Hypersurfaces Moving with Curvature-Dependent Speed Hamilton-Jacobi Equations, Conservation

Rolling ball method for 5-axis surface machiningP.Gray,S.Bedi *,F.IsmailAutomation and Control Group,Department of Mechanical Engineering,University of Waterloo,Waterloo,Ont.,Canada N2L 3G1Received 26June 2001;revised 20October 2001;accepted 24November 2001AbstractCurvature matching for 5-axis surface machining has been plagued by the complexity of the task.As a result the current tool positioning strategies are likewise computationally complicated.Gouging the surface has been the main concern and has presented the greatest dif®culty in the algorithms.Some of the methods perform exhaustive searches of the surface to avoid gouging while others incrementally adjust the tool orientation until gouges are no longer detected.In this paper a new positioning strategy is presented that is simple to implement and is not dif®cult to compute.The rolling ball method rolls a variable radius ball along the tool path and positions the cutting tool to cut the rolling ball.A small region of the ball's surface is used to approximate a small region of the surface being machined.The radius of each ball is computed by checking a grid of points in the area of the surface that the tool casts a shadow for each tool position.A pseudo-radius is computed for each grid point and the most appropriate radius is selected to be the rolling ball's radius.The selection process follows a hierarchy of surface pro®les ranging from convex to concave.Convex,concave,and saddle (mixed)surface regions are all computed in a similar fashion and there are no special cases for which the positioning strategy must be changed to compute a tool position.Local gouge checking is automatically built-in to the positioning computations so that the typical iterative strategy of checking for gouging,then incrementally tilting the tool until no gouges are detected is eliminated.The method is robust and simple to implement and it only requires surface coordinates and surface normals.A simulation of the method and a cutting test were performed and are presented in this document.Crown Copyright q 2002Published by Elsevier Science Ltd.All rights reserved.Keywords :Five-axis;Machining;Surface1.IntroductionSurface machining is performed for a variety of products including dies,molds and turbine blades.Typically ball-nose endmills are used for ®nish milling operations.They are simple to program for three and 5-axis machines as the center of the ball always lies along the normal of the surface at a distance equal to the radius of the ball.Interference is easy to detect with the inside±outside method using the analytic geometric equation of a sphere [1].A ball-nose endmill will not gouge most surfaces as long as the radius of the ball is less than the radius of curvature at the cutter contact point.On the other hand,a bull-nose (toroidal)or corner-radius endmill is preferred for surface machining because higher surface speeds can be achieved at lower rotational speeds leading to higher feedrates which gives higher metal removal rates [2]and an improved surface ®nish.Corner-radius endmills can be used in 3,4,or 5-axis machining but they are more dif®cult to program asmultiple solutions exist for each tool position and interfer-ence with the workpiece can easily occur,often unexpectedly.Five-axis machining is inherently more complicated than 3-axis machining.It is critical to check for interference and travel limitations.The motion of the machine is far less intuitive and the machine itself is less rigid.To position a corner-radiused endmill a cutter contact point (ccp)is selected and,depending on the method,the tool is posi-tioned according to surface characteristics at or around that ccp.Methods vary from simply ®nding a solution to ®nding a tool position and orientation that minimizes the distance between the whole tool and the surface to advanced methods which match the tool to the local shape of the surface.There are three general methods that have been devel-oped.The ®rst development is known as Sturz method.The cutter axis is set to a ®xed angle with respect to the surface normal and feed direction at the ccp.The second,multi-point machining (MPM)pivots the tool about the ccp in an attempt to generate a second ccp for each tool position.The third method,known as the principle axis method (PAM)or curvature matching [3,4]attempts to match the curvature of the surface at the ccp by tilting the tool such0010-4485/03/$-see front matter Crown Copyright q 2002Published by Elsevier Science Ltd.All rights reserved.PII:S0010-4485(02)00056-8*Corresponding author.Tel.:11-519-888-4567x2178;fax:11-519-888-6197.E-mail address:sbedi@uwaterloo.ca (S.Bedi).that the projected effective radius of the tool matches that of the curvature of the surface.Further details of these methods can be found in Refs.[1±5,8,9]but some details are outlined in the body of this paper for completeness.To successfully implement PAM and the Sturz method,each tool position must be checked for local gouging;a secondary process that is more expensive both computation-ally and time-wise than the original positioning strategy.When a gouge is detected,the tool is incrementally inclined further and checked for gouging until the tool clears the part.This increases the apparent curvature of the tool and lifts the tool further away from the surface.This technique of gouge avoidance is only applicable to the local vicinity of the cutting face of the tool and does not check for tool shank and tool holder interference with the part,®xture,or clamp-ing devices.It is,however,a suf®cient check for open face free form surfaces.For MPM,gouge checking and avoid-ance are part of the algorithm and do not need to be performed separately.However,the basic tool position algorithm is very time consuming.PAM uses local curva-ture information generated at one speci®c point on the surface to position the tool.This approximation fails frequently because the tool is a spatial entity and to position it without gouging,regional surface information under the shadow of the tool is needed.MPM on the other hand uses regional surface information via searching for the second ccp.Though MPM generates gouge-free tool positions,it comes at the expense of an extremely complex algorithm.1.1.Local gouge detection versus global interference detectionIt is important at this point to make a distinction between gouge detection and avoidance versus interference detection and avoidance.The purpose of 5-axis surface machining is to match the tool's cutting surface as closely as possible to the model surface so as to cut as wide a strip as possible without gouging the surface.With the attempt to cut a wider strip the area about the ccp is in danger of being gouged.Furthermore,the backside of the tool may also gouge the surface.For these reasons,gouge detection algorithms are critical to any 5-axis tool positioning strategy.Somemethods inherently check for gouging within the algorithm such as MPM and the rolling ball method (RBM)proposed in this paper.Others rely on gouge detection to eliminate gouging by iteratively checking for gouges and tilting the tool incrementally until gouges are no longer detected for each tool position.This is usually performed as a secondary simulation and gouge avoidance algorithm.It is this second-ary simulation and iterative gouge check and avoidance routine that the RBM eliminates in an ef®cient fashion.It is important to note that regardless of the positioning strat-egy all tool paths suffer from the possibility of interference of the tool and tool holder with the work piece and its ®xture.Thus,every tool path must ultimately,in some way,be checked for global interference whether it be a visual check or through simulation software.With compli-cated parts,®xtures,and tool paths this is an essential secondary operation to be performed.For simple operations or when machining open face surfaces it is not as critical.1.2.Research goalsThere are two goals of this work.The ®rst is to simplify the tool positioning computation.The strategy should not be based on optimization and exhaustive search routines because these are slow and unpredictable in the sense that the routines may not converge to a solution.Furthermore,these types of algorithms require special parameters that are dif®cult to set and require a mathematical understanding of the algorithm and parameters.The second goal is to have a built-in gouge checking and avoidance system to eliminate the need for a secondary gouge checking and correction process.It is not the intention of this work to eliminate tool path simulation for detection of global interference of the tool shank and tool holder with the work piece.A clear distinction between local gouge versus global interference detection and avoidance has been made in Section 1.1for the purpose of this work.In this paper a method is presented to match the curvature of a corner-radiused endmill to that of a region of a surface.It is relatively simple to calculate for convex,concave,saddle,and planar surfaces and gouge avoidance is built-in the method.1.3.Introduction to the rolling ball methodThe basic idea of the RBM is to roll a varying radius ball along a tool path on a surface (Fig.1)and position the cutting tool inside the ball (Fig.2).The ball is guaranteed to have one ccp and it approximates the model surface in a speci®ed area around the ccp as a sphere.The radius of the ball is the critical piece of data that must be generated.To evaluate the radius of the ball,the pseudo-curvature of the surface is computed for a grid of points in a small area about the ccp where the tool will be positioned.