The proof of Tchakaloff's Theorem
微分中值定理的证明英文书
微分中值定理的证明英文书《The Proof of the Mean Value Theorem in Differential Calculus》The Mean Value Theorem is a fundamental result in differential calculus, which states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point in the open interval where the derivative of the function is equal to the average rate of change of the function over the closed interval. The theorem has many important applications in mathematics and science, and its proof is a key milestone in the study of differential calculus.To prove the Mean Value Theorem, we first need to consider the function f(x) and its derivative f'(x). Since f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), we can apply the Intermediate Value Theorem to the derivative f'(x) to show that it takes on every value between f'(a) and f'(b) at some point c in the open interval (a, b).Next, we consider the average rate of change of f(x) over the interval [a, b], which is given by the difference in the function values divided by the difference in the input values:Average rate of change = (f(b) - f(a))/(b - a)Now, we can construct a new function g(x) = f(x) - (f(b) - f(a))/(b - a) * x, which represents the difference between f(x) and the line with slope equal to the average rate of change. Notice that g(a) = f(a) and g(b) = f(b), which means g(x) is continuous on the closed interval and differentiable on the open interval.By applying Rolle's Theorem to the function g(x), we find that there exists at least one point d in the open interval (a, b) where g'(d) = 0. Simplifying g'(x) gives us g'(x) = f'(x) - (f(b) - f(a))/(b - a), and therefore g'(d) = f'(d) - (f(b) - f(a))/(b - a) = 0. Rearranging the terms, we get f'(d) = (f(b) - f(a))/(b - a), which is the desired result of the Mean Value Theorem.Therefore, we have successfully proven the Mean Value Theorem for the function f(x) over the interval [a, b] by using the properties of derivatives and applying Rolle's Theorem. This result has significant implications for the behavior of differentiable functions and is a fundamental tool in the study of calculus.。
in the proof of that 句子
in the proof of that 句子经典句子集锦：In the Proof of That1. 有关证明的句子•The proof of a mathematical theorem requires logical reasoning and evidence.•In order to validate a hypothesis, scientists must provide empirical proof.•Legal cases often rely on the presentation of valid proof to support a claim or accusation.•Historical records serve as proof of past events and can provide valuable insights into our collective memory. •Personal anecdotes may not be sufficient proof, but they can offer subjective perspectives and experiences.•In the court of law, the burden of proof lies with the prosecution.•Mathematical proofs provide the necessary steps and logic to demonstrate the validity of a theorem.•The scientific method requires rigorous testing and proof before a hypothesis can be widely accepted.2. 与证明相关的句子目标达到的句子•In the pursuit of truth, one must constantly seek justification and proof for their beliefs.•The accumulation of evidence is essential inestablishing solid proof.•The ultimate goal of a scientific experiment is to obtain conclusive proof that supports or refutes ahypothesis.•Through critical thinking and analysis, one can uncover the proof needed to support their argument.•The process of proving something requires careful examination and evaluation of the available facts.证明难度的句子•Some concepts or ideas are difficult to prove due to their abstract nature or lack of empirical evidence. •Certain claims may require extraordinary proof in order to be widely accepted by the scientific community.•The existence of a higher power or deity is a topic that has been the subject of philosophical debate, as itpresents inherent challenges in terms of proof.不同类型的证明句子•Mathematical proofs rely on logical deductions and mathematical principles to demonstrate the truth of astatement.•Empirical evidence plays a crucial role in scientific proofs, as it relies on observations and experimentation. •Historical documents and records are often used as primary sources of proof to support historical claims. •Personal testimonies and eyewitness accounts can provide individual proof of events or experiences.3. 证明的重要性和局限性证明的重要性•Proof provides a solid foundation upon which knowledge and understanding can be built.•Having proof gives credibility and validity to one’s claims or arguments.•The ability to provide evidence-based proof is essential in various fields, such as science, law, and academia. •Proof allows for replication and verification of results, which is crucial for the advancement of knowledge.证明的局限性•Despite its importance, proof has its limitations and may not always be attainable or conclusive.•Some concepts or phenomena may lie beyond the scope of proof due to inherent complexities or limitations inhuman understanding.•Lack of access to sufficient evidence or data can hinder the ability to provide conclusive proof.•Proof can be influenced by bias, subjectiveinterpretations, or the limitations of existingknowledge and methodologies.证明的局限性（续）•Some truths or beliefs may not be subject to proof, as they are based on personal values, emotions, or faith. •Proof is context-dependent and may vary across different disciplines and fields of study.•The burden of proof can be subjective and may differ depending on the cultural, social, or legal context. •The concept of proof is a human construct and may not hold the same weight or significance in other sentient beings or hypothetical scenarios.结论证明是构建知识和理解的基石，它需要逻辑推理和有效证据的支持。
Analysis and design of associative memories based on stability of cellular neural networks
This article appeared in a journal published by Elsevier.The attached copy is furnished to the author for internal non-commercial research and education use,including for instruction at the authors institutionand sharing with colleagues.Other uses,including reproduction and distribution,or selling or licensing copies,or posting to personal,institutional or third partywebsites are prohibited.In most cases authors are permitted to post their version of thearticle(e.g.in Word or Tex form)to their personal website orinstitutional repository.Authors requiring further informationregarding Elsevier’s archiving and manuscript policies areencouraged to visit:/copyrightAnalysis and design of associative memories based on stability of cellular neural networksQi Han a,n,Xiaofeng Liao b,Tingwen Huang c,Jun Peng a,Chuandong Li b,Hongyu Huang ba School of Electrical and Information Engineering,Chongqing University of Science and Technology,Chongqing401331,Chinab State Key Laboratory of Power Transmission Equipment and System Security,College of Computer Science,Chongqing University,Chongqing400030,Chinac Texas A&M University at Qatar,Doha,P.O.Box23874,Qatara r t i c l e i n f oArticle history:Received6September2011Received in revised form14June2012Accepted17June2012Communicated by J.LiangAvailable online1July2012Keywords:Cellular neural networksAssociative memoriesCloning templatea b s t r a c tIt is well known that the prototype patterns in associative memories can be represented by stableequilibrium points of cellular neural networks(CNNs).Therefore,the stability of equilibrium points ofCNNs is critical in associative memories based on CNNs.In this paper,some criteria about the stabilityof CNNs are established.In fact,these criteria give some constraint conditions for the relationship ofparameters of pared with the previous works,our results relax the conservatism of therelationship of parameters and extend the range of the values of parameters.Two design procedures onthe parameters of CNNs are given to achieve associative memories under our criteria.Finally,anexample is given to verify the theoretical results and design procedures.&2012Elsevier B.V.All rights reserved.1.IntroductionCellular neural networks(CNNs)werefirst introduced in1988[1,2].CNNs can be arranged in matrix and implemented by somesimple analog circuits called cell.Each cell in CNNs is onlyconnected to its neighboring cells.Therefore,CNNs are wellfitfor very large-scale integration implementations due to this localinterconnection property,and have found many applications in avariety of areas,such as image processing[3],pattern recognition[4]and associative memories[5].In this paper,we mainly discussthe application of CNNs in associative memories.At its simplest,an associative memory is a system which stores mappings fromspecific input patterns to specific output patterns.That is to say,asystem which‘‘associates’’two patterns is that when one of twopatterns is presented,the other can be reliably recalled.There aretwo kinds of associative memories:auto-associative memoriesand hetero-associative.In associative memory,the retrievedpattern can be different from the probe in content or format,however,in some other search approaches,these two patternshave to be same or similar,such as similarity search in[6].Since Liu and Michel[7]reported that CNNs are effective as anassociative memories medium,associative memories havereceived a great deal of interest.Next,we would introduceassociative memories according to the time sequence.Sparselyinterconnected neural networks for associative memories werepresented in[8],and sparse synthesis technique was applied tothe design of a class of CNNs.A design algorithm for CNNs withspace-invariant cloning template with applications to associativememories was presented in[9].A synthesis procedure forassociative memories using discrete-time CNNs(DTCNNs)withlearning and forgetting capabilities was presented in[10].Asynthesis procedure of CNNs for associative memories wasintroduced in[11],where the method assured the global asymp-totic stability of the equilibrium point.DTCNNs with a globallyasymptotically stable equilibrium point were designed to behaveas associative memories in[12].In the last ten years,associativememories were realized by local stability of equilibrium points ofCNNs.In[13,14],the number of memory patterns of CNNs whichwere locally exponentially stable was obtained,and the designprocedures of associative memories based on CNNs was given.Adesign method for synthesizing associative memories based ondiscrete-time recurrent neural networks was presented in[15].In[16],a new design procedure for synthesizing associative mem-ories based on CNNs with time delays characterized by input andoutput matrices was introduced.From the above introduction about associative memories,it iseasy to know that stability of CNNs plays an important role inassociative memories.The prototype patterns in associativememories can be represented by stable equilibrium points ofcellular neural networks.Therefore,it is important to study thestability of equilibrium points of CNNs.There have been abundantresearches about stability of CNNs.Some sufficient conditions forContents lists available at SciVerse ScienceDirectjournal homepage:/locate/neucomNeurocomputing0925-2312/$-see front matter&2012Elsevier B.V.All rights reserved./10.1016/j.neucom.2012.06.017n Corresponding author.E-mail address:yiding1981@(Q.Han).Neurocomputing97(2012)192–200CNNs to be stable were obtained by constructing Lyapunov Function [17–24],and these conditions generally made equili-brium point global asymptotically stable.However,some authors presented some conditions which made equilibrium points locally stable,and there generally were multiple equilibrium points [13,14,25–27].In previous papers,the conditions of stability of CNNs are conservative.For example,bias vectors were computed by one of all memory patterns in [7–9];the relations between cloning templates were stronger and closer in [13–16].Therefore,the aim of the paper is to relax conservative relationship among parameters of CNNs.In order to get our theories,we choose the initial states of CNNs are zero.Our theories expend the scope of the values of parameters of CNNs.When the inputs and outputs of a CNN are given,the values of parameters can be obtained by our methods.In fact,if we get appropriate values of parameters of CNNs,we can realize associative memories.Therefore,from the above theoretical analysis,we give design procedures of associa-tive memories based on CNNs.The remaining parts of this paper are organized as follows.In Section 2,a class of CNNs are given.In Section 3,the relationship among parameters of CNNs is given by a theorem and some corollaries.These theories give some methods about how to get the value of parameters A ,D and V of a CNN.In Section 4,two design procedures on associative memories and a flow chart about how to get parameters of a CNN are given.In Section 5,an example is given to verify the theoretical results and design procedures.Some conclusions are finally drawn in Section 6.2.PreliminariesConsider a class of cellular neural networks defined by thefollowing differential equation:_y ij ðt Þ¼Àc ij y ij ðt ÞþX k 2ði ,r Þk ¼k 1ði ,r ÞX l 2ðj ,r Þl ¼l 1ðj ,r Þa kl g i þk ,j þl ðy ðt ÞÞþZ ij Z ij ¼X k 2ði ,r Þk ¼k 1ði ,r ÞX l 2ðj ,r Þl ¼l 1ðj ,r Þd kl u kl þv ij 8>>>>>><>>>>>>:ð1Þwhere y ij ðt ÞA R denotes the states vector,c ij is a positive para-meter,r is positive integer denoting neighborhood radius,A ¼ða kl Þð2r þ1Þð2r þ1Þa 0is intra-neuron connection weight matrix,D ¼ðkl Þð2r þ1Þð2r þ1Þis input cloning template,u kl is the input,v ij is the bias,k 1ði ,r Þ¼max f 1Ài ,Àr g ,k 2ði ,r Þ¼min f N Ài ,r g ,l 1ðj ,r Þ¼min f 1Àj ,Àr g ,l 2ðj ,r Þ¼max M Àj ,r g Èand g ðU Þis the activa-tion function defined by g ðy Þ¼ð9y þ19À9y À19Þ=2g i þk ,j þl ðy ðt ÞÞis output of a cell of CNNs.Outputs of CNNs can be memory patterns of associative memories.The expressions of template A and D are in Appendix A .Let r ¼1and n ¼NM .If the system (1)has N rows and M columns,then it can be put in vector form as _x¼ÀCx þAf ðx ÞþDU þV ð2Þwhere x ¼ðx 1,x 2,...,x n ÞT ¼ðy 11,y 12,...,y 1M ,...,y NM ÞT ,coefficient matrices A and D are obtained through the templates and C ¼diag ðc 1...c n Þ,the input vector U ¼ðu 1,...,u n ÞT ,t V ¼ðv 1,...,v n ÞT and f ðx Þ¼ðg ðy 1Þ,...,g ðy n ÞÞT .The k th cell in Eq.(2)is denoted by O k (k ¼iN þj ,where 1r i r N ,1r j r M ,i denotes i th row and j denotes j th column of the CNN).The expressions of template A and D are in Appendix A .The matrices of A and D are very complex.They are not only sparse matrices,and the posi-tions of their elements in these two matrices have certain rules.Therefore,these two matrices cannot be solved by previousmethods,such as the method in [28].In order to get the values of these two matrices,we need to transform these two matrices in two vectors.The methods will be shown in Section 3.2.Let a ¼ða 1,a 2,...,a n ÞTA U n ¼x i A R n x i ¼1or x i ¼À1,i ¼1,2,È...,n g ,C ða Þ¼x A R n Èx i a i 41,i ¼1,2,...,n g .Then,for x A C ða Þ,the Eq.(2)can be rewritten as _x¼ÀCx þA a þDU þV ð3ÞIf b is an equilibrium point of (3),then we haveb ¼C À1ðA a þDU þV ÞA C ða Þð4ÞLemma 1.[7]Suppose a ¼ða 1,a 2,...,a n ÞT A U n .If b ¼ðb 1,b 2,...,b n ÞT ¼C À1ðA a þDU þV ÞA C ða Þ,then b is an asymptotically stable equilibrium point of (2).Proof.Eq.(3)has a unique equilibrium point at x e ¼C À1ðA a þDU þV Þ,and x e ¼b A C ða Þby assumption.Therefore,this equili-brium is also asymptotical stable,since Eq.(3)has all its n eigenvalues at Àc i ,i ¼1,2,...,n .3.Main resultIn this section,we will give some theories about stability of CNNs firstly.Then,some methods are obtained on the basis of these theories for realizing associative memories based on CNNs.3.1.Stability of CNNsFirst,we give a theorem and some corollaries for achieving associative memories.The Eq.(3)can be rewritten as _xi ¼Àc i x i þX n j ¼1a ij a j þX n j ¼1d ij u j þv i ,i ¼1,2,...,n ð5ÞTheorem 1.In Eq.(5),let x i ð0Þ¼0,i ¼1,2,...,n :(i).If Pn j ¼1a ij a j þP n j ¼1d ij u j þv i 4c i ,then the Eq.(5)convergesto a positive stable equilibrium point,and the value of positive equilibrium point is bigger than 1.(ii).If P n j ¼1a ij a j þP nj ¼1d ij u j þv i o Àc i ,then the Eq.(5)con-verges to a negative stable equilibrium point,and the valuesof negative equilibrium point is less than À1.The proof of Theorem 1see Appendix B1.By the above theorem,we can get the following corollaries immediately.Corollary 1.In Eq.(5),let x i ð0Þ¼0and a ii Z c i ,i ¼1,2,...,n ,(i)If P n j ¼1,j a i a ij a j þP nj ¼1d ij u j þv i 40,then the Eq.(5)con-verges to a positive stable equilibrium point,and the valueof positive equilibrium point is bigger than 1.(ii)If P n j ¼1,j a i a ij a j þP nj ¼1d ij u j þv i o 0,then the Eq.(5)con-verges to a negative stable equilibrium point,and the valueof negative equilibrium point is less than À1.The proof of Theorem 1see Appendix B2.Corollary 2.In Eq.(5),let x i ð0Þ¼0and a ii o c i ,(i)If P n j ¼1a ij a j þP nj ¼1d ij u j þv i 4c i ,then the Eq.(5)convergesto a positive stable equilibrium point,and the value of positive equilibrium point is bigger than 1.Q.Han et al./Neurocomputing 97(2012)192–200193(ii)If P n j ¼1a ij a j þP nj ¼1d ij u j þv i o Àc i ,then the Eq.(5)convergesto a negative stable equilibrium point,and the value of negative equilibrium point is less than À1.The proof of Corollary 2is same to that of Theorem 1.If we choose v i ¼0in Theorem 1,then we can get the following corollary:Corollary 3.In Eq.(5),let x i ð0Þ¼0and v i ¼0,(i)If P n j ¼1a ij a j þP nj ¼1d ij u j 4c i ,then the Eq.(5)converges to apositive equilibrium point,and the value of positive equili-brium point is bigger than 1.(ii)If P n j ¼1a ij a j þP nj ¼1d ij u j o Àc i ,then the Eq.(5)converges to anegative equilibrium point,and the value of negative equili-brium point is less than À1.If we choose v i ¼0in Corollary 1,then we can get the following corollary:Corollary 4.In Eq.(5),let x i ð0Þ¼0,v i ¼0and a ii Z c i ,(i).If P nj ¼1,j a i a ij a j þP nj ¼1d ij u j 40,then the Eq.(5)converges to a positive stable equilibrium point,and the value of positive equilibrium point is bigger than 1.(ii).If Pn j ¼1,j a i a ij a j þP n j ¼1d ij u j o 0,then the Eq.(5)converges toa negative stable equilibrium point,and the value of negative equilibrium point is less than À1.Remark 1.In Theorem 1and its corollaries,though the initial states of a CNN have to return to zero,our methods does not strictly limit the relations between cloning template of the CNN.For example,P n j ¼1ð9a ij þb ij 9Þo c i o 1and P 1i ¼À1P 1i ¼À1ð9a ij þij 9Þo 1in [15,16],however our theories do not exist these limitations.The Theorem 1and its corollaries will be used to realize associative memories based on CNNs in the following.In order to realize associative memories,inputs of the CNN and the memory patterns(outputs of the CNN)should be given.Then,according to the above theories,we train the CNN on the basis of inputs and outputs,and the parameters (weight value)of the CNN can be obtained.Actually,the above theorem and corollaries give some methods or constraints about how to get parameters of a CNN.For example,when initial states of a CNN are zero,the relationship among parameters can be obtained by Theorem 1.If the values of some parameters of a CNN are fixed,the regions of values of other parameters of the CNN can be obtained.If inputs,outputs and parameters A ,C and D are fixed,the region of bias V can be gotten.The main difference between Theorem 1and Corollary 3is whether all bias v i (i ¼1,2,y n )are equal to 0.Similarly,the main difference of Corollary 1and Corollary 4is also whether all bias v i (i ¼1,2,y n )are equal to 0.The main difference between Theorem 1and Corollary 1is that whether there must exists a ii Z c i .Similarly,the main difference of Corollary 3and Corollary 4is also that whether there must exists a ii Z c i .In Appendix C ,we give a lot of notations.These notations will be used in matrix equations of next section.3.2.Associative memoriesNext,we will discuss associative memories based on CNNs by use of the above the theories and notations in Appendix C .In the section,the explanation of new notations can be found in Appendix C .First,we discuss how to get cloning template A and D of a CNN.We can make use of the inputs and outputs of cells in set R to get A and D .Because of the all values of v i (i ¼1,2,y n )of cells in set R are equal to zero,choose that all v i of all cells in set R are equal tozero.Then,by Corollaries 3or 4,the relationship among Pn j ¼1a ij a j ,P nj ¼1d ij u j,c i (i ¼1,y n )can be established.Then,we use Theorem 1or Corollary 1to calculate the regions of v i (i ¼1,2,y n )of cells in set P or Q .Therefore,in order to get parameters A ,D ,C ,and V for a CNN,we divide the question into two cases on the basis of Corollary 3and Corollary 4,respectively.