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Learning Support - Math Classes
Grade 9-12 students who have successfully finished Saxon Algebra II with A or B grade.
Materials Used:
Saxon Advanced Math, 2nd edition and home school packet.
Description:
This challenging course completes the entire text in one year. This pre-calculus course covers topics in advanced algebra, analytic geometry, trigonometry, and calculus. Course focuses on key areas. Most tests online but a few need to be mailed or scanned. Class meets twice weekly.
2 HS credit
Cost: $482.00
Adv Math (Saxon - 2nd Edition) - Year 1
Open To:
Grade 9-12 students who have completed the majority of Saxon Algebra 2.
1 HS credit
Cost: $482.00
Adv Math (Saxon - 2nd Edition) - Year 2
Open To:
Grade 9-12 students who have completed the first half of Advanced Math
1 HS credit
Cost: $482.00
Algebra 1/2 (Saxon - 3rd Edition)
Open To:
Grade 7-12 students. (Placement test may be required.)
Materials Used:
Saxon Algebra 1/2, 3rd edition home school packet from Saxon.
Description:
This course covers topics mastered in a standard Pre-Algebra course. Course focuses on key areas. Students use family site for online tests and quizzes. Class meets once weekly.
1 HS credit if done in Grade 9 or above.
Cost: $300.00
Algebra I (Saxon - 3rd Edition) (2x a week)
Open To:
Grade 7-12 students who have completed a pre-algebra course with a grade of A or B. Note: Please contact us if you would like a Grade 7 student to take this course.
Materials Used:
Saxon Algebra 1, 3rd edition, home school packet from Saxon.
Description:
This course covers topics mastered in a standard Algebra 1 course. Course focuses on key areas. Students use family site for online tests and quizzes. Class meets twice weekly.
1 HS credit if done in Grade 8 or above.
Cost: $482.00
Algebra I (Saxon - 3rd Edition) (4x a week)
Open To:
Students in Grades 7-12 who have completed a pre-algebra course with a grade of A or B.
Note: Please contact us if you would like a Grade 7 student to take this course.
Description: Students watch the lesson using DIVE cd'sAlgebra II (Saxon - 3rd Edition) (2x a week)
Open To:
Grade 8-12 students who have completed the majority of Saxon Algebra 1 with a grade of A or B.
Materials Used:
Saxon Algebra 2, 3rd edition and home school packet from Saxon.
Description:
This course covers topics mastered in a standard Algebra 2 course. This is the necessary course to transition from high school level work to college level work. Course focuses on key areas. Most tests online but a few need to be mailed or scanned. Class meets twice weekly.
1 HS credit
Cost: $482.00
Algebra II (Saxon - 3rd Edition) (4x a week)
Open To:
Students in Grades 8-12 who have completed the majority of Saxon Algebra 1 with a grade of A or B.
Description: Students watch the lesson using DIVE cd's,Geometry (Jacobs - 3rd Edition)
Open To:
Grade 9-12 students who have completed the majority of Saxon Algebra 1 with a grade of C or above.
This course teaches Euclidean geometry; students will discuss and demonstrate proofs in class. Students use family site for online tests and quizzes. Class meets 2x week.
1 HS credit
Cost: $482.00
Math 76 (Saxon - 4th Edition)
Open To:
Students in Grades 5-12.
Materials Used:
Saxon Math 76, 4th edition, home school packet from Saxon.
Description:
This course covers topics mastered in a standard middle school course. Course focuses on key areas including pre algebra skills. Students use family site for online tests and quizzes. Class meets once weekly.
Cost: $300.00
Math 87 (Saxon - 3rd Edition)
Open To:
Students in Grade 6-12
Materials Used:
Saxon Math 87, 3rd edition homeschool packet from Saxon.
Description:
This course covers topics mastered in a standard middle school course. Course focuses on key areas including pre algebra skills. Students use family site for online tests and quizzes. Class meets once week weekly.
This course covers topics mastered in a standard Pre-Calculus course. This is the necessary course to transition from high school level work to college level work. Course focuses on key areas. All tests are graded via mail or uploading feature. Class meets twice weekly.
Cost: $470.00 | 677.169 | 1 |
The course notes can be found at
this webpage.
I contributed to the coursenotes for Ideas in Geometry, working with Bart Snapp to revise some sections in light of our students' responses to the course It is mainly taken by Education and Liberal Arts majors.
Recent Semesters' Teaching
I have taught a wide range of courses using a variety of methods while at the University of Illinois. These include courses based on Lecturing, Group Work, and the use of computers.
Last semester I taught a discussion section of Calculus I in the Merit Workshop style. My students attend a large lecture of Calculus I but also attend
In Spring 2009, I taught a Merit Workshop section of Calculus 1. The students attended a large lecture of Calculus 1 but attendedYou can find the worksheets, exam review sheets, and other information for this course at the Merit Calculus 1 Spring 2009 webpage.
Further information about the Merit Program is available at the UIUC Merit Program website.
In Fall 2008, I taught
Math 181: A Mathematical World . I was the primary instructor for this course, which is aimed at students who will not be majoring in Mathematics, Engineering, or Science. Students encounter a wide variety of different mathematical ideas and explore how these ideas are present in the world around us.
In Spring 2008, I taught Math 119: Ideas in Geometry. The course notes can be found at
this webpage.
I also contributed to the coursenotes for Ideas in Geometry
In Fall 2007, I taught Math 119: Ideas in Geometry (Fall Semester).
My experiences teaching this course this semester lead to the contributions to the coursenotes mentioned above and lead to numerous small changes in how I approached this course in the Spring, as I learned from my experiences and those of my students.
In Spring 2006, I taught a stand-alone Math 220 (Calculus 1 - Spring 2006) class by the
Calculus and Mathematica method.
Students worked in small groups on problems presented through the computer.
The Calculus and Mathematica approach takes advantage of computers (and Mathematica) to allow students to experiment with the ideas of Calculus and to develop an improved conceptual understanding aided by an ability to geometrically visualize these ideas.
My Schedule
Here's my schedule of "fixed" weekly events, but there will occasionally be other meetings, etc., that do not show up on this schedule. If you cannot make it to the regularly scheduled office hours, use the schedule below to find a time to meet that will suit us both. | 677.169 | 1 |
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At Neufeld Learning Systems, we are educators as well as the authors and creators of the Understanding Math programs for grades 4 to 10. Since 1985, we have integrated technology as a learning tool to embrace, enrich and expand the existing mathematics curriculum. We invite you to explore our Web site to find out more about Neufeld Learning Systems and its Understanding Math series of software.
The DIAGNOSYS Home Page Welcome to the www home page for DIAGNOSYS, the expert-system diagnostic test for basic mathematics and other subjects. 14/8/01 DIAGNOSYS v3.51 and MATHINPUT v1.51 include some minor improvements to the design/edit features (edit of correct answers, provision of unit step function for graphs). 26/9/01 DIAGNOSYS v3.53 fixes a minor bug in processing files. Contents ...
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GRACE - Graphical Ruler and Compass Editor Welcome to GRACE, the Graphical Ruler and Compass Editor. GRACE is an interactive ruler and compass construction editor for use in teaching the fundamental concepts of geometry to high school students. With GRACE's intuitive graphical user interface, students can both define geometric constructions, and create formal proofs about their constructions.
Our primary objective at Eduscape is to create learning software for business and education. We feel the Internet, along with your PC, is the ultimate distant learning tool. We use a combination of CDROMs and the Internet to present learning materials. The learning programs use Real Audio sound clips, digital movies, and hyperlinking. These materials allow the learner to interact with the media, ...
Eight hundred schools use Caves software to make maths easy. Mental maths tutors teach numeracy in primary and secondary schools. One hundred and seventy quizzes train pupils in the basic rules of arithmetic. Special needs versions help the partially sighted and those with motor problems.
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Lesson Plans
Are colleges affordable? – Lesson Plan
By Mary F. Klein a mathematics teacher from Urbana, Illinois
Subject(s)
Second-year algebra courses and beyond, high school mathematics
Time
One to two class periods
Objectives
A mathematical model is a method of understanding a complex real world process involving multiple variables. To design a model, important features are identified and quantified symbolically using a number of mathematical tools. Once designed, a mathematical model must be evaluated and critiqued for accuracy, advantages, and limitations.
In this lesson, the students investigate a mathematical model that compares the cost of education to potential earnings in order to decide if the investment is a good one. The students compute the amount to be repaid each year for the life of the loan and then they compare the answer to some probable starting salaries. Finally, the students critique the model discussing which other factors (demand for the job, other costs, etc.) should be taken into consideration.
Overview
This lesson and activity are designed to help students understand the idea of a mathematical model and, at the same time, provide a tool that may be useful to them when they choose a college and evaluate the potential debt that may result.
Procedures
Distribute the Online NewsHour story, 'Paying More For College, Getting Less' for students to read. Briefly discuss the story. Bring up the idea that one goes into debt to finance a house. Does it pay to borrow to finance future earnings? Point out that borrowing for a house or an education can be considered an investment, while borrowing for a car is not an investment since the car depreciates in value. Tell the class that today's goal is to use mathematics to investigate the relationship between how much we borrow and how much we can afford to borrow.
Tell students that the power of mathematics is that it can be used to help us understand our world and that an important tool is the mathematical model. Define 'mathematical model'. Distribute the handout, 'The Mathematical Model,' and the worksheet.
Go over the parts of the model and then work out the details of the first problem on the worksheet.
Point out that since we know the amount of the yearly payment, we can consider the yearly payment to be a constant. Equation D can be simplified to y = (205,800) / x. What sort of graph can we expect? A hyperbola! Graph equation D. (Let 0 < x < $60,000 using increments of $10,000 and let 0 < y < 100% using increments of 10%.)
Discuss how to interpret the graph. (As earnings increase the percent of income used to repay the debt decreases). The financial industry standard is that total debt (including housing) should be less that 36% of gross earnings. Is the value we found for A in fact a reasonable debt ratio for a first year teacher earning $28,132 (the median income for a teacher with a BA in Illinois)? The minimum starting salary for such a teacher is $20,299. Discuss how low that first year starting salary could be and still support a reasonable debt ratio. (Use a graphing utility to plot y = 36. Find the intersection of the graphs.) Would it matter if there were a glut or a shortage of teachers in a particular field?
Assign students to groups and have them present their findings. (To save time, you might assign just 1 or 2 problems to each group.)
Check answers.
Discuss the model itself. Is it a good model? What hidden assumptions does it make? What are its limitations? What would the students change?
Extension Activity
Investigate the effect on the debt ratio if a student charges daily expenses to a credit card and makes only the minimum payment each month.
Borrow a copy of Peterson's Complete Guide to Colleges from a high school counselor. Choose a number of colleges or universities and look up 'the average debt incurred by students the previous year' for each school. Compare the debt ratios for the different schools.
Sources
Robert Andersen, Associate Director of Financial Aid, University of Illinois at Urbana-Champaign
Chris Hopkins, Counselor, Urbana High School
Author Mary F. Klein is a mathematics teacher with experience at the high school, middle school, and junior high school levels. She is the author of several Web-based lessons on mathematics and basketball. She lives with her husband and son in Urbana, Illinois | 677.169 | 1 |
computer vision. • Part I provides an introduction to affine geometry. but in our opinion. We present a coherent view of geometric methods applicable to many engineering problems at a level that can be understood by a senior undergraduate with a good math background. This is also useful to establish the notation and terminology. we provide an introduction to affine geometry. Our main goal is to provide an introduction to the mathematical concepts needed in tackling problems arising notably in computer graphics. In particular. this book should be of interest to a wide audience including computer scientists (both students and professionals). This ensures that readers are on firm grounds to proceed with the rest of the book.Preface
This book is primarily an introduction to geometric concepts and tools needed for solving problems of a geometric nature with a computer. just to mention some key areas. We present some of the main tools used in computer aided geometric design (CAGD). which is a main theme of this book. but our goal is not to write another text on CAGD. In this respect. and engineers interested in geometric methods (for example. Thus. triangulations). This book consists of four parts. books dealing with the relevant geometric material are either too theoretical. In brief. mathematicians. we are flirting with the intuitionist's ideal of doing mathematics from a "constructive" point of view. or else rather specialized and application-oriented. in particular the study of curves and surfaces. we are more interested in dealing with curves and surfaces from an algorithmic point of view . Readers 1
. and motion planning. This material provides the foundations for the algorithmic treatment of polynomial curves and surfaces. geometric modeling. Many problems in the above areas require some geometric knowledge. Although we will touch some of the topics covered in computational geometry (for example. Such a point of view if of course very relevant to computer science. This book is an attempt to fill this gap. mechanical engineers). we are writing about
Geometric Modeling Methods in Engineering
We refrained from using the expression "computational geometry" because it has a well established meaning which does not correspond to what we have in mind.
Riesler [68]. Several sections of this book are inspired by the treatment in one of several of the above texts. there are excellent texts on CAGD.2
CONTENTS proficient in geometry may omit this section. Samuel [69]. we are far from giving a comprehensive treatments of these topics. However. radically changed our perspective on polynomial curves and surfaces. "Blossoming: A connect–the–dots approach to splines" [65]. but as Ramshaw. 36]. 6]. from the short movie Geri's game. We also include a section on subdivision surfaces. although we cover affine geometry in some detail. geometric modeling. Hoschek and Lasser [45]. Farin [32. • Part III deals with an algorithmic treatment of polynomial surfaces (B´zier rectangular e or triangular surfaces). Pedoe [59]. Tisseron [83]. especially the paper by DeRose et al [24] on the animated character Geri. Luckily. For such a treatment. Hilbert and Cohn-Vossen [42]. Berger and Gostiaux [7]. • Part II deals with an algorithmic treatment of polynomial curves (B´zier curves) and e spline curves. just to mention some key areas. We have happily and irrevocably adopted the view that the most transparent manner for presenting much of the theory of polynomial curves and surfaces is to stress the multilinear nature (really multiaffine) of these curves and surfaces. This part has been included to make the material of parts I–III self-contained. 31]. This is why many pages are devoted to an algorithmic treatment of curves and surfaces. This is in complete agreement with de Casteljau's original spirit. including Bartels. computer vision. we only provide a cursory coverage of CAGD methods. and Veblen e and Young [85. Sidler [76]. Pedoe [59]. Our main goal is to provide an introduction to the concepts needed in tackling problems arising in computer graphics. Beatty. 6]. and spline surfaces. and differential calculus. and Piegl and Tiller [62]. Similarly. a great classic. For example. Fiorot and Jeannin [35. On the other hand. one of the most spectacular application of these concepts is the treatment of curves and surfaces in terms of control points. These readers are advised do some extra reading in order to assimilate some basic knowledge of geometry. 86]. • Part IV consists of appendices consisting of basics of linear algebra. do Carmo [26]. a tool extensively used in CAGD. Boehm and Prautzsch [11]. complements of affine geometry. we are more explicit
. readers totally unfamiliar with this material will probably have a hard time with the rest of the book. Samuel [69]. or use it by need . certain technical proofs that were omitted earlier. analysis. we highly recommend Berger [5. Our advice is to use it by need !
Our goal is not to write a text on the many specialized and practical CAGD methods. as attested by several papers in SIGGRAPH'98. and Barsky [4]. Dieudonn´ [25]. we highly recommend Berger [5. Lyle Ramshaw's remarkably elegant and inspirational DEC-SRC Report. an exciting and active area of research in geometric modeling and animation. and motion planning. and Tisseron [83]. As it turns out. and we are happy to thank the authors for providing such inspiration.
Catmull-Clark.
New Treatment. Some great references are Koenderink [46] and Faugeras [33] for computer vision. For further information on these topics.
We will see how this can be done using polynomial curves or surfaces (also called B´zier e curves or surfaces).
The tools and techniques developed for solving the approximation problem will be very useful for solving the other two problems. and solid modeling. by a curve or a surface. Thus. There is also no reasonably thorough textbook presentation of the main surface subdivision schemes (DooSabin. it is very helpful to be aware of efficient methods for numerical matrix analysis. there is no fully developed modern exposition integrating the basic concepts of affine geometry as well as a presentation of curves and surfaces from the algorithmic point of view in terms of control points (in the polynomial case). Strang [81]. New Results
This books provides an introduction to affine geometry.CONTENTS
3
in our use of multilinear tools. background material or rather technical proofs are relegated to appendices. and Metaxas [53] for physics-based deformable models. much of the algorithmic theory of polynomial curves and surfaces is captured by the three words:
Polarize. As the reader will discover. Strang's beautiful book on applied mathematics is also highly recommended as a general reference [80]. Hoffman [43] for solid modeling.
Novelties
As far as we know.
. There are other interesting applications of geometry to computer vision. In fact.
Again. and to numerical methods in matrix analysis. tensorize!
We will be dealing primarily with the following kinds of problems: • Approximating a shape (curve or surface). homogenize. spline curves or surfaces. • Drawing a curve or a surface. Generally. computer graphics. and a technical discussion of convergence and smoothness. we will see how this can be done using spline curves or spline surfaces.
• Interpolating a set of points. readers are referred to the excellent texts by Gray [39].
The material presented in this book is related to the classical differential geometry of curves and surfaces. Loop). and Ciarlet [19]. it is often possible to reduce problems involving certain splines to solving systems of linear equations.
but in the context of wavelets and multiresolution representation.4
CONTENTS
We give an in-depth presentation of polynomial curves and surfaces from an algorithmic point of view. which implies that we give precise definitions and prove almost all of the results in this book. We also urge the reader to write his own algorithms. They provide an attractive alternative to spline surfaces in modeling applications where the topology of surfaces is rather complex. A glimpse at subdivision surfaces is given in a new Section added to Farin's Fourth edition [32]. but we always keep the algorithmic nature of the mathematical objects under consideration in the forefront. all implemented in Mathematica. A a general rule. we provide Mathematica code for most of the geometric algorithms presented in this book. Catmull and Clark [17]. surfaces. etc. Many problems and programming projects are proposed (over 200).
Many algorithms and their implementation
Although one of our main concerns is to be mathematically rigorous. originating in Berger [5]). Some are routine. we give a crash course on discrete Fourier transforms and (circular) discrete convolutions. which yields both geometric and computational insights into the subtle interaction of knots and de Boor control points. and where the initial control polyhedron consists of various kinds of faces. and for this. we are primarily interested in the repesentation and the implementation of concepts and tools used to solve geometric problems. not just triangles or rectangles. and we propose many challenging programming projects. we try to be rigorous. As far as we know. DeRose. some are (very) difficult. we devote a great deal of efforts to the development and implemention of algorithms to manipulate curves. A clean and elegant presentation of control points is obtained by using a construction for embedding an affine space into a vector space (the so-called "hat construction". We even include an optional chapter (chapter 11) covering tensors and symmetric tensors to provide an in-depth understanding of the foundations of blossoming and a more conceptual view of the computational material on curves and surfaces. 28]. which leads very naturally to a presentation of polynomial curves and surfaces in terms of control points (B´zier curves and surfaces). We e present many algorithms for subdividing and drawing curves and surfaces. 29. this is the first textbook presentation of three popular methods due to Doo and Sabin [27. Subdivision surfaces are also briefly covered in Stollnitz. As a matter of fact. Subdivision surfaces form an active and promising area of research. We discuss Loop's convergence proof in some detail. These algorithms were used to prepare most of the illustrations of this book. and Charles Loop [50]. Subdivision surfaces are the topic of Chapter 9 (section 9. and Salesin [79].4).
Open Problems
. triangulations. The approach (sometimes called blossoming) consists in multilinearizing everything in sight (getting polar forms). The continuity conditions for spline curves and spline surfaces are expressed in terms of polar forms. Thus.
and to Marcel Berger.
Acknowledgement
This book grew out of lectures notes that I wrote as I have been teaching CIS510. Chris Croke. Ken Shoemake.CONTENTS
5
Not only do we present standard material (although sometimes from a fresh point of view). and David Harbater. Dimitris Metaxas. and the finite element method is the subject of so many books that we will not even attempt to mention any references. Also thanks to Norm Badler for triggering my interest in geometric modeling. Steve Frye. we have omitted rational curves and rational surfaces. Rich Pito. thus taking the reader to the cutting edge of the field. Ioannis Kakadiaris. I wish to thank some students and colleagues for their comments. Will Dickinson. who screened half of the manuscript with a fine tooth comb. Edith Haber. The first two topics are nicely covered in Hoffman [43]. Charles Erignac. Hartmut Liefke. We also discuss some of the problems with the convergence and smoothness of subdivision surface methods. Gerald Farin. David Jelinek. We also have omitted solid modeling techniques. for sharing some of their geometric secrets with me. DeRose. Bond-Jay Ting. We are also writing a text covering these topics rather extensively (and more). Deepak Tolani. methods for rendering implicit curves and surfaces. and Salesin [79]. and projective geometry. For example. we describe very clearly the problem of finding an efficient way to compute control points for C k -continuous triangular surface splines. but whenever possible. and a more mathematical presentation in Strang [82]. and most of all Raj Iyer. including Doug DeCarlo. Hany Farid. a remarkably clear presentation of wavelets is given in Stollnitz. A good reference on these topics is [31]. many thanks to Eugenio Calabi for teaching me what I know about differential geometry (and much more!).
. the finite elements method. Herman Gluck. Jaydev Desai. Finally. Jeff Nimeroff. Andy Hicks. Ron Donagi.
What's not covered in this book
Since this book is already quite long. we state some open problems. Introduction to Geometric Methods in Computer Science. Dianna Xu. for the past four years. and wavelets.
6
CONTENTS
.
geometry has played a crucial role in the development of many scientific and engineering disciplines such as astronomy. in order to put it on firmer grounds and to obtain more rigorous proofs. Basically. On the other hand. and Ferguson.Chapter 1 Introduction
1. e used geometric methods. and in this century. What happened then is that mathematicians started using more algebra and analysis in geometry. and that this could lead to wrong results or fallacious arguments. and it is very useful from the point of view of applications. among others. geodesy. geometry seems to be making a come-back as a fundamental tool used in manufacturing. It is interesting to observe that most College textbooks of mathematics included a fair amount of geometry up to the fourties. starting in the early sixties. 7
. the type of geometry used is very elegant. computer vision. and motion planning. the intuitions behind these fancy concepts almost always come from shapes that can somehow be visualized. automobile and aircraft manufacturing. ship building. mechanics. what a glorious subject! For centuries. The consequence of the strive for more rigor and the injection of more algebra in geometry is that mathematicians of the beginning of the twentieth century began suppressing geometric intuitions from their proofs. stimulated by computer science. One might say that this is only true of geometry up to the end of the nineteenth century. but even when the objects are higher-dimensional and very abstract. with the advent of faster computers. civil and mechanical engineering. it is a branch of affine geometry. Although not very advanced. the amount of geometry decreases to basically disappear in the seventies. just to mention some key areas. except for the initiated. These methods pioneered by de Casteljau. it was discovered at the end of the nineteenth century that there was a danger in relying too much on visual intuition. Geometry lost some of its charm and became a rather inpenetrable discipline. B´zier. Paradoxically. automobile and plane manufacturers realized that it was possible to design cars and planes using computer-aided methods. After being neglected for decades. Beginning with the fifties.1 Geometric Methods in Engineering
Geometry. balistics. there seems to be an interesting turn of events. computer graphics. architecture. Thus. What makes geometry a unique and particularly exciting branch of mathematics is that it is primarily visual .
8
CHAPTER 1. virtual oceans • Shape reconstruction • Weather analysis
. in particular. since they are the most efficient class of curves and surfaces from the point of view of design and representation. In the next section. The demand for technology using 3D graphics. organs.2
Examples of Problems Using Geometric Modeling
The following is a nonexhaustive listing of several different areas in which geometric methods (using curves and surfaces) play a crucial role. surfaces. We will need to understand better how to discretize geometric objects such as curves. since this material relies on some algebra and analysis (linear algebra. etc) will be required. virtual reality. This book represents an attempt at presenting a coherent view of geometric methods used to tackle problems of a geometric nature with a computer. We believe that this can be a great way of learning about curves and surfaces. in order to make the book entirely self-contained. In turn. etc. affine geometry. in order to gain a deeper understanding of this theory of curves and surfaces. directional derivatives. etc). limbs. we present the underlying geometric concepts in some detail. However.
1. while having fun. Thus. animation techniques. We concentrate on methods for discretizing curves and surfaces in order to store them and display them efficiently. we list some problems arising in computer graphics and computer vision that can be tackled using the geometric tools and concepts presented in this book. INTRODUCTION
We are convinced that geometry will play an important role in computer science and engineering in the years to come. we focus on polynomial curves defined in terms of control points. Furthermore. is increasing fast. • Manufacturing • Medical imaging • Molecular modeling • Computational fluid dynamics • Physical simulation in applied mechanics • Oceanography. we provide some appendices where this background material is presented. and it is clear that storing and processing complex images and complex geometric models of shapes (face. there are plenty of opportunities for applying these methods to real-world problems. and volumes. Our main focus is on curves and surfaces. but our point of view is algorithmic.
Many manufacturing problems involve fitting a surface through some data points. x5 . (wings. Again. x3 . in medical imaging. plane design. engine parts. In computer animation. x7 The previous example suggests that curves can be defined in terms of control points.10 d2 d1 x1 x3 d0 d3 x2
CHAPTER 1. x4 . passing through some intermediate locations. fuselage. one may want to have a person move from one location to another. etc. x1 . one may want to find the contour of some organ. etc). Indeed. in a smooth manner. We could go on and on with many other examples. this problem can be solved using B-splines. x2 . but it is now time to review some basics of affine geometry!
. One may do this by fitting a Bspline curve through the data points. Let us mention automobile design. say the heart. ski boots. x6 .1: A C 2 interpolation spline curve passing through the points x0 . For example. INTRODUCTION
d7
d6 x6 x4 d4 x5 d5
x0 = d−1
x7 = d8
Figure 1. specifying curves and surfaces in terms of control points is one of the major techniques used in geometric design. ship hulls. given some discrete data.
Part I Basics of Affine Geometry
11
.
12
.
The geometric properties of a vector space are invariant under the group of bijective linear maps. rotations. Of course. there are more affine maps than linear maps. As in physics. In particular. and other parts of physics (for example. Roughly speaking. etc.1 Affine Spaces
Geometrically. one is also interested in geometric properties invariant under certain transformations. living in a space consisting of "points. One reason is that the point corresponding to the zero vector (0). One could model the space of points as a vector space. curves and surfaces are usually considered to be sets of points with some special properties. in an intrinsic manner. but not a linear map in general. translations. that is. given an m × n matrix A and a vector b ∈ Rm . are handled in an akward manner. such as parallelism. trajectories. called the origin. dynamics. but this is not very satisfactory for a number of reasons. a rigid motion is an affine map. the set U = {x ∈ Rn | Ax = b} of solutions of the system Ax = b is an affine space. projections. when there is really no reason to have a privileged origin. etc. curves. Affine spaces are the right framework for dealing with motions. surfaces. for example. But the deeper reason is that vector spaces and affine spaces really have different geometries. coordinate systems have to be chosen to finally carry out computations. elasticity). and these two groups are not isomorphic. this is highly desirable to really understand what's going on.Chapter 2 Basics of Affine Geometry
2. 13
. but not a vector space (linear space) in general. Affine spaces provide a better framework for doing geometry. Another reason is that certain notions. plays a special role. Thus. and physical forces. it is possible to deal with points. independently of any specific choice of a coordinate system. affine geometry is crucial to a clean presentation of kinematics. whereas the geometric properties of an affine space are invariant under the group of bijective affine maps. among other things. After all. but one should learn to resist the temptation to resort to coordinate systems until it is really necessary." Typically. Also.
corresponding to linear independence and bases. Note that → − b = a + ab. given a point − → x = (x1 . x3 ). This allows us to define a clean notion of parallelism.14
CHAPTER 2. the standard frame in R3 has origin O = (0. e ). realizing that we are forced to restrict our attention to families of scalars adding up to 1. We take advantage of the fact that almost every affine concept is the counterpart of some concept in linear algebra. − )) e e e3 → → → − consisting of an origin O (which is a point) together with a basis of three vectors (− . Our presentation of affine geometry is far from being comprehensive. We investigate briefly some simple affine maps. 0. b3 − a3 ). Then. x3 ). e e
1 2 3
For example. For more details. b3 ). b2 . and Hilbert and Cohn-Vossen [42]. let us identify points with elements of R3 . but it is easy to infer that a frame is a pair (O. and that the position of this point x is determined with respect to a "frame" in R3 by a vector. Corresponding to linear subspaces. points and vectors are identified. stressing the physical interpretation of the definition in terms of points (particles) and vectors (forces). Then. in the standard frame. Corresponding to linear combinations of vectors. −2 . and it is biased towards the algorithmic geometry of curves and surfaces. and − = (0. b2 − a2 . Curiously. We show that every affine map is completely defined by the image of one point and a linear map. addition being understood as addition in R3 . the vector ab = (b1 − a1 . (−1 . We also define convexity. 0) and the basis of three vectors − = (1. the position of x is the vector Ox = (x1 . The position of a point x is then defined by → → → e1 e2 e3 the "unique vector" from O to x. Next. If so. Corresponding to linear maps. the notion of a → → → frame is rarely defined precisely. which coincides with the point itself. − = (0. we define affine combinations of points (barycenters). − . 0). Certain technical proofs and some complementary material on affine geometry are relegated to an appendix (see Chapter B).
. x2 . the reader is referred to Pedoe [59]. x2 . 0. 1). Samuel [69].
But wait a minute. 1. this definition seems to be defining frames and the position of a point without defining what a point is! Well. we define affine maps as maps preserving affine combinations. such as Greenwood [40]. we introduce affine subspaces as subsets closed under affine combinations. In the standard frame. 6]. the translations and the central dilatations. a3 ) and b = (b1 . we define affine independence and affine frames. 0. Snapper and Troyer [77]. Berger [5. Tisseron [83]. Suppose we have a particle moving in 3-space and that we want to describe the trajectory of this particle. given any two points a = (a1 . we characterize affine subspaces in terms of certain vector spaces called their directions. there is a unique free vector denoted → − → − ab from a to b. a2 . If one looks up a good textbook on dynamics. We begin by defining affine spaces. BASICS OF AFFINE GEOMETRY
Use coordinate systems only when needed!
This chapter proceeds as follows. one finds out that the particle is modeled as a point. 0).
2. where we forget the vector space structure. Furthermore. a3 + v3 ). v2 . although trivial in the case of R3 . but the notion of linear combination of points is frame dependent. AFFINE SPACES → → → of λa + µb with respect to the frame (Ω. there is a unique free vector ab such that → − b = a + ab. (−1 . v3 ). are all that is needed to define the abstract notion of affine space (or affine structure). it is natural that E is a vector space. This action + : R3 × R3 → R3 satisfies some crucial properties. λa3 + µb3 − (λ + µ)ω3 ) which are different from (λa1 + µb1 − ω1 . Intuitively. we can think of the − → elements of E as forces moving the points in E. λa3 + µb3 − ω3 ). a3 ) and any vector − = (v1 . where E is a set of points (with no structure) and − → E is a vector space (of free vectors) acting on the set E. A clean way to handle the problem of frame invariance and to deal with points in a more intrinsic manner is to make a clearer distinction between points and vectors. and the second copy corresponding to free vectors. the action of the force u is to "move" every point a ∈ E to the point a + u ∈ E obtained by the translation corresponding to u viewed as a vector. For example.
.1. v → which can be thought of as the result of translating a to b using the vector − . we discovered a major difference between vectors and points: the notion of linear combination of vectors is basis independent. v we obtain the point → a + − = (a1 + v1 . u v u v
→ − and for any two points a. The basic idea is − → to consider two (distinct) sets E and E . Since − → translations can be composed. some restriction is needed: the scalar coefficients must add up to 1.
17
Thus. λa2 + µb2 − ω2 . − → a + 0 = a. → → → → (a + − ) + − = a + (− + − ). a2 . b. We duplicate R3 into two copies. −3 )) are e e2 e (λa1 + µb1 − (λ + µ)ω1 . The effect − → of applying a force (free vector) u ∈ E to a point a ∈ E is a translation. In order to salvage the notion of linear combination of points. unless λ + µ = 1. a2 + v2 . the first copy corresponding to points. λa2 + µb2 − (λ + µ)ω2. It turns out that the above properties. We can v − is placed such that its origin coincides with a and that its tip coincides with → imagine that v b. − . By this. we mean − → that for every force u ∈ E . where the vector space structure is important. we make explicit the important fact that the vector space R3 acts → on the set of points R3 : given any point a = (a1 . considered as physical particles.
b) of points defines a − → − → unique vector ab in E . The formal definition of an affine space is as follows. since a(a + v) is the unique vector such that a+v = a+ a(a + v).1. → − Thus. Thus. but wearing different sets of glasses.1. E . we can choose to look at the points in E. there is a unique u ∈ E such that a + u = b. a vector space E (of translations. b ∈ E. if we also pick any point a in E. − → The axioms defining an affine space E. then we can recover all the points in E as the translated points a + u for all u ∈ E .
. or we can choose to look at the vectors u in E . It is natural to think of all vectors as having the same origin. or free vectors). − → (AF2) (a + u) + v = a + (u + v). it is assumed that all vector spaces under consideration are defined over the field R of real numbers. it is denoted as dim(E). (AF1) a + 0 = a. for every a ∈ E. E . E . Indeed. BASICS OF AFFINE GEOMETRY
For simplicity. The unique → − − → vector u ∈ E such that a + u = b is denoted by ab. + can be interpreted intuitively as saying − → that E and E are two different ways of looking at the same object. or sometimes by b − a. or a triple E. forgetting the points in E. − → This can be formalized by defining two maps between E and E . + consisting − → of a nonempty set E (of points). − → E . Furthermore.18
CHAPTER 2.
Did you say "A fine space"?
− → Definition 2. It is also assumed that all families of vectors and scalars are finite. An affine space is either the empty set. b = a + v is equivalent to ab = v. v ∈ E . satisfying the following conditions. For simplicity. we also write → − b = a + ab (or even b = a + (b − a)). + is the dimension dim( E ) of the vector space
−− − − −→ a(a + v) = v −− − − −→ −− − − −→ − → for all a ∈ E and all v ∈ E . the second set of glasses depending on the choice of an "origin" in E. − → (AF3) For any two points a. forgetting that every pair (a. the null vector. and every u. The following diagram gives an intuitive picture of an affine space. Note that − → − → The dimension of the affine space E. and an − → action + : E × E → E. point that can be viewed as an origin − → in E. for every a ∈ E.
yields b. − → − → where u ∈ E . E ). without making a commitment to a fixed origin in E. However. + as (E. − → − → which. which.
. and we denote it as Ea . rather than reducing E to some isomorphic copy of Rn . The composition of the first mapping with the second is −− − − −→ u → a + u → a(a + u). as soon as we commit to an origin a in E. we urge the reader to − → think of E as a physical set of points and of E as a set of forces acting on E. points are points. consider the mapping from E to E: u → a + u. → − − → When we identify E to E via the mapping b → ab. these compositions are the identity from E to E and the identity from E to E. Thus.2. The composition of the second with the first mapping is → − → − b → ab → a + ab. we say that we consider E as the vector space obtained by taking a as the origin in E. we can view E as the vector space Ea . we will often denote an affine space E.1. or even as − → E. an − → affine space E. + is a way of defining a vector space structure on a set of points E. Thus. Nevertheless. E . in view of (AF3). The vector space E is called the vector space associated with E. and the mappings are both bijections.2: Intuitive picture of an affine space − → For every a ∈ E. E . where b ∈ E. in view of (AF3). AFFINE SPACES E b=a+u u a c=a+w v w − → E
19
Figure 2. After all. and consider the mapping from E to E : → − b → ab. yields u. and not vectors! For − → − → notational simplicity.
we will denote points as row vectors. 0) and (0. an + un ). 2. . − → − → Any vector space E has an affine space structure specified by choosing E = E . Note that the vector space R is isomorphic to the line of equation x + y = 0 passing through the origin.
. the action of the vector space Rn over the set Rn simply viewed as a set of points.
2. . we will see in section 10. for any two points a = (a1 . and vectors as column vectors. an ) + .1 that it is possible to make sense of linear combinations of points. and even mixed linear combinations of points and vectors. is somewhat confusing. 1 − x) on L and any u ∈ R. points and vectors. The set L is the line of slope −1 passing through the points (1. 3. . since it suggests that points can be substracted (but not added!). We will refer to this affine structure on a vector − → space as the canonical (or natural) affine structure on E . Yet. the unique (vector) u ∈ R such that b = a + u is u = b1 − a1 . In order to distinguish between the double role played by members of Rn . For example. In this respect. . . we will consider n = 1. In most cases. Addition is well− → defined on vectors. the notation b − a for the unique vector u such that b = a + u. not in an obvious fashion). un The affine space An is called the real affine space of dimension n. 1 − x − u).2
Examples of Affine Spaces
Let us now give an example of an affine space which is not given as a vector space (at least. = (a1 + u1 . as in u + v. 1). (x. . Thus. In particular. the vector space Rn can be viewed as an affine space denoted as An . 1 − a1 ) and b = (b1 . . 1 − b1 ) on L. and − → letting + be addition in the vector space E . BASICS OF AFFINE GEOMETRY
One should be careful about the overloading of the addition symbol +. but addition of points a + b does not make sense. It immediately verified that this action makes L into an affine space. is given by u1 . y) satisfying the equation x + y − 1 = 0. . 1 − x) + u = (x + u. .20
CHAPTER 2. the translate a + u of a point a ∈ E by a vector u ∈ E is also well-defined. (a1 . Consider the subset L of A2 consisting of all points (x. The line L can be made into an official affine space by defining the action + : L × R → L of R on L defined such that for every point (x.
it is natural to define the affine combination λa + µb as the point of coordinates (λa1 + µb1 . the definition is intrinsic. For any two points a. let us now consider the new coordinate system with respect to the origin c = (1. Consider R2 as an affine space. −1). 0) and basis vectors and . some extra condition is needed in order for affine combinations to make sense. b ∈ E. Thus. under its natural 1 0 coordinate system with origin O = (0. b2 ). and the coordinates of b are (1. −2). 1) (and the same basis vectors). 1).
. λa2 + µb2 ). BARYCENTERS
23
b c O=d a
Figure 2. the point a + b is the point c = (1.4. the following properties hold: (1) If i∈I λi = 1. then → λi − i = aa
i∈I i∈I
− → λi bai . −1) and b = (2. Given any two 0 1 points a = (a1 .2. 2). then − → → a+ λ − = b+ aa λ ba . Thus.
i i i i i∈I i∈I
(2) If
i∈I
λi = 0. It turns out that if the scalars sum up to 1.4. when a = (−1. we give another example showing what goes wrong if we are not careful in defining linear combinations of points. and let (λi )i∈I be a family of scalars.5: Two coordinates systems in R2 certain readers. a + b corresponds to two different points depending on which coordinate system is used for its computation! Thus. 0) of the first coordinate system.1. Lemma 2. a2 ) and b = (b1 . Given an affine space E. 1). AFFINE COMBINATIONS. let (ai )i∈I be a family of points in E. This time. However. and the point a + b is the point d of coordinates (−1. the coordinates of a are (−2. it is clear that the point d is identical to the origin O = (0. as the following lemma shows. However.
24
CHAPTER 2. for any family of points (ai )i∈I in E.
i∈I
=(
i∈I
− → λi bai
=
i∈I
since
i∈I
λi = 0.
(2) We also have → λi − i = aa
i∈I i∈I
→ → − − λi ( ab + bai ) → − λi ) ab + − → λi bai . which is just a pair (a. it is convenient to introduce the notion of a weighted point. λi ))i∈I . or affine combination) of the points ai assigned the weights λi . we have a+
i∈I
→ λi − i = a + aa
i∈I
→ → − − λi ( ab + bai ) → − λi ) ab + − → λi bai
i∈I i∈I
=a+( → − = a + ab + = b+
i∈I i∈I
− → λi bai since
i∈I
λi = 1
− → λi bai
→ − since b = a + ab. where i∈I λi = 1.
Thus. λ). BASICS OF AFFINE GEOMETRY
Proof. we also say that the point i∈I λi ai is the barycenter of the family of weighted points ((ai .3). aa
. Note that the barycenter x of the family of weighted points ((ai .
i∈I
In dealing with barycenters. The unique point x is called the barycenter (or barycentric combination. and λ ∈ R is a scalar. λi ))i∈I .1. and it is denoted as λi ai . by lemma 2. λi ))i∈I is the unique point such that − = → ax
i∈I
→ λi − i for every a ∈ E. for any family (λi )i∈I of scalars such that i∈I λi = 1. Then. (1) By Chasles' identity (see section 2. where a ∈ E is a point.4. given a family of weighted points ((ai . the point x=a+
i∈I
→ λi − i aa
is independent of the choice of the origin a ∈ E.
Note that e the curve passes through a and d. we can define the curve F : A → A2 such that Such a curve is called a B´zier curve. b. the barycenter is the center of mass of the family of weighted points ((ai . we will see that a polynomial curve can be defined as a set of barycenters of a fixed number of points. d) be a sequence of points in A2 . 2 2
Later on. λi ))i∈I (where the masses have been normalized. b.4. and c. Thus. BARYCENTERS and setting a = x. This observation will be used in Chapter 10. F (t) = (1 − t)3 a + 3t(1 − t)2 b + 3t2 (1 − t) c + t3 d. and we may aa denote it as i∈I λi ai . and negative masses are allowed). AFFINE COMBINATIONS. 1). 4 4 2 The point g1 can be constructed geometrically as the middle of the segment joining c to 1 the middle 2 a + 1 b of the segment (a. but generally not through b and c. 2 2 2 2
The point g2 can be constructed geometrically as the point such that the middle 1 b + 1 c of 2 2 the segment (b. regardless of the value of i∈I λi . the point x is the unique point such that → λ − = 0. xa
i i i∈I
25
In physical terms.1 to define a vector space in which linear combinations of both points and vectors make sense. c) is the middle of the segment (a. 1 . We will see in the next chapter how any point F (t) on the curve can be constructed using an algorithm performing three affine interpolation steps (the de Casteljau algorithm). b). d) are called its control points. and (a. (1 − t)3 a + 3t(1 − t)2 b + 3t2 (1 − t) c + t3 d
.
i∈I i i∈I i i
The figure below illustrates the geometric construction of the barycenters g1 and g2 of the weighted points a. and (c. For example. the vector λ − does not depend on the point a. c. Remarks: (1) For all m ≥ 2. Then. and (a. since 2 g1 = 1 1 1 1 a + b + c. let (a. b. 1 . −1). (b. Observe that (1 − t)3 + 3t(1 − t)2 + 3t2 (1 − t) + t3 = 1. g2 ). is a well-defined affine combination. so that i∈I λi = 1.
since the sum on the left-hand side is obtained by expanding (t + (1 − t))3 = 1 using the binomial formula. 1). since g2 = −a + 2 1 1 b+ c . it is easy to prove that the barycenter of m weighted points can be obtained by repeated computations of barycenters of two weighted points. → (2) When λ = 0. 1 . c.2.
yi ) ∈ U means that
axi + byi = c.5. yi ) ∈ U. As an example. Given any m points (xi . According to definition 2. y) ∈ R2 | ax + by = c}.2.5. the barycenter i∈I λi ai belongs to V .1.1. the empty set is trivially an affine subspace.e. a (linear) subspace can be characterized as a nonempty subset of a vector space closed under linear combinations. the set of solutions of the equation ax + by = c. we get
m m m
a
i=1
λi xi
+b
i=1
λi y i
=
i=1
λi c = c. + .
i=1 i=1
λi y i
=
i=1
λi (xi . In affine spaces. yi ) ∈ U and any m scalars λi such that λ1 + · · · + λm = 1. Given an affine space E. yi ) ∈ U.
. It is natural to define an affine subspace as a subset of an affine space closed under affine combinations.
which shows that
m
m
m
λi xi .5. − → Definition 2.
and if we multiply both sides of this equation by λi and add up the resulting m equations.5
Affine Subspaces
In linear algebra. we claim that
m
i=1
λi (xi . An affine subspace is also called a flat by some authors. a subset V of E is an affine subspace if for every family of weighted points ((ai . i.
and since λ1 + · · · + λm = 1.
Indeed. consider the subset U of A2 defined by U = {(x. λi ))i∈I in V such that i∈I λi = 1. and every intersection of affine subspaces is an affine subspace. we get
m m
(λi axi + λi byi ) =
i=1 i=1
λi c. the notion corresponding to the notion of (linear) subspace is the notion of affine subspace. (xi . AFFINE SUBSPACES
27
2. where it is assumed that a = 0 or b = 0. E .
we get a(x − x0 ) + b(y − y0 ) = 0. Thus. Second. and thus (x. y0 ) is any point in U. y0 ) + U . we claim that − → U = (x0 . It turns out that U is closely related to the subset of R2 defined by − → U = {(x. since the right-hand side of the equation is − → − → null. This line can be identified with a line passing through the origin of A2 . the set of solution of the homogeneous equation ax + by = 0 obtained by setting the right-hand side of ax + by = c to zero. u2 ) ∈ U }. then ax + by = c. y − y0 ) ∈ U . and it is just a usual line in R2 . if (x. Hence. u2 ) ∈ U . BASICS OF AFFINE GEOMETRY
Thus. In fact. y0) + U . and since we also have ax0 + by0 = c. for any m scalars λi . Indeed. line which is parallel to the line U of equation ax + by = c. i. y) ∈ R2 | ax + by = 0}. Now.
. y0 ) + U . − → − → which shows that (x − x0 . (x0 . y) ∈ U. U is an affine subspace of A2 . it is just a usual line in A2 . − → U = (x0 . we also have − → − → U ⊆ (x0 . In fact. y0 ) ∈ U. The above example shows that the affine line U defined by the equation ax + by = c − → is obtained by "translating" the parallel line U of equation ax + by = 0 passing through the origin. given any point (x0 .
− → − → First.
this time without any restriction on the λi . since ax0 + by0 = c and au1 + bu2 = 0 for all (u1 . by subtraction. y0) + U = {(x0 + u1 . yi ) ∈ U . y0 ) + U ⊆ U. where − → − → (x0 . U is one-dimensional. y0 ) + U . y0 + u2 ) | (u1 . if (x0 . y0 ) + U . U is a subspace of R2 . and U = (x0 .28
CHAPTER 2. In fact. the same calculation as above yields that
m
i=1
− → λi (xi .e. y) ∈ (x0 .
an ) in V . . . AFFINE SUBSPACES
29
U U
Figure 2. (an . .
n
x=a+
i=1
→ λi − i aa
is the barycenter of the family of weighted points
n
((a1 . . λn ).2. observe that for every x ∈ E. the subset U of An defined by U = {x ∈ Rn | Ax = c} is an affine subspace of An . Let V be a nonempty subset of E. Furthermore. if we consider the corresponding homogeneous equation Ax = 0. Affine subspaces can be characterized in terms of subspaces of − → E . . This is a general situation. and for any x0 ∈ U. . λn ) of scalars.
i=1
. . for any family (λ1 . the set − → U = {x ∈ Rn | Ax = 0} is a subspace of Rn . 1 −
λi )). .5. Given any m × n matrix A and any vector c ∈ Rm . . For every family (a1 . (a. . . for every point a ∈ V . . λ1 ). it is easy to prove the following fact.7: An affine line U and its direction More generally. we have − → U = x0 + U .
An affine subspace of codimension 1 is called a hyperplane (recall that a subspace F of a vector space E has codimension 1 iff there is some subspace G of dimension 1 such that E = F ⊕ G. lemma 2. the set V of barycenters i∈I λi ai (where i∈I λi = 1) is the smallest affine subspace containing
. By the dimension of the subspace V .2. b. E . be characterized a little later. and thus. V is obtained from U by the translation → − ab.5.5.2. we mean the dimension of V . a line is the set of all points of the form a + λu.5. + an affine structure. such that − = λ ab. →
ab
− → A plane is specified by a point a ∈ E and two linearly independent vectors u. We say that three points a. and V = b + U. i. for a subspace U of E. c are ac → − → distinct. for any a ∈ U and any b ∈ V . for any family (ai )i∈I of points in E.e. An affine subspace of dimension 1 is called a line. µ ∈ R. are linearly dependent. a plane is the set of all points of the form a + λu + µv. Equivalently.5. since U = V . The subspace V associated with an affine subspace V is called the direction of V . then there is a unique λ ∈ R. if the vectors ab and − are linearly dependent. if the vectors ab. It is − → also clear that the map + : V × V → V induced by + : E × E → E confers to V. and an affine subspace of dimension 2 is called a plane. If two of the points a. see Appendix A). We say that four points → − − − → a. when E is the natural affine space associated with a vector space E. v ∈ E . V. a line is specified by a point a ∈ E and a nonzero vector v ∈ E . − → By lemma 2. We say that two affine subspaces U and V are parallel if their directions are identical.3. say a = b.2 shows that every affine subspace of E is of the form u + U. Given an affine space E. we have U = a + U. b. c → − → are collinear . The subspaces of E are the affine subspaces that contain 0. AFFINE SUBSPACES E − → E
31
a V =a+V
V
− → Figure 2. Hyperplanes will ac. d are coplanar . for λ.8: An affine subspace V and its direction V In particular. + . − → Lemma 2. b. → and ad.e. for λ ∈ R. i. the direct sum of F and G. and we define the ac → − ac ratio − = λ. c.
32 (ai )i∈I . etc. + . we have the notion of affine independence. a line specified by two distinct points a and b is denoted as a. the set of all points (1 − λ)a + λb.6
Affine Independence and Affine Frames
Corresponding to the notion of linear independence in vector spaces. b). and assume that there are some scalars (λj )j∈(I−{k}) such that − → a− j λj − → = 0 . Given a family (ai )i∈I of points in an affine space E. b ∈ V . Remark: Since it can be shown that the barycenter of n weighted points can be obtained by repeated computations of barycenters of two weighted points. let (ai )i∈I be a family of points in E. If the family (−→)j∈(I−{i}) is linearly independent for some i ∈ I. First. If V contains at least two points. then (−→)j∈(I−{i}) is linearly a− j a− j ia ia independent for every i ∈ I. then V = ∅. we will reduce the notion of (affine) independence of these points to the (linear) independence of the families (−→)j∈(I−{i}) of vectors obtained by chosing any ai as an origin. BASICS OF AFFINE GEOMETRY
Proof. that is. it is enough to show that V is closed under affine combinations. For example. ka
j∈(I−{k})
Since − → = −→ + −→. If (ai )i∈I is nonempty.1. Proof. the set V contains all barycentric combinations of a and b. Assume that the family (−→)j∈(I−{i}) is linearly independent for some specific i ∈ I. and similarly for planes. the following lemma a− j ia shows that it sufficient to consider only one of these families.6. a− j ia Let k ∈ I with k = i. then the smallest affine subspace containing (ai )i∈I must contain the set V of barycenters i∈I λi ai . the set V contains the line determined by a and b. which is immediately verified. E . If (ai )i∈I is empty. − → Lemma 2. a− j a− i a− j ka ka ia
.
2. because of the condition i∈I λi = 1.
Given a nonempty subset S of E. a nonempty subset V of E is an affine subspace iff for every two points a. and thus. V is an affine subspace iff for any two distinct points a. λ ∈ R. Given an affine space E. b . the smallest affine subspace of E generated by S is often denoted as S .
CHAPTER 2. or even (a. b ∈ V .
2 is called the linear map associated with the affine map f . followed by a rotation of angle π/4. but very technical. − → − → Lemma 2. We simply sketch the main ideas. 0 1
this affine map is the composition of a shear. − → for every a ∈ E and every v ∈ E . c. It can be found in Appendix B. Section − → − → B.13. − → − → The unique linear map f : E → E ′ given by lemma 2. d) is shown in figure 2.7. followed by √ a magnification of ratio 2.1. Let a be any point in E. c. Proof.2. If a linear map f : E → E ′ satisfying the condition of the lemma exists.
. c′ . We then have to check that the map defined in such a way is linear and that its definition does not depend on the choice of a ∈ E. such that f (a + v) = f (a) + f (v). b.13: The effect of an affine map is an affine map. The following lemma shows the converse of what we just showed. Since we can write 1 1 1 3 = √
√
a
2
2 2 √ 2 2
−
√ 2 2
√
2 2
1 2 .7. b′ . b. d′ ).7. The effect of this map on the square (a. The proof is not difficult. AFFINE MAPS c d′
′
39
d
c b′
b a′ Figure 2. there is a unique linear map f : E → E ′ . The image of the square (a. d) is the parallelogram (a′ . Every affine map is determined by the image of any point and a linear map. followed by a translation. Given an affine map f : E → E ′ . this map must satisfy the equation −− − − − − − − −→ f (v) = f (a)f (a + v) − → for every v ∈ E .2.
Since affine → bc maps preserves barycenters. c as −ratio(a.
There is a useful trick to convert the equation y = Ax + b into what looks like a linear equation. and form the (n + 1) × (n + 1)-matrix A b 0 1 so that y = Ax + b is equivalent to y 1 = A b 0 1 x . if f maps any three collinear points to collinear points. b. where say. y. which shows that f (a).
4 5 3 and sin θ = 5 . f (b). where A is a rotation matrix. where a = c. see the problems. Given three collinear points where a. f (c) are collinear in E ′ . we have b = (1 − β)a + βc for some unique β. AFFINE MAPS
43
translation.
. There is a converse to this property. b. It is easy to show that this affine map has a unique fixed point. c. → (1 − β) bc provided that β = 1. Such affine maps are called rigid motions. On the other hand.e. then f is affine. c) = ∞. even though
8 5 3 10
−6 5
2 5
=
2 0 1 0 2
4 5 3 5
−3 5
4 5
. the affine map 8 6 −5 x1 1 x1 5 → 3 + 2 x2 1 x2
10 5
has no fixed point. we agree that ratio(a. c. we have f (c) = (1 − λ)f (a) + λf (b). and b. with a = b. We add 1 as the (n + 1)th component to the vectors x. If f : E → E ′ is a bijective affine map. 1
This trick is very useful in kinematics and dynamics. When b = c. given any three collinear points a. b. The proof is rather long (see Berger [5] or Samuel [69]). since f preserves barycenters. as → − β ab ratio(a. For more
and the second matrix is a rotation of angle θ such that cos θ = on fixed points of affine maps.2. and we define the ratio of the sequence a. The converse states that given any bijective function f : E → E ′ between two real affine spaces of the same dimension n ≥ 2. i. which is simpler to state when the ground field is K = R. b. b. it is clear that affine maps preserve the ratio of three points. that b = c. We warn − → ba our readers that other authors define the ratio of a. c) = =−. c) = − . The trick is to consider an (n + 1) × (n + 1)-matrix. b. c in E.7. c = (1 − λ)a + λb. b.
if f = λ idE . perhaps we use use that! Observe that Ha. Recall that the group of bijective linear maps of the vector space E is denoted as GL(E). Ha. for any affine bijection f ∈ GA(E). Given any affine space E.λ. we have → − → − f (b) = f (a + ab) = f (a) + λ ab
. is a subgroup of GL(E). Note how every line is mapped to a parallel line. Snapper and Troyer use the term dilation for an affine dilatation and magnification for a central dilatation [77]. Remark: The terminology does not seem to be universally agreed upon. The subgroup DIL(E) = L−1 (R∗ idE ) of GA(E) is particularly interesting. Choose some origin a in E. BASICS OF AFFINE GEOMETRY
2. and when λ = 0 and x = a. The subset of all linear maps of the form λ idE . and is denoted as R∗ idE (where λ idE (u) = λu.16 shows the effect of a central dilatation of center d. The kernel of this map is the set of translations on E. Note that Ha.µ = Ha.λ ◦ Ha.1. ax Figure 2. for some λ ∈ R∗ with λ = 1. Since f is affine. called the affine group of E.λ (x) = a + λ− ax. The triangle (a. a dilatation (or central dilatation or homothety) of center a and ratio λ is a map Ha. a direct translation of the French "homoth´tie" [69]. Since e dilation is shorter than dilatation and somewhat easier to pronounce.1 is the identity. Samuel uses homothety for a central dilatation.44
CHAPTER 2. The terms affine dilatation and central dilatation are used by Pedoe [59].8
Affine Groups
We now take a quick look at the bijective affine maps. It is immediately verified that Ha.λ (a) = a. the map f → f defines a group homomorphism L : GA(E) → GL(E). b. Ha. Given any point a ∈ E.λ (x) is on the line defined by → a and x. c) is magnified to the triangle (a′ . and R∗ = R − {0}). When λ = 0. and denoted as GA(E). Then. and any scalar λ ∈ R. Given an affine space E. The elements of DIL(E) are called affine dilatations.λµ . and is obtained by "scaling" − by λ. where λ ∈ R − {0}. b′ . it is clear that Ha. then there is a unique point c ∈ E such that f = Hc.8.λ defined such that → Ha. Proof. When λ = 1.λ = λ idE . It turns out that it is the disjoint union of the translations and of the dilatations of ratio λ = 1. We have the following useful result.λ is an affine bijection. the set of affine bijections f : E → E is clearly a group. c′ ). Lemma 2. for every x ∈ E.
H = Ker f ∗ for some nonnull linear form f ∗ . where x = (x1 . f −1 (0) = a + H is a hyperplane H is E. in the sense that H = f −1 (0) = {x ∈ Am | f (x) = 0}. . Lemma 2. which implies that f is surjective.1. xm ) ∈ Rm . The relationship between affine hyperplanes and affine forms is given by the following lemma. − → where H is a hyperplane of E . since f ∗ is nonnull. xm ). Since f (a + v) = f (v) for all v ∈ E . and since f is a linear form. Unlike linear forms f ∗ . Thus. (b) If H is an affine hyperplane. clearly. for any a ∈ H. by a previous observation. it is possible that f −1 (0) = ∅. we have − → f (a + v) = 0 iff f (v) = 0. we have H = f −1 (0). it is interesting to consider affine forms. f : E → R is not identically null.2. letting f : E → R be the affine form defined such that. It is immediately verified that this map is affine. and we call it the kernel of f . − → Proof. and since by lemma A. its kernel H = Ker f is a hyperplane. and f is non constant.5. there is some λ ∈ R such that g = λf (with λ = 0). and the set H of solutions of the equation λ1 x1 + · · · + λm xm = µ is the null set. .9. . For any other affine form g : E → R such that H = Ker g. The second part of (b) is as in lemma A. we have H = a+ H. it is surjective. . .5.5. .
.1. The following properties hold: (a) Given any nonconstant affine form f : E → R. every hyperplane H ′ parallel to H is defined by a nonconstant affine form g such that g(a) = f (a) − λ. By lemma A. f (a + v) = f ∗ (v). AFFINE HYPERPLANES
47
for all (x1 .9. Ker f is a hyperplane H in E .2. there is a nonconstant affine form f : E → R such that H = Ker f . for all a ∈ E. for an affine form f . there is some a ∈ E such that f (a) = 0. (a) Since f : E → R is nonconstant. (b) For any hyperplane H in E. − → Thus. (c) Given any hyperplane H in E and any (nonconstant) affine form f : E → R such that H = Ker f .5. for which Ker f ∗ is never empty (since it always contains the vector 0). which are just affine maps f : E → R from an affine space to R. then by lemma 2. and thus.1 (c). of the affine map f : Am → R. . . for some λ ∈ R. Let E be an affine space. we also denote f −1 (0) as Ker f . or kernel. (c) This follows easily from the proof of (b). Recall that an (affine) hyperplane is an affine subspace of codimension 1.1. and is left an exercise. Given an affine map f : E → R.
n . m′′ ) be the barycentric coordinates of M. D0 ).
′ ′ ′ Let (D0 . This problem uses notions and results from problems 6. ′ ′′ ′ ′ ′′ Similarly. we will say that the barycentric coordinates (x. If D0 and D0 have a unique intersection point. and P = (p. z) such that x + y + z = 0 is also called a system of barycentric coordinates for the point of normalized barycentric coordinates 1 (x. this −→ − MB −→ − MC is equivalent to −→ − − → NC P A −→ − = −1. the intersection of the two lines D and D ′ is either a point or a vector. let M be ′ this point. BN. z) of a point M. y. wu′ − uw ′. y. any triple (x. Given any affine frame (a.
Prove that m n p m′ n′ p′ = 0 m′′ n′′ p′′
. In view of (a) and (b) of problem 7. ′ as explained at the beginning of the problem. p ) as the ′ intersection of D2 and D2 . n ) as the intersection of D1 and D1 . Then. z). and 8. we can think of it as a point at infinity (in the direction of the line defined by that vector). We call M the intersection of D0 and D0 . In either case. in both cases of barycentric coordinates (vw ′ − wv ′ . CP have a unique intersection point or are parallel iff m′′ np′ − m′ n′′ p = 0. such that the four lines belonging to any union of two of the above pairs are neither parallel not concurrent ′ (have a common intersection point). and P = B. D1 ). this is equivalent to − → −→ − − − → MB NC P A − → −→ − = 1. BASICS OF AFFINE GEOMETRY
When M = C. 7. b. x+y+z With this convention. let M denote a nonnull vector defining the common ′ direction of D0 and D0 .52
CHAPTER 2. it is natural to extend the notion of barycentric coordinates of a point in A2 as follows. D2 ) be three pairs of six distinct lines. uv ′ − vu′ ). m′ . When M = C. (D1 . − − − → MC NA P B (c) Prove Ceva's theorem: the lines AM. c) in A2 . let (m. N = A. and (D2 . y. When the above is a vector. − − → NA P B
Problem 9 (20 pts). are the normalized barycentric coordinates of M. and P = B. where x + y + z = 1. N = A. and if D0 and D0 are parallel. p . define N = (n.
