那么由水平线和视线所形成的的角叫做俯角.(5)如果我们知道一个三角形的两条边的长度和对着其中一条边的角度,我们如何解这个三角形呢?这个问题有一点困难来回答,因为所给的信息可能确定两个三角形,一个三角形或者一个也确定不了.2.32.素数:prime 合数:composite 质因数:prime factor/prime divisor 公倍数:common multiple 正素因子: positive prime divisor 除法算式:division equation 最大公因数:greatest common divisor(G.C.D) 最小公倍数: lowest common multiple(L.C.M) 整除:divide by 整除性:divisibility 过程:process 证明:proof 分类:classification 剩余:remainder辗转相除法:Euclidean algorithm 有限集:finite set 无限的:infinitely 可数的countable 终止:terminate 与矛盾:contrary to3.(1)We need to study by which integers an integer is divisible, that is , what factor it has. Specially, it is sometime required that an integer is expressed as the product of its prime factors.(3)The number 1 is neither a prime nor a composite number;A composite number in addition to being divisible by 1 and itself, can also be divisible by some prime number.(5)The number of the primes bounded above by any given finite integer N can be found by using the method of the sieve Eratosthenes.4.(1)数论中一个重要的问题是哥德巴赫猜想,它是关于偶数作为两个奇素数和的表示.(3)一个数,形如2p-1的素数被称为梅森素数.求出5个这样的数.(5)任意给定的整数m和素数p,p的仅有的正因子是p和1,因此仅有的可能的p和m的正公因子是p和1.因此,我们有结论:如果p是一个素数,m是任意整数,那么p整除m,要么(p,m)=1.2.42.集:set 子集:subset 真子集:proper subset 全集:universe 补集:complement 抽象集:abstract set 并集:union 交集:intersection 元素:element/member 组成:comprise/constitute包含:contain 术语:terminology 概念:concept 上有界:bounded above 上界:upper bound 最小的上界:least upper bound 完备性公理:completeness axiom3.(1)Set theory has become one of the common theoretical foundation and the important tools in many branches of mathematics.(3)Set S itself is the improper subset of S; if set T is a subset of S but not S, then T is called a proper subset of S.(5)The subset T of set S can often be denoted by {x}, that is, T consists of those elements x for which P(x) holds.(7)This example makes the following question become clear, that is, why may two straight lines in the space neither intersect nor parallel.4.(1)设N是所有自然数的集合,如果S是所有偶数的集合,那么它在N中的补集是所有奇数的集合.(3)一个非空集合S称为由上界的,如果存在一个数c具有属性:x<=c对于所有S中的x.这样一个数字c被称为S的上界.(5)从任意两个对象x和y,我们可以形成序列(x,y),它被称为一个有序对,除非x=y,否则它当然不同于(y,x).如果S和T是任意集合,我们用S*T表示所有有序对(x,y),其中x术语S,y属于T.在R.笛卡尔展示了如何通过实轴和它自己的笛卡尔积来描述平面的点之后,集合S*T被称为S和T的笛卡尔积.2.52.竖直线:vertical line 水平线:horizontal line 数对:pairs of numbers 有序对:ordered pairs 纵坐标:ordinate 横坐标:abscissas 一一对应:one-to-one 对应点:corresponding points圆锥曲线:conic sections 非空图形:non vacuous graph 直立圆锥:right circular cone 定值角:constant angle 母线:generating line 双曲线:hyperbola 抛物线:parabola 椭圆:ellipse退化的:degenerate 非退化的:nondegenerate任意的:arbitrarily 相容的:consistent 在几何上:geometrically 二次方程:quadratic equation 判别式:discriminant 行列式:determinant3.(1)In the planar rectangular coordinate system, one can set up aone-to-one correspondence between points and ordered pairs of numbers and also a one-to-one correspondence between conic sections and quadratic equation.(3)The symbol can be used to denote the set of ordered pairs(x,y)such that the ordinate is equal to the cube of the abscissa.(5)According to the values of the discriminate,the non-degenerate graph of Equation (iii) maybe known to be a parabola, a hyperbolaor an ellipse.4.(1)在例1,我们既用了图形,也用了代数的代入法解一个方程组(其中一个方程式二次的,另一个是线性的)。
福尔摩斯的名言英语
福尔摩斯的名言英语大全很荣幸同学们能来关注福尔摩斯的名言英语励志名言句子内容,由为大家搜集整理发布,让我们赶快一起来学习一下吧!一、我追求精神上的兴奋。
I pursue spiritual excitement.二、很多被绞死的人死的冤枉。
(.)Many of the hanged man is dead wrong.三、嫉妒是性格的奇异转换器。
Jealousy is a singular converter character.四、感情作用会影响糊涂的理智。
The emotional qualities will affect to clear reasoning.五、不用动脑筋,我就活不下去。