(i)Corollary 3is used to obtain the relationship among parameters A ,D and C .Furthermore,the values of A and D can be gotten.Let l i 4max 1r i r nc i f g .We choose ^F 0¼0:5^D ,ð9Þand^AG þ^DU ¼^LG ð10ÞIt is obvious that Eq.(9)and Eq.(10)satisfy Corollary 3.FromEq.(9),the sign of Pn j ¼1a ij a j in a cell O i and the sign of output ofthe cell are the same.From Eq.(10),sign of P n j ¼1a ij a j þP nj ¼1d ij u j in a cell O i of a CNN and the sign of output of the cell are thesame.By Corollary 3,we know that sign of Pn j ¼1a ij a j þP n j ¼1d ij u j in a cell of a CNN can determine the sign of output of the cell.(ii)Corollary 4is used to achieve associative memories.Let l i 40.We choose ^F 00¼0:5^D ,ð11Þand~AG þ^DU ¼^LG ð12ÞIt is obvious that Eq.(11)and Eq.(12)satisfy Corollary 4.From Eqs.(9)–(12),we give the methods about computing parameters A and D .However,we find that it is difficult to beobtain ^Aand ^D through Eqs.(9)–(12).For example,in Eq.(10),the values of ^AG ,U and ^LG are known,and the value of ^D is unknown.We notice that there is no method to get the value of ^D.So,Eqs.(9)–(12)are needed to transform.Eq.(9)can be transformed as ^OLA ¼0:5^D ,ð13ÞTherefore,LA ¼0:5pinv ð^O0Þ^D ,ð14Þwhere pinv ðU Þdenotes pseudo inverse of a matrix.Eq.(10)can be transformed as ^O00LA ¼0:5^D ð15ÞTherefore,LA ¼0:5pinv ð^O00Þ^D ð16ÞEq.(11)can be transformed as ^XLD ¼^D À^F 0ð17ÞTherefore,LD ¼pinv ð^XÞð^D À^O 0LA Þð18ÞEq.(12)can be transformed as ^XLD ¼^D À^F 00ð19ÞTherefore,LD ¼pinv ð^XÞð^D À^O 00LA Þð20ÞQ.Han et al./Neurocomputing 97(2012)192–200194After transforming Eqs.(9)–(12),from LA and LD ,it is easy to obtain parameters A and D .Remark 2.When matrices ^O0,^O 00or ^X are irreversible,the values of LA or LD are approximate values.Next,we discuss how to get bias v i for a cell in sets P and Q .In set R ,the value of v i of a cell are equal to zero.However,in sets P and Q ,the values of v i (O i A P ,Q )of a cell are not equal to zero.Therefore,we need to calculate the value.Parameters A and D can be obtained by Corollary 3or 4.The ranges of v i (i ¼1,y n )can be obtained by Theorem 1or Corollary 1.From Theorem 1,we knowthat when v i 4c i ÀP n j ¼1a ij a j ÀP nj ¼1d ij u j ,the Eq.(5)converges to a positive stable equilibrium point,which means that the output of cellO i is positive.Because the sign of outputs of all cells in set P ispositive,we can choose that ^v þ4max 1r i r nf c i ÀPn j ¼1a ij a j ÀP nj ¼1d ij u j g is the bias of all cells in set P .Similarly,^v Ào min 1r i r n Àc i ÀP n j ¼1a ij a j ÀP nj ¼1d ij u j Én is the bias of all cells in set Q .Then,the regions of ^vþand ^vÀcan be get by Theorem 1or Corollary 1.We can consider the regions of ^vþand ^v Àfrom two aspects as follows.(i)If Theorem 1is used to realize associative memories based on a CNN,we can get the following result:If a i ¼1in Eq.(5),we get v i ðl Þ4c i ÀP n j ¼1a ij a l jÀP nj ¼1d ij u l j ¼x 0i ðl Þ.If a i ¼À1in Eq.(5),we get v i ðl Þo c i ÀP n j ¼1a ij a l j ÀP nj ¼1d ij u lj ¼x 0i ðl Þ.Therefore,we choose ^vþZ max1r l r m ,O i A P9f x 0i ðl Þg 9ð21Þas bias of all cells in set P ,and ^vÀr Àmax 1r l r m ,O i A Q9f x 0i ðl Þg 9ð22Þas bias of all cells in set Q .(ii)If Corollary 1is used to realize associative memories based on a CNN,we can get the following results:If a i ¼1in Eq.(5),we get v i ðl Þ4ÀP n j ¼1a ij a l j ÀP n j ¼1d ij u lj ¼x 00i ðl Þ.If a i ¼À1in Eq.(5),we get v i ðl Þo ÀP n j ¼1a ij a lj ÀP nj ¼1d iju lj ¼x 00i ðl Þ.Therefore,we choose ^vþZ max1r l r m ,O i A P9f x 00i ðl Þg 9ð23Þas bias of all cells in set P ,and ^vÀr Àmax 1r l r m ,O i A Q9f x 00i ðl Þg 9ð24Þas bias of all cells in set Q .Remark 3.Cloning template D is computed by the cells in set R .Therefore,the values of D are not be affected by the cells in sets P and Q ,which can make the values of D more accurate than that in [15].Bias vector is computed by the cells in sets P and Q ,which make the outputs of cells in sets P and Q always right in the process of associative memories.Remark 4.In [15],the authors proofed that associative memories based on CNNs are high-capacity (High-capacity means that a large number of memory patterns can be stored by a CNN).Themodel of the paper is similar with that of [15],therefore,we donot proof the problems of high-capacity.4.Design procedure of a CNNIn this section,two design procedures of parameters of a CNN are given by the above theories.(i)If Theorem 1and Corollary 3are chosen to achieve associativememories,we give the following design procedure of para-meters of a CNN.We can use Corollary 3to get parameters A and D in set R and use Theorem 1to get biases v i (i ¼1,y n )for a cell in sets P and Q .Step 1.Denote a matrix G ¼ða 1,a 2,...,a m Þas memory pat-terns of associative memories,where a i is a set of outputs of all cells in a CNN,and m is the number of patterns of associative memories.Denote a input matrixU ¼ðU 1,U 2,...,U mÞwhich is corresponding to the matrix of memory patterns G .Step 2.Divide all cells of the CNN into three sets.If all outputs of a cell in all memory patterns are 1,the cell will be classified as set P .If all outputs of a cell in all memory patterns are À1,the cell will be classified as set Q .If the outputs of a cell in all memory patterns are 1and À1,the cell will be classified as set R .Step 3.Choose biases v i ðO i A R Þare equal to zero.Step 4.Determine all state constants c ij ,1r i r N ,1r j r M .Obtain coefficient matrices C .Step 5.Determine matrix L such that l i 4max 1r i r nf c ig ,and getmatrix L 0.Step pute cloning template A by use of Eq.(14)and the relationship between inputs and outputs of cells in set R .Obtain coefficient matrices A .Step pute cloning template by use of Eq.(18)and the relationship between inputs and outputs of cells in set R .Obtain coefficient matrices D .Step pute x 0i ðl Þ(O i A P ,1r l r m )in set P .Choose ^vþZ max 1r l r m ,O i A P9f x 0i ðl Þg 9in terms of (21),and biases v i ðO i A P Þareequal to ^vþ.Step pute x 0i ðl Þ(O i A Q ,1r l r m )in set Q .Choose^vÀo Àmax 1r l r m ,O i A Q9f x 0i ðl Þg 9in terms of (22),and biases v i ðO i A Q Þare equal to ^vÀ.Step 10.Synthesize the CNN with the connection weight matrices A ,C ,D and bias vector V .From these above ten steps,we give the flow chart as in Fig.1:(ii)If Corollary 1and Corollary 4are chosen to realize associativememories,we give the following design procedure for a CNN:Step 1,Step 2,Step 3and Step 4of (ii)are same to these of (i).Step 5.Determine matrix L such that l i 40,and get matrix L 0.Step 6.Determine a 00such that a 00Z max 1r i r nf c ig holds.Step pute cloning template A from (16)in set R .Obtain coefficient matrices A .Step pute cloning template from (20)in set R .Obtain coefficient matrices D .Step pute x 00i ðl Þ(O i A P ,1r l r m )in set P .Choose^v þ4max 1r l r m ,O i A P9f x 00i ðl Þg 9in terms of (23),and bias v i of allcells in set P is equal to ^vþ.Step pute x 00i ðl Þ(O i A Q ,1r l r m )in set Q .Choose^vÀo Àmax 1r l r m ,O i A Q9f x 00i ðl Þg 9in terms of (24),and bias v i of all cells in set Q is equal to ^vÀ.Q.Han et al./Neurocomputing 97(2012)192–200195Step 11.Synthesize the CNN with the connection weight matrices A ,C ,D and bias vector V .5.Numerical examplesIn this section,we will give some numerical simulations to verify the theoretical results in this paper.Consider the same example introduced in [16].The inputs and the output patterns of a CNN are represented by two pairs of (5Â5)pixel images showed in Figs.1and 2(black pixel ¼1,white pixel ¼À1),where the inputs of the CNN compose the word ‘‘MO’’in Fig.2(a),and the patterns to be memorized to constitute the word ‘‘LS’’in Fig.2(b).We design all parameters of a CNN to realize associative memories for patterns in Fig.2by use of design procedure (i)in Section 4.Step 1.In terms of Fig.2,we get memory patterns G ¼ððÀ1,1,À1,...,À1ÞT ,ðÀ1,1,1,...,À1ÞT Þand inputs matrix U ¼ðð1,À1,À1,...,1ÞT ,ðÀ1,1,1,...,1ÞT Þ.Step 2.From memory patterns,all cells of the CNN can be divided into three sets,P ¼f O 2,O 7,O 12,O 22,O 23,O 24g ,Q ¼f O 1,O 5,O 6,O 8,O 9,O 10,O 11,O 15,O 16,O 18,O 20,O 21,O 25g and R ¼f O 3,O 4,O 13,O 14,O 17,O 19g .Step 3.Choose biases of all cells in set R are equal to zero,namely,v 3¼v 4¼v 13¼v 14¼v 17¼v 19¼0.Step 4.Let c ij ¼1,1r i r N ,1r j r M ,then we can obtain C ¼diag ð1,1,...,1Þn Ân .Step 5.Let L ¼diag ð4,4,...,4Þn Ân ,then we have L 0¼diag 4,4,...,4ðÞnm Ânm .Step 6.From Eq.(14),we get LA ¼ð0,0,0,0,2,0,0,0,0ÞT .Then,we have A ¼diag ð2Þn Ân .Step 7.From Eq.(18),we get LD ¼À1:5682,À2:9394,ðÀ0:6136,1:6515,0:6136,0:1212,0:3864,0:3258,0:25ÞT .Then,we can obtain D .Step 8.Let ^vþ¼20in set P .Therefore,v 2¼v 7¼v 12¼v 22¼v 23¼v 24¼20.Step 9.Let ^vÀ¼À20in set Q .Therefore,v 1¼v 5¼v 6¼v 8¼v 9¼v 10¼v 11¼v 15¼v 16¼v 18¼v 20¼v 21¼v 25¼À20.Step 10.Synthesize the CNN with A ,C ,D and V .Note that the a ii 4c ij in the example,however,in previouspaper [15],Pn j ¼19a ij þij 9o c ij must be satisfied.Therefore,in the paper,though the initial states of a CNN have to return to zero,wereduce many limitations for the relationship among cloning templates.From the above ten steps,we can get a CNN which can realize associative memories from ‘‘MO’’to ‘‘LS’’.In Fig.3,when the inputs of the CNN are ‘‘O’’,we can get time response curves of all cells in the CNN in Fig.3.In Fig.3,we find that states of all cells will be stable after a time.When states of all cells are stable,the value of equilibrium point isx n ¼À22:5671,20:7718,4:1968,3:9240,À19:7945,ðÀ20:2415,18:5446,À26:3927,À29:7259,À27:6048,À20:2415,19:7718,4:7119,3:2877,À24:4686,À20:2415,À3:7271,À18:1355,4:7119,À23:6959,À20:1657,20:3248,26:3700,28:2790,À21:1051ÞT :Therefore,we know that all outputs of the CNN corresponding to the equilibrium point aref ðx n Þ¼À1,1,1,1,À1,À1,1,À1,À1,À1,À1,1,1,1,À1,À1,ðÀ1,À1,1,À1,À1,1,1,1,À1ÞT ,where the outputs of the CNN are same with ‘‘S’’.When the inputs of the CNN are ‘‘M’’,we can get time response curves of all cells in the CNN in Fig.4.Then the value of equilibrium point isx n ¼À20:9309,23:3777,À4:0755,À3:9543,À22:3247,ðÀ23:5141,26:4382,À16:0296,À18:7113,À21:1657,Fig.1.Flow chart about how to get parameters of a CNN by use of Theorem 1and Corollary3.Fig.2.(a)Inputs of a CNN and (b)outputs of the CNN or memory patterns.Q.Han et al./Neurocomputing 97(2012)192–200196。
笛卡尔的本体论之争
笛卡尔的本体论之争首先周一公布2001年6月18日;实质性修改太阳2006年10月15日笛卡尔的本体论（或先验）的论点，既是哲学的一个最迷人，他的理解方面的不足。
论据与魅力源于努力证明神的存在，从简单的处所，但功能强大。
存在是产生立即从清晰和明确的想法是一个无比完美。
讽刺的是，简单的说法也产生了一些误读，加剧了部分由笛卡尔没有一套单一版本。
该声明的论点主要出现在第五沉思。
这种说法因果来得早在接踵而至的一个神的存在，沉思在第三，不同的证据提出问题的两项之间的秩序和关系。
重复笛卡尔哲学原理，包括本体论争论的几个文本等中央。
他还辩解首先由一些主要的知识分子，他在一天，严厉打击反对第二次回复，和第五。
笛卡尔不是第一位哲学家，制订一个本体论的论点。
一个早期版本的说法已大力安瑟伦辩护圣在11世纪，然后圣托马斯阿奎那批评由当代），后来被命名为Gaunilo和尚（安瑟伦（尽管他的言论是针对然而，另一个版本参数）。
阿奎那的批评被视为如此具有破坏性，本体论的争论了数百年死亡。
它的出现，作为一个同时代的惊喜笛卡尔，他应该试图复活它。
虽然他声称没有被证明的熟悉安瑟伦的版本，笛卡尔似乎他自己的工艺参数，以阻止传统的反对。
尽管相似之处，笛卡尔的论点的版本不同于安瑟伦方式在重要的。
后者的版本被认为要从定义这个词的含义“上帝”，上帝是一个被一大于不能设想。
笛卡尔的观点相反，中，主要是基于两个他的哲学的中心原则-天生的思想理论和学说明确的印象和独特的。
他声称不依赖于上帝的任意定义，而是一种天生的想法，其内容是“的。
” 笛卡尔的版本也非常简单。
神的存在是直接从推断的事实，有必要存在的想法是包含在一个清晰而鲜明的超级完美的存在。
事实上，在一些场合，他建议，所谓的本体论“的论调”是不是一个正式的哲学偏见的证据，而是在所有不言而喻的公理直观地掌握了一个心灵的自由。
笛卡尔的本体论的争论相比往往以几何论证，认为有必要存在的想法不能排除再从神比事实平等的角度，其角度，例如两权，可以被排除在一个三角形的想法。
glivenko-cantelli格里文科定理证明
1The Glivenko-Cantelli TheoremLet X i,i=1,...,n be an i.i.d.sequence of random variables with distribu-tion function F on R.The empirical distribution function is the function ofx defined byˆFn(x)=1n1≤i≤nI{X i≤x}.For a given x∈R,we can apply the strong law of large numbers to the sequence I{X i≤x},i=1,...n to assert thatˆFn(x)→F(x)a.s(in order to apply the strong law of large numbers we only need to show that E[|I{X i≤x}|]<∞,which in this case is trivial because|I{X i≤x}|≤1).In this sense,ˆF n(x)is a reasonable estimate of F(x)for a given x∈R. But isˆF n(x)a reasonable estimate of the F(x)when both are viewed as functions of x?The Glivenko-Cantelli Thoerem provides an answer to this question.It asserts the following:Theorem1.1Let X i,i=1,...,n be an i.i.d.sequence of random variables with distribution function F on R.Then,supx∈R|ˆF n(x)−F(x)|→0a.s.(1) This result is perhaps the oldest and most well known result in the very large field of empirical process theory,which is at the center of much of modern econometrics.The statistic(1)is an example of a Kolmogorov-Smirnov statistic.We will break the proof up into several steps.Lemma1.1Let F be a(nonrandom)distribution function on R.For each >0there exists afinite partition of the real line of the form−∞=t0< t1<···<t k=∞such that for0≤j≤k−1F(t−j+1)−F(t j)≤ .1Proof:Let >0be given.Let t0=−∞and for j≥0definet j+1=sup{z:F(z)≤F(t j)+ }.Note that F(t j+1)≥F(t j)+ .To see this,suppose that F(t j+1)<F(t j)+ .Then,by right continuity of F there would existδ>0so that F(t j+1+δ)< F(t j)+ ,which would contradict the definition of t j+1.Thus,between t j and t j+1,F jumps by at least .Since this can happen at most afinite number of times,the partition is of the desired form,that is−∞=t0< t1<···<t k=∞with k<∞.Moreover,F(t−j+1)≤F(t j)+ .To see this, note that by definition of t j+1we have F(t j+1−δ)≤F(t j)+ for allδ>0.The desired result thus follows from the definition of F(t−j+1).Lemma1.2Suppose F n and F are(nonrandom)distribution functions on R such that F n(x)→F(x)and F n(x−)→F(x−)for all x∈R.Thensupx∈R|F n(x)−F(x)|→0.Proof:Let >0be given.We must show that there exists N=N( )such that for n>N and any x∈R|F n(x)−F(x)|< .Let >0be given and consider a partition of the real line intofinitely many pieces of the form−∞=t0<t1···<t k=∞such that for0≤j≤k−1F(t−j+1)−F(t j)≤2.The existence of such a partition is ensured by the previous lemma.For any x∈R,there exists j such that t j≤x<t j+1.For such j,F n(t j)≤F n(x)≤F n(t−j+1)F(t j)≤F(x)≤F(t−j+1),which implies thatF n(t j)−F(t−j+1)≤F n(x)−F(x)≤F n(t−j+1)−F(t j).2Furthermore,F n(t j)−F(t j)+F(t j)−F(t−j+1)≤F n(x)−F(x)F n(t−j+1)−F(t−j+1)+F(t−j+1)−F(t j)≥F n(x)−F(x).By construction of the partition,we have thatF n(t j)−F(t j)−2≤F n(x)−F(x)F n(t−j+1)−F(t−j+1)+2≥F n(x)−F(x).For each j,let N j=N j( )be such that for n>N jF n(t j)−F(t j)>− 2and let M j=M j( )be such that for n>M jF n(t−j )−F(t−j)<2.Let N=max1≤j≤k max{N j,M j}.For n>N and any x∈R,we have that|F n(x)−F(x)|< .The desired result follows.Lemma1.3Suppose F n and F are(nonrandom)distribution functions on R such that F n(x)→F(x)for all x∈Q.Suppose further that F n(x)−F n(x−)→F(x)−F(x−)for all jump points of F.Then,for all x∈R F n(x)→F(x)and F n(x−)→F(x−).Proof:Let x∈R.Wefirst show that F n(x)→F(x).Let s,t∈Q such that s<x<t.First suppose x is a continuity point of F.Since F n(s)≤F n(x)≤F n(t)and s,t∈Q,it follows thatF(s)≤lim infn→∞F n(x)≤lim supn→∞F n(x)≤F(t).Since x is a continuity point of F,lim s→x−F(s)=limt→x+F(t)=F(x),3from which the desired result follows.Now suppose x is a jump point of F .Note thatF n (s )+F n (x )−F n (x −)≤F n (x )≤F n (t ).Since s,t ∈Q and x is a jump point of F ,F (s )+F (x )−F (x −)≤lim inf n →∞F n (x )≤lim sup n →∞F n (x )≤F (t ).Sincelim s →x −F (s )=F (x −)lim t →x +F (t )=F (x ),the desired result follows.We now show that F n (x −)→F (x −).First suppose x is a continuity point of F .Since F n (x −)≤F n (x ),lim sup n →F n (x −)≤lim sup n →F n (x )=F (x )=F (x −).For any s ∈Q such that s <x ,we have F n (s )≤F n (x −),which implies thatF (s )≤lim inf n →∞F n (x −).Sincelim s →x −F (s )=F (x −),the desired result follows.Now suppose x is a jump point of F .By as-sumption,F n (x )−F n (x −)→F (x )−F (x −),and,by the above argument,F n (x )→F (x ).The desired result follows.Proof of Theorem 1.1:If we can show that there exists a set N suchthat Pr {N }=0and for all ω∈N (i)ˆFn (x,ω)→F (x )for all x ∈Q and (ii)ˆFn (x,ω)−F n (x −,ω)→F (x )−F (x −)for all jump points of F ,then the result will follow from an application of Lemmas 1.2and 1.3.For each x ∈Q ,let N x be a set such that Pr {N x }=0and for all ω∈N x ,ˆF n (x,ω)→F (x ).Let N 1= x ∈Q .Then,for all ω∈N 1,ˆF n (x,ω)→F (x )by construction.Moreover,since Q is countable,Pr {N 1}=0.4For integer i ≥1,let J i denote the set of jump points of F of size at least 1/i .Note that for each i ,J i is finite.Next note that the set of all jump points of F can be written as J = 1≤i<∞J i .For each x ∈J ,let M x denotea set such that Pr {M x }=0and for all ω∈M x ,ˆF n (x,ω)−F n (x −,ω)→F (x )−F (x −).Let N 2= x ∈J M x .Since J is countable,Pr {N 2}=0.To complete the proof,let N =N 1∪N 2.By construction,for ω∈N ,(i)and (ii)hold.Moreover,Pr {N }=0.The desired result follows.2The Sample MedianWe now give a brief application of the Glivenko-Cantelli Theorem.Let X i ,i =1,...,n be an i.i.d.sequence of random variables with distribution F .Suppose one is interested in the median of F .Concretely,we will defineMed(F )=inf {x :F (x )≥12}.A natural estimator of Med(F )is the sample analog,Med(ˆFn ).Under what conditions is Med(ˆFn )a reasonable estimate of Med(F )?Let m =Med(F )and suppose that F is well behaved at m in the sense that F (t )>12whenever t >m .Under this condition,we can show usingthe Glivenko-Cantelli Theorem that Med(ˆFn )→Med(F )a.s.We will now prove this result.Suppose F n is a (nonrandom)sequence of distribution functions such thatsup x ∈R |F n (x )−F (x )|→0.Let >0be given.We wish to show that there exists N =N ( )such that for all n >N|Med(F n )−Med(F )|< .Choose δ>0so thatδ<12−F (m − )δ<F (m + )−12,5which in turn implies thatF(m− )<12−δF(m+ )>12+δ.(It might help to draw a picture to see why we should pickδin this way.) Next choose N so that for all n>N,supx∈R|F n(x)−F(x)|<δ.Let m n=Med(F n).For such n,m n>m− ,for if m n≤m− ,thenF(m− )>F n(m− )−δ≥12−δ,which contradicts the choice ofδ.We also have that m n<m+ ,for if m n≥m+ ,thenF(m+ )<F n(m+ )+δ≤12+δ,which again contradicts the choice ofδ.Thus,for n>N,|m n−m|< ,as desired.By the Glivenko-Cantelli Theorem,it follows immediately that Med(ˆF n)→Med(F)a.s.6。
Copernicus' Proof of the Earth's motion
3rd International Conference on Absolutes Houston, Texas, July 16-18, 2007
Actual Scientific Research Model
The following shows the actual scientific investigation model including the publication of scientific papers is also controlled by this model.
“Our Sages have, in this astronomical question, abandoned their own theory in favour of the theory of others. Thus, it is distinctly stated, 'The wise men of the other nations have defeated the wise men of Israel.' ” p163, Moses Maimonides.
[Platos' Timaeus, Heath page 52]
If the Earth was created first before the Sun. So, around what was the earth orbiting?