(Pseudo-curvature is a term that refers to the general radius of a sphere that is ®tted to a region of a surface which approximates the surface region.The term is explained in more detail later.)Thus only a small section of the rolling ball sphere is used toP.Gray et al./Computer-Aided Design 35(2003)347±357348Fig.1.Rolling ball method.approximate the region of the surface that the tool casts a shadow on.These calculations are simple because they only require surface coordinates and surface normals.The `most concave'radius of curvature is used as the radius of the ball.This ensures that the tool will not gouge the surface because it is the smallest ball that will ®t in the area without gouging the surface while maintaining contact at the ccp.In contrast to PAM and other methods that use curvatures computed at a single point on the surface,the RBM uses regional infor-mation around the ccp beneath the tool to determine the curvature of the region instead of the curvature at a single point.Two methods were used to verify this theory.An advanced simulation program was created to check for gouging tool positions and to view the generated tool path.An actual cutting test was performed to physically prove the theory.The sample workpiece was measured with a CMM and the data compared to the CAD model to verify the accuracy of the positioning putation times are also included for different tool dimensions.2.Other 5-axis tool positioning strategiesTool positioning in 5-axis has generally focused on posi-tioning a milling cutter on a surface such that it does not gouge the model surface.There are several ways to accom-plish this,three of which are reviewed here.2.1.Sturz or inclined toolThis is the simplest method for 5-axis positioning.The tool is inclined at a constant angle in the feed direction about the ccp in the plane containing the ccp,the feed direction and the surface normal [5](Fig.3).The inclination angle istypically between 5and 108.Obviously the tool will not be optimally positioned for every point but the gains in metal removal rate are signi®cant over 3-axis surface machining [6].The dif®cult question is:at what angle should the tool be inclined?If the angle is too high then lower gains in metal removal rate will occur.If it is too low the tool may gouge the workpiece.This means that the tool paths generated must be simulated and checked for gouging with the design surface.If any is detected the tool path must be regenerated with a larger inclination angle.2.2.Multi-point machiningIn contrast with the Sturz method,MPM is one of the most complicated 5-axis tool positioning strategies.The objective of MPM is to position the tool such that there are two ccps instead of only one.This is a mathematically dif®cult task that involves searching speci®c areas for poten-tial candidates for the second ccp for each tool position.The tool is forced to maintain contact at the ®rst ccp.It is then rotated about two independent axes until an optimal position is achieved that minimizes the distance between the tool and the second ccp [7,8](Fig.4).The method also fails to converge to a solution in some cases.Due to the complex-ities of the algorithm this method would have limited use in industry.P.Gray et al./Computer-Aided Design 35(2003)347±357349Fig.3.Sturz method (taken with permission from Warkentin et al.[6]).Fig.2.Circular line of contact between a toroidal tool and a sphere for a concave surfaceregion.Fig.4.Multi-point machining.2.3.Principle axis methodPAM is actually a degenerate case of MPM where the two ccps converge to a single ccp.The tool position is selected based on the local principle curvatures of the surface at the ccp.The tool is inclined about the maximum curvature direction by an angle such that the effective projected radius of the tool is equal to the minimum radius of curvature [9](Fig.5).Originally the feed direction was chosen to follow the direction of minimum curvature computed at each ccp.However,the method has been modi®ed to permit tool paths in any direction [1].The inclination angle computed from the maximum curvature is a starting point for an iterative search for an angle that does not cause gouging.Thus each tool position must be incrementally checked for gouging and adjusted until it is gouge-free.The axis about which the tool is lifted is selected in an ad hoc manner or by an optimization method that can be time consuming and temperamental.The secondary step of gouge checking and avoidance makes this technique dif®cult to implement.2.4.DiscussionThe common dif®culty encountered with all the methods is that they are extremely complicated to implement;the calculations and algorithms are dif®cult.Often the methods require data from the surface that is dif®cult to compute such as curvature which requires directional second deriva-tives.The data required may not be available if the surface does not have high enough continuity in all directions or the surface is given as a triangulated set of data points.Some-times these dif®culties can be overcome by using numerical computation methods which are often impractical to imple-ment because of the complexity of implementation,long calculation times,and unreliable results if the solution does not converge.Often special code to detect and account for special cases has to be implemented.Furthermore,for most of the positioning methods,local gouge checking is not part of the tool positioning strategy and the strategy only gives a starting point.A secondary process involving aniterative procedure of gouge checking and tool position adjustment is then performed until the tool is clear of the surface.This is not ef®cient and the computation time for each position is unpredictable and varying.For these reasons industry has not bene®ted from these sophisticated positioning strategies and has only implemented the Sturz method because it is simple,reliable,and robust.3.Rolling ball method 3.1.OverviewThe principle objectives in the development of the RBM was to simplify the computation for 5-axis tool positioning and to make a robust algorithm capable of machining complex surfaces reliably without having to perform a secondary local gouge checking and avoidance procedure.To accomplish this it was determined that the method should use simple algorithms based on regional information of the surface in the area surrounding the tool and is applic-able to all types of surfaces and surface de®nitions.The method is a derivative of PAM and MPM in that it takes the best features of both and combines them into a unique tool positioning strategy that produces gouge-free tool positions.It can be thought of as a ball of varying radius rolling along the surface in the direction of the tool path (Fig.1).The tool is placed inside the rolling ball and main-tains contact at the ball's ccp similar to PAM but it will not gouge the surface and the inclination angle is only computed once.Similar to MPM but in contrast to PAM,the RBM uses regional information around the ccp to compute the curvature of the region below the tool instead of curvature computed at a single point.However,unlike MPM,the RBM is a stable algorithm that is simple to implement and fast to compute.3.2.TheoryRBM uses the principle of MPM to position the cutting tool inside of a rolling sphere.It can be shown that any cylindrical or toroidal tool can be placed on a sphere and have a full circular line of contact with the sphere [10](Figs.2and 6).A modi®cation of the PAM strategy is used for positioning the rolling ball.In PAM the curvature values computed at the ccp are local in nature and do not provide any information about the surrounding area which,in many cases,will cause the tool to gouge the surface.For the RBM,a pseudo-radius of curvature is calculated for each point of a grid of points in the vicinity of the ccp where the tool contacts the surface.The rolling ball radius is selected as one of the pseudo-radii of curvature of the grid of points and used for positioning the tool at the speci®c ccp.Depending on the regional surface characteristics in the area of the grid of points,the rolling ball approximates the surface region as either convex or concave or as a ball of in®nite radius if the surface is a ¯at planar area.Determining the regionalP.Gray et al./Computer-Aided Design 35(2003)347±357350Fig.5.Principle axis method (taken with permission from Warkentin et al.[6]).surface pro®le is automatic in the algorithm because the radius is used to determine the concavity or convexity of the region.Tool positioning for concave regions is identical to tool positioning for convex regions.The identical approach to all surface pro®les makes the RBM very robust and reliable.Saddle regions (mixed concave and convex)are considered to be concave because the RBM positioning strategy will lift the tool further above the surface for concave balls than for convex balls.A hierarchy was devel-oped to determine which of the pseudo-radii computed from the shadow checking area should be used for the rolling ball.The radius is selected as the `most concave'radius computed from the shadow checking area.Fig.7shows this hierarchy structure.3.3.Radius of rolling ballThis is the key element to the RBM.The radius of the ball depends on the surface region around the ccp.When a tool is positioned in 5-axis to touch the surface at the ccp,it may gouge the region of the surface where the tool resides.In order to compute a tool orientation that does not cause gouging,the positioning strategy must take into account the shape of the surface in the shadow of the tool as the tool is being positioned.The shadow region over which the rolling ball radius must be calculated depends on the tool orientation and is not initially known.This means that the shape and size of this region must be estimated in order to position the tool.Though the shape of the region is arbitrary,it must encapsulate all possible shadows that the tool could cast onto the surface once it is positioned.In our implemen-tation we use a circular region.The region is an oversized estimate of the shadow the tool will project onto the surface once it is positioned.This shadow area is divided into a grid of points which are used to determine the rolling ball's radius (Fig.8).A pseudo-radius of curvature is computed for eachshadow area grid point.The pseudo-radius of curvature is de®ned as the radius of the circle whose center lies along the surface normal at the ccp and the circle passes through the ccp and the shadow area grid point (Fig.9)(Eq.