Don't think, I can't live.六、虚心和傲慢都是与事实相背而行的。
Modesty and pride are back and done with facts.七、除去其他的因素,剩下的必是事实了。
To get rid of other factors, the rest must be the truth.八、我一直不作任何例外。
定律没有例外。
I never make any exception. Law is no exception.九、对于装病,我认为我有力量写个专论。
For faking it, I think I have the ability to write a monograph.十、生活之谜是任何大脑也创造不出来的。
The riddle of life is beyond the contrivance of any brain also. 十一、这是一个要抽足三斗烟才能解决的问题。
This is a smoke foot three measures to solve the problem.十二、再也没有比显而易见的事实更虚伪的了。
No more hypocritical than obvious fact.十三、世上的事都是前人做过的,没什么新奇的。
数学专业英语吴炯圻第版
This does not mean that one should not make use of geometry in studying properties of real numbers. On the contrary, the geometry often suggests the method of proof of a particular theorem, and sometimes a geometric argument is more illuminating than a purely analytic proof (one depending entirely on the axioms for the real numbers). 这并不意味着研究实数的性质时不会应用到几何。相 反,几何经常会为证明一些定理提供思路,有时几何 讨论比纯分析式的证明更清楚。
毫无疑问,读者都熟悉通过在直线上描点的方式表 示实数的几何意义。如图2-4-1所示,选择一个点表 示0,在0右边的另一个点表示1。这种做法决定了刻 度。
第12页/共20页
If one adopts an appropriate set of axioms for Euclidean geometry, then each real number corresponds to exactly one point on this line and, conversely, each point on the line corresponds to one and only one real number.
现在我们来定义正整数,就是属于每一个归纳集的实数。
第6页/共20页
Let P denote the set of all positive integers. Then P is itself an inductive set because (a) it contains 1, and (b) it contains x+1 whenever it contains x. Since the members ive set, we refer to P as the smallest inductive set.
洛克菲勒名言名句英文
洛克菲勒名言名句英文第一篇:洛克菲勒名言名句英文相关As a student who is interested in business, I have always been inspired by the wisdom and philosophy of John D. Rockefeller. Throughout his life, Rockefeller was known for his incredible success in the oil industry, but he was also known for the insights that he shared with others. These insights, often captured in his famous quotes, have become an invaluable resource for anyone seeking to succeed in business and in life.One of Rockefeller's most famous quotes is "Don't be afraid to give up the good to go for the great." This quote emphasizes the importance of taking risks and pursuing greatness, even if it means leaving behind something that is merely good or satisfactory. This is a valuable lesson for anyone, but particularly for business leaders. In business, it is easy to settle for mediocrity and to become complacent. However, Rockefeller's quote reminds us that true success requires us to constantly strive for greatness, even if it means giving up something that is comfortable or familiar. Another famous Rockefeller quote is "I believe in the dignity of labor, whether with head or hand; that the world owes no man a living but that it owes every man an opportunity to make a living." This quote speaks to the importance of hard work and the opportunities that are available to those who are willing to work for them. In today's fast-paced world, it is easy to become disillusioned and to feel that success is out of reach. However, Rockefeller's quote reminds us that success is not handed to us, but rather it is something that we must work for. By valuing hard workand seizing opportunities, we can achieve great things.In conclusion, John D. Rockefeller's famous quotes have had a profound impact on generations of business professionals and ordinary people alike. His insights on risk-taking, hard work, and the pursuit of greatness continue to inspire us to this day. As a student who is interested in business, I have found Rockefeller's teachings to be invaluable, and I strive to incorporate his wisdom into my own life and future career.重点分析:本篇文章主要介绍了对于洛克菲勒名言名句的理解和思考。