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The earth does not rotate: “...the world also is stablished, that it cannot be moved,” Psalm 93:1 The earth hangs on nothing: “He stretcheth out of the north over the empty place, and hangeth the earth upon nothing.” Job 26:7 The earth is a sphere: “It is he that sitteth upon the circle of the earth, and the inhabitants are as grasshoppers;” Isaiah 40:22
Conformal restriction the chordal case
a rX iv:mat h /29343v2[mat h.PR]25Apr23Conformal restriction:the chordal case Gregory Lawler ∗Oded Schramm †Wendelin Werner ‡Abstract We characterize and describe all random subsets K of a given simply connected planar domain (the upper half-plane H ,say)which satisfy the “conformal restriction”property,i.e.,K connects two fixed boundary points (0and ∞,say)and the law of K conditioned to remain in a simply connected open subset H of H is identical to that of Φ(K ),where Φis a conformal map from H onto H with Φ(0)=0and Φ(∞)=∞.The construction of this family relies on the stochastic Loewner evolution processes with parameter κ≤8/3and on their distortion under conformal maps.We show in particular that SLE 8/3is the only random simple curve satisfying conformal restriction and relate it to the outer boundaries of planar Brownian motion and SLE 6.Keywords:Conformal invariance,restriction property,random fractals,SLE.MSC Classification:60K35,82B27,60J69,30C99Contents1Introduction3 2Preliminaries7 3Two-sided restriction10 4Brownian excursions15 5Conformal image of chordal SLE19 6Restriction property for SLE8/323 7Bubbles267.1Brownian bubbles (26)7.2Adding a Poisson cloud of bubbles to SLE (29)8One-sided restriction318.1Framework (31)8.2Excursions of reflected Brownian motions (31)8.3The SLE(κ,ρ)process (34)8.4Proof of Theorem8.4 (38)8.5Formal calculations (41)9Equivalence of the frontiers of SLE6and Brownian motion449.1Full plane SLE6and planar Brownian motion (44)9.2Chordal SLE6and reflected Brownian motion (46)9.3Chordal SLE6as Brownian motion reflected on its past hull..499.4Non-equivalence of pioneer points and SLE6 (50)9.5Conditioned SLE6 (52)10Remarks5221IntroductionConformalfield theory has been extremely successful in predicting the ex-act values of critical exponents describing the behavior of two-dimensional systems from statistical physics.In particular,in the fundamental papers [5,6],which were used and extended to the case of the“surface geometry”in[9],it is argued that there is a close relationship between critical planar systems and some families of conformally invariantfields.This gave rise to intense activity both in the theoretical physics community(predictions on the exact value of various exponents or quantities)and in the mathematical community(the study of highest-weight representations of certain Lie alge-bras).However,on the mathematical level,the explicit relation between the two-dimensional systems and thesefields remained rather mysterious.More recently,a one-parameter family of random processes called stochas-tic Loewner evolution,or SLE,was introduced[44].The SLEκprocess is ob-tained by solving Loewner’s differential equation with driving term B(κt), where B is one-dimensional Brownian motion,κ>0.The SLE processes are continuous,conformally invariant scaling limits of various discrete curves arising in the context of two-dimensional systems.In particular,for the models studied by physicists for which conformalfield theory(CFT)has been applied and for which exponents have been predicted,it is believed that SLE arises in some way in the scaling limit.This has been proved for site-percolation on the triangular lattice[46],loop-erased random walks[29] and the uniform spanning tree Peano path[29](a.k.a.the Hamiltonian path on the Manhattan lattice).Other models for which this is believed include the Ising model,the random cluster(or Potts)models with q≤4,and the self-avoiding walk.In a series of papers[23,24,25,26],the authors derived various prop-erties of the stochastic Loewner evolution SLE6,and used them to compute the“intersection exponents”for planar Brownian paths.This program was based on the earlier realization[32]that any conformally invariant process satisfying a certain restriction property has crossing or intersection expo-nents that are intimately related to these Brownian intersection exponents. In particular,[32]predicted a strong relation between planar Brownian mo-tion,self-avoiding walks,and critical percolation.As the boundary of SLE6is conformally invariant,satisfies restriction,and can be well understood,com-putations of its exponents yielded the Brownian intersection exponents(in particular,exponents that had been predicted by Duplantier-Kwon[15,14],3disconnection exponents,and Mandelbrot’s conjecture[34]that the Haus-dorffdimension of the boundary of planar Brownian motion is4/3).Sim-ilarly,the determination of the critical exponents for SLE6in[23,24,25] combined with Smirnov’s[46]proof of conformal invariance for critical per-colation on the triangular lattice(along with Kesten’s hyperscaling relations) facilitated proofs of several fundamental properties of critical percolation [47,28,45],some of which had been predicted in the theoretical physics literature,e.g.,[37,35,36,38,43].The main goal of the present paper is to investigate more deeply the restriction property that was instrumental in relating SLE6to Brownian mo-tion.One of our initial motivations was also to understand the scaling limit and exponents of the two-dimensional self-avoiding walk.Another motiva-tion was to reach a clean understanding of the relation between SLE and conformalfield theory.Consequences of the present paper in this direction are the subject of[18,19].See also[2,3]for aspects of SLE from a CFT perspective.Let us now briefly describe the conformal restriction property which we study in the present paper:Consider a simply connected domain in the complex plane C,say the upper half-plane H:={x+iy:y>0}.Suppose that two boundary points are given,say0and∞.We are going to study closed random subsets K of H such that:•1.The restriction measure Pαexists if and only ifα≥5/8.2.The only measure Pαthat is supported on simple curves is P5/8.It isthe law of chordal SLE8/3.3.The measures Pαforα>5/8can be constructed by adding to thechordal SLEκcurve certain Brownian bubbles with intensityλ,where α,λandκare related byα(κ)=6−κ2κ.4.For allα≥5/8,the dimension of the boundary of K defined under Pαisalmost surely4/3and locally“looks like”an SLE8/3curve.In particular, the Brownian frontier(i.e.,the outer boundary of the Brownian path) looks like a symmetric curve.As pointed out in[30],this gives strong support to the conjecture that chordal SLE8/3is the scaling limit of the infinite self-avoiding walk in the upper half-plane and allows one to recover(modulo this conjecture)the critical exponents that had been predicted in the theoretical physics literature(e.g., [36,16]).This conjecture has recently been tested[21,22]by Monte Carlo methods.Let us also mention(but this will not be the subject of the present paper,see[18,19])that in conformalfield theory language,−λ(κ)is the central charge of the Virasoro algebra associated to the discrete models(that correspond to SLEκ)and thatαis the corresponding highest-weight(for a degenerate representation at level2).To avoid confusion,let us point out that SLE6is not a chordal restriction measure as defined above.However,it satisfies locality,which implies a different form of restriction.We give below a proof of locality for SLE6, which is significantly simpler than the original proof appearing in[23].We will also study a slightly different restriction property,which we call right-sided restriction.The measures satisfying right-sided restriction sim-ilarly form a1-parameter collection P+α,α>0.We present several con-structions of the measures P+α.First,whenα≥5/8,these can be obtained from the measures Pα(basically,by keeping only the right-side boundary). Whenα∈(0,1),the measure P+αcan also be obtained from an appropriately reflected Brownian excursion.It follows that one can reflect a Brownian ex-cursion offa ray in such a way that its boundary will have precisely the law5of chordal SLE8/3.A third construction of P+α(valid for allα>0)is given by a process we call SLE(8/3,ρ).The process SLE(κ,ρ)is a variant of SLE where a drift is added to the driving function.In fact,it is just Loewner’s evolution driven by a Bessel-type process.The word chordal refers to connected sets joining two boundary points of a domain.There is an analogous radial theory,which investigates sets joining an interior point to the boundary of the domain.This will be the subject of a forthcoming paper[31].We now briefly describe how this paper is organized.In the preliminary section,we give some definitions,notations and derive some simple facts that will be used throughout the paper.In Section3,we study the family of chordal restriction measures,and show(1.1).Section4is devoted to the Brownian excursions.We define these measures and use a result of B.Vir´a g (see[49])to show that thefilling of such a Brownian excursion has the law P1.The key to several of the results of the present paper is the study of the distortion of SLE under conformal maps,for instance,the evolution of the image of the SLE path under the mappingΦ(as long as the SLE path remains in H),which is the subject of Section5.This study can be considered as a cleaner and more advanced treatment of similar questions addressed in[23]. In particular,we obtain a new short proof of the locality property for SLE6, which was essential in the papers[23,24,25].The SLE distortion behaviour is then also used in Section6to prove that the law of chordal SLE8/3is P5/8and is also instrumental in Section7,where we show that all measures Pαforα>5/8can be constructed by adding a Poisson cloud of bubbles to SLE curves.The longer Section8is devoted to the one-sided restriction measures P+α. As described above,we exhibit various constructions of these measures and show as a by-product of this description that the two-sided measures Pαdo not exist forα<5/8.A recurring theme in the paper is the principle that the law P of a random set K can often be characterized and understood through the function A→P[K∩A=∅]on an appropriate collection of sets A.In Section9we use this to show that the outer boundary of chordal SLE6is the same as the outer boundary(frontier)of appropriately reflected Brownian motion and the outer boundary of full-plane SLE6stopped on hitting the unit circle is the same as the outer boundary of Brownian motion stopped on hitting the unit circle.6We conclude the paper with some remarks and pointers to papers in preparation.2PreliminariesIn this section some definitions and notations will be given and some basic facts will be recalled.Important domains.The upper half plane{x+iy:x∈R,y>0} is denoted by H,the complex plane by C,the extended complex plane by ˆC=C∪{∞}and the unit disk by U.Bounded hulls.Let Q be the set of all bounded A⊂A∩H and H\A is(connected and)simply connected.We call such an A a bounded hull.The normalized conformal maps g A.For each A∈Q,there is a unique conformal transformation g A:H\A→H with g A(z)−z→0as z→∞. We can then define(as in[23])z(g A(z)−z).(2.1)a(A):=limz→∞First note that a(A)is real,because g A(z)−z has a power series expansion in1/z near∞and is real on the real line in a neighborhood of∞.Also note thata(A)=limy H(i y),(2.2)y→∞where H(z)=Im z−g(z) is the bounded harmonic function on H\A with boundary values Im z.Hence,a(A)≥0,and a(A)can be thought of as a measure of the size of A as seen from infinity.We will call a(A)the half-plane capacity of A(from infinity).The useful scaling rule for a(A),a(λA)=λ2a(A)(2.3) is easily verified directly.Since Im g A(z)−Im z is harmonic,bounded,and has non-positive boundary values,Im g A(z)≤Im z.Consequently,0<g′A(x)≤1,x∈R\A.(2.4) (In fact,g′A(x)can be viewed as the probability of an event,see Proposition 4.1.)7∗-hulls.Let Q∗be the set of A∈Q with0∈A.We call such an A a∗-hull. If A∈Q∗,then H=H\A is as the H in the introduction.The normalized conformal mapsΦA.For A∈Q∗,we defineΦA(z)= g A(z)−g A(0),which is the unique conformal transformationΦof H\A onto Hfixing0and∞withΦ(z)/z→1as z→∞.Semigroups.Let A be the set of all conformal transformationsΦ:H\A→H withΦ(0)=0andΦ(∞)=∞,where A∈Q∗.That is,A={λΦA:λ> 0,A∈Q∗}.Also let A1={ΦA:A∈Q∗}.Note that A and A1are both semigroups under composition.(Of course,the domain ofΦ1◦Φ2isΦ−12(H1) if H1is the domain ofΦ1.)We can consider Q∗as a semigroup with the product·,where A·A′is defined byΦA·A′=ΦA◦ΦA′.Note thata(A·A′)=a(A)+a(A′).(2.5) As a(A)≥0,this implies that a(A)is monotone in A.±-hulls.Let Q+be the set of A∈Q∗with A∩R⊂(0,∞).Letσdenote the orthogonal reflection about the imaginary axis,and let Q−={σ(A):A∈Q+}be the set of A∈Q∗with A∩R⊂(−∞,0).If A∈Q∗,then we can find unique A1,A3∈Q+and A2,A4∈Q−such that A=A1·A2=A4·A3. Note that Q+,Q−are semigroups.Smooth hulls.We will call A∈Q a smooth hull if there is a smooth curveγ:[0,1]→C withγ(0),γ(1)∈R,γ(0,1)⊂H,γ(0,1)has no self-intersections,and H∩∂A=γ(0,1).Any smooth hull in Q∗is in Q+∪Q−. Fillings.If A⊂H such that any path from z to∞inH\A.Similarly, F R H(A)denotes the union of A with the connected components ofH\A which do not intersect[0,∞).Also,for closed A⊂C, F C(A)denotes the union of A with the bounded connected components of C\A.Approximation.We will sometimes want to approximate A∈Q by smooth hulls.The idea of approximating general domains by smooth hulls is standard (see,e.g.,[17,Theorem3.2]).8Lemma2.1.Suppose A∈Q+.Then there exists a decreasing sequence of smooth hulls(A n)n≥1such that A= ∞n=1A n and the increasing sequence Φ′An(0)converges toΦ′A(0).Proof.The existence of the sequence A n can be obtained by various means,for example,by considering the image underΦ−1A of appropriately chosenpaths.The monotonicity ofΦ′An (0)follows immediately from the monotonic-ity of A n and(2.4).The convergence is immediate by elementary properties of conformal maps,sinceΦAnconverges locally uniformly toΦA on H\A. Covariant measures.Our aim in the present paper is to study measures on subsets of H.In order to simplify further definitions,we give a general definition that can be applied in various settings.Suppose thatµis a measure on a measurable spaceΩwhose elements are subsets of a domain D.Suppose thatΓis a set of conformal transformations from subdomains D′⊂D onto D that is closed under composition.We say thatµis covariant underΓ(orΓ-covariant)if for allϕ∈Γ,the measureµrestricted to the setϕ−1(Ω):={ϕ−1(K):K⊂D}is equal to a constant Fϕtimes the image measureµ◦ϕ−1.Ifµis afiniteΓ-covariant measure,then Fϕ=µ[ϕ−1(Ω)]/µ[Ω].Note that a probability measure P onΩisΓ-covariant if and only if for allϕ∈Γwith Fϕ=P[ϕ−1(Ω)]>0,the conditional law of P onϕ−1(Ω)is equal to P◦ϕ−1.Also note that ifµis covariant underΓ,then Fϕ◦ψ=FϕFψfor allϕ,ψ∈Γ, because the image measure ofµunderϕ−1is F−1ϕµrestricted toϕ−1(Ω),sothat the image underψ−1of this measure is F−1ϕF−1ψµrestricted toψ−1◦ϕ−1(Ω).Hence,the mapping F:ϕ→Fϕis a semigroup homomorphism fromΓinto the commutative multiplicative semigroup[0,∞).Whenµis a probability measure,this mapping is into[0,1].We say that a measureµisΓ-invariant if it isΓ-covariant with Fϕ≡1. Chordal Loewner chains.Throughout this paper,we will make use of chordal Loewner chains.Let us very briefly recall their definition(see[23]for details).Suppose that W=(W t,t≥0)is a real-valued continuous function. Define for each z∈g t(z)−W t,g0(z)=z.(2.6) For each z∈Loewner evolution is defined as K t:={z∈κB t,where B is a standard one-dimensional Brownian motion,then the corresponding random Loewner chain is chordal SLEκ(SLE stands for stochastic Loewner evolution).3Two-sided restrictionIn this section we will be studying certain probability measures on a collection Ωof subsets of H.We start by definingΩ.Definition3.1.LetΩbe the collection of relatively closed subsets K of H such that1.K is connected,K is connected.A simple example of a set K∈Ωis a simple curveγfrom0to infinity in the upper half-plane.Ifγis just a curve from zero to infinity in the upper half-plane with double-points,then one can take K=F R H(γ)∈Ω,which is the set obtained byfilling in the loops created byγ.We endowΩwith theσ-field generated by the events{K∈Ω:K∩A=∅},where A∈Q∗.It is easy to check that this family of events is closed underfinite intersection,so that a probability measure onΩis characterized by the values of P[K∩A=∅]for A∈Q∗.Thus:Lemma3.2.Let P and P′be two probability measures onΩ.If P[K∩A=∅]=P′[K∩A=∅]holds for every A∈Q∗,then P=P′.It is worthwhile to note that theσ-field onΩis the same as the Borel σ-field induced by the Hausdorffmetric on closed subsets of2.P is A-covariant.3.There exists anα>0such that for all A∈Q∗,P[K∩A=∅]=Φ′A(0)α.4.There exists anα>0such that for all smooth hulls A∈Q∗,P[K∩A=∅]=Φ′A(0)α.Moreover,for eachfixedα>0,there exists at most one probability measure Pαsatisfying these conditions.Definition3.4.If the measure Pαexists,we call it the two-sided restriction measure with exponentα.Proof.Lemma3.2shows that a measure satisfying3is unique.A probability measure isΓ-covariant if and only if it isΓ-invariant.Therefore,1and2are equivalent.As noted above,any A∈Q∗can be written as A+·A−with A±∈Q±.Using this and Lemma2.1,we may deduce that conditions3and4are also equivalent.SinceΦ′λA(0)=Φ′A(0)for A∈Q∗,λ>0,3together withLemma3.2imply that P isΓ-invariant.BecauseΦ′A1·A2(0)=Φ′A1(0)Φ′A2(0),3also implies that for all A1,A2∈Q∗,P[K∩(A1·A2)=∅]=P[K∩A1=∅]P[K∩A2=∅],which implies1.Hence,it suffices to show that1implies4.Suppose1holds.Define the homomorphism F of Q∗onto the multiplica-tive semigroup(0,1]by F(A)=P[K∩A=∅].We also write F(ΦA)for F(A).Let G t(z)be the solution of the initial value problem∂t G t(z)=2G t(z)H.Note that this function can equivalently be defined as G t(z)= g t(z)−g t(0)=g t(z)+2t,where(g t)is the chordal Loewner chain driven by the function W t=1−2t.Hence,G t is the unique conformal map from H\K t onto H such that G t(0)=0and G t(z)/z→1when z→∞.(Here,K t is the evolving hull of g t.)Also,and this is why we focus on these functions G t,one has G t◦G s=G t+s in H\K t+s,for all s,t≥0.Since F is a homomorphism,11this implies that F (G t )=exp(−2αt )for some constant α≥0and all t ≥0,or that F (G t )=0for all t >0.However,the latter possibility would imply that K ∩K t =∅a.s.,for all t >0.Since t>0K t ={1}and 1/∈H \A and n A n is bounded away from 0and ∞.(Thisis very closely related to what is known as the Carath´e odory topology.)Now assume that A n →A ,where A n ,A ∈Q +.It is immediate that Φ′A n (0)→Φ′A (0),by Cauchy’s derivative formula (the maps may be extended to a neighborhood of 0by Schwarz reflection in the real line).Set A +n =ΦA n (A \A n )and A −n =ΦA (A n \A ).We claim that there is a constant δ>0and a sequence δn →0such thatA +n ∪A −n ⊂{x +iy :x ∈[δ,1/δ],y ≤δn }.(3.3)Indeed,since the map ΦA n ◦Φ−1A converges to the identity,locally uniformlyin H ,it follows (e.g.,from the argument principle)that for every compact set S ⊂H for all sufficiently large n ,S is contained in the image of ΦA n ◦Φ−1A ,which means that A +n ∩S =∅.Similarly,Φ−1A ◦ΦA n converges locally uniformly inthere is someǫ>0such that for infinitely many n P[K∩A−n]>ǫ.There-fore,with positive probability,K intersects infinitely many A−n.Since K is closed and(3.3)holds,this would then imply that P K∩[δ,1/δ] >0,a contradiction.Thus lim sup n→∞F(A n)≤F(A).A similar argument alsoshows that lim inf n→∞F(A n)≥F(A),and so lim n→∞F(A n)=F(A),and the continuity of F is verified.To complete the proof of the Lemma,we now show that A0is dense in Q+.Let A∈Q+.Set A′:=A∪[x0,x1],where x0:=inf(A∩R)and x1:=sup(A∩R).Forδ>0,δ<ΦA(x0)/2,let Dδbe the set of points in H with distance at mostδfrom[ΦA(x0),ΦA(x1)].Let Eδdenote theclosure of A∪Φ−1A (Dδ).It is clear that Eδ→A asδ→0+in the topologyconsidered above.It thus suffices to approximate Eδ.Note thatH withβ(0),β(s)∈R.We may assume thatβis parametrized by half-plane capacity from∞,so that a β[0,t] = 2t,t∈[0,s].Set g t:=gβ[0,t],Φt:=Φβ[0,t]=g t−g t(0),U t:=g t β(t) ,˜Ut:=U t−g t(0)=Φt β(t) ,t∈[0,s].By the chordal version of Loewner’s theorem,we have∂t g t(z)=2Φt(z)−˜U t +2(Φt(z)−˜U t)˜U t,Φ0(z)=z.(3.4)Since˜U t is continuous and positive,there is a sequence of piecewise constant functions˜U(n):[0,s]→(0,∞)such that sup{|˜U(n)t−˜U t|:t∈[0,s]}→0 as n→∞.LetΦ(n)t be the solution of(3.4)with˜U(n)t replacing˜U t.Then, clearly,Φ(n)s(z)→Φs(z)=ΦEδlocally uniformly inΦ′A+(0)are bounded away from zero,but limǫց0Φ′A(0)=0.AsΦA∗−◦ΦA+=ΦA=ΦA∗+◦ΦA−,(3.5) we haveΦ′A∗−(0)→0whenǫ→0.By applying F to(3.5)we getΦ′A∗−(0)α−Φ′A+(0)α=Φ′A∗+(0)αΦ′A−(0)α−.AsΦ′A∗−(0)=Φ′A∗+(0)(by symmetry),this means thatΦ′A∗−(0)α−α−staysbounded and bounded away from zero asǫց0,which givesα−=α.Since every A∈Q∗can be written as A+·A−this establishes4withα≥0.The caseα=0clearly implies K=∅a.s.,which is not permitted.This completes the proof.Let us now conclude this section with some simple remarks:Remark3.6.If K1,...,K n are independent sets with respective laws Pα1,...,Pαn,then the law of thefilling K:=F R H(K1∪...∪K n)of the union of the K j’s is Pαwithα=α1+···+αn becauseP[K⊂Φ−1(H)]=nj=1P[K j⊂Φ−1(H)]=Φ′(0)α1+···+αn.Remark3.7.Whenα<1/2,the measure Pαdoes not exist.To see this, suppose it did.Since it is unique,it is invariant under the symmetryσ: x+iy→−x+iy.Let A={e iθ:θ∈[0,π/2]}.Since K is almost surely connected and joins0to infinity,it meets either A orσ(A).Hence,symmetry implies thatΦ′A(0)α=P[K∩A=∅]≤1/2.On the other hand,one can calculate directlyΦ′A(0)=1/4,and henceα≥1/2.We will show later in the paper(Corollary8.6)that Pαonly exists for α≥5/8.Remark3.8.We have chosen to study subsets of the upper half-plane with the two special boundary points0and∞,but our analysis clearly applies to any simply connected domain O=C with two distinguished boundary points a and b,a=b.(We need to assume that the boundary of O is sufficiently nice near a and b.Otherwise,one needs to discuss prime ends in place of the distinguished points.)For instance,if∂O is smooth in the neighborhood of a and b,then for a conformal mapΦfrom a subset O′of O onto O,we getP[K∩(O\O′)=∅]=(Φ′(a)Φ′(b))α,14where P denotes the image of Pαunder a conformal map from H to O that takes a to0and b to∞.Remark3.9.The proof actually shows that weaker assumptions onΩare sufficient for the proposition.DefineΩb just asΩwas defined,except that Condition1is replaced by the requirements that K=∅andH\{0} such thatγ[0,1]∩R={γ(0),γ(1)}and and with positive P-probability K∩γ[0,1]=∅andγ[0,1]separates K inK and K is unbounded.Thus,P[Ω]=1. Using this fact,Lemma3.2may be applied,giving the remaining implication 3⇒2.4Brownian excursionsAn important example of a restriction measure is given by the law of the Brownian excursion from0to infinity in H.Loosely speaking,this is simply planar Brownian motion started from the origin and conditioned to stay in H at all positive times.It is closely related to the“complete conformal invariance”of(slightly different)measures on Brownian excursions in[32,27].Let X be a standard one-dimensional Brownian motion and Y an inde-pendent three-dimensional Bessel process(see e.g.,[41]for background on three-dimensional Bessel processes,its relation to Brownian motion condi-tioned to stay positive and stochastic differential equations).Let us briefly re-call that a three-dimensional Bessel process is the modulus(Euclidean norm) of a three-dimensional Brownian motion,and that it can be defined as the solution to the stochastic differential equation dY t=dw t+dt/Y t,where w is standard Brownian motion in R.It is very easy to see that(1/Y t,t≥t0)is15a local martingale for all t0>0,and that if T r denotes the hitting time of rby Y,then the law of(Y Tr+t ,t<T R−T r)is identical to that of a Brownianmotion started from r and conditioned to hit R before0(if0<r<R).Note that almost surely lim t→∞Y t=∞.The Brownian excursion can be defined as B t=X t+iY t.In other words, B has the same law as the solution to the following stochastic differential equation:dB t=dW t+i1Proof.LetΦ=ΦA.Suppose that W is a planar Brownian motion and Z is a Brownian excursion in H,both starting at z∈H\A.When Im(z)→∞, Im(Φ−1(z))=Im(z)+o(1).Hence,with a large probability(when R is large),a Brownian motion started from z∈I R(respectively,z∈Φ−1(I R)) will hitΦ−1(I R)(resp.,I R)before R.The strong Markov property of planar Brownian motion therefore shows that when R→∞,P[W hits I R before A∪R]∼P[W hitsΦ−1(I R)before A∪R].But sinceΦ◦W is a time-changed Brownian motion,andΦ:H\A→H,the right-hand is equal to the probability that a Brownian motion started from Φ(z)hits I R before R,namely,Im(Φ(z))/R.Hence,P[Z hits I R before A]=P[W hits I R before A∪R]Im(z)+o(1)when R→∞.In the limit R→∞,we getP[Z⊂H\A]=Im[Φ(z)]Im(z).(4.2)When z→0,Φ(z)=zΦ′(0)+O(|z|2)so thatP[B[0,∞)∩A=∅]=lims→0P[B[s,∞)⊂H\A]=lims→0E Im(Φ(B s))Figure R×[0,1].Using almost the same proof as in Proposition4.1(but keeping track ofthe law of the path),one can prove the following:Lemma4.2.Suppose A∈Q∗and B is a Brownian excursion in H starting at0.Then the conditional law of(ΦA(B(t)),t≥0)given B∩A=∅is the same as a time change of B.Finally,let us mention the following result that will be useful later on. Lemma4.3.Let P x+iy denote the law of a Brownian excursion B starting at x+iy∈y=a(A)yx2+y2 ,and the lemma readily follows.18Figure5.1:The various maps.Using Cauchy’s Theorem,for example,it is easy to see that the second statement of the lemma may be strengthened toy ∞−∞P x+iy B[0,∞)∩A=∅ dx=πa(A),y>sup{Im z:z∈A}.(4.3) 5Conformal image of chordal SLELet W:[0,∞)→R be continuous with W0=0,and let(g t)be the(chordal) Loewner chain driven by W satisfying(2.6).It is easy to verify by differen-tiation and(2.6)that the inverse map f t(z)=g−1t(z)satisfies2f′t(z)∂t f t(z)=−+o(z−1),z→∞,zwhere the coefficient a(t)depends on G and W t.Note that˜g t satisfies the Loewner equation∂t a(t)∂t˜g t(z)=where ˜Wt :=h t (W t ),h t :=˜g t ◦G ◦g −1t =g A t .(This follows from the proof of Loewner’s theorem,because ˜g t (˜K t +δ\˜K t )lies in a small neighborhood of ˜W t when δ>0is small.Also see [23,(2.6)].)The identity (2.5)gives a (g t (K t +∆t \K t ))=2∆t .The image of ˜K t +∆t \˜K t under˜g t is h t g t (K t +∆t \K t ) .The scaling rule (2.3)of a tells us that as ∆t →0+,the half-plane capacity of h t g t (K t +∆t \K t ) is asymptotic to h ′t (W t )2·2∆t .(The higher order derivatives of h t can be ignored,as follows from (2.2).Also see [23,(2.7)].)Hence,∂t a (t )=2h ′t (W t )2.(5.1)Using the chain rule we get[∂t h t ](z )=2h ′t (W t )2z −W t .(5.2)This formula is valid for z ∈H \g t (A )as well as for z in a punctured neighborhood of W t in R .In fact,it is also valid at W t with[∂t h t ](W t )=lim z →W t 2h ′t (W t )2z −W t =−3h ′′t (W t ).Computations of a similar nature appear (in a deterministic setting)in [11].Differentiating (5.2)with respect to z gives the equation[∂t h ′t ](z )=−2h ′t (W t )2h ′t (z )(z −W t )2−2h ′′t (z )2h ′t (W t )−4h ′′′t (W t )κdB tfor some measurable process b t adapted to the filtration of B t which satisfies t 0|b s|ds <∞a.s.for every t >0.20。