(1)).This is similar to the calculation of the osculating plane presented by Farin [11]for three-dimensional spline curves.A poten-tial radius for the ball is computed at each grid point in the shadow-area beneath the tool.Each radius is compared and the ®nal rolling ball radius is selected as the `most concave radius'found in the shadow area checking region.The most concave radius refers to a hierarchy for the radius of curva-ture shown in ®gure (Fig.7).The `most concave radius'is the most concave radius of the shadow region.It is used to ensure that the tool position is matched to the highest protruding point in the shadow region which guarantees no gouging will occur provided the shadow-checking grid is dense enough.It ensures that the tool will be lifted higher than all the other points where potential radii are calculated.r2~E´~E 2£^n´~E 1where,~E is the S Åhadow_checking_pt 2c Åcp,nÃthe Surface normal at ccp and r is the pseudo-radius of curvature.P.Gray et al./Computer-Aided Design 35(2003)347±357351Fig.7.Hierarchy for rolling ball radiusselection.Fig.8.Schematic of the shadow checking area boundary and gridpoints.Fig.6.Toroidal tool positioned on outside of sphere for convex surface regions.This is a simple task when using functional parametricsurfaces where the control points are mapped linearly from the domain to Euclidean space.For functional parametric patches the checking area can be stored as a constant array of D u and D v parameters measured as offsets from the u and v coordinates of the ccp.Unfortunately the algorithm for calculating the shadow area cannot be implemented so easily for surfaces that are not linearly mapped between the domain and Euclidean spaces.A marching method would then be required to outline the boundaries of the shadow checking area.3.4.Positioning the tool inside the rolling ballPositioning the tool inside the rolling ball is a simple matter.Warkentin et al.[10]showed that any cylindrical tool,be it toroidal or ¯at-ended can be positioned in a sphere such that it forms a circular line of contact.For RBM the tool is placed such that it maintains contact at the ccp and has a circular line of contact with the rolling ball (Fig.2)(Eqs.(2)and (3)).The RBM matches convex surface regions by positioning the tool on the outside of the rolling ball (Fig.6).This can only be done using toroidal tools that have a clearance in the bottom center area.a R sphere 2R sphere 2r 22R 2major q 2 Tool center 0 0;0;a Tool center 1 R x ;f Tool center 0Tool center 2^du x ^d v x ^n x ccp x^d u y ^d v y ^n y ccp y ^d u z^d v z ^n z ccp z 0012666666437777775Tool center 1where R sphere is the rolling ball radius,R major the major radius of toroidal tool,r the minor radius of toroidal tool,a the initial z -height of a toroidal tool inside the rolling ball with it's axis aligned with the z -axis,ccp (ccp x ,ccp y ,ccp z ),thecutter contact point coordinate,^du ^d u x ;^d u y ;^d u z ;the tangent to surface in u -parametric direction at ccp,^dv ^dv x ;^d v y ;^d v z ;the tangent to surface in v -parametric direc-tion at ccp,^n^n x ;^n y ;^n z ;the surface normal at ccp,[R x ,f ]the rotation about x -axis by the tool inclination angle f for arolling ball centered at (0,0,0),Tool center 0the tool center position inside rolling ball sphere centered at the origin (0,0,0),Tool center 1the tool center rotated about the x -axis by the inclination angle f such that the tool contacts the bottom of the rolling ball where the z -axis intersects the ball and this will be the ccp,Tool center 2is the tool center trans-formed such that the lowest point of the rolling ball touches the surface at the ccp and the z -axis of the ball is aligned with the surface normal.^Tool axis 00;0;1 3^Tool axis 1 R x ;f ^Tool axis 0^Tool axis 2 ^du x ^d v x ^n x ^d u y ^d v y ^n y ^du z ^dv z ^nz 2666437775^Tool axis 1where Tool axis 0is the tool axis vector inside rolling ballsphere centered at the origin (0,0,0),Tool axis 1the tool axis rotated about the x -axis by the inclination angle f such that the tool contacts the bottom of the rolling ball where the z -axis intersects the ball and this will be the ccp,Tool axis 2is the tool axis rotated such that the lowest point of the rolling ball touches the surface at the ccp and the z -axis of the ball is aligned with the surface normal.3.5.Tool path generationTypical tool paths for surface machining move the tool along constrained x ,y ,or z directions along isoparametric lines or in a spiraling fashion.In the RBM,the tool path can be as complicated or as simple as desired.The tool can turn corners or change feed directions simply by changing the shadow area boundaries and computing the rolling ball radius for the new region.The tool can then be pivoted about the ccp inside the rolling ball sphere.This permits the ability to perform sharp turns,commonly found in zig-zag or spiral surface cutting routines,smoothly without any gouges.3.6.Convex curvature matchingA simple check for convex curvature matching had to be implemented to avoid interference of the protruding surface and the bottom of non-center cutting corner radius endmills (Fig.10).If the convex surface point has too small a radius then the rolling ball may interfere with the center of the cutting tool.To avoid this the minimum allowable convex radius is used as the rolling ball radius instead of the radius computed from the shadow area (Eq.(4)).This effectively lifts the backside of the tool and prevents the center area from interfering with the surface (Fig.11).Rball convex2 Min 2clearance 1R 2major 2r 2minor 2£ r minor 2Min clearance4P.Gray et al./Computer-Aided Design 35(2003)347±357352Fig.9.Schematic of the pseudo-curvature circle.Rball convex is the minimum rolling ball radius for a convex surface region,Min clearance the user-speci®ed minimum safety clearance between the tool center and the surface,R major the major radius of toroidal tool,r minor is the minor radius of toroidal tool.4.Testing the rolling ball methodTo test the RBM a simulator for tool paths was con-structed to check for gouging and an actual cutting test was performed for a sample surface.4.1.Tool path generation and RBM implementation The RBM algorithm described in the previous section was implemented with a few additional items that limit the selection of some of the free parameters.These para-meters and items are discussed in the following sections.4.1.1.Test surfaceThe test surface is a bi-cubic Bezier patch.It was chosen to be a functional patch (to have a linear mapping between the domain space and Euclidian space)to simplify the computation of the shadow area and boundary.The control points were chosen to create a complex and dif®cult surface to machine to demonstrate the robustness of RBM.The bi-cubic surface has convex,concave and saddle (mixed concave and convex)regions,which test all the positioningpossibilities.Fig.12shows the test surface and Table 1is a list of the control points.4.1.2.Tool path and adaptive forward step algorithmFor simplicity tool passes were chosen to follow adjacent isoparametric lines of the test surface.As the tool moves along the tool path its trajectory is a function of the milling machine's kinematics.To minimize deviation of the tool from the planned course,a tool path must contain numerous points.As the distance between points decreases,the controller processing time increases.This can lead to the machine starting and stopping for every move while waiting for the next command.To avoid this a simple adaptive forward step algorithm was used to maximize the distance between ccps.A linear transition is assumed over the rela-tively short distances between two consecutive ccps.The cutter will cut off peaks between ccps in convex sections and will leave material between ccps in concave sections.The adaptive forward step searches for the next ccp along the selected isoparametric line by incrementally marching forward and computing the pseudo-radius of curvature between the current ccp and the next potential ccp.The pseudo-radius of curvature is de®ned in Fig.9and Eq.(1).A straight line cuts the circle of radius equal to the radius of curvature between the current and next potential ccp.TheP.Gray et al./Computer-Aided Design 35(2003)347±357353Fig.11.Rolling ball radius substitution for tool centerclearance.Fig.12.Cutting testsurface.。

Available Face-Changing EffectXiaobai Ai【期刊名称】《现代物理（英文）》【年(卷),期】2018(9)12【摘要】Based on mathematical foresight and beyond the mainstream inertial thinking pattern, the author believes that if neutrinos were really tachyons, the mystery of neutrinos might be solved. Fortunately, the space-like theory of special relativity reveals that there would exist an observable effect i.e. a “face-changing effect”, not oscillation, which was just related to the superluminal motion. As long as the motion velocity of an electron anti-neutrino was greater than c2/v, where v was the instantaneous thermal motion velocity of its mother neutron at the time of β-decay, a corresponding electron neutrino formed from the face-changing would be observed on the journey. Therefore, a special and easy way to judge the physical nature of neutrinos may be suggested the reactor neutrino experimental groups all over the world, in addition to the current studies involving the disappearance mode of , to add a new experimental search after ve in the current, to see whether a few ve neutrinos would exceed the background counting. “Yes” result would reveal the neutrinos being tachyons, and “no” would be not.【总页数】13页(P2193-2205)【关键词】Special;Relativity;in;Space-Like;Region;Tachyon;Face-Changing;Effect;Mother;Neutron;in;β-Decay;Reactor;Neutrino;Experiments 【作者】Xiaobai Ai【作者单位】Shanghai Institute of Applied Physics, The Chinese Academy of Sciences, Shanghai, China【正文语种】中文【中图分类】R73因版权原因，仅展示原文概要，查看原文内容请购买。

Unit 1Science Fiction【话题词汇】1.innovation n. 创新；革新2.hardware n. 硬件；硬设备3.database n. 数据库4.e-book n. 电子书5.password n. 密码6.smartphone n. 智能手机7.automatic adj. 自动的8.install v t. 安装9.upload v t. 上传10.calculate v t. 计算11.browse v t. 浏览12.reliable adj. 可信赖的；可信任的13.portable adj. 可携带的14.digital adj. 数字的；数码的15.intelligent adj. 智能的；有才智的16.manual adj. 手工的；手动的【话题短语】1.log in/on 登陆2.log off/out 退出3.save the data 存储数据4.shut down/off 关闭；关机5.break down 系统瘫痪6.artificial intelligence(AI) 人工智能7.the information superhighway 信息高速公路8.make a breakthrough 突破rmation Technology 信息技术10.remote control 远程遥控【话题佳句】1.In 2015, Chinese scientist Tu Youyou won the Noble Prize in Physiology orMedicine due to the great breakthrough she had made in medicine。

由于在医学方面取得巨大突破，中国科学家屠呦呦2015年获得了诺贝尔生理学或医学奖。

2.Since the first spaceship was launched, some sci-fi novels and the space toys have been launched one after another.自从第一艘宇宙飞船发射以来，一些科幻小说和太空玩具也陆续上市。