C). (c. A′ . C ′ respectively. 1). Let (λ1 . (3. a′′ ). Use the affine frame (A. c′′ a′ ). such that λi = 1. The equation X ⊤ AX = 0 is called the homogeneous equation of the conic. 0). λ ∈ R. 1/3). (6. be the barycentric coordinates of A′ .54
CHAPTER 2. . does it also take the centroid of ∆1 to the centroid of ∆2 ? Justify your answer. B ′ . (ab′′ . P have barycentric coordinates (bc. BASICS OF AFFINE GEOMETRY (a) Letting A = C ⊤ BC. show that the barycenter of any n ≥ 3 points can be determined by repeated computations of barycenters of two points. for
. (0. 1/3). 1 ≤ i ≤ n. (b) Given any affine frame (A. CB ′ ). In this case. with λ1 + · · · + λn = 1. 9)} to the vertices of triangle ∆2 = {(1. if no three of the above points are collinear. . To simplify the notation. b′′ ). We say that a conic of homogeneous equation X ⊤ AX = 0 is nondegenerate if det(A) = 0. a′′ b′′ ). and let E be an affine space over K. and (AB ′ . (b. N. and (c. . From this. b′ .
Prove that A is symmetric. and show that M. Problem 13 (10 pts). λn ) be any n scalars in R. C). . c′′ b). C. Let E be an affine space over R. B ′ . AC ′ ). (b. 4). are collinear in the sense of problem 9. In the case where λ1 + · · · + λn = 1 and λi = 1. and degenerate if det(A) = 0. Problem 15 Extra Credit (20 pts). . 1/3). . then a nondegenerate conic passes through these six points iff the intersection points M. an ) be any n ≥ 3 points in E. . Given any six distinct points A. Prove that a conic containing more than one point is degenerate iff it contains three distinct collinear points. c) is the barycenter of (a. a′′ b′ . prove that any conic passing through A. BA′ ). Show that the barycenter λ1 a1 + · · · + λn an can be obtained by first determining the barycenter b of the n − 1 points a2 . (CA′ . Problem 14 (20 pts). an assigned some appropriate weights. c′ . P (in the sense of problem 7) of the pairs of lines (BC ′ . and that X ⊤ AX is homogeneous of degree 2. B. . and let (a1 . y ∈ V . 1)}. a′ . then V contains the entire line (1 − λ)x + λy. and let (a. N. B. . assume that λ1 = 1. the conic is the union of two lines. (5. Deduce from the above that a nonempty subset V of E is an affine subspace iff whenever V contains any two points x. . prove that the equation of the conic becomes X ⊤ AX = 0. (c) Prove Pascal's Theorem. . Show that this condition does not depend on the choice of the affine frame. and then the barycenter of a1 and b assigned the weights λ1 and λ2 + · · · + λn . If an affine map takes the vertices of triangle ∆1 = {(0. Assume that K is a field such that 2 = 1 + 1 = 0. cb′ . 0). c′′ ). b. C has an equation of the form ayz + bxz + cxy = 0. B. Hint. that det(A) = det(B). The centroid of a triangle (a. B. Show that there must be some i. C ′ . (c′ a. c′ a′ . .
Prove that if λµ = 1 and λ > 0. Show that these are rotations of center c. where a. y) can be represented as the complex number z = x + iy. In the plane A2 .
57
Can you figure out what Sn looks like in general? (you may want to write a computer program). where a.10. or there are infinitely many fixed points. Problem 19 (20 pts). b are complex numbers such that a = 0. with respect to the standard affine frame. Describe what these maps do geometrically.2. b2
Prove that such maps have a unique fixed point c if θ = 2kπ. 0). Prove that these maps are affine and that this set of maps is a group under composition. (c) Consider the set of geometric transformations of the form z → az + b or z → az + b. b2
Prove that such maps have a unique fixed point iff (λ + µ) cos θ = 1 + λµ. a point of coordinates (x. 0)). (2) Consider affine maps of the form x1 x2 → λ cos θ −λ sin θ µ sin θ µ cos θ x1 x2 + b1 . these affine maps are represented by rotation matrices. (a) Prove that these maps are affine. which means that with respect to a frame with origin c (the unique fixed point). The purpose of this problem is to study certain affine maps of A2 . Do you think that Sn has a limit? Problem 18 (20 pts). (1. there is some angle θ for which either there is no fixed point. Describe what these maps do geometrically. (b) Prove that the above set of maps is a group under composition. b are complex numbers such that a = 0. Consider the set of geometric transformations of the form z → az + b. for all integers k. and where z = x − iy if z = x + iy. PROBLEMS Construct S1 starting from the line segment L = ((−1. (1) Consider affine maps of the form x1 x2 → cos θ − sin θ sin θ cos θ x1 x2 + b1 . (3) Prove that the affine map x1 x2 →
8 5 3 10 6 −5 2 5
x1 x2
+
1 1
.
PROBLEMS
59
(ii) Give an algorithm to find the intersection of the plane H and of the triangle determined by a. c.2. Matlab. etc). Mathematica. (iii) (extra credit 20 pts) Implement the above algorithm so that the intersection can be visualized (you may use.
. Maple.10. b.
60
CHAPTER 2. BASICS OF AFFINE GEOMETRY
.
Part II Polynomial Curves and Spline Curves
61
.
62
.
or several pieces joined together? We can model the entire shape S via a single system of equations.e. with some degree of smoothness. This is usually mathematically simpler. and render curves and surfaces. and the shape has some dimension s ≤ n (s = 1 for a curve. or as the zero locus of this function (i. the shape S lives in some object space E of some dimension n (typically. and unsuitable to tackle problems for which it may be necessary to modify certain parts and leave the other parts unchanged. and it works well for very simple shapes. Nevertheless. which is used to determine which points lie on S and which do not. the affine plane A2 . and we will now concentrate on modelling a single curve or surface. we will come back to splines. such an approach is not modular. Later on.. the set of points that are mapped to "zero" under F ). since composite shapes are decomposed into simpler pieces. The two models just mentioned define S either in terms of 63
. s = 2 for a surface). manipulate. and by specifying how these pieces join together.Chapter 3 Introduction to the Algorithmic Geometry of Polynomial Curves
3. we face a number of questions which are listed below. A single piece. we model composite shapes with splines.1 Why Parameterized Polynomial Curves?
In order to be able to design. Oversimplifying a bit. Parametric or Implicit Model? Mathematically. The kind of model that we will adopt consists of specifying a curve or surface S in terms of a certain system of equations. In CAGD jargon. we have to choose some mathematical model for defining curves and surfaces. or the 3D affine space A3 ). possibly composite. Given that we adopt such a model. we view these equations as defining a certain function F . But for complex shapes. a combinatorial explosion usually occurs which makes this approach impractical. and the shape S is viewed either as the range of this function. Furthermore. it is very important to know how to deal with these simpler building blocks effectively. Such design problems are usually handled by breaking the curve or surface into simpler pieces. when a shape S is modeled by a system of equations.
where E is of dimension n ≥ s. the function F : A → A2 defined such that F1 (t) = 2t + 1. a shape S of dimension s specified by a function F : P → E. F2 (t) = t2 . where A denotes the affine line. and in the second case where F : E → P . the parameter space P also has dimension s ≤ n. we say that we have a parametric model . F2 (t) = t − 1. Let us examine each model. represents a straight line in the affine plane. in which case P doesn't have a standard name. INTRODUCTION TO POLYNOMIAL CURVES
Figure 3. where A2 denotes the affine plane). where P is another space called the parameter space (typically P = A for a curve. or a function F : E → P . for some parameter value a ∈ P (possibly many parameter values).
. For example. In the first case where F : P → E.1: A parabola a function F : P → E. Thus. we say that we have an implicit model . represents a parabola in the affine plane. The function F : A → A2 defined such that F1 (t) = 2t. In the parametric model .64
CHAPTER 3. and P = A2 for a surface. Every point lying on the shape S is represented as F (a). is defined as the range F (P ) of the function F .
F (a) = 0}. the space P is a vector space. This would have certain mathematical advantages. In order to avoid such situations. the function F : A2 → A3 defined such that F1 (u. since we are primarily interested in the real part of shapes (curves and surfaces). F3 (u. it possible that F −1 (0) = ∅. as the set of zeros of the function F : S = F −1 (0) = {a | a ∈ E. y) = x2 + y 2 + 1. In the implicit model .66
CHAPTER 3. v) = u3 − 3v 2 u. we could assume that our spaces are defined over the field of complex numbers (or more generally. F2 (u. Of course. F defines the empty curve.
. since the equation x2 + y 2 = −1 has no real solutions. INTRODUCTION TO POLYNOMIAL CURVES
2 y 0 -1 -2 4 1
3 z
2 1 0 -2 -1 0 x 1 2
Figure 3. an algebraically closed field). In this case. v) = v. represents what is known as the monkey saddle. v) = u. where E is of dimension n ≥ s.3: An elliptic paraboloid For a more exotic example. we will make the working assumption that we are only considering functions that define nonempty shapes. a shape S of dimension s specified by a function F : E → P . is defined as the zero locus F −1 (0) of the function F . Thus. if F : A2 → A is the function defined such that F (x. that is. and it has some dimension d ≥ n − s. For example. but does not help us much for visualizing the shape.
The function F : A2 → A defined such that F (x. it is not clear that it can be decided whether an arbitrary algebraic equation has real solutions or not. F2 (x. y ∈ A. Although this is not entirely obvious.5 -1 2 -1 1 0. However.3. As a simple example of a curve defined implicitly. z) = z − xy. the twisted cubic is defined implicitly by the function F : A3 → A2 . defines an ellipse in the affine plane. y) = 2y − x + 3
for all x. defines a straight line (in fact. defined such that F1 (x.5 0 0. y. y) = 2x2 + y 2 − 1. z) = y − x2 . For example. the implicit model is more natural to deal with certain classes of problems. defines the same parabola as the above parametric definition.
for all x.
. for all x. y) = 4y − x2 .5 1
67
1
z
0
-1
-2
Figure 3. These complications are one of the motivations for paying more attention to the parametric model.5 x -0. y. since we immediately see that y = x2 /4 for every point on this parabola.1. The function F : A2 → A defined such that F (x. the same line as defined above parametrically). where implicit equations of shapes are directly available.4: A monkey saddle There are some serious problems with such an assumption. the function F : A2 → A defined such that F (x. WHY PARAMETERIZED POLYNOMIAL CURVES?
y 0 -0. y ∈ A. y ∈ A.
since the function F corresponds to d scalar-valued functions (F1 . . The class of functions definable by polynomials is very well-behaved. using polynomials is rarely a restriction (continuous functions on reasonable domains can
. . . . The hyperbolic paraboloid discussed in the parametric model is defined by the function F (x. we can turn to the class of functions definable by rational funtions (fractions of polynomials). As a matter of fact. y = sin θ. F2 = 0.. dealing with rational fractions turns out to be largely reduced to dealing with polynomials. z) = z − x2 + y 2 . we may want to use a simpler class of functions. From a practical point of view.68
CHAPTER 3. z) = z − 2x2 − y 2 . This last example leads us to another major question:
What Class of Functions for F ? Although trigonometric functions are perfectly fine. and turns out to be sufficient for most purposes. the unit circle is defined implicitly by the equation x2 + y 2 − 1 = 0. y. y. If P has dimension d ≥ n − s. Fd ). For another familiar example. and it has the parametric representation x = cos θ. for computational (algorithmic) reasons.. z) = x2 + y 2 + z 2 − 1. The elliptic paraboloid discussed in the parametric model is defined by the function F (x. defined such that F (x. Certainly. we should only consider continuous functions which are sufficiently differentiable. instead of saying that S is the zero locus of the function F : E → P . . to yield shapes that are reasonably smooth. we usually say that S is defined by the set of equations F1 = 0. which is sufficient for most computational applications. or that S is the zero locus of the above system of equations. Fd = 0. When polynomials are insufficient. INTRODUCTION TO POLYNOMIAL CURVES The unit sphere is defined implicitly by the function F : A3 → A. y.
In general. We will make a very modest use of elementary notions of differential geometry. In the implicit model. since polynomials are differentiable at any order. and is done using the algebraic concept of a resultant. studying shapes S defined by systems of polynomial equations is essentially the prime concern of algebraic geometry. degree depends on how many pieces are involved and how complex each pieces joined small pieces. the function F : P → E provides immediate access to points on the shape S: every parameter value a ∈ P yields a point F (a) lying on the shape S. The discrepancy shows up for m = 2 for polynomial functions. In the parametric model. the choice of piece is. way beyond the scope of this course. Having decided that we will use polynomial (or rational) functions.
1
. and for m = 3 for rational functions. another venerable subject.
Because the study of shapes modelled as zero loci of polynomial equations is quite difficult. Given a point b ∈ E.1.3. Also. where the function F : E → P does not provide immediate access to points in S. For example. there are some subtleties regarding over which field we view the curves or surfaces as being defined. then some unexpected objects show up as the traces of
Actually. What Degree for F ? In most practical applications involving many small together (splines). for a single piece. and an even more modest use of very elementary notions of algebraic geometry. which may be very difficult. there is one more question. venerable. it can be shown that it is always possible to find an algebraic implicit definition G for S. what we use primarily is some elements of multilinear algebra. functions that are only continuous. we will be dealing with polynomial functions (and rational functions). as we shall see very soon. it is often useful to find an implicit definition G : E → P for S. and very difficult subject. it is easier to "carve out" portions of the shape. we will focus primarily on the study of shapes modelled parametrically. by simply considering subsets of the parameter space. WHY PARAMETERIZED POLYNOMIAL CURVES?
69
be approximated by polynomial functions). There are other advantages in dealing with the parametric model. given a parametric definition F : P → E for a shape S. for example. the degree could be quite high.1 This is called implicitization. to determine whether b ∈ S usually requires solving a system of algebraic equations. Nevertheless. the degree m = 2 or m = 3 is sufficient for the However. It should be noted that it is not always possible to go from an implicit algebraic definition G : E → P for S to a parametric algebraic definition F : P → E for S. In the case where F is given by rational functions. Thus. This is not so for the implicit model. Bold readers are urged to consult Fulton [37] or Harris [41]. studying shapes S defined by systems of parametric polynomials (or rational fractions) is in some sense subsumed by differential geometry. It should also be noted that if we consider parametric curves F : A → E defined by functions that are more general than polynomial or rational functions. This is a fascinating. In fact.
The point F of coordinates (a. i. of equation y 2 = 4ax. Polynomials certainly qualify! Before we launch into our study of parametric curves. and we briefly mention the classical bifocal definition. continuous curves F : [0. then the parabola of equation y 2 = 4ax is the set of points in the plane such that MH = MF . of total degree m ≤ 2.. and the line D of equation x = −a (parallel to the y-axis). is called the directrix of the parabola. if H is the intersection of the perpendicular through M to D. x2 y 2 + 2 =1 a2 b (Without loss of generality. with real coefficients. Such an equation defines a (plane) conic (because it is the curve obtained by intersecting a circular cone. For example. and AB denotes the Euclidean length of the line segment AB. with a plane). that is. 0) is called the focus of the parabola. It is very easy to see that straight lines are also definable in the parametric model. we believe that it may be useful to review briefly what kind of plane curves are defined in the implicit model. INTRODUCTION TO POLYNOMIAL CURVES
curves. y) is a polynomial in x. For any point M in the plane.e. For m = 1. where f (x. Hyperbolas. y. we can assume a ≥ b). 1] → A2 whose trace is the entire unit square (including its interior)! See the problems.70
CHAPTER 3. Peano and Hilbert (among others) showed that there are space filling curves. the conics can be classified in three types (after some suitable change of coordinates): Parabolas.2
The above definition of the conics assumes a Euclidean structure on the affine plane. This is a good motivation for assuming that the functions F are sufficiently differentiable. of equation x2 y 2 − 2 = 1.
2
. It is possible to give more geometric characterizations of the conics. y) = 0. Except for some degenerate cases. which defines a straight line. we have the equation ax + by + c = 0. The general curve of degree 2 is defined by an equation of the form ax2 + bxy + cy 2 + dx + ey + f = 0. of equation
The above definitions are algebraic. what sort of curves are defined by polynomial equations f (x. a2 b Ellipses. Let us begin with the parabola.
0) and (c. WHY PARAMETERIZED POLYNOMIAL CURVES? D H M
71
F
Figure 3. let c = a2 − b2 (so that a2 = b2 + c2 ). Then. This is the "gardener's definition" of an ellipse. the parabola of equation y 2 = 4x has focus F = (1. make two knots at distance 2a from each other. the ellipse defined by x2 y 2 + 2 =1 a2 b is the set of points in the plane such that MF + MF ′ = 2a (note that c ≤ a). one can take a string. one can draw the ellipse.5: A Parabola For example. 0).1.
.3. using a stick. and directrix D of equation x = −1: √ In the case of an ellipse. since we are assuming that a ≥ b. Indeed. Each of F. F ′ is a focus of the ellipse. and drive a nail through each knot in such a way that the nails are positioned on the foci F and F ′ . 0). on the ground. Then. Let F and F ′ be the points of coordinates (−c. by moving the stick along the string (perpendicular to the ground) in such a way that the string is kept stiff.
y) + ϕ0 = 0. y) + ϕ2 (x. the reader is referred to standard geometry texts. where each ϕi (x. The curve defined by such an equation is called a (plane) cubic.7: A Hyperbola is called the exentricity of the conic. y) + ϕ1 (x. For more details. for example. 3. we can pick any point of the cubic as the origin. In general. y) + ϕ1 (x. and hyperbolas. On the other hand. WHY PARAMETERIZED POLYNOMIAL CURVES?
73
M
F
F′
Figure 3. 2. y) is a homogenous polynomial of total degree i. We will also characterize those cubics that can be defined parametrically in terms of polynomials. y) is the null polynomial. There is also a monofocal definition of ellipses. things become a lot tougher! The general curve of degree 3 is defined by an equation of the form ϕ3 (x. It can be shown that the cubics that have rational representations are characterized by the fact that ϕ1 (x. We will see very soon that only the parabolas can be defined parametrically using polynomials. y) + ϕ2 (x. y) = 0. Berger or Pedoe. a plane cubic cannot be defined parametrically using rational functions (and a fortiori. and the equation of the cubic becomes ϕ3 (x. with i = 1. all the conics can be defined parametrically using rational fractions.3.
. using polynomials). involving the exentricity. parabolas. Since we assumed that we are only considering nonempty curves.1. and ϕ0 is a scalar. For m = 3. In this case ϕ0 = 0.
We will now try to gain some insight into polynomial curves by determining the shape of the traces of plane polynomial curves (curves living in E = A2 ) of degree m ≤ 3. it is preferable to consider the parameter space R as the affine line A. s]) in E is called the trace of the polynomial curve segment F [r. a polynomial curve is obtained by bending and twisting the affine line A using a polynomial map. It should be noted that if d is the maximum degree of the polynomials F1 . b0 ). many different parametric representations may yield the same trace. and it will be convenient to "raise" the degree of some of these curve segments to a common degree. s) for A with r < s. → → F (t) = a + F (t)− + · · · + F (t)− . will in fact be convenient later on for CAGD applications.1. s] → E of a polynomial curve F : A → E of degree at most m.
We begin with m = 1.
3.
. . and it is only required that d ≤ m. s] of degree (at most) m is the restriction F : [r. . and write F (t) instead of F (t). For example. 1 ≤ i ≤ n. and similarly. e e
0 1 1 n n
where t = (1 − t)0 + t1. and every Fi (X) is a polynomial in R[X] of degree ≤ m. . (−1 . such that for every t ∈ A.74
CHAPTER 3. we will need to join curve segments of possibly different degrees. we will introduce a major technique of CAGD. y(t) = F2 (t) = b1 t + b0 . and we begin with a rigorous definition. As a matter of fact. Otherwise. . For reasons that will become clear later on. This decision.2
Polynomial Curves of degree 1 and 2
x(t) = F1 (t) = a1 t + a0 . . A polynomial curve F of degree ≤ 1 is of the form
If both a1 = b1 = 0. which may seem unusual. 1). the trace of F reduces to the single point (a0 . we view t ∈ A as the real number t ∈ R. Recall that every point t ∈ A is expressed as t = (1 − t)0 + t1 in terms of the affine basis (0. it is now time to focus on the parametric model for defining curves. . a1 = 0 or b1 = 0. INTRODUCTION TO POLYNOMIAL CURVES
After there general considerations. Definition 3. On the way.1. a (parameterized) polynomial curve of degree at most m is a map e en F : A → E. Given any affine space E of finite dimension n and any affine frame → → (a0 . . the set of points F ([r. and we can eliminate t between x and y. sometimes called the geometric curve associated with F : A → E. getting the implicit equation a1 y − b1 x + a0 b1 − a1 b0 = 0. in most cases. Given any affine frame (r. s]. The set of points F (A) in E is called the trace of the polynomial curve F . − )) for E. Fn defining a polynomial curve F of degree m. then d = m is possible. we are more interested in the trace of a curve than in its actual parametric representation. A parameterized polynomial curve is defined as follows.
For simplicity of notation. a (parameterized) polynomial curve segment F [r. blossoming. Intuitively. Also. .
Thus. In fact. with b > 0. The corresponding implicit equation is Y = b 2 X . letting X(u) = x1 (u − µ) − a′0 . The diagram below shows the parabola defined by the following parametric equations F1 (t) = 2t. Y (u) = au2 . and we get a parametric representation of the form X(u) = au. since by an appropriate change of coordinates. the nondegenerate polynomial curves of true degree 2 are the parabolas. In summary. hyperbolas) are precisely the curves definable as rational curves of degree 2 (parametric definitions involving rational fractions of quadratic polynomials).
. ellipses and hyperbolas are not definable as polynomial curves of degree 2. we can change coordinates again (a translation). passing through the origin. and having the Y -axis as axis of symmetry. every parabola can be defined as the parametric polynomial curve X(u) = u. F2 (t) = t2 .76
CHAPTER 3.
Intuitively. it will be shown later that the conics (parabolas ellipses. the previous degenerate case corresponds to
b a2
Conversely. y1 (u − µ) = bu2 + b′0 . every parabola is defined by the implicit equation Y = aX 2 . Finally. INTRODUCTION TO POLYNOMIAL CURVES
Then. Y (u) = y1 (u − µ) − b′0 . it can be shown easily that ellipses and hyperbolas are not definable by polynomial curves of any degree. Y (u) = bu2 .
= ∞. we have a parametric representation of the form x1 (u − µ) = au + a′0 . a2
This is a parabola. On the other hand. with b > 0.
where r = s. we must have t = (1 − λ)r + λs = r + λ(s − r).
3. let us begin with straight lines. Thus.8: A parabola Remark: In the degenerate case of a half line. multilinearize) polynomials. Given any affine frame (r. We now show that there is another way of specifying quadratic polynomial curves which yields a very nice geometric algorithm for constructing points on these curves. x2 (t) = a2 t + b2 . s) for A. FIRST ENCOUNTER WITH POLAR FORMS (BLOSSOMING)
77
Figure 3. where i = 1. F : A → A3 is itself an affine map.3
First Encounter with Polar Forms (Blossoming)
As a warm up.3. The general philosophy is to linearize (or more exactly. In this case. the mismatch goes away. 2. In fact. Observe that each function t → ai t + bi is affine. The parametric definition of a straight line F : A → A3 . every t ∈ A can be writen uniquely as t = (1 − λ)r + λs.3. is of the form x1 (t) = a1 t + b1 .
. 3. the mismatch between the parametric representation and the implicit representation can be resolved. x3 (t) = a3 t + b3 . since every polynomial has roots (possibly complex numbers). if we view our curves as the real trace of complex curves.
at " s−r of the way from F (r)").78 and so.
This means that F (t) is completely determined by the two points F (r) and F (s). For this. and t = 0. and picking b = F (r). maps that are affine in each of their arguments. we have F (t) = F ((1 − λ)r + λs). Substituting the value of λ in the above. then F (0.
t−r which shows that F (t) is on the line determined by F (r) and F (s). We now show how to turn a quadratic polynomial into a biaffine map. since F is affine. we know that −− −→ −− −→ −− −→ bF (t) = (1 − λ)bF (r) + λbF (s). s) for A. and the ratio of interpolation λ. s = 1. that is. by several interpolation steps from some (finite) set of given points related to the curve F . we have −− −→ −−− −− −→ −−− F (r)F (t) = λF (r)F (s). when r ≤ t ≤ s. since F (t) is the barycenter of F (r) and F (s) assigned the weights 1 − λ and λ. λ =
t−r s−r
CHAPTER 3. it is first necessary to turn the polynomials involved in the definition of F into multiaffine maps. at " s−r of the way from t−r F (r)" (in particular. = (1 − λ) F (r) + λ F (s).25) is a fourth of the way from F (0) between F (0) and F (1). INTRODUCTION TO POLYNOMIAL CURVES (and 1 − λ =
s−t ). F (t) is indeed between F (r) and F (s). we have −− −→ −−− F (r)F (t) = t−r s−r −− −→ −−− F (r)F (s).
.
as illustrated in the following diagram. For example. every point F (t) on the line defined by F is obtained by a single interpolation step F (t) = s−t s−r F (r) +
t−r s−r
t−r s−r = 1: 3
F (s). where r = s. for every point b. to determining the point F (t) on a polynomial curve F .25. s−r
Now. in the case of an affine map F : A → A3 . where
We would like to generalize the idea of determining the point F (t) on the the line defined by F (r) and F (s) by an interpolation step. Furthermore.
Thus. if r = 0. given any affine frame (r.
for all X ∈ R. consider the polynomial F (X) = X 2 + 2X − 3.3. but this is not true. x2 ) = f (x2 .3. X). for all X ∈ R. It would be tempting to say that f1 is linear in each of x1 and x2 . FIRST ENCOUNTER WITH POLAR FORMS (BLOSSOMING) F (s)
79
F (t) F (r)
Figure 3. simply form f (x1 . For an arbitrary polynomial F (X) = aX 2 + bX + c
. X). It would be nicer if we could find a unique biaffine function f such that F (X) = f (X. or polar form. such a function should satisfy some additional property. 2
The symmetric biaffine function f (x1 . for all X ∈ R. It turns out that requiring f to be symmetric is just what's needed. x2 . Observe that the function of two variables f1 (x1 . To make f1 (and f2 ) symmetric. in the sense that F (X) = f1 (X. x2 ) + f1 (x2 . x1 ) = x1 x2 + x1 + x2 − 3. due to the presence of the term 2x1 and of the constant −3. and F (X) = f2 (X. We say that a function f of two arguments is symmetric iff f (x1 .9: Linear Interpolation As an example. X). for all x1 . and f1 is only biaffine. x2 ) = x1 x2 + 2x1 − 3 gives us back the polynomial F (X) on the diagonal. of F . x2 ) = f1 (x1 . x2 ) = x1 x2 + x1 + x2 − 3 is called the (affine) blossom. but f1 is also affine in each of x1 and x2 . Note that f2 (x1 . x1 ). x2 ) = x1 x2 + 2x2 − 3 is also biaffine. and of course.
s]. but the shells are not nicely nested. t2 ) = t1 + t2 . t2 ) = −t1 t2 . s = 1.4. t2 ):
. FIRST ENCOUNTER WITH THE DE f (r. t)
f (t. assuming different control points. Note that the shells are nested nicely. Actually. The de Casteljau algorithm can also applied to compute any polar value f (t1 . r) f (t. and the resulting diagram is called a de Casteljau diagram. r) Figure 3. t) determined by λ. The parabola of the previous example is actually given by the parametric equations F1 (t) = 2t. The first polyline is also called a control polygon of the curve.10: A de Casteljau diagram
f (s.3. Each polyline given by the algorithm is called a shell . on the curve F . f2 (t1 . we still obtain two polylines and a de Casteljau diagram. for r = 0. when t is outside [r. s)
with the desired point f (t. F2 (t) = −t2 . The example below shows the construction of the point F (t) corresponding to t = 1/2. s)
f (r.
This example also shows the construction of another point on the curve. The polar forms are f1 (t1 . s) CASTELJAU ALGORITHM
83
f (t.
can be viewed as meaningful labels for the node of de Casteljau diagrams.
. we can say that the polar form f (t1 . t2 ) of the polynomial function defining a parabola gives precisely the intersection point of the two tangents at F (t1 ) and F (t2 ) to the parabola. 2 y = bt1 t2 . The equation of the tangent to the parabola at (x(t). Let us consider the parabola given by x(t) = at y(t) = bt2 . s−r Thus. FIRST ENCOUNTER WITH THE DE CASTELJAU ALGORITHM
85
from the points f (t1 . t1 ) and f (t1 . we solve the system of linear equations ay − 2bt1 x + abt2 = 0 1 ay − 2bt2 x + abt2 = 0. or ay − 2btx + abt2 = 0. Turning this property around. We recommend reading de Casteljau's original presentation in de Casteljau [23]. A nice geometric interpretation of the polar value f (t1 . For this. t2 ). s) computed during the first stage.3. It is in this sense that the term blossom is used: by forming the blossom of the polynomial function F . To find the intersection of the two tangents to the parabola corresponding to t = t1 and t = t2 .4.
Thus. also called blossom values. it does not work in general. we need to look closely at the intersection of two tangents to a parabola. There is a natural generalization of this nice geometric interpretation of polar forms to cubic curves. t2 ) can be obtained. y(t)) is x′ (t)(y − y(t)) − y ′ (t)(x − x(t)) = 0. but unfortunately. a(y − bt2 ) − 2bt(x − at) = 0. r) = f (r. the coordinates of the point of intersection of any two tangents to the parabola are given by the polar forms of the polynomials expressing the coordinates of the parabola. some hidden geometric information is revealed. the ratio of interpolation being t2 − r λ2 = . that is. the polar values f (t1 . 2 and we easily find that x=a t1 + t2 .
86
CHAPTER 3. cubic curves. 0). b1 = (1. we first show that by a change of coordinates
. The polar forms of F1 and F2 are f1 (t1 . let us assume that a3 = 0 or b3 = 0. and b2 = f (1.
Let us now consider m = 3.4). b1 = f (0. b2 ). As in the case of quadratic curves. The de Casteljau algorithm. for curves not contained in a plane (it involves intersecting the osculating planes at three points on the curve).
A polynomial curve F of degree ≤ 3 is of the form
Since we already considered the case where a3 = b3 = 0. F2 (t) = t2 . also has the property that the line determined by f (r. have coordinates: b0 = (0. t2 ) = t1 t2 . b1 ). t) and f (s. 1). to the middle of the segment (b1 . 0). t) is the tangent at F (t) (this will be shown in section 5. f2 (t1 . Consider the parabola given by F1 (t) = 2t.5
Polynomial Curves of Degree 3
x(t) = F1 (t) = a3 t3 + a2 t2 + a1 t + a0 . s) is the tangent at F (s). y(t) = F2 (t) = b3 t3 + b2 t2 + b1 t + b0 . t2 ) = t1 + t2 . INTRODUCTION TO POLYNOMIAL CURVES
and when it does. Example. the point F (1/2) = (1.
3. b2 = (2. 1). The control points b0 = f (0. Let us give an example of the computation of the control points from the parametric definition of a quadratic curve. 1/4) on the parabola. For t = 1/2. in addition to having the nice property that the line determined by F (r) and f (r. and that the line determined by F (s) and f (r. s) is the tangent at F (r). is the middle of the line segment joining the middle of the segment (b0 . 1). that is. 0).
3a X2 = − −b . If b > 0. then Y ′ (X) has two roots. and we get an implicit equation of the form y = a′ x3 + b′ x2 + c′ x + d′ . then Y ′ (0) = 0. with a′ > 0. which means that the origin is an inflexion point. and Y ′′ (0) = 0. where Y = y − c. Its shape will depend on the variations of sign of its derivative Y ′ = 3aX 2 + b. and Y (X) is strictly increasing with X. we can suppress the term b′ X 2 by the change of coordinates b′ x = X − ′. and increasing again when X varies from X2 to +∞. It has a flat S-shape. 3a
Then. and the tangent at the origin is the X-axis. 3a We get an implicit equation of the form y = aX 3 + bX + c. we get the implicit equation Y = aX 3 + bX. If b < 0. the origin is an inflexion point. we can eliminate t between x1 (t) and y1 (t). X1 = + −b . then Y ′ (X) is always strictly positive. Then. with a > 0. the slope b of the tangent at the origin being positive. As in the case of quadratic curves. Case 2. The curve still has a flat S-shape. note that a line parallel to the Y -axis intersects the curve in a single point. and 0 is a double root of Y ′ . In all three cases. This is the reason why we get a parametric representation. with a > 0. the slope b of the tangent at the origin being negative. Y (X) is increasing when X varies from −∞ to X1 . the other case being similar. let us assume that a1 b3 − a3 b1 > 0.
.88
CHAPTER 3. a2 b3 − a3 b2 = 0. The curve has an S-shape. If b = 0. This curve is symmetric with respect to the origin. INTRODUCTION TO POLYNOMIAL CURVES
If a1 b3 − a3 b1 = 0. By one more change of coordinates. decreasing when X varies from X1 to X2 . The following diagram shows the cubic of implicit equation y = 3x3 − 3x. Also. since Y ′′ = 6aX.
b3 b1 b2 y=Y − . we can show the following lemma.5. as it will be clearer in a moment. Lemma 3. b3
Furthermore. the curve defined by the above implicit equation. The origin (X. is equal to the trace of the polynomial curve F .e. 0). Proof. where b3 > 0. We leave it as an exercise. and when b1 > 0. after the translation of the origin given by x=X− b1 a2 . any polynomial curve of the form x(t) = F1 (t) = a2 t2 .
In fact. we have the implicit equation (a2 y − b2 x)2 = a2 x b3 x + b1 a2
2
. y(t) = F2 (t) = b3 t3 + b2 t2 + b1 t. b3
with b3 > 0. Y ) = (0. then the curve defined by the above implicit equation is equal to the trace of the polynomial curve F . y.90
CHAPTER 3. Then. Y ) = (0. if b1 ≤ 0. INTRODUCTION TO POLYNOMIAL CURVES
In terms of the original coordinates x.. b3
and we get the implicit equation a2 b2 a2 Y − X b3 b3
2
+
b1 a2 2 X = X 3. b3
the trace of F satisfies the implicit equation a2 b2 a2 Y − X b3 b3
2
+
b1 a2 2 X = X 3. It is straightforward and not very informative.1. Given any nondegenerate cubic polynomial curve F . 0) is called a singular point of the curve defined by the implicit equation. it is preferable to make the change of coordinates (translation) x=X− b1 a2 . excluding the origin (X. i. b3 b1 b2 y=Y − .
.
which is also the set of points on the curve defined by the implicit equation c(aY − bX)2 + cdX 2 = X 3 . Y ) = (0. with the exception that when d > 0. unless it is a tangent at the origin to the trace of the polynomial curve F (which only happens when d ≤ 0). Y ) = cd. 0) must be excluded from the trace of the polynomial curve.3. 0). we discover a nice parametric representation of the polynomial curve in terms of the parameter m. a of the trace of F (also on the axis of symmetry) is vertical. when d > 0. For every nondegenerate cubic polynomial curve F .1 shows that every nondegenerate polynomial cubic is defined by some implicit equation of the form c(aY − bX)2 + cdX 2 = X 3 . Again. lemma 3. Y ) = (0.5. Y ) = (0. a (X. The reason for choosing the origin at the singular point is that if we intersect the trace of the polynomial curve with a line of slope m passing through the singular point. such that F and G have the same trace. Y1 ) and (X.2.5. Y ) = (0. Lemma 3.5. Y2 ) such that 2b Y1 + Y2 = X. Lemma 3. Y1 ) belongs to the trace of F iff (X. there is some parametric definition G of the form X(m) = c(a m − b)2 + cd. Y2 ) belongs to the trace of F . this mismatch can be resolved if we treat these curves as complex curves. every line of slope m passing through the origin (X. the singular point (X. The tangent at the point bcd (X. with the exception that when d > 0. Since every line of slope m through the origin has
. Y ) = (0. Y (m) = m(c(a m − b)2 + cd). Proof. 0). in the sense that for any two points (X. POLYNOMIAL CURVES OF DEGREE 3
91
Thus. excluding the origin (X. 0) must be excluded from the trace of the polynomial curve.5. The line aY −bX = 0 is an axis of symmetry for the curve. The case where d > 0 is another illustration of the mismatch between the implicit and the explicit representation of curves.1 shows that every nondegenerate polynomial cubic is defined by some implicit equation of the form c(aY − bX)2 + cdX 2 = X 3 . the singular point (X. Furthermore. 0) intersects the trace of the polynomial curve F in a single point other than the singular point (X.
and Y ′ (m) has two roots. which means that the singular point (X. In this case. the polynomial Y ′ (m) has no roots. tangent to the vertical line X = cd at the intersection of this line with the axis of symmetry aY − bX = 0. Y ′ (m) has a double root m0 = 3a . and increases when m varies from m2 to +∞. but between the two vertical lines X = cd and X = cd + cb2 . tangent to the vertical line X = cd at the intersection of this line with the axis of symmetry aY − bX = 0. m2 . that is. the case c < 0 being similar. since d > 0. Y ) = (0. and Y ′ (m) is positive except for m = m0 . 0) is not on the trace of the cubic. 0) is not on the trace of the cubic either. the polynomial Y ′ (m) has two roots m1 . We also get a kind of "humpy" curve. In this case. and only intersecting the X-axis for m = 0. a
. for X = cd + cb2 . 3a b m2 = . decreases when m varies from m1 to m2 . In this case. When b2 − 3d > 0. When b2 − 3d < 0. then Y (m) increases when m varies from −∞ to m1 . and only intersecting the X-axis for m = 0. CLASSIFICATION OF THE POLYNOMIAL CUBICS
93
3.3. and there is an inflexion point for m = m0 . and since we assumed c > 0. and since we assumed c > 0. which are easily computed: m1 = b . The cubic of equation 3(Y − 2X)2 + 3X 2 = X 3 is shown below: Case 3: d = 0 (a cuspidal cubic). The cubic of equation 3(Y − X)2 + 6X 2 = X 3 is shown below: Case 2: b2 ≥ 3d > 0. We get a kind of "humpy" curve. 2b When b2 − 3d = 0.6
Classification of the Polynomial Cubics
We treat the case where c > 0. which means that Y (m) is strictly increasing.6. we must have d > 0. for X = cd + cb2 . the singular point (X. that is. Thus. Y ) = (0. the curve makes two turns. we have b2 − 3d > 0. We also get a kind of "humpy" curve. Case 1: 3d > b2 . the polynomial Y ′ (m) is always positive. Y (m) increases when m varies from −∞ to ∞.
17: Nodal Cubic (d < 0) increases when m varies from −∞ to m1 . Remark: The implicit equation c(aY − bX)2 + cdX 2 = X 3
. from "humpy" to "loopy". The curve also intersects the X-axis for m = 0. that is.6. and increases when m varies from m2 to +∞. CLASSIFICATION OF THE POLYNOMIAL CUBICS
97
Figure 3. tangent to the vertical line X = cd at the intersection of this line with the axis of symmetry aY − bX = 0. having the origin as a double point.3. through "cuspy". The trace of the cubic is a kind of "loopy curve" in the shape of an α. The cubic of equation 3 (Y − X)2 − 3X 2 = X 3 4 is shown below: One will observe the progression of the shape of the curve. decreases when m varies from m1 to m2 . for X = cd + cb2 .
x1 ). there are some cubics that cannot be represented even as rational curves. Y ) = X 3 . We easily verify that f must be given by f (x1 . Note that X 3 itself is a homogeneous polynomial in X and Y of degree 3. which is affine in each argument.
3. X. fractions of polynomials of degree ≤ 3) are exactly those cubics whose implicit equation is of the form ϕ2 (X. cannot be parameterized rationally. To make this statement precise. x2 . Y ) are homogeneous polynomial in X and Y of total degree respectively 2 and 3. x3 . F (X) = aX 3 + bX 2 + cX + d. i. Y ). Using some algebraic geometry. x2 . x2 . x3 ) = f (x1 . Elliptic curves are a venerable and fascinating topic. x2 ) = f (x3 . the polynomial case is obtained in the special case where ϕ3 (X. Such cubics are elliptic curves. 1. 3 3
. x1 . such that f (x1 . Thus. where ϕ2 (X. the singular point is at infinity. Y ) is a homogeneous polynomial in X and Y of total degree 2 (in the case of a degenerate cubic of equation y = aX 3 + bX 2 + cX + d. x3 ) = f (x2 . Y ) and ϕ3 (X. x1 ) = f (x3 . and such that F (X) = f (X. x2 . Y ) = ϕ3 (X.. These cubics have a singular point at the origin. the polar form of F is a symmetric triaffine function f : A3 → A. INTRODUCTION TO POLYNOMIAL CURVES
of a nondegenerate polynomial cubic (with the exception of the singular point) is of the form ϕ2 (X. a function which takes the same value for all permutations of x1 . X). we would like to extend blossoming to polynomials of degree 3.. Furthermore. but definitely beyond the scope of this course! Returning to polynomial cubics. it can be shown that the (nondegenerate) cubics that can be represented by parametric rational curves of degree 3 (i. the cubics defined by the implicit equation Y 2 = X(X − 1)(X − λ). x3 . where λ = 0. For example. for all X ∈ R. inspired by our treatment of quadratic polynomials.e. projective geometry is needed). Given any polynomial of degree ≤ 3. x3 . where ϕ2 (X. x3 ) = ax1 x2 x3 + b x1 x2 + x1 x3 + x2 x3 x1 + x2 + x3 +c + d. we need to define the polar form (or blossom) of a polynomial of degree 3.e. that is.98
CHAPTER 3. x2 ) = f (x2 . x1 . Y ) = X 3 .7
Second Encounter with Polar Forms (Blossoming)
First.
b1.j−1 .8. t) as b0.j .1 ).j−1 and bi+1. the index j denotes the stage of the computation.3 . the algorithm constructs the three polylines (b0 . Note that in bi. b3 ) is also called a control polygon of the curve.j−1 .3 . b2 ). b1. f (t.2 b2. b1. and 0 ≤ i ≤ 3 − j.
called shells. and F (t) = b0.3 .1
2
3
b0.j lies between bi. We have F (t) = b0. (b2 .
As will shall see in section 5. (b1 .3. but the shells are not nicely nested. Then the triangle representing the computation is as follows:
0 b0 = b0.3 b1. r).2 b1. they form the de Casteljau diagram. we still obtain three shells and a de Casteljau diagram. Note that the shells are nested nicely. b1 ).2 )
where 1 ≤ j ≤ 3. When λ is outside [r. The following diagram illustrates the de Casteljau algorithm for computing the point F (t) on a cubic. When r ≤ t ≤ s. b2 ).j : bi.1. and s = 6: The above example shows the construction of the point F (3) corresponding to t = 3.1 . (b2 .1 ) (b0. and the point f (t. The polyline (b0 .j = s−t s−r bi. t.0
Then. (b1 . t. we have the following inductive formula for computing bi.0
1 b0.1
b2 = b2. where r = 0.2 . b3 ) (b0.0 b1 = b1.3 . In this case. and bi.
. each interpolation step computes a convex combination. (b1.2 . b2.j−1 + t−r s−r bi+1. s].1 . and with the point b0.1 b0. SECOND ENCOUNTER WITH THE DE CASTELJAU ALGORITHM
103
the intermediate points f (t. on the curve F . b1 ).0 b3 = b3. geometrically.2 . s) as b0. the above formula generalizes to any degree m. t.
2 ). s.1 . t2 . s. s)
1 f (r. s) f (t1 . b2. t3 ) = b0. t.3 . s)
As above.2 . r). It is possible for some (even all!) of the control points to coincide. it is convenient to denote the intermediate points f (r. f (t1 . t1 ). t1 .2 = b1. b1 ). b1.4. f (t1.2 . s) f (s.j . and f (t1 . the de Casteljau algorithm can also be used to compute any polar value f (t1 . t3 ) as b0. r. b1. s. s). b2 = b3 . SECOND ENCOUNTER WITH THE DE CASTELJAU ALGORITHM
105
The de Casteljau algorithm also gives some information about some of the tangents to the curve. s) as b0. t2 . r) f (r. t2 .8. t2 .2 are computed during the second stage of the de Casteljau algorithm. t) correctly. given by tj − r . Then the triangle representing the computation is as follows:
.2 .
Remark: The above statements only make sense when b0 = b1 . that the tangent at b0 is the line (b0 . t3 ) (which is not generally on the curve). r. All we have to do is to use a different ratio of interpolation λj during phase j.3. t2 . It will be shown in section 5. As in the quadratic case.1 . s. Note that in bi. t1 )
2
3
f (t1 . s)
f (r. s) f (t1 . t3 ) f (t1 . the tangent at b3 is the line (b2 . t2 . where b0.2 . r. and b0. b3 ). but the tangents may not be computed as easily as above. f (r.2 and b1. and the point f (t1 . the intermediate points f (t1 . t2 . s) as b0.3 . t1 . b1. and the tangent at F (t) is the line (b0. s−r
λj =
The computation can also be represented as a triangle:
0 f (r. the index j denotes the stage of the computation. t2 . r.1 . The algorithm still computes f (t. r) f (r.
6) = (6. 0) = (7. 0) = (11.
The second "hump" of the curve is outside the convex hull of this new control polygon. 6). 5) = (8. where b2 ≥ 3d > 0. We leave as an exercise to verify that this cubic corresponds to case 2.3. 4) = (3. 6). the coordinates of the control points are: b0 b1 b2 b3 The curve has the following shape. = (15.
. s = 1: b′0 b′1 b′2 b′3 = (15.9. s = 2. The axis of symmetry is y = 2x. EXAMPLES OF CUBICS DEFINED BY CONTROL POINTS
109
Figure 3. 10) = (3. It is interesting to see which control points are obtained with respect to the affine frame r = 0.22: Bezier Cubic 3 With respect to the affine frame r = 0.
. Consider the cubic. t2 . F3 (t) = t3 . t3 ) = t1 t2 t3 . This curve has some very interesting algebraic properties. 0) 3 3 b3 = (1. defined as follows: F1 (t) = t. 1). known as a twisted cubic (it is not a plane curve). b3 ). b1 . t3 ) = (t1 + t2 + t3 ) 3 1 f2 (t1 . The reader should apply the de Casteljau algorithm to find the point on the twisted cubic corresponding to t = 1/2. 0) 3 2 1 b2 = ( . 0. F2 (t) = t2 . this time. t2 . For example. the coordinates of the control points are: b0 = (0. It can also be shown that any four points on the twisted cubic are affinely independent. s = 1. . 0) 1 b1 = ( . is it possible to find the singular point geometrically? Is it possible to find the axis of symmetry geometrically? Let us consider one more example. Example 6. z − xy = 0. it is the zero locus (the set of common zeros) of the two polynomials y − x2 = 0. 3). With respect to the affine frame r = 0. EXAMPLES OF CUBICS DEFINED BY CONTROL POINTS
115
Challenge: Given a planar control polygon (b0 . t3 ) = (t1 t2 + t1 t3 + t2 t3 ) 3 f3 (t1 . t2 . of a space curve. This is the object of the next two chapters. we need to investigate some of the basic properties of multiaffine maps. and later on. to rational curves and also to polynomial and rational surfaces. We would like to extend polar forms and the de Casteljau algorithm to polynomial curves of arbitrary degrees (not only m = 2. We get the polar forms 1 f1 (t1 . 0.9. 1.3. b2 . In order to do so.
0). = (1. 0).10
Problems
Problem 1 (40 pts). Problem 2 (20 pts). −10). or any other available software in which graphics primitives are available. 10). Problem 3 (30 pts). (2) What are the slopes of the tangents at the origin. = (−6. (1) Find the polar forms of x(t) and y(t). s = 1. Plot the curve as well as possible (choose some convenient value for p). Plot the cubic as well as you can. −6). = (4. 6). y(t) = 9pt(3t2 − 1). Consider (in the plane) the cubic curve F defined by the following 4 control points. = (6. You may use Mathematica. Consider (in the plane) the cubic curve F defined by the following 4 control points. Remark: This cubic is known as the "Tchirnhausen cubic". = (−6. Problem 4 (20 pts). over some interval [r. and F (2/3). 6). INTRODUCTION TO POLYNOMIAL CURVES
3. 6). Write a computer program implementing the the de Casteljau algorithm for cubic curves. Find the control points with respect to the affine frame r = −1. Give a geometric construction of the tangent to the cubic for t = 0.
Use the de Casteljau algorithm to find the coordinates of the points F (1/3).116
CHAPTER 3. s].
. F (1/2). Consider the cubic defined by the equations: x(t) = 9p(3t2 − 1). where p is any scalar. assuming that r = 0 and s = 1: b0 b1 b2 b3 = (6. assuming that r = 0 and s = 1: b0 b1 b2 b3 = (−4. = (−1.
Use the representation in polar coordinates. where p is any scalar.
. over [0. (1 − cos θ)2
(ii) Prove that the tangents to the cubic at N1 . Plot the Bernstein polynomials Bi3 (t). 1 − cos θ sin θ(1 − 2 cos θ) y(θ) = 2p . Problem 10 (20 pts). You can check that you are right if you find that the cubic is defined by: x(θ) = 3p + 6p cos θ . does the curve have points where the curvature is null? What happens when three control points are identical? What if the four control points are identical? Problem 9 (20 pts). (i) Let Hi be the foot of the perpendicular from F to the tangent to the cubic at Ni . Prove that H1 . let ω be the intersection of the normal at M to the parabola with the mediatrix of the line segment F M. Hint. (iii) What is the locus of the center of gravity of this triangle. H3 belong to the parabola of equation y 2 = 4px.118
CHAPTER 3. N3 . INTRODUCTION TO POLYNOMIAL CURVES
Does the curve have a cusp at some control point? Does the curve have an inflexion point at some control point? Assuming that A2 is equipped with its usual Euclidean inner product. N3 intersect in three points forming an equilateral triangle. 1]. Show that when M varies on the parabola. N2 . and let N be the symmetric of F with respect to ω. It is best to use polar coordinates (with pole F = (p. N varies on the cubic. Any line D through F intersects the cubic in three points N1 . You may want to use Mathematica to play around with these curves. defined by the equations: x(t) = 27pt2 . Consider the cubic of problem 3 (except that a new origin is chosen). What is the maximum number of intersection points of two plane cubics? Give the control points of two plane cubics that intersect in the maximum number of points. Hint. Consider the following construction: for every point M on the parabola y 2 = 4px. 0)). H2 . 0). y(t) = 9pt(3t2 − 1). Let F be the point of coordinates (p. and that MN is tangent to the cubic at N. 0 ≤ i ≤ 3. Problem 11 (80 pts) Challenge. N2 .
We conclude by showing that the definition of a polynomial curve in polar form is equivalent to the more traditional definition (definition 3. and some discussion of the affine case can be found in Berger [5] (Chapter 3. The chapter proceeds as follows. and it is not very easily accessible. After a quick review of the binomial and multinomial coefficients. This has applications to curve and surface design. and the de Casteljau algorithm.1. Next.7. there is little doubt that the polar approach led him to the discovery of the beautiful algorithm known as the "de Casteljau algorithm". showing that they subsume the usual multivariate polynomial functions. in the case of polynomial maps between vector spaces. e This material is quite old. Affine polynomial functions h : E → F are described explicitly in the case where E has finite dimension. and more specifically to B´zier curves. Section 3).1 Multiaffine Maps
In this chapter. and that the Bernstein polynomials of degree ≤ m form a basis of the vector space of polynomials of degree ≤ m. but it plays a crucial role in section 10. section 6). multilinear and multiaffine maps are defined. we prove a generalization of lemma 2. It is shown how polynomials in one or several variables are polarized. Polynomial curves in polar form are defined. splines. The uniqueness of the polar form of an affine polynomial function is proved next. and the equivalence between polynomials and symmetric multiaffine maps is established. characterizing multiaffine maps in terms of multilinear maps. Affine polynomial functions and their polar forms are defined.Chapter 4 Multiaffine Maps and Polar Forms
4. which first appeared in 1939. It should be pointed out that de Casteljau pioneered the approach to polynomial curves and surfaces in terms of polar forms (see de Casteljau [23]). in a form usually called "blossoming". A presentation of the polarization of polynomials can be found in Hermann Weyl's The Classical Groups [87] (Chapter I. In fact. The proof can be omitted at first reading. going back as early as 1879. This result is quite technical in nature.1). page 4-6). An equivalent form of polarization is discussed quite extensively in Cartan [16] (Chapter I. 119
. and their characterization in terms of control points and Bernstein polynomials is shown.1. we discuss the representation of certain polynomial maps in terms of multiaffine maps.2.
For n ∈ N. . . n} consisting of k elements is namely k n n n−1 n−1 are also called binomial .120
CHAPTER 4. . and n (read "n choose k") as follows: k ∈ Z. . there are as many subsets of k elements from {1. and there are as many subsets of k of k − 1 elements from {2. there are two kinds of subsets of {1. MULTIAFFINE MAPS AND POLAR FORMS
We first review quickly some elementary combinatorial facts. . . n}. we define k n k 0 0 n k = 0. we define n! (read "n factorial") as follows: 0! = 1. n−1 . Thus. . = 1. n}. = n−1 n−1 . Furthermore. (n + 1)! = (n + 1)n! It is well known that n! is the number of permutations on n elements. k!(n − k)!
for 0 ≤ k ≤ n. . . . . The numbers . n}. when n ≥ 0. + k−1 k if n ≥ 1. . if k ∈ {0. . . n} not containing 1 as there are subsets of k elements from {2. n} having k elements: those containing 1. . . Now. and those not containing 1. = n−k k The binomial coefficients can be computed inductively by forming what is usually called
. . we have the empty set which has only one subset. . . n} containing 1 as there are subsets n−1 . . we can prove by induction that
n is the k number of subsets of {1. which is equal to + k k k−1 k coefficients. when n = 0. . namely itself. . the number of subsets of {1. . . For every n ∈ N. When n ≥ 1. Indeed. n} consisting of k elements. . . because they arise in the expansion of the binomial expression (a + b)n . . . It is easy to see that n n . . /
It is immediately shown by induction on n that n k = n! . . . . namely k−1 elements from {1.
s] as F ([r.1 when E is of finite dimension. Definition 4. it is convenient to denote the polynomial map defining the curve by an upper-case letter. as given in section 3. We define the trace of F as F (A). and E is any affine space (of dimension at least 2). .2. s ∈ A. definition 4.1. It would then be confusing to denote the affine space which is the range of the maps F and f also as F . for example. . we denote it as E (or at least.2. is that formal polynomials can be added and multiplied. and the polar form of F by the same. s]) in polar form of degree m is the restriction F : [r. a (parameterized) polynomial curve segment F ([r. MULTIAFFINE MAPS AND POLAR FORMS
between the polynomial functions of definition 4. and the the trace of F [r. Thus. f is also defined by a polynomial map of polar degree m − k).130
CHAPTER 4.2.4 turns out to be more general and equivalent to definition 3. in the sense that it could be equivalent to a polynomial map of lower polar degree (the symmetric multilinear maps fm .4. F = A3 . Definition 4. s] → E of an affine polynomial curve F : A → E in polar form of degree m. Given any r. Remark: When defining polynomial curves. in which case. Nevertheless.1).
Typically. Although we can make sense of addition as affine combination in the case of polynomial functions with range an affine space. we will allows ourselves the abuse of language where we abbreviate "polynomial curve in polar form" to "polynomial curve". defined by its m-polar form. Also note that we defined a polynomial curve in polar form of degree at most m. we are led to consider symmetric m-affine maps f : Am → F . such as F : A → E. with r < s. .1 (see definition 3. unless confusions arise. f . we can define (parameterized) polynomial curves in polar form. which is some symmetric m-affine map f : Am → E.1. Recall that the canonical affine space associated with the field R is denoted as A. . this generalization of the notion of a polynomial function is very fruitful. and it is also more convenient for the purpose of designing curves satisfying some simple geometric constraints. but lower-case letter.4 is not the standard definition of a parameterized polynomial curve. rather than a polynomial curve in polar form of degree exactly m. where F is any affine space. as the next example will show. s]).2.1 and formal polynomials. fm−1 . and thus. Indeed. where A is the real affine line. A (parameterized) polynomial curve in polar form of degree m is an affine polynomial map F : A → E of polar degree m. fm−k+1 involved in the unique decomposition of f as a sum of multilinear maps may be identically null. For convenience.
. because an affine polynomial map f of polar degree m may end up being degenerate. multiplication does not make any sense. we use a letter different from the letter used to denote the polynomial map defining the curve). the affine space E is the real affine space A3 of dimension 3. However.
s](t)
=
m k
s−t s−r
m−k
t−r s−r
k
are the Bernstein polynomials of degree m over [r. whose polar form f : Am → E satisfies the conditions f (r.m} i∈I I∩J=∅..
Summarizing the above considerations. we have
m m Bk [r.1.3. but we hope that the above considerations are striking motivations for using polar forms. Given any sequence of m + 1 points a0 . Of course. . we showed the following lemma.
and F (t) is given by the formula
m
F (t) =
k=0
m Bk [r. r. Furthermore. |J|=k
s − ti s−r
j∈J
tj − r s−r
ak . .
4. am in some affine space E.4
Uniqueness of the Polar Form of an Affine Polynomial Map
Remarkably. we can prove in full generality that the polar form f defining an affine polynomial h of degree m is unique. 1]). . s].. .
where the polynomials
m Bk [r. s](t) = Bk
t−r s−r
. . s. r = s). . .. the polar form f of F is given by the formula
m
f (t1 . . s) = ak . UNIQUENESS OF THE POLAR FORM OF AN AFFINE POLYNOMIAL MAP 133 The polynomials
m Bk (t) =
m (1 − t)m−k tk k
are the Bernstein polynomials of degree m (over [0. Lemma 4. . tm ) =
k=0 I∪J={1. . . .
m−k k
(where r. . We could use Cartan's proof in the case of polynomial
. .4. It is not hard to show that they form a basis for the vector space of polynomials of degree ≤ m. s](t) ak . . s ∈ A.. Clearly. there is a unique polynomial curve F : A → E of degree m.4. . we will come back to polynomial curves and the de Casteljau algorithm and its generalizations.
x2 ) = 1 4h 2 x1 + x2 2 − h(x1 ) − h(x2 ) . Section B.4. x. Given two affine spaces E and F . section §8. and can be found in Chapter B.1. Lemma 4. and is given by the following expression: f (a1 .1.5. x) for all x ∈ R.4. ..1. the above identity reduces to the (perhaps more familiar) identity f (x1 . 2
used for passing from a quadratic form to a bilinear form. .IV. (1) For every polynomial p(X) ∈ R[X].2. We first show the following simple lemma.1. there is a symmetric m-affine form f : Rm → R. . . and [15] (chapter A. am ) = 1 m! (−1)m−k k m h
i∈H H⊆{1. since it contains 2m − 1 terms. k≥1
Proof. of degree ≤ m. since it applies to arbitrary affine spaces. but we can give a slightly more direct proof which yields an explicit expression for f in terms of h. It is quite technical. If p(X) ∈ R[X] is a homogeneous polynomial of degree exactly m.
. All the ingredients to prove this result are in Bourbaki [14] (chapter A. the polar form is given by the identity f (x1 .I. even of infinite dimension (for example. a formula giving the polar form for a polynomial h(X) of degree 3. the polar form f : E m → F of h is unique. MULTIAFFINE MAPS AND POLAR FORMS
functions. for any polynomial function h of degree m.4. such that p(x) = f (x.
You may also want to try working out on your own. Lemma 4. Note that when h(X) is a homogeneous polynomial of degree 2. x2 ) = 1 [h(x1 + x2 ) − h(x1 ) − h(x2 )] .4.5
Polarizing Polynomials in One or Several Variables
We now use lemma 4. .m} k=|H|.
4. it is often possible to reduce the number of terms.1 is very general. proposition 3).1 to show that polynomials in one or several variables are uniquely defined by polar forms which are multiaffine maps. .134
CHAPTER 4. . you may want to verify that for a polynomial h(X) of degree 2. In particular cases. Before plunging into the proof of lemma 4. section §5.
ai . k
It should be noted that lemma 4. but they are deeply buried! We now prove a general lemma giving the polar form of a polynomial in terms of this polynomial. .4.... Hilbert spaces). The expression of lemma 4. then the symmetric m-affine form f is multilinear.1 is far from being economical. proposition 2).4.
The computation consists of m phases. t) by repeated linear interpolations. s. bm . As we observed. The essence of the de Casteljau algorithm is to compute f (t. . as shown
1
Actually. All that is given is the sequence b0 .. as shown below: 1 f (r.8. When t is outside [r. b1 ). certain properties holding for the curve segment F ([r. corresponding to the parameter value t ∈ A. some of the points bk and bk+1 may coincide. . b1 . but the algorithm still works fine. and the idea is to approximate the shape of the polygonal line consisting of the m line segments (b0 . t) 2 3
The de Casteljau algorithm can also be used to compute any polar value f (t1 . only one interpolation is performed. but it works just as well for any t ∈ A. t. t. t.