公布的英语短语
公布的英语短语“君如察臣婴之言,推君之盛德,公布之於天下,则汤武可为也。
”下面就由店铺为大家带来关于公布的英语短语集锦,希望大家能有所收获。
关于公布的相关短语公布细则 open specification;公布运价 published rate公布开会日期announce the date of the meeting;公布法令promulgate a decree;公布罪状announce sb.'s crimes;公布名单make public a name list公布价格 published price公布系统 public address system公布时间 Apabi released date ; date of publication信息公布 Information released ;成绩公布 announcement of results公布告诉 Releasing Notices充分公布 Full Disclosure速公布 instant-publishing公布计划 unveil a plan关于公布的相关例句1. He will have been heartened by the telephone opinion poll published yesterday.他会为昨天公布的电话民意调查结果感到鼓舞。
2. Police have released an E-fit picture of the suspected gunman.警方已公布持枪嫌犯的电子画像。
3. The Romanian government yesterday unveiled a new, scaled-down security force.罗马尼亚政府昨天对外公布了精简编制后新组建的安全部队的情况。
4. The official reason given for the President's absence was sickness.对外公布的总统未到场原因是身体有恙。
洛必达法则的证明
C A S E 1 f (a) = g(a) = 0. Since g (x) is non-zero near x = a, there is an interval (a, b) such that g (x) is
positive or negative for x ∈ (a, b), and therefore, g(x) is either strictly increasing or strictly decreasing on (a, b) (Figure 1). However, g(a) = 0, so g(x) itself is non-zero for x ∈ (a, b). Thus for all x ∈ (a, b), the hypotheses of Cauchy’s Mean Value Theorem for the interval [a, x] are satisfied and hence, there exists c ∈ (a, x) such that
h(a)
=
h(b)
=
f
(a)g(b) g(b)
− −
f (b)g(a) g(a)
Thus, the hypotheses of Rolle’s Theorem are satisfied and the conclusion holds, namely, there exists c ∈ (a, b) such that h (c) = 0. In other words,
f (x) g(x)
− −
f (a) g(a)
=
f g
(c) (c)
Since f (a) = g(a) = 0, this reduces to
多项式的正交性与其零点有界性的一个注记
其中 Φ0 ( z) = 1 . j = 1, 2, …, n 代入式( 1 ) 当中, 证明: 把 z n, j , j = 1, 2, …, n 是 Φ n 的零点, 由假设: z n, 不难推 j , 假设式 ( 2 ) 成立, 让 εn 代 出式( 2 ) 成立. 反过来, zΦ n -1 ( z) + ε n Φ * 表式( 2 ) 当中的比值. 那么, n -1 ( z) 是首 项 系 数 为 1 , 次 数 为 n 的 多 项 式, 记为 j = 1, 2, …, n 是其所有 Ψ n ( z) . 由式 ( 2 ) 知 z n, j , 零点, 这表明 Ψ n ( z) 和 Φ n ( z) 的零点完全一样, 又两 者 都 是 首 项 系 数 为 1 的 多 项 式, 从而必有 Ψ n ( z) = Φ n ( z) . 所以式 ( 2 ) 对 任 意 的 n ≥ 1 , n ∈N 成立. 证完. 接下来, 我们给出本文的中心结果. 定理 2 设 { Φ n ( z) }" 是次数为 n 的首一复 n =0 n∈N , 多项式序列, 若对每一个 n ≥ 1 , 式( 2 ) 成 * 立, 同时 Φ n ( z) 与 Φ n ( z) 的零点互不相同. 则下 面三个条件相互等价: ( i) 序列 { Φ n ( z) }" 关于唯一的有限的正波 n =0 且测度的支集是无限的. 雷尔测度正交, ( ii) | ε n | < 1 , 其中 - ε n 是式 ( 2 ) 当中的 比值. ( iii ) Φn ( z) 的 零 点 都 在 单 位 圆 内, 即 | zn, j = 1, 2, …, n. j | < 1, 证明: 由( i) 到( ii) 是平凡的, 而由( ii) 到 ( iii ) 是正交多项式的经典结果. 反过来, 由 ( ii ) 到 ( i ) 5]中定理 来自于单位圆上 Favard 定理, 见文献[ 8. 1 , p. 156 和文献[ 6]中定理 8. 3 , p. 140. 因此, 我们只需证明: 如果( iii) 成立, 那么( ii) 成立. j = 1, 2, …, n 代表 Φ n ( z) 的所有零 设 z n, j , 点, 他们当中的某些零点可以相等. 所以若( iii ) 成 j = 1, 2, …, n . 把 z n, j = 1, 立, 则有 | z n, j | < 1 , j , 2, …, n 代入式( 1 ) , 可得 | ε n | = | z n, j | Φ n - 1 ( z n, j) * Φ n - 1 ( z n, j)