《经济学人》杂志原版英文(整理完整版)之欧阳学创编
Digest Of The. Economist.2006(6-7)Hard to digestA wealth of genetic information is to be found in the human gutBACTERIA, like people, can be divided into friend and foe. Inspired by evidence that the friendly sort may help with a range of ailments, many people consume bacteria in the form of yogurts and dietary supplements. Such a smattering of artificial additions, however, represents but a drop in the ocean. There are at least 800 types of bacteria living in the human gut. And research by Steven Gill of the Institute for Genomic Research in Rockville, Maryland, and his colleagues, published in this week's Science, suggests that the collective genome of these organisms is so large that it contains 100 times as many genes as the human genome itself.Dr Gill and his team were able to come to this conclusion by extracting bacterial DNA from the faeces of two volunteers. Because of the complexity of the samples, they were not able to reconstruct the entire genomes of each of the gut bacteria,just the individual genes. But that allowed them to make an estimate of numbers.What all these bacteria are doing is tricky to identify—the bacteria themselves are difficult to cultivate. So the researchers guessed at what they might be up to by comparing the genes they discovered with published databases of genes whose functions are already known.This comparison helped Dr Gill identify for the first time the probable enzymatic processes by which bacteria help humans to digest the complex carbohydrates in plants. The bacteria also contain a plentiful supply of genes involved in the synthesis of chemicals essential to human life—including two B vitamins and certain essential amino acids—although the team merely showed that these metabolic pathways exist rather than proving that they are used. Nevertheless, the pathways they found leave humans looking more like ruminants: animals such as goats and sheep that use bacteria to break down otherwise indigestible matter in the plants they eat.The broader conclusion Dr Gill draws is that people are superorganisms whose metabolism represents an amalgamation of human and microbial attributes. The notionof a superorganism has emerged before, as researchers in otherfields have come to view humans as having a diverse internal ecosystem. This, suggest some, will be crucial to the successof personalised medicine, as different people will have different responses to drugs, depending on their microbial flora. Accordingly, the next step, says Dr Gill, is to see how microbial populations vary between people of different ages, backgrounds and diets.Another area of research is the process by which these helpful bacteria first colonise the digestive tract. Babies acquire their gut flora as they pass down the birth canal and take a gene-filled gulp of their mother's vaginal and faecal flora. It might not be the most delicious of first meals, but it could well be an important one.Zapping the bluesThe rebirth of electric-shock treatmentELECTRICITY has long been used to treat medical disorders. As early as the second century AD, Galen, a Greek physician, recommended the use of electric eels for treating headaches and facial pain. In the 1930s Ugo Cerletti and Lucio Bini, two Italian psychiatrists, used electroconvulsive therapy to treat schizophrenia. These days, such rigorous techniques are practised less widely. But researchers are still investigatinghow a gentler electric therapy appears to treat depression.Vagus-nerve stimulation, to give it its proper name, was originally developed to treat severe epilepsy. It requires a pacemaker-like device to be implanted in a patient's chest and wires from it threaded up to the vagus nerve on the left side of his neck. In the normal course of events, this provides an electrical pulse to the vagus nerve for 30 seconds every five minutes.This treatment does not always work, but in some cases where it failed (the number of epileptic seizures experienced by a patient remaining the same), that patient nevertheless reported feeling much better after receiving the implant. This secondary effect led to trials for treating depression and, in 2005, America's Food and Drug Administration approved the therapy for depression that fails to respond to all conventional treatments, including drugs and psychotherapy.Not only does the treatment work, but its effects appear to be long lasting. A study led by Charles Conway of Saint Louis University in Missouri, and presented to a recent meeting of the American Psychiatric Association, has found that 70% of patients who are better after one year stay better after two years as well.The technique builds on a procedure called deep-brain stimulation, in which electrodes are implanted deep into the white matter of patients' brains and used to “reboot” faulty neural circuitry. Such an operation is a big undertaking, requiring a full day of surgery and carrying a risk of the patient suffering a stroke. Only a small number of people have been treated this way. In contrast, the device that stimulates the vagus nerve can be implanted in 45 minutes without a stay in hospital.The trouble is that vagus-nerve stimulation can take a long time to produce its full beneficial effect. According to Dr Conway, scans taken using a technique called positron-emission tomography show significant changes in brain activity starting three months after treatment begins. The changes are similar to the improvements seen in patients who undergo other forms of antidepression treatment. The brain continues to change over the following 21 months. Dr Conway says that patients should be told that the antidepressant effects could be slow in coming.However, Richard Selway of King's College Hospital, London, found that his patients' moods improved just weeks after the implant. Although brain scans are useful indetermining the longevity of the treatment, Mr Selway notes that visible changes in the brain do not necessarily correlate perfectly with changes in mood.Nobody knows why stimulating the vagus nerve improves the mood of depressed patients, but Mr Selway has a theory. He believes that the electrical stimulation causes a region in the brain stem called the locus caeruleus (Latin, ironically, for “blue place”) to flood the brain with norepinephrine, a neurotransmitter implicated in alertness, concentration and motivation—that is, the mood states missing in depressed patients. Whatever the mechanism, for the depressed a therapy that is relatively safe and long lasting is rare cause for cheer. The shape of things to comeHow tomorrow's nuclear power stations will differ from today'sTHE agency in charge of promoting nuclear power in America describes a new generation of reactors that will be “highly economical” with “enhanced safety”, that “minimise wastes” and will prove “proliferation resistant”. No doubt they will bake a mean apple pie, too.Unfortunately, in the world of nuclear energy, fine words are not enough. America got away lightly with its nuclearaccident. When the Three Mile Island plant in Pennsylvania overheated in 1979 very little radiation leaked, and there were no injuries. Europe was not so lucky. The accident at Chernobyl in Ukraine in 1986 killed dozens immediately and has affected (sometimes fatally) the health of tens of thousands at the least. Even discounting the association of nuclear power with nuclear weaponry, people have good reason to be suspicious of claims that reactors are safe.Yet political interest in nuclear power is reviving across the world, thanks in part to concerns about global warming and energy security. Already, some 441 commercial reactors operate in 31 countries and provide 17% of the planet's electricity, according to America's Department of Energy. Until recently, the talk was of how to retire these reactors gracefully. Now it is of how to extend their lives. In addition, another 32 reactors are being built, mostly in India, China and their neighbours. These new power stations belong to what has been called the third generation of reactors, designs that have been informed by experience and that are considered by their creators to be advanced. But will these new stations really be safer than their predecessors?Clearly, modern designs need to be less accident prone.The most important feature of a safe design is that it “fails safe”. Fo r a reactor, this means that if its control systems stop working it shuts down automatically, safely dissipates the heat produced by the reactions in its core, and stops both the fuel and the radioactive waste produced by nuclear reactions from escaping by keeping them within some sort of containment vessel. Reactors that follow such rules are called “passive”. Most modern designs are passive to some extent and some newer ones are truly so. However, some of the genuinely passive reactors are also likely to be more expensive to run.Nuclear energy is produced by atomic fission. A large atom (usually uranium or plutonium) breaks into two smaller ones, releasing energy and neutrons. The neutrons then trigger further break-ups. And so on. If this “chain reaction” can be controlled, the energy released can be used to boil water, produce steam and drive a turbine that generates electricity. If it runs away, the result is a meltdown and an accident (or, in extreme circumstances, a nuclear explosion—though circumstances are never that extreme in a reactor because the fuel is less fissile than the material in a bomb). In many new designs the neutrons, and thus the chain reaction, are kept under control by passing them through water to slow themdown. (Slow neutrons trigger more break ups than fast ones.) This water is exposed to a pressure of about 150 atmospheres—a pressure that means it remains liquid even at high temperatures. When nuclear reactions warm the water, its density drops, and the neutrons passing through it are no longer slowed enough to trigger further reactions. That negative feedback stabilises the reaction rate.Can business be cool?Why a growing number of firms are taking global warming seriouslyRUPERT MURDOCH is no green activist. But in Pebble Beach later this summer, the annual gathering of executivesof Mr Murdoch's News Corporation—which last year led to a dramatic shift in the media conglomerate's attitude tothe internet—will be addressed by several leading environmentalists, including a vice-president turned climatechangemovie star. Last month BSkyB, a British satellite-television company chaired by Mr Murdoch and run by hisson, James, declared itself “carbon-neutral”, having taken various steps to cut or offset its discharges of carboninto the atmosphere.The army of corporate greens is growing fast. Late lastyear HSBC became the first big bank to announce that itwas carbon-neutral, joining other financial institutions, including Swiss Re, a reinsurer, and Goldman Sachs, aninvestment bank, in waging war on climate-warming gases (of which carbon dioxide is the main culprit). Last yearGeneral Electric (GE), an industrial powerhouse, launched its “Ecomagination” strategy, aiming to cut its output ofgreenhouse gases and to invest heavily in clean (ie, carbon-free) technologies. In October Wal-Mart announced aseries of environmental schemes, including doubling the fuel-efficiency of its fleet of vehicles within a decade.Tesco and Sainsbury, two of Britain's biggest retailers, are competing fiercely to be the greenest. And on June 7thsome leading British bosses lobbied Tony Blair for a more ambitious policy on climate change, even if that involvesharsher regulation.The greening of business is by no means universal, however. Money from Exxon Mobil, Ford and General Motorshelped pay for television advertisements aired recently in America by the Competitive Enterprise Institute, with thedaft slogan “Carbon dioxide: they call it pollution; we call it life”. Besides, environmentalist critics say, some firmsa re engaged in superficial “greenwash” to boost the image ofessentially climate-hurting businesses. Take BP, themost prominent corporate advocate of action on climate change, with its “Beyond Petroleum” ad campaign, highprofileinvestments in green energy, and even a “carbon calculator” on its website that helps consumers measuretheir personal “carbon footprint”, or overall emissions of carbon. Yet, critics complain, BP's recent record profits arelargely thanks to sales of huge amounts of carbon-packed oil and gas.On the other hand, some free-market thinkers see the support of firms for regulation of carbon as the latestattempt at “regulatory capture”, by those who stand to profit from new rules. Max Schulz of the ManhattanInstitute, a conservative think tan k, notes darkly that “Enron was into pushing the idea of climate change, becauseit was good for its business”.Others argue that climate change has no more place in corporate boardrooms than do discussions of other partisanpolitical issues, such as Darfur or gay marriage. That criticism, at least, is surely wrong. Most of the corporateconverts say they are acting not out of some vague sense of social responsibility, or even personal angst, butbecause climate change creates real business risks and opportunities—from regulatory compliance to insuringclientson flood plains. And although these concerns vary hugely from one company to the next, few firms can besure of remaining unaffected.Testing timesResearchers are working on ways to reduce the need for animal experiments, but new laws mayincrease the number of experiments neededIN AN ideal world, people would not perform experiments on animals. For the people, they are expensive. For theanimals, they are stressful and often painful.That ideal world, sadly, is still some way away. People need new drugs and vaccines. They want protection fromthe toxicity of chemicals. The search for basic scientific answers goes on. Indeed, the European Commission isforging ahead with proposals that will increase the number of animal experiments carried out in the EuropeanUnion, by requiring toxicity tests on every chemical approved for use within the union's borders in the past 25years.Already, the commission has identified 140,000 chemicals that have not yet been tested. It wants 30,000 of theseto be examined right away, and plans to spend between €4 billion-8 billion ($5 billion-10 billion) doing so. Thenumberof animals used for toxicity testing in Europe will thus, experts reckon, quintuple from just over 1m a yearto about 5m, unless they are saved by some dramatic advances in non-animal testing technology. At the moment,roughly 10% of European animal tests are for general toxicity, 35% for basic research, 45% for drugs andvaccines, and the remaining 10% a varietyof uses such as diagnosing diseases.Animal experimentation will therefore be around for some time yet. But the hunt for substitutes continues, and lastweekend the Middle European Society for Alternative Methods to Animal Testing met in Linz, Austria, to reviewprogress.A good place to start finding alternatives for toxicity tests is the liver—the organ responsible for breaking toxicchemicals down into safer molecules that can then be excreted. Two firms, one large and one small, told themeeting how they were using human liver cells removed incidentally during surgery to test various substances forlong-term toxic effects.PrimeCyte, the small firm, grows its cells in cultures over a few weeks and doses them regularly with the substanceunder investigation. The characteristics of the cells are carefully monitored, to look for changes in theirmicroanatomy.Pfizer,the big firm, also doses its cultures regularly, but rather than studying individual cells in detail, it counts cellnumbers. If the number of cells in a culture changes after a sample is added, that suggests the chemical inquestion is bad for the liver.In principle, these techniques could be applied to any chemical. In practice, drugs (and, in the case of PrimeCyte,food supplements) are top of the list. But that might change if the commission has its way: those 140,000screenings look like a lucrative market, although nobody knows whether the new tests will be ready for use by2009, when the commission proposes that testing should start.Other tissues, too, can be tested independently of animals. Epithelix, a small firm in Geneva, has developed anartificial version of the lining of the lungs. According to Huang Song, one of Epithelix's researchers, the firm'scultured cells have similar microanatomy to those found in natural lung linings, and respond in the same way tovarious chemical messengers. Dr Huang says that they could be used in long-term toxicity tests of airbornechemicals and could also help identify treatments for lung diseases.The immune system can be mimicked and tested, too. ProBioGen, a company based in Berlin, is developinganartificial human lymph node which, it reckons, could have prevented the near-disastrous consequences of a drugtrial held in Britain three months ago, in which (despite the drug having passed animal tests) six men sufferedmultiple organ failure and nearly died. The drug the men were given made their immune systems hyperactive.Such a response would, the firm's scientists reckon, have been identified by their lymph node, which is made fromcells that provoke the immune system into a response. ProBioGen's lymph node could thus work better than animaltesting.Another way of cutting the number of animal experiments would be tochange the way that vaccines are tested, according to CoenraadHendriksen of the Netherlands Vaccine Institute. At the moment, allbatches of vaccine are subject to the same battery of tests. DrHendriksen argues that this is over-rigorous. When new vaccine culturesare made, belt-and-braces tests obviously need to be applied. But if abatch of vaccine is derived from an existing culture, he suggests that itneed be tested only to make sure it is identical to the batch from which itis derived. That would require fewer test animals.All this suggests that though there is still some way to go before drugs,vaccines and other substances can be tested routinely oncells ratherthan live animals, useful progress is being made. What is harder to see ishow the use of animals might be banished from fundamental research.Anger managementTo one emotion, men are more sensitive than womenMEN are notoriously insensitive to the emotional world around them. At least, that is the stereotype peddled by athousand women's magazines. And a study by two researchers at the University of Melbourne, in Australia,confirms that men are, indeed, less sensitive to emotion than women, with one important and suggestiveexception. Men are acutely sensitive to the anger of other men.Mark Williams and Jason Mattingley, whose study has just been published in Current Biology, looked at the way aperson's sex affects his or her response to emotionally charged facial expressions. People from all cultures agreeon what six basic expressions of emotion look like. Whether the face before you is expressing anger, disgust, fear,joy, sadness or surprise seems to be recognised universally—which suggests that the expressions involved areinnate, rather than learned.Dr Williams and Dr Mattingley showed the participants intheir study photographs of these emotional expressions inmixed sets of either four or eight. They asked the participants to look for a particular sort of expression, andmeasured the amount of time it took them to find it. The researchers found, in agreement with previous studies,that both men and women identified angry expressions most quickly. But they also found that anger was morequickly identified on a male face than a female one.Moreover, most participants could find an angry face just as quickly when it was mixed in a group of eightphotographs as when it was part of a group of four. That was in stark contrast to the other five sorts of expression,which took more time to find when they had to be sorted from a larger group. This suggests that something in thebrain is attuned to picking out angry expressions, and that it is especially concerned about angry men. Also, thishighly tuned ability seems more important to males than females, since the two researchers found that men pickedout the angry expressions faster than women did, even though women were usually quicker than men to recognizeevery other sort of facial expression.Dr Williams and Dr Mattingley suspect the reason for this is that being able to spot an angry individual quickly hasasurvival advantage—and, since anger is more likely to turn into lethal violence in men than in women, the abilityto spot angry males quickly is particularly valuable.As to why men are more sensitive to anger than women, it is presumably because they are far more likely to getkilled by it. Most murders involve men killing other men—even today the context of homicide is usually aspontaneous dispute over status or sex.The ability to spot quickly that an alpha male is in a foul mood would thus have great survival value. It would allowthe sharp-witted time to choose appeasement, defence or possibly even pre-emptive attack. And, if it is right, thisstudy also confirms a lesson learned by generations of bar-room tough guys and schoolyard bullies: if you wantattention, get angry. The shareholders' revoltA turning point in relations between company owners and bosses?SOMETHING strange has been happening this year at company annual meetings in America:shareholders have been voting decisively against the recommendations of managers. Until now, mostshareholders have, like so many sheep, routinely voted in accordance with the advice of the peopletheyemploy to run the company. This year managers have already been defeated at some 32 