. s) f (r. Since f is symmetric. s]) is contained in the convex hull of the control polygon (or control polyline) determined by the control points b0 . t) is computed by extrapolation. . In phase m (the last phase). s. we usually say that F (t) = f (t. where in phase j.1 The curve goes through the two end points b0 and bm . . We already discussed the de Casteljau algorithm in some detail in section 3. even outside [r. . bm of m + 1 control points. s) f (s. are arbitrary: f (t. . What's good about the algorithm is that it does not assume any prior knowledge of the curve. . All we have to do is to use tj during phase j. (bm−1 . bm ). t2 . . t. but not through the other control points.144
´ CHAPTER 5. s]. s) f (r. We strongly advise our readers to look at de Casteljau's original presentation in de Casteljau [23]. (b1 . . and thus. The following diagram shows an example of the de Casteljau algorithm for computing the point F (t) on a cubic. s. s) The above computation is usually performed for t ∈ [r. we can think of its m arguments as a multiset rather than a sequence. we can write the arguments in any order we please. b1 . and t. r) f (t. using the fact that f is symmetric and m-affine. . . such as the fact that the curve segment F ([r. s]) may not hold for larger curve segments of F . where r. s]. t. and the result is the point F (t) on the curve. . r. r. m + 1 − j interpolations are performed. the computation of the point F (t) on a polynomial cubic curve F can be arranged in a triangular array. b2 ). s) f (t. t3 ) (which is not generally on the curve). s]. r. s. Let us review this case (polynomial cubic curves). . s) f (t. POLYNOMIAL CURVES AS BEZIER CURVES
in the more general case. . r) f (r. . t) f (r.
q (* w.q_List. Thus. Note that the parameter t can take any value in A. s]. This can be advantageous for numerical stability. The result is the point F (t). s] (t). . The function decas computes the point F (t) on a polynomial curve F specified by a control polygon cpoly (over [r.r) p + (t . . s]. affine basis [r. [r.t. we have an explicit formula giving any point F (t) associated with a parameter t ∈ A. . . but when we refer to the B´zier e curve segment over [r. . These functions all use a simple function lerp performing affine interpolation between two points p1 and p2. .5. and not necessarily only in [r. s]. as B b0 .t_] := (s . . s] . We will sometimes denote the B´zier curve determined by the sequence of control points b0 .t)/(s . with respect to the affine frame [r.r.
m−i i
where f is the polar form of the B´zier curve. s]. . . s]. (* Performs general affine interpolation between two points p. and we give below several versions in Mathematica. bm . the de Casteljau algorithm provides an iterative method for computing F (t). . for a value t of the parameter. . . The output is a list consisting of two sublists. .r)/(s . . on the unique polynomial curve F : A → E of degree m determined by the sequence of control points b0 . . s. s]. in the sense that f (r. .r) q. *)
. or B[r. s]. s](t) bk . . s](t)
=
m k
s−t s−r
m−k
t−r s−r
k
are the Bernstein polynomials of degree m over [r. and defined with respect e to the interval [r. . s].3.r_. s]). without actually using the Bernstein polynomials. r . The function badecas simply computes the point F (t) on a polynomial curve F specified by a control polygon cpoly (over [r. the first one being the shells of the de Casteljau diagram. . . we are assuming that t ∈ [r. s]). bm . . but also the shells of the de Casteljau diagram. and the second one being F (t) itself. s) = bi . THE DE CASTELJAU ALGORITHM
149
By lemma 4. . bm . [r. and interpolating value t *) lerp[p_List. The de Casteljau algorithm is very easy to implement. bm . .
where the polynomials
m Bk [r.1. The point F (t) is given by the formula
m
F (t) =
k=0
m Bk [r. and the point e corresponding to the parameter value t as B b0 .s_. .1.
and also Then. This is because. Qµ = (1 − µ)B + µC. we claim that Qλ = (1 − λ)B + λC. h(bm ). b = {}. What is not entirely obvious.
(1 − µ)Pλ + µQλ = (1 − λ)Pµ + λQµ . u We now consider a number of remarkable properties of B´zier curves. [r. [r. . . . images e of the original control points b0 .
153
Remark: The version of the de Casteljau algorithm for computing polar values provides a geometric argument for proving that the function f (t1 . bb[[1]]]. . Thus. res := Append[lseg. .j + 1} ]. h(bm ). . bm . and let Pλ = (1 − λ)A + λB. . s] ) = B [h(b0 ). bm . . obtaining a curve h(F ). Since the algorithm proceeds by affine interpolation steps.1. Let A. {j. is a symmetric multiaffine map. . m . This means that if an affine map h is applied to a B´zier curve F e specified by the sequence of control points b0 . . bb = b. we only need to show that f (t1 . bm are obtained by computing affine combinations. . m} ]. . . interpolating first with respect to λ. . 1. . .
. res ) ]. . 1. . C be three points. B. . Pµ = (1 − µ)A + µB. yields the same result a interpolating first with respect to µ. . and then with respect to µ. points on a B´zier curve specified by a sequence of control points e b0 . and that affine maps preserve affine combinations. Since every permutation is a product of transpositions. {i. tm ) remains invariant if we exchange any two arguments. and then with respect to λ. THE DE CASTELJAU ALGORITHM . s]] . e Affine Invariance. is symmetry. then the B´zier curve F ′ determined by the sequence of control points h(b0 ). The above result is closely related to a standard result known as Menela¨ s's theorem. . tm ) that it computes. . . . . . . This can be expressed as h(B b0 . it is clear that it yields a multiaffine map. . . bm . . . since both of these points coincide with the point R = (1 − λ)(1 − µ)A + (λ + µ − 2λµ)B + λµC. We can establish this fact. .
which is obvious. is identical to the curve h(F ). as follows.5.
[s. . r + h] (t) = B b0 . . . . . . . . and B b0 . all affine combinations involved in computing a point F (t) on the curve segment are convex barycentric combinations. is contained within the convex hull of the control points b0 . .
. s]. bm . and left to the reader. bm . bm and defined over the interval [r. s] (t) = B bm . [0. over [0.
This fact is basically obvious. bm . . [r. s] (r < s). . over the interval [r. . . bm . [r. . bm . who can also verify the following two properties: B b0 . bm . . This is usually a lot cheaper than moving e every point around. s] (t) = B b0 . . The segment of B´zier curve F (t) specified by the sequence of e control points b0 . The B´zier curve F (t) specified by the see quence of control points b0 . bm . 1].154
´ CHAPTER 5. .4: Symmetry of biaffine interpolation
C
This property can be used to save a lot of computation time. . . . [0. 1] t−r r−s . r] (t). . [r. is the same as the B´zier e t−r curve F r−s . . when r ≤ t ≤ s. . bm . . b0 . . . . . . Invariance under affine parameter change. . . This is because. This can be expressed as B b0 . and then contruct the new B´zier curve. . . Convex Hull Property. . . . . we simply apply the affine map to the control points. If we want to move a curve using an affine map. with the same sequence of control points b0 . h] (t − r). rather than applying the affine map to all the points computed on the original curve. POLYNOMIAL CURVES AS BEZIER CURVES B
Pµ R Pλ
Qλ
Qµ
A Figure 5. . . .
THE DE CASTELJAU ALGORITHM
155
This property can be used to determine whether two B´zier curve segments intersect e each other. . . passes trough the points b0 and bm . is to observe that the curve F (r + s − t). r) and b1. the tangent to the B´zier curve at the point b0 is the line determined by b0 e and b1 . if some control point bi is moved a little bit. s] (t). over the interval [r. r. Determination of tangents. If the points b0 . r + s − tm ). . .4. . . tm ) = f (r + s − t1 . the tangent at the point bm is the line determined by bm−1 and bm (provided that these points are distinct). b0 . m−1 = f (t. The B´zier curve F (r + s − t) specified by the sequence of control points e b0 . is equal to the B´zier curve F ′ (t) specified by the e sequence of control points bm . .
m−1 m−1
given by the de Casteljau algorithm. bm . . . . s. . bm . bm are collinear (belong to the same line). . . Given a B´zier curve F specified by a sequence of control points e b0 .5. . . . e This is obvious since the points on the curve are obtained by affine combinations. Similarly. the curve is most affected i around the points whose parameter value is close to n . Another way to see it. . . s]. . Linear Precision. s. t. . . over the interval [r. . . This can be expressed as B b0 . . s). . .1. . s) = f (s. whose polar form is g(t1 . . s] (r + s − t) = B bm . This is because it can be shown i that the Bernstein polynomial Bin reaches its maximum at t = n . . . it is indeed F ′ (t). . . It will be shown in section 5. . . Symmetry. . . . [r. . . that when b0 and b1 are distinct. over the interval [r. [r. Pseudo-local control. then the B´zier curve determined by the sequence of control points b0 . . . . bm . . . satisfies the conditions g(r. where f is the polar form of the original curve F (t) specified by the sequence of control points b0 . . bm . . . m−1 = f (t. . . t. . . . The B´zier curve F (t) specified by the sequence of control points e b0 . . . . bm . . . . s]. . . . r) = bm−i . is determined by the two points b0. bm is that same line. . . . s]. the tangent at the current point F (t) determined by the parameter t. . Furthermore. This can be seen easily using the symmetry of the Bernstein polynomials. b0 . r. . Endpoint Interpolation. . .
.
m−i i m−i i
and since this curve is unique.
bm−k−j. . .1 . . . [r. .. . . s] and the hyperplane H.j . m− k . bm . .0 b0.m−k . Observe that the two diagonals b0. .j−1 bi+1.0 bm−1. .j−1 .
5. .. As we will see. b0.. and an interval [r. b0.1 bm−k. .0 .m .156
´ CHAPTER 5. . the number of intersections between the B´zier curve segment F [r. POLYNOMIAL CURVES AS BEZIER CURVES
Variation Diminishing Property. bm−j+1. b0. s]. . subdivision can be used to approximate a curve using a polygon.. . . is less than or equal e to the number of intersections between the control polygon determined by b0 . . . and the computation can be conveniently represented in the following triangular form: 0 b0. . .j . .1 bm. b0. bi. . . over [r.. and the hyperplane H. bk. We will prove the variation diminishing property as a consequence of another property. . . . . . for every t ∈ A.j 1 . and the convergence is very fast. we saw how the de Casteljau algorithm gives a way of computing the point B b0 . . bm−1. .m−k b0. . s] (t) = b0. .j .j . . bm−j.
.j−1 .2
Subdivision Algorithms for Polynomial Curves
We now consider the subdivision method. . . Given a sequence of control points b0 .0 Let us now assume that r < t < s. .m
. m
. j −1 j . b0.0 . for every hyperplane H. .. Given a B´zier curve F specified by a sequence of e control points b0 . bm . .j−1 bm−k−1.0 ..j−1 b0. the subdivision property. . bm−k−j+1. . bm . .1 b1. . . As a consequence..m on the B´zier e curve. a convex control polygon corresponds to a convex curve. . . bm . s]. bi.
Then. using b.r. m = Length[pola]. Given any two adjacent edges (a. we now have the three edges (a. wrt original frame (s.r. pt}. b]. c). b] ]. c). if we contruct the points b′ = (1−λ)a+λb and b′′ = (1−λ)b+λc. b] different from the interval [r. a_.b] . r. m} ]. (b′ ..a] . s. where 0 ≤ λ ≤ 1. to get B[a. b′ ).r. b′′ ). (* (* (* (* Computes the control polygon wrt new affine frame (a. pol1 = subdecas[poly. npoly. Observe that the subdivision process only involves convex affine combinations. [r. s].2. we get the control polygon B[r. We can now prove the variation diminishing property. npoly[[1]] ) ]. [r. [a. pol2 = {}. r] using b.mu) r + mu t. . s]. and then we reverse this control polygon and subdivide again w. b. t) Returns control poly (f(a. b_] := Block[ {poly = {cpoly}. s] using the parameter a. pt].. pola = pol1[[1]]. a. SUBDIVISION ALGORITHMS FOR POLYNOMIAL CURVES
167
The convergence is indeed very fast.t. b)) *) *) *) *)
newcpoly[{cpoly__}... (* Print[" npoly: ". assuming a = r. we subdivide w. a].. npoly = subdecas[pol2. m. pol2 = Prepend[pol2. b] of a polynomial curve given by a control polygon B over [r.. Indeed.lambda) r + lambda t and that b = (1 . r.
.t. we observe that a hyperplane intersects the new polygon in a number of points which is at most the number of times that it intersects the original polygon. c) of the control polygon. {i. When r = a. i. Do[ pt = pola[[i]]. 1. pol1. b) Assumes that a = (1 . s_. s.b] over a new affine frame [a. b) and (b. pol2.t. This immediately follows by convexity. r. a. s] over which the original control polygon is defined. The above function can be used to render curve segments over intervals [a. npoly] *) npoly = subdecas[poly. by subdividing once w. Another nice application of the subdivision method is that we can compute very cheaply the control polygon B[a.5. r_. . b) and (b. and (b′′ . ( If[a =!= r. pola. and modify the control polygon so that instead of having the two edges (a.
g as defined above is clearly m + 1-affine and symmetric. . . . Indeed. . This can be done very easily in terms of polar forms. the following notation is often used. . For example. . . tm+1 ) = 1 m+1 f (ti1 . . . t3 ) + f (t2 . t3 ) = f (t1 . and it is an 0 m+1 easy exercise to show that the points b1 are given in terms of the original points bi . for example. we have g(t1. if F is defined by the polar form f : Am → E. . . . . . . . with b1 = b0 . t) = F (t). . . t2 ) + f (t1 . . tm+1 ) = m+1
m+1
f (t1 .
i=1
where the hat over the argument ti indicates that this argument is omitted. . it is the desired polar form. . degree raising. and thus. b1 . and b1 0 m+1 = bm . POLYNOMIAL CURVES AS BEZIER CURVES
by induction. . 3
If F (and thus f ) is specified by the m + 1 control points b0 . . For example.
. . . . ti . it is sometimes necessary to view F as a curve of polar degree m + 1. . We consider one more property of B´zier curves. say 3. is specified by m + 2 control points b1 . Instead of the above notation. we can show that a hyperplane intersects any polygon obtained by subdivision in a number of points which is at most the number of times that it intersects the original control polygon. and since it is unique. . in the sense that g(t. tm+1 ). t2 . e a hyperplane intersects the B´zier curve in a number of points which is at most the number e of times that it intersects its control polygon. . .<im ≤m+1
Indeed.. Given a B´zier curve e e F of polar degree m. . 1 g(t1 . in order to use such a system on a curve of lower degree. by the j equations: m+1−i i bi−1 + bi . Or a system may only accept curves of a specified degree.
m+1 m
is necessarily g(t1 . the polar form g : Am+1 → E that will yield the same curve F . . and specified by a sequence of m + 1 control points b0 . it is equal to F on the diagonal. . then F considered of degree m + 1 (and thus g). a curve of degree 2. bm . bm . . .
1≤i1 <. .. t3 ) . Since these polygons obtained via subdivision converge to the B´zier curve. if f is biaffine. . . .168
´ CHAPTER 5. tim ) . certain algorithms can only be applied to curve segments of the same degree. t) = f (t. . b1 = i m+1 m+1 where 1 ≤ i ≤ m. it may be necessary to raise the (polar) degree of the curve.
. It can be shown that the control polygons obtained by successive degree raising. uk+m). u5. u3 )
2
3
f (t1 . u4 . u3). but of the form f (uk+1. we consider the following four control points: f (u1 . we say that this sequence is progressive iff the inequalities indicated in the following array hold: u1 = u2 = = u3 = = = u4 u5 u6 Then. u6 )
. u2 . t2 . . t2 . by sliding a window of length 3 over the sequence. and so on. Now. for reasons that will become clear shortly. u4) f (t1 . we can compute any polar value f (t1 . u4 . u4. u6 .5. u5 . . f (u4. u5 . t2 . it is convenient to consider control points not just of the form f (r m−i si ). t2 . this convergence is much slower than the convergence obtained by subdivision. u2m of length 2m. f (u2 . f (u3. . . u2 . u5 . u6 . Given a sequence u1 . u6 ). and it is not useful in practice. However. u2. . u4 . u5 ). u3 .3. where the ui are real numbers taken from a sequence u1 . u3 . using the following triangular array obtained using the de Casteljau algorithm:
0 f (u1. u3. u2 . Let us begin with the case m = 3. u5 ) f (u4. u3 . u3. u4). converge to the original curve segment. t3 ) f (t1 . This explains the name "progressive case". u3 ) f (u2. t3 ) from the above control points. from left to right. u4 ) f (t1 . u4 . u4 )
1 f (t1 . u3) f (t1 . .3
The Progressive Version of the de Casteljau Algorithm (the de Boor Algorithm)
We now consider one more generalization of the de Casteljau algorithm that will be useful when we deal with splines. u5 )
f (u3.
5. Such a version will be called the progressive version. When dealing with splines. u2. . satisfying certain inequality conditions. THE PROGRESSIVE VERSION OF THE DE CASTELJAU ALGORITHM
169
One can also raise the degree again. Observe that these points are obtained from the sequence u1 .
At stage 1, we can successfully interpolate because u1 = u4 , u2 = u5 , and u3 = u6 . This corresponds to the inequalities on the main descending diagonal of the array of inequality conditions. At stage 2, we can successfully interpolate because u2 = u4 , and u3 = u5 . This corresponds to the inequalities on the second descending diagonal of the array of inequality conditions. At stage 3 we can successfully interpolate because u3 = u4 . This corresponds to the third lowest descending diagonal. Thus, we used exactly all of the "progressive" inequality conditions.
Note that the reason why the interpolation steps can be performed is that we have the inequalities uj = um+1 , uj+1 = um+2 , . . . , ui+j = um+i+1 , . . . , um = u2m−j+1, which corresponds to the j-th descending diagonal of the array of progressive inequalities, counting from the main descending diagonal. In order to make the above triangular array a bit more readable, let us define the following points bi,j , used during the computation: bi,j = f (t1 . . . tj ui+j+1 . . . um+i ), for 1 ≤ j ≤ m, 0 ≤ i ≤ m − j, with bi,0 = f (ui+1 , . . . , um+i ), for 0 ≤ i ≤ m. Then, we have the following equations:
bi,j =
um+i+1 − tj um+i+1 − ui+j
bi,j−1 +
tj − ui+j um+i+1 − ui+j
bi+1,j−1 .
The progressive version of the de Casteljau algorithm is also called the de Boor algorithm. It is the major algorithm used in dealing with splines. One may wonder whether it is possible to give a closed form for f (t1 , . . . , tm ), as computed by the progressive case of the de Casteljau algorithm, and come up with a version of lemma 4.3.1. This turns out to be difficult, as the case m = 2 already reveals!
The coefficients of f (u1, u2 ) and f (u3 , u4 ) are symmetric in t1 and t2 , but it is certainly not obvious that the coefficient of f (u2 , u3 ) is symmetric in t1 and t2 . Actually, by doing more calculations, it can be verified that t1 − u1 u3 − u1 is symmetric. These calculations are already rather involved for m = 2. What are we going to do for the general case m ≥ 3? u3 − t2 u3 − u2 + t2 − u2 u3 − u2 u4 − t1 u4 − u2
We can still prove the following theorem generalizing lemma 4.3.1 to the progressive case. The easy half follows from the progressive version of the de Casteljau algorithm, and the converse will be proved later.
In this section, it is assumed that E is some affine space An , with n ≥ 2. Our intention is to give the formulae for the derivatives of polynomial curves F : A → E in terms of control points. This way, we will be able to describe the tangents to polynomial curves, as well as the higher-order derivatives, in terms of control points. This characterization will be used in the next section dealing with the conditions for joining polynomial curves with C k -continuity. A more general treatment of (directional) derivatives of affine polynomial functions F : Am → E will be given in section 10.5, as an application of the homogenization of an affine space presented in chapter 10. In this section, we decided to go easy on our readers,
5.4. DERIVATIVES OF POLYNOMIAL CURVES
175
and proofs are omitted. Such proofs are easily supplied (by direct computation). Our experience shows that most readers are happy to skip proofs, since they can find them later in chapter 10. In this section, following Ramshaw, it will be convenient to denote a point in A as a, to distinguish it from the vector a ∈ R. The unit vector 1 ∈ R is denoted as δ. When dealing → − with derivatives, it is also more convenient to denote the vector ab as b − a.
Given a polynomial curve F : A → E, for any a ∈ A, recall that the derivative DF (a) is the limit F (a + tδ) − F (a) lim , t→0, t=0 t if it exists.
Recall that since F : A → E, where E is an affine space, the derivative DF (a) of F at a − → is a vector in E , and not a point in E. Since coefficients of the form m(m−1) · · · (m−k + 1) occur a lot when taking derivatives, following Knuth, it is useful to introduce the falling power notation. We define the falling power mk , as mk = m(m − 1) · · · (m − k + 1), for 0 ≤ k ≤ m, with m0 = 1, and with the convention that mk = 0 when k > m. The falling powers mk have some interesting combinatorial properties of their own. The following lemma giving the k-th derivative Dk F (r) of F at r in terms of polar values, can be shown. Lemma 5.4.1. Given an affine polynomial function F : A → E of polar degree m, for any r, s ∈ A, with r = s, the k-th derivative Dk F (r) can be computed from the polar form f of F as follows, where 1 ≤ k ≤ m: Dk F (r) = mk (s − r)k
i=k
i=0
k (−1)k−i f (r, . . . , r, s, . . . , s). i
m−i i
A proof is given in section 10.5. It is also possible to obtain this formula by expressing F (r) in terms of the Bernstein polynomials and computing their derivatives. If F is specified by the sequence of m + 1 control points bi = f (r m−i s i ), 0 ≤ i ≤ m, the above lemma shows that the k-th derivative Dk F (r) of F at r, depends only on the k + 1 control points b0 , . . . , bk In terms of the control points b0 , . . . , bk , the formula of lemma 5.4.1 reads as follows: i=k mk k k D F (r) = (−1)k−i bi . k i (s − r)
i=0
176
´ CHAPTER 5. POLYNOMIAL CURVES AS BEZIER CURVES
In particular, if b0 = b1 , then DF (r) is the velocity vector of F at b0 , and it is given by DF (r) = − m −→ m (b1 − b0 ). b0 b1 = s−r s−r
This shows that when b0 and b1 are distinct, the tangent to the B´zier curve at the point e b0 is the line determined by b0 and b1 . Similarly, the tangent at the point bm is the line determined by bm−1 and bm (provided that these points are distinct). More generally, the tangent at the current point F (t) defined by the parameter t, is determined by the two points b0, m−1 = f (t, . . . , t, r) and b1, m−1 = f (t, . . . , t, s),
m−1 m−1
given by the de Casteljau algorithm. It can be shown that DF (t) = m (b1,m−1 − b0,m−1 ). s−r
More generally, if b0 = b1 = . . . = bk , and bk = bk+1 , it can be shown that the tangent at the point b0 is determined by the points b0 and bk+1 . In the next section, we use lemma 5.4.1 to give a very nice condition for joining two polynomial curves with certain specified smoothness conditions. This material will be useful when we deal with splines. We also urge the readers who have not yet looked at the treatment of derivatives given in section 10.5 to read chapter 10.
5.5
Joining Affine Polynomial Functions
The weakest condition is no condition at all, called C −1 -continuity. This means that we don't even care whether F (q) = G(q), that is, there could be a discontinuity at q. In this case, we say that q is a discontinuity knot. The next weakest condition, called C 0 -continuity, is that F (q) = G(q). In other words, we impose continuity at q, but no conditions on the derivatives. Generally, we have the following definition.
When dealing with splines, we have several curve segments that need to be joined with certain required continuity conditions ensuring smoothness. The typical situation is that we have two intervals [p, q] and [q, r], where p, q, r ∈ A, with p < q < r, and two affine curve segments F : [p, q] → E and G : [q, r] → E, of polar degree m, that we wish to join at q.
which we already proved, we have established the conditions of the lemma. Another way to state lemma 5.5.2 is to say that the curve segments F ([p, q]) and G[q, r]) join with continuity C k at q, where 0 ≤ k ≤ m, iff their polar forms f : Am → E and g : Am → E agree on all multisets of points that contain at least m−k copies of the argument q. Thus, the number k is the number of arguments that can be varied away from q without disturbing the values of the polar forms f and g. When k = 0, we can't change any of the arguments, and this means that f and g agree on the multiset q, . . . , q ,
m
i.e., the curve segments F and G simply join at q, without any further conditions. On the other hand, for k = m − 1, we can vary m − 1 arguments away from q without changing the value of the polar forms, which means that the curve segments F and G join with a high degre of smoothness (C m−1 -continuity). In the extreme case where k = m (C m -continuity), the polar forms f and g must agree when all arguments vary, and thus f = g, i.e. F and G coincide. We will see that lemma 5.5.2 yields a very pleasant treatment of parametric continuity for splines. The following diagrams illustrate the geometric conditions that must hold so that two segments of cubic curves F : A → E and G : A → E defined on the intervals [p, q] and [q, r], join at q with C k -continuity, for k = 0, 1, 2, 3. Let f and g denote the polar forms of F and G.
Problem 17 (20 pts). .186
´ CHAPTER 5. for each r ≥ 1. After a number of subdivision steps. or a controlled displacement.
Remark: For any t ∈ [0. . there is some i such that i/(m + r) is closest to t. Write a computer program implementing a version of the de Casteljau algorithm applied to curves of degree four modified as exlained below. Experiment with various ways of perturbing the middle control point. bm+r ) be the control polygon for the curve F obtained after r steps of degree elevation. . Then. 1]. During a subdivision step. Let F be a polynomial curve defined by its control polygon B = (r) (r) (b0 . allow the middle control point (b2 ) to be perturbed in each of the two subpolygons. as r → ∞. You may want to write computer programs to draw such diagrams. this convergence is very slow. . Let B(r) = (b0 .
As a consequence. POLYNOMIAL CURVES AS BEZIER CURVES
for curves of degree 6. 1]. Problem 16 (30 pts). .
. Prove that
m
bi =
j=0
(r)
m j
r i−j m+r i
bj . . it can be shown (using Stirling's formula) that lim r i−j m+r i
m i/(m+r)→t
i/(m+r)→t
= tj (1 − t)m−j .
This means that the control polygons B(r) converge towards the curve segment F [0. . bm ). However. you should get "fractal-style" curves with C 1 -continuity. . lim
(r) bi
=
j=0
m bj Bj (t) = F (t). and is not useful in practice. by a random displacement.
When m is large. For the above reasons. and briefly present the more traditional approach to B-splines in terms of basis functions. but they are also unsatisfactory in a number of ways: 1. we present a class of spline curves known as B-splines and explain how to use knots to control the degree of continuity between the curve segments forming a spline curve. We show how knot sequences and sequences of de Boor control points can be used to specify B-splines and we present the de Boor algorithm for B-splines. 2. Moving any control point will affect the entire curve.1 Introduction: Knot Sequences. Given a control polygon containing m sides (m + 1 vertices). 3. In order to 187
.Chapter 6 B-Spline Curves
6. We also discuss interpolation using B-splines. We know that it takes m(m+1) 2 steps to compute a point on the curve. people thought about segmenting a curve specified by a complex control polygon into smaller and more manageable segments. and solving such a system can be impractical when the degree of the curve is large. we will have to compute control points from points on the curve. this may be too costly. This is the idea behind splines. If we are interested in interpolation rather than just approximating a shape. In this chapter. Each segment of the curve will be controlled by some small subpolygon of the global polygon. and the segments will be joined together in some smooth fashion. in the sense that moving a control point only affects a small region of the curve. and thus impractical. as well as the useful knot insertion algorithm. the degree of the polynomial curve determined by these control points is m. de Boor Control Points
Polynomial curves have many virtues for CAGD. It would be desirable to have better local control. This leads to systems of linear equations.
. Extending our sequence a bit. obtaining the sequence . 1. . [2. 6. 3]. . . 3] is equal to the first control point of the curve segment F [3. For the time being. 3. we are just focusing on a particular subsequence of some bi-infinite sequence. 3. [3. the above knot sequence becomes . [2. 9. 3]. 9]. 9. we could have the intervals [1. 3. . 15. 6]. we can assume that we are dealing with infinite sequences of contiguous intervals. . We also allow collapsing several consecutive intervals. [1. . 6]. 2]. 3. 6]. s] as in the case of polynomial curves. 5. and so on. 9]. . 6. we will assume that the fourth control point of the curve segment F [1. 12. 2. . 1 and 9). 1. . . For example. . . and that such flexibility is in fact useful. where the sequences extend to infinity in both directions. F [6. [6. 3. say to . . 6] is equal to the first control point of the curve segment F [6. 5]. 3. [5. . 1. The number of consecutive occurrences of a knot is called its multiplicity. We will see that it is useful to collapse intervals. F [2. . 6) than just contact. 6. 5] to [3. Such a sequence is called a knot sequence. 3. . that the fourth control point of the curve segment F [5. . 1. 6. 9]. . Since we want to join these cubic curve segments. 3 has multiplicity 3. . It is also possible to handle cyclic sequences. When looking at a finite sequence such as . B-SPLINE CURVES
achieve this. . we may also want to be more careful about endpoints (in this case. 2. . if we want to collapse the interval [3. . 3]. 9. F [5. . . Why did we choose a nonuniform spacing of the intervals? Because it would not really simplify anything. Let us denote the corresponding cubic curve segments as F [1. 5. 3]. that the fourth control point of the curve segment F [2. rather than a single interval [r. . Usually. we can also collapse [2. as in the following sequence: . Clearly. [5.188
CHAPTER 6. 2] is equal to the first control point of the curve segment F [2. 3. 9. we would need four control points for each of these intervals. This can be achieved. . as we shall see later. and if we want to use cubic segments. . 3. 3]. For example. 9]. . F [3. and finally. . . 5]. 2]. 3] to [3. in the above sequence. . [6. we will want better continuity at the junction points (2. 3. 5]. This can be captured by a knot sequence by letting a knot appear more than once. Note that it is slightly more economical to specify the sequence of intervals by just listing the junction points of these intervals. For example. 5]. [3. it becomes necessary to consider several contiguous parameter intervals. Thus.
. as we will explain shortly. since this is a way to lower the degree of continuity of a join. 10. . . 2]. 3]. .
3. 10. 6. We will also see how to extend the de e Casteljau algorithm in order to compute points on any curve segment directly from the de Boor control points. it does not make sense for the knot multiplicity to exceed m + 1. 1. 6. 3]. multiplicity m + 1 corresponds to a discontinuity at the control point associated with that multiple knot). F4 . 3. 9. F [2. F [5. and get several multiple knots. and to find more convenient control points than the B´zier control points e of the curve segments (in our example. . F6 . F [6. F5 . given the knot sequence . 12] to [10. . which turn out to be more convenient (and more economical) than the B´zier control points of the curve segments. we will denote F [1. 1. 3. 2]. e which is called the de Boor algorithm. Since we are now using knot sequences to represent contiguous intervals. For example. 3. The problem now is to find a convenient way of specifying the degree of continuity that we want at each join. The more mathematical presentation is to consider the polar form fi of each curve segment Fi associated with an interval [ui . INTRODUCTION: KNOT SEQUENCES. and we index each curve segment by the index (position) of the last occurrence of the knot corresponding to the left of its interval domain. . . . 3. F [3. F8 . . F [5. 10. . . 9]. . DE BOOR CONTROL POINTS
189
we can collapse several intervals. we could compute the B´zier control points using this algorithm. 3]. . where m is the degree of each curve segment (in fact. Knots of multiplicity 1 are called simple knots. 6. The above sequence has a triple knot 3. . 1. uk+m . However. we have curve segments: F1 . . there is a nice answer to both questions. F [1. . and thus. simply as F1 . F [3. 15. 9.1. . F [2. . 10]: . as shown by collapsing [10. . 9. 3. 10. As we will see. 10. F [6. 3. . 9]). we allow knot multiplicity at most m (although it is easy to accommodate knots of multiplicity m + 1). without having to first compute the B´zier control points for the given e curve segment. Fortunately. We simply consider the subsequence of strictly increasing knots. . . 5]. There are several possible presentations. . F9 . ui+1 ]. 6]. F4 . If the degree of all curve segments is ≤ m. and a double knot 10. 2. and F5 . . and knots of multiplicity greater than 1 are called multiple knots.6. . and to figure out the conditions that C n -continuity at ui+1 impose on the polar forms fi and fi+1 . as in the sequence . 2]. F2 . it turns out that we are led to consider sequences of consecutive knots of length m of the form uk+1. 5. Note that we need to take care of multiple knots. 5]. Lemma 5. We now have to explain how the de Boor control points arise. . F3 . . 15. and for simplicity.2 will yield a very pleasant answer to the problem of continuity of joins and this answer will also show that there are natural control points called de Boor points. we can simplify our notation of curve segments by using the index (position) of the knot corresponding to the beginning of an interval as the index of the curve segment on that interval. 16 .5.
. 6].
dj+1. d2. d0. uk+m+j+1].1 is 4 7−3 = . where 0 ≤ j ≤ m − 2. 2 . The interpolation ratio associated with the point d0. 11 − 6 5 2 7−5 = .3 is
. At every round. 2 . m obtained during the m-th round is a point on the curve segment. we have d0. the middle interval [uk+m. .1 is
the interpolation ratio associated with the point d0. 9−6 3 1 7−6 = . uk+m+j+1] onto the line segment (dj. and we repeat the procedure. the intersection of the m original intervals [uk+j+1.1 is 7−5 2 1 = = . we get a point dj. 9−5 4 2 7−6 1 = . 1 ). dm−2. and d0. and the starting knot of the leftmost interval used during this round is the (right) successor of the starting knot of the leftmost interval used at the previous round. 1 .3 : 777. . uk+m+1 ]. the immediate successor of the starting knot uk+1 of the leftmost interval used at the previous stage.1 : 789.2 is
the interpolation ratio associated with the point d1. 8−6 2
and the interpolation ratio associated with the point d0. The above round gives us a new control polygon determined by the m − 1 points d0. uk+m+1]. 1 . For example. 2 on (dj.1 : 567. so that at the m-th round. d1. and that the leftmost interval [uk+2. 8−5 3
the interpolation ratio associated with the point d2. we only have one interval. . d1. 1 ). d1. These points are also shown as labels of polar values.2 is 7−6 1 = . uk+m+1] starts at knot uk+2 . . Figure 6. where 0 ≤ j ≤ m − 1. The point d0. dj+1. 2 .2 : 677.194
CHAPTER 6. Note that each interval [uk+j+2.2 illustrates the computation of the point corresponding to t = 7 on the spline of the previous figure. 8−3 5
the interpolation ratio associated with the point d1. uk+m+j+1] now consists of m − 1 consecutive subintervals.1 : 678.2 : 778. and for t ∈ [uk+m . the number of consecutive intervals affinely mapped onto a line segment of the current control polygon decreases by one. B-SPLINE CURVES
and we map affinely each of the m − 1 intervals [uk+j+2.
11]. d8 . to distinguish between real numbers in R and points in A. . . we will denote knots as points of the real affine line A. 9]). 5]. A knot sequence is a bi-infinite nondecreasing sequence uk k∈Z of points uk ∈ A (i. If we "squeeze" any of these intervals. corresponding to the intervals [3. 6]. 2.4). 3. 9] (previously [8. we are also shrinking the corresponding curve segment to a single point. The bookkeeping which consists in labeling control points using sequences of m consecutive knots becomes clearer: it is used to keep track of the consecutive intervals (on the knot line) and how they are mapped onto line segments of the current control polygon. 8]) will be lower. meaning that the tangents may not agree.2
Infinite Knot Sequences. 5] this will have the effect that the corresponding curve segment shrinks to a single point. 9. 11] (previously [9. 9]) will join with even less continuity. . say [5. 8]. 5. 5] and [5. and [9. This time. 1.3 shows the construction of the B´zier control points of the five B´zier segments forming e e this part of the spline. 8] (which used to be [6. [6. [5. say [5. If we also squeeze the interval [5. and in fact. d6 . now that 5 is a triple knot. . the curve segments corresponding to the intervals [3. Thus. d2 . 11]) may not even join at the point associated with the knot 5. 8. d3 . d5 . B-SPLINE CURVES
We recognize the progressive version of the de Casteljau algorithm presented in section 5. d7 . We can now provide a better intuition for the use of multiple knots. 9] (previously [8. 5] and [5. . we see how the knot multiplicity can be used to control the degree of continuity of joins between curve segments. Open B-Spline Curves
We begin with knot sequences.3. and the (portion of a) sequence of control points . 8] to the empty interval. 6. in the sense that the curve segments associated with [3. .
6. . and now. and the result will be that the degree of continuity of the junction between the curve segments associated with [3. . as u ∈ A (as explained in section 5. d1 . 15. Definition 6. we may even have a discontinuity at the parameter value 5.196
CHAPTER 6. 6]. such that every knot in the sequence has finitely many
. Figure 6. 11.e. . we may not even have C 1 -continuity. . .2. we can see that these control points determine five curve segments. . to the empty interval [5. d4 . 14.1. [8. . As usual. 5] and [5. Going back to the previous example of the (portion of a) knot sequence . The extreme is to squeeze one more interval. We can now be more precise and prove some results showing that splines are uniquely determined by de Boor control points (given a knot sequence). . uk ≤ uk+1 for all k ∈ Z). 9].
2. A knot of multiplicity 1 is called a simple knot. A knot uk in a knot sequence uk k∈Z has multiplicity n (n ≥ 1) iff it occurs exactly n (consecutive) times in the knot sequence. a piecewise polynomial curve of degree m based on the knot sequence uk k∈Z is a function F : A → E. A knot uk of multiplicity m + 1 is called a discontinuity (knot). Thus. F agrees with a polynomial curve Fi on the interval [ui . if ui+1 is a discontinuity knot. = ui+n < ui+n+1 ). We can now define spline (B-spline) curves. ui+1 [. ui+1 [ agrees with a polynomial curve Fi of polar degree m. . A knot sequence uk k∈Z is uniform iff uk+1 = uk +h. for a knot sequence of degree of multiplicity at most m + 1. it is more convenient to index the curve segment on the interval [ui .
. Definition 6.1.198
CHAPTER 6. i. then we have C −1 -continuity. for any two consecutive distinct knots ui < ui+1 . we must have uk ≤ uk+1 for all k ∈ Z.e. A spline curve F of degree m based on the knot sequence uk k∈Z is a piecewise polynomial curve F : A → E. there are at most m + 1 occurrences of identical knots in the sequence. so that the junction knot is indeed ui+1 . where n is the multiplicity of the knot ui+1 (1 ≤ n ≤ m + 1). ui+1 [ by the last occurrence of the knot p = ui in the knot sequence. such that. such that.2. Remarks: (1) Note that by definition. Given any natural number m ≥ 1. The set F (A) is called the trace of the spline F . with associated polar form fi . ui+n+1 [. if ui+1 is a knot of multiplicity n. B-SPLINE CURVES
occurrences. rather than its first occurrence. = uk+n . . where E is some affine space (of dimension at least 2). the next distinct knot being ui+n+1 (since we must have ui+1 = . then the following condition holds: 1. in the sense of definition 5. for every two consecutive distinct knots ui < ui+1 . Thus. . Thus. for some fixed h ∈ R+ . then 1 ≤ n ≤ m + 1. if uk+1 = uk+2 = . The restriction of F to [ui . . the following condition holds: 2. and for every k ∈ Z.5. and any knot sequence uk k∈Z of degree of multiplicity at most m + 1. that is. and with a polynomial curve Fi+n on the interval [ui+1 . a knot sequence has degree of multiplicity at most m + 1 iff every knot has multiplicity at most m + 1. a knot of multiplicity m + 1. The curve segments Fi and Fi+n join with continuity (at least) C m−n at ui+1 . in particular. Given any natural number m ≥ 1. and Fi (ui+1 ) and Fi+n (ui+1 ) may differ.
that every knot has multiplicity ≤ m. we have F (ui+1 ) = Fi (ui+1 ) = Fi+n (ui+1 ). we want the spline function F to be defined at ui+1 .
.
However. to have C m continuity at all joins: this is the case when F is a polynomial curve! In this case. We could have instead used intervals ]ui+1 . Since F agrees with Fi on [ui . Practically. We could also have used a more symmetric approach. when ui+1 is a knot of multiplicity m + 1. which amounts to say that each join has continuity C −1 (but may be better). t>ui+1 t→ui+1 . agrees with a polynomial curve Fi+m+1 . ui+m+2 [. t<ui+1
lim
F (t) = Fi (ui+1 ). agrees with a polynomial curve Fi+m+1 .e. then clause (1) can be simplified a little bit: we simply require that the restriction of F to the closed interval [ui . although possibly discontinuous. ui+1 ] agrees with a polynomial curve Fi of polar degree m. F (ui+1 −) =
t→ui+1 .
Thus. we would have F (ui+1 −) = and F (ui+1 +) =
t→ui+1 . this would complicate the treatment of control points (we would need special control points corresponding to the "dicontinuity values" F (ui+1 )). is equal to the limit of Fi (t) when t approaches ui+1 (from below). when ui+1 has multiplicity n ≤ m. In this case. ui+m+2 [. INFINITE KNOT SEQUENCES. However.2. OPEN B-SPLINE CURVES
199
(2) If we assume that there are no discontinuities. and thus. open on the right. the limit F (ui+1 −) of F (t) when t approaches ui+1 from below. ui+m+2 ]. t<ui+1
lim
F (t) = Fi (ui+1 ). (3) The number m + 1 is often called the order of the B-spline curve. This is achieved by requiring that the restriction of F to the interval [ui+1 . This ensures that F (ui+1 ) = Fi+m+1 (ui+1 ). For the curious reader. open on the left. are called a discontinuity of the first kind (see Schwartz. there is no advantage in viewing F as a spline. where we require that the restriction of F to the open interval ]ui+1 . (4) Note that no requirements at all are placed on the joins of a piecewise polynomial curve.
lim
F (t) = Fi+m+1 (ui+1 ). and by giving an arbitrary value to F (ui+1 ). ui+1 [. since we have at least C 0 -continuity. or for a spline curve. i. we mention that the kind of discontinuity arising at a discontinuity knot ui+1 . and one can safely ignore these subtleties. [71]). (5) It is possible for a piecewise polynomial curve. The first option seems the one used in most books. when ui+1 has multiplicity m + 1. discontinuities are rare anyway.6.
such that dk = fi (uk+1 . are called de Boor points. . . there exists a unique spline curve F : A → E. the multiset {ui+1 . uk+m ) = fj (uk+1. . Since we have ui−m+1 ≤ .2. . Proof. . OPEN B-SPLINE CURVES
207
Consider i < j such that k ≤ i < j ≤ k + m. the sequence is progressive. . .4.3.2. for any bi-infinite sequence dk k∈Z of points in some affine space E. . . . . . . fi and fj agree on {uk+1 .1. there is a unique polynomial curve Fi of polar degree m. . Then. .
. for all k. where ui < ui+1 and k ≤ i ≤ k + m.2. . and they play an important role for splines. .2. where i − m ≤ k ≤ i. ≤ ui < ui+1 ≤ . Theorem 6. Let i. . . . . .6. Basically. . Then. . Assume that such a spline curve F exists. . This is confirmed formally by the following e theorem. we need to show the existence of such a spline curve. uk+m ) = dk . . . which can be viewed as a generalization of lemma 4. . The polar values fi (uk+1 . For this. uk+m }. . . .4 uniquely determine every curve segment Fi .2. . they play for splines the role that B´zier control points play for polynomial curves. . centered at the interval [ui . . Since the requirements of theorem 6. ≤ ui+m .4. and consider the sequence of 2m knots ui−m+1 . ui+m . . uj } is a submultiset of the multiset {uk+1. The proof is due to Ramshaw. uk+m ). . if it exists. we know that fi (uk+1 . we have shown that the spline curve F is unique. Let ui < ui+1 be two consecutive distinct knots.2. be such that i < j ≤ i + m. ui < ui+1 and uj < uj+1.3. uk+m ). and any knot sequence uk k∈Z of degree of multiplicity at most m + 1. uj }.2. . uk+m ) (where k ≤ i ≤ k + m). j. . We must show that the polar forms fi and fj agree on all multisets of m elements from A (supermultisets) containing the multiset of intervening knots {ui+1 . . such that the following condition holds: dk = fi (uk+1 . uk+m }.3. for all k. We will use lemma 6. . Given any m ≥ 1. From the requirements of theorem 6. .3. i. we have to show that the curve segments Fi fit together in the right way to form a spline with the required continuity conditions. INFINITE KNOT SEQUENCES. ui+1 ]. . and by lemma 6. . by theorem 5. . Thus.
. . . . . and we will often do so. . B-SPLINE CURVES
for k.3. vl ) = fj (v 1 . uj+1 . . implies that gi = gj . . . . uk+m ). is defined by a unique symmetric multiaffine map. . . when the index i is clear from the context. although for every knot ui such that ui < ui+1 . .. for all v 1 . On the other hand. ui+m−1 . . . . . . . ui . . . . uj ). . uj }. . ui+m is progressive. which by theorem 5. ui . is equivalent to proving that gi = gj . corresponding to control points. . . . ui+1 [. . it is important to realize that F itself is generally not the polar form of some (unique) symmetric multiaffine map. . tm ).2. Given a knot ui in the knot sequence. . the inequality k ≤ i ≤ k + m can be interpreted in two ways. ui ui+1 . . . . . . . . and think of the inequalities as i − m ≤ k ≤ i. uj ). Nevertheless. . . ui+1 . . cut the middle j − i columns out of this parallelogram. . . . . . . vl ∈ A. ui+1 . . then gi and gj agree on the l + 1 sequences associated with the progressive sequence uj−m+1 . the curve segment Fi that is the restriction of F to [ui . . ui+m−1 ui+m
Now. . the theorem tells us which curve segments Fi of the spline F are influenced by the specific de Boor point dk : the de Boor point dk influences at most m + 1 curve segments.
. uk+m ). with j − m ≤ k ≤ i. If we think of k as fixed. omitting the subscript i. . . since each element of the left half is strictly less than each element of the right half (since ui < ui+1 ≤ uj ). tm ) as f (t1 . . and let gi : Al → E and gj : Al → E denote the symmetric multiaffine functions of l arguments defined as follows: gi (v1 . We can only say that F is a piecewise polar form. . . In this case. . . . . . . . . . vl ) = fi (v1 .. . . ui+m . v l . . . . ui ui+1 ui+1 . . . v l . and collapse the remaining two triangles to form a smaller parallelogram. . . This does not depend on the knot multiplicity. and since fi and fj agree on the previous parallelogram. Given a spline curve F : A → E. . . Let l = m − j + i. uj+1. ui ui+1 uj−m+2 . . and thus. uj uj+1 . the theorem tells us which de Boor points influence the specific curve segment Fi : there are m+1 de Boor points that influence the curve segment Fi . . uj uj+1 . that fi = fj . We may even denote polar values fi (t1 . . . simply as f (uk+1 . . This is achieved when all the knots are simple.208
CHAPTER 6. we can consider i as fixed. This means that fi and fj agree on the rows of the following parallelogram: uj−m+1 uj−m+2 . uj uj+1 . . . . . . . . . . . it is convenient to denote polar values fi (uk+1 . . . uj . . Proving that fi and fj agree on all supermultisets of {ui+1 . such that ui < ui+1 . Note that the sequence of 2l knots uj−m+1 . and gj (v1 .
tm ). . Given a knot sequence uk k∈Z of degree of multiplicity at most m + 1. provided that this common value makes sense. . um ) as the common value fS (u1 . . . we define fI (u1 . we have to deal with the two end knots. . Given a nonempty open interval I. . Another reason for letting the end knots have multiplicity m + 1 is that it is needed for the inductive definition of B-spline basis functions (see section 6. . um }. validity intervals can be inferred automatically. um }. FINITE KNOT SEQUENCES. .j} (ui+1 . ui+m ) (the de Boor points) are always well defined. . um }. q[. . . Under some reasonable conventions. . the polar forms fi .uj+1 [ (ui+1 .
6. . . where tj−i+1 . ui+1 [ ∩ I = ∅}. A reasonable method is to assume that the end knots have multiplicity m + 1. . . These adjustments are minor. . . . we will next consider what kinds of adjustments are needed to handle finite knot sequences and infinite cyclic knot sequences. agree on the multiset {u1 . where i ∈ S. This way the first curve segment is unconstrained at its left end. . omitting validity intervals in polar forms.6.ui+m+1 [ (ui+1 . We have to be a little careful to deal with knots of multiplicity ≥ 2. . the terms f]ui . the interested reader is referred to Ramshaw [65]. Definition 6. . and wild overloading. Actually. . . The interval I is called a validity interval .3
Finite Knot Sequences. uj . FINITE B-SPLINE CURVES
209
A clean way to handle this overloading problem is to define the notion of a validity interval . that is. . . . For details. for any multiset of arguments {u1 . if we know that for some nonempty set S of knots. . We prefer using the notation fI . . when i < j ≤ i + m. . Then. . It is possible to introduce conventions for overloading the notation. Given any natural numbers m ≥ 1 and N ≥ 0. Having considered the case of bi-infinite knot sequences.3. . . um ) as fS (u1 . . it is quite complex. um ). where S = {i | ]ui . In particular. . . . where I is an open interval ]p. According to this convention. um ). and the last curve segment is unconstrained at its right end.1. . tame overloading. tj−i+1. for any multiset of arguments {u1 . . .3.7). a finite knot sequence of degree of multiplicity at most m + 1 with N intervening knots is any finite nondecreasing
. . uj . . Finite B-Spline Curves
In the case of a finite knot sequence. tm ) is well defined iff ui < uj+1 . . . . . . we denote this common value fi (u1 . f]ui . multiplicity m will give the same results. tj−i+1 . but multiplicity m + 1 allows us to view a finite spline curve as a fragment of an infinite spline curve delimited by two discontinuity knots. . the common value of fi and fj is denoted as f{i. . . Ramshaw introduced two such conventions. tm are arbitrary. Although the notation fS is helpful. . ui < ui+1 and uj < uj+1. .
Figure 6.18: A cyclic knot sequence Definition 6.4.2. Given any natural number m ≥ 1, given any cyclic knot sequence uk k∈Z of period L, cycle length N, period size T , and of degree of multiplicity at most m, a piecewise polynomial curve of degree m based on the cyclic knot sequence uk k∈Z is a function F : A → E, where E is some affine space (of dimension at least 2), such that, for any two consecutive distinct knots ui < ui+1 , if ui+1 is a knot of multiplicity n, the next distinct knot being ui+n+1 , then the following condition holds: 1. The restriction of F to [ui , ui+1 ] agrees with a polynomial curve Fi of polar degree m, with associated polar form fi , and Fi+N (t + T ) = Fi (t), for all t ∈ [ui , ui+1 ]. A spline curve F of degree m based on the cyclic knot sequence uk k∈Z or for short, a closed (or cyclic) B-spline, is a closed piecewise polynomial curve F : A → E, such that, for every two consecutive distinct knots ui < ui+1 , the following condition holds: 2. The curve segments Fi and Fi+n join with continuity (at least) C m−n at ui+1 , in the sense of definition 5.5.1, where n is the multiplicity of the knot ui+1 (1 ≤ n ≤ m). The set F (A) is called the trace of the closed spline F . Remarks:
. um+k ) = dk . and during round j. 0 ≤ k ≤ m − j.
The computation proceeds by rounds.j =
um+k+1 − t bk. for 0 ≤ k ≤ m.
Recall that F (t) = f (t. Its relationship to spline curves is that. b1. . we translate all knots by I − m.j .3. B-SPLINE CURVES
6. in case of a finite spline).j are computed.
As in section 5. since the indexing will be a bit more convenient. . . . um+k+1 − uk+j
where
bk. . and a sequence dk of control points (corresponding to the nature of the knot sequence).3. .230
CHAPTER 6.
j
for 1 ≤ j ≤ m. Such a computation can be conveniently represented in the following trian-
. uI+1 ] for which uI ≤ t < uI+1 . where 1 ≤ k ≤ m + 1.j = f (t uk+j+1 . the points b0. starting from the m + 1 control points indexed by the sequences uI−m+k . bm−j. . we just have to find the interval [uI .j−1. um+1 ] is the middle of the sequence u1 . .j−1 + um+k+1 − uk+j
t − uk+j bk+1. u2m of length 2m. . we presented the progressive version of the de Casteljau algorithm. or cyclic). . finite.j . in order to compute the point F (t) on the spline curve F determined by uk and dk .5
The de Boor Algorithm
In section 5. . . Indeed. . and with bk. For the general case. . t) is computed by iteration as the point b0.0 = f (uk+1 . . in this case [um . given any parameter t ∈ A (where t ∈ [u0 . . . . . um+k ). uN +1 ]. and then to apply the progressive version of the de Casteljau algorithm. . . let us assume for simplicity that I = m.m determined by the inductive computation
bk. uI+k−1 . given a knot sequence uk (infinite.
the effect will be to increase the degree of multiplicity of w by 1.j−1
endfor endfor. for j := 1 to m − r do for i := I − m + j + 1 to I + 1 − r do di. um+i−j − ui−1
begin I = max{k | uk ≤ t < uk+1 }. for i := I − m + 1 to I + 1 − r do di.0 = di−1 . if t = uI then r := multiplicity(uI ) else r := 0 endif. i The de Boor algorithm can be described as follows in "pseudo-code":
um+i−j − t di−1.0 := di−1 endfor. .j =
where 1 ≤ j ≤ m − r. and in the latter case. . The knot w may be new or may coincide with some existing knot of multiplicity r < m. and r = 0 when uI < t < uI+1 (1 ≤ r ≤ m). If I is the largest knot index such that uI ≤ w < uI+1 . Note that other books often use a superscript for the "round index" j. . B´zier control points associated with the curve segments forming a spline curve. when I − m + 1 ≤ i ≤ I + 1 − r.j−1.j−1 + um+i−j − ui−1
t − ui−1 di. I − m + j + 1 ≤ i ≤ I + 1 − r.m−r end
6. uI+k ) associated with the sequences
. The point F (t) on the spline curve is dI+1−r. This is the de Boor algorithm. and write our di.6.j−1 +
t−ui−1 um+i−j −ui−1
di. and with di. F (t) := dI+1−r.6.m−r . . THE DE BOOR ALGORITHM AND KNOT INSERTION
233
di.6
The de Boor Algorithm and Knot Insertion
The process of knot insertion consists of inserting a knot w into a given knot sequence without altering the spline curve. where r is the multiplicity of the knot uI when t = uI . inserting the knot w will affect the m − 1 − r control points f (uI−m+k+1 . and even e for computing a point on a spline curve. Knot insertion can be used either to construct new control points.j :=
um+i−j −t um+i−j −ui−1
di−1.j as dj .
to raise the original multiplicity r of the knot t to m (again. and bi+2 . are 4 7−3 = . dI+1−r. is described in Cohen. 1.0
endfor return dI−m+2. and Riesenfeld [20].6. 6. 4 4 and b′2i+2 = 1 3 bi + bi+1 . It is also possible to formulate a version of knot insertion where a whole sequence of knots. and d3. is inserted. 9.0 +
w−ui−1 um+i−1 −ui−1
di. create two new control points according to the formulae b′2i+1 = 3 1 bi + bi+1 . in the spirit of our treatment of single knot insertion. . Lyche. 8. more illuminating) presentation of this algorithm in terms of polar forms. bi+1 . two new control points b′2i+1 and b′2i+2 are created according to the formulae b′2i+1 = 1 1 bi + bi+1 . The interpolation ratios associated with the points d1. . bi+1 ).1 end
Note that evaluation of a point F (t) on the spline curve amounts to repeated knot insertions: we perform m − r rounds of knot insertion. + bi+2 . . An amusing application of this algorithm is that it yields simple rules for inserting knots at every interval midpoint of a uniform cyclic knot sequence.1 . 15. It is rather straightforward to give an equivalent (and in fact. d2. one may consult Risler [68] or Barsky [4]). 9−5 4 2 7−6 1 = .1 . consists in inserting the knot t = 7 three times. 11 − 6 5
Evaluation of the point F (7) on the spline curve above. 14. 11.23 illustrates the process of inserting the knot t = 7. For instance. THE DE BOOR ALGORITHM AND KNOT INSERTION for i := I − m + 2 to I + 1 − r do di. 2 2 and b′2i+2 = 1 6 1 bi + bi+1 . . . 8−3 5 7−5 2 1 = = . as opposed to a single knot.1 . 5. 8 8 8
. in the case of a quadratic B-spline curve specified by a closed polygon of de Boor control points. for any three consecutive control points bi . . 4 4
This is Chaikin's subdivision method [18].1 .6. r = 0 if t is distinct from all existing knots). For a cubic B-spline curve specified by a closed polygon of de Boor control points. 2. 3. . . . in the knot sequence . Such an algorithm often coined "Olso algorithm". We leave the formulation of such an algorithm as an exercise to the reader (for help.1 :=
um+i−1 −w um+i−1 −ui−1
235
di−1. for every edge (bi . Figure 6. . .
Given a finite knot sequence u0 . u1 . Endpoint interpolation. and we learned that the current point F (t) on a spline curve F can be expressed as an affine combination of its de Boor control points. In the traditional theory of B´zier curves. m where 0 ≤ i ≤ N + m. since it consists in affine interpolation steps. a B´zier curve e e is viewed as the result of blending together its B´zier control points using the Bernstein e
. uN +1 . Variation diminishing property. the resulting spline curve reproduces the straight line. The coefficients of the de Boor points can be viewed as real functions of the real parameter t. . uN −1 . a spline curve passes through the first and the last de Boor control points. If one of de Boor control points is changed. where u0 and uN +1 are of multiplicity m + 1. An easy way to prove this property is to use knot insertion.7. . u1 . This is a variation diminishing property. This property is inherited from the corresponding property of the B´zier e segments. . Affine invariance. this affects m+1 curve segments. . the control polygon consists of B´zier subpolygons. Every point on the spline curve lies in the convex hull of no more than m + 1 nearby de Boor control points. . In case of a finite knot sequence u0 . uN −1 . if we read off control points on this line at the Greville abscissas. Strong convex hull property. . At that stage. Insert every knot until it has multiplicity m. where u0 and uN +1 are of multiplicity m + 1. given a straight line of the form l(u) = au + b. Linear Precision. given the N + m + 1 knots (known as Greville abscissas) 1 ξi = (ui + ui+1 + · · · + ui+m−1 ). which we now briefly e review. The spline curve is not intersected by any straight line more often that is the control polygon. uN . u2 .7
Polar forms of B-Splines
Our multiaffine approach to spline curves led us to the de Boor algorithm. .
6. uN +1 . uN . u2 . . for which the variation diminishing e property has been established. Local control.6. POLAR FORMS OF B-SPLINES
237
Spline curves inherit a number of the properties of B´zier curves.
.k is the segment forming the spline curve Bj. for every t ∈ [uk . B-splines are usually not defined in terms of the de Boor algorithm. we know that Fk (t) =
j
Bj.k (t) f]uj . which may cause a slight confusion with the (normalized) B-splines Bj.m+1.k (t) dj =
j
Bj. but the notation Bj. In the traditional theory. We define B-splines as follows.m+1 is well established.k (t) f]uj . thus showing the equivalence with the traditional approach. we will only consider an infinite knot sequence uk k∈Z where every knot has multiplicity at most m + 1. . . since the adaptations needed to handle a finite or a cyclic knot sequence are quite trivial. Some authors use the notation Njm . uk+1[. .7. .
we get fk (t1 . tm ) f]uj .m+1 which are piecewise polynomial functions. the normalized B-splines Bj. .m+1 : A → A) of order m + 1. Remark: The word "curve" is sometimes omitted when referring to B-spline curves. we tried to be careful in keeping the word "curve" whenever necessary.m+1 is actually of degree ≤ m. to avoid ambiguities. where dk = f]uk . . and defined by the de Boor control points dk k∈Z . . .m+1 : A → A (or Nj. such that δi.uk+m+1 [ (uk+1 . a spline curve can be viewed as the result of blending together its de Boor control points using certain piecewise polynomial functions called B-splines as the weights. . .k (t1 .j .238
CHAPTER 6. . Remark: The normalized B-spline Bj. or by a formula involving so-called divided differences. . is the unique spline curve whose de Boor control points are the reals xi = δi. Definition 6. uj+m ). and δi.uj+m+1 [ (uj+1 . For simplicity. but instead in terms of a recurrence relation.j = 0 otherwise. B-SPLINE CURVES
basis polynomials as the weights. uj+m).m.j = 1 iff i = j.m+1. We will now show briefly how the multiaffine approach to spline curves leads to the recurrence relation defining B-splines. uk+1 [. for every k such that uk < uk+1 .m+1.uj+m+1 [ (uj+1 . tm ) =
j
bj. . . where δi.
. uk+m ). where every knot has multiplicity at most m+1.j is the Kronecker delta symbol. and not curves in the traditional sense.1. Thus.uj+m+1 [ (uj+1 . and it would perhaps make more sense to denote it as Bj. .m+1. the j-th normalized (univariate) B-spline Bj. . uj+m ). . .m+1. . .
where Bj. . any B-spline curve) F : A → E over the knot sequence uk k∈Z (where E is an arbitrary affine space).m+1. Given any spline curve (really. .m+1 over [uk . If we polarize both sides of the equation Fk (t) =
j
Bj. Given an infinite knot sequence uk k∈Z. Thus. Similarly.k are indeed the weights used for blending the de Boor control points of a spline curve. B-splines are piecewise polynomial functions that can be viewed as spline curves whose range is the affine line A. .
. . .m+1. .uj+m+1 [ (uj+1 . . . tm }. . then we have a computation triangle for fk and for fk+n . . k]. . we obtain another interpretation for bj. . Recall that the computation can be arranged into a triangle. . for j outside [k − m.k (t1 .k .k (t1 .k is null. if tk+1 = .
If we label the intermediate nodes as we do this top-down computation. . . . Recall that the above sum only contains at most m + 1 nonzero factors. then the bottom row node yk gives the value bj. . We discover that the triangles used in the computation of the polar values fk (t1 .k . k]. m−j
. . tm ). . and it is just the de Boor algorithm. and the bottom row consists of the values yk = fk (t1 . we have computed the influence of the chosen j-th de Boor control point on all segments fk of the spline. tm ). .m+1. An interesting phenomenon occurs if we fix the multiset {t1 . . uj+m ). The first method proceeds from the top-down. tm } and we compute fk (t1 .6. . Intuitively.m+1.m+1. . but there are no triangles for fk+1 through fk+n−1 . . Fixing an index j. . .k (t1 . Our goal is to find a recurrence formula for bj. .k is the polar form of Bj.7. tm ) associated with the normalized B-spline Bj. tm ) (where uk < uk+1).k (t1 . . . . . The nodes of the above lattice are labeled to reflect the computation of the polar value fk (t1 . tm ). . . The de Boor algorithm provides a method for computing the polar value fk (t1 .j for the top nodes. Thus. . .m+1.m+1. this lattice will have triangular notches cut out of the bottom of it. POLAR FORMS OF B-SPLINES
239
where bj. . and the bottom row with the polar values yk . It is easily seen that the label assigned to every node is the sum of the products of the edge-weights along all descending paths from xj to that node. . . More specifically. The top row consists of the de Boor control points dj = f]uj . the polynomial Bj. . . tm ) overlap in large parts. uj+m). forming a kind of lattice with m + 1 rows. whose heights correspond to knot multiplicity. since Fk is influenced by the m + 1 de Boor control points dj only when j ∈ [k − m. .uj+m+1 [ (uj+1 . .24 shows a portion of this lattice in the case of a cubic spline. tm ) is the sum of the products of the edge-weights along all descending m paths from xj to yk (there are such paths). . . . = tk+n . . . . as specified in the de Boor algorithm. It is interesting to draw the same lattice in the case where the object space is the affine line A.m+1. . and this is what generates a triangular hole of height n at the bottom of the lattice. . . if we let xi = δi.m+1. where we let k vary. If there are multiple knots. . . . . We now show how this second lattice can be used to compute the polar value bj. but this time. and bj. at the fixed multiset {t1 . except that we label the top row with the de Boor points xi (which are real numbers). omitting the node labels. . Figure 6. tm ) from the de Boor control points f]uj . where each subsequent row is computed from the row above by taking affine combinations controlled by one of the arguments ti .m+1 . tm ).