小学上册英语第1单元期中试卷考试时间：100分钟（总分:100）B卷一、综合题(共计100题共100分)1. 选择题：What do we call the act of planting seeds?A. GerminationB. SowingC. CultivatingD. Harvesting答案: B2. 填空题：The invention of the airplane changed the way people _____.3. snorkeling) allows you to explore underwater ecosystems. 填空题：The ____4. 选择题：Which beverage is made from leaves?A. JuiceB. CoffeeC. TeaD. Soda答案:C5. 填空题：My cat likes to _______ (抓老鼠).6. 选择题：What is the name of the largest ocean?A. AtlanticB. IndianC. ArcticD. Pacific答案: DThis ball is my favorite ________.8. 听力题：A space station orbits the Earth to conduct ______.9. 填空题：I often create stories about friendships between my ________ (玩具名).10. 填空题：The __________ (古代文明) had their own writing systems.11. 选择题：What do we call the large body of saltwater that covers most of the Earth?A. RiverB. LakeC. OceanD. Pond答案: C. Ocean12. 填空题：The ________ (兔子) hops around quickly and loves to eat carrots.13. 听力题：Chemical bonds are formed when atoms ______ with each other.14. 听力题：The __________ can help scientists understand earth's history.15. 选择题：What is 30 15?A. 10B. 15C. 20D. 25答案:B16. 选择题：What do you call a person who studies the human mind?A. PsychologistB. SociologistC. TherapistD. Psychiatrist答案: AThe celestial sphere is useful for visualizing ______.18. 填空题：The ________ was a key event in the history of civil rights in America.19. 听力题：The atomic number of an element indicates the number of _______ it has.20. 选择题：What do we call the process of a seed developing into a plant?A. GerminationB. PhotosynthesisC. PollinationD. Fertilization答案:A21. collect _________. (坚果) 填空题：Squirrel22. 填空题：I like to eat ______.23. 听力题：A ____ can fly high in the sky and has a sharp beak.24. (66) is known for its ancient ruins. 填空题：The ____25. 填空题：The ________ was a significant cultural exchange in history.26. 选择题：What is 7 + 3?A. 9B. 10C. 11D. 1227. 选择题：What do you call a baby cat?A. PuppyB. KittenC. CubD. Calf答案: BThe sun is ___ in the sky. (high)29. 听力题：The density of an object determines whether it will ______ or sink in water.30. 填空题：My pet is very ______.31. 听力题：A suspension differs from a solution because the particles can ______.32. 填空题：A ____(map) shows different features of an area.33. 填空题：The _____ (大雁) flies south for the winter.34. 听力题：Mars has the largest volcano, called ______.35. 听力题：The ant climbs up the ____.36. 听力题：His favorite food is ________.37. 听力填空题：I think technology is amazing. It has changed the way we live and communicate. I’m particularly interested in __________ and how it can improve our lives.38. 填空题：In _____ (意大利), you can find delicious pizza.39. 听力题：Telescopes help us see objects that are very _______ away.40. 听力题：The chemical symbol for aluminum is ______.41. 听力题：The outer layer of the Earth is called the ______.42. 听力题：The cat is on the ___. (roof)Which animal is known as the ship of the desert?A. CamelB. HorseC. DonkeyD. Elephant答案:A. Camel44. 填空题：The __________ is a major city known for its cultural significance. (伦敦)45. 填空题：I enjoy hiking in the mountains to see stunning ______ (风景).46. 填空题：The __________ was a time of great exploration and discovery. (大航海时代)47. 选择题：What is the name of the first person to walk on the moon?A. Neil ArmstrongB. Buzz AldrinC. Michael CollinsD. Yuri Gagarin48. 听力题：We make _____ (饼干) together.49. 听力题：The chemical symbol for silicon is ______.50. 选择题：How many months are there in a year?A. 10B. 12C. 11D. 9答案:B51. 选择题：How many days are in March?A. 28B. 30C. 31D. 32The car is ___. (fast)53. 选择题：Which animal is known for its ability to fly?A. FishB. BirdC. CatD. Dog答案:B54. 填空题：A plant’s _____ (生命周期) includes germination, growth, and reproduction.55. 听力题：The _____ (clock) shows the time.56. 填空题：I want to _______ (成为) a scientist.57. 选择题：What is the capital of Paraguay?A. AsunciónB. Ciudad del EsteC. EncarnaciónD. Concepción答案: A58. 选择题：What is the name of the famous poet known for his sonnets?A. Robert FrostB. William WordsworthC. William ShakespeareD. John Keats答案: C59. 听力题：A sound's timbre is related to its ______.60. 听力题：I have a _____ (拼图) to solve.61. 听力题：The ____ has shiny scales and swims in rivers.The girl is wearing a ________ dress.63. 听力题：A chemical change alters the identities of the ______.64. 听力题：We have _____ (two/four) eyes.65. 听力题：The Sun rotates on its axis every ______ days.66. 填空题：The __________ is a large desert in the southwestern United States. (莫哈维沙漠)67. 选择题：What do you call a baby walrus?A. CalfB. PupC. KitD. Cub68. 听力题：The girl is very ________.69. 填空题：My sister's favorite activity is _______ (名词). 她总是 _______ (动词).70. 听力题：In a chemical equation, the reactants are shown on the _______.71. 听力题：The cake is ______ with frosting. (decorated)72. 选择题：What is the capital of Thailand?A. HanoiB. BangkokC. JakartaD. Manila73. 填空题：I planted a _____ in my backyard.74. 填空题：My brother is my best _______ because we share everything.Which holiday comes in December?a. Thanksgivingb. Halloweenc. Christmasd. New Year答案:c76. 填空题：We visit ______ (艺术馆) to see paintings.77. 选择题：What do we call the act of making a choice?A. Decision-makingB. PlanningC. OrganizingD. Arranging答案:A78. 选择题：What do bees produce?A. MilkB. HoneyC. EggsD. Butter答案:B79. 选择题：What is 3 + 4?A. 5B. 6C. 7D. 880. 填空题：The _____ (jasmine) flower smells sweet at night.81. 听力题：She is a friendly ________.82. 听力题：A butterfly starts as a ______.83. 听力题：The temperature at which a substance changes from solid to liquid is called ______.What do you call a young frog?A. TadpoleB. KitC. PupD. Calf85. 填空题：I enjoy playing ________ (运动) with my classmates.86. 填空题：The ______ (松鼠) gathers food for the winter.87. 听力题：A __________ is formed through the accumulation of organic materials.88. 填空题：The first successful face reconstruction was performed in ________.89. 选择题：What is the name of the famous battle fought in 1066?A. Battle of HastingsB. Battle of WaterlooC. Battle of GettysburgD. Battle of Agincourt答案: A90. 填空题：A rabbit's teeth never stop ______ (生长).91. 选择题：What is the capital of Sweden?A. StockholmB. GothenburgC. UppsalaD. Malmo92. 选择题：What is the color of grass?A. BlueB. GreenC. YellowD. Red答案:BWhat do we call the study of the universe?A. AstronomyB. AstrologyC. CosmologyD. Geology答案: A94. 选择题：What is the main reason we have seasons on Earth?A. Distance from the sunB. Tilt of the Earth's axisC. Cloud coverD. Ocean currents95. 听力题：A __________ is a mixture where the different parts are not visible.96. 选择题：What do we call the scientific study of life?A. BiologyB. ChemistryC. PhysicsD. Geology答案: A. Biology97. 填空题：A ladybug has a red ______ (外表) with black spots.98. 选择题：Which animal is known as the king of the jungle?A. ElephantB. LionC. TigerD. Bear答案:B99. 选择题：What is the opposite of hot?A. ColdB. WarmC. CoolD. Boiling答案:A100. 填空题：He is a pilot, ______ (他是一名飞行员), flying passengers safely.。

a r X i v :m a t h /0501095v 1 [m a t h .P R ] 6 J a n 2005The divergence of fluctuations for the shape on first passage percolation Yu Zhang ∗,Department of Mathematics,University of Colorado Abstract Consider the first passage percolation model on Z d for d ≥2.In this model we assign independently to each edge the value zero with probability p and the value one with probability 1−p .We denote by T (0,v )the passage time from the origin to v for v ∈R d and B (t )={v ∈R d :T (0,v )≤t }and G (t )={v ∈R d :ET (0,v )≤t }.It is well known that if p <p c ,there exists a compact shape B d ⊂R d such that for all ǫ>0tB d (1−ǫ)⊂B (t )⊂tB d (1+ǫ)and G (t )(1−ǫ)⊂B (t )⊂G (t )(1+ǫ)eventually w.p.1.We denote the fluctuations of B (t )from tB d and G (t )by F (B (t ),tB d )=inf l :tB d 1−l t F (B (t ),G (t ))=inf l :G (t ) 1−l t .The means of the fluctuations E [F (B (t ),tB d ]and E [F (B (t ),G (t ))]have been conjec-tured ranging from divergence to non-divergence for large d ≥2by physicists.In this paper,we show that for all d ≥2with a high probability,the fluctuations F (B (t ),G (t ))and F (B (t ),tB d )diverge with a rate of at least C log t for some constant C .The proof of this argument depends on the linearity between the number of pivotal edges of all minimizing paths and the paths themselves.This linearity is also indepen-dently interesting.Key words and phrases:first passage percolation,shape,fluctuations.AMS classification:60K 35.1Introduction of the model and results.We consider Z d,d≥2,as a graph with edges connecting each pair of vertices x=(x1,···,x d)and y=(y1,···,y d)with d(x,y)=1,where d(x,y)is the distance between x and y.For anytwo vertex sets A,B⊂Z d,the distance between A and B is also defined byd(A,B)=min{d(u,v):u∈A and v∈B}.(1.0) We assign independently to each edge the value zero with a probability p or one with prob-ability1−p.More formally,we consider the following probability space.As sample space,we takeΩ= e∈Z d{0,1},points of which are represented as configurations.Let P=P p be the corresponding product measure onΩ.The expectation with respect to P is denotedby E=E p.For any two vertices u and v,a pathγfrom u to v is an alternating sequence(v0,e1,v1,...,e n,v n)of vertices v i and edges e i in Z d with v0=u and v n=v.Given a pathγ,we define the passage time ofγasT(γ)=ni=1t(e i).(1.1)For any two sets A and B we define the passage time from A to B asT(A,B)=inf{T(γ):γis a path from some vertex of A to some vertex in B}, where the infimum takes over all possiblefinite paths.A pathγfrom A to B with t(γ)= T(A,B)is called the route of T(A,B).If we focus on a special configurationω,we may write T(A,B)(ω)instead of T(A,B).When A={u}and B={v}are single vertex sets, T(u,v)is the passage time from u to v.We may extend the passage times over R d.If x and y are in R d,we define T(x,y)=T(x′,y′),where x′(resp.,y′)is the nearest neighbor of x(resp.,y)in Z d.Possible indetermination can be dropped by choosing an order on the vertices of Z d and taking the smallest nearest neighbor for this order.In particular,the point-point passage time wasfirst introduced by Hammersley and Welsh (1965)as follows:a m,n=inf{t(γ):γis a path from(m,···,0)to(n,···0)}.By Kingman’s subadditive argument we havelimn→∞1where p c=p c(d)is the critical probability for Bernoulli(bond)percolation on Z d and the non-random constantµp is called the time constant.Given a unit vector x∈R d,by the same argument in(1.2)1limn→∞1+xtχ(2)t ⊂B(t)⊂tB21+xtχ(2)t ⊂B(t)⊂G(t)are close to one or zero for large x or for small x≥0.In words,thefluctuations of B(t) diverge with a rate t1/3.There have been varying in discussions about the nature of the fluctuations of B(t)for large d ranging from the possible independence ofχ(d)on d(Kar-dar and Zhang(1987))through the picture ofχ(d)decreasing with d but always remaining strictly positive(see Wolf and Kertesz(1987)and Kim and Kosterlitz(1989))to the possi-bility that for d above some d c,χ(d)=0and perhaps thefluctuations do not even diverge (see Natterman and Renz(1988),Haplin-Healy(1989)and Cook and Derrida(1990)).Mathematicians have also made significant efforts in this direction.Before we introduce mathematical estimates,let us give a precise definition of thefluctuation of B(t)from some set.For a connected setΓof R d containing the origin,letΓ+l={v∈R d:d(v,Γ)≤l}andΓ−l={v∈Γ:d(v,∂Γ)≥l}.Note thatΓ−l⊂ΓandΓ⊂Γ+l.Note also thatΓ−l might be empty even thoughΓis non-empty.We say B(t)has afluctuation fromΓifF(B(t),Γ)=F(B(t)(ω),Γ)=inf{l:Γ−l⊂B(t)(ω)⊂Γ+l}.