companies,including household names such as Boeing, ExxonMobil and General Motors.This shareholders' revolt has focused entirely on one issue: the method by which members of the boardof directors are elected. Shareholder resolutions on other subjects have mostly been defeated, as usual.The successful resolutions called for directors to be elected by majority voting, instead of by thetraditional method of “plurality”—which in practice meant that only votes cast in favour were counted,and that a single vote for a candidate would be enough to get him elected.Several companies, led by Pfizer, a drug giant, saw defeat looming and pre-emptively adopted a formalmajority-voting policy that was weaker than in the shareholder resolution. This required any director whofailed to secure a majority of votes to tender his resignation to the board, which would then be free todecide whether or not to accept it. Under the shareholder resolution, any candidate failing to secure amajority of the votes cast simply would not be elected. Intriguingly, the shareholder resolution wasdefeated at four-fifths of the firms that adopted a Pfizer-style majority voting rule, whereas itsucceedednearly nine times out of ten at firms retaining the plurality rule.Unfortunately for shareholders, their victories may prove illusory, as the successful resolutions were all“precatory”—meaning that they merely advised management on the course of action preferred byshareholders, but did not force managers to do anything. Several resolutions that tried to imposemajority voting on firms by changing their bylaws failed this year.Even so, wise managers should voluntarily adopt majority voting, according to Wachtell, Lipton, Rosen &Katz, a Wall Street law firm that has generally helped managers resist increases in shareholder power butnow expects majority voting eventually to “become universal”. It advises th at, at the very least,managers should adopt the Pfizer model, if only to avoid becoming the subject of even greater scrutinyfrom corporate-governance activists. Some firms might choose to go further, as Dell and Intel have donethis year, and adopt bylaws requiring majority voting.Shareholders may have been radicalised by the success last year of a lobbying effort by managersagainst a proposal from regulators to make it easier for shareholders to put up candidates in boardelections. It remains to be seen if they willbe back for more in 2007. Certainly, some of the activistshareholders behind this year's resolutions have big plans. Where new voting rules are in place, they plancampaigns to vote out the chairman of the compensation committee at any firm that they think overpaysthe boss. If the 2006 annual meeting was unpleasant for managers, next year's could be far worse.Intangible opportunitiesCompanies are borrowing against their copyrights, trademarks and patentsNOT long ago, the value of companies resided mostly in things you could see and touch. Today it liesincreasingly in intangible assets such as the McDonald's name, the patent for Viagra and the rights toSpiderman. Baruch Lev, a finance professor at New York University's Stern School of Business, puts theimplied value of intangibles on American companies' balance sheets at about $6 trillion, or two-thirds ofthe total. Much of this consists of intellectual property, the collective name for copyrights, trademarksand patents. Increasingly, companies and their clever bankers are using these assets to raise cash.The method of choice is securitisation, the issuing ofbonds based on the various revenues thrown off byintellectual property. Late last month Dunkin' Brands, owner of Dunkin' Donuts, a snack-bar chain, raised$1.7 billion by selling bonds backed by, among other things, the royalties it will receive from itsfranchisees. The three private-equity firms that acquired Dunkin' Brands a few months ago have used thecash to repay the money they borrowed to buy the chain. This is the biggest intellectual-propertysecuritisation by far, says Jordan Yarett of Paul, Weiss, Rifkind, Wharton & Garrison, a law firm that hasworked on many such deals.Securitisations of intellectual property can be based on revenues from copyrights, trademarks (such aslogos) or patents. The best-known copyright deal was the issue in 1997 of $55m-worth of “Bowie Bonds”supported by the future sales of music by David Bowie, a British rock star. Bonds based on the films ofDreamWorks, Marvel comic books and the stories of John Steinbeck have also been sold. As well asDunkin' Brands, several restaurant chains and fashion firms have issued bonds backed by logos andbrands.Intellectual-property deals belong to a class known as operating-asset securitisations. These differ fromstandard securitisations of future revenues, such as bonds backed by thepayments on a 30-yearmortgage or a car loan, in that the borrower has to make his asset work. If investors are to recoup theirmoney, the assets being securitised must be “actively exploited”, says Mr Yarett: DreamWorks mustcontinue to churn out box-office hits.The market for such securitisations is still small. Jay Eisbruck, of Moody's, a rating agency, reckons thataround $10 billion-worth of bonds ar e outstanding. But there is “big potential,” he says, pointing out thatlicensing patented technology generates $100 billion a year and involves thousands of companies.Raising money this way can make sense not only for clever private-equity firms, but also for companieswith low (or no) credit ratings that cannot easily tap the capital markets or with few tangible assets ascollateral for bank loans. Some universities have joined in, too. Yale built a new medical complex withsome of the roughly $100m it raised securitising patent royalties from Zerit, an anti-HIV drug.It may be harder for investors to decide whether such deals are worth their while. They are, after all,highly complex and riskier than standard securitisations. The most obvious risk is that the investorscannot be sure that the assets will yield。
the nature of scientific reasoning
本次翻译练习的难度比较大，文章出自北京师范大学研究生英语阅读与翻译课程所用的授课材料，作者布洛诺夫斯基是英国著名的数学家和散文家，剑桥大学数学博士。
这篇文章从科学发展史的角度出发，论述的问题主要是科学并不排斥想象力和创造力。
因此标题翻译成“科学理性的本质”或“科学推理的本质”是比较恰当的。
要翻译好这篇文章不仅应在在宏观的层面牢牢把握文章的主旨，也需要从微观的角度考虑作者使用的语言在语法和修辞上的特点，这样才能在理解的基础上恰当的表达。
当然，这篇文章相对于大家目前的英语水平，在理解和表达两个方面都具有不小的挑战性。
下面通过对这次翻译比较好的赵新平同学作业的点评，来分段落说一说这篇文章究竟有哪些细节部分需要注意，以及相应的翻译策略。
1What is the insight in which the scientist tries to see into nature? Can it indeed be called either imaginative or creative? To the literary man the question may seem merely silly. He has been taught that science is a large collection of facts; and if this is true, then the only seeing which scientists need to do is, he supposes, seeing the facts. He pictures them, the colorless professionals of science, going off to work in the morning into the universe in a neutral, unexposed state. They then expose themselves like a photographic plate. And then in the darkroom or laboratory they develop the image, so that suddenly and startlingly it appears, printed in capital letters, as a new formula for atomic energy.原译：什么是洞察力？科学家一直试图弄清它的本质。
Proof of the Minimax Theorem
Proof of the Minimax TheoremMichael A.GoodrichSeptember17,2007The proof of the minimax theorem follows the format given in Luce and Raiffa[2],but has been modified to use our class terminology.Changes in the tutorial since it wasfirst posted in Fall2007are highlighted in red. FormalismsWefirst present the properties of a two-person,zero-sum game.To help you through these properties,I will group the properties together in a way that makes sense to me.I hope it will help you. Formalization of a2Person Zero-Sum Game1.There are two players,P1and P2.2.P1has a set A={a1,a2,...,a m}of m pure strategies(or actions).3.P2has a set B={b1,b2,...,b n}of n pure strategies(or actions).4.Each player has a utility for each(a i,b j)pair of actions.The utility for P1is denoted U1(a i,b j)andthe utility for P2is denoted U2(a i,b j).Since this is a zero-sum game,U1(a i,b j)=−U2(a i,b j)for all i and j.To minimize the number of subscripts we will carry around,let M(a i,b j)=U1(a i,b j) denote the mutual utility for the game.In essence,these properties state the number of players,the set of possible strategies available to each player,and the payoffs that occur when the game is played using pure strategies.Payoffs for Mixed StrategiesThe next properties identify what happens when players use mixed strategies.5.Each player can use a mixed strategy by creating a probability mass function and playing each purestrategy with afixed probability.Let p i denote the probability that player1will play action a i,and let q j denote the probability that player2will play action b j.Since p and q are probabilities,they must satisfy(a)∀i p i≥0,∀j q j≥0.(b) mi=1p i=1,nj=1q j=1.A mixed strategy that uses a particular probability mass function is denoted p=(p1,p2,...,p m)where p i=P r(a i)is the probability that action a i will be played;similarly,for player2q= (q1,q2,...,q n).6.For each randomized strategy pair(p,q),the payoff M(p,q)is defined to beM(p,q)=mi=1nj=1p i M(a i,b j)q j.We denote the payoff when player1uses pure strategy a i and player2uses mixed strategy q asM(a i,q)=nj=1M(a i,b j)q jwith similar notation for M(p,b j).7.In much the same way that A and B denote the set of pure strategies available to players1and2,respectively,we use P and Q to denote the set of all mixed strategies available to players1and2, respectively.Maximum SecurityNow that we have a language for expressing expected payoff for mixed strategies,we can select which of the possible mixed strategies are minimax or maximin strategies.We do this with the introduction of the notions of security and regret.In this section,we talk about maximizing security,and in the next section we discuss minimizing regret.8.Player1’s objective is to select a randomized strategy p from P so as to maximize M(p,q).At thesame time,player2’s objective is to select a randomized strategy q from Q so as to maximize its payoff,which is equivalent to minimizing M(p,q).The rules of the game require that each player choose its strategy in complete ignorance of the opponent’s selection.9.For each mixed strategy p belonging to P,player1’s security level is defined to bev1(p)=minqM(p,q).SinceM(p,q)=nj=1M(p,b j)q jis a weighted sum of the n payoffs M(p,b j),it is minimized when all of the weight is assigned to the least of these(do you see why?look at how we compute the maximin mixed strategy in the lecture notes.)v1(p)=min[M(p,b1),M(p,b2),...,M(p,b n)].You can think of v1(p)as the payoff that player1will receive if player2knows that P1will do p.(Why?Because if player2knows this,then it can choose its best response.)We can define v2(q) for player2in a similar way(but using maximums instead of minimums since high M means low payoff to player2).10.By assumption,player1wants to maximize its security level,so P1must choose a strategy p∗suchthatv1(p∗)≥v1(p)∀p∈P.Let v1denote this maximal security level(i.e.,v1=v1(p∗)).Thenv1(p)≥v1(p)(1)v1=maxpfor all other mixed strategies.We also know thatM(p∗,q)≤M(p∗,q)∀q∈Q.(2)v1=minqInequality(1)means that the strategy that produces v1is superior to all other strategies(in terms of maximizing security level).Inequality(2)means that v1is the worst(minimum payoff)that player 1can expect given that player1plays p∗.If player2doesn’t choose wisely then player1will get more than v1.If player1doesn’t play p∗then player1can receive a lower payoff.The strategy p∗is called the maximin strategy.Minimum RegretIn this section,we repeat the analysis from the previous section but apply it to the minimizing player (player2).11.Because the game is zero-sum,we know that when player2maximizes its security level then itminimizes player1’s payoff.If player2uses strategy q,player1cannot obtain a return greater thanM(p,q).v2(q)=maxpThe value v2is sometimes called regret,which kind of indicates that it is the negation of security.Just like player1tries to maximizes security,player2tries to minimize regret.Definev2(q),q∗=arg minqand definev2=v2(q∗)≤v2(q)∀q∈Q.(3) By repeating the analysis that we did for player1but with player2in mind,we learn thatv2≥M(p,q∗)∀p∈P.(4) The strategy q∗is called the minimax strategy.Relation Between Minimax and Maximin ValuesAs you may have guessed,there is a relationship between the minimax and maximin values for a zero-sum game.In this section,we develop this relationship.12.Putting these pieces together,we learn that if player1uses the maximin strategy,it is guaranteed atleast v1units of(security)payoffv1(p)≤v1≤M(p∗,q)∀q∈Q.(Note that v1(p)is a function of p which means that security is a function of what player1plans on doing.)Similarly,if player2uses the minimax strategy,it is guaranteed no more than v2units of (regret)loss,which is tantamount to guaranteeing that player1can receive no more than v2units of payoffM(p,q∗)≤v2≤v2(q)∀q∈Q.(Again,note that v2(q)is a function of q which means that regret is a function of what player2 plans on doing.)Thus,M(p,q∗)≤v2∀p∈PM(p∗,q∗)≤v2v1≤M(p∗,q)∀q∈Qv1≤M(p∗,q∗)v1≤v2.(5)Equilibrium ValueNow that we have established a relationship between the minimax and maximin value,we can formally define what an(Nash)equilibrium value for a game is.13.A pair of strategies(p ,q )is said to be in equilibrium if p is good against q and vice versa,meaningM(p,q )≤M(p ,q )≤M(p ,q).In words,these two strategies are in equilibrium if neither player has an incentive to change(can increase its payoff by unilaterally changing its behavior).To help understand this,it is useful to recall that a Nash equilibrium is a solution pair such that no player has an incentive to unilaterally change his or her action.Whew!That’s quite a bit of information,but its nothing more than a formalism of the concepts we’ve been talking about informally.Since one of my goals for this course is to help you get confident about reading the literature,I want you to practice putting your thoughts into a concise,mathematical language.A Big Picture MomentIt is sometimes useful when wrestling with concepts like these to take a moment and get the big picture. These notes aim at making clear three things:the formalization of a two person zero-sum game,theproof of the minimax theorem for two person games,and the relationship between minimax values and equilibrium values.You have justfinished slogging your way through the formalization of a two person zero-sum game.Before proceeding to the other two objectives,it is important that you understand the following:•The minimax theorem wasfirst proved by John von Neumann(I’ll try tofind the reference when I get a chance).This theorem states that for zero sum games there exists a unique value for this game.This value is the minimax value for the minimizing agent and the maximin value for the maximizing agent.•It just so happens that the pair of solutions that generate the maximin value and the minimax value for this game form a Nash equilibrium.This holds because the Nash equilibrium corresponds to the minimax/maximin solution pair for zero-sum games.Thus,for two-person zero-sum games,the presence of a unique value for a game implies that a Nash equilibrium exists for that game.•John Nash[1]showed that every game,not just two-person zero-sum games,has at least one equi-librium value.Since the minimax/maximin solution pair corresponds to this equilibrium,we will use a variant of Nash’s more general proof as the basis for the proof of the minimax theorem.•For non-zero sum games,the joint solution produced by players individually choosing a minimax strategy does not necessarily correspond to an equilibrium solution—consider the Battle of the Sexes game.A Useful TheoremWe now turn to an interesting theorem that applies only to zero-sum games.The theorem states that (a)if an equilibrium exists then the maximin value v1equals the minimax value v2,(b)if v1=v2then there exists a real number v and a pair of strategies p∗and q∗such that the payoffs for these strategies are bounded by v,and(c)if the conditions just mentioned hold then the game has an equilibrium.The theorem does not state that each zero-sum,two-person game satisfies any of these conditions;it only says that if it does then the maximin and minimax solutions are equilibrium solutions.Although we will state the theorem,we will omit the proof of this theorem(it follows almost immediately from our problem specification above)so that we can concentrate on the proof of the minimax theorem.In essence,this theorem states that we can use Nash’s proof of the existence of an equilibrium in general-sum games to show that the minimax solution pair exists and is in equilibrium for zero-sum games.The great thing about this theorem is that it establishes an equivalence between the minimax value,an equilibrium pair,and a concept called the value of the game.Can you think of a way to use the value of the game to create an algorithm thatfinds the equilibrium solution?I should point out that many useful theorems have a form similar to this one;they show that several different conditions are equivalent.These theorems are very serviceable because they allow us to tie a handful of mixed ideas into an equivalence.The proof of these theorems usually follows a procedure wherein we show that condition1implies condition2,condition2implies condition3,and so on until we can show that condition implies condition1.Theorem1For two-person,zero-sum games as we have presented them,each of the following three conditions implies the other two.1.An equilibrium pair exists.2.v1=maxp minqM(p,q)=minqmaxpM(p,q)=v2.3.There exists a real number v,a mixed strategy p∗,and a mixed strategy q∗such that(a)iM(a i,b j)p∗i≥v for j=1,2,...,n(b)jM(a i,b j)q∗j≤v for i=1,2,...,m.Note that condition3(3a)says that the average loss for player2using any pure strategy is no less than v. Similarly,condition3(3b)says that the average payoff for player1using any pure strategy is no greater than v.This value v is known as the value of a game in normal form.The Minimax TheoremWe now present Nash’s proof of the Minimax Theorem.Although there are several proofs of the theorem, Nash’s is fairly easy to understand.Additionally,although we only give the proof for the specification of a zero-sum game given above,the proof extends to N-person general-sum games as well.This means that every general-sum,N-person game has at least one Nash equilibrium.A simple interpretation of the steps in the proof goes something like this:•We are going to make up a transformation that maps mixed strategies to new mixed strategies.•We are going to show that this transformation is special:if the underlying game has an equilibrium then this transformation has afixed point,and if the transformation has afixed point only then the game has an equilibrium.•Since the transformation is continuous and since probability spaces are“topologically close”to a multi-dimensional sphere(we’ll discuss this in class),we know by Brouwer’sfixed point theorem that the transformation has afixed point.•Since thefixed point is,in essence,indistinguishable from the equilibrium point,it follows that the game has an equilibrium point.Theorem2For every two-person,zero-sum game,there exists an equilibrium strategy.Proof:Consider a transformation T that maps mixed strategy pairs(p,q)into mixed strategy pairs T(p,q)=(p ,q ).What we’ll show is that this transformation T has the following two properties:1.p∗and q∗are optimal(i.e.,maximin and minimax)strategies if and only if T(p∗,q∗)=(p∗,q∗).(Any point which is mapped to itself under a transformation is called afixed point of this transfor-mation.For example,consider the transformation T: +→ +defined as T(x)=x2.The value of x=1is afixed point since T(1)=12=1.)2.T,defined below,has at least onefixed point.In essence,c i(p,q)represents the improvement to player1for switching from strategy p to strategy a i. Similarly,d j(p,q)represents the improvement to player2for switching from strategy q to strategy b j.We will now construct a transformation that satisfies our two properties.To do this,define T as follows.Letc i(p,q)=M(a i,q)−M(p,q)if M(a i,q)−M(p,q)>0 0otherwise;d j(p,q)=M(p,q)−M(p,b j)if M(p,q)−M(p,b j)>0 0otherwise;Using the notation T(p,q)=(p ,q ),we definep i =p i+c i(p,q)1+mk=1c k(p,q)q i =q i+d i(p,q)1+nk=1d k(p,q).Before proceeding,we need to check and see if this is indeed a transformation from a probability to another probability.Thus,we want to show that T:P×Q→P×Q(i.e.that the things produced by the transformations are probabilities),so we need to show that p and q are probability mass functions. Clearly,since c i and d i are both positive then p i and q i are also both nonnegative.(Recall that probabilities cannot be negative.)The next requirement is that these probabilities sum to one.Sincemi=1pi=mi=1p i+c i(p,q)1+mk=1c k(p,q)=mi=1p i+mi=1c i(p,q)1+mk=1c k(p,q)=1+mi=1c i(p,q)1+mk=1c k(p,q)=1.Now that we know that the transformation is valid,we want to start the proof.Thefirst step is the ”If”step.When you look carefully at the statement of thefirst property of the transformation,you’ll see an if and only if A then B statement.Most of you remember this,but just in case you don’t,the way you establish such a statement isfirst showing the if A then B part and then showing the if B then A part. We’ll start by showing that if p∗and q∗are optimal then T(p∗,q∗)=(p∗,q∗).Observe that c i(p,q)is a measurement of the amount that a i is better than p(if at all)as a response against q.Similarly,d i(p,q)is a measurement of the amount that b i is better than q(if at all)as a response against p.When p∗and q∗are optimal,it follows that c i(p∗,q∗)=0for all i(can you see why?)so p∗i=p i for all i.Similary,q∗i=q i. Thus,T(p∗,q∗)=(p∗,q∗).Turning to the only if portion of the proof,suppose that(p,q)is afixed point.We need to show that (p,q)is optimal.Wefirst show that there exists an i such that both p i>0and c i(p,q)=0.Since,by definition(in class,ask me about how convex combinations can be graphically depicted),M(p,q)=mi=1p i M(a i,q)we conclude that M(p,q)<M(a i,q)cannot be true for all i such that p i>0.To see this,we’ll do a mini-proof within this proof.Suppose that,∀i such that p i>0,M(p,q)<M(a i,q).ThenM(p,q)=mi=1p i M(a i,q)>mi=1p i M(p,q)=M(p,q)m i=1p i=M(p,q).This is a contradiction,which means that for at least one p i>0it must follow that M(p,q)≥M(a i,q). This ends our mini-proof.We will now use this to show thatfixed points are optimal points.But this implies that,for this i,0≥M(a i,q)−M(p,q)so,by definition of c i,c i(p,q)=0for this i. For this i,the fact that(p,q)is afixed point implies thatp i=p i+01+mk=1c k(p,q).Since p i>0for this i,it follows that mk=1c k(p,q)=0.But the terms c k are all non-negative(bydefinition),so they must all equal0.This,in turn,means that M(p,q)≥M(a i,q)for all a i.Since this is true regardless of q,it follows that no other pure or mixed strategy has higher payoff for all q.Thus, M(p,q)≥M(p ,q)for all p and for all q so p is good against all q.Similarly,we can show that q is good against all p,so when(p,q)is afixed point of the transformation T then it is also optimal.The only thing we have to do now to complete the proof is show that the transformation T has a fixed point.This existence follows from the Brouwerfixed-point theorem.We won’t show how thefixed point theorem applies,but you might want to think about it a little bit.I’ll paraphrase Luce and Raiffa’s statement of the theorem.The theorem says that any continuous transformation that maps a point of a spheroid(or something topologically“close”to a spheroid)in afinite dimensional Euclidean space into another point of the sphereoid has at least onefixed point.The space P×Q is topologically“close”to a spheroid and our transformation is continuous,so we know that afixed point exists.Any optimal point is afixed point,and anyfixed point is an optimal point so we know that every zero-sum,two-player game has a mixed strategy equilibrium point.Yippee!We did it.References[1]Jr.J.F.Nash.The bargaining problem.Econometrica,18:155–162,1950.Reprinted in Classics inGame Theory,H.W.Kuhn,ed.[2]R.D.Luce and H.Raiffa.Games and Decisions.John Wiley,New York,1957.。
爱德华 诺顿 洛伦茨
“蝴蝶效应”是指在一个动力系统中，初始条件下微小的变化能带动整个系统的长期而巨大的连锁反应。这 是一种混沌现象，“一只蝴蝶在巴西轻拍翅膀，会使更多蝴蝶跟著一起振翅。最后将有数千只的蝴蝶都跟著那只 蝴蝶一同挥动翅膀，其所产生的飓风可以导致一个月后在美国得州发生一场龙卷风。”
罗伦兹是个兴趣十分广泛的人，他喜欢越野滑雪、徒步旅行，去世前的一周，他还参加过一次徒步活动。
研究成就
混沌理论
蝴蝶效应