.k of the homogenized version bj. This justifies the introduction of end knots u0 and uN +1 of multiplicity m + 1. Intuitively. .k () = δj.m+1. tm − uj bj.j .k (t1 .m (t). uj+1 [ 0 otherwise. t − uj uj+m+1 − t Bj. . . the bottom-up approach chooses a spline segment fk and computes how this segment is influenced by all de Boor control points. the sequence of tensors (b⊙ j. taking linear combinations as specified by the edge-weights. . and we let yi = δi. and Cox: 1 if t ∈ [uj . the above recurrence equations show the necessity for the knots u−m and uN +m+1 . but under the symmetric path interpretation. de Boor. . . Thus. . If we consider the tensored version b⊙ j. This time.k . we get the standard recurrence relation defining B-splines.242
CHAPTER 6. tm ) = bj.k for the bottom nodes. B-SPLINE CURVES
The symmetry of this second approach suggests a bottom-up computation. tm−1 ). . . .k . .m+1.m+1 is null outside [uj . . tm ): We have bj.m. . ti+m ) = δi. . the above equations show that the sequence of symmetric tensors (b⊙ j. .m+1 (t) = uj+m − uj uj+m+1 − uj+1 Bj. .m.k )k−m≤j≤k are linearly independent.m+1. it is easy to see that the label of the top row node xj is the polar value bj.1.k (t1 . Bj.k of bj. As a consequence. and drop the subscript k.k )k−m≤j≤k is the dual basis of the basis of symmetric tensors (uj+1 · · · uj+m )k−m≤j≤k .k (ti+1 .
. Remark: In the case of a finite spline based on a finite knot sequence uk −m≤k≤N +m+1 . .m+1 . uj+m+1 [.m+1. tm ). Note that the combinations involved are no longer affine (as in the top-down approach).m+1. we choose a fixed k with uk < uk+1 . . The bottom-up approach yields a recurrence relation for computing bj.m+1.m+1. . tm−1 ) + uj+m − uj uj+m+1 − tm bj+1.k (t1 . uj+m+1 − uj+1 If we set all ti to t. the polynomials (Bj. . .m+1. One will observe that bj. We then assign labels to the nodes in a bottom-up fashion.k (t1 . . A nice property about this recurrence relation is that it is not necessary to know the degre m ahead of time if we want to compute the B-spline Bj. the dual space of
∗
A.1 (t) = It is easily shown that Bj.m+1.m+1.k (t1 .m (t) + Bj+1.k )k−m≤j≤k forms a basis of the vector space
m m
A . . due to Mansfield.
1. 2)
g(0. 2) = We note that the control polygon of G is "closer" to the curve. 1. 2. 2) and f (1. 2. 2. . where vk < vk+1 . 2)
g(1. which consists of the knots in the sequence uk k∈Z . This is a general phenomenon. 2) + f (1.
. ti . B-SPLINE CURVES
g(0. since if the left-hand side is well defined. 1. 2)
g(0. Then. vk+1 ]. 1. 0. is related to the polar form fk of the original spline curve F of degree m by the identity 1 gk (t1 . . . then the right-hand side is also well defined. with the multiplicity of each knot incremented by 1. 1. 0. In the general case.27 shows the de Boor control points of the quartic spline curve G and its associated control polygon. 1. since the multiplicity of every knot goes down by 1. 1.
i=1
where the hat over the argument ti indicates that this argument is omitted. 1) g(1. 2. It can be shown that as the degree is raised. . 0. .1. 1) + 2f (0. We observe that the above expression is indeed well defined. where each knot is of multiplicity at most m + 1. 4 The point g(0. given a knot sequence sequence uk k∈Z. 2.244 g(0. 1. 1). 1. . 0)
g(2. 2) . 0. 1. . . given a spline curve F of degree m based on this knot sequence. . tm+1 ). Figure 6. the control polygon converges to the spline curve. 1)
g(1. the polar form gk of the spline curve segment G of degree m + 1 over the interval [uv . 1. 1. . . in bold lines. . 2) is easily seen to be the middle of the line segment between f (0. it is first necessary to form a new knot sequence vk k∈Z . tm+1 ) = m+1
i=m+1
fk (t1 . 1)
CHAPTER 6. 0. 2)
Figure 6.27: Degree raising of a cubic spline as a quartic G and
2f (0. 1. g(0. 1. 1. as in section 5.
. we can try to find the de Boor control points of a C 2 cubic spline curve F based on the finite knot sequence u0 . uN −1 . 0 ≤ i ≤ N − 1. . we can add various "end conditions". In order to solve the above problem. since Lagrange interpolants tend to oscillate in an undesirable manner. uN −2 . it is not possible to obtain a more explicit formula. . . and we consider one of the most common problems which can be stated as follows: Problem 1: Given N +1 data points x0 .vk+m+1 [ (vl+1 . There are a number of interpolation problems. . Thus.8
Cubic Spline Interpolation
We now consider the problem of interpolation by smooth curves. . v l+m+2 ] = [uk . . . . . v l+i . the de Boor control points of the spline curve G of degree m + 1 are given by the formula g]vl . for a uniform knot sequence. xN . 0 ≤ i ≤ N. the above problem has two degrees of freedom. . Cubic spline curves happen to do very well for a large class of interpolation problems. we only come up with N − 1 equations expressing x1 . We note that we are looking for a total of N + 3 de Boor control points d−1 . In general. The control points d0 and d7 = dN were chosen arbitrarily. . . Actually. which amounts to the assumption that the de Boor control points d0 and dN are known. u0 . dN . . . find a C 2 cubic spline curve F . vl+m+1 ) = 1 m+1
i=m+1
f]uk . . and the last control point dN +1 coincides with xN . xN −1 in terms of the N + 1 unknown variables d0 .vl+m+2 [ (v l+1 . we are looking for N + 1 de Boor control points d0 . . dN . . .28 shows N + 1 = 7 + 1 = 8 data points. we can
. Figure 6. . To remove these degrees of freedom. and where the hat over the argument v l+i indicates that this argument is omitted. Unlike the problem of approximating a shape by a smooth curve. Thus. . . u2 . e
6.
i=1
where [v l . dN +1 . u0 . . . uN . . For example. . and it is under-determined. . . for all i. . using the de Boor evaluation algorithm. Thus. . interpolation problems require finding curves passing through some given data points and possibly satisfying some extra constraints. u1 .6. v l+m+1 ). . and a sequence of N +1 knots u0 . with ui < ui+1 for all i. uN . . since the first control point d−1 coincides with x0 . . However. . . we turn to spline curves. It is well known that Lagrange interpolation is not very satisfactory when N is greater than 5. uN . . . . as in the case of B´zier curves. and a C 2 cubic spline curve F passing through these points. such that F (ui ) = xi . CUBIC SPLINE INTERPOLATION
245
as we move from F to G. uk+m+1 ].8. uN . . .
We also show how to compute a new control net with respect to a new rectangle (or new triangle) from a given net. i2 )) for P is chosen. i1 = . it is easy to use the algorithms developed for polynomial curves. and the other one for triangular nets.Chapter 7 Polynomial Surfaces
7. (−1 . things are more tricky. we investigate the possibility of defining polynomial surfaces in terms of polar forms. and i2 = . . − )) be an affine frame for E. We begin with the traditional definition of polynomial surfaces. we will also denote the affine plane as P. . This is one of the many indications that dealing with surfaces is far more complex than dealing with curves. However. and let (Ω1 . there are two natural ways to polarize a polynomial surface. As we shall see. After a quick review of the traditional parametric definition in terms of polynomials. To reduce the amount of superscripts. in the case of triangular nets. one for rectangular nets. The de Casteljau algorithm splits into two versions. we take up the study of polynomial surfaces. typically. and that the affine plane is denoted as A2 . Intuitively. Because the polynomials involved contain two variables. this depends on whether we decide to tile the parameter plane with rectangles or with triangles. and the second approach yields total degree surfaces. there are two natural ways to polarize a polynomial surface. The first approach yields bipolynomial surfaces (also called tensor product surfaces). We assume that − − → → some fixed affine frame (O. Let E be some affine space of finite dimension 0 1 → → n ≥ 3. In the case of rectangular nets.1 Polarizing Polynomial Surfaces
In this chapter. e en 261
. the canonical affine frame − → − → 1 0 where O = (0. Bipolynomial surfaces are completely determined by rectangular nets of control points. and we give an efficient method for performing subdivision. 0). . ( i1 . . We show how these versions of the de Casteljau algorithm can be turned into subdivision methods. Recall that the affine line is denoted as A. and total degree surfaces are completely determined by triangular nets of control points.
7.
. First.p}. POLARIZING POLYNOMIAL SURFACES Note that we are intentionally denoting A × ···× A×A × ···× A
p q
263
as (A) × (A) . . instead of Ap × Aq . However. q ... . if m is such that F is a polynomial surface of total degree m. . .. and to polarize the polynomials in both variables simultaneously. |I|=h J⊆{1. since A × A is isomorphic to P = A2 .q}. 2 : F1 (U. we can view F as a polynomial surface F : P → E.. . up . . Note that in this case. where h ≤ p and k ≤ q. Using linearity. since F (u. v)).. vq ) = 1 p h q k
I⊆{1. . which is symmetric in all of its m arguments. v1 . . it is enough to explain how to polarize a monomial F (u. V ) = U 2 + V 2 + UV + 2U + V − 1 F2 (U. . . we get an m-multiaffine map f : P m → E. . The advantage of this method is that it allows the use of many algorithms applying in the case of curves. (u. . This way. v) = f (u. v) in P. |J|=k i∈I
the surface F is indeed a map F : P → E. namely as the coordinates of a point (u.
p
p
q
We get what are traditionally called tensor product surfaces. In some sense. Consider the following surface viewed as a bipolynomial surface of degree 2. v).
We will present both methods.
Example 1. V ) = UV + U + V + 1. in which we polarize separately in u and v. v) = f ((u. . . . v).
m
the surface F is really a map F : A × A → E. and investigate appropriate generalizations of the de Casteljau algorithm. We begin with the first method for polarizing.. .
p q
The second way to polarize..
ui
j∈J
vj
. since e F (u. we consider several examples to illustrate the two ways of polarizing..1. to avoid the confusion between the affine space Ap and the cartesian product A × · · · × A. . . v. . Indeed. with parentheses around each factor A. u. is to treat the variables u and v as a whole. this method is the immediate generalization of B´zier curves. V ) = U − V + 1 F3 (U. v) of the form uh v k with respect to the bidegree p. It is easily seen that f (u1 . .
defined by an implicit equation of the form z= x2 y 2 + 2. in more details. the algebraic surfaces defined by implicit equations of degree 2 that are also defined as polynomial surfaces of degree 2 are. except for some degenerate cases. z = v2.5. a b or an elliptic paraboloid . q . we go back to bipolynomial surfaces. First. An implicit equation for this surface is z = (y − x2 )2 . either an hyperbolic paraboloid . which is of degree 4. and total degree surfaces. defined by an implicit equation of the form x2 y 2 z = 2 − 2. defined by an implicit equation of the form y 2 = 4ax. a2 b
or a parabolic cylinder . applying lemma 4.2
Bipolynomial Surfaces in Polar Form
Given a bipolynomial surface F : P → E of degree p. which together. and then with respect to V . we get polar forms fi : (A)p × (A)q → A. V ) defining F . we leave as an exercise to show that.270
CHAPTER 7. and it is easy to see that this is the smallest degree. define a (p + q)-multiaffine map f : (A)p × (A)q → E.
.
7. We will now consider bipolynomial surfaces. first with respect to U. It should be noted that parametric polynomial surfaces of degree 2 may correspond to implicit algebraic surfaces of degree > 2. y = u2 + v. POLYNOMIAL SURFACES
Generally. as shown by the following example: x = u. where E is of dimension n.1 to each polynomial Fi (U.
q -symmetric multilinear map. . V1 . . .2. . Vq ) is symmetric in its first p-arguments. q in polar form. .
p q
for all u. v ∈ R.1 is that it does not depend on the dimension of E.2. v.
. such that there is some multiaffine map f : (A)p × (A)q → E. . . v).1. a bipolynomial surface of degree p. and thus. and with F (u. . and symmetric in its last q-arguments. It is easily verified that the set of all symmetric q-linear maps h : (A)q → E forms a vector space SML((A)q .
p q
for all u. . . v) = f (u. . and with F (u. and can be safely omitted by other readers. is a map F : A × A → E. . . A). . We also say that f is p. . it is natural to propose the following definition. Remark: It is immediately verified that the same polar form is obtained if we first polarize with respect to V and then with respect to U. Let f : (A)p × (A)q → E be its homogenized p. Up . . u. . q -symmetric. v ∈ A. v). . q -symmetric multilinear map f : (A)p × (A)q → E is in bijection with the symmetric p-linear map g : (A)p → SML((A)q . The advantage of definition 7. Let F : A × A → E be a bipolynomial surface of degree p. and symmetric in its last q-arguments. E). . the p. q . Given any affine space E of dimension ≥ 3. and let f : (A)p × (A)q → E be its polar form.2. . . u.7. Remark: This note is intended for those who are fond of tensors. E). Definition 7. . By analogy with polynomial curves. v) = f (u. v. . . which is symmetric in its first p-arguments. The trace of the surface F is the set F (A. BIPOLYNOMIAL SURFACES IN POLAR FORM
271
such that f (U1 .
j . 0 .q . . v) for all v ∈ A. is a curve on the bipolynomial surface F . where 0 ≤ i ≤ p − 1. Similarly. bp. 0 . when we fix the parameter v. j . s2 − v s2 − r2
j−1 bi∗. q . the corners of the surface patch. defined such that Fv (u) = F (u. and bp. s1 ]. RECTANGULAR SURFACE PATCHES 275 as a rectangular grid of (p + 1)(q + 1) points in A × A.2. [r 1 .
7. and then compute bp . e The de Casteljau algorithm can be generalized very easily to bipolynomial surfaces. j )(i. bi. and F ([r1 . THE DE CASTELJAU ALGORITHM. Such curves are called isoparametric curves. . s2 ]) is the trace of the rectangular patch. j )(i. k +
v − r2 s2 − r2
j−1 bi∗. By lemma 7. s1 ]. . b0. the images of the line segments [r 1 . bp∗ . and together.q
Given a rectangular control net N = (bi. and 0 ≤ j ≤ q − 1. can be viewed as an image of the rectangular grid p. s2 ]. and [r 2 . . r2 ]. the map Fv : A → E. called the boundary curves of the rectangular surface patch. k+1 . is connected to the three points bi+1. .j)∈
. bj k = i∗. j .3
The de Casteljau Algorithm for Rectangular Surface Patches
p. . . pq quadrangles are obtained in this manner. we first compute the points bj k . v) for which the parameters u. are B´zier e curve segments. Remark: We can also consider curves on a bipoynomial surface. e e In particular. s2 ]. but of degree p + q. . When we fix the parameter u. The control net N = (bi. When we fix u. contains the four control points b0. where b0 j = bi. v satisfy the inequalities r1 ≤ u ≤ s1 and r2 ≤ v ≤ s2 . For every i.j)∈ p. q . Note that there is a natural way of connecting the points in a control net N : every point bi. defined by the constraint u + v = λ. [r2 . [s1 . for some λ ∈ R.q in the affine space E. .2. j+1 .7. bi. or rectangular B´zier patch. and when we fix v. we can first compute the points
b0∗ . 0 . and i∗. . we obtain a B´zier curve of degree p. and bi+1. such a control net N determines a unique bipolynomial surface F of degree p. with 0 ≤ i ≤ p. They are B´zier curves. q . the map Fu : A → E. by applying the de Casteljau algorithm to the control 0∗ points b0∗ . where bi∗ is obtained by applying the de Casteljau algorithm to the B´zier control points e bi. we obtain a B´zier curve of degree q. The e surface F (or rectangular patch) determined by a control net N . bp∗ . is a curve on the bipolynomial surface F . .3.
. they form a polyhedron which gives a rough approximation of the surface patch. with 0 ≤ i ≤ p. is called a rectangular (surface) patch. defined such that Fu (v) = F (u. Generally. . j+1 . v) for all u ∈ A. i∗. q . The portion of the surface F corresponding to the points F (u.
y). r. f (u. x. it is clear that the rectangular surface patch defined by a control net N is contained inside the convex hull of the control net. x) is denoted as rrs. v 1 . 3 . v. There are other interesting properties of tangent planes. for j := 1 to q do for k := 0 to q − j do
−u bj := ss11−r1 i∗ endfor endfor. endfor. u. F (u. x. u. y). that will be proved at the end of
From the above algorithm. v) := bp 0∗ end
j−1 −v bj k := ss22−r2 bi∗. It is assumed that (r. x. for instance. y). u.
. bi∗ = bq . v 3 ) is denoted as u1 u2 u3 . s. u. i∗. and then.7. v). endfor endfor. v. j i∗. u. f (r. and that the image by an affine map h of a bipolynomial surface defined by a control net N is the bipolynomial surface defined by the image of the control net N under the affine map h. k + i∗. x). are computed first. y. for i := 0 to p do b0 = bi∗ i∗ endfor. the polar value f (u1 . y. using these points (shown as square dark dots) as control point. x. u. xxx. u. k+1
u−r1 s1 −r1
j−1 bi+1∗
The following diagram illustrates the de Casteljau algorithm in the case of a bipolynomial surface of degree 3. u3 . s) and (x. It is also clear that bipolynomial surfaces are closed under affine maps. y. u. f (u. The computation shows that the points f (u. u. RECTANGULAR SURFACE PATCHES 277 begin for i := 0 to p do for j := 0 to q do b0 j := bi. v) (shown as a round dark dot) is computed. not all points have been labeled.0 endfor. v1 v2 v3 . and for simplicity of notation. The diagram shows the computation of the point F (u. x. Since the figure is quite crowded. x. y) have been chosen as affine bases of A. THE DE CASTELJAU ALGORITHM. and f (u. v 2 . for j := 1 to p do for i := 0 to p − j do j−1 bi∗ +
v−r2 s2 −r2
j−1 bi∗. u2 . x. the point f (u.3. u.
0 . which determine another tangent line (unless these points are all collinear). q . v) = f ((u. b1. the four tangent planes at the corner points are known. V1 ). Definition 7. . we get polar forms fi : P m → A. a). 0 . such that there is some symmetric multiaffine map f : P m → E. which together. 1 . which determine one tangent line. s. where E is of dimension n. such that F (u. where λi + µi + νi = 1. By analogy with polynomial curves.7. v)). are not collinear. v) is determined by the points bp−1 .4
Total Degree Surfaces in Polar Form
Given a surface F : P → E of total degree m. Vi ).1 is that it does not depend on the dimension of E. v). λm r + µm s + νm t)
. Thus. applying lemma 4.
m
Note that each f ((U1 . . b∗1 . q . . TOTAL DEGREE SURFACES IN POLAR FORM
279
this chapter. .4.5. The trace of the surface F is the set F (P). and bp. . V ) defining F . . is a map F : P → E. . t) ∈ P. it is also natural to propose the following definition. . we can expand f (a1 . the tangent plane to the surface at F (u. (u. am ) = f (λ1 r + µ1 s + ν1 t.1 to each polynomial Fi (U.
7. a surface of total degree m in polar form. For example. and 1 ≤ i ≤ m. We now go back to total degree surface. for any ai = λi r + µi s + νi t. bp−1 . Given any affine space E of dimension ≥ 3. . 0 . then the plane that they define is the tangent plane to the surface at the corner point b0.4.1. and with F (a) = f (a. . . Vm )) is multiaffine and symmetric in the pairs (Ui . . if the control points b0. and b0. and the 1∗ 0∗ q−1 q−1 points b∗0 . in general. . . . bp. Given any barycentric affine frame (r.4. (Um . 0 . . define an m-multiaffine and symmetric map f : P m → E.
m
for all a ∈ P. A similar property holds for the other corner control points b0. . The advantage of definition 7. Also. . . with respect to both U and V . .
u2. 0 and b0. for example. A similar property holds for the other corner control points b0. 0. that will be proved at the end of this Chapter.284 ttt
CHAPTER 7.
There are other interesting properties of tangent planes. points are all collinear). For example. 0. the three tangent planes at the corner points are known. s) is denoted as rrs. and it consists of three shells. 0. m. 0. bm−10 . 0. and for simplicity of notation.2: The de Casteljau algorithm for polynomial surfaces of total degree 3 base of P. Also. The diagram shows the computation of the point F (u). 0 . 0. r. and bm−1. u3 ) is denoted as u1 u2 u3 . if the control points bm. 0 . 1. bm−1. Given a polynomial surface F : P → E. 1. then the plane that they define is the tangent plane to the surface at the corner point bm. 0 . the polar value f (u1 . 0. Thus. and bm−11 (unless these 1. 0. in general. f (r. 1 . the tangent plane to the surface at F (a) is determined by the points bm−10 . are not collinear. each one obtained via two-dimensional interpolation steps. m . POLYNOMIAL SURFACES
ttu rtt tuu rrt rtu tsu tss tts
uuu rrr rru ruu rst suu ssu sss
rsu
rrs
rss
Figure 7. by considering points a ∈ P of barycentric
.
and they are called the boundary curves of the e triangular patch. it is shown that if F : E → E is an affine polynomial function of polar degree − → → m.6. the directional derivative Du F (a) is given by → u Du F (a) = mf (a. q . − ). or ν constant. . . and if − is any nonnull vector in E .
p−1 q−1
f : (A)p × (A)q → E
This directional derivative is also denoted as ∂2F → → (a. We conclude this chapter by considering directional derivatives of polynomial surfaces. when represented as a bipolynomial surface of degree p. . µ. b) is given by → → u v Du Dv F (a. requires a control net of (p + 1)(q + 1) control points. 0. b. b) = p q f (a. and when represented as a surface of total degree m. It is interesting to note that the same polynomial surface F . and let f : (A)p × (A)q → E be its polar form.5. This way. a. DIRECTIONAL DERIVATIVES OF POLYNOMIAL SURFACES
285
coordinates (0. 1 − µ). requires a control net of (m+1)(m+2) 2 points. we get a curve on the surface F passing through F (r) and F (t). we can consider curves on a surface F of total degree m. Let be its homogenized p. . q -symmetric multilinear map.
m−1
Now. Using some very similar calculations. − ∈ R. we also get a B´zier curve e of degree m.6
Directional Derivatives of Polynomial Surfaces
In section 10. b). . for any two points a. Finally. − ). for any u a ∈ E. lies on a line. Other isoparametric curves are obtained by holding µ constant. 0).
7. q . . a. and similarly. → → it is easily shown that for any two nonnull vectors − . u v the directional derivative Du Dv F (a. . considering points of barycentric coordinates (λ. b. they form a polynomial curve called an isoparametric curve. obtained by holding λ constant. . . considering points of barycentric coordinates (λ. − . determined by the constraint that the parameter point a = λr + µs + νt ∈ P. . let F : A × A → E be a bipolynomial surface of degree p. These curves are B´zier curves of degree m. 1 − λ. we get a curve on the surface F passing through F (s) and F (t). . 1 − λ). b ∈ A.7. we get a curve on the surface F passing through F (r) and F (s). If we consider the points F (λr + µs + νt) on the surface F . ∂− ∂− u v
. where E and E are normed affine spaces. More generally. .
j + 1]]. {j. m .m .m]. and ∆art. dd. getting new control nets N ars. A naive method would require twelve calls. j. {a__}] := Block[ {cc = {net}. st. (m = mm. dd = Join[dd. we give a method for subdividing a reference triangle using four calls to the de Casteljau algorithm in its subdivision version. 1. (* polar value by de Casteljau for triangular patches poldecas3[{net__}. 0. In fact. it seems natural to subdivide ∆rst into the three subtriangles ∆ars. i. Do[ pt = barcor[[l. SUBDIVISION ALGORITHMS FOR POLYNOMIAL SURFACES
The following function computes a polar value. res ) ]. ∆ast. m . If[m . res = dd ]. and a list a of mm points given by their barycentric coordinates. and repeat this process recursively. pt]. net0 = convtomat[dd. res}.l]. 0. m. N ast and N art using the functions described earlier.3]] * net0[[i + 1. because no progress is made on the edges rs. m} ]. Do[ row = {}. net0. row = Append[row. pt.2]] * net0[[i + 1. Do[ dd = {}.1]] * net0[[i + 2. and thus such a triangulation does not converge to the surface patch. where a = (1/3. and we will propose a rather efficient method which yields a very regular triangulation. j + 2]] + barcor[[l. 1/3) is the center of gravity of the triangle ∆rst. 1/3. k. mm_.
. *)
If we want to render a triangular surface patch F defined over the reference triangle ∆rst. Therefore. this process does not yield a good triangulation of the surface patch. row. net0 = convtomat[cc. j + 1]] + barcor[[l. given a control net net. However. we need to subdivide the triangle ∆rst in such a way that the edges of the reference triangle are subdivided. {l. {i. l.302
CHAPTER 8.i .l} ]. There are many ways of performing such subdivisions.l} ]. row]. in order to compute triangulations that converge to the surface patch. and tr.l =!= 0. barcor = {a}.
as shown in the diagram below: The first step is to compute the control net for the reference triangle ∆bat. and Kahman [10]. and ∆sca. Farin. Boehm and Farin [9]. 1/2). which is more expensive than the standard 3-dimensional version. 1/2. This can be done using two steps. split the triangle ∆rst into the two triangles ∆art and ∆ars. the nets N bat. This algorithm is also sketched in B¨hm [12] (see pages o 348-349). SUBDIVISION ALGORITHMS FOR TRIANGULAR PATCHES s
303
sca a bac abt crb c
t
b
r
Figure 8. The subdivision strategy that we will follow is to divide the reference triangle ∆rst into four subtriangles ∆abt. 1/2. and c = (1/2. 0). where a = (0. and present a variant of it below. B¨hm [12].2: Subdividing a reference triangle ∆rst
Subdivision methods based on a version of the de Casteljau algorithm which splits a control net into three control subnets were investigated by Goldman [38]. 1/2.
. and they use a version of the de Casteljau algorithm computing a 5-dimensional simplex of polar values. Using the function sdecas3. 1/2). 1/2. and N bar are obtained. where a = (0. which turns out (0. It was brought to our attention by Gerald Farin (and it is mentioned at the end of Seidel's paper [75]) that Helmut Prautzsch showed in his dissertation (in German) [63] that regular subdivision into four subtriangles can be achieved in four calls to the de Casteljau algorithm. and Filip [34]). N ast. we need the barycentric coordinates of b with respect to the triangle ∆art. 1/2). Then. In the first step. rt and rs respectively. and we throw away N ast (which is degenerate anyway). 1/2)). we split ∆art into the two triangles ∆bat and ∆bar. 1/2. Using the function sdecas3 (with a = (0. and we throw away N brt. ∆crb. ∆bac. are the middle points of the sides st. b = (1/2. 1/2) is the middle of st. 0. N brt. For this. the nets N art. o However. and N ars are obtained. We rediscovered this algorithm. and Seidel [75] (see also Boehm.8. some of these authors were not particularly concerned with minimizing the number of calls to the de Casteljau algorithm.1.
and N cba are obtained. N bar and N ars from N rst
cas a c
bat bar t b r
Figure 8. and we throw away N car.4: Computing the net N cas from N ars We will now compute the net N cas from the net N ars. we need the barycentric coordinates of c with respect to the triangle ∆ars. N car. We can now compute the nets N cbr and N cba from the net N bar. Finally. we apply transposej to the net N bat to get the net N abt. For this. 1/2. SUBDIVISION ALGORITHMS FOR POLYNOMIAL SURFACES s
a
ars
bat bar t b r
s Figure 8. we need the barycentric coordinates of c with respect to the reference triangle ∆bar which turns out to be (−1. 1). For this. 1/2). the snet N cbr. 1. transposek to N cba
. Using the function sdecas3. Using the function subdecas3sa. the net N cas is obtained.304
CHAPTER 8. which turns out to be (0.3: Computing the nets N bat.
N bac.5: Computing the nets N cbr and N cba from N bar
sca a bac abt crb c
t
b
r
Figure 8. and transposek twice to N cas to get the net N sca.
. using four calls to the de Casteljau algorithm. Thus. Remark: For debugging purposes. and ∆sca to get the net N bac. N abt is blue. we assigned different colors to the patches corresponding to N abt. we obtained the nets N abt. ∆bac. and N sca. and we found that they formed a particularly nice pattern under this ordering of the vertices of the triangles. In fact.6: Subdividing ∆rst into ∆abt. The subdivision algorithm just presented is implemented in Mathematica as follows. transposej followed by transposek to the net N cbr to get the net N crb. N bac. N crb. N crb is green. N bac is red. and N sca. SUBDIVISION ALGORITHMS FOR TRIANGULAR PATCHES s
305
cas a cba bat cbr c
t
b
r
s Figure 8. N crb. ∆crb.1.8. and N sca is yellow.
e.e. 0. {i. In order to render the surface patch. that is.
. i}. or color the polygons as desired. {4.
(*
performs n subdivision steps using itersub4. where each net is a list of points.. i}. (l = Length[cnet]. debug_] := Block[ {newnet = {net}. splitting into four patches
itersub4[{net__}. mainsubdecas4[cnet[[i]]. debug]. lnet ) ]. lnet = {}. polygons are considered nontransparent. m_. n} ]. 2}. i. Do[ newnet = itersub4[newnet. {6. 0. l. and the rendering algorithm automatically removes hidden parts. l} ]. i. 0}. debug]] . m_. debug_] := Block[ {cnet = {net}.308
CHAPTER 8. m. SUBDIVISION ALGORITHMS FOR POLYNOMIAL SURFACES *)
(* using subdecas4. n_. m. 1. {2. 1. ( newnet = {newnet}. Indeed. This is very crucial to understand complicated surfaces. 2}. 0. It is also very easy to use the shading options of Mathematica. The best thing to do is to use the Polygon construct of Mathematica. 0}. recursively splits into 4 patches *)
rsubdiv4[{net__}. Do[ lnet = Join[lnet. to join the control points in a net by line segments. {i. it is necessary to triangulate each net. and is left as an exercise. The subdivision method is illustrated by the following example of a cubic patch specified by the control net net = {{0. This can be done in a number of ways.. The function rsubdiv4 creates a list of nets. newnet ) ]. 0.
{4. 2. ν) with respect to ∆rst. {5.
This surface patch looks like a sink! Another pleasant application of the subdivision method is that it yields an efficient method for computing a control net N abc over a new reference triangle ∆abc. we need to review how a change of reference triangle is performed. ν3 ). µ. In the second step. Given any arbitrary point d. c. In this case. SUBDIVISION ALGORITHMS FOR POLYNOMIAL SURFACES {1.
−1 ′ λ λ1 λ2 λ3 λ µ′ = µ1 µ2 µ3 µ ν ν1 ν2 ν3 ν′
. by inverting a matrix. using subdecas3ra. we can compute the new control net N abc over the new reference triangle ∆abc. ν) of d with respect to ∆rst. 2}. 4. (λ2 . we easily get ′ λ λ1 λ2 λ3 λ µ = µ1 µ2 µ3 µ′ ν′ ν ν1 ν2 ν3
and thus.312
CHAPTER 8. and then the control net N abt over the reference triangle ∆abt. and coordinates (λ′ . given a reference triangle ∆rst and a control net N over ∆rst. ν2 ). Such an algorithm is useful if we wish to render a surface patch over a bigger or different reference triangle than the originally given one. from a control net N over an original reference triangle ∆rst. Let ∆rst and ∆abc be two reference triangles. µ′ . and let (λ1 . 2}. In the first step. {3. and then the control net N abc over the reference triangle ∆abc. µ. since d = λr + µs + νt = λ′ a + µ′ b + ν ′ c and a =λ1 r + µ1 s + ν1 t. 6. b. ν1 ). Now. 2}. c =λ3 r + µ3 s + ν3 t. 2. we compute the control net N cab using subdecas3ta. be the barycentric coordinates of a. if d has coordinates (λ. using three subdivision steps as explained below. {3. b =λ2 r + µ2 s + ν2 t. µ3. ν ′ ) with respect to ∆abc. with respect to ∆rts. {2. -8}. µ′ . µ1 . In the third step. Before discussing such an algorithm. 2.
Thus. and (λ3 . ν ′ ) of d with respect to ∆abc can be computed from the coordinates (λ. µ2 . 4. the coordinates (λ′ . we compute the control net N ast over the reference triangle ∆ast. 0}}. using transnetj. using transnetk. this is easily done using determinants by Cramer's formulae (see Lang [47] or Strang [81]). we compute the control net N bat using subdecas3sa. 2}.
and then N abc using transnetk.11: Case 1a: a ∈ st. we need the coordinates of b with respect to the reference triangle ∆ast. some adaptations are needed.
First. Case 2a: s = a (and thus.
Case 1a: b ∈ at. and then N abs using transnetj. and in the third step. /
First. b ∈ at / / Note that in the second step. and then go back to case 1. Finally. The implementation in Mathematica requires some auxiliary functions. / Compute N ast using subdecas3ra. Next. In general. Case 1b: b ∈ at. This can be easily done by inverting a matrix of order 3.314
CHAPTER 8. SUBDIVISION ALGORITHMS FOR POLYNOMIAL SURFACES s
b a
t
r
Figure 8. The function collin checks whether three points in the plane are collinear.
In this case. and then go back to case 1. First compute N art using transnetj. ∆rst = ∆rat. we need the coordinates of c with respect to the reference triangle ∆abt. as explained earlier. Case 2b: a ∈ st and s = a. and implemented in Mathematica. We used the strategy explained below. and b does not belong to at.
. and then N abt using transnetj. then compute N bas using subdecas3ra. a ∈ st). compute N bat using subdecas3sa. compute N tas from N ast using transnetk twice. Compute N ars using subdecas3ta. compute N abc using subdecas3ta. One should also observe that the above method is only correct if a does not belong to st. Case 1: a ∈ st. compute N cab using subdecas3ta.
1. newnet ) ]. {0. Another nice application of the subdivision algorithms is an efficient method for computing the control points of a curve on a triangular surface patch. We used the triangles ref trig1 = ((−1. 0}. 1/3. (1. {0.1. 0}. 1. 0.
319
As an example of the use of the above functions. It is easily shown that the monkey saddle is specified by the following triangular control net monknet over the standard reference triangle ∆rst. 0}. newnet = subdecas3ta[netb. where r = (1. {2/3. 0}. m. 0}. b. 0). m] ]. −1. 1. What we need is to compute the control points di = f (a. (1. −1)) and ref trig2 = ((1. we get the following picture. (−1. . {2/3. . 0}. newnet = transnetk[newnet. . . 1). monknet = {{0. defined by the equations x = u. b). 2/3. 0}. 0). Using newcnet3 twice to get some new nets net1 and net2. and then subdividing both nets 3 times. (−1. 0. 3). 1. −1. 0. 0. Note that z is the real part of the complex number (u + iv)3 . y = v. z = u3 − 3uv 2. 1). a. . {1/3. b. 1). 2/3.
m−i i
. 0. {1. s = (0. where the curve is the image of a line in the parameter plane. 0}. c]. 1/3. specified by two points a and b. {1/3. 1/3. 1}}. 3)) with newcnet3. {1/3. . . . −1). SUBDIVISION ALGORITHMS FOR TRIANGULAR PATCHES nc = solve3[a. 1. {0.8. -1}. nc]. ss. we can display a portion of a well known surface known as the "monkey saddle". −1. 0. and t = (0.
15: Case 1: r ∈ ab / where m is the degree of the surface. . Case 1: r ∈ ab. 1. . 0 .8. More precisely. . and we compute N tba using newcnet3. SUBDIVISION ALGORITHMS FOR TRIANGULAR PATCHES s b
321
t
r
a
Figure 8. m−i . Case 2a: r ∈ ab and a ∈ rt. Indeed. We could compute these polar values directly. we must have t ∈ ab. and the control points (d0 . m−i−1 . but there is a much faster method.
. / We compute N rba using newcnet3. if r does not belong to the line (a. Case 2b: r ∈ ab and a ∈ rt. /
The corresponding Mathematica program is given below. b). / In this case. . bi.1. bi. . since r ∈ ab. 0. we simply have to compute N rba using newcnet3. assuming that the surface is defined by some net N over the reference triangle ∆rts. . assuming the usual representation of a triangular net as the list of rows bi. we have the following cases. . . dm ) are simply the bottom row of the net N rba. We compute N sba using newcnet3. m−i.
The final picture (corresponding to 3 iterations) is basically as good as the triangulation shown earlier, and is obtained faster. Actually, it is possible to convert a triangular net of degree m into a rectangular net of degree (m, m), and conversely to convert a rectangular net of degree (p, q) into a triangular net of degree p + q, but we will postpone this until we deal with rational surfaces.
for all i, j, where 0 ≤ i ≤ p and 0 ≤ j ≤ q. Prove that the rectangular surface defined by N is equal to F (we say that rectangular patches have bilinear precision). Problem 2 (20 pts). Give a method for recursively subdividing a triangular patch into four subpatches, using only three calls to the de Casteljau algorithm. Show the result of performing three levels of subdivision on the orginal reference triangle (r, s, t). Problem 3 (20 pts). Investigate the method for recursively subdividing a triangular patch into six subpatches, using four calls to the de Casteljau algorithm as follows: first apply the subdivision version of the de Casteljau algorithm to the center of gravity of the reference triangle, and then to the middle of every side of the triangle. Show the result of performing three levels of subdivision on the orginal reference triangle (r, s, t). Problem 4 (40 pts). Implement your own version of the de Casteljau algorithm splitting a triangular patch into four triangles, as in section 8.1. Use your algorithm to draw the surface patch over [−1, 1] × [−1, 1] defined by the following control net domenet3 = {{0, 0, 0}, {3/4, 3/2, 3/10}, {-1/4, -1/2, 3/10}, {1/2, 1, 0}, {3/2, 0, 3/10}, {1/2, 1/2, 1/2}, {5/4, -1/2, 3/10}, {-1/2, 0, 3/10}, {1/4, 3/2, 3/10}, {1, 0, 0}}; Problem 5 (40 pts). Implement your own version of the de Casteljau algorithm splitting a rectangular patch into four rectangles, as in section 8.2. Problem 6 (40 pts). Given a surface specified by a triangular net, implement a program drawing the u-curves and the v-curves of the patch over [a, b] × [c, d], using the function curvcpoly.
Show that this map behaves like a global deformation of the original rectangular grid p,q . Show how to use such maps to globally deform a B´zier curve specified by its control points e (c0 , . . . , cm ) (where each ci is inside the grid defined by p,q ).
First. we will not be able to propose a nice scheme involving control points. the parameter space is the affine line A. we will basically only consider subdivisions made of rectangles or of (equilateral) triangles. n meeting with C n−1 continuity. we will see that there is a nice scheme involving de Boor control points. and even if we just want to subdivide the plane into convex regions. However. To the best of our knowledge. In the case of a curve. in the line of de Boor control points. but on the positive side. and the only reasonable choice is to divide the affine line into intervals. where the edges are line segments. In fact. This is not quite as good as in the triangular case.4). we will find that C n continuity is only possible if m ≥ 2n + 2. We will then consider spline surfaces of degree m based on a rectangular subdivision of the plane. Next. there is a tremendous variety of ways of doing so. We will discover that C n continuity is only possible if 2m ≥ 3n + 2.1 Joining Polynomial Surfaces
We now attempt to generalize the idea of splines to polynomial surfaces. for bipolynomial splines of bidegree n. we will find necessary and sufficient conditions on polar forms for two surface patches to meet with C n continuity. the parameter space is the affine plane P. We will find necessary conditions on spline surfaces of degree m = 3n + 1 to meet with C 2n continuity.Chapter 9 Polynomial Spline Surfaces and Subdivision Surfaces
9. this is far more subtle than it is for curves. As we shall see. This time. we will restrict our attention to subdivisions of the plane into convex polygons. We conlude this chapter with a section on subdivision surfaces (Section 9. in the case of a surface. We also need to decide what kind of continuity we want to enforce. Subdivision surfaces provide an attractive alternative to spline surfaces in modeling applications where the topology of surfaces is rather complex. but unfortunately. we will first consider parametric continuity. we will take a closer look at spline surfaces of degree m based on a triangular subdivision of the plane. and where the initial control polyhedron consists 337
. As in the case of curves. Thus. and to view the curve as the result of joining curve segments defined over these intervals. finding such a scheme is still an open problem.
. . Geri. → → for all −1 . which. . uk ∈ P. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES of various kinds of faces.1.2. . a. . uk ∈ P. FA and FB join with C k continuity along (r. where it is tangent plane continuous. for any point a ∈ P. Recall that lemma B. ak+1. . uk . . iff FA and FB agree to kth order for all a ∈ (r. s). Lemma 9. Given two polynomial surface FA and FB of degree m. . . . iff fA (u1 . . . We present three subdivision schemes due to Doo and Sabin [27. since it is always possible to convert a rectangular net to a triangular net. . s). It is also easier to deal with bipolynomial surfaces than total degree surfaces.
m−k m−k
for all u1. . .4. s). . . . FA and FB agree to kth order at a iff their polar forms fA : P m → E and fB : P m → E agree on all multisets of points that contain at least m − k copies of a. 28]. FA and FB join with C k continuity along (r. Dui F (a) = Du1 .5 tells us that for any a ∈ (r. iff fA (u1 . The idea is to start with a rough polyhedron specified by a mesh (a collection of points and edges connected so that it defines the boundary of a polyhedron). we can prove the following crucial lemma. 29. am ) = fB (u1 . . . . recall from section 11. uk . This is not a real restriction. Using this fact.1 that we say that F and G agree to kth order at a. s) be the line segment along which they are adjacent (where r. . In this section. . am ∈ (r. . s) be the line segment along which they are adjacent (where r. . . am ). and we concentrate on the more difficult case. s ∈ P are distinct vertices of A and B). and to apply recursively a subdivision scheme. and Charles Loop [50]. . . we restrict our attention to total degree polynomial surfaces. Let A and B be two adjacent convex polygons in the plane. . . a) = fB (u1 . .1. a. . s ∈ P are distinct vertices of A and B). s). Catmull and Clark [17]. Given two polynomial surface F and G of degree m. . and for this. . s). . uk . and let (r. Let A and B be two adjacent convex polygons in the plane. that is. where 0 ≤ i ≤ k. notably. uk . a computer model of a character from the short movie Geri's game. . . ak+1 . . s) iff their polar forms fA : P m → E and fB : P m → E agree on all multisets of points that contain at least m − k points on the line (r. and let (r. Dui G(a). . and all ak+1 . in the limit. . we give a crash course on discrete Fourier transforms and (circular) discrete convolutions. Subdivision schemes typically produce surfaces with at least C 1 -continuity.338CHAPTER 9.1. A number of spectacular applications of subdivision surfaces can be found in the 1998 SIGGRAPH Conference Proceedings. iff Du1 .
. . . u u Definition 9. that is. . −i ∈ R2 . . . except for a finite number of so-called extraordinary points. yields a smooth surface (or solid). Given two polynomial surfaces FA and FB of degree m. not just triangles or rectangles. We discuss Loop's convergence proof in some detail. . . . for all u1 . . . . . a).
The proof is more complicated than it might appear. This means that given any four
. and B and C join with C n continuity along (s.2. A and B join with C n continuity along (s. which means that changing some control points in a small area does not affect the entire spline curve. More precisely. In the case of spline curves. each with n shells. n = 3. First. then fA and fB are completely determined. We assume that the parameter plane has its natural euclidean structure.
9. p). For simplicity. C n continuity is ensured by the overlaping of m − n + 1 pairs of de Casteljau diagrams.2. or triangles. in the case of C 3 continuity. recall that it was possible to achieve C m−1 continuity with curve segments of degree m. all the control points agree.2
Spline Surfaces with Triangular Patches
In this section.9. Also.1. we will prove that if 2m ≤ 3n + 1. there is no reference triangle containing all of these three edges! Lemma 9. In the case of surfaces. t). if the parameter plane is divided into equilateral triangles. Surface splines consisting of triangular patches of degree m ≥ 1 joining with C n continuity cannot be locally flexible if 2m ≤ 3n + 1.4: Constraints on triangular patches Finally. In general. The difficulty is that even though A and D join with C n continuity along (s. if fC and fD are known. q). which means that fA = fB . we will consider surface patches of degree m joining with the same degree of continuity n for all common edges. the situation is not as pleasant. given any 4 adjacent patches as shown in the figure below. i. spline curves have local flexibility.. We now investigate the realizability of the continuity conditions in the two cases where the parameter plane is subdivided into rectangles. SPLINE SURFACES WITH TRIANGULAR PATCHES p t B
343
C
A q D
s
Figure 9. then it is generally impossible to construct a spline surface.e. we study what happens with the continuity conditions between surface patches.
y. q1 . → Given a polynomial surface F : P → E of degree m. w) contains h polar values. v. but there is an equation relating them. z. s) and (s. if 0 ≤ l1 ≤ h. s). This time. v) can be computed from the equation obtained by letting k = l1 + 1. all the other polar values in the diamond (t. q2 . the polar value fA (sh−l1 t2h+2 q l1 ) along (u. the map u − → → Du F : P → E defined by the directional derivative of F in the fixed direction − . l = l1 . x. the proof proceeds by induction on h. p2 . and j = 2h + 1 − l1 . where 1 ≤ l1 ≤ h. we get a strictly smaller diamond of h2 polar values contained in (t. is a u
. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES since all the other polar values involved in this equation are inside (s. Generally. u. which implies that n = 2h and m = 3h + 1. j1 !j2 !j3 !
since all the other polar values involved are inside (s. and in fact. the argument for the previous case can be used. but only C 2h−1 -continuity between A and B. for any vector − ∈ R2 . the problem remains to actually find a method for contructing spline surfaces when 2m = 3n + 2. q1 . but it is not practical. both polar values fA (t2h+1 q h ) and fB (t2h+1 ph ) are undetermined. y. q2 . and thus. so that the diamond (t. Knowing that we must have 2m ≥ 3n + 2 to have local flexibility. (−1)i3
i1 +i2 +i3 =l1 +1
(l1 + 1)! fA (sh−l1 +i1 t2h+1−l1 +i2 q l1 +i3 ) = i1 !i2 !i3 ! (−1)j3
j1 +j2 +j3 =l1
l1 ! fB (sh−l1 +j1 t2h+1−l1 +j2 pl1 +1+j3 ). w) can be computed using only C n−1 -continuity constraints. Thus. w). Instead of presenting this method. i = h − l1 . this argument uses C 2h continuity between A and D and between B and C.
2
In both cases. u. to find any reasonable scheme to constuct triangular spline surfaces. When 2m = 3n + 2. Such a method using convolutions is described by Ramshaw [65]. the polar value associated with u is fA (t2h+1 q h−1 ) and the polar value associated with w is fB (t2h+1 ph−1 ). fA (t2h+2 q h ) can be computed. s). and thus there is at most one degree of freedom for both patches FA and FB . p1 . v. and we use the induction hypothesis to compute these polar values. s) and (s. we attempt to understand better what are the constraints on triangular patches when n = 2N and m = 3N + 1. y. Thus. Again. p1 . z. x. v. u. The key is to look at "derived surfaces". p2 . After a similar computation to determine the polar values fB (sh−l1 t2h+2 pl1 ). We now consider the case n = 2h and m = 3h. y. note that C n -continuity was only needed to compute the polar values associated with u and w.346CHAPTER 9. and that once they are determined.
. Lemma 9.3. . and since F and G agree on L. .2. for → → − → any three vectors − . . then Du F and Du G also join with C n u continuity along L. → → Proof. . parallel to the three directions of the edges of triangles in the α γ → → − → → − triangular grid. Dun Du G meet with C 0 continuity. . we can compute Du F (a) by evaluating F at a and at points near a on L.9. → → → → α β γ 0 We have the following lemma. Dun F and Du Du1 . the following lemmas show that if F and G join with → C n continuity along a line L and if − is parallel to L. u u then it is clear that the derived surfaces Du1 . If a ∈ L. if F and G meet → with C n continuity along a line L. We can now derive necessary conditions on surfaces F and G of degree 3n + 1 to join → → − → with C 2n continuity. −n be any vectors in R2 . Lemma 9. . and such that
. . Given two triangular surfaces F : P → E and G : P → E.2. . − . we will get the same value for Du F (a) and Du G(a).4.2. . parallel to the three directions of α γ − +− +− =− . . . β . Dun G meet with C 0 conti→ nuity along L. then Du F and Du G also meet u n with C continuity along L. . − . then Du F and Du G also u 0 meet with C continuity along L. and if − is parallel to L. β . . Dun Du F and Du1 . Given a spline surface F : P → E of degree 3n + 1 having C 2n continuity. Since the various directional derivatives commute. by lemma 9. Let −1 . Taking the derivative in the direction − . the derived surface α γ
the edges of triangles in the triangular grid.3. Consider three vectors − . Given two triangular surfaces F : P → E and G : P → E. which means that Du F and Du G also meet with C n continuity along L. . Du1 . if F and G meet → with C 0 continuity along a line L.2. Proof. . Dun F and Du1 . Dun G also meet with C 0 continuity. Lemma 9.2. If F and G meet with C n continuity along L.6: A stripe in the parameter plane for triangular patches
347
polynomial surface of degree m − 1. SPLINE SURFACES WITH TRIANGULAR PATCHES − → β B − → α A C − → γ Figure 9. and if − ∈ R2 is parallel to L. Given two triangular surfaces F : P → E and G : P → E. and such that − + β + − = 0 .2. for every triangle A. called a derived surface of F . the derived surfaces u Du Du1 .
{β ⊙− γ
349
which is impossible since such a tensor has order at least 3n + 3. This is one of the outstanding open problems for spline surfaces. it is easy to show that each patch is defined by 3(n + 1)2 − 2 control points. → {− α γ n+1 − → →n+1 ⊙ η | η ∈ (P)⊙(n−1) }. Such spline surfaces do exist. Some interesting related work on joining triangular patches with geometric continuity (G1 -continuity) can be found in Loop [51]. as discussed very lucidly by Ramshaw [65]. the derived surface Dn+2 Dn FA is the same in any stripe in the β α − . →n+1 →n+1 ⊙ − {− α β ⊙ η | η ∈ (P)⊙(n−1) }. if the parameter plane is divided into rectangles. Next we will see that we have better luck with rectangular spline surfaces. the derived surface Dn+2 Dn F is the same in any stripe in the direction − . it is easy to see that these subspaces are pairwise disjoint. if the three subspaces above had a →n+1 − n+1 →n+1 ⊙ − nontrivial intersection. As a α γ consequence. we will consider surface patches
. We can also consider surface splines of degree 3n + 3 with C 2n+1 continuity. satisfying the independent conditions − → Dn+2 Dn FA = Dn+2 Dn FA = Dn+2 Dn FA = 0 . In summary. to the best of our knowledge. no nice scheme involving de Boor control points is known for such triangular spline surfaces.
However.
It can be shown that if we consider surface splines of degree 3n + 3 with C 2n+1 continuity. we were led to consider surface splines of degree 3n + 1 with C 2n continuity. Unfortunately.
9.9. α γ α γ β β Each patch is then defined by 3(n + 1)2 control points. and their existence can be shown using convolutions.3. →n+1 ⊙ − n+1 ⊙ η | η ∈ (P)⊙(n−1) }. then for every triangle A. they would contain a tensor of the form − α β ⊙→ γ ⊙ δ. reasoning on dimensions. satisfying the independent conditions − → Dn+1 Dn+1 FA = Dn+1 Dn+1 FA = Dn+1 Dn+1 FA = 0 . and the conditions are indeed independent. For example. SPLINE SURFACES WITH RECTANGULAR PATCHES correspond to three subspaces of (P)⊙(3n+1) .3
Spline Surfaces with Rectangular Patches
We now study what happens with the continuity conditions between surface patches. β α γ α γ β Each patch is then defined by 3(n + 1)2 − 2 control points. For simplicity. → → direction γ α γ A β − → and the derived surface Dn+2 Dn FA is the same in any stripe in the direction β .
B. Since A and B meet with C n continuity along (x. fA (xi y m−i−j z j ) = fB (xi y m−i−j z j ). z). and so either i ≤ n or j ≤ n. More precisely. if fB and fD are known. we must have m ≥ 2n + 2. We shall consider the case of rectangular spline surfaces of degree 2n meeting with C n−1 continuity. Surface splines consisting of rectangular patches of degree m ≥ 1 joining with C n continuity cannot be locally flexible if m ≤ 2n + 1. by lemma 9. since A and D join with C n continuity along (y. Furthermore. Similarly. D as in the previous figure. there is at most one free control point for every two internal adjacent patches. First. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES x w
B y
A z
C
D
Figure 9. However. for all j ≤ n. then fA is completely determined. When m = 2n + 2. As opposed to the triangular case. Lemma 9. i + j ≤ m ≤ 2n + 1. the proof is fairly simple. for all i ≤ n. which shows that fA (xi y m−i−j z j ) is completely determined. we have fA (xi y m−i−j z j ) = fD (xi y m−i−j z j ).350CHAPTER 9.1. the only control point which is not determined is fA (xn+1 z n+1 ).3. One can prove using convolutions (see Ramshaw [65]) that such spline
. in order to have rectangular spline surfaces with C n continuity. when m = 2n + 2.7: Constraints on rectangular patches of degree m joining with the same degree of continuity n for all common edges. then fA is completely determined. Take ∆xyz as reference triangle. given any 4 adjacent patches as shown in the figure below.1. y). then it is generally impossible to construct a spline surface. if fB and fD are known. This means that given any three adjacent patches A. Proof.2. Thus. we will prove that if m ≤ 2n + 1.
we will look for necessary conditions in terms of derived surfaces. A surface of degree 2n is specified by (2n+2)(2n+1) control points. and any vertical vector β .
n−1
Lemma 9. Thus. we find 2 that each rectangular patch is determined by (n + 1)2 control points.3. The following lemma is the key result. for every rectangle A. setting it to zero corresponds to (n+1)n 2 constraints.2.9. Proof. we will be successful in finding a nice class of spline surfaces specifiable in terms of de Boor control points. n . This time.3. so they must be identical.3 n + 1 times. In view of lemma 9. then u n+1 n+1 Du F = Du G. we deduce that Dn+1 F and Dn+1 G meet with u u C n−1 continuity along L.3. it makes sense to look for rectangular spline surfaces of degree 2n with continuity C n−1 satisfying the constraints − → Dn+1 FA = Dn+1 FA = 0 α β
n+1 for all rectangles A. we discover that bipolynomial spline surfaces of bidegree n.2. Lemma 9. Applying lemma 9. we have a total of (n + 1)n constraints.3. Instead. An immediate consequence of lemma 9. the derived α → surface Dn+1 FA is the same in any stripe in the direction − . n are an answer to our quest. In fact.3. − → → for any horizontal vector − . but the construction is not practical. Since Dα FA has degree n − 1. Given a spline surface F : P → E of degree 2n having C n−1 continuity. since each rectangle is the product of two intervals.3. we can easily adapt what we have done for spline curves to bipolynomial spline surfaces. Proof. and subtracting the (n + 1)n constraints. as in the case of triangular spline surfaces.3. and the derived surface Dn+1 FA α α β − → is the same in any stripe in the direction β . in the present case of rectangular spline surfaces. We can now derive necessary conditions on surfaces F and G of degree 2n to join with C continuity. note that a surface of degree 2n such that
− → Dn+1 FA = Dn+1 FA = 0 α β is equivalent to a bipolynomial surface of bidegree n. Furthermore. we can do this for bipolynomial spline surfaces of
. Given two triangular surfaces F : P → E and G : P → E of degree 2n.2. However. and if − is parallel to L. → if F and G meet with C n−1 continuity along a line L. But these surfaces have degree at most n − 1. SPLINE SURFACES WITH RECTANGULAR PATCHES
351
surfaces exist. and thus.
In summary. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES bidegree p. and leave this topic as an interesting project. tl+1 ]. . where sk < sk+1 and tl < tl+1 . where k − p ≤ i ≤ k and l − q ≤ i ≤ l.l has the (p + 1)(q + 1) de Boor control points xi. in the limit. and two patches adjacent in the v-direction meet with C q−r continuity. Furthermore. two subdivision schemes for surfaces were proposed by Doo and Sabin [27. The challenge of finding such a scheme for triangular spline surfaces remains open. . In 1978.e.
9. q basically reduces to the study of spline curves.. The patches of the spline surface have domain rectangles of the form Rk. The main difference between the two schemes is the following. Since the study of bipolynomial spline surfaces of bidegree p. and by Catmull and Clark [17]. we will not elaborate any further. 29.e. . where r is the multiplicity of the knot tj that divides them. One of the major advantages of such a scheme is that it applies to surfaces of arbitrary topology. The idea is to start with a rough polyhedron specified by a mesh (a collection of points and edges connected so that it defines the boundary of a polyhedron). the process of recursively inserting a knot at the midpoint of every interval in a cyclic knot sequence. and a knot sequences (tj ) along the v-direction. sk+1 ] × [tl . . After one round of subdivision the Doo-Sabin scheme produces a mesh whose vertices all have the same degree 4. si+p . where r is the multiplicity of the knot si that divides them.4
Subdivision Surfaces
An alternative to spline surfaces is provided by subdivision surfaces. we were able to generalize the treatment of spline curves in terms of knot sequences and de Boor control points to bipolynomial spline surfaces. who (in 1974) defined a simple subdivision scheme applying to curves defined by a closed control polygon [18]. a "good" subdivision scheme produces large portions of spline surfaces. The patch defined on the rectangle Rk. and to apply recursively a subdivision scheme. contrary to the case of triangular spline surfaces. The idea of defining a curve or a surface via a limit process involving subdivision goes back to Chaikin. 28]. . yields a smooth surface (or solid). i. .j = f (si+1 . except for a finite number of so-called extraordinary points. Given a knot sequences (si ) along the u-direction. which.j . . .352CHAPTER 9..l = [sk . and that it is not restricted to a rectangular mesh (i. in the case of rectangular spline surfaces. tj+1 . Soon after that. The progressive version of the de Casteljau algorithm can be generalized quite easily. Two patches adjacent in the u-direction meet with C p−r continuity. and most faces are rectangular. a mesh based on a rectangular grid). we have de Boor control points of the form xi. The reader should have no trouble filling in the details. Riesenfeld [67] realized that Chaikin's scheme was simply the de Boor subdivision method for quadratic uniform B-splines. q . except for faces
. tj+q ).
For every edge E common to two faces F1 and F2 . In Loop's scheme. Then. large regions of the mesh define biquadratic B-splines. we will restrict ourselves to a brief description of the Doo-Sabin method.
. A new E-face is created as follows. and wF1 . After one round of subdivision. However. a nontrivial problem. Catmull-Clark method. The limit surface is C 1 -continuous except at extraordinary points. subdivision hit the big screen with Pixar's "Geri's game". in 1998. also referred to as extraordinary points. During every round of the subdivision process. wF2 of the other end vertex w of E are connected to form a rectangular face. it is not until roughly 1994 that subdivision surfaces became widely used in computer graphics and geometric modeling applications.8. and it turns out that these faces shrink and tend to a limit which is their common centroid. and most vertices have degree 4. The centroid of each nonrectangular face is referred to as an extraordinary point. Charles Loop in his Master's thesis (1987) introduced a subdivision scheme based on a mesh consisting strictly of triangular faces [50]. Every vertex v of the current mesh yields a new vertex vF called image of v in F . the number of nonrectangular faces remains constant. except for original vertices whose degree is not equal to six. Since 1994. every triangular face is refined into four subtriangles. This process is illustrated in figure 9.9. referred to as extraordinary points. The limit surface is C 2 -continuous except at extraordinary points. who also proposed a method for improving the design of boundary curves. Most vertices have degree six. and V -faces. and it is obtained by connecting the image vertices of the boundary vertices of F in F . image vertices are connected to form three kinds of new faces: F -faces. and Loop method. after one round of subdivision. Large regions of the mesh define triangular splines based on hexagons consisting of 24 small triangles each of degree four (each edge of such an hexagon consists of two edges of a small triangle). Although such subdivision schemes had been around for some time.4. On the other hand. The Doo-Sabin scheme is described very clearly in Nasri [56]. refinements of previous subdivision schemes and new subdivision schemes have been proposed. the CatmullClark scheme is closer to a process of face shrinking. An F -face is a smaller version of a face F . the four image vertices vF1 . so is the new F -face. Note that if F is an n-sided face. vF2 of the end vertex v of E. The limit surface is C 2 -continuous except at extraordinary points. Although both schemes can be viewed as cutting-off corners. referring the reader to the SIGGRAPH'98 Proceedings and Course Notes on Subdivision for Modeling and Animation. for every face F having v as a vertex. except for vertices arising from original nonrectangular faces and from vertices of degree not equal to four. Furthermore. not unlike a sculptor at work. Several years later. new vertices and new faces are created as follows. Large regions of the mesh define bicubic B-splines. SUBDIVISION SURFACES
353
arising from original vertices of degree not equal to four and from nonrectangular faces.9. Due to the lack of space. E-faces. as illustrated in figure 9. the Catmull-Clark scheme produces rectangular faces.