(1.10) If we setΓ=tB d for d=2,the conjecture in(1.8)is equivalent to ask ifF(B(t),tB2)≈t1/3with a high probability.When p≥p c(d),B d is unbounded and so is B(t).Also,when p=0,there are no fluctuations so we require in this paper that0<P(t(e)=0)=p<p c(d).(1.11) The mathematical estimates for the upper bound of thefluctuation F(B(t),Γ),when Γ=tB d andΓ=G(t),are quite promising.Kesten(1993)and Alexander(1993,1996) showed that for p<p c(d)and all d≥2,there is a constant C such that√F(B(t),tB d)≤Ct)B d,where log denotes the natural logarithm.In this paper C and C i are always positive con-stants that may depend on p or d,but not t,and their values are not significant and change from appearance to appearance.On the other hand,the estimates for the lower bound of thefluctuations are quite un-satisfactory.Under(1.11)it seems that the only result for all d≥2(see Kesten(1993)) isF(B(t),tB d)≥a non-zero constant eventually w.p.1.(1.13)For d=2,Piza and Newman(1995)showed thatF(B(t),tB d)≥t1/8and F(B(t),G(t))≥t1/8eventually w.p.1.(1.14) Clearly,one of most intriguing questions in thisfield is to ask if thefluctuations of B(t) diverge as some statistical physicists believed to be true while others did not.In this paper we answer the conjecture affirmatively to show that thefluctuations of B(t)always diverge for all d≥2.We can even tell that the divergence rate for B(t)is at least C log t.Theorem1.If0<p<p c,for all d≥3,t>0and any deterministic setΓ,there exist positive constantsδ=δ(p,d)C1=C1(p,d)such thatP(F(B(t),Γ)≥δlog t)≥1−C1t−d+2−2δlog p.Remark 1.If we setΓ=tB d orΓ=G(t),together with(1.14),thefluctuations F(B(t),tB d)and F(B(t),G(t))are at leastδlog t with a large probability.Also,it follows from this probability estimate thatE(F(B(t),Γ))≥C log t(1.15) for a constant C=C(p,d)>0.Remark2.We are unable to estimate whether F(tB d,G(t))diverges even though we believe it does.The proof of Theorem1is constructive.In fact,if F(B(t),Γ)≤δlog t,then we can construct t d−1+2δlog p zero paths from∂B(t)toΓ+δlog t.For each such path,we can use the geometric property introduced in section2to show that the path contains a pivotal edge defined in section3.Therefore,we can construct about t d−1−2δlog t pivotal edges.However, in section3,we can also show the number of pivotal edges is of order t.Therefore,for a suitableδwe cannot have as many pivotal edges as we constructed.The contradiction tells us that F(B(t),Γ)≥δlog t.With these estimates for pivotal edges in section3,we can also estimate the number of the total vertices in all routes.This estimate is independently interesting.For a connected setΓ⊂R d withα1tB d⊂Γ⊂α2tB d for some constants0<α1<1<α2,letRΓ= γt,whereγt is a route for T(0,∂Γ).Theorem2.If0<p<p c(d),for all d≥2and t>0,there exists C(p,d,α1,α2)such thatE(|RΓ|)≤Ct,where|A|denotes the cardinality of A for some set A.Remark3.Kesten(1986)showed that there exists a route for T(0,Γ)with length Ct for some positive constant C.Theorem2gives a stronger result with the number of ver-tices in all routes for T(0,∂Γ)also in order t.For quite some time,the author believed that the routes of T(0,∂Γ)resembled a spiderweb centered at the origin so the number of the vertices in the routes should be of order t d.However,Theorem2negates this assumption.Remark4.Clearly,there might be many routes for the passage time T(0,∂Γ).As a consequence of Theorem2,each route contains at most Ct vertices.Specifically,let M x,t=sup{k:there exists a route of TΓ(0,∂Γ)containing k edges}.If0<p<p c,for all d≥2and t>0,there exists a positive constant C=C(p,d)such thatE(M x,t)≤Ct.(1.16) Remark5.We may also consider Theorem2for a point-point passage time.LetR x,t= γn,whereγn is a route for T(0,xt)for a unit vector x.We may use the same argument of Theorem2to show if0<p<p c(d), for all d≥2and t>0,there exists a positive constant C(p,d)such thatE(|R x,t|)≤Ct.(1.17) Remark6.The condition p>0in Theorem2is crucial.As p↓0,the constant C in Theorem2,(1.16)and(1.17)may go to infinity.When p=0,all edges have to take value one.If we takeΓas the diamond shape with a diagonal length2t both in vertical and horizontal directions,it is easy to say that all edges inside the diamond belong to RΓso|RΓ|=O(t d).(1.18) This tells us that Theorem2will not work when p=0.2Geometric properties of B(t)In this section we would like to introduce a few geometric properties for B(t).We let B′(t)be the largest vertex set in B(t)∩Z d and G′(t)be the largest vertex set in G(t)∩Z d.Similarly, given a setΓ⊂R d,we also letΓ′be the largest vertex set inΓ.It is easy to see thatΓ′⊂Γ⊂{v+[−1/2,1/2]d:v∈Γ′}.(2.0)As we mentioned in the last section,both B(t)and G(t)arefinite as well as B′(t)and G′(t). We now show that B′(t)and G′(t)are also connected.Here a set A is said to be connected in Z d if any two vertices of A are connected by a path in A.Proposition1.B′(t)and G′(t)are connected.Proof.Since T(0,0)=0≤t,0∈B′(t).We pick a vertex v∈B′(t),so T(0,v)≤t. This tells us that there exists a pathγsuch thatT(γ)≤t.Therefore,for any u∈γ,T(0,u)≤t so u∈B′(t).This implies thatγ⊂B′(t),so we know B′(t)is connected.The same argument shows that G′(t)is connected.2Given afinite setΓof Z d we may define its vertex boundary as follows.For each v∈Γ, v∈Γis said to be a boundary vertex ofΓif there exists u∈Γbut u is adjacent to v. We denote by∂Γall boundary vertices ofΓ.We also let∂oΓbe all vertices not inΓ,but adjacent to∂Γ.With these definitions,we have the following Proposition.Proposition2.For all v∈∂B′(t),T(0,v)=t and for all u∈∂o B′(t)T(0,u)=t+1.Proof.We pick v∈∂B′(t).By the definition of the boundary,v∈∂B′(t)so T(0,v)≤t. Now we show T(0,v)≥t for all v∈∂B′(t).If we suppose that T(0,v)<t for some v∈∂B′(t),then T(0,v)≤t−1,since T(0,v)is an integer.Note that t(e)only takes zero and one values so there exists u∈∂o B(t)and u is adjacent to v such that T(0,u)≤t.This tells us that u∈B′(t).But we know as we defined that∂o B′(t)∩B(t)=∅.This contradiction tells us that T(0,v)≥t for all v∈∂B′(t).Now we will prove the second part of Proposition2.We pick a vertex u∈∂o B′(t).Since u is adjacent to v∈B′(t),T(0,u)≤1+T(0,v)≤1+t.On the other hand,any path from0to u has to pass through a vertex of∂B′(t)before reaching∂o B′(t).We denote the vertex by v.As we proved,T(0,v)=t.The passage time of the rest of the path from v to u has to be greater or equal to one,otherwise,u∈B′(t). Therefore,any path from0to u has a passage time larger or equal to t+1,that isT(0,u)≥t+1.Therefore,T(0,∂o B′(t))=t+1.2Given afixed connected setκ=κt containing the origin,define the event{B′(t)=κ}={ω:B′(t)(ω)=κ}.Proposition3.The event that{B′(t)=κ}only depends on the zeros and ones of the edges ofκ∪∂oκ.Proposition2for d=2has been proven by Kesten and Zhang(1998).In fact,they gave a precise structure of B′(t).We may adapt their idea to prove Proposition2for d≥3by using the plaquette surface(see the definition in section12.4Grimmett(1999)).To avoid the complicated definition of the plaquette surface,we would rather give the following direct proof.Proof of Proposition3.LetκC denote the vertices of Z d\κand{ω(κ)}= edge inκ{0,1}and{ω(κC)}= edge inκC{0,1},where edges inκare the edges whose two vertices belong toκ∪∂oκand the edges inκC are the other edges.For eachω∈Ω,we may rewriteωasω=(ω(κ),ω(κC)).Suppose that Proposition3is not true,so the zeros and ones inω(κC)can affect the event {B′(t)=κ}.In other words,there exist two differentω1,ω2∈Ωwithω1=(ω(κ),ω1(κC))andω2=(ω(κ),ω2(κC))such thatB′(t)(ω1)=κbut B′(t)(ω2)=κ.(2.4) From(2.4)there are two cases:(a)there exists u such that u∈B′(t)(ω2),but u∈κ.(b)there exists u such that u∈κ,but u∈B′(t)(ω2).If(a)holds,T(0,u)(ω2)≤t.(2.5) There exists a pathγfrom0to u such thatT(γ)(ω2)≤t.(2.6)Since u∈κ,any path from0to u has to pass through∂oκ=∂o B′(t)(ω1).Letγ′be the subpath ofγfrom0to∂oκ=∂o B′(t)(ω1).Then by Proposition2,T(γ′)(ω1)≥t+1.Note that the zeros and ones in bothω2=(ω(κ),ω2(κC)andω1=(ω(κ),ω1(κC)are the same sot+1≤T(γ′)(ω1)=T(γ′)(ω2)≤T(γ)(ω2).(2.7) By(2.6)and(2.7)(a)cannot hold.Now we assume that(b)holds.Since any path from0to u has to pass through ∂o B′(t)(ω2),by Proposition2,T(0,u)(ω2)≥t+1.(2.8) But since u∈κand B′(t)(ω1)=κ,there exists a pathγinside B′(t)(ω1)from0to u such thatT(γ)(ω1)≤t.Therefore,T(0,u)(ω2)≤T(γ)(ω2)≤t,(2.9) sinceγ⊂κand the zeros and ones in bothω2=(ω(κ),ω2(κC)andω1=(ω(κ),ω1(κC)are the same.The contradiction of(2.8)and(2.9)tells us that(b)cannot hold.23The linearity of a number of pivotal edgesIn this section we will discuss afixed value0<p<p c and afixed interval open I p⊂(0,p c) centered at p.First we show that the length of a route from the origin to∂B′(t)is of order t.Lemma1.For a small interval I p⊂(0,p c)centered at p,there exist positive constants α=α(I p,d),C1=C1(I p,d)and C2=C2(I p,d)such that for all t and all p′∈I p, P(∃a routeγfrom the origin to∂B′(t)with|γ|≥αt)≤C1exp(−C2t). Proof.By Theorem5.2and5.8in Kesten(1986)for all p′∈I p and for all t there exist C3=C3(I p,d)and C4=C4(I p,d)such thatP(2t/3B d⊂B(t))≤C3exp(−C4t)and P(B(t)⊂3t/2B d)≤C3exp(−C4t).(3.0) If we put these two inequalities from(3.0)together,we have for all p′∈I p and all t, P(t/2B d)⊂B′(t)⊂2tB d)for all large t)≥1−C3exp(−C4t).(3.1)On the event of{a routeγfrom the origin to∂B(t)with|γ|≥αt}∩{t/2B d)⊂B(t)⊂2tB d}we assume that there exists a routeγfrom the origin to some vertex u∈∂B′(t)witht/2≤d(0,u)≤2t and|γ|≥αtsuch thatt=T(0,∂B′(t))=T(γ)=T(0,u).Therefore,by(3.1)P(∃a routeγfrom the origin to∂B′(t)with T(γ)≤t,|γ|≥αt)≤ t/2≤d(0,u)≤2t P(∃a routeγfrom the origin to u with T(γ)≤t,|γ|≥αt)+C3exp(−C4t).Proposition5.8in Kesten(1986)tells us that there exist positive constantsβ(I p,d),C5= C5(I p,d)and C6=C6(I p,d)such that for all p′∈I p and tP(∃a self avoiding pathγfrom(0,0)to y contains n edges,but with T(γ)≤βn)≤C5exp(−C6n),(3.2) where n is the largest integer less than t.Together with these two observations if we take a suitableα=α(I p,d),we have for all p′∈I pP(∃a routeγfrom the origin to∂B′(t)with|γ|≥αt)≤(2t)d C5exp(−C6t). Lemma1follows.2To show Theorems we may concentrate to a“regular”set satisfying(3.1).Here we give the following precise definition.Given a deterministic connectedfinite setΓ=Γt⊂R d,Γis said to be regular if there exists t such that1/2tB d⊂Γ⊂2tB d.