1963年罗伦兹提出了“混沌理论”，这一理论拥有巨大的影响力，其主要精神是，在混沌系统中，初始条件 的微小变化，可能造成后续长期而巨大的连锁反应。此理论最为人所知的论述之一是“蝴蝶效应”：“一只蝴蝶 在巴西轻拍翅膀，会使更多蝴蝶跟着一起振翅，最后将有数千只的蝴蝶都跟着那只蝴蝶一同挥动翅膀，结果可以 导致一个月后在美国德州发生一场龙卷风。”
罗伦兹发现“混沌理论”颇具戏剧性效果，也可以算是混混沌沌中发现的。1961年，冬季的一天，罗伦兹在 电脑上进行关于天气预报的计算。为了考察一个很长的序列，他走了一条捷径，没有令计算机从头运行，而是从 中途开始。他把上次的输出直接打入作为计算的初值，然后他穿过大厅下楼，去喝咖啡。一小时后他回来时，发 生了出乎意料的事。
谢谢观看
在《混沌学传奇》等书中皆有这样的描述：“1961年冬季的一天，罗伦兹在计算机上进行关于天气预报的计 算。为了考察一个很长的序列，他走了一条捷径，没有令计算机从头运行，而是从中途开始。他把上次的输出直 接打入作为计算的初值，然后他穿过大厅下楼，去喝咖啡。一小时后他回来时，发生了出乎意料的事，他发现天 气变化同上一次的模式迅速偏离，在短时间内，相似性完全消失了。进一步的计算表明，输入的细微差异可能很 快成为输出的巨大差别。
宇宙探索编辑部 小节 英文名
宇宙探索编辑部小节英文名Cosmic Exploration Editorial BoardThe Cosmic Exploration Editorial Board is a prestigious organization dedicated to the advancement of space exploration and discovery. As the leading authority in the field, the board plays a crucial role in shaping the direction of humanity's journey into the unknown.Composed of renowned scientists, astronauts, and visionary thinkers, the Cosmic Exploration Editorial Board is responsible for evaluating proposals, approving missions, and guiding the global community in its pursuit of unlocking the secrets of the universe. With a keen eye for innovation and a deep respect for the challenges that lie ahead, the board ensures that every step taken by space agencies and private enterprises is thoughtful, strategic, and aligned with the broader goals of the scientific community.At the heart of the board's mission is a steadfast commitment to expanding our understanding of the cosmos. Through rigorous analysis and thorough deliberation, the members of the Cosmic Exploration Editorial Board carefully assess the feasibility and potential impact of various space exploration initiatives. From thedevelopment of cutting-edge technologies to the planning of ambitious interplanetary expeditions, the board's decisions carry immense weight and shape the trajectory of humanity's cosmic journey.One of the board's primary responsibilities is the evaluation of research proposals submitted by scientists and research institutions around the world. These submissions cover a wide range of topics, from the study of exoplanets and the search for extraterrestrial life to the development of advanced propulsion systems and the exploration of the solar system. The board's thorough review process ensures that the most promising and impactful projects are selected for funding and implementation.In addition to evaluating research proposals, the Cosmic Exploration Editorial Board also plays a crucial role in the planning and execution of space missions. Working closely with space agencies and private space companies, the board provides guidance and oversight to ensure that every mission is properly conceived, resourced, and executed. This includes the careful consideration of mission objectives, the selection of appropriate technologies and instruments, and the coordination of international collaborations.The board's expertise extends beyond the technical aspects of space exploration, as they also consider the broader societal and ethicalimplications of humanity's activities in space. Questions of planetary protection, the responsible use of resources, and the potential impact on extraterrestrial environments are all carefully deliberated by the board, ensuring that the pursuit of scientific knowledge is balanced with a deep sense of responsibility and stewardship.As the world's leading authority on space exploration, the Cosmic Exploration Editorial Board plays a critical role in shaping the future of humanity's cosmic endeavors. Through their tireless efforts, the board ensures that the most promising and impactful projects are selected, the most ambitious missions are executed, and the pursuit of scientific knowledge is balanced with a deep sense of responsibility and ethical consideration.In the face of the countless challenges and complexities that come with the exploration of the unknown, the Cosmic Exploration Editorial Board stands as a beacon of expertise, vision, and unwavering commitment to the advancement of human knowledge and the expansion of our cosmic horizons. As we continue to push the boundaries of what is possible, the board's guiding hand will remain a vital force in charting the course of our celestial journey.。
证明哥德巴赫猜想的原文(英文)并附译后汉语
此文发表在：Advances in Theoretical and Applied Mathematics (A TAM), ISSN 0793-4554, V ol. 7, №4, 2012, pp.417-424Proving Goldbach’s Conjecture by Two Number Axes’ Positive Half Lines which Reverse from Each Other’s DirectionsZhang TianshuNanhai west oil corporation,China offshore Petroleum,Zhanjiang city, Guangdong province, P.R.ChinaEmail: tianshu_zhang507@;AbstractWe know that every positive even number 2n(n≥3) can express in a sum which 3 plus an odd number 2k+1(k≥1) makes. And then, for any odd point 2k+1 (k≥1)at the number axis, if 2k+1 is an odd prime point, of course even number 3+(2k+1) is equal to the sum which odd prime number 2k+1 plus odd prime number 3makes; If 2k+1 is an odd composite point, then let 3<B<2k+1, where B is an odd prime point, and enable line segment B(2k+1) to equal line segment 3C. If C is an odd prime point, then even number 3+(2k+1) is equal to the sum which odd prime number B plus odd prime number C makes. So the proof for Goldbach’s Conjecture is converted to prove there be certainly such an odd prime point B at the number axis’s a line segment which take odd point 3 and odd point 2k+1 as ends, so as to prove the conjecture by such a method indirectly.KeywordsNumber theory, Goldbach’s Conjecture, Even number, Odd prime number, Mathematical induction, Two number axes’ positive half lines which reverse from each other’s direction, OD, PL, CL, and RPL.Basic ConceptsGoldbach’s conjecture states that every ev en number 2N is a sum of two prime numbers, and every odd number 2N+3 is a sum of three prime numbers, where N≥2.We shall prove the Goldbach’s conjecture thereinafter by odd points at two number axes’ positive half lines wh ich reverse from each other’s directions and which begin with odd point 3.First we must understand listed below basic concepts before the proof of the conjecture, in order to apply them in the proof.Axiom．Each and every even number 2n (n≥3) can express in a sum which 3 plus each odd number 2k+1 (k≥1) makes.Definition 1．A line segment which takes two odd points as two ends at the number axis’s positive half line which begins with odd point 3 is called an odd distance. “OD” is abbreviated from “odd distance”.The OD between odd point N and odd point N+2t is written as OD N(N+2t), where N≥3, and t≥1.A integer which the length of OD between two consecutive odd points expresses is 2.A length of OD between odd point 3 and each odd point is unique.Definition 2．An OD between odd point 3 and each odd prime point at the number axis’s positive half line which begins with odd point 3, otherwise called a prime length. “PL” is abbreviated from “prime length”, and “PLS”denotes the plural of PL.An integer which each length of from small to large PL expresses be successively 2, 4, 8, 10, 14, 16, 20, 26. . .Definition 3．An OD between odd point 3 and each odd composite point at the number axis’s positive half line which begins with odd point 3, otherwise called a composite length. “CL” is abbreviated from “composite length”.An integer which each length of from small to large CL expresses be successively 6, 12, 18, 22, 24, 30, 32. . .We know that positive integers and positive integers’ points at the number axis’s positive half line are one-to-one correspondence, namely each integer’s point at the number axis’s positive half line represen ts only a positive integer. The value of a positive integer expresses the length of the line segment between point 0 and the positive integer’s point here. When the line segment is longer, it can express in a sum of some shorter line segments; correspondingly the positive integer can also express in a sum of some smaller integers.Since each and every line segment between two consecutive integer’s points and the line segment between point 0 and point 1 have an identical length, hence when use the length as a unit to measure a line segment between two integer’s points or between point 0 and any integer’s point, the line segment has some such unit length, then the integer which the line segment expresses is exactly some.Since the proof for the conjecture relate merely to positive integers which are not less than 3, hence we take only the number axis’s positive half line which begins with odd point 3. However we stipulate that an integer which each integer’s point represents expresses yet the length of the line segment between the integer’s point and point 0. For example, an odd prime value which the right end’s point of any PL represents expresses yet the length of the line segment between the odd prime point and point 0 really.We can prove next three theorems easier according to above-mentioned some relations among line segments, integers’ points and integers.Theorem 1．If the OD which takes odd point F and odd prime point P S as two ends is equal to a PL, then even number3+F can express in a sum of two odd prime numbers, where F>P S.Proof．Odd prime point P S represents odd prime number P S, it expresses the length of the line segment from odd prime point P S to point 0.Though lack the line segment from odd point 3 to point 0 at the number axis’s positive half line which begins with odd point 3, but odd prime point P S represents yet odd prime P S according to above-mentioned stipulation;Let OD P S F=PL 3P b, odd prime point P b represents odd prime number P b, it expresses the length of the line segment from odd prime point P b to point 0. Since PL 3P b lack the segment from odd point 3 to point 0, therefore the integer which the length of PL 3P b expresses is even number P b-3, namely the integer which the length of OD P S F expresses is even number P b-3. Consequently there is F=P S+(P b-3), i.e. 3+F=odd prime P S + odd prime P b .Theorem 2．If even number 3+F can express in a sun of two odd prime numbers, then the OD which takes odd point 3 and odd point F as ends can express in a sun of two PLS, where F is an odd number which is more than 3.Proof．Suppose the two odd prime numbers are P b and P d, then there be 3+F= P b+P d.It is obvious that there be OD 3F=PL 3P b + OD P b F at the number axis’s positive half line which begins with odd point 3.Odd prime point P b represents odd prime number P b according to above-mentioned stipulation, then the length of line segment P b(3+F) is precisely P d, nevertheless P d expresses also the length of the line segment from odd prime point P d to point 0. Thereupon cut down 3 unit lengths of line segment P b(3+F), we obtain OD P b F; again cut down 3 unit lengths of the line segment from odd prime point P d to point 0, we obtain PL 3P d, then there be OD P b F=PL 3P d.Consequently there be OD 3F=PL 3P b + PL 3P d.Theorem 3．If the OD between odd point F and odd point 3can express in a sum of two PLS, then even number 3+F can express in a sum of two odd prime numbers, where F is an odd number which is more than 3. Proof．Suppose one of the two PLS is PL 3P S, then there be F＞P S,and the OD between odd point F and odd prime point P S is another PL. Consequently even number 3+F can express in a sum of two odd prime numbers according to theorem 1.The ProofFirst let us give ordinal number K to from small to large each and every odd number 2k+1, where k≥1,then from small to large each and every even number which is not less than 6is equal to 3+(2k+1).We shall prove this conjecture by the mathematical induction thereinafter.1．When k=1, 2, 3 and 4, we getting even number be orderly 3+(2*1+1)=6=3+3, 3+(2*2+1)=8=3+5, 3+(2*3+1)=10=3+7 and 3+(2*4+1)=12=5+7. This shows that each of them can express in a sum of two odd prime numbers.2．Suppose k=m, the even number which3 plus №m odd number makes, i.e. 3+(2m+1) can express in a sum of two odd prime numbers, where m≥4.3．Prove that when k=m+1, the even number which 3 plus №(m+1) odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers.Proof．In case 2m+3 is an odd prime number, naturally even number 3+（2m+3）is the sum of odd prime number 3 plus odd prime number 2m+3 makes.When 2m+3 is an odd composite number, suppose that the greatest odd prime number which is less than 2m+3 is P m, then the OD between odd prime point P m and odd composite point 2m+3 is either a PL or a CL. When the OD between odd prime point P m and odd composite point 2m+3 is a PL, the even number 3+（2m+3）can express in a sum of two odd prime numbers according to theorem 1.If the OD between odd prime point P m and odd composite point 2m+3 is a CL, then we need to prove that OD 3（2m+3）can express in a sum of two PLS, on purpose to use the theorem 3.When OD P m(2m+3) is a CL, from small to large odd composite number 2m+3 be successively 95, 119, 125, 145. . .First let us adopt two number axes’ positive half lines which reverse from each other’s directions and which begin with odd point 3.At first, enable end point 3 of either half line to coincide with odd point 2m+1 of another half line. Please, see first illustration:3 5 7 2m-3 2m+12m+1 2m-3 7 5 3First IllustrationSuch a coincident line segment can shorten or elongate, namely end point 3 of either half line can coincide with any odd point of another half line.This proof will perform at some such coincident line segments. And for certain of odd points at such a coincident line segment, we use usually names which mark at the rightward direction’s half line.We call PLS which belong both in the leftward direction’s half line and in a coincident line se gment “reverse PLS”. “RPLS” is abbreviated from “reverse PLS”, and “RPL” denotes the singular of RPLS.The RPLS whereby odd point 2k+1 at the rightward direction’s half line acts as the common right endmost point are written as RPLS2k+1,and RPL2k+1 denotes the singular, where k>1.This is known that each and every OD at a line segment which takes oddpoint 2m+1 and odd point 3 as two ends can express in a sum of a PL and a RPL according to preceding theorem 2 and the supposition of №2 step of the mathematical induction.We consider a PL and the RPL2k+1 wherewith to express together the length of OD 3(2k+1)as a pair of PLS, where k >1. One of the pair’s PLS is a PL which takes odd point 3 as the left endmost point, and another is a RPL2k+1 which takes odd point 2k+1 as the right endmost point. We consider the RPL2k+1 and another RPL2k+1 which equals the PL as twin RPLS2k+1.For a pair of PLS, the PL is either unequal or equal to the RPL2k+1. If the PL is unequal to the RPL2k+1, then longer one is more than a half of OD 3(2k+1), yet another is less than the half. If the PL is equal to the RPL2k+1, then either is equal to the half. A pair of PLS has a common end’s point.Since each of RPLS2k-1 is equal to a RPL2k+1,and their both left endmost points are consecutive odd points, and their both right endmost points are consecutive odd points too. So seriatim leftwards move RPLS2k+1to become RPLS2k-y, then part left endmost points of RPLS2k+1 plus RPLS2k-y coincide monogamously with part odd prime points at OD 3(2k+1), where y=1, 3, 5, ...Thus let us begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, where y= 1, 3, 5, 7 ... ỹ ...Suppose that y increases orderly to odd number ỹ,and∑ part left endmostpoints of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously, then there are altogether(ỹ+3)/2 odd points at OD (2m-ỹ)(2m+1), and let μ=(ỹ+3)/2.Let us separate seriatim OD 3(2m-y+2) (y=1, 3, 5,…) from each coincident line segment of two such half lines, and arrange them from top to bottom orderly. After that, put an odd prime number which each odd prime point at the rightward direction’s half line expresses to on the odd prime point, and put another odd prime number which each left endmost point of RPLS2m-y+2 at the leftward direction’s half line expresses to beneath the odd prime point.For example, when 2m+3=95, 2m+1=93, 2m-1=91 and 2m-ỹ =89，μ=3. For the distributer of odd prime points which coincide monogamously with left endmost points of RPLS95, RPLS93, RPLS91, RPLS89 and RPLS87, please see second illustration:OD 3(95)19 31 37 61 67 7979 67 61 37 31 19OD 3(93)7 13 17 23 29 37 43 53 59 67 73 79 83 8989 83 79 73 67 59 53 43 37 29 23 17 13 7 OD 3(91) 5 11 23 41 47 53 71 83 8989 83 71 53 47 41 23 11 5 OD 3(89)13 19 31 61 73 7979 73 61 31 19 13OD 3(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 8383 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7Second IllustrationTwo left endmost points of twin RPLS2m-y+2 at OD 3(2m-y+2) coincide monogamously with two odd prime points, they assume always bilateral symmetry whereby the centric point of OD 3(2m-y+2)acts as symmetriccentric. If the centric point is an odd prime point, then it is both the left endmost point of RPL2m-y+2 and the odd prime point which coincides with the left endmost point, e.g. centric point 47 of OD3(91) in above-cited that example.We consider each odd prime point which coincides with a left endmost point of RPLS2m-ỹ alone as a characteristic odd prime point, at OD 3(2m+1). Thus it can seen, there is at least one characteristic odd prime point at OD 3(2m-ỹ) according to aforesaid the way of making things, e.g. odd prime points 19, 31 and 61 at OD 3(89) in above-cited that example.Whereas there is not any such characteristic odd prime point in odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹ+2.In other words, every characteristic odd prime point is not any left endmost point of RPLS2m+1plus RPLS2m-1 ... plus RPLS2k-ỹ+2.Moreover left endmost points of RPLS2m-y are №1 odd points on the lefts of left endmost points of RPLS2m-y+2monogamously, where y is an odd number≥1.Consequently, №1 odd point on the left of each and every characteristic odd prime point isn’t a ny left endmost point of RPLS2m-1plus RPLS2m-3…plus RPLS2m-ỹ. — {1}Since each RPL2m-y+2 is equal to a PL at OD 3(2m-y+2).In addition, odd prime points which coincide monogamously with left endmost points of RPLS2m+1 plus RPLS2m-1 ... plus RPLS2m-ỹare all odd prime points at OD 3(2m+1).Hence considering length, at OD 3(2m+1) RPLS whose left endmost points coincide monogamously with all odd prime points are all RPLS at OD 3(2m+1), irrespective of the frequency of RPLS on identical length. Evidently the longest RPL at OD 3(2m+1)is equal to PL 3P m.When OD P m(2m+3) is a CL, let us review aforesaid the way of making thing once again, namely begin with odd point 2m+1, leftward take seriatim each odd point 2m-y+2 as a common right endmost point of RPLS2m-y+2, and∑ part left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincide just right with all odd prime points at OD 3(2m+1) monogamously.Which one of left endmost points of RPLS2m-y+2(1≤y≤ỹ+2) coincides first with odd prime point 3? Naturally it can only be the left endmost point of the longest RPL P m whereby odd prime point P m acts as the right endmost point.Besides all coincidences for odd prime points at OD 3(2m+1) begin with left endmost points of RPLS2m+1, whereas left endmost points of RPLS2m-ỹare final one series in the event that all odd prime points at OD 3(2m+1) are coincided just right by left endmost points of RPLS.Therefore odd point 2m-ỹ as the common right endmost point of RPLS2m-ỹcannot lie on the right of odd prime point P m, then odd point 2m-ỹ-2 can only lie on the left of odd prime point P m . This shows that every RPL2m-ỹ-2 at OD 3(2m-ỹ-2) is shorter than PL 3 P m.In addition, №1odd point on the left of a left endmost point of each and every RPL2k-ỹ is a left endmost point of RPLS2k-ỹ-2.Therefore each and every RPL2m-ỹ-2can extend contrary into at least one RPL2m-y+2, where y is a positive odd number ≤ỹ+2.That is to say, every left endmost point of RPLS2m-ỹ-2 is surely at least one left endmost point of RPLS2k-y+2.Since left endmost points of RPLS2m-ỹ-2 lie monogamously at №1 odd point on the left of left endmost points of RPLS2m-ỹ including characteristic odd prime points.Consequently, №1 odd point on the left of each and every characteristic odd prime point is surely a left endmost point of RPLS2m-y+2, where 1≤y≤ỹ+2.—— {2}So we draw inevitably such a conclusion that №1 odd point on the left of each and every characteristic odd prime point can only be a left endmost point of RPLS2m+1 under these qualifications which satisfy both above-reached conclusion {1}and above-reached conclusion {2}.Such being the case, let us rightwards move a RPL2m+1 whose left endmost point lies at №1odd point on the left of any characteristic odd prime point to adjacent odd points, then the RPL2m+1 is moved into a RPL2m+3. Evidently the left endmost point of the RPL2m+3is the characteristic odd prime point, and its right endmost point is odd point 2m+3.So OD 3(2m+3) can express in a sum of two PLS, and the commonendmost point of the two PLS is exactly the characteristic odd prime point. Thus far we have proven that even if OD P m(2m+3) is a CL, likewise OD 3(2m+3) can also express in a sum of two PLS.Consequently even number which 3 plus №(m+1)odd number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime numbers according to aforementioned theorem 3.Proceed from a proven conclusion to prove a larger even number for each once, then via infinite many an once, namely let k to equal each and every natural number, we reach exactly a conclusion that every even number 3+(2k+1) can express in a sum of two odd prime numbers, where k≥1.To wit every even number 2N can express in a sum of two odd prime numbers, where N>2.In addition let N =2, get 2N=4=even prime number 2+even prime number 2. Consequently every even number 2N can express in a sum of two prime numbers, where N≥2.Since every odd number 2N+3 can express in a sum which a prime number plus the even number makes, consequently every odd number 2N+3 can express in a sum of three prime numbers, where N≥2.To sum up, we have proven that two propositions of the Goldbach’s conjecture are tenable, thus Goldbach’s conjecture holds water.附，翻译成汉语：利用互为反向数轴的正射线证明哥德巴赫猜想张天树Tianshu_zhang507@摘要我们知道,依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.于是,对于数轴上的任意一个奇数点2k+1（k≧1）,如果2k+1是一个奇素数点,当然,偶数3+(2k+1）可等于奇素数2k+1 与奇素数 3 之和;如果2k+1是一个奇合数点,那么,取3<B<2k+1,这里,B是一个奇素数点,且使线段B(2k+1)等于线段3C. 如果C是一个奇素数点,那么, 偶数3+（2k+1）等于奇素数B与奇素数C之和.于是对哥德巴赫猜想的证明就变换成了去证明：在数轴的以奇数点3和2k+1为端点的线段上，总是存在着这样的奇素数点B，以此方法来间接地证明哥德巴赫猜想.关键词数论、哥德巴赫猜想、数学归纳法、偶数、奇素数、互为反方向的数轴的正射线、奇距、素长、合长和反向素长.基本慨念哥德巴赫猜想表为:每一个偶数2N都是两个素数之和,每一个奇数2N+3都是三个素数之和,这里N≧2.本文将用互为反方向的两条数轴的从奇数点3开始的正方向射线上的奇数点来证明这个猜想.在证明这个猜想之前，我们先要熟知下述的基本慨念，以便在证明的过程中应用它们.公理.依次增大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一个奇数2k+1（k≧1）之和.定义1. 在数轴的从奇数点3开始的正方向射线上，以任意两个奇数点为端点的线段,称为这两个奇数点之间的距离，简称奇距．我们用符号“OD”表示奇距.奇数点N与奇数点N+2t之间的奇距,写成OD N(N+2t)，这里N≧3,t≧1. 相邻两个奇数点之间的奇距是2，奇数点3与各个奇数点之间的奇距长度,都是唯一的.定义2.在数轴的从奇数点3开始的正方向射线上，以任意一个奇素数点和奇数点3为端点的奇距，也被称为这个奇素数点的素长，并用符号“PL”表示一条素长，和符号“PLS”表示至少两条素长，即素长的复数.从小到大的素长依次是: 2、4、8、10、14、16、20、26. . . . . .定义3.在数轴的从奇数点3开始的正方向射线上，以任意一个奇合数点和奇数点3为端点的奇距，又被称为这个奇合数点的合长，并用符号“C L”表示一条合长.从小到大的合长依次是: 6、12、18、22、24、30、32. . . . . .我们知道，在数轴的正方向射线上的整数点与正整数是一一对应的.也就是说，在数轴的正方向射线上，每一个整数点只代表一个整数，这个整数的值在这里表示这个整数点距0点的线段长度. 当这条线段较长时，它可表为若干条线段之和. 因此，这个整数也可表为若干个整数之和. 且因为0点与1点、以及相邻两个整数点之间的线段长度都是相等的,当我们把这些相等线段每条的长度作为1个长度单位去度量任意一个整数点距0点的线段或任意两个整数点之间的线段长度时,该线段有多少个这样的长度单位，它就代表多大的整数.因为本文的证明仅仅用到不小于3的整数，所以，我们只取数轴的从奇数点3开始的正方向射线，但我们规定：在这射线上的每个整数点代表的整数值仍然是指这个整数点距0点的线段长度. 例如，在这射线上任意一条素长的右端点所代表的奇素数值，实际上是指这个奇素数点距0点的线段长度.根据以上所述，我们很容易地证明以下三条定理:定理1.如果以奇数点F和奇素数点P S 为端点的奇距等于一条素长，那么，偶数3+F可表为两个奇素数之和，这里F＞P S.证明：奇素数点P S代表奇素数P S，P S的值表示奇素数点P S到0点的长度.在数轴的从奇数点3开始的正方向射线上虽然缺少从奇数点3到0点的一段，但是按照前面的规定，奇素数点P S仍然代表奇素数P S；令OD P S F=PL 3P b，奇素数点P b代表奇素数P b，P b的值表示奇素数点P b 到0点的长度. 但是，PL 3P b缺少从奇数点3到0点的一段，因此，PL 3P b 的长度表示的整数是偶数P b-3，即OD P S F的长度表示的整数是偶数P b-3. 所以, F=P S+(P b-3)，即3+F=奇素数P S+奇素数P b.定理2. 如果偶数3+F可表为两个奇素数之和，那么，以奇数点F和奇数点3为端点的奇距可表为两条素长之和，这里F是一个大于3的奇数.证明：假设这两个奇素数为P b和P d，则有3+F= P b+P d. 显然，在数轴的从奇数点3开始的正方向射线上，OD 3F=PL 3P b + OD P b F. 根据前面的规定，奇素数点P b代表奇素数P b，那么，线段P b(3+F)的长度就是P d，该长度是指从奇素数点P d到0点的线段长度.在线段P b(3+F)上去掉3个单位长度，得到OD P b F；在从奇素数点P d到0点的线段上去掉3个单位长度，得到PL 3P d，于是，OD P b F=PL 3P d.所以, OD 3F=PL 3P b + PL 3P d.定理3.如果奇数点F与奇数点3之间的奇距可表为两条素长之和，那么，偶数3+F可表为两个奇素数之和，这里F是一个大于3的奇数.证明：假设这两条素长中的一条是PL 3P S，那么，F＞P S，且奇数点F与奇素数点P S 之间的奇距是另一条素长. 根据定理1，偶数3+F可表为两个奇素数之和.证明首先,让我们以自然数k给每一个依次增大的奇数2k+1编上序号，这里k≧1,那么，每一个不小于6的依次增大的偶数等于3+(2k+1). 在下文中, 我们将运用数学归纳法来证明这个猜想.1.当k= 1、2、3和4时，我们依次得到的偶数: 3+（2×1+1）=6=3+3，3+（2×2+1）=8=3+5，3+（2×3+1）=10=3+7或5+5，3+（2×4+1）=12=5+7都可表为两个奇素数之和.2.假设当k=m时，3加上第m个奇数所得的偶数、即3+（2m+1）可表为两个奇素数之和，这里m≧4.3.