10. the new V -face is also n-sided. if vF is the image of v in the face F containing v.4. Another rule is
n
vi =
j=1
αij wj . vertices of degree n ≤ 2. if i = j. If v has degree n. and the image vF of v in F as the midpoint of c and v (if F has n sides. and vi is the image of wi in F .. create the new vertex 1 3 vF + v. SUBDIVISION SURFACES F1 vF1 v
355
F2
vF2
vF3
F3
Figure 9.
k=1
because it is the real part of the sum of the n-th roots of unity. cos(2π(i − j)/n) = −1. j ≤ n and n ≥ 3 is the number of boundary edges of F . The scheme just described applies to surfaces without boundaries. 1/n). v′ = Various rules are used to determine the image vertex vF of a vertex v in some face F . 4 4 Another method was proposed by Nasri [56].10: Vertices of a new V -face A new V -face is obtained by connecting the image vertices vF of a given vertex v in all the faces adjacent to v. with αij =
n+5 4n 3+2 cos(2π(i−j)/n) 4n
if i = j. since
n
cos(2πk/n) = 0.e. for every i. where the v's are the vertices of F ). One way to handle boundaries is to treat them as quadratic B-splines.
j=i
.
where the wj are the vertices of the face F .9. A simple scheme used by Doo is to compute the centroid c of the face F . For every boundary vertex v. provided that v has degree n ≥ 3.
where 1 ≤ i. Special rules are needed to handle boundary edges. i. 1 ≤ i ≤ n. the centroid of F is the barycenter of the weighted points (v. This process is illustrated in figure 9. Note that the above weights add up to 1. and thus.
The subdivision rules are modified to allow for nonuniform knot spacing. if F1 and F2 are the two faces sharing E as a common edge. by Sederberg. edge points. this method consists in subdividing every face into smaller rectangular faces obtained by connecting new face points. if F denotes the average of the new face points of all (old) faces adjacent to v and E denotes the average of the midpoints of all (old) n edges incident with v. Given a vertex v (an old one). . new edges points are denoted as hollow round points. i. vE = v + w + vF1 + vF2 . where vF1 and vF2 are the centroids of F1 and F2 . We now turn to a description of the Catmull-Clark scheme. to model clothes or human skin. This can be achieved by eigenvalue analysis. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES Observe that after one round of subdivision. the idea is to analyze the iteration of subdivision around extraordinary points. and new vertex points are denoted as hollow square points. the new edge point vE is the average of the four points v. n 1 vF = vi . or cusps. Figure 9. These matters were investigated by Doo and Sabin [28] and by Peters and Reif [60]. vF1 . and vertex points. all vertices have degree four. or better. vn .11 shows this process. Zheng. and Sabin [74]. The version presented in Catmull and Clark [17] is as follows.
.e. darts. In fact.e. New face points are denoted as solid square points.356CHAPTER 9. However. It is also easy to check that these faces shrink and tend to a limit which is their common centroid. The Doo-Sabin method has been generalized to accomodate features such as creases. 4
The computation of new vertex points is slightly more involved. and what kind of smoothness is obtained at extraordinary points. . n i=1 Given an edge E with endpoints v and w. Roughly. i. vF2 . it is not obvious that such subdivision schemes converge. Such features are desirable in human modeling. called "NURSS" by the authors. Given a face F with vertices v1 . n n n
New faces are then determined by connecting the new points as follows: each new face point vF is connected by an edge to the new edge points vE associated with the boundary edges E of the face F . using discrete Fourier transforms. Note that only rectangular faces are created. each new vertex point v ′ is connected by an edge to the new edge points vE associated with all the edges E incident with v. w. and the number of nonrectangular faces remains constant. the new face point vF is computed as the centroid of the vi . . for example. . there are several different versions. Sewell. Unlike the previous one. the new vertex point v ′ associated with v is v′ = 1 2 n−3 F+ E+ v.
Then for any three consecutive control points pl . n n n
Doo and Sabin analyzed the tangent-plane continuity of this scheme using discrete Fourier transforms [28].11: New face point. It is also possible to accomodate boundary vertices and edges. two new control points pl+1 and pl+1 are i i+1 i+2 2i+1 2i+2
. and pl of a boundary curve. A more general study of the convergence of subdivision methods can be found in Zorin [89] (see also Zorin [88]). SUBDIVISION SURFACES
357
Figure 9. 4 2 4
but it was observed that the resulting surfaces could be too "pointy" (for example. Another version studied by Doo and Sabin is v′ = 1 1 n−2 F+ E+ v. edge points.9. and the number of extraordinary points (vertices of degree different from four) remains constant. and using rules for knot insertion at midpoints of intervals in a closed knot sequence. and vertex points An older version of the rule for vertex points is v′ = 1 1 1 F + E + v. all faces are rectangular. We have only presented the Catmull-Clark scheme for surfaces without boundaries. and C 1 -continuity is investigated by Peters and Reif [60]. pl . The tangent-plane continuity of various versions of Catmull-Clark schemes are also investigated in Ball and Storry [3] (using discrete Fourier transforms). Boundaries can be easily handled by treating the boundary curves a cubic B-splines.4. Observe that after one round of subdivision. starting from a tetrahedron).
since exactly two triangles ∆prs and ∆qrs share the edge (rs). The method of DeRose Kass.12: Loop's scheme for computing edge points created according to the formulae pl+1 = 2i+1 1 l 1 l p + p . 8 8 8 i+2
DeRose. Loop's method only applies to meshes whose faces are all triangles. and the midpoint of the edge (rs). 8 8 8 8
as illustrated in figure 9. For every edge (rs). This new scheme was used in modeling the character Geri in the short film Geri's game. One of the techniques involved is called eigenbasis functions. we compute the new edge point ηrs as the following convex combination: ηrs = 1 3 3 1 p + r + s + q. as opposed to parametric models which obviously do. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES
p
F
ηrs
G
q
r Figure 9. . . For any vertex v of degree n. This corresponds to computing the affine combination of three points assigned respectively the weights 3/8. . using rules to determine new edge points and new vertex points. 3/8. Before presenting Loop's scheme. which were also studied by Zorin [89]. let us mention that a particularly simple subdivision method termed "midedge subdivision" was discovered by Peters and Reif [61]. if p0 . pn−1 are the other endpoints of all (old) edges
. also allows semi-sharp creases in addition to (infinitely) sharp creases. in which the Loop scheme was extended to allow (infinitely) sharp creases. Their work is inspired by previous work of Hoppe et al [44]. However. except that DeRose et al's method applies to Catmull-Clark surfaces.s 358CHAPTER 9. and Truong [24]. + pl . have generalized the Catmull-Clark subdivision rules to accomodate sharp edges and creases. Loop's method consists in splitting each (triangular) face into four triangular faces. A common criticism of subdivision surfaces is that they do not provide immediate access to the points on the limit surface. and Truong [24]. it was shown by Stam [78] that it is in fact possible to evaluate points on Catmull-Clark subdivision surfaces. and 2/8: the centroids of the two triangles ∆prs and ∆qrs. Kass. Unlike the previous methods. 2 i 2 i+1 and pl+1 = 2i+2 1 l 6 l 1 pi + pi+1 . .12.
Observe that after one round of subdivision. the new vertex point v ′ associated with v is
n−1
v = (1 − αn )
′
i=0
1 pi n
+ αn v. Loop's method was first formulated for surfaces without boundaries. in order to insure convergence and better smoothness at extraordinary points. we will present its main lines. except for vertices coming from orginal vertices of degree different from six.13.
. Large regions of the mesh define triangular splines based on hexagons consisting of small triangles each of degree four (each edge of such an hexagon consists of two edges of a small triangle). Vertices of degree different from six are called extraordinary points.9. all vertices have degree six.4. Boundaries can be easily handled by treating the boundary curves a cubic B-splines. where hollow round points denote new edge points. Since the principles of Loop's analysis are seminal and yet quite simple. but in some cases. Loop's method is illustrated in figure 9. He proves convergence of extraordinary points to a limit. In his Master's thesis [50]. Loop rigorously investigates the convergence and smoothness properties of his scheme. The limit surface is C 2 -continuous except at extraordinary points.
where αn is a coefficient dependent on n. He also figures out in which interval αn should belong. as in the Catmull-Clark scheme. but such vertices are surrounded by ordinary vertices of degree six.13: Loop's scheme for subdividing faces incident with v. tangent plane continuity is lost at extraordinary points. Loop determined that the value αn = 5/8 produces good results [50]. Thus. ordinary points have a well defined limit that can be computed by subdividing the quartic triangular patches. and hollow square points denote new vertex points. SUBDIVISION SURFACES
359
v1
η2
η3
v2 η1
v3
Figure 9.
cn−1 ) are then given by the formulae ck = 1 1 f (k) = n n
n−1
fj e−i2πjk/n . where the Fourier transform f of the function f is given by f (x) =
∞
f (t)e−ixt dt. .
j=0
Note the analogy with the continuous case. . . which means that the transform of a convolution should be the product of the transforms. fn−1 ) ∈ Cn as f = Fn f. Loop defines the discrete Fourier transform as Fn f . Inspired by the continuous case. . and following Strang [82]. . it is natural to define the discrete Fourier transform f of a sequence f = (f0 .
. 0 ≤ k ≤ n − 1. The drawback of this choice is that the convolution rule has an extra factor of n.
and the Fourier coefficients of the Fourier series f (x) = are given by the formulae ck =
∞ k=−∞
−∞
ck eikx
1 2π
π
f (x)e−ikx dx. c ⋆ d = c d.e. We will come back to this point shortly. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES rule (discrete or not!) must be defined in such a way that they form a harmonious pair. the Fourier coefficients c = (c0 . .362CHAPTER 9. . . In view of the formula Fn Fn = Fn Fn = n In . which causes problem with the convolution rule. as
n−1
f (k) =
j=0
fj e−i2πjk/n
for every k.
−π
Remark: Others authors (including Strang in his older book [80]) define the discrete Fourier 1 transform as f = n Fn f . We also define the inverse discrete Fourier transform (taking c back to f ) as c = Fn c. i. or equivalently. where the multiplication on the right-hand side is just the inner product of c and d (vectors of length n).
8 Since the absolute value of the cosine is bounded by 1. 8
all we have to prove is that Rl⋆ tends to the null sequence as l goes to infinity. and we get 0 if j = 0. i
l→ ∞
Therefore. and so. applying the Fourier transform in its cosine form and the convolution rule.
j
we get P =
l
5 M− A 8
⋆ P 0 + Ql . He proves that q l (and thus v l ) has the limit (1 − βn )q 0 + βn v 0 .366CHAPTER 9. namely. Loop gives explicit formulae for the limit of extraordinary points. we have Rl⋆ = (R)l . letting R= M− 5 A .
n−1
j=0
M−
5 A 8
l⋆
= 0. Since both M and A are even sequences. However. the faces surrounding extraordinary points converge to the same limit as the centroid of these faces.
n
At this stage. where βn = 3 . this is easy to do.
l→ ∞
and consequently that
l→ ∞
lim pl = lim q l . 0 ≤ j ≤ n − 1.
lim Rl⋆ = lim Rl⋆ = 0n .
where cn⋆ stands for the n-fold convolution c ⋆ · · · ⋆ c. and thus which proves that
l→ ∞
l→ ∞
lim (R)l = 0n . 11 − 8αn
. (R)j = 3 1 + 4 cos (2πj/n) if j = 0. we just have to compute the discrete Fourier transform of R. POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES Taking advantage of certain special properties of M and A. 5 1 ≤ (R)j ≤ 8 8 for all j.
Loop also discusses curvature continuity at extraordinary points. there are problems with curvature continuity at extraordinary points (the curvature can be zero). general approaches to study the properties (convergence. For example. such as. Another phenomenon related to the eigenvalue distribution of the local subdivision matrix is the fact that the mesh may be uneven. for a vertex of degree three (n = 3). Stam [78] also implemented a method for computing points on Catmull-Clark surfaces. and it is nontrivial. especially after 1996. 3 αn = + 8 3 1 + cos (2π/n) 8 4
2
. they have their problems too. it is possible to find a formula for the tangent vector function at each extraordinary point. and code simplicity. If αn is chosen in the correct range. we advise our readers to consult the SIGGRAPH Proceedings and Course Notes. there is convergence. certain triangles being significantly larger than others near extraordinary points. He proposes the following "optimal" value for αn . Loop's scheme was extended to accomodate sharp edges and creases on boundaries. the values α3 = 5/8 is outside the correct range.9. uniformity of representation.
. In summary. but we hope that we have at least given pointers to the most important research directions. The implementation of the method is discussed. At extraordinary points. Although subdivision surfaces have many attractive features. Again. arbitrary topology of the mesh. as Loop first observed experimentally. Their method makes use of Loop's scheme.4. The related issue of adaptive parameterization of surfaces is investigated in Lee et al [49].
Note that α6 = 5/8 is indeed this value for regular vertices (of degree n = 6). We conclude this section on subdivision surfaces by a few comments. Extraordinary points of large degree may exhibit poor smoothness. αn = 5/8 yields βn = 1/2. namely 5 11 − < αn < . There are many other papers on the subject of subdivision surfaces. He proves that tangent plane continuity is insured if αn is chosen so that 1 3 1 − cos (2π/n) < αn < + cos (2π/n) . see Hoppe et [44]. 4 4 4 For instance. Loop also investigates the tangent plane continuity at these limit points. SUBDIVISION SURFACES
367
The bounds to insure convergence are the same as the bounds to insure convergence of v l to q l . numerical stability. First. 8 8 In particular. except at a finite number of extraordinary points. The phenomenon of "eigenvalue clustering" can also cause ripples on the surface. and there is a range of values from which αn can be chosen to insure tangent plane continuity. but his study is more tentative. and we apologize for not being more thorough. smoothness) of subdivision surfaces have been investigated in Reif [66] and by Zorin [89]. Loop proves that his subdivision scheme is C 2 -continuous.
the derived surface Dn+2 Dn FA is the same in any α β → stripe in the direction − . . Show that the number of independent conditions is generally 2m + 1. . Show that the other 4 interior control points can be found by computing the corner twists of each patch (twist vectors are defined in section 7. Show that the number of conditions required for two triangular patches of degree m to meet with C 1 -continuity is 3m + 1. Problem 3 (30 pts).. 1. 0.8. the derived surface Dn+2 Dn FA is the same in any stripe in the γ γ β − . Problem 5 (40 pts). where r = −1. . Implement the interpolation method proposed in problem 4. We would like to find a rectangular bicubic C 1 -continuous B-spline surface F interpolating the points xi. 1 and s = −1. vj ) = xi. 0. j . 0≤j≤N be a net of data points. Show that the number of independent conditions is generally 3m. vj ) to be the bilinear interpolant of the four bilinear patches determined by the nine points xi+r. e Problem 4 (30 pts). (ii) However. . Let u0 . Experiment with various methods for determining corner twists. and let (xi. Formulate a de Boor algorithm for rectangular B-spline surfaces. Compute the Bessel twists. j )0≤i≤M. . accounting for 12 control points per patch. uM and v0 . (2) Prove that the conditions − → Dn+2 Dn FA = Dn+2 Dn FA = Dn+2 Dn FA = 0 . Show that the number of conditions required for two triangular patches of degree m to meet with C 2 -continuity is 6m − 2. prove that for every triangle A. Problem 2 (20 pts). j . (iii) Various methods exist to determine twist vectors.5
Problems
Problem 1 (30 pts). Use it to convert a B-spline surface into rectangular B´zier patches. each patch requires 16 control points. and the derived surface Dn+2 Dn F is the same in any stripe in the direction → direction α α γ A − → β.e. i. . Formulate a knot insertion algorithm for rectangular B-spline surfaces. uN be two knot sequences consisting of simple knots. (i) Using the method of section 6. α β γ α γ β
.6). POLYNOMIAL SPLINE SURFACES AND SUBDIVISION SURFACES
9. such that F (ui . (1) If we consider surface splines of degree 3n + 3 with C 2n+1 continuity. Problem 6 (20 pts). show that the control points on the boundary curves of each rectangular patch can be computed. . . j+s . One method (Bessel twist) consists in estimating the twist at (ui .368CHAPTER 9.
Let us first explain how to distinguish between points and vectors practically. Actually. at least mathematically. However. it would also be advantageous if we could manipulate points and vectors as if they lived in a common universe. It turns out that E is characterized by a universality property: affine maps to vector spaces extend uniquely to linear maps.1 The "Hat Construction". It is often more convenient. Such an homogenization of an affine space and its associated vector space is very useful to define and manipulate rational curves and surfaces. A disadvantage of the affine world is that points and vectors live in disjoint universes. Similarly. curves and surfaces live in affine spaces. Next. the linearization of multiaffine maps is used to obtain formulae for the directional derivatives of polynomial maps. linear maps). or the directional derivatives of polynomial surfaces. It is shown how affine frames in E become bases in E. affine combinations. 371
. we will show that such a procedure can be put on firm mathematical grounds. the construction of a vector space E in which − → both E and E are embedded as (affine) hyperplanes is described. (vector spaces. multiaffine maps extend to multilinear maps. these formulae lead to a very convenient way of formulating the continuity conditions for joining polynomial curves or surfaces. rather than affine objects (affine spaces. First. affine maps). using what amounts to a "hacking trick". such a treatment will be given elsewhere. In turn. or Homogenizing
For all practical purposes. It also leads to a very elegant method for obtaining the various formulae giving the derivatives of a polynomial curve. This chapter proceeds as follows. to deal with linear objects. As a consequence. Then. affine maps between affine spaces E and F extend to linear maps between E and F . linear combinations. using perhaps an extra bit of information to distinguish between them if necessary. Such an "homogenization" (or "hat construction") can be achieved.Chapter 10 Embedding an Affine Space in a Vector Space
10.
y)-plane. 0) whose fourth coordinate is 0. Clearly. translations and dilatations. in which E is embedded as a hyperplane passing through the origin. −2 ). satisfies a universal property with respect to the extension of affine maps to linear maps. u3 ) over the basis (−1 . Given an affine space (E. the vector space E is unique up to isomorphism. and its actual construction is not so important. can we make sense of such elements. and that we have some → → v affine frame (a0 . −2 . → → − every − ∈ E induces a mapping tu : E → E. every point x ∈ E is v v → represented by its coordinates (x1 . and vectors) are finite. in which E corresponds to the (x. it is assumed that all vector spaces are defined over the field R of real numbers. x2 . the set of translations is a vector space isomorphic u − → → → to E . x2 . Of course. called
. In the case of an affine space E of dimension 2. we are opening the door for strange elements such as (x1 . 1). Berger explains the construction in terms of vector fields. and E corresponds to the plane of equation z = 1. E ) is embedded in − → a vector space E. We will also show that the homogenization E of − → an affine space (E. We prefer a more geometric and simpler description in terms of simple geometric transformations. y)-plane. 0. for some linear form ω : E → R. As usual. Remark: Readers with a good knowledge of geometry will recognize the first step in embedding an affine space into a projective space. The construction of the vector space E is presented in some details in Berger [5]. −2 . Thus. and of such a construction? The − → answer is yes. and defined such that u → tu (a) = a + − . This "programming trick" works actually very well. The question is. We will present a construction in which an affine space (E. E ).λ : E → E. x3 . Ramshaw explains the construction using the symmetric tensor power of an affine space [65]. E ). we define the mapping Ha. we can think of E as the vector space R3 of − → dimension 3. it is quite useful to visualize the space E. u2. x3 . 1) whose fourth coordinate is 1. x3 ). parallel to the (x. We begin by defining − → two very simple kinds of geometric (affine) transformations. called a translation. −2 )). v v v → → − → → v A vector − ∈ E is also represented by its coordinates (u1 . With respect to this affine frame. As a consequence.372
CHAPTER 10. where → → → a = a0 + x1 −1 + x2 −2 + x3 −3 . (−1 . and E itself is embedded as an affine hyperplane. v3 . x2 . However. EMBEDDING AN AFFINE SPACE IN A VECTOR SPACE
Assume that we consider the real affine space E of dimension 3. Given any point a and any scalar λ ∈ R. and passing through the point on the z-axis of coordinates (0. where the fourth coordinate is neither 1 nor 0. u v v → One way to distinguish between points and vectors is to add a fourth coordinate. for simplicity. and that all families of scalars (points. and that vectors are represented by (row) vectors (v1 . we will use the same notation − for both the vector − and the translation u u tu . defined as ω −1 (1). and to agree that points are represented by (row) vectors (x1 . v2 . for every a ∈ E. 5). The extension to arbitrary fields and to families of finite support is immediate.
it will turn out that barycenters show up quite naturally too! In order to "add" two dilatations Ha1 . since → a1 . λ1 + a2 . We would like − → to give this set the structure of a vector space. a translation − can be viewed as the constant xa. When λ = 0. THE "HAT CONSTRUCTION".1−λ as a. We have Ha. note that Ha. λ1 (x) = x + λ1 − 1 and xa → a2 . λ . consisting of the union of the (disjoint) sets of translations and dilatations of ratio λ = 1. λ can be viewed as → → the vector field x → λ− Similarly. where λ = 0.λ (a) = a. λ2 (x) = x + λ2 − 2 . Thus. Thus. each a. let us see the effect of such a dilatation on a point x ∈ E: we have → → → → Ha. and when λ = 0 and x = a. Ha.0 is the constant affine map sending every point to a. in such a way that both E and E can be naturally embedded into E. xa
if we want to define a1 . OR HOMOGENIZING
373
dilatation (or central dilatation.λ (x) = a + λ− ax. and since a is a fixed-point.λ1 and Ha2 . λ (x) = x + λ− xa.λ is never the identity. and is obtained by "scaling" − by λ. We prefer our view in terms of geometric transformations.1. since → → → → λ1 − 1 + λ2 − 2 = λ1 − 1 − λ1 − 2 = λ1 − →.1−λ (x) = Hx. it turns out that it is more convenient to consider dilatations of the form Ha. and defined such that → Ha.10.1−λ (x) = a + (1 − λ)− = a + − − λ− = x + λ− ax ax ax xa.0 (x) = a for all x ∈ E. Then.λ(a). let us denote Ha.λ2 . To see this. and Ha.1−λ .λ (x) is on the line → defined by a and x. The effect is a uniform dilatation ax (or contraction. − → (2) Berger defines a map h : E → E as a vector field . In this case. u − . we see that we have to distinguish between two cases: (1) λ1 + λ2 = 0.λ is never a translation. In fact. For simplicity of notation. we have → a. we could define E as the (disjoint) union of these two → vector field x → u vector fields. Then. xa xa xa xa a− 1 2a
. if λ < 1). or homothety) of center a and ratio λ. Ha. for every x ∈ E. We now consider the set E of geometric transformations from E to E. Ha. λ2 . Remarks: (1) Note that Ha. If we assume λ = 1.
u u → → and since we are identifying a + − with a + − .10. we make a few observations. for λ = 1. when viewing λ as a point in both A and A. THE "HAT CONSTRUCTION". From u u → lemma 10. u u → → we will refrain from writing λa + − as λa + − . The elements of A are called B´zier sites.5. 1 . by e Ramshaw. However. 1 as a. We can now justify rigorously the programming trick of the introduction of an extra coordinate to distinguish between points and vectors. which we will call a weighted point. Thus. and any family (λi )i∈I of scalars in R. we go one step further. a little bit like an infinite stack of very thin pancakes! There are two privileged − → pancakes: one corresponding to E. and denote → → notation. and write 1a just as a. First.1. when we consider the homogenized version A of the affine space A associated with the field R considered as an affine space. When we want to be more precise. and i( E ) and E . and one corresponding to E .1.1.
i∈I
where
−−→ −− λi ai =
i∈I i∈I
− → λi bai
. in view of the fact that → → a + − . OR HOMOGENIZING
377
for each λ. and simply λ. λ + − = a + λ−1 − . 1 = a. in the simplified u u − = a + − . λi =
i∈I
−− − −→ λi ai . However. which can be denoted as 2. λ . since u u → → a. 1 + − . 2+3 denotes 2 the middle point of the segment [2. Then. since it is not a barycentric combination. 3]. we may also write a. As an example. it is easily shown by induction on the size of I that the following holds: (1) If
i∈I
λi = 0. 2 . λ . the expression 2 + 3 denotes the real number 5. we will identify j(E) and E. then ai . Given any family (ai )i∈I of points in E. in A. for λ = 0. for every a ∈ E. the expression 2 + 3 makes sense: it is the weighted point 2. when viewing λ as a vector in R and in A. we write λ for λ. the above reads as a + u u → → a + − as a + − . every element of E can be written uniquely as − + λa. In particular. We u also denote λa + (−µ)b as λa − µb. − → − → From now on. and 2 + 3 does not make sense in A.5. because we find it too confusing. 1 (under the injection j). in A. We will also write λa instead of a.
When i∈I λi = 1. where the occurrences of − (if any) are replaced by −. When i∈I λi = 0. which. which. or (2) If
i∈I
λi = 0. we have the following slightly more general property. EMBEDDING AN AFFINE SPACE IN A VECTOR SPACE
for any b ∈ E. Grassmann. By induction on the size of I and the size of J. → vj
where
−−→ −− λi ai =
i∈I i∈I
− → λi bai
for any b ∈ E. for any family (ai )i∈I of points in E. λi =
i∈I i∈I
λi
i∈I
λi
ai .
Proof. by lemma 2. then
ai .5. λi . then ai . . − → Lemma 10. for − → → any family (λ ) of scalars in R. λi +
i∈I j∈J
− = → vj
−−→ −− λi ai +
i∈I j∈J
−. .1. with I ∩ J = ∅. is a vector independent of b. we allow ourselves to write λ1 a1 + · · · + λn an . we can make sense of i∈I ai . it is a point. λi belongs to the hyperplane ω (1).1.
i∈I
λi
. by lemma 2. This will be convenient when dealing with derivatives in section 10.
i i∈I j j∈J
the following properties hold: (1) If
i∈I
λi = 0.
Thus. . is a vector independent of b. λi +
− = → vj
j∈J i∈I
λi
i∈I
i∈I
λi
ai +
j∈J
− → vj
i∈I
λi
. the linear combination of points i∈I λi ai is a vector. n}. and that in E.2. the element −1 i∈I ai . we see how barycenters reenter the scene quite naturally. and thus.1. Operations on weighted points and vectors were introduced by H. regardless of the value of i∈I λi . The above formulae show that we have some kind of extended barycentric calculus.4. or (2) If
i∈I
λi = 0. where some of the occurrences of + can be replaced by − . In fact. as λ1 a1 + · · · + λn an .
i∈I
λi
. in his book published in 1844!
. then ai . and when I = {1.378
CHAPTER 10. and any family (− ) v of vectors in E . E ). Given any affine space (E. which is left as an exercise.4. .
t→0. EMBEDDING AN AFFINE SPACE IN A VECTOR SPACE
10. since we view A as embedded as a line in A) as a. we will write λ1 a1 + · · · + λn an . . However.
lim
F (a + tδ) − F (a) . . In this section. and we write simply a + 1 − a.1. t = 0. or as δ. Remark: when we write a + 1−a. such that λ1 + · · · + λn = 0. with t ∈ R. t when t → 0. − → The vector 1 of A will be denoted as 1 . Note that − → δ = 1 = a + 1 − a. as suggested in section 10. we have F (a + tδ) − F (a) = F (a + tδ) − F (a). it will be convenient to denote a point in A (and in A. it is also more convenient to denote the vector ab as b − a. remember that such combinations are vectors in E (and in E). For any a ∈ A. and λ1 . of its m-polar form f : Am → E. for any a ∈ R. One of the major benefits of homogenization is that the derivatives of an affine polynomial function F : A → E can be obtained in a very simple way from the homogenized version f : (A)m → E. as − → However. . to distinguish it from the vector a ∈ R (and a ∈ A. . we assume that E is a normed affine space. following Ramshaw. since we view R as embedded in A). t
if it exists. we need to see what is the limit of F (a + tδ) − F (a) . and thus. . since F agrees with F on A. . Osculating Flats
In this section.388
CHAPTER 10. given a1 . λn ∈ R. but we prefer to be less pedantic. the derivative DF (a) is the limit.
. an ∈ E. . .5
Differentiating Affine Polynomial Functions Using Their Homogenized Polar Forms. we mean a + 1 −a in A. When dealing with → − derivatives. In this section. t=0
λ1 a1 + · · · + λn an .
. . = bk . t. the above reasoning can be used to show that the tangent at the point b0 is determined by the points b0 and bk+1 .m−1 ). . . . . s). t. . then DF (r) is the velocity vector of F at b0 . s−r
m−1 m−1
and since f agrees with f on Am . the formula of lemma 10. .10. . . δ) =
m−1
s−r s−r
m (f (t. m−1 = f (t. if b0 = b1 .m−1 − b0.5. . Similarly. we have basically done all the work already. t. k i (s − r) i=0 In particular. . . t. s)). . . . This shows that when b0 and b1 are distinct. t. the tangent to the B´zier curve at the point b0 is the line determined by b0 and b1 . and it is given by DF (r) = − m m −→ (b1 − b0 ). b0 b1 = s−r s−r
the last expression making sense in E. . bk In terms of the control points b0 . . . . . if b0 = b1 = . we have justified the claims about tangents to B´zier curves made in section 5. . r) and b1. More generally. Let us assume that E and E are normed affine spaces. . note that since δ= and DF (t) = mf (t. bk . However. . it will be necessary to generalize the above results to directional derivatives. .2 reads as follows: i=k mk k k D F (r) = (−1)k−i bi . e Similarly. 2 (s − r) (s − r)2
the last expression making sense in E. s−r
Thus. In order to see that the tangent at the current point F (t) defined by the parameter t. Recall from
. Later on when we deal with surfaces.5.1. the acceleration vector D2 F (r) is given by D2 F (r) = − −→ − m(m − 1) m(m − 1) −→ (b0 b2 − 2b0 b1 ) = (b2 − 2b1 + b0 ). . we have DF (t) = m (b1. r) − f (t. . . e the tangent at the point bm is the line determined by bm−1 and bm (provided that these points are distinct). .
m−1 m−1
given by the de Casteljau algorithm. . DIFFERENTIATING AFFINE POLYNOMIAL FUNCTIONS
391
control points b0 . is determined by the two points b0. and consider a map F : E → E. and bk = bk+1 . m−1 = f (t.
Lemma 10.5.3 is a generalization of lemma 10.5.1 to any domain E which is a normed affine space. We are going to make use of this lemma to study local approximations of a polynomial map F : E → E, in the neighborhood of a point F (a), where a is any point in E. In order to be sure that the polar form f : E m → E is continuous, let us now assume that E is of finite dimension, → → Since by lemma 10.5.3, the directional derivatives D . . . D F (a) exist for all − , . . . , − u u
u1 uk 1 k
Of course, for k > m, the derivative Dk F (a) is the null k-linear map. Remark: As usual, for k = 0, we agree that D0 F (a) = F (a). We could also relax the condition that E is of finite dimension, and assume that the polar form f is a continuous map. Now, let a be any point in E. For any k, with 0 ≤ k ≤ m, we can truncate the Taylor expansion of F : E → E at a at the (k + 1)-th order, getting the polynomial map Gk : E → E a defined such that, for all b ∈ E, Gk (b) = F (a) + a → − → − 1 1 1 D F (a)( ab) + · · · + Dk F (a)( ab k ). 1! k!
The polynomial function Gk agrees with F to kth order at a, which means that a Di Gk (a) = Di F (a), a
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CHAPTER 10. EMBEDDING AN AFFINE SPACE IN A VECTOR SPACE
for all i, 0 ≤ i ≤ k. We say that Gk osculates F to kth order at a. For example, in the case a of a curve F : A → E, for k = 1, the map G1 is simply the affine map determined by the a tangent line at a, and for k = 2, G2 is a parabola tangent to the curve F at a. a
k As pointed out by Ramshaw, it is tempting to believe that the polar form ga of Gk is a simply obtained from the polar form f of F , by fixing m − k arguments of f at the point a, more precisely, if k ga (b1 , . . . , bk ) = f (b1 , . . . , bk , a, . . . , a), m−k
for all b1 , . . . , bk ∈ E. Unfortunately, this is false, even for curves. The problem is a silly one, it has to do with the falling power mk . For example, if we consider a parabola F : A → A2 , it is easy to see from Taylor's formula, that
1 Gb (a) = 2f (a, b) − f (b, b), 1 and Ga (b) = 2f (a, b) − f (a, a),
which means that f (a, b) is both the middle of the two line segments (f (a, a), G1 (b)) and a (f (b, b), G1 (a)), which happen to be tangent to F at F (a) and F (b). Unfortunately, it is not b true that G1 (b) = G1 (a) = f (a, b). a b It is possible to fix this problem and to find the relationship between the polar forms f k and ga , but this is done most conveniently using symmetric tensors, and will be postponed until section 11.1 (see lemma B.4.3). The ennoying coefficients mk can also be washed out, if we consider the affine subspaces spanned by the range of Gk , instead of osculating curves or surfaces. a Definition 10.5.4. Given any two normed affine spaces E and E, where E is of finite dimension, for any polynomial map F : E → E of degree m, for any a ∈ E, for any k, with 0 ≤ k ≤ m, the polynomial map Gk : E → E is defined such that, for all b ∈ E, a Gk (b) = F (a) + a → − → − 1 1 1 D F (a)( ab) + · · · + Dk F (a)( ab k ). 1! k!
We say that Gk osculates F to kth order at a. The osculating flat Osck F (a) is the affine a subspace of E generated by the range of Gk . a If F : A → E is a curve, then we say that F is nondegenerate iff Osck F (a) has dimension k for all a ∈ A. In such a case, the flat Osc1 F (a) is the tangent line to F at a, and Osc2 F (a) is the osculating plane to F at a, i.e., the plane determined by the point F (a), the velocity vector D1 F (a), and the acceleration vector D2F (a). The osculating plane is the usual notion used in differential geometry. The osculating plane to the curve F at the point F (a) is the limit of any plane containing the tangent line at F (a) and any other point F (b) on the curve F , when b approaches a.
It is not too difficult to show that there are k(k + 3) (k + 1)(k + 2) −1 = 2 2 partial derivatives in the above expression, and we say that the surface F is nondegenerate iff Osck F (a) has dimension k(k+3) for all a ∈ P. For a nondegenerate surface, Osc1 F (a) is 2 the tangent plane to F at a, and Osc2 F (a) is a flat of dimension 5, spanned by the vectors 2 2 ∂F ∂F ∂2F (a), ∂x2 (a), ∂ F (a), ∂ F (a), and ∂x1 ∂x2 (a). The flat Osc3 F (a) is a flat of dimension 9, and ∂x1 ∂x2 ∂x2 1 2 we leave as an exercise to list the vectors spanning it. Thus, plane curves are degenerate in the above sense, except lines, and surfaces in A3 are degenerate in the above sense, except planes. There is a simple relationship between osculating flats and polar forms, but it is much more convenient to use tensors to prove it, and we postpone the proof until section 11.1 (see lemma B.4.4). Let us simply mention a useful corollary. Given a polynomial map F : E → E of degree m with polar form f : E m → E, the affine subspace spanned by the range of the multiaffine map (b1 , . . . , bm−k ) → f (a, . . . , a, b1 , . . . , bm−k ),
k
is the osculating flat Oscm−k F (a). This leads to a geometric interpretation of polar values. We note in passing that the geometric interpretation of polar forms in terms of osculating flats, was investigated by S. Jolles, as early as 1886. Let F : A → E be a nondegenerate curve of degree 3 (and thus, a space curve). By the previous corollary, the polar value f (r, s, t) is the unique intersection of the three osculating planes Osc2 F (r), Osc2 F (s), and Osc2 F (t). The polar value f (s, s, t) is the intersection of the tangent line Osc1 F (s) with the osculating plane Osc2 F (t). More generally, given a nondegenerate curve or surface F : E → E, the polar value f (b1 , . . . , bm ) is the unique intersection of the osculating flats corresponding to all of the distinct points a that occur in the multiset {b1 , . . . , bm }.
This interpretation is so nice that one wonders why it was not chosen as a definition of polar forms. Unfortunately, this idea does not work as soon as F : A → E is degenerate,
396
CHAPTER 10. EMBEDDING AN AFFINE SPACE IN A VECTOR SPACE
which happens a lot. Indeed, B´zier points are usually not affinely independent. Thus, we e had to use a more algebraic definition. In the case of surfaces, osculating flats intersect more often that we would expect. For example, given a nondegenerate cubic surface, we know that it lies in an affine space of dimension 9. But for each a ∈ P, the osculating flat Osc2 F (a) has dimension 5. In general, three 5-flats in a 9-space do not intersect, but any three 5-flats Osc2 F (a), Osc2 F (b), and Osc2 F (c), intersect at the polar value f (a, b, c). Readers who would like to have an in-depth understanding of the foundations of geometric design and a more conceptual view of the material on curves and surfaces are urged to read the next chapter on tensors. However, skipping this chapter will only have very minor consequences (basically, ignoring the proofs of a few results).
This chapter is not absolutely essential and can be omitted by readers who are willing to accept some of the deeper results without proofs (or are willing to go through rather nasty computations!). On the other hand, readers who would like to have an in-depth understanding of the foundations of computer-aided geometric design and a more conceptual view of the material on curves and surfaces, should make an effort to read this chapter. We hope that they will find it rewarding! First, tensor products are defined, and some of their basic properties are shown. Next, symmetric tensor products are defined, and some of their basic properties are shown. Symmetric tensor products of affine spaces are also briefly discussed. The machinery of symmetric tensor products is then used to prove some important results of CAGD. For example, an elegant proof of theorem 5.3.2 is given. We have seen that multilinear maps play an important role. Given a linear map f : E → → F , we know that if we have a basis (−i )i∈I for E, then f is completely determined by its u − ) on the basis vectors. For a multilinear map f : E n → F , we don't know if there → values f ( ui is such a nice property, but it would certainly be very useful. In many respects, tensor products allow us to define multilinear maps in terms of their action on a suitable basis. Once again, as in section 10.1, we linearize, that is, we create a new vector space E ⊗ · · · ⊗ E, such that the multilinear map f : E n → F is turned into a linear map f⊗ : E ⊗ · · · ⊗ E → F , which is equivalent to f in a strong sense. If in addition, f is symmetric, then we can define a symmetric tensor product E ⊙ · · · ⊙ E, and every symmetric multilinear map f : E n → F is turned into a linear map f⊙ : E ⊙ · · · ⊙ E → F , which is equivalent to f in a strong sense. Tensor products can be defined in various ways, some more abstract than others. We tried to stay down to earth, without excess! 397
What is important about lemma 11.1.3 is not so much the construction itself, but the fact that a tensor product with the universal property with respect to multilinear maps stated in that lemma holds. Indeed, lemma 11.1.3 yields an isomorphism between the vector space of linear maps L(E1 ⊗ · · · ⊗ En ; F ), and the vector space of multilinear maps L(E1 , . . . , En ; F ), via the linear map − ◦ ϕ defined by h → h ◦ ϕ,
However, there vectors are not linearly independent. This situation can be fixed when considering bases, which is the object of the next lemma. − → Lemma 11.1.4. Given n ≥ 2 vector spaces E1 , . . . , En , if ( uk )i∈Ik is a basis for Ek , 1 ≤ i k ≤ n, then the family of vectors − → − → (u11 ⊗ · · · ⊗ unn )(i1 ,...,in )∈I1 ×...×In i i
Of course, this is just what we wanted! Tensors in E1 ⊗ · · · ⊗ En are also called n-tensors, → → → and tensors of the form −1 ⊗ · · · ⊗ −n , where −i ∈ Ei , are called simple (or decomposable) u u u n-tensors. Those n-tensors that are not simple are often called compound n-tensors. → → We showed that E ⊗ · · · ⊗ E is generated by the vectors of the form − ⊗ · · · ⊗ − . u u
1 n 1 n
K) is naturally isomorphic to ( n E)∗ . Those → → → u1 un ui symmetric n-tensors that are not simple are often called compound symmetric n-tensors. Indeed. using polynomials. . it turns out that we can easily construct symmetric tensors powers of affine spaces from what we have done so far. As a matter of fact.3. F ). its inverse is the map f → f⊙ . since this might also be helpful to those readers who wish to study Ramshaw's paper [65] carefully. an )). with f (a1 . we now briefly discuss symmetric tensor powers of affine spaces.3 yields an isomorphism between the vector space of linear maps L( n E. Thus. an ) = f⊙ (ϕ(a1 . A symmetric n-th tensor power (or tensor product) of an affine space E. as affine maps f⊙ : E → F . . or for short f = f⊙ ◦ ϕ.
The last identity shows that the "operation" ⊙ is commutative. for every symmetric multilinear map f ∈ S(E n . the vector space of symmetric multilinear forms S(E n . this yields a (noncanonical) isomorphism between the vector space of symmetric multilinear forms S((E ∗ )n .3. the map − ◦ ϕ is bijective. .
407
where h ∈ L( n E. .3. . . Definition 11.2. for every affine space F and for every symmetric multiaffine map f : E n → F .
. such that. .
11. u u
Remark: As in the case of general tensors. for all a1 . F ). and since by lemma 11. and the vector space of symmetric multilinear maps S(E n . . Thus. where − ∈ E. F ). . . and tensors of the form − ⊙· · ·⊙ − . there is a unique affine map f⊙ : S → F . is an affine space S. When the space E has finite dimension.3
Affine Symmetric Tensor Products
Ramshaw gives a construction for the symmetric tensor powers of an affine space.
Lemma 11. h ◦ ϕ is clearly symmetric multilinear. AFFINE SYMMETRIC TENSOR PRODUCTS for all permutations π on n elements. where n ≥ 2. are called simple (or decomposable) symmetric n-tensors. Actually. F ) such that f = f⊙ ◦ ϕ. It is also possible to define tensor products of affine spaces. Symmetric tensors in n E are also called symmetric n-tensors. via the linear map −◦ϕ defined by h → h ◦ ϕ.11. together with a symmetric multiaffine map ϕ : E n → S.1. there is a unique linear map f⊙ ∈ L( n E. and symmetric tensor powers of affine spaces. The motivation is to be able to deal with symmetric multiaffine maps f : E m → m F . we can view the → → symmetric tensor −1 ⊙ · · · ⊙ −n as a multiset. . . an ∈ E.2. K) and n E. F ).
Lemma 11.− → − ⊙ ··· ⊙ − → → → → u1 un u1 un from (E)n to n E. where a1 . . where a1 . is defined by f⊙ (a1 ⊙ · · · ⊙ an ) = f (a1 . . . but with affine spaces and affine and multiaffine maps. . In order to prove the existence of such tensor products. . are as follows: (1) The notion of affine space freely generated by a set I makes sense. . It turns out that if we first homogenize the affine space E. getting the vector space E. If ϕ(a1 . it is easy to show that symmetric tensor powers are unique up to isomorphism. . and behave well. whose proof is left as an exercise. on the generators a1 ⊙ · · · ⊙ an of
n
E. − − → → Then. is a symmetric n-th tensor power for E. . Given an affine space − → − → E and a subspace F of E .. . then the symmetric tensor power E of the affine space E already sits inside n n E consists of all affine E as an affine space! As a matter of fact.
However. is an affine space. Indeed. . determined such that → → − − a ≡F b iff ab ∈ F . . an are points in E. and for every symmetric multiaffine map f : E n → F . . the unique affine map n f⊙ : E → F such that f = f⊙ ◦ ϕ. . ..2. and then construct the symmetric tensor power n n E. an ∈ E. the symmetric tensor power n E is spanned by the tensors a1 ⊙· · ·⊙an . TENSOR PRODUCTS AND SYMMETRIC TENSOR PRODUCTS Equivalently. . the affine space combinations of simple n-tensors of the form a1 ⊙ · · · ⊙ an . . it is identical to the construction of the free vector space K (I) . defined as the restriction to E n of the multilinear map − . we can proceed in at least two ways. . where a1 . .408 CHAPTER 11. an ). and together with the multiaffine map ϕ : E n → n E..3. makes what we just claimed more precise. it can be shown that E/ ≡F is an affine space with associated vector space E / F . an affine congruence on E is defined as a relation ≡F on E.. the subset n E of the vector space n E consisting of all affine combinations of simple n-tensors of the form a1 ⊙ · · · ⊙ an . there is a unique affine map f⊙ such that the following diagram commutes: E nC
/ S CC CC f⊙ C f CC!
ϕ
F
As usual.
. . Given an affine space E. The first method is to mimick the constructions of this section. The following lemma. the reasons why tensor products of affine spaces can be constructed. an are points in E. (2) Quotients of affine spaces can be defined. In fact. there is a more direct approach. an ) is denoted as a1 ⊙· · ·⊙an . .
an ) = h(a1 · · · an ). for any n ≥ 2. . . .4 for symmetric tensor powers. From lemma 11. an ∈ E. The existence of the symmetric tensor power E m justifies the multiplicative notation f (a1 · · · am ) for the polar value f (a1 . we define the set JM of functions η : {1. by lemma A. . it is sufficent to show that every linear map n h: E → F is uniquely determined by the tensors of the form an .
409
Let E and E be two affine spaces. . since (I) i∈I M(i) = i∈dom(M ) M(i). . We let dom(M) = {i ∈ I | M(i) = 0}. am ) = f⊙ (a1 ⊙ · · · ⊙ am ).
It is worth noting that the affine tensor power n E is spanned not only by the simple n-tensors a1 · · · an .2. . am ∈ E.1. u u u u n generate E. that is. . Following Ramshaw. . This is because Ramshaw defines the hat construction as a special case of his construction of the affine symmetric tensor power. . and dom(M) is finite. and write tensors a1 ⊙ · · · ⊙ am simply as a1 · · · am . which is a finite set. this is a consequence of lemma 4. and this is quite easy to show. am ): view the notation f (a1 · · · am ) as an abbreviation for f⊙ (a1 ⊙ · · · ⊙ am ). .11. for any multiset M ∈ N(I) . such that M(i) = 0 for finitely many i ∈ I. . Indeed. note that the sum i∈I M(i) makes sense.1. as follows: JM = {η | η : {1. and we simply followed an approach closer to traditional linear algebra. . However. . . we can associate a unique affine map f⊙ : E m → E.2. and that the set of all multisets over I is denoted as N(I) . . since f is symmetric multiaffine and h = f⊙ is the unique affine map such that f (a1 .2.4.4. .5. Left as an exercise. for all a1 . . where a1 . we call the affine map f⊙ . −n ∈ E.3. . We personally favor the reading n E. For this. By lemma 11. . for all a1 . i ∈ dom(M).4
Properties of Symmetric Tensor Products
→ → → → Going back to vector spaces. but Ramshaw favors the reading n E. . . . Then. . the vectors −1 ⊙ · · · ⊙ −n . if we let f = h ◦ ϕ. .3. . . we just have to show that a symmetric multiaffine map is uniquely determined by its behavior on the diagonal. . given any polynomial affine map F : E → E of degree m with polar form f : E m → E. PROPERTIES OF SYMMETRIC TENSOR PRODUCTS Proof. an ∈ E. such that f (a1 . . |η −1 (i)| = M(i). . recall that a (finite) multiset over a set I is a function M : I → N. The answer is that that they are isomorphic. whereas Ramshaw followed an approach in which affine spaces played a more predominant role.
11. one may wonder what is the relationship between E and n E. the two approaches are equivalent.
n Given an affine space E. In the end. n} → dom(M). omitting the symbol ⊙. where −1 . again. We will prove a version of lemma 11. . . n} → dom(M). . For every multiset M ∈ N .
i∈I
. but they are not linearly independent. . but also by the simple n-tensors of the form an . drop the subscript ⊙ from f⊙ . M(i) = n}. the affine blossom of F . . where a ∈ E. .
we have avoided talking about algebras (even though polynomial rings are a prime example of algebra). and if these tensors are linearly independent in n A. We could also have defined S(E) from T(E). First. multisets of size n + 1 consisting of points u1 . is an interesting object. then S(E) is obtained as the quotient of T(E). .4. called the symmetric tensor algebra of E. This elegant proof is due to Ramshaw.5
Polar Forms Revisited
When E = A. Very elegant. we have n + 1 tensors uk+1 ⊙ · · · ⊙ un+k .412 CHAPTER 11.2.1). by saying that an algebra over a field K (or a ring A) consists basically of a vector space (or module) structure. The generalization to the symmetric tensor product f1 ⊙ · · · ⊙ fn of n ≥ 3 linear maps fi : E → F is immediate. and 0 E = K). When E is of finite dimension n.
11. Then.
2
. . by the subspace of T(E) generated by all vectors in T(E). given any progressive sequence u1 . is that S(E) provides an intrinsic definition of a polynomial ring (algebra!) in any set I of variables. it can be shown to correspond to the ring of polynomials with coefficients in K in n variables (this can be seen from lemma 11. . and left to the reader. which is easily done. . . . then they form a basis of n A. we can easily verify that (f ′ ◦ f ) ⊙ (g ′ ◦ g) = (f ′ ⊙ g ′ ) ◦ (f ⊙ g). we give a short proof of the hard part of theorem 5.
Let me tease the reader anyway. but this would also require defining a multiplication operation (⊗) on T(E). we recall theorem 5. un+1 in A can be viewed as symmetric tensors u1 ⊙ · · · ⊙ un+1 . Actually. . When this operation is associative and has an identity. . an algebra essentially has both a vector space (or module) structure. corresponds to the ring of polynomials in infinitely many variables in I. we would have to define a multiplication operation on T(E). → → → → of the form − ⊗ − − − ⊗ − .
(with 1 E = E. and we will continue to do so. u2n .2 What's nice about the symmetric tensor algebra. → When E is of infinite dimension and (−i )i∈I is a basis of E.3.3. If this is done. and a ring structure. Remark: The vector space
m
S(E) =
m≥0
E.2. together with a multiplication operation which is bilinear. where 0 ≤ k ≤ n. the vector space S(E) (also u denoted as Sym(E)). As a reward to the brave readers who read through this chapter. However. TENSOR PRODUCTS AND SYMMETRIC TENSOR PRODUCTS If we also have linear maps f ′ : E ′ → E ′′ and g ′ : E ′ → E ′′ . isn't it? u v v u We can finally apply this powerful machinery to CAGD.
A family (ai )i∈I is finite if I is finite. by Strang [81]. projectives. wi = ui if i ∈ I. e e e e G´om´trie 1 and 2. we say that a family (ai )i∈I has finite support iff ai = 0 for all i ∈ I − J. by Mac Lane and Birkhoff [52]. and wk = v. We recommend the following excellent books for an extensive treatment of linear algebra. the notation and terminology is a bit strange! The text Geometric Concepts for Geometric Design. polynomials. We begin by reviewing some basic properties of vector spaces. and euclidiennes. Alg`bre Lin´aire et G´om´trie El´mentaire by Dieudonn´ [25]. In fact. Algebra. The first two chapters of Strang [81] are highly recommended. The appendices contain no original material. 15]. We can deal with an arbitrary set X by viewing it as the family (Xx )x∈X corresponding to the identity function id : X → X. Alg`bre. 425
. a subfamily of (ui )i∈I is a family (uj )j∈J where J is any subset / of I. by e e e e Tisseron [83]. by Van Der Waerden [84]. Given a set A. by Bourbaki [14. by Bertin [8]. Algebra.Appendix A Linear Algebra
A. by Berger [5. Given a family (ui )i∈I . (ai )i∈I = ∅. there is probably more material than really needed. Alg`bre lin´aire et g´om´trie classique. we denote as (ui )i∈I ∪k (v) the family (wi )i∈I∪{k} defined such that. and geometry: Linear Algebra and its Applications. except perhaps for the presentation and the point of view. Algebra. by Lang [47]. e e e e e e Another useful and rather complete reference is the text Finite-Dimensional Spaces. 6]. by Walter Noll [57]. recall that a family (ai )i∈I of elements of A is simply a function a : I → A. by Michael Artin [1]. by Boehm and Prautzsch [11] is also an excellent reference on geometry geared towards computer aided geometric design. We agree that when I = ∅. G´om´tries affines. But beware.1 Vector Spaces
The purpose of the appendices is to gather results of Linear Algebra and Analysis used in our treatment of the algorithmic geometry of curves and surfaces. and we advise our readers to proceed "by need". Given a family (ui )i∈I and any element v. where J is a finite subset of I (the support of the family). If A is ring with additive identity 0. where k is any index such that k ∈ I. e Algebra 1.
and all λ. The subspace { 0 } will be denoted as 0 (with a mild abuse of notation). Letting λ = µ = 0. − ∈ F . → → A family (−i )i∈I is linearly dependent iff some −j in the family can be expressed as a u u linear combination of the others vectors in the family. u
i∈I
→ Equivalently. → Given a vector space E. → → − λi −i = 0 implies that λi = 0 for all i ∈ I.1. Indeed.1. we stipulate that − = 0 .
. the subset V of E consisting of the v − → → null vector 0 and of all linear combinations of (−i )i∈I is easily seen to be a subspace of v E. and motivate the following definition. a subset F of E is a linear subspace (or subspace) → → → → of E iff F is nonempty and λ− + µ− ∈ F for all − .2. given a family (−i )i∈I . − → − → we see that every subspace contains the vector 0 . µ ∈ K.1.A. u j
The notion of a subspace of a vector space is defined as follows. the family ∅ is linearly independent. Subspaces having such a "generating family" play an important role. VECTOR SPACES
427
→ Definition A. a family (−i )i∈I is linearly dependent iff there is some family (λi )i∈I of scalars u in K such that → → − λi −i = 0 and λj = 0 for some j ∈ I. Given a vector space E. and that any intersection of subspaces is a subspace. We say that a family (−i )i∈I is linearly independent v u iff for every family (λi )i∈I of scalars in K. A vector − ∈ E is a linear combination of a v − ) of elements of E iff there is a family (λ ) of scalars in K such that → family ( ui i∈I i i∈I − = → v
i∈I
→ λi −i . Definition A. u v u v It is easy to see that a subspace F of E is closed under arbitrary linear combinations of vectors from F . u
→ → − → When I = ∅. Let E be a vector space. u − = → uj
i∈(I−{j})
i∈I
which implies that
→ −λ−1 λi −i .3. there is some family (λi )i∈I of scalars in K such that → → − λi −i = 0 and λj = 0 for some j ∈ I. u
i∈I
We agree that when I = ∅.
If a subspace V of E is generated by a finite family (− ) . if B does not generate E.4. by lemma / −) → ′ A. If µ = 0. We begin with a crucial lemma. Given any finite family S = (−i )i∈I generating a vector space E and any u −) → linearly independent subfamily L = ( uj j∈J of S (where J ⊆ I). In particular. We prove the above result in the case where a vector space is generated by a finite family. Consider the set of linearly independent families B such that L ⊆ B ⊆ S. B is a basis of E such that L ⊆ B ⊆ S. every basis of E has n elements. it has some maximal element.428
APPENDIX A. we have λ − = 0 . Thus. Assume that µ− + i∈I λi −i = 0 . and the integer n is called the dimension of the vector space E. with p ∈ H. we only prove the theorem for finitely generated vector spaces. there is some family (λ ) of → → vectors vi v i i∈I scalars in K such that − = → → λi −i . v v
i∈I
→ → We also say that the elements of (−i )i∈I are generators of V and that V is spanned by (−i )i∈I . Given a vector space E and a subspace V of E. if v ui i∈I ui i∈I k → v − to the family (− ) is linearly independent (where k ∈ I). v u → → then µ has an inverse (because K is a field).
. then the family (− ) ∪ (− ) obtained → → → E. and since the family (− ) is linearly independent. for any family (λi )i∈I of scalars in K. a family (−i )i∈I of v − ∈ V spans V or generates V iff for every − ∈ V . µ = 0. Since this → set is nonempty and finite. LINEAR ALGEBRA
→ Definition A. Indeed. v v − ) .5. → Theorem A. showv u − is a linear combination of (− ) and contradicting the hypothesis. Then. It is a standard result of linear algebra that every vector space E has a basis. A family ( ui i∈I called a basis of V .1. → → by adding v ui i∈I / → → → − Proof. and since L ⊆ B ⊂ B ′ ⊆ S. I and J have the same cardinality. Thus. and thus we have − = − i∈I (µ−1 λi )−i . Proof. and that → → for any two bases (−i )i∈I and (−j )j∈J . this contradicts the maximality of B. there is a basis B of E such that L ⊆ B ⊆ S. We claim that u → B generates E. Thus.6. but the proof is more sophisticated for vector spaces that do not have a finite set of generators (it uses Zorn's lemma). we u u have λi = 0 for all i ∈ I. if E u v has a finite basis of n elements.
i∈I i i i i∈I
Note that the proof of the above lemma holds for arbitrary vector spaces.1. we say → → or generated by ( vi i∈I vi i∈I − ) that spans V and is linearly independent is → that V is finitely generated . → → ing that v ui i∈I − → → → But then. not only for finitely generated vector spaces. say B = (−h )h∈H . the family B = ( uh h∈H∪{p} is linearly independent. The next theorem also holds in general. → Lemma A.1.1. Given a linearly independent family (−i )i∈I of elements of a vector space u − ∈ E is not a linear combination of (− ) .5. then there is some −p ∈ S that is not u a linear combination of vectors in B (since S generates E).
−p is a u → linear combination of (−i )i∈(I−{p}) ∪ (−→)l∈L . for any family B = (−i )i∈I of vectors of E. we can replace L by (L − {p}) ∪ {p } where p′ does not belong to I ∪ L. The following lemma giving useful properties characterizing a basis is an immediate consequence of theorem A. Appendix §1 (pp. Then. Let u vρ(l) − = → up
i∈(I−{p})
→ λi −i + u
l∈L
λl −→ vρ(l)
(1)
If λl = 0 for all l ∈ L. Given a vector space E.1. there exists a set L and an injection ρ : L → J → combination of ( vj j∈J → → such that L ∩ I = ∅. By the induction u hypothesis. and the families (−i )i∈(I−{p}) ∪(−→)l∈L and (−j )j∈J generate the same subspace u vρ(l) v ′ of E.A. §5 (pp. When m = 0.6 also holds for vector spaces that are not finitely generated.1. and the families (−i )i∈I ∪ (−→)l∈L and (−j )j∈J generate u vρ(l) v the same subspace of E. The existence of such a maximal family can be shown using Zorn's lemma. VECTOR SPACES
429
Theorem A.1.6 should consult either Lang [47].1. Readers who want to know more about Zorn's lemma and the proof of theorem A. Given a vector space E. 588-589).
. where p is any member of I. (3) B is a minimal generating family of E. m ≤ n. we can always assume that L ∩ I = ∅.6. In particular. → Lemma A.8.1. In this case. → → |L| = n−m. we have
i∈(I−{p})
→ → u → − u λi −i − −p = 0 . where |J| = n. (2) B is a maximal linearly independent family of E. §2 (pp. Appendix 2. → → Thus.1. or Artin [1]. |L| = n − m. 139-140). and let ( vj j∈J be any finite family such that every −i is a linear u − ) . Since −p is a linear combination of (−j )j∈J and u v −) → −→) −) → → the families ( ui i∈(I−{p}) ∪ (vρ(l) l∈L and ( vj j∈J generate the same subspace of E. Assume |I| = m + 1. the v following properties are equivalent: (1) B is a basis of E. and u the lemma holds trivially with L = J (ρ is the identity).
→ Proof. If p ∈ L.7. let (−i )i∈I be any finite linearly independent family u −) → → in E. the family (−i )i∈I is empty. → Lemma A. The following replacement lemma shows the relationship between finite linearly independent families and finite families of generators of a vector space. the problem is to garantee the existence of a maximal linearly independent family B such that L ⊆ B ⊆ S. Consider → the linearly independent family (−i )i∈(I−{p}) . and replace ρ by the injection ρ′ which agrees with ρ on L − {p} and such that ρ′ (p′ ) = ρ(p). We proceed by induction on |I| = m. 878-884) and Chapter III. there exists a set L and an injection ρ : L → J such that L ∩ (I − {p}) = ∅. where |I| = m.
say u l = q. we must have I = L. Let (−i )i∈I be a basis of E. since otherwise. v → → we obtain |J| ≤ |I|. Thus. and assume that I is infinite. one can prove that lemma A. and by lemma A.1.1. vi = 0}. this would contradict the fact that ( → u
i i∈I
is linearly independent.1. → Theorem A.1. for every two bases → → (− ) and (− ) u v of E. A symmetric argument yields |J| ≤ |I|.6 and lemma A. since I = j∈J Lj with J infinite and the Lj finite. Thus. Since λq = 0. This → → can be shown as follows. Actually. lemma A. and vice-versa. we have −→ = v− ρ(q) → → (−λ−1 λi )−i + λ−1 −p + u q q u
i∈(I−{p}) l∈(L−{q})
(−λ−1 λl )−→ vρ(l) q
(2)
→ → We claim that the families (−i )i∈(I−{p}) ∪ (−→)l∈L and (−i )i∈I ∪ (−→)l∈(L−{q}) generate u vρ(l) u vρ(l) the same subset of E. we obtain the following fundamental theorem. let Lj ⊆ I be the finite set → Lj = {i ∈ I | −j = v Let L =
j∈J
→ vi −i . λl = 0 for some l ∈ L. Let E be a finitely generated vector space. J must be infinite. let (−j )j∈J be a generating family of u v E.8 implies that |I| ≤ |J|. The first part follows immediately by applying theorem A. LINEAR ALGEBRA
→ contradicting the fact that (−i )i∈I is linearly independent.8 implies theorem A. and thus.1.9. Assume that (−i )i∈I and (−j )j∈J are bases of E.6 when the vector space if finitely generated. and vρ(q) ui i∈I vρ(l) l∈(L−{q}) .9 also holds for vector spaces that are not finitely generated.1. Any family (−i )i∈I generating u −) → E contains a subfamily ( uj j∈J which is a basis of E. Furthermore. |I| = |J| for any two bases (− ) and (− ) u v of E.
i i∈I j j∈J
Proof.8 together. |I| ≤ |J|. the families − ) ∪ (−→) → −) → ( ui i∈I vρ(l) l∈(L−{q}) and ( vj j∈J generate the same subspace of E.1.6 with L = ∅ and → → → → S = (−i )i∈I . we say that E is of infinite dimension. The dimension of a vector space
. Indeed. Putting theorem A. For every j ∈ J. the dimension |I| being a cardinal number which depends only on the vector space E. because the Lj are finite. If (−j )j∈J is also a basis.
i i∈I j j∈J
When |I| is infinite. u
i∈I
→ → Lj . by a → standard result of set theory. But then. and the lemma holds for L − {q} and the restriction of the injection ρ : L → J to L − {q}. Remark: Theorem A. since L ∩ I = ∅ and |L| = n − m imply that (L − {q}) ∩ I = ∅ and |L − {q}| = n − (m + 1). by a symmetric argument. since otherwise (−i )i∈L u u −) would be another basis of E. Since (−i )i∈I is a basis of E. we have |I| = |J| = n.5.430
APPENDIX A. by vρ(l) ρ(q) −→ is a linear combination of (− ) ∪ (−→) − → (1).1. the second family is obtained from the first by replacing −→ by − . I would be finite. and − is a linear combination of (− ) → → → v− up up ui i∈(I−{p}) ∪ (−→)l∈L . Since (−i )i∈I is linearly u u v u −) → independent and ( vj j∈J spans E. Furthermore.1. by (2).
there would be a family (µi )i∈I of scalars not all null such that → → − µi −i = 0 u
i∈I
and µj = 0 for some j ∈ I. we can form a vector space K (I) which. Clearly. λi = µi for all i ∈ I. since the family u v − ) generates E. Thus dim(K) = 1. contradicting the assumption that (λi )i∈I is the unique family → → such that − = i∈I λi −i .1. let (−i )i∈I be a family of vectors in E. → → Let (−i )i∈I be a basis of a vector space E. → → v ui i∈I i∈I λi ui → Proof. the family (λ ) → v v i i∈I of scalars such that i∈I λi ui − = → − is unique iff (− ) is linearly independent. Let u − ∈ E. → Lemma A.A. for any vector − ∈ E. The converse is shown by contradiction.1. If (µi )i∈I is another family of u − = → − → scalars in K such that v i∈I µi ui . assume that (−i )i∈I is linearly independent. then every family (a) where a ∈ K and a = 0 is a basis. For any vector − ∈ E. VECTOR SPACES
431
E is denoted as dim(E). If (−i )i∈I was linearly u dependent. v u → → If (−i )i∈I is a basis of a vector space E. Then. → ( ui i∈I Given a field K and any (nonempty) set I. there is a family (λ ) of scalars K such that → (u
i i∈I i i∈I
− = → v
i∈I
→ λi −i . if (vi )i∈I is the unique u v family of scalars in K such that − = → → v vi −i .