(3.3) For a regular setΓwe denote byTΓ(0,∂Γ)=inf{T(γ):γ⊂Γ′is a path from the origin to some vertex of∂Γ′}.Now we try to compute the derivative of ETΓ(0,∂Γ)in p for a regular setΓ.As a result,we haveETΓ(0,∂Γ)= i≥1P(TΓ((0,0),∂Γ)≥i).Note that t(e)takes zero with probability p and one with probability1−p.But we know that the standard Bernoulli random variable takes zero with probability1−p and one with probability p.Hence we have to define an increasing event(see the definition in section2of Grimmett(1999))in reverse.An event A is said to be increasing if1−I A(ω)≤1−I A(ω′)wheneverω≤ω′,where I A is the indicator of A.Note thatΓis afinite set,so dETΓ(0,∂Γ)= i≥1dP(TΓ(0,∂Γ)≥i)dp=− i≥1 e∈ΓP({TΓ(0,∂Γ)≥i}(e)),(3.6)dpwhere{TΓ(0,∂Γ)≥i}(e)is the event that e is a pivotal for{TΓ(0,∂Γ)≥i}.In fact,given a configurationω,e is said to be a pivotal edge for{TΓ(0,∂Γ)(ω)≥i}if t(e)(ω)=1andTΓ(0,∂Γ)(ω′)=i−1where w′is the configuration that t(b)(ω)=t(b)(ω′)for all edges b∈Γexcept e and t(e)(ω′)=0.The event{TΓ(0,∂Γ)≥i}(e)is equivalent to the event that there exists a route of TΓ(0,∂Γ)with passage time i passing through e and t(e)=1.With this observation, dETΓ(0,∂Γ)=E(KΓ).(3.7)dpNow we give an upper bound for E(KΓ)by giving an upper bound for−dETΓ(0,∂Γ)We show that the size of NΓcannot be more than Ct for some constant C.Lemma2.For a regular setΓand the interval I p,there exist positive constants C i= C i(I p,d)(i=1,2,3)such that for all p′∈I p and tP(NΓ≥C1t)≤C2exp(−C3t).Proof.We follow the proof of Theorem8.2in Smythe and Wierman(1979).Letω+r denote the time state of the lattice obtained by adding the r to t(e)for each edge e.It follows from the definitions of the passage time and NΓTΓ(0,∂Γ)(ω+r)≤TΓ(0,∂Γ)(ω)+rNΓ(ω).(3.8) If we take a negative r in(3.8),we haveNΓ(ω)≤TΓ(0,∂Γ)(ω+r)−TΓ(0,∂Γ)r≤−T(L)µr.(3.10)If we can show that for some r<0,there exist constants C4=C4(I p,d)and C5=C5(I p,d) such that for all p′∈I p and all tP(TΓ(0,∂Γ)(ω+r)≤0)≤C4exp(−C5t),(3.11) then by(3.9)and(3.10),Lemma2holds.Therefore,to show Lemma2,it remains to show (3.11).Note thatΓis afinite connected set so for eachωthere exists x=x(ω)∈∂Γsuch thatTΓ(0,∂Γ)(ω+r)=TΓ(0,x)(ω+r)≥T(0,x)(ω+r).Since x∈∂ΓandΓis regular,thent/2≤d(0,x)≤2t.We haveP(TΓ(0,∂Γ)(ω+r)≤0)≤P(T(0,x(ω))(ω+r)≤0)≤ t/2≤d(0,y)≤2t P(T(0,y)(ω+r)≤0).(3.12)Therefore,by(3.2)and(3.12)we takeβand|r|small with r<0andβ>|r|>0to obtain for all p′∈I pt/2≤d(0,y)≤2t P(T(0,y)(ω+r)≤0)≤ t/2≤d(0,y)≤2t P(∃a self avoiding pathγfrom0to y contains n edges,but with T(γ)(ω)≤2βn)≤C6t d exp(−C7t),(3.13) where n is the largest integer less than t.Therefore,(3.11)follows from(3.13).2 With Lemma2we are ready to give an upper bound for−dETΓ(0,∂Γ)dp≤Ct.Proof.We assign s(e)≥t(e)either zero or one independently from edge to edge with probabilities p−h or1−(p−h)for a small number h>0,respectively.With this definition, P(s(e)=1,t(e)=0)=P(s(e)=1)−P(s(e)=1,t(e)=1)=P(s(e)=1)−P(t(e)=1)=1−(p−h)−(1−p)=h.(3.14) Letγt be a route for T tΓ(0,Γ)with time state t(e)and letγs be a route for T sΓ(0,Γ)with time state s(e).Here we pickγt such that|γt|=NΓ.For each edge e∈γt,if t(e)=1,then s(e)=1.If t(e)=0but s(e)=1,we just add one for this edge.Therefore,T sΓ(0,Γ)≤T(γt)+ e∈γt I(t(e)=0,s(e)=1).(3.15)Clearly,γt may not be unique,so we select a route from theseγt in a unique way.We still writeγt for the unique route without loss of generality.By(3.15)and this selectionET sΓ(0,Γ)≤ET(γt)+ β e∈βP(t(e)=0,s(e)=1,γt=β),(3.16)where thefirst sum in(3.16)takes over all possible pathsβfrom0to∂Γ′.Let us estimateβ e∈βP(t(e)=0,s(e)=1,γt=β).SinceΓis regular,the longest path from(0,0)to∂Γ′is less than(2t)d.By Lemma2,there exist C1=C1(I p,d),C2(I p,d)and C3(I p,d)such thatβ e∈βP(t(e)=0,s(e)=1,γt=β)≤ |β|≤C1t e∈βP(t(e)=0,s(e)=1,γt=β)+C2t d exp(−C3t).(3.17)Note that the value of s(e)may depend on the value of t(e),but not the other values of t(b) for b=e,so by(3.14)P(t(e)=0,s(e)=1,γt=β)=P(s(e)=1|t(e)=0,γt=β)P(t(e)=0,γt=β)≤P(s(e)=1|t(e)=0)P(γt=β)=hp−1P(γt=β).(3.18) By(3.18)we have|β|≤C1 e∈βP(t(e)=0,s(e)=1,γt=β)≤ |β|≤C1t e∈βhp−1P(γt=β)≤C1htp−1.(3.19) By(3.17)and(3.19)there exists C4=C4(I p,d)such thatE(T sΓ(0,∂Γ))≤E(T tΓ(0,∂Γ)+C4th.(3.20) If we setf(p)=E(TΓ(0,∂Γ))for time state t(e)with P(t(e)=0)=p,then by(3.20)−d f(p)−h≤C4t.(3.21) Therefore,we have−dETΓ(0,∂Γ)dp≤C4t.(3.22). Therefore,Lemma3follows from(3.22).2Together with(3.7)and Lemmas3,we have the following proposition.Proposition4.If0<p<p c,then for a regular setΓthere exists a constant C=C(p) such thatEKΓ≤Ct.4Proof of Theorem1In this section,we only show Theorem1for d=3.The same proof for d>3can be adapted directly.Given afixed setΓ⊂R3defined in section2,Γ′⊂Z3is the largest vertex set in Γ,whereΓ′⊂Γ⊂{v+[1/2,1/2]3:v∈Γ′}.(4.0) Suppose that there exists a deterministic setΓsuch thatF(B(t),Γ)≤δlog t.(4.1) (4.1)means thatΓ−δlog t⊂B(t)⊂Γ+δlog t,whereΓ+l={v∈R3:d(v,Γ)≤l}andΓ−l={v∈Γ:d(v,∂Γ)≥l}.Wefirst show that ifΓ+δlog t does not satisfy the regularity condition in(3.3),then the probability of the event in(4.1)is exponentially small.We assume thatΓ+δlog t⊂2tB d.(4.2)IftF(B(t),Γ)≤δlog t withδlog t<B d.(4.4)2To see(4.4),note that3tB(t)⊂'%&%'$S mt ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡∂(Γ+δlog t )′∂B ′(t )Figure 1:The graphs:S mt ,∂B ′(t ),Γ+δlog t ,the cylinder T s i with the center at L s i ,pivotal edge e v i ,zigzag path γv i from v i to ∂(Γ+δlog t )′.Now we focus on Γ+δlog t satisfying (3.3).We need to show that under (4.1)there are of order t 2disjoint zero paths from ∂B (t )to Γ+δlog t .To accomplish this,let S mt denote a sphere with the center at the origin and a radius tm for small but positive number m .Then by (3.1)for a suitable m >0P (S mt ⊂B (t )⊂2tB d )≥1−C 1exp(−C 2t ).(4.9)Here we select the sphere S mt without a special purpose since the sphere is easy to describe.For each s ∈∂S mt ,let L s be the normal line passing though s ,that is the line orthogonal to the tangent plane of S mt at s .We denote the cylinder with center at L s by (see Fig.1)T s (M )={(x,y,z )∈R 3:d ((x,y,z ),L s )≤M }for some constant M >0.Now we work on the regular polyhedron with ct 2faces embedded on S mt ,where ct 2is an integer and c =c (m,M )is a small number such that the radius of each face of the regular polyhedron is larger than M .We denote the center of each face in the regular polyhedron by {s i }ct 2i =1.By this construction,we have at least ct 2disjoint cylinders {T s (M )}.We denote them by {T s i (M )}ct 2i =1.For each s i ,we may take M large such that there exists a path γs i ⊂Z d ∩T s i (M )from some vertex of Z 3in S mt to ∞.To see the existence of such a path if L s i is the ray going along the coordinate axis,we just use L s i as the path.If it is not,we can construct a zigzag path in Z d next to L s i from s i to ∞(see Figure 1).In fact,we may take M =2to keepour zigzag path inside T s i (M ).For simplicity,we use T s i to denote T s i (2).There might be many such zigzag paths,so we just select one in a unique manner.We denote by u i ∈Z d with d (u i ,s i )≤2the initial vertex in γs i .Since γs i is next to L s i ,for any point x on the ray L s i ,there is v in γs i with d (v,y )≤2.Furthermore,by a simple induction we conclude thatthe number of vertices from u i to v along γs i is less than 2d (s i ,x ).(4.10)(4.10)tells us that the length of γs i is linear to the length of L s i .Since γs i is from S mt to∂(Γ+δlog t )′,it has to come outside of B ′(t )from its inside.Let v i be the vertex in ∂o B ′(t )andγv i be the piece of γs i outside B ′(t )from v i to ∂(Γ+δlog t )′(see Figure 1).On F (B (t ),Γ)≤δlog tfor a regular Γ,we know B ′(t )⊂(Γ+δlog t )′.Therefore,by our construction (see Figure 1)γv i ⊂(Γ+δlog t )′.(4.11)Also,by our special construction in (4.10),we have|γv i |≤2δlog t.(4.12)When B ′(t )=κfor a fixed vertex set κ,then γv i is a fixed path from ∂o κto ∂(Γ+δlog t )′with a length less than 2δlog t .Therefore,on B ′(t )=κP (γv i is a zero path)≥p 2δlog t .(4.13)We say T s i is good if there exists such a zero path γv i .On B ′(t )=κ,let M (Γ,κ)be the number of such good cylinders T s i .By (4.13),we haveEM (Γ,κ)≥(ct 2)p 2δlog t =ct 2−2δlog p .(4.14)On B ′(t )=κ,note that the event that T s i is good depends on zeros and ones of the edges inside T s i ,but outside of ∂o κ.Note also that T s i and T s j are disjoint for i =j ,so by a standard Hoeffding inequality there exist C 1=C 1(p,d )and C 2=C 2(p,d )such thatP (M (Γ,κ)≤ct (2−2δlog p )/2)≤C 1exp(−C 2ct −2+2δlog p ).(4.15)We denote byD (Γ,κ)={M (Γ,κ)≥ct (2+2δlog p )/2}.Note that D (Γ,κ)only depends zeros and ones outside ∂o κ.(4.16)By Proposition 3and (4.16),{B ′(t )=κ}and D (Γ,κ)are independent.(4.17)By Proposition 2,any route from (0,0,0)to v i in B ′(t )∪∂o B ′(t )has a passage time t +1.We just pick one from these routes and denote it by γ(0,v i ).On F (B (t ),Γ)≤δlog t if T s i is good,there exists a zero path γv i from v i to ∂(Γ+δlog t )′.This implies that there exists a pathγ(0,∂Γ+δlog t )=γ(0,v i )∪γv ifrom (0,0,0)to ∂(Γ+δlog t )′with a passage time t +1and the path passes through the edgeadjacent v i between ∂B (t )and ∂o B (t ).On the other hand,note that any path from theorigin to ∂(Γ+δlog t )′has to pass through ∂o B ′(t )first,so by Proposition 2it has to spend atleast passage time t +1.Therefore,if we denote by e v i the edge adjacent v i from ∂B (t )to ∂o B (t ),then the path γ(0,∂Γ+δlog t )with passage time T ((0,0,0),∂Γ+δlog t )passes through e v i and t (e v i )=1.By (4.11)and B ′(t )⊂(Γ+δlog t )′,the path γ(0,∂Γ+δlog t )has to stay inside (Γ+δlog t )′.These observations tell us that e v i is a pivotal edge for T Γ+δlog t((0,0),Γ+δlog t ).Therefore,on F (B (t ),Γ)≤δlog t if T s i is good,T s i contains at least one pivotal edge for T Γ+δlog t ((0,0),Γ+δlog t ).(4.18)With these preparations we now show Theorem 1.Proof of Theorem 1.If Γ+δlog t is not regular,1/2tB d ⊂Γ+δlog t or Γ+δlog t ⊂2tB d ,by (4.8)there are C 1=C (p,d )and C 2(p,d )such thatP p (F (B (t ),Γ)≤δlog t )≤C 1exp(−C 2t ).