证明:当k=m+1时，3加上第m+1个奇数所得的偶数、即3+（2m+3）也可表为两个奇素数之和.证明如果2m+3是一个奇素数，当然，偶数3+（2m+3）可表为奇素数3与奇素数2m+3之和.当2m+3是一个奇合数时，假设小于2m+3的最大奇素数为P m，那么，奇合数点2m+3与奇素数点P m之间的奇距或是一条素长，或是一条合长.当奇合数点2m+3与奇素数点P m之间的奇距是一条素长时，根据定理1，偶数3+（2m+3）可表为两个奇素数之和.如果奇合数点2m+3与奇素数点P m之间的奇距是一条合长，我们则需要证明OD 3（2m+3）可表为两条素长之和，以便应用定理3.当OD P m(2m+3)是一条合长时，2m+3从小到大的值依次是95、119、125、145. . . . . .首先，让我们采用两条互为相反方向的数轴的从奇数点3开始的正方向射线，最初,让一条射线上的奇数点3与另一条射线上的奇数点2m+1重合. 请参看第一图：3 5 7 2m-3 2m+12m+1 2m-3 7 5 3第一图这两条射线互相重合的线段能够缩短或伸长，也就是说，一条射线的端点3能够重合另一条射线的任意一个奇数点.本文的整个证明将在这样互相重合、且可伸缩的线段上施行，并且，对这线段上互相重合后的一个奇数点，我们主要使用它在向右方向射线上的名称.我们把既属于向左方向射线，又在两条射线互相重合的线段上的素长称为“反向素长”. 一条反向素长用符号“RPL”表示，它的复数表为“RPLS”. 以向右方向射线上的奇数点2k+1作为共同右端点的至少两条反向素长被写作RPLS2k+1,以及RPL2k+1表示其中的一条，这里k＞1.根据前列的定理2和数学归纳法中第二步的假设，我们知道，在以奇数点2m+1和奇数点3为端点的线段上的各条奇距，都能够表为一条素长与一条反向素长之和. 我们把表示OD 3(2k+1)为两条素长之和的两条素长看作是一对素长，这里k＞1. 在这对素长中，一条是以奇数点3为左端点的素长，另一条则是以奇数点2k+1为右端点的反向素长. 我们把这对素长中的反向素长和在长度上等于这对素长中素长的另一条以奇数点2k+1为右端点的反向素长看作是孪生的反向素长.对于共表OD 3(2k+1)长度的一对素长，如果它们不等，那么，较长的一条比OD 3（2k+1）的一半长，而另一条则比这一半短. 如果这两条素长相等，那么，每一条都等于OD 3（2k+1）的一半.共表OD 3（2k+1）长度的一对素长，它们有一个共同的端点.因为在长度上每一条RPL2k-1 等于一条RPL2k+1. 并且，它们的左端点是相邻的奇数点，右端点也是相邻的奇数点. 于是,逐点向左移动RPLS2k+1使之成为RPLS2k-y,那么，RPLS2k+1和RPLS2k-y 的部分左端点就一对一地重合OD 3（2k+1）上的奇素数点，这儿y=1, 3, 5, …...因此,让我们在数轴的从奇数点3开始的正方向射线上，从奇数点2m+1开始，向左依次取每一个奇数点2m-y+2作为反向素长的共同右端点，这里y=1、3、5、7、...ỹ、...假设y依次增大到奇数ỹ，且RPLS2m+1、RPLS2m-1 ... RPLS2m-ỹ+2和RPLS2m-ỹ的部份左端点一对一地恰好重合完OD 3（2m+1）上的全部奇素数点，那么，在OD（2m- ỹ）（2m+1）上共有（ỹ+3）/2个奇数点，并且令μ =（ỹ+3）/2.然后，我们按照从长到短的次序，从两条射线的重合段中依次分离出OD 3（2m-y+2），并把它们从上到下依次排列起来，还把向右方向射线上表示奇素数点的奇素数放在这个奇素数点的上方，把重合这奇素数点的表示反向素长左端点的奇素数放在它的下方.例如:当2m+3=95时，2m+1=93, 2m-1=91和2m-ỹ=89，μ=3. 对于OD 3(95)、OD 3(93)、OD 3(91) 、OD 3(89) 和OD 3(87)上反向素长左端点重合奇素数点的分布情况，如下图：OD 5(95)19 31 37 61 67 7979 67 61 37 31 19OD 5(93) 7 13 17 23 29 37 43 53 59 67 73 79 83 8989 83 79 73 67 59 53 43 37 29 23 17 13 7OD 5(91)5 11 23 41 47 53 71 83 8989 83 71 53 47 41 23 11 5OD 5(89) 13 19 31 61 73 7979 73 61 31 19 13OD 5(87) 7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 8383 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7第二图在OD 3（2m-y+2）上，孪生反向素长的两个左端点一对一重合的两个奇素数点，以该奇距的中点为对称中心左右对称. 如果该奇距的中点是一个奇素数点，那么,它既是反向素长的左端点，又是这个左端点重合的奇素数点，例如，在上面引用例子中，OD 3（91）的中点47.在OD 3（2m+1）上，我们仅仅把唯一由RPLS2m-ỹ的左端点重合的奇素数点，即在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的奇素数点中没有的奇素数点看作是特有的奇素数点. 那么，根据上述的作法，在RPLS2m-ỹ的左端点一一重合的奇素数点中，有至少一个特有的奇素数点.例如，上面所举那个例子中，在OD 3(89)上的奇素数点19、31和61.因为在RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点一一重合的奇素数点中，没有一个特有的奇素数点. 换言之，每一个特有的奇素数点都不是RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ+2的左端点.又RPLS2m-y+2的左端点左边的第一个奇数点是RPLS2m-y的左端点，这里y是≧1的正奇数.所以,在每一个特有奇素数点左边的第一个奇数点不是RPLS2m-1、RPLS2m-3、...和RPLS2m-ỹ的左端点.--- {1}因为在OD 3(2m-y+2)上，每一条反向素长等于一条素长. 加之，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.因此，仅以长度而言，在OD 3(2m+1)上，其左端点一对一地重合完全部奇素数点的反向素长是OD 3(2m+1)上全部的反向素长，而不与具有同一长度的反向素长的条数有关.显然，在OD 3(2m+1)上的全部反向素长中，最长的反向素长等于PL 3P m.当奇合数点2m+3与奇素数点P m之间的奇距是一条合长时，让我们回顾前述的作法，即从2m+1开始，向左依次取每一个奇数点作为反向素长的共同右端点，并且，RPLS2m+1、RPLS2m-1、...和RPLS2m-ỹ的左端点一对一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.究竟哪一条反向素长的左端点首先重合奇素数点3呢？显然，它仅仅能够是以奇素数点P m为右端点的最长的反向素长的左端点.又从RPLS2m+1的左端点开始，RPLS2m-ỹ的左端点是重合完OD 3（2m+1）上全部奇素数点的最后一组.所以，作为RPLS2m-ỹ的共同右端点的奇数点2m-ỹ不能位于奇素数点P m的右边，于是，奇数点2m-ỹ-2位于奇素数点P m的左边. 由此可知，在OD 3（2m- ỹ-2）上的每一条反向素长都比PL 3P m短.又因为在RPLS2m-ỹ的左端点左边的第一个奇数点全部是RPLS2m-ỹ-2的左端点.所以，每一条RPL2m-ỹ-2都能反向延长成为至少一条RPL2m-y+2. 那就是说，RPLS2m-ỹ-2的每个左端点都一定是RPLS2m-y+2的至少一条的左端点，这里y是≦ỹ+2的正奇数.因为RPLS2m-ỹ-2的左端点一对一地全部在RPLS2m-ỹ左端点（包括特有奇素数点）左边的第一个奇数点上，于是，在特有奇素数点左边的第一个奇数点一定是RPLS2m-y+2的左端点，这里y是≦ỹ+2的正奇数. ---{2}所以,在既要满足结论{1}、又要满足结论{2}的情况下，我们只能够得出: 在特有奇素数点左边的第一个奇数点是RPLS2m+1的左端点.既然如此，让我们向右移动任意一条这样的PLS2m+1到相邻的奇数点，那么,这条PLS2m+1被移动成了一条PLS2m+3. 显然，这条PLS2m+3的左端点重合这个特有奇素数点，而它的右端点是奇数点2m+3.这样，OD 3（2m+3）可以被表为两条素长之和，这两条素长的共同端点就是这个特有奇素数点.到此为止，我们已经证明了: 当OD P m（2m+3）是一条合长时，OD 3（2m+3）也能表为两条素长之和.那么，根据定理3，3加上第m+1个奇数所得的偶数,即3+(2m+3)可以表示成两个奇素数之和.每次都从已证得的结论出发，就可以推出:当k等于每一个自然数时，每一个偶数3+(2k+1)都可以表示成两个奇素数之和. 即当N>2时，每一个偶数2N都可以表示成两个奇素数之和.又当N=2时,有2N=4=偶素数2+偶素数2.所以,每一个偶数2N都可以表示成两个素数之和，这里N≧2.因为每一个奇数2N+3等于偶数2N加上奇素数3，因此，每一个奇数2N+3都可以表示成三个素数之和，这里N≧2.综上所述，我们已经证明了哥德巴赫猜想的两个命题都是站得住脚的，因此，哥德巴赫猜想成立.。
反证法英语作文
反证法英语作文In the realm of mathematics a common method used to prove a statement is the method of contradiction also known as reductio ad absurdum. This method involves assuming the opposite of what youre trying to prove and then showing that this assumption leads to a contradiction thereby proving the original statement must be true. Lets explore this concept in the context of an English essay.Title The Power of Proof by ContradictionIntroductionThe method of contradiction is a powerful tool in the mathematicians arsenal. It is a technique that has been used for centuries to prove the validity of various mathematical theorems and propositions. This essay will delve into the essence of the method of contradiction its historical significance and how it can be applied in solving complex problems.Historical BackgroundThe method of contradiction has its roots in ancient Greek philosophy particularly in the works of the philosopher Aristotle. He used this method to establish the principles of logic and reasoning. Over time this method has been refined and adapted by mathematicians to prove a wide range of mathematical statements.Understanding the MethodAt its core the method of contradiction involves four main steps1. Assume the opposite of the statement you want to prove.2. Deduce logical consequences from this assumption.3. Show that these consequences lead to a contradiction or an absurdity.4. Conclude that the original assumption must be false and therefore the statement you wanted to prove is true.Examples of ApplicationLets consider a classic example from geometry proving that the sum of the angles in a triangle is always 180 degrees. Using the method of contradiction one would assume thatthe sum of the angles in a triangle is not 180 degrees. By examining the implications of this assumption one would find that it leads to a contradiction with the properties of a straight line and the parallel postulate thus proving the original statement to be true.Advantages of the MethodThe method of contradiction offers several advantagesIt can be used to prove a statement without directly constructing a solution.It is particularly useful when the direct approach is difficult or not feasible.It encourages critical thinking by challenging the reader to consider the opposite of what is being proven.Challenges and LimitationsHowever the method of contradiction is not without its challenges. It requires a deep understanding of the subject matter to identify the correct assumptions and to deduce the logical consequences that lead to a contradiction. Additionally it can be less intuitive than direct proof methods making it harder for some to grasp.ConclusionThe method of contradiction is a testament to the power of logical reasoning in mathematics. It allows us to prove statements that might otherwise be elusive. By embracing this method we not only solve mathematical problems but also strengthen our analytical skills and our ability to think critically.ReflectionIn conclusion the method of contradiction is a valuable tool in the field of mathematics. It challenges us to think beyond the obvious and to explore the implications of our assumptions. As we apply this method we gain a deeper understanding of the principles that govern the world around us and we develop a more profound appreciation for the beauty of mathematical proof.。
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.。
有关于科幻小说的作文英语
有关于科幻小说的作文英语Science fiction, a genre that has captivated the minds of readers for centuries, is not merely a tale of the unknownbut a mirror reflecting our deepest fears, aspirations, andthe potential future of humanity. It is a literary devicethat transcends the boundaries of time and space, inviting us to ponder upon the possibilities that lie beyond the realm of our current understanding.In the realm of science fiction, authors weave intricate narratives that often begin with a 'what if' scenario. This simple yet powerful question opens the door to a world where technology, space travel, alien life, and dystopian societies are not just figments of imagination but plausible realities. The genre has given birth to some of the most iconic works of literature, such as George Orwell's "1984," Isaac Asimov's "Foundation" series, and Ray Bradbury's "Fahrenheit 451."One of the most significant aspects of science fiction is its ability to critique society. By placing readers in a worldthat is both familiar and alien, authors can address contemporary issues without the constraints of reality. For instance, "1984" explores the dangers of totalitarianism,while "Fahrenheit 451" delves into the perils of censorship and the suppression of knowledge.Moreover, science fiction serves as a catalyst for innovation. Many technological advancements we take for granted todaywere once the stuff of science fiction. The concept of smartphones, for example, can be traced back to the'communicator' devices found in Star Trek. Similarly, virtual reality and artificial intelligence, now integral parts of our technological landscape, were once the preserve of science fiction narratives.The genre also encourages a sense of wonder and curiosity about the universe. Stories of interstellar travel and extraterrestrial life spark our imagination and inspire us to explore the cosmos. They challenge us to consider our placein the universe and question the very nature of existence.However, science fiction is not without its darker themes. It often warns us of the potential consequences of unbridled technological advancement and the ethical dilemmas that may arise. Works like "Brave New World" by Aldous Huxley and "Do Androids Dream of Electric Sheep?" by Philip K. Dick force us to confront the existential questions that come with creating life-like artificial beings or the pursuit of pleasure at the expense of individuality and freedom.In conclusion, science fiction is a genre that enriches our lives by expanding our horizons and challenging our perceptions. It is a medium through which we can explore the complexities of the human condition and the vastness of the universe. As we continue to push the boundaries of what is possible, science fiction remains a vital tool for envisioning the future and understanding the present. It is a genre that will undoubtedly continue to evolve, reflectingthe ever-changing landscape of human imagination and technological progress.。
10.1007%2Fs11537-013-1280-5
Japan.J.Math.8,147–183(2013)DOI:10.1007/s11537-013-1280-5About the Connes embedding conjectureAlgebraic approachesNarutaka Ozawa?Received:28December2012/Accepted:15January2013Published online:20March2013©The Mathematical Society of Japan and Springer Japan2013Communicated by:Yasuyuki KawahigashiAbstract.In his celebrated paper in1976,A.Connes casually remarked that anyfinite von Neu-mann algebra ought to be embedded into an ultraproduct of matrix algebras,which is now known as the Connes embedding conjecture or problem.This conjecture became one of the central open problems in thefield of operator algebras since E.Kirchberg’s seminal work in1993that proves it is equivalent to a variety of other seemingly totally unrelated but important conjectures in the field.Since then,many more equivalents of the conjecture have been found,also in some other branches of mathematics such as noncommutative real algebraic geometry and quantum infor-mation theory.In this note,we present a survey of this conjecture with a focus on the algebraic aspects of it.Keywords and phrases:Connes embedding conjecture,Kirchberg’s conjecture,Tsirelson’s prob-lem,semi-pre-C -algebras,noncommutative real algebraic geometryMathematics Subject Classification(2010):16W80,46L89,81P151.IntroductionThe Connes embedding conjecture([Co])is considered as one of the most im-portant open problems in thefield of operator algebras.It asserts that anyfi-nite von Neumann algebra is approximable by matrix algebras in a suitable sense.It turns out,most notably by Kirchberg’s seminal work([Ki1]),that the N.O ZAWAResearch Institute for Mathematical Sciences,Kyoto University,Kyoto606-8502,Japan(e-mail:)?Partially supported by JSPS(23540233)and by the Danish National Research Foundation (DNRF)through the Centre for Symmetry and Deformation.148N.Ozawa Connes embedding conjecture is equivalent to a variety of other important con-jectures,which touches most of the subfields of operator algebras,and also someother branches of mathematics such as noncommutative real algebraic geome-try([Sm])and quantum information theory.In this note,we look at the alge-braic aspects of this conjecture.(See[BO,Ki1,Oz1]for the analytic aspects.)This leads to a study of the C -algebraic aspect of noncommutative real alge-braic geometry in terms of semi-pre-C -algebras.Specifically,we treat someeasy parts of Positivstellensätze of Putinar([Pu]),Helton–McCullough([HM]),and Schmüdgen–Bakonyi–Timotin([BT]).We then treat their tracial analogueby Klep–Schweighofer([KS]),which is equivalent to the Connes embeddingconjecture.We give new proofs of Kirchberg’s theorems on the tensor productC F d˝B.`2/and on the equivalence between the Connes embedding conjec-ture and Kirchberg’s conjecture.We also look at Tsirelson’s problem in quantuminformation theory([Fr,J+,Ts]),and prove it is again equivalent to the Connesembedding conjecture.This paper is an expanded lecture note for the author’slecture for“Masterclass on sofic groups and applications to operator algebras”(University of Copenhagen,5–9November2012).The author gratefully ac-knowledges the kind hospitality provided by University of Copenhagen duringhis stay in Fall2012.He also would like to thank Professor Andreas Thom forvaluable comments on this note.2.Ground assumptionWe deal with unital -algebras over k2f C;R g,and every algebra is assumedto be unital,unless it is clearly not so.The unit of an algebra is simply denotedby1and all homomorphisms and representations between algebras are assumedto preserve the units.We denote by i the imaginary unit,and by the complexconjugate of 2C.In case k D R,one has D for all 2k.3.Semi-pre-C -algebrasWe will give the definition and examples of semi-pre-C -algebras.Recall thata unital algebra A is called a -algebra if it is equipped with a map x!xsatisfying the following properties:(i)1 D1and.x / D x for every x2A;(ii).xy/ D y x for every x;y2A;(iii). x C y/ D x C y for every x;y2A and 2k.The sets of hermitian elements and unitary(orthogonal)elements are writtenrespectively asA h WD f a2A W a D a g and A u WD f u2A W u u D1D uu g:About the Connes embedding conjecture149 Every element x2A decomposes uniquely as a sum x D a C b of an hermitian element a and a skew-hermitian element b.The set of hermitian elements is an R-vector space.We say a linear map'between -spaces is self-adjoint if ' D',where' is defined by' .x/D'.x / .We call a subset A C A h a -positive cone(commonly known as a quadratic module)if it satisfies the following:(i)R 01 A C and a C b2A C for every a;b2A C and 2R 0; (ii)x ax2A C for every a2A C and x2A.For a;b2A h,we write aÄb if b a2A C.We say a linear map'be-tween spaces with positivity is positive if it sends positive elements to positive elements(and often it is also required self-adjoint),and a positive linear map' is faithful if a 0and'.a/D0implies a D0.Given a -positive cone A C, we define the -subalgebra of bounded elements byA bdd D f x2A W9R>0such that x xÄR1g:This is indeed a -subalgebra of A.For example,if x is bounded and x xÄR1,then x is also bounded and xx ÄR1,because0ÄR 1.R1 xx /2D R1 2xx C R 1x.x x/x ÄR1 xx : Thus,if A is generated(as a -algebra)by S,then S A bdd implies A D A bdd.Definition.A unital -algebra A is called a semi-pre-C -algebra if it comes together with a -positive cone A C satisfying the Combes axiom(also called the archimedean property)that A D A bdd.Since hÄ.1C h2/=2for h2A h,one has A h D A C A C for a semi-pre-C -algebra.We define the ideal of infinitesimal elements byI.A/D f x2A W x xÄ"1for all">0gand the archimedean closure of the -positive cone A C(or any other cone)by arch.A C/D f a2A h W a C"12A C for all">0g:The cone A C is said to be archimedean closed if A C D arch.A C/.A C -alge-bra A is of course a semi-pre-C -algebra,with a zero infinitesimal ideal and an archimedean closed -positive coneA C D f x x W x2A g:If A B.H/(here B.H/denotes the C -algebra of the bounded linear opera-tors on a Hilbert space H over k),then one also hasA C D f a2A h W h a ; i 0for all 2H g:150N.Ozawa Note that the condition a being hermitian cannot be dropped when k D R.Itwill be shown(Theorem1)that if A is a semi-pre-C -algebra,then A=I.A/is a pre-C -algebra with a -positive cone arch.A C/.Definition.We define the universal C -algebra of a semi-pre-C -algebra Aas the C -algebra C u.A/together with a positive -homomorphismÃW A!C u.A/which satisfies the following properties:Ã.A/is dense in C u.A/andevery positive -representation of A on a Hilbert space H extends to a -representation N W C u.A/!B.H/,i.e., D N ıÃ.In other words,C u.A/isthe separation and completion of A under the C -semi-normsup fk .a/k B.H/W a positive -representation on a Hilbert space H g: (We may restrict the dimension of H by the cardinality of A.)We emphasize that only positive -representations are considered.Everypositive -homomorphism between semi-pre-C -algebras extends to a posi-tive -homomorphism between their universal C -algebras.It may happen that A C D A h and C u.A/D f0g,which is still considered as a unital(?)C -algebra.Every -homomorphism between C -algebras is automatically pos-itive,has a norm-closed range,and maps the positive cone onto the positivecone of the range.However,this is not at all the case for semi-pre-C -algebras,as we will exhibit a prototypical example in Example1.On the other hand,we note that if A is a norm-dense -subalgebra of a C -algebra A such thatarch.A C/D A\A C,then every positive -representation of A extends to a-representation of A,i.e.,A D Cu .A/.(Indeed,if x2A has k x k A<1,then1 x x2A C and hence k .x/k<1for any positive -representation ofA.)It should be easy to see that the following examples satisfy the axiom of semi-pre-C -algebras.Example1.Let be a discrete group and kŒ be its group algebra over k:.f g/.s/DXt2f.st 1/g.t/and f .s/D f.s 1/ for f;g2kŒ :The canonical -positive cone of kŒ is defined as the sums of hermitian squares,kŒ C Dn n Xi D1 i i W n2N; i2kŒo:Then,kŒ is a semi-pre-C -algebra such that C u.kŒ /D C ,the full group C -algebra of ,which is the universal C -algebra generated by the unitary representations of .There is another group C -algebra.Recall that the left regular representation of on`2 is defined by .s/ıt Dıst for s;t2 , or equivalently by .f/ D f for f2kŒ and 2`2 .The reducedAbout the Connes embedding conjecture151 group C -algebra C r of is the C -algebra obtained as the norm-closure of .kŒ /in B.`2 /.The group algebra kŒ is equipped with the corresponding -positive conek rŒ C D f f2kŒ W9f n2kŒ C such that f n!f pointwise gD f f2kŒ W f is of positive type g;and the resultant semi-pre-C -algebra k rŒ satisfies C u.k rŒ /D C r .Indeed, if f2kŒ \ 1.C r C/,then for D .f/1=2ı12`2 ,one has f D and f is the pointwise limit of n n2kŒ C,where n2kŒ are such that k n k2!0.On the other hand,if f is of positive type(i.e.,the kernel .x;y/!f.x 1y/is positive semi-definite),then f D f and h .f/Á;Ái 0 for everyÁ2`2 ,which implies .f/2C r C.It follows that k rŒ C D arch.kŒ C/if and only if is amenable(see Theorem1).Example2.The -algebra kŒx1;:::;x d of polynomials in d commuting hermitian variables x1;:::;x d is a semi-pre-C -algebra,equipped with the -positive conekŒx1;:::;x d C D -positive cone generated by f1 x2i W i D1;:::;d g:One has C u.kŒx1;:::;x d /D C.Œ 1;1 d/,the algebra of the continuous func-tions onŒ 1;1 d,and x i is identified with the i-th coordinate projection.Example3.The -algebra k h x1;:::;x d i of polynomials in d non-commuting hermitian variables x1;:::;x d is a semi-pre-C -algebra,equipped with the -positive conek h x1;:::;x d i C D -positive cone generated by f1 x2i W i D1;:::;d g: One has C u.k h x1;:::;x d i/D C.Œ 1;1 / C.Œ 1;1 /,the unital full free product of d-copies of C.Œ 1;1 /.Example4.Let A and B be semi-pre-C -algebras.We denote by A˝B the algebraic tensor product over k.There are two standard ways to make A˝B into a semi-pre-C -algebra.Thefirst one,called the maximal tensor product and denoted by A˝max B,is A˝B equipped with.A˝max B/C D -positive cone generated by f a˝b W a2A C;b2B C g: The second one,called the minimal tensor product and denoted by A˝min B, is A˝B equipped with.A˝min B/C D.A˝B/h\.ÃA˝ÃB/ 1..C u.A/˝min C u.B//C/: (See Theorem14for a“better”description.)One has C u.A˝˛B/D C u.A/˝˛C.B/for˛2f max;min g.The right hand side is the C -algebra maximal u152N.Ozawa (resp.minimal)tensor product(see[BO,Pi1]).For A1 A2and B1 B2, one has.A1˝min B1/C D.A1˝min B1/\.A2˝min B2/C;but the similar identity need not hold for the maximal tensor product. Example5.The unital algebraic free product A B of semi-pre-C -algebras A and B,equipped with.A B/C D -positive cone generated by.A C[B C/;is a semi-pre-C -algebra,and C u.A B/D C u.A/ C u.B/,the unital full free product of the C -algebras C u.A/and C u.B/.The following is very basic(cf.[Ci]and Proposition15in[Sm]).Theorem1.Let A be a semi-pre-C -algebra andÃW A!C u.A/be the uni-versal C -algebra of A.Then,one has the following.kerÃD I.A/,the ideal of the infinitesimal elements.A h\à 1.C.A/C/D arch.A C/,the archimedean closure of A C.uAlthough it follows from the above theorem,we give here a direct proof of the fact that arch.A C/\. arch.A C// I.A/.Indeed,if h2Ä1and "1<h<"1for"2.0;1/,then one has0Ä.1C h/." h/.1C h/D".1C h/2 h h.2C h/hÄ.4"C"/1 .2 "/h2; which implies h2<5"1.We postpone the proof and give corollaries to this theorem.4.PositivstellensätzeWe give a few results which say if an element a is positive in a certain class of representations,then it is positive for an obvious reason.Such results are re-ferred to as“Positivstellensätze.”Recall that a C -algebra A is said to be resid-uallyfinite dimensional(RFD)iffinite-dimensional -representations separate the elements of A,i.e., .a/ 0for allfinite-dimensional -representations implies a 0in A.All abelian C -algebras and full group C -algebras of residuallyfinite amenable groups are RFD.Moreover,it is a well-known result of Choi that the full group C -algebra C F d of the free group F d of rank d is RFD(see Theorem26).In fact,finite representations(i.e.,the unitary repre-sentations such that .F d/isfinite)separate the elements of C F d([LS]). However,we note that the full group C -algebra of a residuallyfinite group need not be RFD([Be1]).We also note that the unital full free products of RFDAbout the Connes embedding conjecture153 C -algebras is again RFD([EL]).In particular,C u.k h x1;:::;x d i/is RFD.The results mentioned here have been proven for complex C -algebras,but they are equally valid for real cases.See Sect.7.Theorem1,when combined with resid-ualfinite dimensionality,immediately implies the following Positivstellensätze (cf.[Pu,HM]).Corollary2.The following are true.Let f2kŒ h.Then, .f/ 0for every unitary representation if and only if f2arch.kŒ C/.The full group C -algebra C of a group is RFD if and only if the fol-lowing statement holds.If f2kŒ h is such that .f/ 0for every finite-dimensional unitary representation ,then f2arch.kŒ C/.Let f2kŒx1;:::;x d h.Then,f.t1;:::;t d/ 0for all.t1;:::;t d/2Œ0;1 d if and only if f2arch.kŒx1;:::;x d C/.(See Example2.)Let f2k h x1;:::;x d i h.Then,f.X1;:::;X d/ 0for all contractive her-mitian matrices X1;:::;X d if and only if f2arch.k h x1;:::;x d i C/.(See Example3.)In some cases,the -positive cones are already archimedean closed.We will see later(Theorem26)this phenomenon for the free group algebras kŒF d . 