. But then. and assume that − = → → − . if the field K itself is viewed as a vector space. u
with λj = λj +µj since µj = 0. u
A very important fact is that the family (λi )i∈I is unique. that u → is. − = → v
i∈I
→ → − λi −i + 0 = u
i∈I
→ λi −i + u
i∈I
→ µi −i = u
i∈I
→ (λi + µi )−i . we must have λi − µi = 0 for all i ∈ I. u
i∈I
→ each vi is called the component (or coordinate) of index i of − with respect to the basis v −) . then we have → → − (λi − µi )−i = 0 .10. First. in some sense. u
i∈I
→ and since (−i )i∈I is linearly independent. Given a vector space E. is the standard vector space of dimension |I|.
or linear map. n}. is e e
i i∈I i i j i i
clearly a basis of the vector space K (I) . Linear maps formalize the concept of linearity of a function. a linear map between E and F is a function f : E → F satisfying the following two conditions: → → → → f (− + − ) = f (− ) + f (− ) x y x y − ) = λf (− ) → → f (λ x x → → for all − . In fact. dim(K (I) ) = |I|. Definition A. x y → for all λ ∈ K. given any family (λi )i∈I of scalars in K.11.2.1. Given a → family (−i )i∈I of vectors in E. we get f ( 0 ) = 0 . addition and multiplication by a scalar are well defined. Further−→ − −→ − → → more.432
APPENDIX A. Given two vector spaces E and F . LINEAR ALGEBRA
Definition A.
A. we have u f(
i∈I
→ λi −i ) = u
i∈I
→ λi f (−i ). − ∈ E. is clearly an injection. because families with finite support are considered. such that ι(i) = −i for every i ∈ I.
2
Where K I denotes the set of all functions from I to K. When I = {1. . The → function ι : I → K (I) . Thus.2 We define addition and multiplication by a scalar as follows: (λi )i∈I + (µi )i∈I = (λi + µi )i∈I .2
Linear Maps
A function between two vector spaces that preserves the vector space structure is called a homomorphism of vector spaces. K (I) = K I . but dim(K I ) is strictly greater when I is infinite.2. and λ · (µi )i∈I = (λµi )i∈I . but this is false when I is infinite. . . K (I) is a vector space. It is immediately verified that. − ∈ E.1. e
When I is a finite set. x
− → − → − → → → Setting − = − = 0 in the first identity.1.
. Given a field K and any (nonempty) set I. u
→ u The above identity is shown by induction on the size of the support of the family (λi −i )i∈I . . we denote K (I) as K n . the family (− ) of vectors − defined such that (e ) = 0 if j = i and (e ) = 1. let K (I) be the subset of the cartesian product K I consisting of families (λi )i∈I with finite support of scalars in K. using the properties of definition A. The basic property of x y linear maps is that they transform linear combinations into linear combinations.
the following properties hold: u → → (1) The family (−i )i∈I generates E iff for every family of vectors (−i )i∈I in F . since f = f ◦ ι. Since (− ) → → → → E. there is a unique linear map f : K (I) → F such that → f (−i ) = f (i) for every i ∈ I.5. (1) If there is any linear map f : E → F such that f (−i ) = −i for all i ∈ I. there is some vector w
. and (f (i))i∈I is a family e of vectors in F .2.2. there is at u v − ) = − for all i ∈ I. → for every i ∈ I. we must have → f (− ) = x
i∈I
→ xi f (−i ) = u
i∈I
→ xi −i . However. since u v − ) generates E. LINEAR MAPS
435
Lemma A. and by lemma A. v
→ This shows that f is unique if it exists. such that f = f ◦ ι.3. → → there is some linear map f : E → F such that f ( ui vi
→ xi −i . given any family → (−i )i∈I of vectors in E. Since F is nontrivial. which proves the existence and uniqueness of a linear map f e such that f = f ◦ ι. u v − ) = − for all i ∈ I. for any vector space F .2. and for any function f : I → F . → → most one linear map f : E → F such that f ( ui vi
ι
→ → Proof. every vector − ∈ E can written as some linear combination → → ( ui i∈I x − = → x
i∈I
→ → (2) The family (−i )i∈I is linearly independent iff for every family of vectors (−i )i∈I in F . Given any two vector spaces E and F . there is a unique linear map f : K (I) → F . The following simple lemma will be needed later when we study spline curves.4. we must have → e f (i) = f (ι(i)) = f (−i ). as in the following diagram: I CC / K (I) CC CC f C f C! C F Proof. Conversely. If such a linear map f : K (I) → F exists. Given any set I. u
and by linearity.2. with F nontrivial. the family (−i )i∈I is a basis of K (I) . assume that (−i )i∈I does not generate u − ∈ F such that − = − . Lemma A. there is some some vector y y 0 ui i∈I − ∈ E which is not in the subspace generated → does not generate E.A.
(−i )i∈I is linearly independent.1.2. u
i∈I
→ By the assumption. Conversely. defining f : E → F to be the constant linear → family in F such that vi 0 − → − → → map with value 0 .1.
Lemma A. and since f = g. the relation ≡M is an equivalence relation with the following two congruential properties: → → → → → u → → → 1. and f (− ) = − . Given any vector space E and any subspace M of E. there is a basis ( ej j∈I0∪J of E. and let M be any subspace of E. for j ∈ I − {i}. we have a linear map such that f (−i ) = 0 for all i ∈ I. then −1 + −2 ≡M −1 + −2 .
A. By definition of the basis (−j )j∈I0 ∪J of E. there is a unique linear map g : E → F such that g(− ) = − .6. we would get → → y u 0
i j
− → 0 = fi (
i∈I
→ λi −i ) = u
i∈I
→ → λi fi (−i ) = λi − . → → → → u v u v
M
We have the following simple lemma. → (2) if the family (−i )i∈I is linearly independent. and by lemma A.1. The subspace M induces a relation → → ≡M on E. there is some linear map fi : E → F . The next section may be omitted until needed. this implies λi = 0. for some j0 ∈ J. and u v u v u v v
. Then. − ∈ E. for all w y e − → → → j ∈ (I0 ∪ J) − {j0 }. for all i ∈ I0 . for every i ∈ I. Letting (−i )i∈I be the e u w ej v → − = − for all i ∈ I. we have. such that fi (−i ) = u → − .3. u y
→ → − → and since − = 0 . and g(−j ) = 0 . for every i ∈ I. and they are needed to define tensor products. If −1 ≡M −1 and −2 ≡M −2 . Thus.2. defined as follows: For all − . there is some family
i i∈I
(λi )i∈I of scalars (not all zero) such that
→ → − λi −i = 0 . u v − ≡ − iff − − − ∈ M. Then. there is a linearly independent subfamily (−i )i∈I0 of (−i )i∈I u u u −) → generating the same subspace. assume that (− ) u is linearly dependent.3.3. and − = −0 . then the conclusion follows by lemma u → A. y u Although in this course. By lemma u − → → → → A. this contradicts the fact that there is at most one such map.6 again. By lemma A. g(−i ) = 0 for all e u i ∈ I. LINEAR ALGEBRA
→ → → by (−i )i∈I . → → → → → such that −i = −i .436
APPENDIX A.3
Quotient Spaces
Let E be a vector space. which are useful to provide nice conceptual proofs of certain properties of splines. they are fundamental in algebra. we will not have many occasions to use quotient spaces.
For every i ∈ I. defined such that. let (Ei )i∈I be a family of vector spaces. and let f : E → F be a linear map. (
i∈I
hi ) ◦ ini = hi . such that f ◦ s = idF .4. We also have injection maps ini : Ei → where fi = x. (f ( ui i∈I vj j∈J − = f (− ). Let I be any nonempty set. The (external) direct sum i∈I Ei of the family (Ei )i∈I is defined as follows:
i∈I Ei consists of all f ∈ i∈I Ei . defined such that. Definition A. for all i ∈ I. First. Let (−i )i∈I be a basis of E. then. let us recall the notion of the product of a family (Ei )i∈I . By lemma A. and for every family (hi )i∈I of linear maps hi : Ei → G. DIRECT SUMS
441
We now give the definition of a direct sum for any arbitrary nonempty index set I.1. then i∈I Ei = ∅. there is a surjective linear map r : F → E called a retraction. by lemma u − )) is linearly independent in F .3. we denote . for every i ∈ I. It is one of the many versions of the axiom of choice.
We also have the following basic lemma about injective or surjective linear maps. ini (x) = (fi )i∈I .
. is the set of all functions f : I → i∈I Ei . its product i∈I Ei . there is a basis (− ) → → A. and let (Ei )i∈I be a family of vector spaces. is often denoted as (fi )i∈I .A. and fj = 0. we find the vector space K of definition A. and let G be any vector space.
→ Proof. Lemma A. Let E and F be vector spaces. for all i ∈ I. πi ((fi )i∈I ) = fi . Let I be any nonempty set.1. and r(− ) = − for all → → → → r : F → E can be defined such that r( vi ui vj w
Lemma A. f (i) ∈ Ei . when i∈I Ei as E (I) Ei = K. Given a family of sets (Ei )i∈I .
such that. and addition and multiplication by a scalar are defined as follows:
(fi )i∈I + (gi )i∈I = (fi + gi )i∈I . such that r ◦ f = idE . for every i ∈ I. We now define direct sums.4. In particular.2.2. If f : E → F is injective.7.3.
i∈I
Ei .4. then. If f : E → F is surjective. λ(fi )i∈I = (λfi )i∈I .4.6. which have finite support. where I ⊆ J. By theorem A. if Ei = ∅ for every i ∈ I. The direct sum i∈I Ei is a vector space.8. there is a unique linear map hi :
i∈I i∈I
Ei → G. that. A member f ∈ i∈I Ei .
(I) Remark: When Ei = E. for all j ∈ (I − {i}).9. for all i ∈ I. we have the projection πi : i∈I Ei → Ei .4. for all i ∈ I. there is an injective linear map s : F → E called a section. such that.
The following lemma is an obvious generalization of lemma A. and where vi ui − ) = − . a linear map → → of F . Since f : E → F is an injective linear map.11.2.
the reader is urged to consult Bourbaki ([14] Chapter III. tensor algebras. very extensive!) of tensors.B. etc.
. OSCULATING FLATS REVISITED
465
geometry. Chapter IV). Determinants can also be defined in an intrinsic manner. For an extensive treatment (in fact.4. The exterior algebra (E) can also be obtained as the quotient of T(E). by the subspace of T(E) generated by → → all vectors in T(E). of the form − ⊗ − (this requires defining a multiplication operation u u (⊗) on T(E)). and [15].
466
APPENDIX B. COMPLEMENTS OF AFFINE GEOMETRY
.
71. z. First.1.1. (D3) d(x. y) + d(y. called a metric. z)| ≤ d(x. We recommend the following texts for a thorough treatment of topology and analysis: Topology. 73]. A metric space is a set E together with a function d : E × E → R+ . y) − d(y. Next. assigning a nonnegative real number d(x. and the Analysis Courses Analyse I-IV. z).Appendix C Topology
C. Undergraduate Analysis. normed vector spaces are defined. Recall that R+ = {x ∈ R | x ≥ 0}. y. y. the length of any side is bounded by the sum of the lengths of the other two sides. by Munkres [54]. We begin with metric spaces. y) to any two points x. as |x| = a2 + b2 . 72. y) = d(y. and we first review these notions. Let us give some examples of metric spaces. condition (D3) expresses the fact that in a triangle with vertices x. Definition C. Most spaces considered in this book have a topological structure given by a metric or a norm. and for a complex √ number x = a + ib. or distance. 467
. Recall that the absolute value |x| of a real number x ∈ R is defined such that |x| = x if x ≥ 0.1 Metric Spaces and Normed Vector Spaces
This appendix contains a review of basic topological concepts. z ∈ E: (D1) d(x. From (D3). z) ≤ d(x. by Schwartz [70. y) ≥ 0. by Lang [48]. a First Course. (symmetry) (positivity) (triangular inequality)
Geometrically. y) = 0 iff x = y. and satisfying the following conditions for all x. we immediately get |d(x. The chapter ends with the definition of a normed affine space. and d(x. metric spaces are defined. x). |x| = −x if x < 0. y ∈ E. z). and their basic properties are stated. Closed and open sets are defined. (D2) d(x.
b ∈ R such that a < b. a + ρ[. For example. with ρ > 0. and d(x. b]. A subset X of a metric space E is bounded if there is a closed ball B(a. One should be aware that intuition can be midleading in forming a geometric image of a closed (or open) ball. We have the Euclidean metric d2 (x. In E = R3 with the Euclidean metric. yn ). an open ball of center a and radius ρ is the set of points inside the disk of center a and radius ρ. For this. Then.
We will need to define the notion of proximity in order to define convergence of limits and continuity of functions. for every a ∈ E. closed on the right)
]a. [a. x) ≤ ρ} is called the closed ball of center a and radius ρ. and d(x. an open ball of center a and radius ρ is the set of points inside the sphere of center a and radius ρ. ]a. ρ) ∪ S(a. xn ) and (y1 . x) < ρ} S(a. b] = {x ∈ R | a < x ≤ b}. ρ) = {x ∈ E | d(a. x) = ρ}
Let E = [a. x) = 0. . (closed interval) (open interval) (interval closed on the left.
Example 3: For every set E. and a closed ball of center a and radius ρ ≥ 1 consists of the entire space! Clearly. and the set
is called the sphere of center a and radius ρ. we can define the discrete metric. b[ = {x ∈ R | a ≤ x < b}. . an open ball of center a and radius ρ is the open interval ]a − ρ. Definition C. defined such that d(x. y) = |x − y|.
is called the open ball of center a and radius ρ. Given a metric space E with metric d. and d(x. . ρ) such that X ⊆ B(a. the set B0 (a. y) = |x1 − y1 |2 + · · · + |xn − yn |2
1 2
. .
. not +∞). It should be noted that ρ is finite (i. ρ) = {x ∈ E | d(a. we introduce some standard "small neighborhoods". b]. we define the following sets: [a. This is the so-called natural metric on R.e. excluding the boundary points on the circle. . ρ). ρ). . for every ρ ∈ R. a closed ball of center a and radius ρ < 1 consists only of its center a. TOPOLOGY
Example 1: Let E = R. y) = |x − y|. In E = R2 with the Euclidean metric. excluding the boundary points on the sphere. B(a. b] = {x ∈ R | a ≤ x ≤ b}. In E = R with the distance |x − y|. b[ = {x ∈ R | a < x < b}. . . the set B(a. open on the right) (interval open on the left. y) = 1 iff x = y. if d is the discrete metric. d) is a metric space.1.2. ρ) = {x ∈ E | d(a. the absolute value of x − y.468
APPENDIX C.
the distance between the points (x1 . ρ) = B0 (a. ([a. Example 2: Let E = Rn (or E = Cn ). Example 4: For any a.
1. . A subset U ⊆ E is an open set in E iff either U = ∅. and of a topological space. defined such that. Let E be a metric space with metric d. .
∞
Some work is required to show the convexity inequality for the Euclidean norm. ρ) ⊆ U. defined such that. We may use the notation B(ρ) and B0 (ρ). . The following lemma is easy to show. given n intervals [ai . In fact.470 → we have the Euclidean norm − . every open ball of center a is contained in E. x
2
APPENDIX C.5. xn ) ∈ E | ai < xi < bi . there is some open ball B0 (a. The set E itself is open. or for every a ∈ U. since for every a ∈ E. .1 A subset F ⊆ E is a closed set in E iff its complement E − F is open in E. 1 ≤ i ≤ n}. ≤ n x
2
In a normed vector space. ρ) such that. but this can be found in any standard text. B0 (a. bi ].4. . Definition C. In E = Rn . . Note that the Euclidean distance is the distance associated with the Euclidean norm. it is easy to show that the open n-cube {(x1 . TOPOLOGY
− → x → and the sup-norm − x
∞
2
= |x1 |2 + · · · + |xn |2
1 2
. we define a closed ball or an open ball of radius ρ as a closed − → ball or an open ball of center 0 . it is possible to find a metric for which such open n-cubes are open balls! Similarly. with ai < bi . 1 ≤ i ≤ n} is an open set. we can define the closed n-cube {(x1 . We will now define the crucial notions of open sets and closed sets. x ∞ √ − → .
. − → x = max{|xi | | 1 ≤ i ≤ n}. xn ) ∈ E | ai ≤ xi ≤ bi . x ≤n − ∞ √ − ≤ n → . . → → Lemma C. which is a closed set.1. . The following inequalities hold for all − ∈ Rn (or − ∈ Cn ): x x − → x − → x − → x
∞ ∞ 2
→ ≤ − x → x ≤ − → x ≤ −
1 2 1
→ . The open sets satisfy some important properties that lead to the definition of a topological space.
1
Recall that ρ > 0.
.
which is the product topology.2
Continuous Functions. the family of open sets is not closed under infinite intersections. x) ≤ η. . i. The reader is referred to standard texts on topology. . which is not open. If each (Ei . a ∈ Ua . . Limits
If E and F are metric spaces defined by metrics d1 and d2 .5 satisfies the following properties: (O1) For every finite family (Ui )1≤i≤n of sets Ui ∈ O. One can also verify that when Ei = R. . LIMITS
471
Lemma C. i ) is a normed vector space. and Ua ∩ Ub = ∅. b)/2 works too). and E ∈ O. . there is some η > 0. . .2.
C. The above lemma leads to the very general concept of a topological space. we have under arbitrary unions. but n Un = {0}. the topology product on Rn is the standard topology induced by the Euclidean norm. the family O of open sets defined in definition C. we must have Ua ∩ Ub = ∅. i. the same set of open sets. (O2) For every arbitrary family (Ui )i∈I of sets Ui ∈ O. letting ρ = d(a.
2 n
1 2
x1
+ · · · + xn
1
. for every x ∈ E.6. . O is closed
= x1 =
1
+ · · · + xn
2 1
n
.
= max{ x1
. . xn )
1 2 ∞ i∈I
Ui ∈ O.
It is easy to show that they all define the same topology. +1/n[.1. such that. For the last point. xn
n }. Proof. O is closed under finite intersections.1. By the triangle inequality. b)/3 (in fact ρ = d(a. for any two distinct points a = b in E. such as Munkres [54]. there are three natural norms that can be defined on E1 × · · · × En : (x1 . (O3) ∅ ∈ O. f (x)) ≤ ǫ. with the standard topology induced by |x − y|.C. b ∈ Ub . One should be careful that in general. It is straightforward. each Un is open. . . Given a metric space E with metric d. . For example. letting Un =] − 1/n. then d2 (f (a). f is continuous at a iff for every ǫ > 0. ρ). in R under the metric |x − y|. that is.
. i. there exist two open sets Ua and Ub such that. . . CONTINUOUS FUNCTIONS. we have U1 ∩ · · · ∩ Un ∈ O. xn ) (x1 . xn ) (x1 .e. ∅ and E belong to O. we can pick Ua = B0 (a. . Furthermore. . if d1 (a.e.e. ρ) and Ub = B0 (b.
The following lemma is useful for showing that real-valued functions are continuous. we say that (E. E ). if E and F are normed vector spaces defined by norms → continuous at − iff a → for every ǫ > 0. Given a normed affine space. Proof. Left as an exercise. a) ≤ ǫ. If E is a topological space.472
APPENDIX C. such that. such that. u u
. λf . b) = ab . x a
C. there is some n0 ≥ 0. there is some η > 0. . we consider normed affine spaces. TOPOLOGY
1
Similarly. b + − ) = d(a. for every ǫ > 0.3. |x − y|) the reals under the standard topology. then
→ → f (− ) − f (− ) x a
2
≤ ǫ. Given an affine space (E. such that. for all n ≥ n0 . E ) where the associated − → space of translations E is a vector space equipped with a norm.1. b). for any a ∈ E. and (R. there is some n0 ≥ 0.3
Normed Affine Spaces
− → For geometric applications. − → − → Definition C. where the space of translations E is a − → − → vector space over R or C. and f /g is continuous at a if g(a) = 0. x if − −− → → x a
1
and
2. d(xn .1. Using lemma C. if f and g are continuous at a. for every − ∈ E. −n − − ≤ ǫ. we can show easily that every real polynomial function is continuous. a sequence (xn )n∈N converges to some a ∈ E iff When E is a normed vector space with norm a ∈ E iff Finally. we will need to consider affine spaces (E. defined such that → − d(a.2. Lemma C. f · g. there is a natural metric on E itself. When E is a metric space with metric d. that is → → d(a + − . E ) is a normed affine space iff E is a normed vector space with norm .2. for any two functions f : E → R and g : E → R. Observe that this metric is invariant under translation. for all n ≥ n0 .
f is
≤ η. then f + g.1. for any λ ∈ R. a sequence (xn )n∈N converges to some
→ → for every ǫ > 0. are continuous at a.
In fact. y). → h(x) = a + λ− ax. a1 ) ⊕ · · · ⊕ (Em . and each Ei is also a normed affine space with norm i . we make (E1 . Of course. b) → ab from E × E to E is continuous. . . am ) into a normed affine space. If an affine space E is a finite direct sum (E1 . defined such that. under the same norm. We are now ready to define the derivative (or differential) of a map between two normed affine spaces. am ).C. This will lead to tangent spaces to curves and surfaces (in normed affine spaces). h(y)) = λd(x. .
Similarly. . .3. . Rn is a normed affine space under the Euclidean metric. by giving it the norm (x1 . NORMED AFFINE SPACES
473
Also. is we consider the map h : E → E. the map − → a + − is a homeomorphism from E u u u to Ea . and similarly for the map − → − → → → → a → a + − from E × E to E. then d(h(x). . → − − → Note that the map (a. xn ) = max( x1
1
. xn
n ). the finite product E1 × · · · × Em is made into a normed affine space.
. a1 ) ⊕ · · · ⊕ (Em . . and it is also complete. for every fixed a ∈ E and λ > 0.
474
APPENDIX C. TOPOLOGY
.
Definition D. in some small open set U ⊆ A containing a).1. dx If f ′ (a) exists for every a ∈ A. and let a ∈ A. A thorough treatment of differential calculus can be found in Munkres [55]. is that locally around a (that is. The main idea behind the concept of the derivative of f at a. including the chain rule. or Df . we define directional derivatives and the total derivative of a function f : E → F between normed affine spaces. we will use linear maps! Let us now review the formal definition of the derivative of a real-valued function. denoted as f ′ (a). This limit is denoted as f ′ (a). h→0. First. For any function f : A → R. and consider any a ∈ A. Of course. Basic properties of derivatives are shown. or 475 x → f (a) + f ′ (a)(x − a). where A is a nonempty open subset of R. Cartan [16]. h∈U h df (a). We first review the notion of derivative of a realvalued function whose domain is an open subset of R. the function f is approximated linearly by the map Part of the difficulty in extending this idea to more complex spaces. We show how derivatives are represented by Jacobian matrices. Lang [48]. Next. and Avez [2].Appendix D Differential Calculus
D. Let A be any nonempty open subset of R. h = 0}. Schwartz [71]. the map df a → f ′ (a) is denoted as f ′ . In this case. Total Derivatives
This appendix contains a review of basic notions of differential calculus. we say that f is differentiable on A. dx where U = {h ∈ R | a + h ∈ A. is to give an adequate notion of linear approximation. Let f : A → R. the derivative of f at a ∈ A is the limit (if it exists) f (a + h) − f (a) lim . we review the definition of the derivative of a function f : R → R.1. or Df (a). or .1 Directional Derivatives.
.
U is indeed open and the definition makes sense. − → Since F is a normed affine space. DIFFERENTIAL CALCULUS
Note that since A is assumed to be open. is the limit f ′ (a− ). where E and F are normed affine spaces. We would like to extend the notion of derivative to functions f : A → F . then it is continuous at a. that this quantity is a vector and not a point. h < 0}. Thus. f (a + h) − f (a) lim . f (a + h) = f (a) + f ′ (a) · h + ǫ(h)h. The first difficulty is to make sense of the quotient f (a + h) − f (a) . the unique vector translating f (a) to f (a + h ). we say that the derivative of f at a on the left. and A is some nonempty open subset of E. and if they are equal. We can also define f ′ (a) as follows: there is some function ǫ(h). in defining derivatives. and
h→0. then f is continuous on the left at a (and similarly on the right). We should note however. If f is differentiable on A. For example. A − {a} is also open. But now. where ǫ(h) is defined for all h such that a + h ∈ A. such that. the vector ab will be denoted as b − a. Remark: A function f has a derivative f ′ (a) at a iff the derivative of f on the left at a and the derivative of f on the right at a exist.1. −− − − −→ −−−−− − → − → it is notationally more pleasant to denote f (a)f (a + h ) as f (a + h ) − f (a). in the → − rest of this chapter. Also. it will be notationally convenient to assume that − → the vector space associated with E is denoted as E . and since the function h → a + h is continuous and U is the inverse image of A − {a} under this function. how do we define the quotient by a vector? Well. h∈U
lim
ǫ(h) = 0. then f is continuous on A. making sense of f (a + h ) − f (a) is easy: we can define −− − − −→ −−−−− − → − → this as f (a)f (a + h ). and that the vector space associated − → with F is denoted as F .1 has a derivative f ′ (a) at a. we don't!
. If a function f as in definition D. h∈U h where U = {h ∈ R | a + h ∈ A. if the derivative of f on the left at a exists. The composition of differentiable functions is differentiable. whenever a + h ∈ A. and derivative of f at a on the right.
Remark: We can also define the notion of derivative of f at a on the left. Nevertheless. h If E and F are normed affine spaces.476
APPENDIX D. if it exists. h→0.
u viewed as a vector). the existence and value of a directional derivative is independent of the choice of norms in E and F .1. which is that their definition is not sufficiently uniform. and let f : A → F be any function. and yet not be continuous at a. where t ∈ R u (or C) (more exactly. denoted as D f (a). Two functions may have all directional derivatives in some open sets. curve which lies on the image f (E) of E under the map f : E → F (more exactly. in the sense of definition D.2.2 makes sense. and the directional derivative Du f (a). t
→ → where U = {t ∈ R | a + t− ∈ A. let A be a nonempty open − → − → → subset of E. For any a ∈ A. Definition D. Let E and F be two normed affine spaces.e. when E has dimension ≥ 2. the vector − . let A be a nonempty open subset of E. is the limit (if it u
u
exists)
t→0. We can consider the vector f (a + t− ) − f (a). DIRECTIONAL DERIVATIVES.1. the real number 1. s] in A containing a.t. the points → of the form a + t− form a line segment).D. defines the direction of the tangent line at a to this curve. Definition D. and yet their composition may not. The directional derivative is sometimes called the Gˆteaux derivative. This curve (segment) is defined by the map t → f (a + t− ). directional derivatives present a serious problem.r. For any a ∈ A. a function can have all directional derivatives at a. However. and that the image of this line defines a curve in u F . Let E and F be two normed affine spaces. provides a suitable generalization of the notion of derivative. we say that f is differentiable
. and since A − {a} is open. also denoted as f ′ (a). As u a consequence. the inverse image U u of A − {a} under the above map is open. This leads us to the following definition. t∈U
lim
→ f (a + t− ) − f (a) u .1.r. Since the notion of limit is purely topological. for any − = 0 in E . The idea is that in E. u u → Since the map t → a + t− is continuous. a → → − In the special case where E = R and F = R. Thus. t = 0}). where t ∈ R (or t ∈ C). all nonnull vectors − share something in common.3. there is no reason to believe → that the directional derivatives w. and let f : A → F be any function. and we let − = 1 (i. u → f (a + t− ) − f (a) u t → makes sense. TOTAL DERIVATIVES
477
→ → − A first possibility is to consider the directional derivative with respect to a vector − = 0 u − → → in E . Indeed. the derivative D1 f (a). from u R to F (more exactly from [r. it is immediately verified that D1 f (a) = f ′ (a). as long as they are equivalent norms. When E = R (or E = C) and F is any normed vector space. we introduce a more uniform notion. for t in some small closed interval [r. t = 0} (or U = {t ∈ C | a + t− ∈ A. s] to F ). the points of the form a + t− form a line. u → the directional derivative of f at a w.1. and the definition of the limit in definition D. Now. a small → curve segment on f (A)).1.t.1.
h∈U
− → − → − → → − Note that for every h ∈ U. when a + h ∈ A. DIFFERENTIAL CALCULUS
− → − → − → at a ∈ A iff there is a linear continuous map L : E → F and a function ǫ( h ). and it is called the Fr´chet derivative. then f is − → − → → continuous at a. u
. In fact. and let f : A → F be any function. where ǫ( h ) is defined for every h such that a + h ∈ A and
h→0.
− → − − → − → − → → where U = { h ∈ E | a + h ∈ A. and we will assume this from now on. The following lemma shows that our new definition is consistent with the definition of the directional derivative. as long as they are equivalent norms.478
APPENDIX D. it does no harm to − → − → assume that ǫ( 0 ) = 0 . since h = 0 . For any a ∈ A. ǫ( h ) is uniquely determined since − → − → − → f (a + h ) − f (a) − L( h ) ǫ( h ) = .
h→0. let A be a nonempty open subset of E.
→ Du f (a) = Df (a)(− ). or f ′ (a). Lemma D. Let E and F be two normed affine spaces. and since A is open in E. Again. However. or total e derivative. Note that the continuous linear map L is unique. such that − → − → − − → → f (a + h ) = f (a) + L( h ) + ǫ( h ) h − → − → − → − → for every a + h ∈ A. The linear map L is denoted as Df (a). h = 0 }. if it exists. the − → inverse image U of A − {a} under the above map is open in E . and u
u
furthermore. Since the notion of limit is purely topological. and f has a directional derivative D f (a) for every − = 0 in E .4. if Df (a) is defined.1. h∈U
lim
− → − → ǫ( h ) = 0 . locally around a. the next lemma implies this as a corollary. and it makes sense to say that − → − → lim ǫ( h ) = 0 . − → h
− → and that the value ǫ( 0 ) plays absolutely no role in this definition. or total differential . or Dfa . − → − → − → Since the map h → a + h from E to E is continuous. of f at a. or differential . The condition for f to be differentiable at a amounts to the fact that the right-hand side of the above expression − → − − → − → → − → approaches 0 as h = 0 approaches 0 . we note that the derivative Df (a) of f at a provides an affine approximation of f . or dfa . or derivative. or df (a). the existence and value of a derivative is independent of the choice of norms in E and F .
. For any two normed affine spaces E and F . if they exist. are called the partial derivatives of f with → → respect to the frame (a0 . Indeed. when E is of finite dimension. we have u u → → → → f (a + t− ) = f (a) + tL(− ) + ǫ(t− )|t| − . for any frame (a0 .1. called the derivative of f on A. the variable xj really has nothing to do with the formal definition.6. . −n )) of E. we get a map Df : A → L( E . or ∂xj ∂f (a) for a partial derivative. F ). (−1 . and → → − → − thus. u u the directional derivatives Duj f (a). then − → Df (a) = f . . This is just another of these situations where tradition is just too hard to overthrow! The notation We now consider a number of standard results about derivatives. .D.
Definition D. The partial derivative Duj f (a) is also denoted as u u ∂f (a). it is easily shown that every linear map is continuous. It is important to note that the derivative Df (a) of f at a is a continuous linear map − → − → from the vector space E to the vector space F . . we have a + t− ∈ A. Also. ǫ( h ) h approaches 0 . −n )) for E. and letting h = t− . then Df (a) = 0. This way.
definition of partial derivatives. . we can define the directional derivatives with respect to the vectors in the → → basis (− . . for |t| ∈ R small enough (where t ∈ R or u − → → → t ∈ C). Given two normed affine spaces E and F . for every frame (a0 . −n ) u u u u − → is a basis of E . and this assumption is then redundant. The uniqueness of L follows from lemma D. .1. If f : E → F is a continuous affine map. for every a ∈ E. (−1 .4. → → → → When E is of finite dimension n. . we can also do it for an infinite frame). for every a ∈ E. if f : E → F is a constant function. (−1 .
.1. . as follows. is a "logical obscenity". −n )). and also denoted as df . u u u u → f (a + t− ) − f (a) u |t| → → → u = L(− ) + ǫ(t− ) − . . . and not a function from the affine space E − − → → to the affine space F . . . For any − = 0 in E . Lemma D. for every function f : E → F . although customary and going back to ∂xj Leibnitz. if E is of finite dimension → → n. f is continuous at a. u u t t and the limit when t = 0 approaches 0 is indeed Du f (a). . − ) (actually.5. . where (−1 . DIRECTIONAL DERIVATIVES. . we obtain the u u
1 n
and for t = 0. If Df (a) exists for every a ∈ A. for every a ∈ E. . . If h = 0 approaches 0 . TOTAL DERIVATIVES
479
− → − − → → − → − → − → Proof. ∂j f (a).1. . the linear map associated with f . since L is continuous.
y) = 4 if (x. defined such that. f (0. As a u matter of fact. There are functions such that all the partial derivatives exist at some a ∈ A. where 0 < λ < 1. 0). and the open segment ]a. if f : A → F is differentiable at every − → point of the open segment ]a. if a function f : A → F is differentiable at a ∈ A. 0) = 0. if Df (0) existed. However. y) = (0. We first state a lemma which plays an important role in the proof of several major results of differential calculus. f (x.486
APPENDIX D. and when we approach the origin on this parabola. then its Jacobian matrix is well defined. Du f (0) exists for all − = 0 . and
x∈]a. but yet. Let E and F be two normed affine spaces. there are sufficient conditions on the partial derivatives for Df (a) to exist. It turns out that the continuity of the partial derivatives on A is a necessary and sufficient condition for Df to exist and to be continuous on A. where 0 ≤ λ ≤ 1. For example. if the closed segment [a. the limit is 1 . given any two points a. 0) = 0. then f defines a − − → → function Df : A → L( E . the closed segment [a. Lemma D. b ∈ E. when 2 2 in fact.2. given an open subset A of E. On the other hand. λ ∈ R. If E is an affine space (over R or C). let A be an open subset of E. then
− → − → f (a + h ) − f (a) ≤ M h .a+h[
max
Df (x) ≤ M. the function f is not continuous at (0. and we would have → Du f (0) = Df (0)(− ) = αh + βk. a + h [. namely continuity of the partial derivatives. b[ is the set of all points a + λ(b − a). For example. we have For any − = 0 . the function is not differentiable at a. a + h ] is contained in A. y) = 1 . − → − → and let f : A → F be a continuous function on A.
− → → Thus.2.
One should be warned that the converse is false. F ). x + y2 → h → − → . letting − = u u k − → − → → h2 k f ( 0 + t− ) − f ( 0 ) u = 2 4 . b] is the set of all points a + λ(b − a). and not even continuous at a. it would be u a linear map Df (0) : R2 → R represented by a row matrix (α β). consider the function f : R2 → R. Given any a ∈ A and any h = 0 in − → − → E . 0). t t h + k2
so that Du f (0) =
0
h2 k
if k = 0 if k = 0. λ ∈ R. and x2 y f (x.
. If f is differentiable on A. on the parabola y = x2 . DIFFERENTIAL CALCULUS
Given two normed affine spaces E and F of finite dimension. but the explicit formula for Du f (0) is not linear. f (0.
for some M ≥ 0.
for all j.2 can be used to show the following important result. −n )) is a frame of E.a+h[ Df (x) − L .3 holds. or a C 0 -function on A. an ). F ) is defined and continuous on A iff every partial derivative ∂fi ∂j fi (or ) is defined and continuous on A. Since the existence of the derivative on an open set implies continuity.D. we say that a function f : A → F is of class C 0 on A. if L : E → F is a continuous linear map. It is also possible to define higher-order derivatives. . and using the more general partial derivatives Dj f introduced before lemma D. iff Df exists and is continuous on A. and where (a0 . . ∂xj → → As a corollary. or E = E1 × · · · × En . where M = maxx∈]a.2. Lemma D. (−1 . − )) is a frame of F . 1 ≤ i ≤ m. j. For a complete treatment of higherorder derivatives and of various versions of the Taylor formula. for all i. and an open subset A of E. . . F ) is defined and continuous on A iff ∂f every partial derivative ∂j f (or ) is defined and continuous on A. . or a C 1 -function on A. Definition D.1. see Lang [48]. if F is of finite dimension m. Given two normed affine spaces E and F . and (b0 .4. given any open subset A of E. the v vm − − → → derivative Df : A → L( E .14. Given two normed affine spaces E and F . a1 ) ⊕ · · · ⊕ (En . JACOBIAN MATRICES − → − → As a corollary. We say that f : A → F is of class C 1 on A. (−1 . Lemma D. . It is easy to show that the composition of C 1 -functions (on appropriate open sets) is a C 1 -function. . 1 ≤ j ≤ n.3.3 gives a necessary and sufficient condition for the existence and continuity of the derivative of a function on an open set.2. assuming that E = (E1 . ∂xj Lemma D. iff f is continuous on A. .2. then − → − → − → f (a + h ) − f (a) − L( h ) ≤ M h . 1 ≤ j ≤ n.2. where E is of finite dimension → → n. a C 1 -function is of course a C 0 -function.
487
The above lemma is sometimes called the "mean-value theorem".2.2. given any u u − − → → function f : A → F .3 gives a necessary and sufficient condition for a function f to be a C 1 -function (when E is of finite dimension). It should be noted that a more general version of lemma D. Lemma D.2. the derivative Df : A → L( E .
. | 677.169 | 1 |
Real Analysis (Oxford Graduate Texts in Mathematics)
by Fon-Che Liu
Real Analysis is indispensable for in-depth understanding and effective application of methods of modern analysis. This concise and friendly book is written for early graduate students of mathematics or of related disciplines hoping to learn the basics of Real Analysis with reasonable ease. The essential role of Real Analysis in the construction of basic function spaces necessary for the application of Functional Analysis in many fields of scientific disciplines is demonstrated with due explanations and illuminating examples.
After the introductory chapter, a compact but precise treatment of general measure and integration is taken up so that readers have an overall view of the simple structure of the general theory before delving into special measures. The universality of the method of outer measure in the construction of measures is emphasized because it provides a unified way of looking for useful regularity properties of measures. The chapter on functions of real variables sits at the core of the book; it treats in detail properties of functions that are not only basic for understanding the general feature of functions but also relevant for the study of those function spaces which are important when application of functional analytical methods is in question. This is then followed naturally by an introductory chapter on basic principles of Functional Analysis which reveals, together with the last two chapters on the space of p-integrable functions and Fourier integral, the intimate interplay between Functional Analysis and Real Analysis. Applications of many of the topics discussed are included to motivate the readers for further related studies; these contain explorations towards probability theory and partial differential equations | 677.169 | 1 |
Omtale
11+ Maths Revision Guide
Secure the top marks in 11+ independent school entrance exams and Pre-Tests leading to a better chance at getting into the school of choice with this essential revision guide. Complete coverage of the ISEB 11+ Maths syllabus and stretching extra content ensures that every topic is thoroughly revised ahead of the exams. - This book covers everything required for the 11+ Maths exam - Prepares pupils for a wide range of independent school exams and pre-tests with challenging extension material - Consolidates revision with all the key information in one place - Features helpful insight in to the exams, with examples, practical tips and advice - Tests understanding and technique with timed, levelled exam-style questions Also available for 11+ Maths preparation: - 10-Minute Maths Tests Workbook Age 8-10 9781471829611 - 10-Minute Maths Tests Workbook Age 9-11 9781471829635 - Mental Arithmetic Workbook Age 8-10 9781471829505 - Mental Arithmetic Workbook Age 9-11 9781471829628 Revision Guides, Workbooks and Practice Papers are also available for English, Science, Verbal Reasoning and Non-Verbal Reasoning on | 677.169 | 1 |
97805217715 Beginner's Guide to MATHEMATICA , Version 4
This update of Jerry Glynn and Theodore Gray's hugely successful textbook covers not only the basic Mathematica features, but also the new features of Mathematica Version 4. The book teaches new Mathematica users some of the important basics of the latest release of this powerful software tool: using the typesetting features, programming palettes, defining functions, creating graphs and notebooks, and applying useful problem solving techniques. Using their skills as Mathematica experts and teachers, the authors provide a brisk but careful tutorial for the Mathematica novice. From the fundamentals of installing and running Mathematica on your computer, through to tips on how to get the most from the advanced programming features, the presentation maintains its concise and knowledgeable tone, providing indexes for both concepts and Mathematica function names. This book will be a valuable tool for both students and individual Mathematica users | 677.169 | 1 |
Mathematics Pocket Notes (Iseb Revision Guides)
This pocket-sized book is suitable for pupils aged 13+ (Year 8 upwards) and distills the key information required for pupils approaching Maths Common Entrance and other entrance exams at 13+. The material is arranged clearly in bullet point format in the following topics: Number; Calculation; Problem Solving; Algebra; Shape, Space and Measures; Handling Data. Ideal for quick revision, this small booklet is perfect for any pupil revising for entrance exams at 13+.
- Endorsed by ISEB
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Also available from Galore Park
- Mathematics ISEB Revision Guide
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- Maths Practice Exercises 13+ Level 3
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"synopsis" may belong to another edition of this title.
Product Description:
Mathematics Pocket Notes
Book Description:
Ideal for any pupil aged 13+ (Year 8 upwards), this pocket book distills the key information required for pupils approaching Common Entrance and entrance exams at 13+.
Book Description Galore Park Publishing Ltd, 20041150537620 | 677.169 | 1 |
When you read the textbook introduction pages to the four chapters that we cover in this course (Chap
are introduced to a number of famous mathematicians. Sir Isaac Newton and Gottfried Leibnitz are the
mathematicians who are credited with the creation of
Pied Piper
Introduction:
Pied Piper is currently in a position to increase sales and gain a competitive advantage over the
competition by being the first to contract with Citrix and allowing them to handle all
telecommunication needs. This will allow the3.1 IS Business Foundation
Pied Piper has joined with a company, Citrix, who is going to provide many different enterprise
business products to support our organization and business processes. One of the applications we
are going to use is Citrix NetscaleCin3:9:6:8:2
Problem
If
+1.5 , find
f(g(x) and
decimal places.
f(x)=[a][b]x+x2 and
f(g([d]) and
g(f([d]) by first determining
g(f(x) before evaluating for
Example: if (3)=6, (9)=2, and (6)=8,
Example: if
f(x)=62x+x2 and
,
, and
g(x)=xx[c]
, then
[d] . Rou
Flag this Question
Question 125 pts
Read pages 655, 700, 742, and 768 in the textbook and answer the following question:
The most prolific mathematician of all time was
Leonard Euler
Blaise Pascal
Isaac Barrow
Gottfried Leibnitz
Isaac Newton
Pierre de Fer
I would recommend this course taught because I believe Mr. Foroughi is the best professor to explain mathematical steps to college students. I think Mr. Foroughi can explain mathematical steps to a deaf person. He is that good. He breaks the steps down really well.
Course highlights:
I learnt everything about the course from my excellent professor. Although the professor gave lot of assignment and tests, all efforts paid off with excellent results for the class.
Hours per week:
9-11 hours
Advice for students:
I would recommend taking Mr. Bahram Foroughi as your class instructor for this course.
Course Term:Fall 2016
Professor:Bahram Foroughi
Course Required?Yes
Course Tags:Math-heavyGo to Office HoursMany Small Assignments
Feb 23, 2017
| Would recommend.
This class was tough.
Course Overview:
This course is extremely hard if you don't have calculus down. Don't even bother with the book, read other equations online, or get a tutor, and/or use Khan academy.
Course highlights:
That I would never want to be a mathematician. However, you truly appreciate the you can use calculus in pretty much every single every day activity or movement.
Hours per week:
6-8 hours
Advice for students:
The text book wasn't necessary. There are tons of examples and guides online that you can source your information from, along with asking the professor. | 677.169 | 1 |
Analytical Geometry
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analytic geometry, also known as coordinate geometry.Analytic Geometry is a branch of algebra that is used to model geometric objects - points, (straight) lines, and circles being the most basic of these | 677.169 | 1 |
Downloads Basic College Mathematics (4th Edition) e-book
Description:
Elayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. Basic College Mathematics, Fourth Edition was written to help readers effectively make the transition from arithmetic to algebra. The new edition offers new resources like the Student Organizer and now includes Student Resources in the back of the book to help students on their quest for success.
The E-Book Basic College Mathematics (4th Edition) by Elayn Martin-Gay is available at the next formats: fb2, pdf, mobi. | 677.169 | 1 |
Algebra provides a systematic way to represent mathematical
relationships and analyze change. Students will make connections
and build relationships between algebra and arithmetic, geometry,
and probability and statistics. Connections will also be made to
other subject areas through practical applications. Students are
required to use algebra as a tool for representing and solving a
variety of practical problems. Tables and graphs will be used to
interpret algebraic expressions, equations, and inequalities and to
analyze behaviors of functions.
Graphing calculators, computers, and other appropriate
technology tools will be used to assist in teaching and learning.
Students must take the Virginia SOL test for Algebra I | 677.169 | 1 |
Groups and Symmetry A Guide to Discovering Mathematics
ISBN-10: 0821804502
ISBN-13: 9780821804506 Mathematical World Series features well-written, challenging works that capture the fascination and usefulness of mathematics.In most mathematics textbooks, the most exciting part of mathematics - the process of invention and discovery - is completely hidden from the reader. The aim of Groups and Symmetry is to change all that. By means of a series of carefully selected tasks, this book leads readers to discover some real mathematics. There are no formulas to memorize; no procedures to follow. The book is a guide: its job is to start you in the right direction and to bring yo back if you stray too far.Discovery is left to | 677.169 | 1 |
This book is intended for an introductory course in number theory, primarily taken by mathematics majors. The author wrote the text with four main goals in mind. He wrote a book that is both easy to read and easy to teach from. The author also aims to ease students into using proofs, and to develop a self-confidence in mathematics surrounding the difficulty of mathematical proof. Although the main users of this text will be mathematics students, a large audience could easily use the book. | 677.169 | 1 |
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Algebraic Expressions
Divided into 3 sections: (a) Algebraic notation and simplifying by collecting like terms and multiplying/dividing expressions.
(b) Substituting values, including negative numbers, into algebraic expressions, with consideration of BIDMAS.
(c) Forming algebraic expressions from potentially worded information, particularly appreciating it as the first step in solving equations (the latter which isn't covered till the 'Equations' topic). | 677.169 | 1 |
Part of a project funded by the National Science Foundation to improve the quality of mathematics and science teaching in grades K-12, this new guide models the student-centered approach recommended by the National Council of Teachers of Mathematics to teach introductory statistics. Provides future middle grade mathematics teachers with a strong foundation, connecting the mathematics they are learning with the mathematics they will be teaching. Gives enhanced meaning to formulas via a visual or geometric approach. Uses numerous illustrations to assist readers in making explicit connections between a typical college elementary statistics course and the statistical concepts taught by middle school teachers. For anyone interested in introductory statistics.. PAPERBACK. Book Condition: New. 01314492221449222 | 677.169 | 1 |
Calculus
Calculus, Second Edition discusses the techniques and theorems of calculus. This edition introduces the sine and cosine functions, distributes ?-? material over several chapters, and includes a detailed account of analytic geometry and vector analysis.
This book also discusses the equation of a straight line, trigonometric limit, derivative of a power function, mean value theorem, and fundamental theorems of calculus. The exponential and logarithmic functions, inverse trigonometric functions, linear and quadratic denominators, and centroid of a plane region are likewise elaborated. Other topics include the sequences of real numbers, dot product, arc length as a parameter, quadric surfaces, higher-order partial derivatives, and Green's theorem in the plane.
This publication is a good source for students learning calculus | 677.169 | 1 |
Plotting Real-Life Trig Applications
Students work with a program similar to the graphing calculator called TI Interactive. They examine real-life applications, graph the data, and look at the trigonometric functions as the result using the computer program to help with the steps and to visualize the relationships. | 677.169 | 1 |
Understanding LTE with MATLAB: From Mathematical Modeling to Simulation and PrototypingOverviewThis book examines the Physical Layer (PHY) of the LTE standards by incorporating three conceptual elements: an overview of the theory behind key enabling technologies; a concise discussion regarding standard specifications; and the MATLAB® algorithms needed to simulate the standard.
The use of MATLAB®, a widely used technical computing language, is one of the distinguishing features of this book. Through a series of MATLAB® programs, the author explores each of the enabling technologies, pedagogically synthesizes an LTE PHY system model, and evaluates system performance at each stage. Following this step-by-step process, readers will achieve deeper understanding of LTE concepts and specifications through simulations.
Key Features:
• Accessible, intuitive, and progressive; one of the few books to focus primarily on the modeling, simulation, and implementation of the LTE PHY standard • Includes case studies and testbenches in MATLAB®, which build knowledge gradually and incrementally until a functional specification for the LTE PHY is attained • Accompanying Web site includes all MATLAB® programs, together with PowerPoint slides and other illustrative examplesRelated Subjects
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This is truly a superb text book. I was pleasantly surprised to find that it is an excellent combination of the description of the global industry standard, 4G LTE, and the practical implementation details that are needed to implement the standard within the MATLAB environment. The text was written at just the right level, with all of the MATLAB code provided within the book so that the user could further explore either the standard as written or explore user-defined deviations from the standard. I particularly enjoyed the fact that the MATLAB code was so well documented, and the ease of its use added further to the value of the text book. Amidst the jungle of books that are available on the topic of 4G LTE, I think that this text book is truly top-notch because the user can gain first hand knowledge of the standard. It is a jewel, which is well worth its purchase price. Very well written and perfectly suited for the academician or industry specialist who seeks to learn more about 4G LTE within a MATLAB software simulation environment. Well done! | 677.169 | 1 |
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Amit Jaiswalashish vaidya
Good practice. Why don't you keep options visible when some1 is solving sum on digital notepad as, in GMAT, keeping an eye on options is important even while solving sums. It's an improvement i would suggest.
Shivangi Sharma
Brilliant Application I don't think if I could ask for more! :) This is a brilliant application. Thank you for designing such a brilliant and terrific application! :)
Carlos Kan
Wrong answers Some are wrong. But, disregarding it, rest it s very helpful
Sergey Davydov
Mistakes The answer of the FIRST question is wrong, the quantity of revolutions is not measured in meters
lavanya tamarapalli
This app is a great help for math practice but found few incorrect answers as well..hope they will be corrected for future referers
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Amit Jaiswal August 21, 2015Good practice. Why don't you keep options visible when some1 is solving sum on digital notepad as, in GMAT, keeping an eye on options is important even while solving sums. It's an improvement i would suggest. | 677.169 | 1 |
Algebra 1 Second Semester Quick Review Spring 2010
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this file type before downloading and/or purchasing.
121 KB|2 pages
Product Description
As the second semester wound down, there were still several topics for which I felt my students could benefit from additional review. I wrote this worksheet during "dead week" to make sure my students saw the following topics before their eyes at least one last time before the final: exponents, fractions, factoring trinomials, expanding/multiplying polynomials vs. factoring out the GCF, solving literal equations and equation with fractional coefficients, the fundamental counting principle, and basic statistics | 677.169 | 1 |
1. Prerequisites *
At the start of this course the student should have acquired the following competences: an active knowledge of
Dutch
specific prerequisites for this course
The competences of the courses:
mathematics with technological and economical applications (1HI/B)
general economy (1HI/B)
principles of engineering (1HI/B)
2. Learning outcomes *
You can deal with, visualize and interpret data using Matlab software
You can deal with concrete situations related to mathematics, economics and engineering with lots of data using Matlab and give an explanation to your conclusions
You can think about creative solutions within the boundaries of the Matlab software: you know the limitations of the package and are able to deal with this
You can explain in a clear way your way of thinking and working
3. Course contents *
This course is an introduction to the calculation software Matlab in an integrated manner. You will be encouraged to discover this software and to deal with it in a self-discovery way. Throughout concrete problems related to other subjects in the curriculum with an emphasis on mathematics, economics and engineering, you will become proficiant in the basic functions of the calculation software.
You learn to deal with data, matrix operations, charts in 2D and 3D, functions, While & if then else-structures in the calculation package. You also get acquainted with Matlab toolboxes on optimisation, finance and differential equations.
For mathematics, we will rely on functions, eigenvalues and -vectors, optimization techniques, differential equations and line integrals. In engineering concepts such as modeling, vector fields, waves and energy will play a role. Economically will the behaviour of economic variables such as stocks or prices be at the heart of the problems.
4. Teaching method and planned learning activities
Class contact teaching
Lectures
Seminars/Tutorials
Laboratory sessions
Personal work
Exercises
Assignments Individually
5. Assessment method and criteria
Continuous assessment
Assignments
(Interim) tests
Participation in classroom activities
6. Study material *
6.1 Required reading
Published on Blackboard.
6.2 Optional reading
The following study material can be studied voluntarily :
7. Contact information *
In addition to the workshops, there are instructions to continue to practice at home yourself. For questions you can always contact the teacher or assistants. | 677.169 | 1 |
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Pre-Algebra Riddles is a diverse and uniquely illustrated collection of fifty challenging riddles, teasers, ciphers, and cryptograms. The book is designed for the Middle School student, but it's blend of arithmetic, algebra, and geometry puzzles that can also be used for enrichment and remediation at the lower high school levels. This book was published by Incentive Publications - a leader in publishing workbooks for Middle School subjects and students.
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Gem Mathematics Basic Facts has been extensively revised and updated to give explanations of key terms and concepts that school and college students will encounter when working for GCSE, Standard Grade or other qualifications at this level.
The extensive revision brings the book in line with the most recent changes in the National Curriculum and examination syllabuses and with new initiatives in teaching.
All the entries have been rewritten and the book contains many worked examples and diagrams, essential for understanding mathematical concepts.
Alphabetical arrangements of entries encourages quick reference and a system of cross-references supports each entry so that it can be put in context.
Well over 1.5 million copies of the series sold. It is established as an invaluable tool for reference and for exam revision.
"synopsis" may belong to another edition of this title.
From the Back Cover:
• Includes the key terms and concepts you need to know, all clearly defined with the help of diagrams and worked examples • Ideal for exam revision and quick reference • The Basic Facts series includes Physics, Chemistry, Biology, Science, Business Studies, Geography, and Twentieth-Century History23358
Book Description Paperback. Book Condition: Good. The book has been read but remains in clean condition. All pages are intact and the cover is intact. Some minor wear to the spine. Bookseller Inventory # GOR00182884221543VG | 677.169 | 1 |
articles for middleschool
... Older Children
Chinese articles for MiddleSchool students 1
6.99
Assorted articles book 1 for Children in Middleschool which is in simplified Chinese characters ... Details...
The MiddleSchool Mathematician, Revised with CD
... correlated to middleschool math skills identified by the NCTM. Use them for lesson starters, review, specific skill lessons, or practice. How exciting to find activities that appeal to the middleschool student's ...
Basicconcepts of quantum mechanics
... physics. In the second chapter, based upon the concept of amplitude probability, various questions on physics of microphenomena ... ). The simplest quantum mechanical systems ---microobjects with two basic states--- are analyzed in detail. In the third ... | 677.169 | 1 |
fresh reader-friendly design, MATHEMATICS FOR ELECTRICITY AND ELECTRONICS, 4E is more current, comprehensive, and relevant than ever before. Packed with practical exercises and examples, it equips learners with a thorough understanding of essential algebra and trigonometry for electricity and electronics technology, while helping them improve critical thinking skills. Well-illustrated information sharpens the reader's ability to think quantitatively, predict results, and troubleshoot effectively, while drill and practice sets reinforce comprehension. To ensure mastery of the latest ideas and technology, the text thoroughly explains all mathematical concepts, symbols, and formulas required by future technicians and technologists. In addition, a new homework solution offers a wealth of online resources to maximize study efforts as well as provides an online testing tool for instructors. | 677.169 | 1 |
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Honors Pre-Calculus with Trigonometry
This course focuses on six critical areas in greater depth; (1) the complex number system; (2) arithmetic with polynomials and rational expressions; (3) operations with functions; (4) conic sections; (5) trigonometric functions; and (6) applications of trigonometry. Students also have the opportunity to study sequences and series, vectors, limits, and derivatives as an introduction to calculus. Suggested Calculator: Scientific or Graphing Calculator (TI-83 or TI-84) | 677.169 | 1 |
Graph Transforms
In this graph transforms learning exercise, 10th graders solve and complete 15 various types of problems. First, they evaluate each function and sketch the graph. Then, students determine why two transformation illustrated are identical for the given function. | 677.169 | 1 |
The Official SAT Question of the Day
Wednesday, October 19, 2011
Graphs of Functions and Systems of Equations
Precalculus will be moving beyond their introduction to functions and function notation from Algebra II and into a in-depth development of transformations. We will be applying transformations (translations, reflections, dilations, etc.) to many of the parent functions included in the image above (we will save trigonometry for the Spring!). Understanding transformations is a crucial link between the algebraic form of a function and its graphical behavior. We will spend a considerable amount of time exploring this topic and working toward mastery. You all will continue to complete the lengthy linear programming problem set and for that reason I have a great deal of information to post--I will post the homework, reference materials, and a YouTube video that goes over the basics of linear programming from the previous week on linear programming as well as the upcoming information on transformations! Enjoy.
Algebra 2 will be moving into systems of equations through the lens of coordinate geometry. Below is a student work example of some of the resources I will be using and what we will be exploring. The purpose is for students to have a concrete link between their algebraic worlds and geometric worlds as well as to ground the solving of systems in the graphical world and slowly move to more abstract manipulation. | 677.169 | 1 |
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Too many students end their study of mathematics before ever taking an algebra course. Others attempt to study algebra, but are unprepared and cannot keep up. Key to Algebra was developed with the belief that anyone can learn basic algebra if the subject is presented in a friendly, non-threatening manner and someone is available to help when needed. Some teachers find that their students benefit by working through these book before enrolling in a regular algebra course; others use them as supplemental help and review. In books 1-4, students study multiplying, division, integers, area, perimeter, exponents, distributive principle, equations, polynomials, quadratic equations and more. This answer key provides brief notes to the teacher and gives the answers to the workbook problems. Student pages are reduced and overlaid with the correct answers. Accompanies Key to Algebra Book 1, Key to Algebra Book 2, Key to Algebra Book 3, and Key to Algebra Book 4.
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Q:
Is KTA algebra appropriate to use at high school level I teach a modified algebra 2 class. Students are 11th and 12th grade special ed. Would this be an appropriate curriculum for high school students struggling with algebra 2
A:
Yes, this could be used as a course for high school students struggling in Algebra. | 677.169 | 1 |
Integrals
Integrals
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The integral is an important concept in mathematics. Integration is one of the two main operations in calculus, with its inverse(almost), differentiation, being the other. Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral | 677.169 | 1 |
Problem-solving technique: different ways to come up
with the same answer
Second Semester
2011-12-04, Session 11
Math Auction
2011-12-11, Session 12
Winter Olympiad
2011-01-09, Session 13
Olympiad review. Euler graphs
2011-01-23, Session 14
Problem-solving technique: representation of numebrs
2011-01-30, Session 15
Binary numbers
2011-02-06, Session 16
Binary numbers: how big are they?
2011-02-13, Session 17
Geometry 1, Math Kangaroo prep
2011-02-27, Session 18
Geometry 2, Math Kangaroo prep
2011-03-04, Session 19
Geometry 3, Math Kangaroo prep
2011-03-11, Session 20
Geometry 4
Third Semester
2011-03-19, Session 21
Spring Olympiad
2011-03-26, Session 22
Complex logical statements
2011-04-02, Session 23
Invariant
2011-04-16, Session 24
Math Auction
2011-04-23, Session 25
Colors as invariants
2011-04-30, Session 26
Geometry 5
2011-05-06, Session 27
Strategic games and wining positions
2011-05-13, Session 28
Strategic games and wining positions
2011-05-20, Session 29
Games: solving backward
2011-06-05, Session 30
End-of-the-year Math Domino | 677.169 | 1 |
Description
Do you want to plot a function and you do not have a graphical calculator at hand? That is not a problem if you have a smart phone and GraphIT. GraphIT can almost plot any function you want. And that is even faster than the graphical calculator. If you are a student or teacher, then you do not want to miss this App.