(4.19)Now we only need to focus on a regular Γ+δlog t .P p (F (B (t ),Γ)≤δlog t )= ΓP (F (B (t ),Γ)≤δlog t,B ′(t )=κ),(4.20)where the sum takes over all possible sets κ.For each fixed κ,by (4.17)and (4.15)there exist C 3=C 3(p,d )and C 4=C 4(p,d )such thatκP (F (B (t ),Γ)≤δlog t,B ′(t )=κ)≤κP (F (B (t ),Γ)≤δlog t,B ′(t )=κ,D (Γ,κ))+C 3exp(−C 4t 2+2δlog p ).By (4.18)κP (F (B (t ),Γ)≤δlog t,B ′(t )=κ,D (Γ,κ))≤ κPK Γ+δlog t ≥ct 2+2δlog pWe combine(4.20)-(4.21)together to haveP(F(B(t),Γ)≤δlog t)≤P KΓ+δlog t≥ct2+2δlog p.(5.3)k dFor each cube B k(u),it has at most4d neighbor cubes,where we count its diagonal neigh-bor cubes.We say these neighbors are connected to B k(u)and denote by¯B k(u)the vertex set of B k(u)and all of its neighbor cubes.Note that,by the definition,RΓis a connected set that contains the origin,soˆRΓ(k)is also connected in the sense of the connection of two of its diagonal vertices.IfΓis regular,then|RΓ|≥t/2.By(5.3),we haveP(|RΓ|≥Mt)= m≥Mt/(2k d)P(|RΓ|≥MKΓ,|ˆRΓ(k)|=m).(5.4)We say a cube B k(u)for u∈ˆRΓ(k)is bad,if there does not exist an edge e∈¯B k(u)∩RΓsuch that t(e)=1.Otherwise,we say the cube is good.Let B k(u)be the event that B k(u) is bad and let DΓbe the number of bad cubes B k(u)for u∈ˆRΓ.If B k(u)occurs,there is a zero pathγk⊂RΓfrom∂B k(u)to∂¯B k(u).By Theorem5.4 in Grimmett(1999),there exist C1=C1(p,d)and C2=C2(p,d)such that forfixed B k(u)P(B k(u))≤C1exp(−C2k).(5.5) On{|ˆRΓ(k)|=m,|RΓ|≥MKΓ},if2(4k)d<M,we claimDΓ≥mpivotal edges.Therefore,4d2KΓ>mM|ˆRΓ(k)|4d2≥As we mentioned,ˆRΓis connected,so there are at most(4)dm choices for all possibleκm. With this observation and(5.11)we haveP(|RΓ|≥MKΓ)= m≥Mt/(2k d)(4)dm m m m/2 exp(−C2km/2)≤C1 m≥Mt/k d m[4d2exp(−C2k/2)]m.(5.12) We choose k large to make4d2exp(−C2k/2)<1/2.By(5.12),there are C3=C3(p,d)and C4=C4(p,d)such thatP(|RΓ|≥MKΓ)≤C3exp(−C4t).(5.13) Therefore,by(5.2)note that there are at most t2d vertices onΓ,so there exists C5=C5(p,d) such thatE|RΓ|=C3t2d exp(−C4t)+MCt≤C5t.(5.14) Theorem2follows from(5.14).。

练习1I、在下列每个句子的空白处填上适当的冠词（如果必要的话），然后将句子译成汉语：1. There has been _____ ever greater interest in this subject.2. The power rating is the maximum power the resistor can safely dissipate without too great _____ rise in temperature.3. Its primary disadvantage is _____ increase in noise.4. _____ successful design of the equipment requires _____ detailed knowledge of the performance specifications.5. In _____ Bohr model of the hydrogen atom, _____ single electron revolves around _____ single proton in a circle of radius R.6. The unit of frequency is _____ hertz.【7. If _____ voltage is applied across _____ circuit, _____ electric current will flow in _____ circuit.8. _____ Fig. 5-1 shows _____ Oersted’s experiment.9. We should use _____ 18-volt battery here.10. _____ machine is _____ device for transmitting force to accomplish _____ definite purpose.11. _____ hydraulic press will be considered in _____ Chapter 14.12. _____ study of fluids in motion is one of _____ more difficult branches of mechanics because of _____ diversity of phenomena that mayoccur.13. It is easy to determine _____ value of _____ parameter μ.14. By _____ Eq. (2-1) we have _____ following relation.\15. It is necessay to use _____ S-shaped tube here.16. The authors work at _____ University of Texas at _____ Arlinton.17. This is _____ R-bit transformer.18. _____ XOR gate must be used here.II、将下列句子译成英语，注意正确地使用冠词：1、这是一个h参数（parameter）。

关于Gauss—Kronecker曲率为零的极小超曲面的注记作者：马红娟赵秀兰郑喜英来源：《湖南师范大学学报·自然科学版》2013年第02期摘要研究了R4中满足Gauss-Kronecker曲率为零的极小超曲面.Hasanis猜想：R4中Gauss-Kronecker曲率恒为零的极小超曲面是R3中极小曲面与实数直线的黎曼乘积.对于上述猜想，Hasanis等人给出了部分证明，得到了一个定理，本文利用具体例子说明该定理中的部分条件是不必要的，并得到分类定理.关键词Gauss-kronecker曲率；主曲率；极小超曲面中图分类号O186文献标识码A文章编号1000-2537（2013）02-0013-03超曲面的Gauss-Kronecker曲率是一个重要的几何不变量.因此研究具有常Gauss-Kronecker 曲率的黎曼流形，尤其是空间形式中具有常Gauss-Kronecker曲率的超曲面是有意义的.关于Qn+1中极小超曲面Mn，Dajczer和Gromoll[1]利用S4中一类极小曲面构造S4中具有零Gauss-Kronecker曲率的极大超曲面M3，de Almeida和Brito[2-3]对S4中具有零Gauss-Kronecker曲率的紧致极小超曲面进行分类，成庆明[4]研究S4[5]和H41（-1）[6]中具有零Gauss-Kronecker曲率的极大类空间超曲面，Hasanis等人把结果推广到R4[7]，H4[8]和S4[9]中的完备极小超曲面.对于猜想[7]：R4中Gauss-Kronecker曲率恒为零的极小超曲面是R3中极小曲面与实数直线的黎曼直积，给出了如下定理.定理1[7]设M3是数量曲率有下界的完备可定向三维黎曼流形，f：M3→R4是满足Gauss-Kronecker曲率恒为零，第二基本形式恒不为零的等距极小浸入，则f（M3）可分解为黎曼直积L2×R，L2是R3中高斯曲率有下界的完备极小超曲面.上述分类定理对数量曲率和第二基本形式加了很强的条件.如果f：M3→R4是完备极小超曲面，f（M3）可分解为L2×R，其中L2是R3中的完备极小曲面，那么M3的数量曲率一定有下界吗？本文通过具体例子说明上述定理中的部分条件是不必要的，并给出分类定理.参考文献：[1]DAJCZER M， GROMOLL D. Gauss parametrizations and rigidity aspects of submanifolds[J]. J Differential Geom， 1985，22（1）：1-12.[2]ALMEIDA S， BRITO F. Minimal hypersurfaces with constant Gauss-Kronecker curvature[J]. Math Z， 1987，195（8）：99-107.[3]ALMEIDA S， BRITO F. Closed hypersurfaces of S4（1） with two constant symmetric curvature[J]. Ann Fac Sci Toulouse Math， 1997，6（6）：187-202.[4]CHENG Q M， SUH Y J. Maximal space-like hypersurfaces in H41（-1） with zero Gauss-Kronecker curvature[J]. Korean Math Soc， 2006，43（1）：147-157.[5]CHENG Q M. Hypersurfaces with constant quasi-Gauss-Kronecker curvature in S4（1）[J]. Adv in Math， 1993，22（2）：125-132.[6]CHENG Q M. Complete space-like submanifolds in a de Sitter space with parallel mean curvature vector[J]. Math Z， 1991，206（3）：333-339.[7]HASANIS T， SAVAS-HALILAJ A， VLACHOS T. Minimal hypersurfaces with zero Gauss-Kronecker curvature[J]. Ill J Math， 2005，49（2）：523-529.[8]HASANIS T， SAVAS-HALILAJ A， VLACHOS T. Complete minimal hypersurfaces in the hyperbolic space H4 with vanishing Gauss-Kronecker curvature[J]. Trans Am Math Soc， 2007，359（6）：2799-2818.[9]HASANIS T， SAVAS-HALILAJ A， VLACHOS T. Complete minimal hypersurfaces in S4 with zero Gauss-Kronecker curvature[J]. Math Proc Camb Philos Soc， 2007，142（1）：125-132.[10]彭家贵，陈卿. 微分几何[M]. 北京：高等教育出版社， 2002.[11]忻元龙. 极小曲面的Bernstein型定理与Gauss映照的值分布[J]. 数学进展， 1989，18（4）：402-411.[12]辛萍芳. 极小曲面的构造[J]. 湖北师范学院自然科学学报， 2002，22（4）：31-34.[13]吴金文. 数量曲率刻划的球面中紧致极小子流形[J]. 湖南师范大学自然科学学报，2001，24（2）：12-14.（编辑沈小玲）。

信阳高中北湖校区2023-2024学年高一下期05月测试（二）英语试题分值：150分时间：120分钟注意事项：1.答题前请填好自己的姓名，班级，考号等信息2.请将答案正确填写在答题卡上一、单词汉译英、英译汉(共20小题；每小题1.5分，共30分)1.g adj.全球的；全世界的2.a adv.每年；一年一度地3.m n.大多数4.d v.消失，不见5.o adj.普通的；平常的6.c adj.当前的，现在的7.f adj.以前的，从前的8.o vt.克服(困难);控制情感9.s n.继续生存，幸存10.p n.建议，提议11.solution n.12.bother vi&vt.13.firm adj.14.involve vt.15.profit n.16.contented adj.17.determination n.18.slim adj.19.destination n.20.indicate vt.二、阅读理解(共两节，共20小题，每小题2.5分，满分50分)第一节(共15小题；每小题2.5分，共37.5分)AGreen Line Performing Arts CenterUpcoming Programs:First Monday Jazz Series:CrosswindMonday Feb 4|7-9 pmFirst Monday Jazz is a free monthly event showcasing local Chicago jazz artists.Green Light Series:South Side Story TimeSunday Feb 17|10-11 amBring your kids to listen,learn,sing,dance,and interact!South Side Story Time is a small gathering that focuses on readings for its young attendees along with the chance for their parents to socialize.Open to families with children of all ages.Sistergirls and Freedom Fighters:Stories in Celebration of Women's Power and Grace Wednesday Feb 20|7-8 pmJoin us for an evening of storytelling featuring the dynamic singing,in the Spirit by Emily Hooper Lansana.This performance will highlight a range of stories that demonstrate women's creative and political genius from folk heroines to Nobel Peace Prize winners.Family Saturdays:Art togetherEvery 1st Saturday of the Month|3-5 pmExplore your child's artistic curiosity with hands-on art workshops designed to stimulate creativity and play.These interdisciplinary workshops are exciting for the entire family,offering activities from music to arts and e to learn something new!Appropriate for families with children.Registration(注册)is encouraged.Follow Arts +Public Life on Facebook for more event details and a full list of all upcomingperformances.21.When can people enjoy jazz in February?A.At 8 pm every first Monday. C.At 7 pm every first Wednesday.B.At 4 pm every other Saturday.D.At 9 pm every other Monday.22.What do we know about Sistergirls and Freedom Fighters?A.It lasts two hours.C.It is a prize-winning performance. B.It is organized by Emily.D.It celebrates women's achievements.23.What do Green Light Series and Family Saturdays have in common?A.Activity typesC.Target participants.B.Event frequencyD.Registration requirements. BI used to believe that only words could catch the essence of the human soul.The literary(文学的)works contained such distinct stories that they shaped the way we saw the world.Words were what composed the questions we sought to uncover and the answers to those questions themselves.Words were everything.That belief changed.In an ordinary math class,my teacher posed a simple question:What's 0.99 rounded to the nearest whole number?Easy.When rounded to the nearest whole number,0.99 =1.Somehow,I thought even though 0.99 is only 0.01 away from 1,there's still a 0.01 difference.That means even if two things are only a little different,they are still different,so doesn't that make them completely different?My teacher answered my question by presenting another equation(等式):1=0.9,which could also be expressed as 1=0.99999...repeating itself without ever ending.There was something mysterious but fascinating about the equation.The left side was unchangeable,objective:it contained a number that ended.On the right was something endless,number repeating itself limitless times.Yet,somehow,these two opposed things were connected by an equal sign.Lying in bed,I thought about how much the equation paralleled our existence.The left side of the equation represents that sometimes life itself is so unchangeable and so clear.The concrete,whole number of the day when you were born and the day when you would die.But then there is that gap in between life and death.The right side means a time and space full of limitless possibilities,and endless opportunities into the open future.So that's what life is.Objective but imaginative.Unchangeable but limitless.Life is an equation with two sides that balances itself out.Still,we can't ever truly seem to put the perfect words to it.So possibly numbers can express ideas as equally well as words can.For now,let's leave it at that:1=0.99999...and live a life like it.What does the author emphasize about words in paragraph 1?A.Their wide variety. C.Their expressive power.B.Their literary origins D.Their distinct sounds.25.What made the author find the equation fascinating?A.The repetition of a number.B.The difference between the two numbers.D.The way two different numbers are