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ANOTHER SIMPLE PROOF OF A THEOREM OF CHANDLER DAVIS IGOR RIVIN Abstract.In 1957,Chandler Davis proved that unitarily invari-ant convex functions on the space of hermitian matrices are pre-cisely those which are convex and symmetrically invariant on the set of diagonal matrices.We give a simple perturbation theoretic proof of this result.(Davis’argument was also very short,though based on completely different ideas).Consider an orthogonally invariant function f defined on the set of n ×n symmetric matrices.Such a function has to factor through the spectrum:(1)f (M )=g ◦λ(M ),where g is a symmetric function:(2)g (λ1,...,λn )=g (λσ(1),...,λσ(n )),for any permutation σ.In the sequel we shall further assume that f is a C 2convex function,and under this assumption we shall show that such functions are pre-cisely those decomposing as per Eq.(1),with convex g.The argument and the statement are identical for unitarily invariant functions of Her-mitian matrices;in that setting the theorem was proved in [Davis57],by a completely different argument (Davis made no regularity assump-tion,but this is easily dispensed with (see Remark 0.2)).
To show this,let M =P +tQ,and let ˜f P,Q (t )=f (M ).It is enough
to show that for any symmetric P,Q,
d 2˜f P,Q
Key words and phrases.convexity,symmetric,hermitian,orthogonally invariant.The author was partially supported by the Department of Mathematical Sciences of the National Science Foundation.He would like to thank Omar Hijab for helpful conversations.
1
2IGOR RIVIN
d˜f
∂λi ˙λi .
(3)
d2˜f
∂λi∂λj ˙λ
i
˙λ
j
+
n i=1∂g ∂λj
−∂g
λj−λi
.
The result then follows from the Lemma below.
Lemma0.1.Let f be a convex function such that f(x,y)=f(y,x), where y=x.Then
∂f
∂y
∂x −
∂f
∂x
−
∂f
ANOTHER SIMPLE PROOF OF A THEOREM OF CHANDLER DAVIS3
Remark0.2.Since smooth symmetric convex functions are dense in
the set of all symmetric functions,we have actually shown the result
for all convex symmetric functions,as follows:
•We start with a symmetric locally integrable function g on the
set of diagonal matrices.
•The function g can be extended(by orthogonal invariance)to
the set of all symmetric matrices,to obtain a function f.
•Convolving f with a C∞Gaussian approximation D n to the
delta function,then averaging over the orthogonal orbits we ob-
tain a convex orthogonally invariant function f n,such that the
f−f n ≤1/n,where denotes the sup-norm.The restric-
tion of f n to the diagonal matrices is smooth and convex(since
convolving with a positive kernel and averaging both preserve
convexity)function g n.
•We apply the smooth argument above to g n and f n,to show that
f n is a convex function.
•Since f is a sup-norm limit of convex functions f n,the result
follows.
Remark0.3.The computation above can also be extended to certain
infinite dimensional situations.In particular,it is not hard to see that
if P has a compact resolvent and Q is bounded then if the function
g is the the spectral zeta function or the logarithm of the determinant
of the operator.In particular,the regularized log det of the Laplacian
on a Riemann surface is convex with respect to bounded perturbations,
as are special values of the spectral zeta function.This is very much
not true for arbitrary perturbations:in particular,when P=Q=∆,
the convexity h log det depends on the sign of the spectral zeta function
at0,which is the same as the sign of the Euler characteristic,so that
log det is not convex with respect to the“diagonal”perturbation of the Laplacian whenever the Riemann surfvace has negative Euler charac-teristic.
References
[Davis57] C.Davis(1957)On invariant convex functions,Arch.Math,1957. [Kato80]T.Kato(1980).Perturbation Theory for Linear Operators,
Springer-Verlag Berlin-Heidelberg-New York.
Mathematics Department,Temple University,1805N Broad St,Philadel-phia,PA19122
Mathematics Department,Princeton University,Princeton,NJ08544 E-mail address:rivin@。