Highlights of GraphIT: 1- Plotting functions 2 Plotting Integral of functions 3 Plotting first drevative of functions 4 you can reference function from another function: In other words you can transform or stretch functions 5 Determie Zero points 6 Determine extreme points 7 Determine Intersection points of two graphs 8 Calculating definite integral between 2 points 9 You can even indirectly determine inflection points 10 The App determine the y scale automatically for you, if you wish 11 Soft key board to edit functions quickly 12 Very intuitive: you can move the cursror by touching the screen. 13 Determining value of the functions and its first drevative 14 takes care of asymptotic lines and domain of the function 14 Smart parser to parse the mathematical formulas | 677.169 | 1 |
A computer algebra system that includes many tools for mathematical calculations, graphing, working with data from the cloud wolfram cloud or databases. It is widely used in scientific, engineering, mathematical and computer fieldsA computer algebra system that includes many tools for mathematical calculations, graphing, working with data from the cloud wolfram cloud or databases. It is widely used in scientific, engineering, mathematical and computer fields.
A computer algebra system that includes many tools for mathematical calculations, graphing, working with data from the cloud wolfram cloud or databases. It is widely used in scientific, engineering, mathematical and computer fields.
Wolfram Mathematica 11 is the latest version of Wolfram Mathematica, the most powerful software for modern technical computing (symbolic, numeric, graphic). Wolfram Mathematica offers a set of tools for general computing both numeric, symbolic, as well as visualization. Mathematica combines powerful computing software with a convenient user interface. It features high-performance symbolic and numeric computation, 2D and 3D data visualization, and programming capabilities. In addition, It also seamlessly integrates general computing engine, documentation system, graphics system, programming language, and advanced connectivity to other applications. | 677.169 | 1 |
Mathematics Department
Welcome to the Math Department Page! I hope this page provides students, staff, and the community with some helpful information and links.
Please do not hesitate to contact me with any questions regarding mathematics curriculum and instruction.
Due to changes in the math curriculum, the high school mathematics pathways for incoming ninth graders for the 2016-2017 school year has been modified. Here is the pathways chart that Mrs. Naber referenced at the Board meeting in April 2015. Click here to view the entire presentation.
What does math look like for the 2016-2017 8th grade students?
As you are aware, this year is the second year of a new mathematics pathway, which begins in 8th grade and continues into 9th & 10th grades. Algebra 1 & 2 have been combined to create a two‐year algebra sequence. Students will learn the entire Common Core Algebra 1 curriculum within the two years, as well as the non-trigonometric topics from Common Core Algebra 2. The topics from Algebra 1 and Algebra 2 were shifted in order to appropriately align with one another. This will allow students to go deeper into these algebraic topics each year. The Common Core Algebra 1 Regents exam will be administered in June of the second year of Algebra. It is important to note that by reorganizing the curriculum to go deeper into fewer topics each year, not all of the Common Core Algebra 1 topics will be taught in the first year. Therefore, students will not be prepared to take the Common Core Algebra 1 Regents Examination until the end of the second year. All other courses would culminate in a departmentally‐designed final examination. Following the Algebra 1 and Algebra 2 courses, students will take Geometry/Trigonometry. By the end of three years, students will have completed the entire Common Core Algebra 1, Geometry, and Algebra 2 standards. They are simply being reorganized to maximize learning and promote mastery.
It is important to note that the year-end grade earned by your child in this year's Algebra 1 course will appear on his/her high school transcript. The Common Core Algebra 1 Regents Exam score will appear on the high school transcript next to your child's 9th grade Algebra 2 grade.
Also, all of the 8th Grade students will take the 8th Grade State Exam in May. | 677.169 | 1 |
Journey into Mathematics: An Introduction to Proofs
The book is designed for self study or "bridge courses" that prepare students finishing calculus for first courses in real variables and abstract algebra.
Prompting students to do mathematics, not merely read about it, this interesting and uniquely enjoyable text prepares students for reading and writing proofs by having them do just that at the outset. Complete proofs are given from the start and coverage begins with elementary mathematics to allow students to focus on the writing and reading of proofs without the distraction of absorbing new ideas simultaneously. | 677.169 | 1 |
This book, first published in 2004, uses the problem of exact evaluation of definite integrals as a starting point for exploring many areas of mathematics, including analysis, number theory, algebra and combinatorics. This will be a guided tour of exciting discovery for undergraduates and their teachers in mathematics, computer science, physics, and engineering.
"Sinopsis" puede pertenecer a otra edición de este libro.
Críticas:
'I recommend this book highly as a source of rewarding projects for undergraduates (and others) to home their analytic skills and gain an appreciation for this area of mathematics. The authors clearly had great love for the material and their enthusiasm comes through in an infectious manner.' SIAM Review
'The authors have managed to write a very readable account about integrals, accessible even to advanced undergraduates. Some of the topics of the book could be used for undergraduate reading and research projects. This way the book could serve as a 'springboard to many unexpected investigations and discoveries in mathematics.' Zentralblatt MATH
Reseña del editor:
Descripción 2004. Hardback. Estado de conservación: NEW. 9780521791861 This listing is a new book, a title currently in-print which we order directly and immediately from the publisher. Nº de ref. de la librería HTANDREE0741544
Descripción Cambridge University Press1791861
Descripción Cambridge University Press1791861
Descripción Cambridge University Press 2004-06-211528 The | 677.169 | 1 |
Description of Advanced Mathematics Programs and Curriculum
France: Description of the Advanced Mathematics Programs and Curriculum
The Grade 11 and 12 scientific track offers robust mathematical knowledge and skills to students aiming for careers in science, technology, engineering, and mathematics (STEM). The mathematics curriculum is meant to develop students' scientific thinking and strengthen their interest in and affinity for scientific research. Together with introducing new mathematical knowledge and content, the curriculum targets developing students' skills and mathematical faculties in these areas:
Mathematical activities assigned to students both in class and for homework are focused on intra-mathematical or contextually diverse problem solving situations. Students are trained in:
Searching for information, experimenting, and modeling, all using technology
Choosing and executing calculation techniques
Implementing algorithms
Reasoning, proving, and validating results
Explaining an answer, communicating a result
The mathematical content is organized in three parts: Analysis, Geometry, Probability and Statistics. About half of class time should be devoted to Analysis, one quarter to Geometry, and the last quarter to Probability and Statistics. The topics included in each content area are listed below.
Differentiation: calculating derivatives, including the derivatives of common functions, derivatives of sums, products, and quotients of functions, and applications of derivatives, including the relationship between the intervals over which a function increases or decreases and the value of its derivative on those intervals and function extrema
Sine and cosine functions
Exponential functions
Natural logarithms
Integration on an interval, including the relationship between the definite integral and the area under a curve, notation, the antiderivative of a function, linearity, and the additive property of definite integrals
Geometry
Complex numbers, including the algebraic form, conjugate, geometric representation, and polar form of a complex number; the sum, product, and quotient of complex numbers, complex solutions to quadratic equations
Euclidean vectors, including the characterization of a line and a plane, scalar product, coordinates, equation of a plane
Trigonometry, including trigonometric functions defined on the unit circle, radian, the sine and cosine of supplementary and complementary angles
Algorithms fit naturally in all mathematical fields. Algorithmic problem solving in each content area is situated in contexts related to academic subject areas and contexts from real life. Students learn how to implement elementary instructions, loops, and conditional instructions as well as to implement validation and control steps in their programs.
Students learn how to use formal mathematical notation (e.g., for functions, derivatives, and integrals) as well as notation for number sets and intervals.
Students learn elements of formal logic, such as the logical operators for "and" and "or"; the concepts of the contrapositive, the converse and the negative of a conditional statement; logical equivalence; types of arguments, such as the counterexample, the logical disjunction, and the contrapositive; and proof by contradiction. | 677.169 | 1 |
KEY BENEFIT Emphasizing physical interpretations of mathematical solutions, this book introduces applied mathematics and presents partial differential equations. KEY TOPICS Leading readers from simple exercises through increasingly powerful mathematical techniques, this book discusses hear flow and vibrating strings and membranes, for a better understand of the relationship between mathematics and physical problems. It also emphasizes problem solving and provides a thorough approach to solutions. The third edition of , Elementary Applied Partial Differential Equations; With Fourier Series and Boundary Value Problems has been revised to include a new chapter covering dispersive waves. It also includes new sections covering fluid flow past a circular cylinder; reflection and refraction of light and sound waves; the finite element method; partial differential equations with spherical geometry; eigenvalue problems with a continuous and discrete spectrum; and first-order nonlinear partial differential equations. An essential reference for any technical or mathematics professional.
Boundary Value Problems is the leading text on boundary value problems and Fourier series. The author, David Powers, (Clarkson) has written a thorough, theoretical overview of solving boundary value problems involving partial differential equations by the methods of separation of variables. Professors and students agree that the author is a master at creating linear problems that adroitly illustrate the techniques of separation of variables used to solve science and engineering. * CD with animations and graphics of solutions, additional exercises and chapter review questions * Nearly 900 exercises ranging in difficulty * Many fully worked examples
Published by McGraw-Hill since its first edition in 1941, this classic text is an introduction to Fourier series and their applications to boundary value problems in partial differential equations of engineering and physics. It will primarily be used by mathematics students with a background in ordinary differential equations and advanced calculus. There are two main objectives of this text. The first is to introduce the concept of orthogonal sets of functions and representations of arbitrary functions in series of functions from such sets. The second is a clear presentation of the classical method of separation of variables used in solving boundary value problems with the aid of those representations.
This significantly expanded fourth edition is designed as an introduction to the theory and applications of linear PDEs. The authors provide fundamental concepts, underlying principles, a wide range of applications, and various methods of solutions to PDEs. In addition to essential standard material on the subject, the book contains new material that is not usually covered in similar texts and reference books. It also contains a large number of worked examples and exercises dealing with problems in fluid mechanics, gas dynamics, optics, plasma physics, elasticity, biology, and chemistry; solutions are provided.
DIFFERENTIAL EQUATIONS WITH BOUNDARY-VALUE PROBLEMS, 8th Edition strikes a balance between the analytical, qualitative, and quantitative approaches to the study of differential equations. This proven and accessible text speaks to beginning engineering and math students through a wealth of pedagogical aids, including an abundance of examples, explanations, Remarks boxes, definitions, and group projects. Written in a straightforward, readable, and helpful style, the book provides a thorough treatment of boundary-value problems and partial differential equations. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version. | 677.169 | 1 |
Stability, Instability and Chaos
By providing an introduction to nonlinear differential equations, Dr Glendinning aims to equip the student with the mathematical know-how needed to appreciate stability theory and bifurcations. His approach is readable and covers material both old and new to undergraduate courses. Included are treatments of the Poincaré-Bendixson theorem, the Hopf bifurcation and chaotic systems. The unique treatment that is found in this book will prove to be an essential guide to stability and chaos | 677.169 | 1 |
calculus, including: Limits of a function Derivatives of a function Monomials and polynomials Calculating maxima and minima Logarithmic differentials Integrals Finding the volume of irregularly shaped objects By breaking down challenging concepts and presenting clear explanations, you'll solidify your knowledge base--and face calculus without fear!
Calculus 1 & 2 covers differentiation and integration of functions using a guided and an analytical approach. All the normally difficult to understand topics have been made easy to understand, apply and remember. The topics include continuity, limits of functions; proofs; differentiation of functions; applications of differentiation to minima and maxima problems; rates of change, and related rates problems. Also covered are general simple substitution techniques of integration; integration by parts, trigonometric substitution techniques; application of integration to finding areas and volumes of solids. Guidelines for general approach to integration are presented to help the student save trial-and-error time on examinations. Other topics include L'Hopital's rule, improper integrals; and memory devices to help the student memorize the basic differentiation and integration formulas, as well as trigonometric identities; differentiation and integration of hyperbolic functions. This book is one of the most user-friendly calculus textbooks ever published
Calculus is a difficult branch of mathematics that focuses on the study of change and often deals with the application of complex equations. Due to its complex nature, calculus contains dozens of terms that can be near impossible to remember. A reference guide for calculus will greatly benefit students and mathematicians alike by providing a quick tool with which to study or look up terms on a whim. Without a reference, most people attempting a calculus equation are flying blind.
James Stewart's CALCULUSCALCULUS I WITH PRECALCULUS, developed for one-year courses, is ideal for instructors who wish to successfully bring students up to speed algebraically within precalculus and transition them into calculus. The Larson Calculus Two primary objectives guided the authors in writing this book: to develop precise, readable materials for students that clearly define and demonstrate concepts and rules of calculus and to design comprehensive teaching resources for instructors that employ proven pedagogical techniques and saves the instructor time. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version. | 677.169 | 1 |
Algebra 2 and Trig Questions & Answers
Algebra 2 and Trig Flashcards
Algebra 2 and Trig Advice
Algebra 2 and Trig Advice
Showing 1 to 3 of 3
It is an easy course if you do all the work and pay attention in class.
Course highlights:
The highlights were learning more advance algebra and trig that helps in precalculus and calculus.
Hours per week:
0-2 hours
Advice for students:
Always do the practice problems and go into the math lab if you need help.
Course Term:Fall 2013
Professor:Gutter
Course Tags:Math-heavyGreat Intro to the SubjectParticipation Counts
Oct 03, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
I enjoyed this course because (mostly) my teacher was great at explanation and I felt I really understood it. If you have taken geometry and algebra 1, the course is very doable and not excrutiating in the slightest. Sure, some tests required studying, but this is not unusual where math classes are concerned.
Course highlights:
I learned about factoring in the first part of the year, which helps now in precalc and set a base for the math course. I also learned about SOH CAH TOA which proved really useful in AP physics (which I was taking at the time), but also was a solid foundation for basic trig and math.
Hours per week:
3-5 hours
Advice for students:
Study up on the topics right away if you don't understand them when they are taught to you, because most serve as bases for future learnings and are necessary to start or calculate solutions to more complex problems.
I would absolutely recommend this course to any student, even though its an elective. It rounds ones already existing knowledge of Algebra, and is a great reward in itself knowing that many colleges require for this course to be taken, but if it is taken in high school, it makes first semester of college that much easier.
Course highlights:
Yes this class is an elective, and being an art major I wasn't really looking forward to it. But to my surprise, since the first day of this course, I realized that as difficult as some of the subjects we learn are, thanks to Mr.McEntee, they aren't impossible to understand. In fact, they can be enjoyable. I honestly never thought I would ever understand functions as easily as I do now. All the algebra that we have learned in the past years is, in most cases, finally explained more or is expanded to more calculations than I thought existed. To make such a huge change in my eyes about math, Mr.MacEntee has definitely proven himself to be a great teacher.
Hours per week:
3-5 hours
Advice for students:
Definitely worth signing up for. In the long run, its a good credit to have under your belt. Even if math isn't a strong point for them, if the teacher knows how to teach the subject and is understanding towards the students, it shouldn't be too difficult. | 677.169 | 1 |
Careers in Mathematics
The UC Davis Internship and Career Center(ICC) helps forge connections between the university and the workplace. No matter what your field of study, you can benefit from personalized career counseling and from internships that will expose you to new working environments and help develop your employment skills.
Relative to K-12 teaching, the Mathematics and Science Teaching Program at UC Davis (MAST) provides information about multiple pathways to a teaching credential. Their academic advisers will show you how to explore your interest in teaching /and/ earn academic credit. They also provide information about scholarships and financial aid for prospective teachers. Check out their web site, or email them at MAST@ucdavis.edu.
Learn about actuary careers, there are several resources. The Society of Actuaries maintains a career web page for applicants, interns and recruiters interested in risk assessment. The actuarial profession also operates some web sites, such as Be An Actuary and Actuaries: Risk is Opportunity. The Department of Mathematics in at least one university (Austin) has an actuarial program, and the curriculum may be of interest.
Learn about careers in statistics at the American Statistical Association(ASA) website. This site provides general examples of statistical careers and also describes several fields that a statistician may work in, such as law, forestry, education, medicine, and the social sciences.
Read a wide variety of career profiles of actual people at the Mathematical Association of America(MAA) website. Most of those profiled use mathematics on a daily basis; others rely on the general problem solving skills acquired in their mathematics courses. Includes diverse careers such as teacher, biostatistician, systems engineer, and aerospace mathematician.
Career profiles of actual people working in a mathematical field. Each person's profile includes question and answer forum so that you can ask them your own specific questions. Don't forget to check out the archived profiles as well.
An index of mathematical applications provided by applied mathematicians working in industry and in government. See examples of how you can use your knowledge of Laplace transforms, ordinary differential equations, linear algebra, or other topics to investigate a real-life problem. | 677.169 | 1 |
handyCalc Calculator HandyCalc is a powerful calculator with automatic suggestion and solving which makes it easier to learn and useAlgeo Algeo is a graphing calculator. It has some notable features: calculate derivatives, definite integrals and Taylor-series of functions.
What's New in 0.7 | 677.169 | 1 |
Pre GED Mathematics
3 DVDs ΓÇó 5 hrs. 14 mins.
free study guide download
item #VAI1337 ΓÇó price $79.95
isbn 9781573851336
upc 600459133795
GUARANTEED in stock!
A SOLID FOUNDATION IN BASIC GED SKILLS!
Easy to follow lessons in algebra, geometry, and coordinate geometry
Proven strategies for solving word problems
Make common sense work for you on multiple choice questions
Convert the skills you already have into the math techniques you need
How to apply mathematics to real life situations
How to avoid the math mistakes that are tested on the GED
Its no secret that math can be extremely tough. In order to pass the GED Mathematics Test, you11 need to demonstrate your knowledge of: number operations; probability; statistics; data analysis; algebra; geometry; and coordinate geometry. Since many test takers find themselves most unprepared for algebra, geometry, and coordinate geometry, this review covers these areas exclusively. This program will teach you the basics, including: solving equations; word problems; factoring; radicals; angle relationships; similar triangles; Pythagorean theorem; area; volume; distance; slope; and much more. (Be sure you can handle the basic lessons presented here before you view our GED Mathematics course.) | 677.169 | 1 |
Calculators in the new A levels for Mathematics (from 2017)
Ofqual's subject-level conditions and requirements for Mathematics and Further Mathematics state that calculators used must include the following features:
an iterative function
the ability to compute summary statistics and access probabilities from
standard statistical distributions
the ability to perform calculations with matrices up to at least order 3 x 3 (FM only)
For the 2017 A levels students will require a calculator that can calculate Binomial and Normal probabilities directly from values. The minimum standard for this is an advanced scientific calculator, such as the Casio 991EX ClassWiz or the TI-30X Pro; however, graphical calculators have this facility along with the additional advantage of being able to plot the graphs of functions. Graphing tools should be permeating the study of A level and integrating graphical calculators into the teaching and learning is a very effective way of achieving this. | 677.169 | 1 |
Me being a student majoring in journalism, it is rather odd for me to be recommending a math class to someone. However, I do feel that I should recommend a math class to someone because it is necessary to take one. Math is always an argued class that people say they don't need in their life but people actually do need basic math skills in life. Depending on their major and what they want to do with their life, they don't need algebra.
Course highlights:
I learned some new math equations this semester and I learned some easier ways to solve problems. The best thing I learned was to create a song to help me remember the Quadratic Formula. I parodied "Happy Birthday" to help remember it and it did and still does help me solve problems when I need to use the Quadratic Formula.
Hours per week:
3-5 hours
Advice for students:
My professor did not make every homework assignment due for the next class, but I still did the assignments for one reason: to make sure I understood what I was studying and what I was learning. My advice is to do every assignment and get as much help as possible because one simple mistake can put you in trouble. | 677.169 | 1 |
About this product
Description
Description
Providing an introduction to modern algebra, this book aims to have the reader learn to work with mathematics through reading, writing, speaking, and listening. It proves only a few of the theorems. Most proofs are left as exercises, and these exercises can form the core of a course based on this book. | 677.169 | 1 |
Textbooks Collection Effects Of The Use Of Technology In Mathematics Instruction On Student Acheivement, Ron Y. Myers
FIU Electronic Theses and Dissertations
The purpose of this study was to examine the effects of the use of technology on students' mathematics achievement, particularly the Florida Comprehensive Assessment Test (FCAT) mathematics results. Eleven schools within the Miami-Dade County Public School System participated in a pilot program on the use of Geometers Sketchpad (GSP). Three of these schools were randomly selected for this study. Each school sent a teacher to a summer in-service training program on how to use GSP to teach geometry. In each school, the GSP class and a traditional geometry class taught by the same teacher were the study participants. Students' mathematics ...Determining The Effects Of Technology On Children, Kristina E. Hatch
Senior Honors Projects
Determining the Effects of Technology on Children
Kristina Hatch
Faculty Sponser: Timothy Henry, Computer Science and Statistics
Technology has become an essential part of Americans' daily lives, affecting our communications, mail, relationships, the management of our bills and finances. As we have become more immersed in the benefits and capabilities of these constantly developing technologies, children as well as adults have become avid users. Laptops and cell phones are specially developed for preteens. Software and game companies have been targeting children in their game development. Video games have become common entertainment for children as young as four. Children today can ...
Faculty Work: Comprehensive List
Algebra deals with more than computations such as addition or exponentiation; it also studies relations. Calculus touches on this a bit with locating extreme values and determining where functions increase and decrease; and in elementary algebra you occasionally "solve" inequalities involving the order relations of < or ≤ , but this almost seems like an intrusion foreign to the main focus, which is making algebraic calculations.
Relational ideas have become more important with the advent of computer science and the rise of discrete mathematics, however. Many contemporary mathematical applications involve binary or n-ary relations in addition to computations. We began discussing this topic in the last chapter when we introduced equivalence relations. In this chapter we will explore other kinds ofActive Calculus, Matthew Boelkins, David Austin, Steven Schlicker
Open Textbooks
Active Calculus is different from most existing calculus texts in at least the following ways: the text is free for download by students and instructors in .pdf format; in the electronic format, graphics are in full color and there are live html links to java applets; the text is open source, and interested instructors can gain access to the original source files upon request; the style of the text requires students to be active learners — there are very few worked examples in the text, with there instead being 3-4 activities per section that engage students in connecting ideas, solving problems ...
Australian Institute for Innovative Materials - Papers
Presently, lithium-ion batteries (LIBs) are the most promising commercialized electrochemical energy storage systems. Unfortunately, the limited resource of Li results in increasing cost for its scalable application and a general consciousness of the need to find new type of energy storage technologies. Very recently, substantial effort has been invested to sodium-ion batteries (SIBs) due to their effectively unlimited nature of sodium resources. Furthermore, the potential of Li/Li+ is 0.3 V lower than that of Na/Na+, which makes it more effective to limit the electrolyte degradation on the outer surface of the electrode.[1] Nevertheless, one major obstacleVertex Weighted Spectral Clustering, Mohammad Masum
Electronic Theses and Dissertations
Spectral clustering is often used to partition a data set into a specified number of clusters. Both the unweighted and the vertex-weighted approaches use eigenvectors of the Laplacian matrix of a graph. Our focus is on using vertex-weighted methods to refine clustering of observations. An eigenvector corresponding with the second smallest eigenvalue of the Laplacian matrix of a graph is called a Fiedler vector. Coefficients of a Fiedler vector are used to partition vertices of a given graph into two clusters. A vertex of a graph is classified as unassociated if the Fiedler coefficient of the vertex is close to ...
Dissertations
In the field of chemistry there is a growing demand for small molecule organocatalysts such as amino acids, more specifically proline and its analogues, which could catalyze various key chemical reactions in the synthesis of several biologically important molecules. Even though natural proline is reported to catalyze various chemical reactions, its use as organocatalyst is limited mainly due to the solubility issues in the reaction media and high catalyst loadings, which is not very ideal for bulk scale manufacturing. To address these limitations we planned to develop unnatural analogues of proline that could catalyze the reactions with lower catalyst loadings ...
Number Theory
Wave Propagation Inside Random Media, Xiaojun Cheng
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
This thesis presents results of studies of wave scattering within and transmission through random and periodic systems. The main focus is on energy profiles inside quasi-1D and 1D random media.
The connection between transport and the states of the medium is manifested in the equivalence of the dimensionless conductance, g, and the Thouless number which is the ratio of the average linewidth and spacing of energy levels. This equivalence and theories regarding the energy profiles inside random media are based on the assumption that LDOS is uniform throughout the samples. We have conducted microwave measurements of the longitudinal energy profiles ...
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
Exoplanet direct detections are reaching the temperature regime of cool brown dwarfs, motivating further understanding of the coolest substellar atmospheres. These objects, T and Y dwarfs, are numerous and isolated in the field, thus making them easier to study in detail than objects in companion systems. Brown dwarf spectral types are derived from spectral morphology and generally appear to correspond with decreasing mass and effective temperature (Teff). However, spectral subclasses of the colder objects do not share this monotonic temperature correlation, indicating that secondary parameters (gravity, metallicity, dust) significantly influence spectral morphology. These secondary atmospheric parameters can provide insight ...
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
We develop new upper bounds for several effective differential elimination techniques for systems of algebraic ordinary and partial differential equations. Differential elimination, also known as decoupling, is the process of eliminating a fixed subset of unknown functions from a system of differential equations in order to obtain differential algebraic consequences of the original system that do not depend on that fixed subset of unknowns. A special case of differential elimination, which we study extensively, is the question of consistency, that is, if the given system of differential equations has a solution. We first look solely at the ``algebraic data" of ...
Diophantine Approximation And The Atypical Numbers Of Nathanson And O'Bryant, David Seff
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
For any positive real number $\theta > 1$, and any natural number $n$, it is obvious that sequence $\theta^{1/n}$ goes to 1. Nathanson and O'Bryant studied the details of this convergence and discovered some truly amazing properties. One critical discovery is that for almost all $n$, $\displaystyle\floor{\frac{1}{\fp{\theta^{1/n}}}}$ is equal to $\displaystyle\floor{\frac{n}{\log\theta}-\frac{1}{2}}$, the exceptions, when $n > \log_2 \theta$, being termed atypical $n$ (the set of which for fixed $\theta$ being named $\mcA_\theta$), and that for $\log\theta$ rational, the number of atypical $n ...
Rewriting Methods In Groups With Applications To Cryptography, Gabriel Zapata
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
In this thesis we describe how various rewriting methods in combinatorial group theory can be used to diffuse information about group elements, which makes it possible to use these techniques as an important constituent in cryptographic primitives. We also show that, while most group-based cryptographic primitives employ the complexity of search versions of algorithmic problems in group theory, it is also possible to use the complexity of decision problems, in particular the word problem, to claim security of relevant protocols.
Fabrication And Applications Of Multifunctional Superhydrophobic Surfaces Based On Surface Chemistry And Morphology, Yang Liu
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
Superhydrophobic surfaces are gaining great interests in both fundamental researches and technological applications, because of their unique non-wetting and self-cleaning properties. By mimicking the hierarchical surface structure of the natural superhydrophobic surface, i.e. lotus leaf, numerous artificial surperhydrophobic surfaces were developed. However, the challenge is how to fabricate superhydrophobic surfaces by a scalable and economical method. To address this challenge, our group has developed methodologies that enable the fabrication of superhydrophobic surfaces in inexpensive and potentially scalable ways, such as lamination and 3-D printing. To expand on applications, we also combined other desired functionalities into the superhydrophobic surfaces.
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
We highlight some of our research done in the fields of quantum optical interferometry and quantum state engineering. We discuss the body of work for which our research is predicated, as well as discuss some of the fundamental tenants of the theory of phase estimation. We do this in the context of quantum optical interferometry where our primary interest lies in the calculation of the quantum Fisher information as it has been shown that the minimum phase uncertainty obtained, the quantum Cramer-Rao bound, is saturated by parity-based detection methods. We go on to show that the phase uncertainty one obtains ...
Turaev Surfaces And Toroidally Alternating Knots, Seungwon Kim
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
In this thesis, we study knots and links via their alternating diagrams on closed orientable surfaces. Every knot or link has such a diagram by a construction of Turaev, which is called the Turaev surface of the link. Links that have an alternating diagram on a torus were defined by Adams as toroidally alternating. For a toroidally alternating link, the minimal genus of its Turaev surface may be greater than one. Hence, these surfaces provide different topological measures of how far a link is from being alternating.
First, we classify link diagrams with Turaev genus one and two in terms ...
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects ...
Joint Laver Diamonds And Grounded Forcing Axioms, Miha Habič
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
In chapter 1 a notion of independence for diamonds and Laver diamonds is investigated. A sequence of Laver diamonds for κ is joint if for any sequence of targets there is a single elementary embedding j with critical point κ such that each Laver diamond guesses its respective target via j. In the case of measurable cardinals (with similar results holding for (partially) supercompact cardinals) I show that a single Laver diamond for κ yields a joint sequence of length κ, and I give strict separation results for all larger lengths of joint sequences. Even though the principles get strictly ...
Travel Mode Identification With Smartphone Sensors, Xing Su
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
Personal trips in a modern urban society typically involve multiple travel modes. Recognizing a traveller's transportation mode is not only critical to personal context-awareness in related applications, but also essential to urban traffic operations, transportation planning, and facility design. While the state of the art in travel mode recognition mainly relies on large-scale infrastructure-based fixed sensors or on individuals' GPS devices, the emergence of the smartphone provides a promising alternative with its ever-growing computing, networking, and sensing powers. In this thesis, we propose new algorithms for travel mode identification using smartphone sensors. The prototype system is built upon the ...
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
We study the geometry of fully augmented link complements in the 3-sphere by looking at their link diagrams. We extend the method introduced by Thistlethwaite and Tsvietkova to fully augmented links and define a system of algebraic equations in terms of parameters coming from edges and crossings of the link diagrams. Combining it with the work of Purcell, we show that the solutions to these algebraic equations are related to the cusp shapes of fully augmented link complements. As an application we use the cusp shapes to study the commensurability classes of fully augmented links. | 677.169 | 1 |
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Overview - This volume concentrates on the structure of Boolean algebras and rings as developed through simpler algebraic systems. The algebra of logic and set theory appears as applications or illustrations throughout, and numerous problems form an integral part of the text.Read more...
This volume concentrates on the structure of Boolean algebras and rings as developed through simpler algebraic systems. The algebra of logic and set theory appears as applications or illustrations throughout, and numerous problems form an integral part of the text. No prior knowledge of Boolean algebra is necessary. The text begins with a consideration of some concepts of intuitive logic. Subsequent chapters introduce Boolean functions, basic ordered sets, and a succession of algebraic systems. An examination of the principle of duality leads to a final chapter on applications, including electrical networks and computer design. Suitable for courses in computer design, circuit analysis, and switching circuits, this volume can also serve as a reference for professionals. | 677.169 | 1 |
sábado, 27 de dezembro de 2014
The book outlines the basics of descriptive geometry in direct connection with the basics of technical drawing and sketching; fundamentals of engineering drawing, rules for the implementation of schemes; are elements of the construction and topographic drawing.
LINK
sexta-feira, 26 de dezembro de 2014
This tutorial, together with manuals TI Trofimova "Physics for technical training areas" (qualification "Bachelor"), "The course of Physics", "Physics in tables and formulas" and "physics course. Waves »TI Trofimova and AV Firsov is a single teaching kit in physics for students of technical colleges. About half of the problems are given with detailed solutions and explanations are provided for the remaining independent decision. This makes it possible to use this guide as a book of problems for schools. The book consists of seven chapters, covering all sections of physics course for engineering specialties of higher education.
LINK ORIGINAL
The collection contains more than 1,000 questions and challenges for all sections of university course in general physics.Each section presents the basic physical laws and relations necessary for solving problems.
The new edition replaced the outdated definitions and terms, introduced the International System of Units (SI).All tasks are provided with answers and the most difficult - the instructions, comments to the decision.
The collection includes tasks for all sections of the course of physics taught in secondary vocational schools.In the first part of the book offers solutions to typical problems and examples of recording solutions.
This will help students develop independent skills in solving problems in physics.The second part contains tasks for self-determination and the answers to them.These tasks can be used for the preparation and conduct of examinations, as well as the repetition of the material.
Collection may be useful to students of secondary schools, lyceums and gymnasiums.
quinta-feira, 25 de dezembro de 2014
Collection of tasks can serve as a guide for self-training for the Olympiad in mathematics. The collection consists of objectives proposed in recent years in the city of Samara Mathematical Olympiads: sammatah, University Nayanova, competitions and SSU SamGGU for graduates.
These formula are related to the solid geometry dealing with the 2-D & 3-D figures in the space & miscellaneous articles in Trigonometry & Geometry. These are useful the standard formula to be remembered for case studies & practical applications. Although, all the formula for the plane figures (i.e. planes bounded by the straight lines only) can be derived by using standard formula of right triangle that has been explained in details in "HCR's Theory of Polygon" published with International Journal of Mathematics & Physical Sciences Research in Oct, 2014.
And the analysis of oblique frustum of right circular cone has been explained in his research paper 'HCR's Infinite-series' published with IJMPSR
HCR's Hand Book.pdf
Adobe Acrobat Document [4.2 MB]
sábado, 13 de dezembro de 2014
Explaining Logarithms
by Dan Umbarger - Brown Books Publishing Group , 2006
Description:
These materials show the evolution of logarithmic ideas over 350 years. I do believe that a quick review of mathematics as it was practiced for hundreds of years would be helpful for many students in understanding logarithms as they are still used today. I see three potential audiences for this material: 1.) students who have never studied logarithms, 2.) students who have studied logarithms but who did not master the concepts or have forgotten key ideas, or 3.) summer school reading for students taking calculus in the fall.
LINK ORIGINAL
LINK DIRECT | 677.169 | 1 |
August 4, 2009
I guess I am getting a little anxious, but school will be back in session within two weeks. Everyone is gathering supplies, new clothes and backpacks. But what about lesson materials to use before and during the school year? Although some students enjoy Math/Algebra/Geometry, many find these topics difficult. This is where Geometry Stash can help. It is geared solely towards putting geometry formulas in one organized package so that you can readily find the formulas you need to study for an exam, or complete your homework assignment. The features of this app include: landscape and portrait mode, search to narrow down results, descriptive, visual diagrams with each item, and shake to hide/show diagram area. Just enter the first few letters of the formula you are researching, and bingo, Geometry Stash will display it for you along with a handy example and description of how to work this formula. All you need to do is drag your finger to get around the display. I love the simple organization of this app, and it will be a huge help to tutors, teachers and students alike. Don't let your Math assignments get you down when you have Geometry Stash to help you through.
"Access the most commonly-used theorems, postulates and corollaries through Geometry Stash's fluid, clean interface. Drag your finger up and down the list to navigate its contents. Touch the lower textured button to hide/show the diagram and adjust viewing space."
--Note: This app requires OS 3.0 or later. Rated 4+. --Tip: Geometry Stash will be offered at no charge until the end of August. --Languages: English --Developer: Anson Liu
Note: Prices, Ratings and Versions are subject to change. Please visit the iTunes App Store for more details, and to download: GEOMETRY STASH
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--More Cool, Fun, Tech-Savvy Stuff!--
--SMS Acronyms: HAGO - Have A Good One
--Tech Jargon: cipher - A procedure that transforms data between plaintext and ciphertext; a crypto algorithm. Basically a "cipher" is any method of encrypting text to conceal its readability and meaning. Its origin is the Arabic "sifr," meaning empty or zero. The term cipher is sometimes used as a synonym for ciphertext (which means "encrypted text"), but it actually describes the method of encryption rather than the result. "Plain text" is what you have before encryption, and "ciphertext" is the encrypted result. | 677.169 | 1 |
Bach/Leitner's progressive text lays a solid foundation for elementary algebra that carefully addresses student needs. The authors' clear, non-intimidating, and humorous style reassures math-anxious readers. Unlike workbook-format Prealgebra texts that stress competence at procedures, this text emphasizes understanding and mastery through careful step-by-step explanations that strengthen students' long-term abilities to conceptualize and solve problems. The text's innovative sequencing builds students' confidence with arithmetic operations early on before extending the basic concepts to algebraic expressions and equations. The authors' unusually thorough introduction to variables eases students through the crucial transition from working with numbers. Throughout the text, interesting applied examples and exercises and math-appreciation features highlight key concepts at work in a wide variety of real-world contexts.
"synopsis" may belong to another edition of this title.
About the Author:
Daniel Bach earned his B.A and M.A degrees in Mathematics from the University of California at Berkeley, specializing in algebraic number theory. He is currently a popular math instructor at Diablo Valley College, having previously taught the subject at Mills College and UC-Berkeley. Dan was head tutor for the Minority Engineering Students Asociation and a Math Specialist at Black Pine Circle Day School, where he coauthored several California Junior High Math Competitions during the 1980s. Dan is the author and programmer of "Dan's Basic Math Clinic," an interactive multimedia courseware in arithmetic, prealgebra, and beginning algebra. In the 1990s, Dan was awarded an NSF developer's grant to write a series of notebooks Precalculus and Mathematica, for the Interactive Math Textbook Project. Since 1997, he has maintained an active website with free math lessons, a weekly contest, feature pages, and an international puzzle-problem community. He is an avid coffee drinker, runner, and bicyclist and enjoys creating 3D animations in his hard-to-find spare time.
Patricia Leitner received her B.A. in Applied Mathematics and her M.A. in Mathematics from the University of California at Berkeley. She has taught Mathematics at the University of California at Berkeley and at City College in San Francisco and currently teaches at Diablo Valley College in Pleasant Hill, California. Patricia has focused her career on making Mathematics understandable and accessible to every student. Many have attributed their success in Math to her teaching style. She was awarded the Nikki Kose Memorial Teaching Prize by the UC-Berkeley, and she has also served on the Mathematical Association of America's National Committee on Two-Year Colleges. Patricia helped design and implement a state-of-the-art computer lab at Diablo Valley College and has written a series of Computer Calculus Tutorials. She has also been a principal developer of the college's self-paced Algebra program and recently created a video series aimed at helping learning-disabled students grasp the concepts of algebra. Patricia has traveled extensively, particularly in France, where she has fun subjecting the locals to her suboptimal French. Most recently, she has begun a second career writing fiction. Patricia currently lives in the San Francisco Bay Area where she enjoys long walks, fitness swimming, and relaxing in cafes with a good book.
Book Description McGraw-Hill Science/Engineering/Math. Paperback. Book Condition: VERY GOOD. little to no wear, pages are clean. The cover and binding are crisp with next no creases. Bookseller Inventory # 2749030631
Book Description McGraw-Hill Science/Engineering/Math. Book Condition: Good. 0073101575 May have signs of use, may be ex library copy. Book Only. Used items do not include access codes, cd's or other accessories, regardless of what is stated in item title. Bookseller Inventory # Z0073101575Z3
Book Description Blacklick, Ohio, U.S.A.: McGraw-Hill Science Engineering, 2004. Soft cover. Book Condition: Good. Dust Jacket Condition: Good. 3rd Edition. Good condition, used in outstanding condition. Book is the US version. Textbook only, accessories such as CD, infotrac, web access codes, may not be included2173977433 | 677.169 | 1 |
Simpson's Rule
In this integral learning exercise, students use the Simpsons's Rule for evaluating integrals. They use different methods to calculate integrals and compare the results. This eight-page learning exercise contains approximately 16 multi-step problems. | 677.169 | 1 |
College Mathematics
In the previous post, we have learned the graphical representation of domain and range. The domain of the function is the shadow or projection of the graph of to the x-axis (see red segment in the figure below). The range of is the projection of the graph of to the y-axis (see green segment in […] Continue reading…
The domain of a function is the set of x-coordinates of the points in the function. The range of the function f is the set of y-coordinates of the points in the function. So if we have a function f with points (-3, -2), (-1, 3), (2, 3), and (5,4), then the domain of the […] Continue reading…
I bought this book a year ago as a refresher of Calculus and as of now, I am almost finished reading it. I think what separates this book from the rest are the numerous worked examples (well, 1000 of them) with detailed solutions and explanations. Additional pointers and explanations in layman's words are provided as notes. […] Continue reading… | 677.169 | 1 |
Integration Applications
Given equations that define a region in the plane students are asked to find its area and the volume of the solid formed when the region is revolved around a line or used as a base of a solid with regular cross-sections. This standard application of the integral has appeared every year since 1969 on…
The Free-response Questions The free-response questions fall into 10 general categories or types. The multiple-choice questions fall largely into the same categories plus some straight-forward questions asking students to find limits, derivatives, and integrals. Often two or more type are combined into one question. The types are the following. Rate and Accumulation Linear motion Graph Analysis…
Density, as an application of integration, has snuck onto the exams. It is specifically not mentioned in the " Curriculum Framework" chapter of the new Course and Exam Description. There is an example (#12 p. 58) in the AB sample exam question section of Course and Exam Description.The first time this topic appeared was in…
The sixth in the Graphing Calculator / Technology Series Both graphing calculators and CAS calculators allow students to evaluate definite integrals. In the sections of the AP Calculus that allow calculator use students are expected to use their calculator to evaluate definite integrals. On the free-response section, students should write the integral on their paper,…
Sometimes I think textbooks are too rigorous. Behind every Riemann sum is a definite integral. So, authors routinely show how to solve an application of integration problem by developing the method starting from the Riemann sum and proceeding to an integral that give the result that is summarized in a "formula." There is nothing wrong with…
Today's question is not a good questions. It's a bad question. But sometimes a bad question can become a good one. This one leads first to a discussion of units, then to all sorts of calculus. Here's the question a teacher sent me this week taken from his textbook: The normal monthly rainfall at the | 677.169 | 1 |
Description
Provides step-by-step recipes for constructing and analyzing models
Interesting biological applications
Explores classical models in ecology and evolution
Questions at the end of every chapter
Primers cover important mathematical topics
Exercises with answers
Appendixes summarize useful rules
Labs and advanced material available
About the authors
Sarah P. Otto is Professor of Zoology at the University of British Columbia. Troy Day is Associate Professor of Mathematics and Biology at Queen's University
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amarcobio
LibraryThingSimilarMathematical models are the decisive tool to explain and predict phenomena in the natural and engineering sciences. With this book readers will learn to derive mathematical models which help to understand real world phenomena. At the same time a wealth of important examples for the abstract concepts treated in the curriculum of mathematics degrees are given. An essential feature of this book is that mathematical structures are used as an ordering principle and not the fields of application.
Methods from linear algebra, analysis and the theory of ordinary and partial differential equations are thoroughly introduced and applied in the modeling process. Examples of applications in the fields electrical networks, chemical reaction dynamics, population dynamics, fluid dynamics, elasticity theory and crystal growth are treated comprehensively.
Mathematics for the Life Sciences provides present and future biologists with the mathematical concepts and tools needed to understand and use mathematical models and read advanced mathematical biology books. It presents mathematics in biological contexts, focusing on the central mathematical ideas, and providing detailed explanations. The author assumes no mathematics background beyond algebra and precalculus. Calculus is presented as a one-chapter primer that is suitable for readers who have not studied the subject before, as well as readers who have taken a calculus course and need a review. This primer is followed by a novel chapter on mathematical modeling that begins with discussions of biological data and the basic principles of modeling. The remainder of the chapter introduces the reader to topics in mechanistic modeling (deriving models from biological assumptions) and empirical modeling (using data to parameterize and select models). The modeling chapter contains a thorough treatment of key ideas and techniques that are often neglected in mathematics books. It also provides the reader with a sophisticated viewpoint and the essential background needed to make full use of the remainder of the book, which includes two chapters on probability and its applications to inferential statistics and three chapters on discrete and continuous dynamical systems.
The biological content of the book is self-contained and includes many basic biology topics such as the genetic code, Mendelian genetics, population dynamics, predator-prey relationships, epidemiology, and immunology. The large number of problem sets include some drill problems along with a large number of case studies. The latter are divided into step-by-step problems and sorted into the appropriate section, allowing readers to gradually develop complete investigations from understanding the biological assumptions to a complete analysis introductory textbook on mathematical biology focuses on discrete models across a variety of biological subdisciplines. Biological topics treated include linear and non-linear models of populations, Markov models of molecular evolution, phylogenetic tree construction, genetics, and infectious disease models. The coverage of models of molecular evolution and phylogenetic tree construction from DNA sequence data is unique among books at this level. Computer investigations with MATLAB are incorporated throughout, in both exercises and more extensive projects, to give readers hands-on experience with the mathematical models developed. MATLAB programs accompany the text. Mathematical tools, such as matrix algebra, eigenvector analysis, and basic probability, are motivated by biological models and given self-contained developments, so that mathematical prerequisites are minimal. Integrates deterministic and stochastic approaches Teaches skills in model construction, analysis, inference, and interpretation Features numerous exercises and their detailed elaborations Motivated by real-world applications throughout
Written for social science students who will be working with or conducting research, Mathematics for Social Scientists offers a non-intimidating approach to learning or reviewing math skills essential in quantitative research methods. The text is designed to build students' confidence by presenting material in a conversational tone and using a wealth of clear and applied examples. Author Jonathan Kropko argues that mastering these concepts will break students' reliance on using basic models in statistical software, allowing them to engage with research data beyond simple software calculationsSelf-contained and suitable for undergraduate students, this text offers a working knowledge of calculus and statistics. It assumes only a familiarity with basic analytic geometry, presenting a coordinated study that develops the interrelationships between calculus, probability, and statistics. Starting with the basic concepts of function and probability, the text addresses some specific probabilities and proceeds to surveys of random variables and graphs, the derivative, applications of the derivative, sequences and series, and integration. Additional topics include the integral and continuous variates, some basic discrete distributions, as well as other important distributions, hypothesis testing, functions of several variables, and regression and correlation. The text concludes with an appendix, answers to selected exercises, a general index, and an index of symbols.
Dynamic Models in Biology offers an introduction to modern mathematical biology. This book provides a short introduction to modern mathematical methods in modeling dynamical phenomena and treats the broad topics of population dynamics, epidemiology, evolution, immunology, morphogenesis, and pattern formation.
Primarily employing differential equations, the author presents accessible descriptions of difficult mathematical models. Recent mathematical results are included, but the author's presentation gives intuitive meaning to all the main formulae. Besides mathematicians who want to get acquainted with this relatively new field of applications, this book is useful for physicians, biologists, agricultural engineers, and environmentalists.
Key Topics Include:
Chaotic dynamics of populationsThe spread of sexually transmitted diseasesProblems of the origin of lifeModels of immunologyFormation of animal hide patternsThe intuitive meaning of mathematical formulae explained with many figuresApplying new mathematical results in modeling biological phenomena
Miklos Farkas is a professor at Budapest University of Technology where he has researched and instructed mathematics for over thirty years. He has taught at universities in the former Soviet Union, Canada, Australia, Venezuela, Nigeria, India, and Columbia. Prof. Farkas received the 1999 Bolyai Award of the Hungarian Academy of Science and the 2001 Albert Szentgyorgyi Award of the Hungarian Ministry of Education.
A 'down-to-earth' introduction to the growing field of modern mathematical biologyAlso includes appendices which provide background material that goes beyond advanced calculus and linear algebraBIOCALCULUS: CALCULUS, PROBABILITY, AND STATISTICS FOR THE LIFE SCIENCES shows with a sound knowledge of mathematics, an understanding of the importance of mathematical arguments, and a clear understanding of how these mathematical concepts and techniques are central in the life sciencesed to download a file and transfer it to your device. Please follow the detailed Help center instructions to transfer the files to supported eReaders. | 677.169 | 1 |
Homework Package - Multi-step Linear Equations
Be sure that you have an application to open
this file type before downloading and/or purchasing.
389 KB|9 pages
Product Description
This homework packet includes 7 half-sheet homeworks about equations. They can include 2-step or multi-step equations, most with variables on both sides of the equality. One sheet asks students to represent and solve equations with algebra tiles and another asks students to represent an equation on a balance. One sheet has 4 word problems.
Each sheet has two homeworks on it to conserve paper. There are also two homeworks that are front and back.
You can also use these sheets as exit passes or do nows (bell work/ringers). | 677.169 | 1 |
...
Show More range of upper level students. The author has learned through many years of teaching that the best way to present theoretical concepts is to take advantage of the precision and clarity of mathematical language. In a way that is accessible to students still learning this language, he presents the necessary mathematical tools gently and gradually which provides discussion and examples that make the language intelligible | 677.169 | 1 |
1-Month Free Trial
Math can be hard, but it doesn't have to be. I make math courses to guide you through everything from math's Fundamentals, all the way through advanced Calculus, and I walk with you along the way.
Because you're here, I want to offer you a special 1-month free trial of Calculus Expert.
The trial gives you access to every course in my course library, featuring over 100 hours of video tutorials, plus lecture notes and tons of quiz questions. Together, the resources inside these courses will be your success roadmap: a step-by-step path to crushing your math class.
This special trial is completely free, and it's for people like you who are interested in doing better in their math class without having to search everywhere to get help.
If you stick around after the trial, the monthly membership is just $27/month (less than just one hour of tutoring), which gives you access to the courses, quizzes, and support which I'll explain more about below.
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Get ready to pinch yourself, because math really can be this easy.
I've helped students who would inevitably fail squeeze by with a grade 10% higher than the one they needed to pass. And people who failed high school math get accepted at prestigious universities for degrees in mathematics.
I don't say stuff like that to brag, just to be totally honest with the fact that math is possible to figure out, and even if you feel like you're cursed in that area, there is still hope for you.
Even if you're not a math whiz, math can still be easy for you.
Yes, even if you're in a Calculus III class and everyone else in your class is failing miserably.
Whatever reason you're taking math classes for—to become a scientist, doctor, or just complete the math requirements for your liberal arts degree—there's one fact you can't run away from: you've got to pass this class or change your major.
And if you change your major, you'll also have to change your life plans.
Which is so not fair.
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Because your dream isn't a partial derivative. And it shouldn't be as hard as one, either.
I have students from all walks of life studying inside Calculus Expert:
Traditional, four-year college students who just need to pass a class to graduate on time
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And here's how Calculus Expert has helped them:
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Every single math course I've ever made
The fundamentals of middle school math all the way to advanced calculus. Access to every course and lesson you need to ace your class.
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3.
Open Q&A sections
Still don't understand something? I check into the Q&A sections of the courses at least once every weekday. Ask away. I'm happy to help.
And here's what you get inside of every single lesson:
Videos
Watch over my shoulder as I solve problems for every single math issue you'll encounter in class. We start from the beginning. I explain the setup and why I set it up that way, the steps I take and why I take them, how to work through the fuzzy middle parts, and how to simplify the answer when you get it.
Notes
The notes section of each lesson is where you find the most important things to remember. It's like Cliff Notes for books, but for math. Everything you need to know to pass your class and nothing you don't.
Quizzes
When you think you've got a good grasp on a topic, you can test your knowledge without affecting your grade by taking one of our quizzes. If you pass, wonderful. If not, you can review the videos and notes again, or ask me for help in the Q&A section.
Below each lesson there's a place for you to ask me questions. Anything goes, as long as it's math related. There's even a function for you to upload a picture of your work to show me exactly what you're getting hung up on.
Plus, I've made all of the videos, notes, and quizzes specifically for non-math minded people | 677.169 | 1 |
A N G L E S & P O L Y G O N S
Year 9 Mathematics
A profusion of diagrams, and lots of practice questions make these six lessons a joy to teach.
I use the topic as an excuse to pack in algebra in a meaningful and purposeful context.
Answers to all exercises are included, but not to the concluding test. | 677.169 | 1 |
AS Level Mathematics/English
Why should I study an additional AS?
To complement your three A Levels, you can also study either AS English or Maths. This is in order to can gain an understanding of these core subject areas without the need to complete a full A Level. Universities will only make offers based on the completion of three A Levels.
There are three modular examinations, two in Pure Mathematics and one in Applications
Pure Mathematics develops and expands on the work in algebra, geometry, graphs, coordinates and trigonometry first encountered in the GCSE course. Calculus is introduced. There is less numerical work; a genuine feel for algebra and a real familiarity with all its techniques are vital for success.
Applications is designed to encourage a mathematically analytical approach to practical situations and to develop mathematical models in both Statistics and Mechanics. | 677.169 | 1 |
Introduction to Galois Theory
Introduction to Galois Theory
Introduction to Galois Theory
Higher School of Economics
About this course: A very beautiful classical theory on field extensions of a certain type (Galois extensions) initiated by Galois in the 19th century. Explains, in particular, why it is not possible to solve an equation of degree 5 or more in the same way as we solve quadratic or cubic equations. You will learn to compute Galois groups and (before that) study the properties of various field extensions.
We first shall survey the basic notions and properties of field extensions: algebraic, transcendental, finite field extensions, degree of an extension, algebraic closure, decomposition field of a polynomial.
Then we shall do a bit of commutative algebra (finite algebras over a field, base change via tensor product) and apply this to study the notion of separability in some detail.
After that we shall discuss Galois extensions and Galois correspondence and give many examples (cyclotomic extensions, finite fields, Kummer extensions, Artin-Schreier extensions, etc.).
We shall address the question of solvability of equations by radicals (Abel theorem). We shall also try to explain the relation to representations and to topological coverings.
Finally, we shall briefly discuss extensions of rings (integral elemets, norms, traces, etc.) and explain how to use the reduction modulo primes to compute Galois groups.
PREREQUISITES
A first course in general algebra — groups, rings, fields, modules, ideals. Some knowledge of commutative algebra (prime and maximal ideals — first few pages of any book in commutative algebra) is welcome. For exercises we also shall need some elementary facts about groups and their actions on sets, groups of permutations and, marginally,
the statement of Sylow's theorems.
ASSESSMENTS
A weekly test and two more serious exams in the middle and in the end of the course. For the final result, tests count approximately 30%, first (shorter) exam 30%, final exam 40%.
There will be two non-graded exercise lists (in replacement of the non-existent exercise classes...)
We introduce the basic notions such as a field extension, algebraic element, minimal polynomial, finite extension, and study their very basic properties such as the multiplicativity of degree in towers.
6 videos
Graded: Quiz 1
WEEK 2
Week 2
We introduce the notion of a stem field and a splitting field (of a polynomial). Using Zorn's lemma, we construct the algebraic closure of a field and deduce its unicity (up to an isomorphism) from the theorem on extension of homomorphisms.
5 videos
Graded: QUIZ 2
WEEK 3
Week 3
We recall the construction and basic properties of finite fields. We prove that the multiplicative group of a finite field is cyclic, and that the automorphism group of a finite field is cyclic generated by the Frobenius map. We introduce the notions of separa...
6 videos, 1 reading
Graded: QUIZ 3
WEEK 4
Week 4
This is a digression on commutative algebra. We introduce and study the notion of tensor product of modules over a ring. We prove a structure theorem for finite algebras over a field (a version of the well-known "Chinese remainder theorem").
6 videos
Graded: QUIZ 4
WEEK 5
Week 5
We apply the discussion from the last lecture to the case of field extensions. We show that the separable extensions remain reduced after a base change: the inseparability is responsible for eventual nilpotents. As our next subject, we introduce normal and Gal...
6 videos
Graded: Graded assignment 1
Graded: QUIZ 5
WEEK 6
Week 6
We state and prove the main theorem of these lectures: the Galois correspondence. Then we start doing examples (low degree, discriminant, finite fields, roots of unity).
6 videos
Graded: QUIZ 6
WEEK 7
Week 7
We continue to study the examples: cyclotomic extensions (roots of unity), cyclic extensions (Kummer and Artin-Schreier extensions). We introduce the notion of the composite extension and make remarks on its Galois group (when it is Galois), in the case when t...
7 videos, 1 reading
WEEK 8
Week 8
We finally arrive to the source of Galois theory, the question which motivated Galois himself: which equation are solvable by radicals and which are not? We explain Galois' result: an equation is solvable by radicals if and only if its Galois group is solvabl...
6 videos
Graded: QUIZ 8
WEEK 9
Week 9.
We build a tool for finding elements in Galois groups, learning to use the reduction modulo p. For this, we have to talk a little bit about integral ring extensions and also about norms and traces.This week, the final graded assignment is given.
6 videos
Graded: Graded assignment 2 (final exam)
Graded: QUIZ 93 out of 5 of 65 ratings
SD
Work hard, learn more.
HC
An intriguing course.
Uncommonly advanced material.
Rare content on Coursera.
Several comments, in general the course is well conducted,it is not easy to follow, but this is natural as it a very abstract subject. The exercise sets (peer-reviewed) where very instructive, but I think that the quizzes where a little too easy in comparison. I missed a little a more intuitive view, a broader view of the subject as well as a review of more advanced applications and developments, such as infinite Galois groups, etc.
R
It is a rare online course of advanced pure mathematics. Overall it is very good. It is not recommended to take other courses in parallel with this course as it will consume lots of time on external notes and references. I would recommend National Research University to open more similar courses on abstract algebra, like Groups, Rings, Fields and Modules as a bridge program to this course, or an extension of this course to Lie Algebra or Representation Theory. | 677.169 | 1 |
The Precalculus CLEP study guide and flashcards will teach you everything you need to know to pass the test. You've probably already covered a few of main concepts in College Algebra. This CLEP study guide will give you detailed examples and exercises to teach and test your skills. You don't need a separate textbook with this or any other CLEP study guide we offer. This study guide also includes sample test questions to teach you the material. This is one of the harder of all the CLEP math tests you can take. We recommend that if you need to take a math CLEP test, you first take the College Mathematics CLEP.
$35.97
Credits: 3
Difficulty: 5
Number of Questions: 48
Included in the Precalculus CLEP Study Guide
Functions
Graphing Functions
Analytic Geometry
Logarithms
Functions as Models
Trigonometry
Sequences and Series
Polynomials
Linear Equations
Equations, Inequalities and Their Graphs
Quadratic Equations
Quadratic Formula
Graphing Quadratic Equations
Sample Questions
Answer Key
Test Taking Strategies
What Your Score Means
Testimonials for Pass Your Class - Precalculus CLEP Study Guide
I just wanted to let you know that I used your College Mathematics study guide and passed my test. I had not picked up a math book since high school, 12 years ago. Thank you for the great guide. I am now studying for my Natural Sciences and US History I & II tests. I will be using your guides for those tests as well.
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I have bought two separate materials form passyourclass and passed both exams. –Maynard | 677.169 | 1 |
Grade 8 Math Connects Suggested Course Outline for Schooling at Home 132 lessons I. Introduction: (1 day) Look at p. 1 in the textbook with your child and learn how to use the math book effectively. DO:
Solving With Variables On Both Sides Free PDF ebook Download: Solving With Variables On Both Sides Download or Read Online ebook solving equations with variables on both sides in PDF Format From The Best
What a freshman should know to take Geometry in the 9 th Grade at St. Francis Algebra in 9 th Grade Algebra 1 H. Algebra 1 A Grade and Recommendation Geometry H. Geometry Algebra Trig / Pre-Calc A Grade,
Course Title: Honors Algebra Course Level: Honors Textbook: Algebra Publisher: McDougall Littell The following is a list of key topics studied in Honors Algebra. Identify and use the properties of operations
Course Title: Math A Elementary Algebra Unit 0: Course Introduction In this unit you will get familiar with important course documents, such as syllabus and communication policy. You will register on MyMathLab
Kuta Free PDF ebook Download: Kuta Download or Read Online ebook ratio and proportion kuta in PDF Format From The Best User Guide Database Worksheet by Kuta Software LLC 1) Two similar figures are in a
New York State Mathematics Content Strands, Grade 7, Correlated to Glencoe MathScape, Course 2 and The lessons that address each Performance Indicator are listed, and those in which the Performance Indicator
Warm Up Free PDF ebook Download: Warm Up Download or Read Online ebook geometry warm up questions in PDF Format From The Best User Guide Database Set your class on the right path with these warm-up problems,
2012-2013 Math Content PATHWAY TO ALGEBRA I Unit Lesson Section Number and Operations in Base Ten Place Value with Whole Numbers Place Value and Rounding Addition and Subtraction Concepts Regrouping ConceptsAnswer to the most important question! What should My Students for their Grade Level? H O M E S C H O O L Math Lesson Plan 6 th Grade Math Why This Lesson Plan? This lesson plan answers the most important
Prep for Intermediate Algebra This course covers the topics outlined below, new topics have been highlighted. You can customize the scope and sequence of this course to meet your curricular needs. Curriculum
Grade 8 Correlated Curriculum The following table shows how the Grade 8 JUMP Math curriculum covers all of the Grade 8. All of the lessons in the Grade 8 Teacher s Guide and Assessment & Practice Books
Unit 1: Numeracy and Expressions (Approximate number of days - Extended: 11 days Algebra: 11 days) (FYI, last day of 1 st quarter is Day 42, October 30) Connections to Previous Learning: Students use the
Prep for Calculus This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet curricular
ALGEBRA 2/TRIGONOMETRY Mathematics Prince George s County Public Schools SY 2011-2012 Prerequisites: Geometry with a grade of B or higher Credits: 1.0 Math, Merit CLASS MEETS: Every other day for 90 minutes
Point Pleasant Boro High School Mathematics Department Geometry Summer Review Packet * Welcome to Geometry! This summer review assignment is designed for ALL students enrolled in Geometry for the 2016-2017
Lyman Memorial High School Pre-Calculus Prerequisite Packet Name: Dear Pre-Calculus Students, Within this packet you will find mathematical concepts and skills covered in Algebra I, II and Geometry. These | 677.169 | 1 |
ISBN-10: 0486425657
ISBN-13: 9780486425658 text explores the translation of geometric concepts into the language of numbers in order to define the position of a point in space (the orbit of a satellite, for example). The two-part treatment begins with discussions of the coordinates of points on a line, coordinates of points in a plane, and the coordinates of points in space. Part 2 examines geometry as an aid to calculation and the necessity and peculiarities of four-dimensional space. Written for systematic study, it features a helpful series of "road signs" in the margins, alerting students to passages requiring particular attention, and an abundance of ingenious problems-with solutions, answers, and hints-promote habits of independent work. 1967 | 677.169 | 1 |
Mathematics coursework tasks | canadianbioceutical.com
Data Handling Cycle
Be it your mathematics assignment or any other mathematics project.This Math Homework Help Reference Guide gives you all the basic math skills you need to succeed.The site includes lessons, formulas, online calculators and homework help.
Statistics Maths Coursework Ideas
Internal Assessment IB Math Studies Project
Covering pre-algebra through algebra 3 with a variety of introductory and advanced lessons.
PSAT Math Questions
Discursive Essay Translation
mathematics t coursework stpm 2016 introduction
We are math teachers and professors who believe that immediate help with assigned homework improves math learning.
Do My Statistics Homework. How it works. 1. Upload your homework. 2. We can help you do well in your online math, statistics,.If you need help in college algebra, you have come to the right place.