equal.C.The question the teacher raised.26.Which of the following can replace the underlined word paralleled"in paragraph 6?A.Measured.posed.C.Mirrored.D.Influenced.27.What does the author think of“1=0.9”?A.They uncover the meaning of life.C.They are useful in learning maths.B.They are reliable in expressing ideas.D.They give limited possibilities.CDuring space missions,astronauts can experience a loss of some of the inner structural support in their bones.For trips in space lasting at least six months,that loss is equal to about 20 years of aging.Luckily,a new study finds a year back on Earth rebuilds half of the strength lost in the affected bone.Exercise scientist Leigh Gabel was part of a team that tracked 17 astronauts,each of whom had spent four to seven months in space.The team measured the 3-D structure of bone.They focused on the structure of the tibia(胫骨)and the lower-arm bone.The researchers took images of the bones3times—before spaceflight and again six months and one year later when the astronauts returned home from space.From these pictures,Gabel's team calculated an astronaut's bone strength and density(密度)at each of those times.What did hey find through comparison of the pictures?Astronauts in space for less than six months regained their preflight(起飞前的)bone strength after a year back on Earth.But those who stayed in space longer suffered permanent bone loss in their tibias.That loss was equal to a decade of aging.The lower arm bones showed almost no loss.Tat was likely because these aren’t weigh-bearing bones on Earth,Gabel explains.In fact,those arms can get bigger workout in space than on Earth as astronauts use them to move around their craft by pushing off handles and doors.“Increasing weight-lifing exercises in space could help alleviate (解)bone loss in the legs,”says Steven Boyd,also in exercise scientist.“With longer spaceflight,we can expect bigger bone loss-and probably a bigger problem with recovery,”says physiologist Laurence Vico.“Space agencies should also consider other bone health measures,such as nutrition,to reduce bone loss and increase bone formation.”28.Why did Gabel's team take pictures at different times?A.To offer evidence for their predictions.B.o show their respect for the astronauts.C.To compare the changes in the astronauts' bones.D.To find out the proper length of staying in space for astronauts.29.Which of the following may Gabel agree with?A.A spaceflight in less than half a year does no harm to astronauts.B.Astronauts coming back from a spaceflight look much older.C.Astronauts can aybid bone loss through doing enough exercise.D.There is almost no difference in the astronauts' lower-arm bones.30.What can we conclude from this text?A.People on Earth never suffer bone loss.B.The finding of the new study is good news for astronauts.C.The 17 astronauts knew the bone loss before their spaceflights.D.Astronauts can regain their bone strength as soon as they return to Earth.31.What does the text mainly talk about?A.ow to prevent bone loss in space.B.Problems faced by astronauts in space.C.Astronauts' suffering from bone loss in space.D.Astronauts' contributions to the development of science.DThe innovation is part of research into brain-computer interfaces(接口)to help improve the lives of people with disabilities.The researchers included machine learning capabilities with their brain-computer interface,making it a one-size-fits-all solutionTypically,these devices require extensive adjustment for each user-every brain is different,both for healthy and disabled users and that has been a major obstacle to mainstream adoption.This new solution can quickly understand the needs of an individual subject and self-adjust through repetition.That means multiple patients could use the device without needing to adjust it for better functioning.In a clinical setting,the subjects wear a cap packed with electrodes(电极)that is connected to a computer.The electrodes gather data by measuring electrical signals from the brain,and the decoder translates the brain signals into commands,which are used to control devices or applications.In this case,the decoded signals are translated into actions in a car racing game and a simpler task involving balancing a digital bar.Subjects were trained simultaneously for both the simpler bar game and the more complex car racing game,thus improving their brain function.This project used 18 subjects with no motor disabilities.As they continue down this road,they will test this on people with motor disabilities to apply it to larger groups in clinical settings.“To achieve this,we need to improve our technology to make it easier to use,'Millan.a professor in the University of Texas says.On the side of translating the research,Millan and his team continue to work on a wheel chair that users can drive with the brain-computer interface.At the South by Southwest Conference and Festivals this month,the researchers showed off another potential use of the technology:controlling two rehabilitation(康复)robots for the hand and arm.This was not part of the new paper but a sign of where this technology could go in the future.“We'll continue down this path wherever it takes us in the pursuit of helping people.”Millan says.32.Who will benefit the most from the Brain-Computer Interface innovation?A.People with physical disabilitiesB.Businessmen selling wheelchairsC.Students seeking academic improvement.D.Athletes seeking performance enhancement.33.What is the main advantage of the new solution?A.Faster adjustment process C.I proved mainstream adoption.B.Decreased need for adjustingD.Quick understanding of commands34.What is the function of electrodes?A.Translating signals C.Controlling devicesB.Measuring brain signals D.Enhancing brain function.35-What is the probable focus of Millan and his team's future work?A.Exploring hand and arm rehabilitation.B.Showcasing technology at conferences.C.Translating and publishing research papers.D.Enhancing brain-computer interface technology.第二节七选五(共5小题；每小题2.5分，共12.5分)When I graduated from high school,I wanted to major in comparative literature.But,once I found out my fiends were going into“real”majors,like marketing,nursing,and engineering,II forced myself to believe that figured I needed to do the same to ensure a good career. 36 I would enjoy(it and succeed in the future,but eventually I exhausted myself understanding theeconomy and trying new marketing ing out of this experience,I realize it is OK tobe different from others and to study things like classics,art history and other majors offered in the College of Humanities(人文学科)!The worries most people have about a Humanities degree and finding a career afterwards are Actually ideas discussed in Humanities classes,which that the majors are too abstract. 37 are occasionally different from what people commonly believe,offer a broad perspective(视角).How could one effectively be an unbiased(无偏见的)writer with only a knowledge of the popular opinion of society?How could one speak persuasively with closed minds?Only seeing the world through a single perspective leads to missed learning,missed friendships,and missed growth! 38STEM(science,technology,engineering,and mathematics)graduates learn actual skills in their studies,while humanities majors learn "soft skills"like communicating effectively through writing critically and speaking persuasively,synthesizing(综合)ideas through gathering and interpreting information,and developing cultural awareness.Do those soft skills sound useless and inapplicable Produce ideas?Encounter people from other cultures?Every day.to you?Think of it.39 Every SECOND of every day.So why not master these skills?40 The job market is quietly creating thousands of openings a week for people who can bring a humanist's grace to our rapidly evolving high-tech future.Your skills will be valuable to any workplace you hope to be in.Stick to your dream major with all your energy,no matter what other people think.A.So one will not obtain any practical skills.B.How often do you communicate with others?C.How much progress can you make in a certain time?D.So I changed my mind and chose Business Management as my major.E.If you choose a major in the College of Humanities,you will be needed.F.The number of students majoring in the humanities has declined by about half.G.With a broader perspective,we will be more open-minded and less limited in what we can become!第三节完形填空(共15小题，每小题1分，满分15分)In 1980,Foley was a new nurse,working the night shift in a maternity ward(产房).One night,a baby girl was born with a 41 congenital(天生的)disorder that affected her brain.She was 42 to die soon.As Foley had seen in similar cases,babies in this 43 were often placed in a bassinet(摇篮)and received little 44 until they died.But Foley said Nancy Allspach,a colleague of hers,had a 45 approach.“She went into the nursery multiple times through the shift and 46 that baby,”Foley recalled.“She put her 47 right down next to the baby,and she talked to the baby.And she even 48 the baby a bottle and rocked her in the big rocking chair.And she 49 the baby as thoughshe were her own.”Foley 50 remembers how the baby was treated,and how the mother was treated,too.Over her nearly 50-year 51 as a registered nurse,Foley has 52 many babies and children who died later.And looking back,she said she always tied to treat both the children and their parents with compassion(同情)—the way she 53 from Nancy Allspach."I never forgot the importance of touching and being 54 to the baby or the child and also the parents,”she said.“Because,at that moment,what they need is 55 .And Nancy taught me that.'41. A. fancy42. A. encouraged43. A. atmosphere44. A. attention45. A. passive46. A. spoiled47. A. face48. A. showed49. A. noticed50. A. clearly51. A. request52. A. met with53. A. heard54. A. polite55. A. compassionB. severeB. forcedB. pointB. commitment B. differentB. rankedB. footB. boughtB. treatedB. hardlyB. ambitionB. looked atB. learnedB. familiarB. destinationC. mildC. considered C. conceptC. contribution C. fatalC. held C. earC. drankC. defeated C. normally C. planC. cared for C. received C. similarC. exampleD. joyfulD. managed D. condition D. improvement D. broadD. setD. backD. fedD. keptD. carefully D. careerD. picked up D. borrowed D. closeD. competition第Ⅱ卷(非选择题共55分)第一部分语法填空(共10小题,每小题1.5分;共15分)阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式(注意:答案请写在答题卡上,写在本卷无效)。