5.Eidelheit–Kakutani separation theoremThe most basic tool in functional analysis is the Hahn–Banach theorem.In this note,we will need an algebraic form of it,the Eidelheit–Kakutani separation theorem.We recall the algebraic topology on an R-vector space V.Let C V be a convex subset.An element c2C is called an algebraic interior point of C if for every v2V there is">0such that c C v2C for all j j<". The convex cone C is said to be algebraically solid if the set Cıof algebraic interior points of C is non-empty.Notice that for every c2Cıand x2C, one has c C.1 /x2Cıfor every 2.0;1 .In particular,CııD Cıfor every convex subset C.We can equip V with a locally convex topology,called the algebraic topology,by declaring that any convex set that coincides with its algebraic interior is open.Then,every linear functional on V is continuous with respect to the algebraic topology.Now Hahn–Banach separation theorem reads as follows.Theorem3(Eidelheit–Kakutani([Ba])).Let V be an R-vector space,C an algebraically solid cone,and v2V n C.Then,there is a non-zero linear func-tional'W V!R such that'.c/:'.v/Äinfc2CIn particular,'.v/<'.c/for any algebraic interior point c2C.154N.Ozawa Notice that the Combes axiom A D A bdd is equivalent to that the unit1is an algebraic interior point of A C A h and arch.A C/is the algebraic closure of A C in A h.(This is where the Combes axiom is needed and it can be dispensed when the cone A C is algebraically closed.See Sect.3.4in[Sm].)Let A be a semi-pre-C -algebra.A unital -subspace S A is called a semi-operator system.Here,a -subspace is a subspace which is closed under the -operation. Existence of1in S ensures that S C D S\A C has enough elements to span S h.A linear functional'W S!k is called a state if'is self-adjoint, positive,and'.1/D1.Note that if k D C,then S is spanned by S C and every positive linear functional is automatically self-adjoint.However,this is not the case when k D R.In any case,every R-linear functional'W S h!R extends uniquely to a self-adjoint linear functional'W S!k.We write S.S/ for the set of states on S.Corollary4.Let A be a semi-pre-C -algebra.Let W A be a -subspace and v2A h n.A C C W h/.Then,there is a state'on A such that'.W/D f0g and'.v/Ä0.(Krein’s extension theorem)Let S A be a semi-operator system.Then every state on S extends to a state on A.Proof.Since A C C W h is an algebraically solid cone in A h,one mayfind a non-zero linear functional'on A h such that'.v/Äinf f'.c/W c2A C C W h g:Since'is non-zero,'.1/>0and one may assume that'.1/D1.Thus the self-adjoint extension of'on A,still denoted by',is a state such that'.v/Ä0 and'.W h/D f0g.Let x2W.Then,for every 2k,one has'.x/C. '.x// D'.. x/C. x/ /D0:This implies'.W/D f0g in either case k2f C;R g.For the second assertion,let'2S.S/be given and consider the coneC D f x2S h W'.x/ 0g C A C:It is not too hard to see that C is an algebraically solid cone in A h and v…C for any v2S h such that'.v/<0.Hence,one mayfind a state N'on A such that N'.C/ R 0.In the same way as above,one has that N'is zero on ker', which means N'j S D'.About the Connes embedding conjecture155 6.GNS constructionWe recall the celebrated GNS construction(Gelfand–Naimark–Segal construc-tion),which provides -representations out of states.Let a semi-pre-C -algebra A and a state'2S.A/be given.Then,A is equipped with a semi-inner product h y;x i D'.x y/,and it gives rise to a Hilbert space,which will bedenoted by L2.A;'/.We denote by O x the vector in L2.A;'/that correspondsto x2A.Thus,h O y;O x i D'.x y/and k O x k D'.x x/1=2.The left multiplica-tion x!ax by an element a2A extends to a bounded linear operator '.a/on L2.A;'/such that '.a/O x D ca x for a;x2A.(Observe that a aÄR1implies k '.a/k2ÄR.)It follows that 'W A!B.L2.A;'//is a positive -representation of A such that h '.a/O1;O1i D'.a/.If W A!B.H/is a positive -representation having a unit cyclic vector,then'.a/D h .a/ ; i is a state on A and .x/ !O x extends to a uni-tary isomorphism between H and L2.A;'/which intertwines and '.Sinceevery positive -representation decomposes into a direct sum of cyclic repre-sentations,one may obtain the universal C -algebra C u.A/of A as the closureof the image under the positive -representationM '2S.A/ 'W A !BM'2S.A/L2.A;'/Á:We also make an observation that.A˝min B/C in Example4coincides withc2.A˝B/h W .'˝ /.z cz/ 0for all'2S.A/;2S.B/;z2A˝B:7.Real versus complexWe describe here the relation between real and complex semi-pre-C -algebras. Because the majority of the researches on C -algebras are carried out for com-plex C -algebras,we look for a method of reducing real problems to complex problems.Suppose A R is a real semi-pre-C -algebra.Then,the complexifica-tion of A R is the complex semi-pre-C -algebra A C D A R C i A R.The -algebra structure(over C)of A C is defined in an obvious way,and.A C/C is defined to be the -positive cone generated by.A R/C:.A C/C Dn n Xi D1z i a i z i W n2N;a i2.A R/C;z i2A Co:(This is a temporary definition,and the official one will be given later.See Lemma11.)Note that A R\.A C/C D.A R/C.The complexification A C has an involutive and conjugate-linear -automorphism defined by x C i y!156N.Ozawa x i y ,x;y 2A R .Every complex semi-pre-C -algebra with an involutive and conjugate-linear -automorphism arises in this way.Lemma 5.Let R W A R !B R be a -homomorphism between real semi-pre-C -algebras (resp.'R W A R !R be a self-adjoint linear functional).Then,the complexification C W A C !B C (resp.'C W A C !C )is positive if and only if R (resp.'R )is so.Proof.Weonly prove that 'C is positive if 'R is so.The rest is trivial.Let b D P i z i a i z i 2.A C /C be arbitrary,where a i 2.A R /C and z i D x i C i y i .Then,b D P i .x i a i x i C y i a i y i /C i P i .x i a i y i y i a i x i /.Since x i a i y i y i a i x i is skew-hermitian,one has 'R .x i a i y i y i a i x i /D 0,and 'C .b/D'R .P i x i a i x i C y i a i y i / 0.This shows 'C is positive.We note that if H C denotes the complexification of a real Hilbert space H R ,then B .H R /C D B .H C /.Thus every positive -representation of a real semi-pre-C -algebra A R on H R extends to a positive -representation of its complexification A C on H C .Conversely,if is a positive -representation of A C on a complex Hilbert space H C ,then its restriction to A R is a positive -representation on the realification of H C .The realification of a complex Hilbert space H C is the real Hilbert space H C equipped with the real inner product h Á; i R D <h Á; i .Therefore,we arrive at the conclusion that C u .A R /C D C u .A C /.We also see that .R Œ /C D C Œ ,.A R ˝B R /C D A C ˝B C ,.A R B R /C D A C B C ,etc.8.Proof of Theorem 1We only prove the first assertion of Theorem 1.The proof of the second is very similar.We will prove a stronger assertion thatk Ã.x/k C u.A /D inf f R >0W R 21 x x 2A C g :The inequality Ätrivially follows from the C -identity.For the converse,as-sume that the right hand side is non-zero,and choose >0such that 21 x x …A C .By Corollary 4,there is '2S.A /such that '. 21 x x/Ä0.Thus for the GNS representation ',one hask '.x/k k '.x/O 1k D '.x x/1=2 :It follows that k Ã.x/k .About the Connes embedding conjecture157 9.Trace positive elementsLet A be a semi-pre-C -algebra.A state on A is called a tracial state if .xy/D .yx/for all x;y2A,or equivalently if is zero on the -subspace K D span f xy yx W x;y2A g spanned by commutators in A.We denote by T.A/the set of tracial states on A(which may be empty).Associated with 2T.A/is afinite von Neumann algebra. .A/00; /,which is the von Neumann algebra generated by .A/ B.L2.A; //with the faithful normal tracial state .a/D h a O1;O1i that extends the original .Recall that afinite von Neumann algebra is a pair.M; /of a von Neumann algebra and a faithful normal tracial state on M.The following theorem is proved in[KS]for the algebra in Example3and in[JP]for the free group algebras,but the proof equally works in the general setting.We note that for some groups ,notably for D SL3.Z/([Be2]),it is possible to describe all the tracial states on kŒ . Theorem6([KS]).Let A be a semi-pre-C -algebra,and a2A h.Then,the following are equivalent..1/ .a/ 0for all 2T.A/..2/ . .a// 0for everyfinite von Neumann algebra.M; /and every positive -homomorphism W A!M..3/a2arch.A C C K h/,where K h D K\A h D span f x x xx W x2A g. Proof.The equivalence.1/,.2/follows from the GNS construction.We only prove.1/).3/,as the converse is trivial.Suppose a C"1…A C C K h for some">0.Then,by Corollary4,there is 2S.A/such that .K/D f0g (i.e., 2T.A/)and .a/Ä "<0.10.Connes embedding conjectureThe Connes embedding conjecture(CEC)asserts that anyfinite von Neumann algebra.M; /with separable predual is embeddable into the ultrapower R! of the hyperfinite II1-factor R(over k2f C;R g).Here an embedding means an injective -homomorphism which preserves the tracial state.We note that if is a tracial state on a semi-pre-C -algebra A andÂW A!N is a -preserving -homomorphism into afinite von Neumann algebra.N; /,thenÂextends to a -preserving -isomorphism from .A/00onto the von Neumann subalgebra generated byÂ.A/in N(which coincides with the ultraweak clo-sure ofÂ.A/).Hence,.M; /satisfies CEC if there is an ultraweakly dense -subalgebra A M which has a -preserving embedding into R!.In particular, CEC is equivalent to that for every countably generated semi-pre-C -algebra A and 2T.A/,there is a -preserving -homomorphism from A into R!.We will see that this is equivalent to the tracial analogue of Positivstellensätze in158N.Ozawa Corollary 2.We first state a few equivalent forms of CEC.We denote by tr the tracial state 1N Tr on M N .k /.Theorem 7.For a finite von Neumann algebra .M; /with separable predual,the following are equivalent..1/.M; /satisfies CEC,i.e.,M ,!R !..2/Let d 2N and x 1;:::;x d 2M be hermitian contractions.Then,forevery m 2N and ">0,there are N 2N and hermitian contractions X 1;:::;X d 2M N .k /such thatj .x i 1 x i k / tr .X i 1 X i k /j <"for all k Äm and i j 2f 1;:::;d g ..3/Assume k D C (or replace M with its complexification in case k D R ).Letd 2N and u 1;:::;u d 2M be unitary elements.Then,for every ">0,there are N 2N and unitary matrices U 1;:::;U d 2M N .C /such thatj .u i u j / tr .U i U j /j <"for all i;j 2f 1;:::;d g .In particular,CEC holds true if and only if every .M; /satisfies condition .2/and/or .3/.The equivalence .1/,.2/is a rather routine consequence of the ultra-product construction.For the equivalence to (3),see Theorem 27.Note that the assumption k D C in condition (3)is essential because the real analogue of it is actually true ([DJ]).Since any finite von Neumann algebra M with sep-arable predual is embeddable into a II 1-factor which is generated by two her-mitian elements (namely .M R/N ˝R ),to prove CEC,it is enough to verify the conjecture (2)for every .M; /and d D 2.We observe that a real finite von Neumann algebra .M R ; R /is embeddable into R !R (i.e.,it satisfies CEC)if and only if its complexification .M C ; C /is embeddable into R !C .The “only if”di-rection is trivial and the “if”direction follows from the real -homomorphism M N .C /,!M 2N .R /,a C i b ! a b b a .A complex finite von Neumann al-gebra .M; /need not be a complexification of a real von Neumann algebra,but M ˚M op is (isomorphic to the complexification of the realification of M ).Therefore,M satisfies CEC if and only if its realification satisfies it.For a finite von Neumann algebra .M; /and d 2N ,we denote by H d .M /the set of those f 2k h x 1;:::;x d i h such that .f.X 1;:::;X d // 0for all hermitian contractions X 1;:::;X d 2M .Further,let H d D\M H d .M /and H fin d D \NH d .M N .k //D H d .R/:Notice that H d D arch .k h x 1;:::;x d i C C K h /(see Example 3and Theorem 6).About the Connes embedding conjecture 159Corollary 8([KS]).Let k 2f C ;R g .Then one has the following.Let .M; /be a finite von Neumann algebra with separable predual.Then,M satisfies CEC if and only if H fin d H d .M /for all d .CEC holds true if and only if H fin d D arch .k h x 1;:::;x d i C C K h /for all/some d 2.Proof.It is easy to see that condition (2)in Theorem 7implies H fin d H d .M /.Conversely,suppose condition (2)does not hold for some d 2N ,x 1;:::;x d 2M ,m 2N ,and ">0.We introduce the multi-index notation.For i D .i 1;:::;i k /,i j 2f 1;:::;d g and k Äm ,we denote x i D x i 1 x i k .It may happen that i is the null string ;and x ;D 1.Then,C D closure f .tr .X i //i W N 2N ;X 1;:::;X d 2M N .k /h ;k X i k Ä1gis a convex set (consider a direct sum of matrices).Hence by Theorem 3,there are 2R and ˛i 2k such that <X i ˛i .x i /< Äinf 2C <X i˛i i :Replacing ˛i with .˛i C ˛ i/=2(here i is the reverse of i ),we may omit <from the above inequality.Further,arranging ˛;,we may assume D 0.Thus f D P i ˛i x i belongs to H fin d ,but not to H d .M /.This completes the proof of the first half.The second half follows from this and Theorem 6.An analogue to the above also holds for C ŒF d .Corollary 9([JP]).Let k D C .The following holds.Let .M; /be a finite von Neumann algebra with separable predual.Then,M satisfies CEC if and only if the following holds true:If d 2N and ˛2M d .C /h satisfies that tr .P ˛i;j U i U j / 0for every N 2N and U 1;:::;U d 2M N .C /u ,then it satisfies .P ˛i;j u i u j / 0for every uni-tary elements u 1;:::;u d 2M .CEC holds true if and only if for every d 2N and ˛2M d .C /h the follow-ing holds true:If tr .P ˛i;j U i U j / 0for every N 2N and U 1;:::;U d 2M N .C /u ,then P ˛i;j s i s j 2arch .C ŒF d C C K h /,where s 1;:::;s d are the free generators of F d .11.Matrix algebras over semi-pre-C -algebrasWe describe here how to make the n n matrix algebra M n .A /over a semi-pre-C -algebra A into a semi-pre-C -algebra.We note that x D Œx j;i i;j for x D Œx i;j i;j 2M n .A /.We often identify M n .A /with M n .k /˝A .There。
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.。
Secrets of the Universe
The secrets of the universe have captivated the human mind for centuries. From ancient civilizations gazing at the stars to modern scientists probing the depths of space, the quest to unravel the mysteries of the cosmos has been a driving force for exploration and discovery.One of the most profound secrets of the universe is the nature of dark matter and dark energy. These elusive substances make up the vast majority of the universe, yet their true identities remain a mystery. Dark matter, which does not emit, absorb, or reflect light, has only been detected through its gravitational effects on visible matter. Dark energy, on the other hand, is thought to be responsible for the accelerated expansion of the universe, yet its origin and properties are still not fully understood.Another enigma is the origin of the universe itself. The Big Bang theory, which suggests that the universe began as a hot, dense state and has been expanding ever since, is the leading explanation forthe universe's birth. However, what triggered the Big Bang and what came before it are questions that continue to baffle scientists.The search for extraterrestrial life is another captivating secretof the universe. With the discovery of thousands of exoplanets inour galaxy, the possibility of finding life beyond Earth has never been greater. Scientists are studying the conditions necessary for life to exist and developing new technologies to search for signs of life on distant worlds.The nature of black holes is yet another intriguing puzzle. These cosmic entities, formed from the remnants of massive stars, possess gravitational forces so strong that nothing, not even light, can escape their grasp. The inner workings of black holes and the mysteries of what lies beyond the event horizon continue to confound researchers.As our understanding of the universe deepens, new secrets and mysteries continue to emerge. From the nature of dark matter anddark energy to the search for extraterrestrial life, the secrets of the universe are a source of endless fascination and inspiration for scientists and curious minds alike. The quest to unlock thesesecrets drives humanity's exploration of the cosmos and fuels our insatiable curiosity about the world beyond our own.。
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a rX iv:mat h /52473v2[mat h.NA ]4Ma y25THE PROOF OF TCHAKALOFF’S THEOREM CHRISTIAN BAYER,JOSEF TEICHMANN Abstract.We provide a simple proof of Tchakaloff’s Theorem on the exis-tence of cubature formulas of degree m for Borel measures with moments up to order m .The result improves known results for non-compact support,since we do not need conditions on (m +1)st moments.In fact we reduce the classical assertion of Tchakaloff’s Theorem to a well-known statement going back to F.Riesz.We consider the question of existence of cubature formulas of degree m for Borel measures µ,i.e.a measure defined on the Borel σ-algebra,where moments up to degree m exist:Definition 1.Let µbe a positive Borel measure on R N and m ≥1such that R N x k µ(dx )<∞for 0≤k ≤m holds true.A cubature formula of degree m for µis given by an integer k ≥1,points x 1,...,x k ∈supp µ,weights λ1,...,λk >0such that R N P (x )µ(dx )=k i =1λi P (x i )for all polynomials on R N with degree less or equal m ,where supp µdenotes the closed support of the measure µ,i.e.the complement of the biggest open set O ⊂R N with µ(O )=0.Cubature formulas of degree m have been proved to exist for Borel measures µ,where the (m +1)st moments exist,see [1]and [6].The result in the case of compact supp µis classical,and due to Tchakaloff(see [10]),hence we refer to the assertion as Tchakaloff’s Theorem.We collect some basic notions and results from convex analysis,see for instance [9]:fix N ≥1,for some set S ⊂R N the convex hull of S ,i.e.the smallest convexset in R N containing S ,is denoted by conv(S ),the (topological)closure of conv(S )is denoted byC \int(C )be a boundary point.There is a linear2CHRISTIAN BAYER,JOSEF TEICHMANNfunctional l y and a real numberβy such that the hyperplane defined by l y=βy contains y,and C is contained in the closed half-space l y≤βy.Hyperplanes and half-spaces with this property are called supporting hyperplanes and supporting half-spaces,respectively.Moreover,cone(A).We also introduce the notion of the relative interior of a convex set C:a point x∈C lies in the relative interior ri(C)if for every y∈C there isǫ>0such that x−ǫ(y−x)∈C.In particular we have that the relative interior of a convex set C coincides with the relative interior of its closureK\int(K)in the boundary of K=cone(A).Then all linear functionals l y:R N→R corresponding to the supporting half-spaces l y≤0at y are certainly integrable and we havel y(E)= R N l y(x)µ(dx)≤0,consequently E∈THE PROOF OF TCHAKALOFF’S THEOREM3 Hence E∈cone(A).In particular E∈cone(A),since the interior lies in the convex cone hull of A.If thefirst condition is not satisfied, we obtain that E is an interior point of cone(A)in an affine subspace of R N(where thefirst condition is satisfied),but then E lies in the relative interior of cone(A)in R N,which is the desired result.Corollary1.Letµbe a positive Borel measure on R N concentrated in A⊂R N, i.e.µ(R N\A)=0,such that thefirst moments exist,i.e.R Nx µ(dx)<∞.Then there exist an integer1≤k≤N,points x1,...,x k∈A and weights λ1,...,λk>0such thatR N f(x)µ(dx)=ki=1λi f(x i)for any monomial f on R N of degree1.Proof.The corollary follows immediately from Theorem1and Caratheodory’s The-orem(see[9],Th.17.1and Cor.17.1.2). Corollary2.Letµbe a positive measure on the measurable space(Ω,F)concen-trated in A∈F,i.e.µ(Ω\A)=0,andφ:Ω→R N a Borel measurable map. Assume that thefirst moments ofφ∗µexist,i.e.R Nx φ∗µ(dx)<∞.Then there exist an integer1≤k≤N,pointsω1,...,ωk∈A and weights λ1,...,λk>0such thatΩφj(ω)µ(dω)=ki=1λiφj(ωi)for1≤j≤N,whereφj denotes the j-th component ofφ.Remark1.In other words,A∈F such thatµ(Ω\A)=0correspond to B⊂φ(Ω) such thatφ∗µ(R N\B)=0.Remark2.Note thatµ(Ω)=∞is also possible,since we only speak about inte-grability of N measurable functionsφ1,...,φN.If we haveµ(Ω)<∞,we could addφN+1=1,and we obtain in particular k′i=1λ′i=µ(Ω)(with possibly different number1≤k′≤N+1of points x′i and weightsλ′i).In the setting of Theorem1assume thatµis a probability measure on R N.Then –by the previous consideration–E= R N xµ(dx)lies in the convex hull conv(A). This fact is well-known infinancial mathematics,since it means that the price range of forward contracts is given by the relative interior of the convex hull of the no-arbitrage bounds of the(discounted)price process(see for instance[2],Th.1.40).The result is also well-known in thefield of geometry of the moment problem,seefor instance[4].As mentioned therein,the result for compactly supported measures essentially even goes back to F.Riesz,see[8].4CHRISTIAN BAYER,JOSEF TEICHMANNProof.We solve the problem with respect toφ∗µon R N and obtain1≤k≤N, y1,...,y k∈φ(A)andλ1,...,λk>0such thatR N f(y)(φ∗µ)(dy)=ki=1λi f(y i)for all polynomials f of degree1.Thus we obtain pointsω1,...,ωk withφ(ωi)=y i for1≤i≤k,furthermoreR Nf(y)(φ∗µ)(dy)= Ω(f◦φ)(ω)µ(dω)by definition,hence the result.In an adequate algebraic framework the previous Theorem1yields all cubature results in full generality,and even generalizes those results(see[1],[6]and[7]for related theory and interesting extensive references).For this purpose we consider polynomials in N(commuting)variables e1,...,e N with degree function deg(e i):=k i for1≤i≤N and integers k i≥1.Hence,wecan associate a degree to monomials e i1...e ilwith(i1,...,i l)∈{1,...,N}l forl≥0(note that the monomial associated to the empty sequence is by convenience 1),namelydeg(e i1···e il)=lr=1k ir.We denote by A N deg≤m the vector space of polynomials generated by monomials of degree less or equal m,for some integer m≥1.We define a continuous map φ:R N→A N deg≤m,viaφ(x1,...,x N)= l≥0(i1,...,i l)∈{1,...,N}l, l r=1k i r≤m x i1···x ile i1···e il.Continuity is obvious,since we are given monomials in each coordinate.φis even an embedding and a closed map.The following example shows the relevant idea in coordinates,since for N=1 and deg(e1)=1we obtain A1deg≤m=R m+1.Example 1.Fix m≥1.Thenφ(x)=(1,x,x2,...,x m)is a continuous map φ:R1→R m+1.Given a positive Borel measureµon R1such that moments up to degree m exist,i.e.R|x|kµ(dx)<∞for0≤k≤m,thenφ∗µadmits moments up to degree1.Hence we conclude that there exist1≤k≤m+1,points x1,...,x k and weightsλ1,...,λk>0such thatR N P(x)µ(dx)=ki=1λi P(x i)for all polynomials P of degree less or equal m.THE PROOF OF TCHAKALOFF’S THEOREM 5Theorem 2.Given N ≥1and degree function deg and m ≥1.Fix a finite,positive Borel measure µon R N concentrated in A ⊂R N ,i.e.µ(R N \A )=0,such that R N |x i 1···x i l |µ(dx )<∞for (i 1,...,i l )∈{1,...,N }lwith l r =1k i r ≤m .Then there exist an integer 1≤k ≤dim A N deg ≤m ,points x 1,...,x k ∈A and weights λ1,...,λk >0such thatR N P (x )µ(dx )=k i =1λi P (x i )for P ∈A N deg ≤m .Proof.The measure φ∗µadmits first moments by assumption,hence we conclude by Corollary 2. Remark 3.Tchakaloff’s Theorem is a special case of the above theorem with A =supp µ.Remark 4.Fix a non-empty,closed set K ⊂R N .We note that a finite sequence of real numbers m i 1...i l for (i 1,...,i l )∈{1,...,N }l with lr =1k i r ≤m representsthe sequence of moments of a Borel probability measure µwith support supp µ⊂K ,where moments of degree less or equal m exist,if and only if l ≥0 (i 1,...,i l )∈{1,...,N }l , l r =1k i r≤m m i 1...i l e i 1···e i l ∈conv φ(K ).The argument in one direction is that any element of conv φ(K )is represented as expectation with respect to some probability measure with support in K ,for instance the given convex combination.The other direction is Tchakaloff’s Theorem in the general form of Theorem 2.Consequently we have a precise geometric character-ization of solvability of the Truncated Moment Problem for measures with support in K .Notice that one can often describe conv φ(K )by finitely many inequalties.References[1]Raul E.Curto and Lawrence A.Fialkow,A Duality Proof of Tchakoloff’s Theorem ,J.Math.Anal.Appl.269(2),519–532,2002.[2]Hans F¨o llmer and Alexander Schied,Stochastic Finance ,Walter de Gruyter,Studies in Math-ematics 27,Berlin,2002.[3]Samuel Karlin and Lloyd S.Shapley,Geometry of moment spaces ,Mem.Amer.Math.Soc.12(1953).[4]Johannes H.B.Kemperman,On the sharpness of Tchebychefftype inequalities ,Indag.Math.27,554–601(1965).[5]Johannes H.B.Kemperman,Geometry of the moment problem ,Proc.Symp.Appl.Math.37(1987),AMS,pp.16-52.[6]Mihai Putinar,A Note on Tchakaloff’s Theorem ,Proc.Amer.Math.Soc.125(8),2409–2414,1997.[7]Bruce Reznick,Sums of even powers of real linear forms ,Mem.Amer.Math.Soc.Volume96(463),1992.[8]Frigyes Riesz,Sur certains syst`e mes singuliers d’´e quations int´e grales ,Ann.Sci.´Ecole Norm.Sup.(3),15,69–81(1911).[9]R.Tyrrell Rockafellar,Convex Analysis ,Princeton University Press,1972.[10]V.Tchakaloff,Formules de cubature m´e canique `a coefficients non n´e gatifs ,Bull.Sci.Math.81,123–134,1957.6CHRISTIAN BAYER,JOSEF TEICHMANNTechnical University of Vienna,e105,Wiedner Hauptstrasse8-10,A-1040Wien,Aus-triaE-mail address:cbayer@fam.tuwien.ac.at,jteichma@fam.tuwien.ac.at。