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If you are stuck with your math assignment or writing math papers is not your forte, our educational resource is for you.Algebra, math homework solvers, lessons and free tutors online.Pre-algebra, Algebra I, Algebra II, Geometry, Physics.So-called math-anxious parents who provided frequent help on homework actually hurt their children by passing on their anxiety, a study found.
A resource provided by Discovery Education to guide students and provide Mathematics Homework help to students of all grades.Second grade math builds upon the skills learned in the first grade.To help parents address questions children may have about Eureka Math at home, we provide Tip Sheets for each module for Grades K through 8. | 677.169 | 1 |
Monday, October 27, 2008
MathML Strengthens Digital Math Texts Digital texts are moving beyond simply putting pages on a computer screen. In Kentucky, a small pilot study is demonstrating the benefits of this textbook technology for students with different learning styles. Instead of re-creating a complex math problem as a static image file, digital texts that use math markup language, or MathML, are able to speak words and equations while highlighting corresponding elements on a computer screen. MathML-enabled digital texts helped the study's very small cohort of students struggling with printed text outperform peers who used traditional print texts. Many students said that before MathML, they'd see a problem but not know how to say it. Hearing the formula and how to say it was a big help to these students.
Wednesday, October 22, 2008
So I assigned the classic fence-against-the-river question to my group (you have 200m of fencing, what's the largest rectangular area you can contain where one side is not needed against the river). We had just started quadratics so they have no real background in it at all. One student solved it thusly:
1) He made a table of values for a few fences:
Then, he turned to Maple. We had done a unit on linear systems and did a few questions on finding lines of best fit... so he used the Curve Fitting Tool to find the best fit parabola, then found the vertex of the parabola and hence the best area.
Then he found the vertex of the parabola by zooming:
Almost every other student did it the traditional way of setting up the equation, graphing and finding the vertex. I thought it was a nice bit of synthesis from the previous unit to approach it in this fashion!
Monday, October 20, 2008
Well, we're into quadratics now. The one section is doing really well incorporating Maple into their learning process -- they fluidly jump back and forth between Maple & OneNote. The other class more stubbornly holds on to doing everything by hand/calculator. I keep struggling to move this section to a more exploratory model but they will have none of it.
Maple still perplexes us from time-to-time... the switching between independent & dependent variables based upon alphabetical order is one of the more irritating ones. It will graph it in the desired order but as soon as you change the window (axes, zoom, etc) it switches the order! I've gotten around it by choosing my variables appropriately but when they experiment the students don't always think to (nor should they have to).
Tuesday, October 14, 2008
Well, we've used the Curve Fitting Tool for the past couple of classes. Lesson learned: it's best to be descriptive in your variables because using x and y is not conducive to the CFT (it wants to use x as the regression variable). We're on to quadratics now; I wish the spreadsheet tool was a little more flexible than it is and followed Excel notation ... having to add the tilde before the cell reference is clumsy. I also wonder if there's a way to have the column filled by a list variable? Oh... and results on the survey coming up.
Wednesday, October 8, 2008
Things that immediately pops up.... Having them able to use the TEXT to describe their variables and present their solutions neatly. We focused so much on the math work that I didn't do enough on using the text component to explain the setup and solution. Sigh.... only so many hours in a class is one excuse, but it is an excuse, I should have done it. Goal for the next unit. And another thing... having them use the HELP function in case they got stuck during a test. They are also spending a long time (relatively speaking) per question on the Maple as opposed to the written portion. They finished up the written part (3 algebraic solutions, 1 graphical, 2 analysis questions) in about a half hour, then took another half hour to finish up 1 graphical & 1 algebraic question in Maple. Something is up there... going to survey them to see what they thought.
Wednesday, October 1, 2008
Well, we did a bit of direct instruction (not Direct Instruction, which is completely different... kinda like variables in poorly designed programming languages) where, in order to lead them through using Maple to get a least squares line for a set of data, I had to go through with them step-by-step. For one, I didn't plan enough time to write out instructions & film them ... 2nd, I wanted to be able to discuss WHY we were doing each step with the students.
It went VERY well... in fact, I had a number of students say "I really like using Maple" and most were actively engaged. I did have one student get frustrated and say "I hate Maple" but a caveat: he's not the most focused student and tends to make sloppy errors. His adjacent seatmate helped him make corrections.
They also wanted to experiment a bit... one student didn't want to write all the years 1989-1992 so wanted to write that Year:=1989..1992. I said there was a way to do that so I'll have to work that idea in later.
I'm really hoping that they are open to experimentation... they seem okay with the "button pushing" process but, of course, I really want them to focus on mathematics problem solving. | 677.169 | 1 |
This
Solve equations, draw graphs, and play with quadratics in this interactive course! We live in a world of numbers. You see them every day: on clocks, in the stock market, in sports, and all over the news. Algebra is all about figuring out the numbers you don't see.
In this course you will examine real world problems -- rescue the Apollo 13 astronauts, stop the spread of epidemics, and fight forest fires -- involving differential equations and figure out how to solve them using numerical methods | 677.169 | 1 |
Prerequisite: This course is offered to 8th graders from the Catholic partner schools in the Northland. Students are admitted by merit. A math exam is given in the month of May. Enrollment will be limited to 25 students. (Students not currently attending one of the Northland partner schools may call the principal for more information.)ALGEBRA I TEST-OUT OPTION
8th Grade Students who successfully pass the final exam for Algebra I may enroll in Geometry.
ALGEBRA I (B)
Grade Level: 9 Course: 1 Year Credit: 1
Prerequisite: Placement based on HSPT and 8th grade teacher recommendations
Students in this course will, at a controlled pace, acquire the algebraic skills necessary for success in future math courses. Topics covered include: solving and graphing (one and two variable) linear equations and inequalities, some problem solving, factoring, solving systems of linear equations, rational expressions, simplifying radicals, and an introduction to linear and quadratic functions. This course is the foundation for further mathematical understanding and success.
AL and an introduction to linear and quadratic functions. This course is the foundation for further mathematical understanding and success.
HONORS ALGEOMETRY
Grade Level: 9,10 Course: 1 Year Credit: 1 Prerequisite: Successful Completion of Algebra I or Algebra I Test Out Option
This course will include a study of: polygons, congruency and classification of triangles, deductive proofs, circles and their angles, right triangles, areas and volumes of plane and solid figures, basic transformations, basic constructions, ratio and proportions of similar triangles, and coordinate geometry. Mathematical vocabulary will be stressed and students will review algebra skills.
GEOMETRY B
Grade Level: 10
Prerequisite: Successful Completion of Algebra I (B) or Algebra I Course: 1 Year Credit: 1
This course is the next sequential course for students who were enrolled in Algebra 1B. This course will include a study of: polygons, congruency and classifications of triangles, circles and their angles, right triangles, areas and volumes of some solid figures, basic transformations, ratio and proportions of similar triangles, and coordinate geometry. Mathematical vocabulary will be stressed and students will spend a significant amount of time reviewing algebra skills.
This course will include an intense study of: polygons, congruency and classifications of triangles, deductive proofs, circles and their angles, right triangles, the Law of Sines and Cosines, areas and volumes of plane and solid figures, basic transformations, constructions of geometric figures, ratio and proportions of similar triangles, and coordinate geometry. In addition, students will develop visual skills, expand their mathematical vocabulary and review algebra skills necessary for success in Honors Algebra II. Taught primarily through lecture, topics will be reinforced with daily assignments.
ALGEBRA II
ALGEBRA II
Grade Level: 10,11,12
Credit: 1
Course: 1 Year
Prerequisite: Successful Completion of Algebra I
This course offers a review of algebra followed by the study of functions (linear, quadratic, higher order polynomials, absolute values and logarithmic) and their graphs. Students will also study complex fractions, exponents, nth roots matrices and variation. Upon successful completion, this course prepares students for Trigonometry or College Algebra. Other students will be encouraged to enroll in Introduction to College Algebra.
Note: Algebra II does not provide adequate preparation for Pre-Calculus.
This course begins with a review of Algebra 1 Honors topics, followed by the study of linear functions, complex numbers, quadratic equations and functions, variation, conic sections, exponential and logarithmic functions, sequences and series, permutations and combinations, and matrices and determinants. This course is required for students who wish to take ACCP Pre-Calculus. Students will be expected to keep up with the daily assignments and must have the ability to work independently outside the classroom.
This course will begin with a review of Algebra 2 to provide the student with a solid foundation for college level mathematics courses. Students will then progress to more advanced concepts such as basic trigonometric functions, triangle trigonometry, conic sections, exponential and logarithmic functions, and sequences and series. Students will also be introduced to topics from Probability, Statistics, and College Algebra. This course will also focus on ACT preparation prior to all national testing dates.
This course addresses fundamentals of basic algebra. Topics include the real number system, basic operations of algebra, linear and quadratic equations, inequalities, functions and graphs, and systems of equations. Additional considerations include exponents and logarithms, and fundamental techniques of counting. This course is intended for students needing a thorough study of algebra in preparation for college level math courses for a liberal arts or business major. Students may elect to sign up for advanced college credit through Park University.
This course will begin with a review of functions and their graphs then move into an extensive study of the six trigonometric functions. This will include the unit circle, graphing the six functions, identities, formulas, proofs, solving equations, inverse functions, solving right and non-right triangles, and an introduction to polar coordinates. This course is primarily taught through lectures, small group activities, and problems involving real-life situations. This course is an alternative to College Algebra and is designed for students who may have an interest in a math related field at the college level.
*Graphing calculators are required. (TI-83 or TI-84 recommended)
A consideration of those topics in algebra and trigonometry necessary for Calculus. Topics include mathematical analysis of the line, the conic sections, exponential and logarithmic functions, circular functions, polynomial and rational functions, mathematical induction, and theory of equations. Students may elect to sign up for advanced college credit through Park University.
This course is intended for students who have a thorough knowledge of college preparatory mathematics, including algebra, axiomatic geometry, and analytical geometry (rectangular and polar coordinates, equations and graphs of lines and conics). Topics covered are limits, derivatives, and integrals of a wide variety of functions and their applications to curve sketching, maximum and minimum problems, area and volume. Students may elect to sign up for advanced college credit through Rockhurst University (ACCP). | 677.169 | 1 |
In the past few years, numerous national organizations and funding
agencies, including the MAA, AAAS, NIH and NSF, have identified the
need
to increase the mathematical training of biology students and to
increase the awareness of biological applications among mathematics
students. This workshop will focus on the development of courses
and materials that integrate mathematics and biology in ways that
benefit students from both disciplines.
The expected audience is faculty in the mathematical sciences.
This workshop will be run in collaboration with a National
Computational
Science Institute (NCSI) workshop for biology faculty and will feature
crossover sessions, discussions, and plenary lectures.
Participants will:
• Learn about the recent developments in the fields
of mathematical and computational biology;
• Discuss educational trends and strategies for
implementing changes;
• Work through existing and develop new materials for
use in the classroom;
• Learn about high powered software programs that
ease the learning curve; and
• Learn about resources available to become more
involved in mathematical and computational biology.
Upon completing the workshop, participants will develop a syllabus,
curricular materials, or a project plan for an interdisciplinary
activity using appropriate technologies such as a specialized software
package, interactive simulation, or laboratory exercise. | 677.169 | 1 |
Mr. Myrup's Math
Parents and students please take this survey to help Mr. Myrup become a better teacher!
Math seems to be one of the least popular school subjects among most secondary students. Many find it boring or "too difficult" to suit them. My goal is to help all students improve their confidence in their mathematical capabilities as well as their perception of math. Math is not a road block for graduation, but a stepping stone to bigger and greater things.
Math has been shown to be beneficial to all who pursue it. It opens doors to better careers and financial opportunities. Studying math opens the mind to new ways of thinking and problem solving that help in all aspects of life. It increases the chances of success in other school subjects, college, job searching and performance, financial and domestic stability, and even artistic appreciation. Math makes possible all the scientific and technological advances that we take for granted.
The best way to learn and succeed in math is try and try and try. Everyone experiences difficulties in math. It requires you to think and exercise your mind. This does not come easily nor should it. As we "work out" our intellectual capabilities we will discover that we are smarter and better than we thought we were. The hard work and failure along the way only makes the results more rewarding when we achieve success.
"For the things of this world cannot be made known without a knowledge of mathematics" - Roger Bacon
Secondary Math II
Students in Secondary Math II will focus on quadratic expressions, equations, and functions, extend the set of rational numbers to the set of complex numbers, link probability and data through conditional probability and counting methods, study similarity and right triangle trigonometry, and study circles with their quadratic algebraic representations.
Secondary Math III
Students in Secondary Math 3 will deepen their knowledge and study of rational, radical, polynomial, trigonometric, exponential, and logarithmic functions. They will also study various probability and statistics topics. A focus will be placed on mathematical modeling. | 677.169 | 1 |
GEOMETRY FORMULA HANDBOOK
Be sure that you have an application to open
this file type before downloading and/or purchasing.
330 KB
Product Description
This handbook was written for high school students, and consists of the most common geometry formulas. The book serves as an extra homework helper; it is also a perfect tool for teachers, and students who are studying for major exams such as the SATs and ACTs. In the book the formulas are listed in chronological order. There are formulas for every scenario, and each formula has an example of a problem and its solution, for easier understanding. This handbook is the only tool you'll need while studying, because it contains everything required to grasp basic testing content in a short amount of time. Anyone who has fundamental knowledge of geometry will benefit from this book. It is the best way to prepare for exams without being overwhelmed with too much information and confused by the language of geometry | 677.169 | 1 |
PreCalculus With Limits
ISBN-10: 0470904127
ISBN-13: 9780470904121 Engineers looking for an accessible approach to calculus will appreciate Precalculus, 2nd Edition. The book offers a clear writing style that helps reduce any math anxiety they may have while developing their problem-solving skills. It incorporates Parallel Words and Math boxes that provide detailed annotations which follow a multi-modal approach. Your Turn exercises reinforce concepts by allowing them to see the connection between the exercises and examples. A five-step problem solving method is also used to help engineers gain a stronger understanding of word problems | 677.169 | 1 |
Readers who use this text are motivated to learn mathematics. They become more confident and are better able to appreciate the beauty and excitement of the mathematical world.
That′s why the new Ninth Edition of Musser, Burger, and Peterson′s best–selling textbook focuses on one primary goal: helping students develop a true understanding of central concepts using solid mathematical content in an accessible and appealing format. The components in this complete learning program––from the textbook, to the eManipulative activities, to the online problem–solving tools and the resource–rich website––work in harmony to help achieve this goal.
Book Description John Wiley and Sons Ltd, United Kingdom, 2010. Loose-leaf. Book Condition: New. 9th. 254 x 196 mm. Language: English .
Book Description John Wiley and Sons Ltd, United Kingdom, 2010. Loose-leaf. Book Condition: New. 9th. 254 x 196 mm. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. | 677.169 | 1 |
Favorit Book The Mathematics Calendar 2015 Read Now
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Now more than ever The Mathematics Calendar reminds us how mathematics describes nature, impacts the sciences, is essential to architecture, influences the arts, is inseparable from music, exercises and tantalizes the mind with its puzzles and problems, stimulates and creates new technologies, and reveals the multi-dimensions of our world and universe through its ever evolving ideas and insights.The 2015 calendar includes twelve new fascinating math topics illustrating the incredible influence of mathematics on our lives. Each day of every month has a problem, whose solution is the date. The brain teaser lies in figuring how to arrive at the answer, and possibly discovering more than one method of solving the date s problem. For each month, the problems range from arithmetic to calculus.Each month's text, photos and graphics have a wealth of information and are even sprinkled with a bit of humor. The twelve topics feature exciting, historic and current math ideas and topics. Theoni Pappas is committed to demystifying mathematics. The Mathematics Calendar has given thousands of people a new perspective about math —it can be fun, fascinating & intriguing.
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Where to Ski and Snowboard 2015 is the only annually updated guide to ski resorts and includes over 1,100 resorts spanning Europe to North America. With over 550 colour graphics and 150 panoramic mountain maps, this practical guide makes for an accessible and engaging read.
Popular Book Topics in Recreational Mathematics 5/2015 (Volume 5) Premium Book Online
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This book is the fifth in our series of books in recreational mathematics. It is designed to be a continuation of the work of our predecessors Martin Gardner and Joseph S. Madachy. There is a complete index to "Recreational Mathematics Magazine�? that was published in the early 1960's by Joe Madachy. A problem column is included as well as research papers and the editors encourage people to contribute items for future issues.
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Foothill College
Math2B-02[40169]
Linear Algebra
SPRING 2017
Basics of the course
contact
Instructor: Park Lee, Young Hee, Ph.D.
Office: 4111
This course is UC/CSU transferable, and required
for the AS degree in Mathematics and a restricted
support course
FOOTHILL COLLEGE
SPRING 2017
MATH2B LINEAR ALGEBRA
Suggested Homework Assignments
These problems are neither collected nor graded. Successful students do the reading and at least
some of the homework problems before the lecture on that topic. We will disc
Math 10 - Lab 2 Descriptive StatisticsDe A
Chapter 10 Homework Solutions
1- When a plane crosses a double cone connected at their vertex, it traces
the graph of conics (ellipse, parabola, and hyperbola). Find a relationship
between the eccentricity and the angle of the
plane with the horizon.
L
Le
1.7 Scientific Notation
Objectives:
Convert between scientific and decimal notations.
Perform calculations with scientific notation.
1/9
Scientific Notation
Definition
A number in scientific notation is expressed in the form
a 10n
where 1 |a| < 10 and n i | 677.169 | 1 |
Core Florida Math for College Readiness Florida Math for College Readiness provides a fourth-year math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness
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From SAT TEST MARCH 8, 214 Summary of Results Page 1 of 1 Congratulations on taking the SAT Reasoning Test! You re showing colleges that you are serious about getting an education. The SAT is one indicator
04 Mathematics CO-SG-FLD004-03 Program for Licensing Assessments for Colorado Educators Readers should be advised that this study guide, including many of the excerpts used herein, is protected by federalProgram Description Successful completion of this maj will assure competence in mathematics through differential and integral calculus, providing an adequate background f employment in many technologicalTitle: Another Way of Factoring Brief Overview: Students will find factors for quadratic equations with a leading coefficient of one. The students will then graph these equations using a graphing calculator
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< P1-6 photo of a large arched bridge, similar to the one on page 292 or p 360-361of the fish book> Maximum or Minimum of a Quadratic Function 1.3 Some bridge arches are defined by quadratic functions.
Factoring Methods When you are trying to factor a polynomial, there are three general steps you want to follow: 1. See if there is a Greatest Common Factor 2. See if you can Factor by Grouping 3. See ifEXCEL SPREADSHEET MANUAL to accompany CALCULUS FOR THE LIFE SCIENCES GREENWELL RITCHEY LIAL Paula Grafton Young Salem College J. Todd Lee Elon College Boston San Francisco New York London Toronto Sydney
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Use the Distributive Property to factor each polynomial. 1. 1b 15a The greatest common factor in each term is 3.. 14c + c The greatest common factor in each term is c. 3. 10g h + 9gh g h The greatest common
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CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.
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9 Areas and Perimeters This is is our next key Geometry unit. In it we will recap some of the concepts we have met before. We will also begin to develop a more algebraic approach to finding areas and perimeters.
TI-Nspire Technology Version 3.2 Release Notes Release Notes 1 Introduction Thank you for updating your TI Nspire products to Version 3.2. This version of the Release Notes has updates for all of the following
Calculus This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike License To view a copy of this license, visit or send a letter
Associate of Applied Science General Business McLENNAN COMMUNITY COLLEGE 2015-2016 Degree Description This program provides an opportunity for students to study the fundamentals of business operations. | 677.169 | 1 |
Mathematics is one of the most important subject which not only decides the careers of many a young students but also enhances their ability of analytical and rational thinking. It is a common belief that Mathematics is a difficult and dry subject, on the contrary, if it is presented in a systematic and illustrative manner it becomes very easy to understand, even for a beginner.
This edition has been revised as per the CCE guidelines based on the latest syllabus prescribed by the CBSE. The entire syllabus has been divided into two terms. Term-1 consists of Chapters 1 to 12 and Term-2 consists of Chapters 13 to 25. Formative Assessments are given at the end of each chapter and Summative Assessments have been given at the end of each term.
Dr. R.D. Sharma is currently working as Head of Deptt. (Science and Humanities) under the Directorate of Technical Education, Delhi. A Ph.D in Mathematics, he is a double gold medalist, ranking first in the order of merit in both B.Sc (Hons.) and M.Sc Examinations from the University of Rajasthan, Jaipur. He has undergone rigorous training from IIT, Kharagpur in computer-oriented mathematical methods and has a long experience of teaching post-graduate, graduate and engineering students.
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It is a nice book which helps in the preparation of the board exams - Class IX. Most of the concepts which we learn in Class IX become the basic fundamentals for Class X CBSE. It is really competitive and complicated. It has a wide range of theory, examples and exercises both for Sa's and Fa's. So, I prefer all of IX- CBSE Students to buy this book.
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If you are not good at maths and your concepts are not clear this the book to buy! It is filled with great examples and twisted question that really really make you think and use your brain! the only reason I did well in my first Summative is because i solved questions from RD Sharma!!! Genius Guy!! If You solve tHE whole book Your Good result is guaranteed!!!
this is a very helpful book for the students of class 9 and will help them to clear their concepts.It contains formative assessment papers after each chapter and summative assessment papers after the completion of certain chapters. | 677.169 | 1 |
Minute 'Minders -- Algebra, Trigonometry, and Calculus Bundle
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PRODUCT DESCRIPTION
The "Minute 'Minders" series is designed to be a set of worksheets that can be used to review prerequisite concepts in a given class. They are meant to be used as quick warm-ups (each a single page) to refresh students' memories about concepts they are already familiar with, but in our teaching experience they may have forgotten.
The Algebra Minute 'Minder has seven worksheets on the following topics | 677.169 | 1 |
Math Word Problems
Beschreibung
Beschreibung
Presents easy-to-use review of the basics of solving math word problems. This title helps you to: discover the six basic steps for solving word problems; translate English-language statements into equations and then solve them; solve geometry problems involving single and multiple shapes; and, work on proportion and percent problems.
Inhaltsverzeichnis
Introduction.Why You Need This Book.How to Use This Book.Visit Our Web Site.Chapter 1: Translating Expressions.Keywords of Basic Mathematical Operations.Addition keywords.Subtraction keywords.Multiplication keywords.Division keywords.Keywords That Indicate a Change in Order.Basic turnaround words.Addition turnaround words.Subtraction turnaround words.Multiplication turnaround words.Division turnaround words.Chapter 2: Inserting Parentheses.Translating a Comma and IS.Leading Keywords.Adjacent Keywords.Turnaround Words.Chapter 3: Simplifying Expressions.Distributive Property of Multiplication over Addition.Combining Like Terms.Distributing a Negative Number.Chapter 4: Equations.Keywords Indicating Equality.Solving Simple Linear Equations.Defining an algebraic expression.Defining an algebraic equation.Four steps for solving simple linear equations.Checking Solutions and Translations.Checking solutions.Checking your translation.Chapter 5: Geometry Problems.Triangles.Perimeter of a triangle.Area of a triangle.Sum of the measures of the angles.Rectangles and Squares.Perimeter of a rectangle.Area of a rectangle.Perimeter of a square.Area of a square.Multiple Shapes.Other Easily Visualized Word Problems.Chapter 6: Proportions and Percents.Proportion Problems.Proportion Word Problems.Percent Problems.Percent Word Problems.Chapter 7: Summation Problems Using the Board Method.Simple Board Problems.Consecutive Integers.Consecutive Even and Odd Integers.Other Problems that Can Be Represented with a Board.Chapter 8: Solving Word Problems Using Tried-and-True Methods.Polya's Four-Step Process.Expanding on Polya's four steps.The five steps of the tried-and-true methods.Consecutive Integers.Consecutive Even and Odd Integers.Other Problems that Cannot Be Represented with a Board.Chapter 9: Systems of Equations Using the Board Method.Money Problems.Investment Problems.Mixture Problems.Distance Problems.Chapter 10: Systems of Equations Using Tried-and-True Methods.Money Problems.Investment Problems.Mixture Problems.Distance Problems.Chapter 11: Common Errors.Variable Omission.Variable Reversal.Methods for Avoiding Common Errors.A firm foundation.Errors in translation.Errors in solving an equation.Errors in answering the question.CQR Review.CQR Resource Center.Glossary.Index.
Portrait
Karen Anglin, a mathematics instructor at Blinn College in Brenham, Texas, since 1990, regularly presents workshops to teachers on best practices for teaching math word problems. She holds an MS in Statistics and a BS in Mathematics from Texas A&M University. | 677.169 | 1 |
beginning teachers understand student thinking in calculus
Slides from a talk given by Tom Judson and me at the MAA Session on Research in Undergraduate Mathematics Education.
Abstract: "Every teacher of calculus encounters various degrees of student understanding. To be a successful teacher, it is essential to understand student misconceptions and to make clear explanations to one's students. Our project is concerned with how new teachers develop their ability to understand student thinking. We conducted individual interviews with graduate students teaching calculus for the first time. We interviewed each graduate student before and after their first teaching assignment. The interviews were transcribed and coded for analysis. We will present the results of our findings in this talk. Our hope is to provide information to that will be useful in developing more effective teaching training programs for graduate students who will teach undergraduate mathematics."
2.
Calculus and Linear Algebra Classes
Instruction for calculus and linear algebra is done
in sections of 25–30 students by teaching fellows
(TFs).
TFs are graduate students, postdocs, and regular
faculty.
A faculty member acts as the course coordinator
for all sections and writes a common syllabus.
Students have common homework assignments
and common exams.
Thursday, January 21, 2010
3.
Preservice Training for Graduate Students
Graduate students are supported for their
first year and have no teaching duties.
Graduate students attend a one-semester
teaching seminar where they learn
speaking skills, pedagogical mechanics,
and have some opportunities to work with
actual calculus students.
Thursday, January 21, 2010
4.
The Apprenticeship
Each graduate student is required to apprentice
under an experienced coach.
The apprentice attends the coach's class for several
weeks and holds office hours.
The apprentice teaches the coach's class three times.
At the end of the apprenticeship, the graduate
student will be put in the teaching lineup with the
coach's approval or the coach will recommend
additional training for the graduate student.
Thursday, January 21, 2010
5.
Mathematical Knowledge for Teaching
Common Content Knowledge (CCK)—Formal
mathematical knowledge that mathematicians have
developed through study and/or research.
Pedagogical Content Knowledge (PCK)—
Knowledge used to follow student thinking and
problem solving strategies in the classroom.
Specialized Content Knowledge (SCK)—
Mathematical knowledge that is used in the
classroom but has not been developed in formal
courses.
Thursday, January 21, 2010
6.
l'Hôpital's Rule is a consequence of the
Cauchy Mean Value Theorem or Taylor's
Theorem (CCK).
Students armed with the sledgehammer
of l'Hôpital's Rule will use it on limits
which are not in indeterminate form and
arrive at wrong answers (PCK).
Thursday, January 21, 2010
7.
ples in [18]. Ma examined the complex mathematical k
elementary school teachers. For example, Ma posed t
Specialized Content Knowledge
to both American and Chinese teachers.
Students performed the following multiplication
123
Liping Ma gives the
following example: × 645
Suppose that a student 615
performs the following 492
multiplication. What
would you say to the 738
student? 1845
What would you say to these students?3
Thursday, January 21, 2010
8.
Participants
We interviewed seven graduate students
before and after their first teaching
assignments.
The graduate students were from Asia,
eastern Europe, and the U.S.
Both men and women were represented.
Thursday, January 21, 2010
9.
Pre-Teaching Interview
"Can you talk a little bit about your
background, and how you got here?"
"Can you tell us about your career plans
and how you see teaching as part of those
plans?"
Each participant was given four questions
involving different calculus scenarios.
Thursday, January 21, 2010
10.
All of the TFs planned a research career or
saw research as a strong component of their
future career.
All thought teaching was important. Those
planning an academic career thought that
teaching would be an important duty.
Several looked forward to the teaching.
All had some idea of the need for PCK in
the classroom.
Thursday, January 21, 2010
11.
graphics.nb
3. The graph of f (x), given below, is made up of straight lines and a semicircle.
f HxL
4
2
x
-5 -3 -1 1 3 5
-2
-4
We define the function F (x) by
x
F (x) = f (t) dt
0
One of your students understands that F (2) = 4 but believes that F (−2) is undefined.
What would you say to the student?
Thursday, January 21,often
4. Students 2010 have difficulty working in three dimensions. One of your students comes
to you and asks how to match each of the following equations with the appropriate
12.
Several participants gave an explanation
by appealing to signed area.
"You could say, why do we have this rule
in the first place? One reason for it is that
we want the Fundamental Theorem of
Calculus to hold."
No one gave an explanation using the
integral as net change without some
prompting.
Thursday, January 21, 2010
13.
Post-Teaching Interview
"Now that you've had a chance to work
with students, has your view of teaching
changed at all?"
"What surprised you about teaching?
What happened that you didn't think would
happen?"
We asked four more questions involving
different calculus scenarios.
Thursday, January 21, 2010
14.
View of Teaching
"I've always thought that the professor
doesn't like to have all that many
questions. And it just sounds silly
sometimes. And then when I taught, I
realized that even the serious questions,
I really wanted those questions. ... It was
a very different perspective that I got."
Thursday, January 21, 2010
15.
"It went great. I really loved it. I mean, I thought I'd
like teaching, but it went better than I expected. I was
nervous, but only for the first couple of classes. Then I
really became comfortable with them. ... They asked a
lot of questions. They are pretty demanding. They
really want to know things. And you can't just get
away with stuff with them. There will definitely be at
least one person who has something to say, you know.
So I thought that was great. But I realized how much I
love questions. I mean, whenever they were a little
tired and they weren't asking so many questions, I felt
sad, you know? It feels great when they have
questions and you feel that they understand
everything."
Thursday, January 21, 2010
16.
What Surprised Them
"I was surprised at how heterogeneous
the students were that I had in terms of
mathematical ability. Some of them had
trouble understanding that x/2 and (1/2)
x were equal to one another, and others
were well over prepared for the class.
They'd taken calculus in high school."
Thursday, January 21, 2010
17.
"When I was teaching, students would really
ask me sometimes some questions that I would
never expect. I saw at first, for example, for
log x times a constant. Everyone knows the
derivative is 1/x times the constant. Then, I
put some kind of extra constant, then people
are very confused ... I think this is should be
kind of easy and obvious to me, but it's really
not obvious to the students. It's a little bit
surprising to me, so I really have to know
what students are really thinking about."
Thursday, January 21, 2010
18.
A: So I felt that I could assume that this is well-known to
students, so I can just move faster when deriving or finding
[something on the] blackboard. But then—Well, since students
always ask the question, but why the equation is true or ... how
could I get second line from first line like that? So after that I
found I need to be more careful and I needed to be prepared.
Q: Do you think it's that these basic facts about algebra and
trigonometry is that they don't know them, or that they just lack
the necessary fluency?
A: Oh, it's just lack.
Q: Lack fluency?
A: Yeah. They're just slow, yeah. ... if I just do it line by line
slowly
Thursday, January 21, 2010
19.
"There are some things I guess
everybody could use help with. They
have trouble doing derivatives that
involve recursing more than twice. If
they need to use the product rule alone,
that's fine. If they need to use the
product rule on the chain rule, that's
fine. But if you need to use the product
rule, the chain rule, and something
else..."
Thursday, January 21, 2010
20.
∞
∞ ∞
ak ≤ an+1 + a(x) dx ≤ a(x) dx.
k=n+1 n+1 n
What would you say to the student?
3. Consider the following problem. Let
x 1/x
1 1
F (x) = 2
dt + 2
dt,
0 1+t 0 1+t
where x = 0.
(a) Show that F (x) is constant on (−∞, 0) and constant on (0, ∞).
(b) Evaluate the constant value(s) of F (x).
What sort of difficulties would would a student encounter when trying
to solve this problem? What would you say to the student?
4. Students often have difficulty working in three dimensions. One of
your students comes to you and asks contour plots. If the contour plot
of f (x, y) is given below, at which of the labelled points is |∇f | the
Thursday, January 21, 2010 smallest? What would you say to this student? What
greatest? The
21.
Two TFs found at least three different solutions
to the problem.
Students will integrate 1/(1 + t2) and then get
stuck.
Students will be able to differentiate the first
term using the Fundamental Theorem of
Calculus but will have difficulty differentiating
the second term.
No one mentioned that students will have
difficulty with locally constant functions.
Thursday, January 21, 2010
22.
Conclusions
Pedagogical content knowledge comes with
teaching experience. It is difficult to "teach"
PCK.
Pre and inservice training should train TFs to
look for PCK and provide in depth examples.
TFs should have opportunities to work with
real students BEFORE they enter the
classroom as the primary instructor.
Thursday, January 21, 2010
23.
Acknowledgements
Thanks to our participants and
colleagues.
Thanks to the generous support of the
Educational Advancement Foundation.
Thursday, January 21, 2010 | 677.169 | 1 |
Zach Wissner-Gross
Zach is the CEO of School Yourself. He completed his doctorate in Physics at Harvard, where he won multiple teaching awards, including Harvard's White Award for Excellence in Teaching. He is a Hertz Fellow, and has authored papers in neuroscience, biophysics, and biotechnology, as well as three interactive math textbooks with the School Yourself team.
Customize your search:
Solve equations, draw graphs, and play with quadratics in this interactive course! We live in a world of numbers. You see them every day: on clocks, in the stock market, in sports, and all over the news. Algebra is all about figuring out the numbers you don't see.
Measure angles, prove geometric theorems, and discover how to calculate areas and volumes in this interactive course! More than 2000 years ago, long before rockets were launched into orbit or explorers sailed around the globe, a Greek mathematician measured the size of the Earth using nothing more than a few facts about lines, angles, and circles | 677.169 | 1 |
I'm certainly no expert either but the way I understand it, it's one of the main branch of mathematics, where by that I mean that it pops up everywhere. ODE, PDE, geometry, you name it. Maybe not group theory itself but a closely related ramification such as modules or whatnot.
For instance, I'm reading the PhD thesis of a student at my uni at the moment and he introduces a homological method to prove new existence and multiplicity theorems in critical point theory.
As always, maybe if you asked specific questions, you'd be more likely too get an answer. I think matt grime is an algebraist but I haven't seen him in a while.
Functional analysis is basically a cross between analysis and algebra. In complex analysis, you study automorphisms of the complex plane. Anything that requires Hilbert space methods (e.g., in PDE, harmonic analysis, etc.) requires algebra. The modern approach to differential geometry is all algebra. How far are you in your career as an analyst that you haven't seen algebra pop up anywhere?
I know though that group theory is needed to prove certain theorems with regards to lattices in solid state physics. I know it is also quite important for symbolic root finding which is very relevant in computer algebra systems. | 677.169 | 1 |
2017-2018 Catalog
Mathematics (MAT)
Courses
MAT-006. Elements of Algebra. 0 Credits.
LECT 2 hrs, RECI 1 hr Elements of Algebra integrates the fundamental operations of arithmetic and introductory Algebra. It is intended for students whose placement examination indicates a need for a review of arithmetic and basic Algebra skills. Topics include operations on whole numbers, fractions, decimals, percent and signed numbers, linear equations and inequalities in one variable, operations on polynomials, factoring, integer exponents, and graphing. The course incorporates a Support Lab where students will receive personal assistance with problems or questions assigned as homework to supplement the lectures. Prerequisites: Appropriate score on a placement test Additional Fees: Course fee applies.
MAT-007. Foundations of Algebra. 0 Credits.
LECT 2 hrs This course integrates selected topics of arithmetic and introductory algebra, including operations on whole numbers, fractions, decimals, percent and signed numbers, linear equations and inequalities in one variable, operations on polynomials, factoring, integer exponents, and graphing. Students are required to complete a series of laboratory assignments, which are designed to reinforce concepts based on the placement test results. Prerequisites: Appropriate score on a placement test.
MAT-009. Basic Mathematics Ia. 0 Credits.
LECT 1 hr Three (3) hours per day for one week. This is an intensive one-week review of topics typically found on the computation placement test. A passing grade satisfies the Basic Mathematics requirement. Prerequisites: Appropriate score on a placement test.
MAT-010. Basic Algebra 1A. 0 Credits.
LECT 1 hr This is an intensive review of topics typically found on the basic algebra placement test. A passing grade satisfies the Basic Algebra requirement. Prerequisites: Appropriate score on a placement test.
MAT-011. Basic Mathematics I. 0 Credits.
LECT 3 hrs A preparatory course designed for students who need additional practice and review in arithmetic.
MAT-014. Basic Algebra I. 0 Credits.
LECT 3 hrs A preparatory course in elementary algebra which includes rational numbers, polynomials, algebraic operations, first-degree equations, graphing, systems of linear equations, problem solving and an introduction to the quadratic equations. Prerequisites:MAT-009 or MAT-011 and permission of department chair.
LECT 3 hrs An intensive course designed to prepare students for mathematics courses such as Calculus with Applications to Business and Economics and Precalculus. It covers selected algebra topics including exponents; rational expressions; polynomials, radicals, relations and functions; exponential and logarithmic functions, systems of equations. Prerequisites:MAT-016 or MAT-060 (grade C or better) or equivalent.
MAT-113. Applied Calculus. 4 Credits.
LECT 4 hrs A study of topics which provides a basis for continuing courses in mathematics and the physical sciences. This course includes trigonometric, exponential and logarithmic functions; analytic geometry; differentiation and integration. Prerequisites:MAT-110 or MAT-123 or equivalent.
MAT-117. Mathematical Analysis for Business and Economics. 3 Credits.
LECT 3 hrs Mathematical topics used in business and economics with emphasis on applications. Covered are polynomials, linear and quadratic models, systems of equations, matrix algebra, and linear programming including the Simplex Method. Prerequisites:MAT-016, MAT-060 (grade of C or better) or equivalent.
MAT-118. Calculus With Application to Business And Economics. 3 Credits.
LECT 3 hrs A course covering functions, derivatives and integration, with special consideration of applications to the business and economics areas. Partial differentiation is introduced. Prerequisites:MAT-110 (grade of C or better) or equivalent.
MAT-120. Mathematics for the Liberal Arts. 4 Credits.
LECT 4 hrs A course addressed to liberal arts students. Topics include the history of mathematics, probability, statistics, geometry, number theory, algebra, graphs and functions, and a choice of selected topics. Prerequisites:MAT-007, MAT-014, MAT-050 or equivalent.
LECT 4 hrs The fundamental principles of statistical methods. Descriptive statistics, correlation, regression, probability, binomial and normal distributions, sampling, hypothesis testing, confidence intervals and ethical issues in statistics are included. An introduction to the use of statistical software to analyze data will be emphasized. Prerequisites:MAT-016, MAT-060 or MAT-120 or equivalent.
MAT-131. Analytic Geometry and Calculus I. 4 Credits.
LECT 4 hrs The first semester of a three-semester sequence. Analytic geometry in the plane, differentiation and applications, and integration are covered. Prerequisites:MAT-123 (grade of C or better) or equivalent.
MAT-132. Analytic Geometry and Calculus II. 4 Credits.
LECT 4 hrs A continuation of Analytic Geometry and Calculus I, which covers the calculus of inverse trigonometric functions, methods of integration, analytic geometry in the plane including polar coordinates and conic sections, hyperbolic functions, sequences and series, and parametric equations. Prerequisites:MAT-131 (grade of C or better) or equivalent.
MAT-140. Math for Radiographers. 1 Credit.
LECT 1 hr This course discusses the math skills that are crucial in the healthcare environment. It teaches the basis measurements, calculations, percents, ratios, and proportions, scientific notation, metric conversions, basis algebraic principles and basic geometric principles used in Radiology. It reviews whole numbers, fractions, decimals and exponents. Radiology units and numeric prefixes are also discussed. Prerequisites:MAT-016 or MAT-060 and admission to the Radiography program Corequisites:RAD-100, RAD-104 and RAD-107.
MAT-183. Honors Probability and Statistics. 4 Credits.
LECT 4 hrs An introduction to the principles of statistical methods. The course will integrate spreadsheet software to cover such topics as descriptive statistics, correlation, regression, probability, binomial and normal distributions, sampling, elementary hypothesis testing and confidence intervals. This course will also cover ethical issues in statistics. Comprehensive case studies will be covered throughout the semester. An introduction to the use of statistical software to analyze large data sets will be emphasized. Prerequisites: Permission of department chair or honors advisor.
MAT-210. Probability and Statistics II. 4 Credits.
LECT 4 hrs This course is a continuation of statistical analysis from Probability and Statistics. Techniques for collection and analysis of data emphasizing estimation and hypothesis testing, analysis of variance and regression analysis are included. Also included are nonparametric testing and an introduction to multiple regression. A focus on analyzing large data sets using statistical software. Prerequisites:MAT-124 or MAT-130 or MAT-183 or equivalent (grade of C or better).
MAT-225. Discrete Mathematics. 4 Credits.
LECT 4 hrs This is a 4-credit course in discrete mathematics. It is offered to math & computer science majors in their first two years of study. The course outline shows it is an exposition of real-world and modern mathematics. Discrete Mathematics covers a breadth of unique topics in number theory, graph theory, set theory, probability and statistics, and propositional logic. Prerequisites:MAT-131.
MAT-228. Linear Algebra. 3 Credits.
LECT 3 hrs Selected topics including matrices and determinants, vectors and vector spaces, linear transformations, eigenvalues and eigenvectors, with applications from a variety of disciplines. Prerequisites:MAT-132 (grade of C or better) or equivalent.
MAT-230. Calculus III. 4 Credits.
LECT 4 hrs A continuation of Analytic Geometry and Calculus II which includes analytic geometry in three dimensions, functions of several variables, partial derivatives, multiple integrals, vectors and an introduction to vector analysis. Prerequisites:MAT-132 (grade of C or better) or equivalent.
MAT-232. Differential Equations. 3 Credits.
LECT 3 hrs Ordinary differential equations and methods of solution. Introduction to classical equations and their solutions, with some applications to geometry, physics and engineering. Prerequisites:MAT-132 (grade of C or better) or equivalent.
MAT-270. Numbers and Operations for Middle Grades. 3 Credits.
LECT 3 hrs This course prepares middle-grades mathematics teachers with a concrete understanding of numbers, number systems, operations with fractions, decimals and percent; there is special consideration to ratios, proportions, factors and multiples and including instructional techniques and calculator-structured lessons. Prerequisites: Permission of department chair and Elementary School or N-2 subject matter endorsement.
MAT-271. Algebra for Middle Grades. 3 Credits.
LECT 3 hrs This course explores topics from pre-algebra and algebra. The course prepares middle-grades mathematics teachers with a concrete understanding of patterns, relationships and functions, polynomials, algebraic operations, first degree equations, graphing and systems of linear equations and linear inequalities and including instructional techniques and calculator-structured lessons. Prerequisites: Permission of department chair and Elementary School or N-2 subject matter endorsement.
MAT-272. Mathematics for Middle Grades. 3 Credits.
LECT 3 hrs This course explores topics including history of mathematics, algebra, probability and statistics while infusing instructional techniques and uses of technology. Prerequisites: Permission of department chair and Elementary School or N-2 subject matter endorsement.
MAT-273. Statistics for Middle Grades. 3 Credits.
LECT 3 hrs An introduction to statistical methods and reasoning as applied to practical problems. Topics include collecting and summarizing data, histograms and other types of graphs, descriptive statistics, normal distributions, sampling, surveys, use of computers in statistics and interpretation of data. Prerequisites: Permission of department chair and Elementary School or N-2 subject matter endorsement.
MAT-274. Geometry for Middle Grades. 3 Credits.
LECT 3 hrs This course includes topics in geometry and measurements with use of Geometer Sketchpad Software. Formulas for perimeter, area, and volume for polygons and polyhedrons, properties of parallel lines and perpendicular lines, fundamental topics of measurements, measurement instruments, measurement errors are covered while infusing instructional techniques. Prerequisites: Permission of department chair and Elementary School or N-2 subject matter endorsement. | 677.169 | 1 |
Piecewise Functions Doodle Notes
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Product Description
Piecewise Functions for Algebra: "doodle notes" -
When students color or doodle in math class, it activates both hemispheres of the brain at the same time. There are proven benefits of this cross-lateral brain activity:
- new learning
- relaxation (less math anxiety)
- visual connections
- better memory & retention of the content!
Students fill in the sheets, answer the questions, and color, doodle or embellish. Then, they can use it as a study guide later on.
** This resource is also available as part of a DISCOUNTED bundle: Functions Bundle | 677.169 | 1 |
97803879627pects of Calculus (Undergraduate Texts in Mathematics)
This book is intended for students familiar with a beginner's version of differential and integral calculus stressing only manipulation offormulas and who are now looking for a closer study of basic concepts combined with a more creative use of information. The work is primarily aimed at students in mathematics, engineering, and science who find themselves in transition from elementary calculus to rigorous courses in analysis. In addition, this book may also be of interest to those preparing to teach a course in calculus. Instead of exposing the reader to an excess of premature abstractions that so easily can degenerate into pedantry, I felt it more useful to stress instruc tive and stimulating examples. The book contains numerous worked out examples and many of the exercises are provided with helpful hints or a solution in outline. For further exercises the interested reader may want to consult a problem book by the author entitled Problems and Propositions in Analysis (New York: Marcel Dekker, 1979). For the history of calculus I recommend the book by C. B. Boyer, The Concepts of the Calculus (New York: Dover, 1949 | 677.169 | 1 |
Plan
• Deliver the new GCSE maths curriculum with confidence with a detailed introduction to support you in interpreting the significant change • Help students achieve a smooth transition from KS3 using our thorough explanations of the new assessment objectives • Plan ahead with detailed, practical schemes of work for 2, 3 and 5 year teaching • Pick up and teach with detailed lesson plans – perfect for cover lessons, NQTs and full of ideas for more experienced teachers. New features include: - literacy prompts to help students ask the right questions - extra teacher support on the new, harder topics from AS level and IGCSE • Link the course together with all relevant resources and curriculum and specification references included • Access answers to the accompanying student book
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Product Description:
Plan and deliver the new AQA specification with this pick-up-and-go Teacher Pack. It is perfectly ma.... HarperCollins Publishers. Paperback. Book Condition: new. BRAND NEW, AQA GCSE Maths Higher Teacher Pack (4th Revised edition), Rob Ellis, Kath Hipkiss, Colin Stobart, B9780008113919 HARPER COLLINS, 2015. Paperback. Book Condition: NEW. 9780008113919 This listing is a new book, a title currently in-print which we order directly and immediately from the publisher. Bookseller Inventory # HTANDREE0984109 | 677.169 | 1 |
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Basic calculators are useful for anyone that wants to make fast, accurate calculations, and some models come with additional features and mathematical functions. Basic calculators are available in handheld... | 677.169 | 1 |
Alg1.M.A.S.ID.A.02: I can find the mean, median, interquartile range and standard deviation of a data set and use the appropriate measures to interpret and compare sets of data and graphs. (15 days)
Benchmark 4:
Alg1.M.F.IF.A.01: I can determine the domain and range of a graph, set of points and table, and determine whether or not the given data represents a function. (5 days)
Additional Standards:
Alg1.M.N.RN.B.03: Explain why the sum or product of two rational numbers are rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational. (Month 1)
Alg1 1)
Alg1.M.A.CED.A.01:Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. (Month 2)
Alg1.M.F.IF.C.09)
Alg1.M.F.IF.A.01:Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). (Month 2 & 7)
Alg1 domain for the function. (Month 2, 7, & 9)
Alg1.M.F.IF.A.02:Use function notations, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. (Month 2, 7, & 9)
Alg1.M.F.BF.A.01:Write a function that describes a relationship between two quantities. (Month 2)
Alg1.M.F.IF.C.07:Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. (Month 2 & 8)
Alg1.M.S.ID.C.07:Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. (Month 3)
Alg1.M.S.ID.B.06:Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or chooses a function suggested by the context. Emphasize linear, quadratic, and exponential models. b. Informally assess the fit of a function by plotting and analyzing residuals. (Month 3)
Alg1.M.N.Q.A.03:Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. (Month 3)
Alg1.M.A.CED.A.03:Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non‐viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. (Month 4)
Alg1.M.A.REI.C.05:Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. (Month 4)
Alg1.M.REI.D.11:Explain why the x‐coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions. (Month 4)
Alg1.M.A.REI.D.10:Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). (Month 4)
Alg1.M.A.SSE.A.02: Use 5 and 7)
Alg1.M.A.CED.A.04:Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm's law V = IR to highlight resistance R. (Month 5 & 7)
Alg1.M.F.LE.A.03:Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. (Month 5)
Alg1.M.F.LE.B.05:Interpret the parameters in a linear or exponential function in terms of a context. (Month 5)
Alg1 5, 7, & 9)
Alg1.M.F.IF.C.08: Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. (Month 7)
Alg1.M.A.SSE.B.03: Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. (Month 7)
Alg1.M.A.SSE.B.04: Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. (Month 7)
Alg1.M.S.ID.A.03:Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). (Month 8) | 677.169 | 1 |
Algebra in Business and Science Algebra is used repeatedly in business and science for professions ranging from accountants to astronomers and physicists. The fundamental algebraic equation in accounting is assets = liabilities + capital. Accountants use this equation to balance the books. They also use algebra to calculate journal entries for interest payments and depreciation (Moore, n.d.). To analyze and work with numbers, accountants also need to understand positive and negative numbers, fractions and decimals. Other businesses that use algebra include retail stores, car dealerships and restaurants to name just a few. Each of these types of businesses sells a product and needs to calculate percentage discounts, sales or meals tax and profit margins. For science, algebra is used in astronomy, physics and chemistry. Astronomers and physicists use algebra to research and understand the universe. They observe, measure, interpret and develop theories to explain what is happening in the universe (US Bureau of Labor, 2009). One way that chemists use algebra is when mixing together chemicals. They need to calculate the correct amount of each substance to obtain the desired result. Algebraic Concepts in Everyday Life
The following are examples that I learned from the course discussion board of different ways that algebra can solve everyday life problems. ∙ To determine each person's share of the cost for a limo ride, you would add a tip to the cost of the limo and divide by the number of riders. For example, if there were 10 kids riding in a limo and the cost of the limo was $600 (including a 20% tip), each person would pay $60. ∙ To split the cost of a restaurant bill, you would add a tip to the bill total and divide by the number of diners splitting the bill. ∙ To calculate a batting average in baseball use the following formula: Batting Average = # of hits / #of at bats. ∙ To calculate the cubic feet of a refrigerator when the height in cubic...
YOU MAY ALSO FIND THESE DOCUMENTS HELPFUL
...Algebra in the RealWorld and EverydayLife
Hal Hagood
u07a2
Table of Contents
Page Number
Table of Contents …………………………………………………………………… 2
Introduction ………………………………………………………………………… 3
Ways That Algebra Affects Business or Science ………………………………….. 5
How Algebraic Concepts Can Solve Everyday Problems in Life …………………. 6
WaysAlgebra Can Solve Everyday Problems in Business or Science ……………. 9
A Surprising Finding About How Algebra Affects Personal Life, Business, and Science 10
References ………………………………………………………………………….. 14
Introduction
Algebra, some of us fear it while some of us embrace it, algebra is not "arithmetic with letters" it is better described as a way of thinking. At its most fundamental level, arithmetic and algebra are two different forms of thinking about numerical issues. Many of these examples have been taken from our classroom discussions while others are examples I have discovered in my own research for this paper, several examples of each will be cited.
"Let's start with arithmetic. This is essentially the use of the four numerical operations addition, subtraction, multiplication, and division to calculate numerical values of various things. It is the oldest part of...
...AND
ALGEBRA: ALGEBRAIC EXPRESSIONS
This resource may be copied in its entirety, but is not to be used for commercial purposes without permission from the Centre for Education in Mathematics and Computing, University of Waterloo.
Play the Late Delivery game first! Levels 1 and 2 are recommended. Click on or go to for the link. 1. a. Write each of the following expression as a single number. i. 20 + 5 ii. 15 ÷ 3
iii. 11 × 9
b. For each question in (a) write 3 equivalent expressions using 3 different operations. 2. Match up the equivalent expressions below:
4+3 1+ 2 6×2+2 16 ÷ 2
55 − 52
2×2×2 49 ÷ 7 7×2
Did You Know?
A cheetah can run 76 km/h. The fastest human can only run about 37 km/h!
3.
The scale balances because the mass on the left side is equal to the mass on the right side. A number sentence can be written to describe the picture: 10 + 10 + 1 = 10 + 5 + 5 + 1 or 2 × 10 + 1 = 2 × 5 + 10 + 1 . a. Draw 3 different combinations of masses on a scale that would balance.
b. Write a number sentence to describe each of the 3 new combinations.
4.
a. Balance the scale using a combination of 10 g, 5 g and 1 g weights. Assume you have many different masses. Compare your solution with your classmates.
b. Write an equivalent expression, which is different than the masses in the picture, to describe the total mass in each pan.
Expectations:Why is algebra important in the realworld? The first reason algebra is important in the realworld is because people use algebra every day in their jobs. Having the ability to learn and do algebra will probably help you exceed into the job you want to do one day. Most people do not realize that algebra is used almost every day in adult life. Some examples of obvious jobs that use algebra are engineers, mathematicians, teachers and scientists. I believe everyone uses some sort of algebra in their jobs or their daily life adventure whether they know it or not.
Algebra is a huge part of our lives. Whether we drive a car and need to calculate the distance and mileage of a trip. Working out equations of how many miles per gallon your car will get is also a very important daily issue. What is you are working in a retail store as a cashier, would you use algebra? You would use algebra to determine coins and bills in change you use. In that case, the different bills and coins are your x's and your y's. Most people want think of that as algebra but that's what it is.
There are some jobs that you would not think that would use algebra. Algebra can be used to determine what dimensions to use to make a box...
...Cami Petrides
Mrs. Babich
Algebra Period 4
April 1, 2014
Extra Credit Project
12. When you flip a light switch, the light seems to come on almost immediately, giving the impression that the electrons in the wiring move very rapidly.
Part A: In reality, the individual electrons in a wire move very slowly through wires. A typical speed for an electron in a battery circuit is 5.0x10 to the -4th meters per second. How long does it take an electron moving at that speed to travel a wire 1.0 centimeter, or 1.0x10 to the -2nd?
Part B: Electrons move quickly through wires, but electric energy does. It moves at almost the speed of light, 3.0x10 to the 8th meters per second. How long would it take to travel 1.0 centimeters at the speed of light?
Part C: Electrons in an ordinary flashlight can travel a total distance of only several centimeters .suppose the distance an electron can travel in a flashlight circuit is 15 centimeters, or 1.5x10 to the -1st meter. The circumference of the earth is about 4.0x10 to the 7th meters. How many trips around the earth could a pulse of electric energy make at the speed of light in the same time an electron could travel through 15 centimeters of a battery circuit in 5.0x10 to the -4th meters per second?
For part A, the first step is to put (5.0) to the 10th to the -4th. The numerator would be (0.00050) if someone were trying to put 5.0x10 to the -4th in the form it's supposed to be in. For the second scientific...
...challenges and move...
...Contracts: How They Effect EverydayLife
BUS 311 – Business Law I
Prof. Katheryne Rogers
January 6, 2012
Most people in society think that contracts are pointless and unnecessary. Contractual law is not high on society's list of things to study. What society does not realize is that contracts bind a majority of the decisions that they make on a systematic basis. In this paper I will make evident the effectiveness of contracts and how they are such an immense aspect in society's everydaylife.
This paper will demonstrate the practicality of contracts and how they influence and add to everydaylife. I am also going to show what would happen if we did not have any contract regulatory resolutions made. Without contracts; the law would not be able to make a verdict as to who to rule for if something were to come up; which transpires daily all over the world. The definition of a contract is basically an agreement between two or more persons. This explanation integrates countless things. Almost everyone in today's society is a consumer. Just by being a consumer you make choices based on issues that have been put in place by contracts.
To show how crucial contracts are; a consumer awakens in his home; which he purchased from K.B Homes. K.B home is a company that is well known for designing homes. The consumer's house was built by contractors. By differentiation a...
...Stress and everydayLife
Everyone has used the word "stress" before, what is it, and why is it a commonality upon the population in the western world. The word stress is defined "Physical, mental, or emotional strain or tension". 1 Has Selye, a pioneering endocrinologist, coined the original definition of the word stress, in the mid-twentieth century. 2 But why do we get stressed? It seems that the modern worlds business culture is the main feed for stress. The long hours working in a office, not knowing if your going to be fired or not, or even having a very disruptive supervisor can bring common stress to everybody; the main source of stress can revolve around the ever so threatening "deadlines" that can make or break people. We all know what it feels like to be stressed, and all the immediate effects of being stressed, but why do we keep on doing this to ourselves. 3
The stock market is a perfect example on how today's business culture inflicts stress to the business men and women out there on the floor. The stock exchange where people can loose all their money in a blink of an eye or become a millionaire in the same amount of time. There are countless accounts of people getting trampled on because of 2000 other people wanted to get their money. Even though this is a risky market for money people still flood the floors every day of the week. We have to stand back and remember October 29, 1929 Black Tuesday,... | 677.169 | 1 |
Math 7 Curriculum
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Product Description
Math 7 Curriculum – by All Things Algebra® and Lindsay Perro
What is currently included in this curriculum?
So far, NOTHING! We opted to release this early for those that wish to pre-order the Math 7 units instead of purchasing them separately. To be notified when materials are released, follow All Things Algebra on TpT to get the new product emails. When you see a Math 7 unit has been added, simply re-download the curriculum to access the new material.
What will this curriculum contain?
This bundle will contain warm-ups, notes, homework assignments, quizzes, unit tests, a midterm test, end of year review materials, and a final exam for Math 7. This curriculum will follow the same layout and framework as all other courses in the All Things Algebra® Curriculum.
What units and topics will this curriculum cover?
Estimated completion dates for each unit are given below. Please also see the preview for a copy of the curriculum map so you can compare the topics with your own curriculum and standards.
What grade level(s) is this curriculum designed for??
This curriculum is intended for 7th grade students or advanced 6th grade students. It will work well for anyone taking Math 8 the following year.
What standards are covered by this curriculum?
Because the standards in each state can differ, it is impossible to create a one-size fits all curriculum. Please see the preview for a complete list of topics included so that you can compare to your own curriculum and standards.
How does this curriculum differ from the All Things Algebra® Pre-Algebra Curriculum?
This curriculum will focus entirely on seventh grade skills. Many topics included in the Pre-Algebra Curriculum, specifically eighth grade objectives, will not included in this curriculum. Pacing will be slower in certain sections and the complexity will be toned down. The Pre-Algebra Curriculum is intended for students taking Algebra 1 the following year. This is more of a "pure" Math 7 Curriculum intended for students taking Math 8 the following year. Everything will be rewritten to better suit the needs of students who are on grade-level. Though both bundles will be similar in topics, they will not have the exact same content. There will be no duplication of problems. If you find yourself using the Pre-Algebra Curriculum and wishing you had something scaled down to use with your Math 7 students, then this curriculum would be a good fit for you.
How does this curriculum differ from Lindsay Perro's Math 7 Curriculum?
The topics and sequencing for the All Things Algebra® Math 7 curriculum is completely different from that of Lindsay Perro. The curriculum is also drastically different from a style standpoint. You may find one suits your needs more than the other, or you may find that they work very well in conjunction with each other | 677.169 | 1 |
Put
• Develop problem solving skills with 40 totally new and inspiring topics, some familiar and some unfamiliar, grouped into three sections: beginner, improver, advanced • Build, apply and secure the functional and process skills that are integral to the 2010 GCSE Maths Specifications • Challenge students at all levels with open questions that can be approached in different ways • Encourage discussion and collaboration between students to encourage team work and responsibility • Make mathematics relevant and useful, with scenarios such as Coastguard Search and Rescue, and Investigating Design and Cost in Tea Bag Production • Promote self-assessment with 'How did you find these tasks?' questions at the end of each topic. Paperback. Book Condition: new. BRAND NEW, GCSE Maths Functional Skills: Student Book: Edexcel and AQA, Andrew Bennington, Andrew Manning, Naomi Norman, 'How did you find these tasks?' questions at the end of each topic. Bookseller Inventory # B9780007410064 | 677.169 | 1 |
Physics
This qualification is designed to prepare students for university study in physics or engineering.
Each module covers a topic that can be exemplified in many different applications, bringing the subject to life.
Students must carry out 10 required practical activities in the course. The practicals are not constrained, allowing each school to choose resources and equipment that best suit their circumstances.
Practicals in this specification include the Young modulus of a metal and investigations into interference effects, transformers and determination of g. Exam questions will be asked on the practicals, but there is no coursework or practical exam.
Physics is fundamentally a mathematical subject, and this is reflected in the requirement for students to develop a wide range of mathematical skills.
Examinations will include both shorter questions for quickly checking understanding and longer extended calculations, which allow students to demonstrate Physics Physics | 677.169 | 1 |
Algebra
and why it is important, Algebra Expressions, problems and history
What is Algebra meaning and use
Algebra is a branch
of mathematics that deals with properties of operations and the
structures these operations are defined on. Elementary Algebra that
follows the study of arithmetic is mostly occupied with operations on
sets of whole and rational numbers and solving first and second order
equations. What puts elementary algebra a step ahead of elementary
arithmetic is a systematic use of letters to denote generic numbers.
Mastering of
elementary algebra which is often hailed as a necessary preparatory
step for the study of Calculus, is as often an insurmountable block
in many a career. However, the symbolism that is first introduced in
elementary algebra permeates all of mathematics. This symbolism is
the alphabet of the mathematical language.
The word "algebra"
is a shortened misspelled transliteration of an Arabic title al-jebr
w'al-muqabalah (circa 825) by the Persian mathematician known as
al-Khowarismi. The al-jebr part means "reunion of broken parts",
the second part al-muqabalah translates as "to place in front
of, to balance, to oppose, to set equal." Together they describe
symbol manipulations common in algebra: combining like terms, moving
a term to the other side of an equation, etc.
In its English usage
in the 14th century, algeber meant "bone-setting," close to
its original meaning. By the 16th century, the form algebra appeared
in its mathematical meaning. Robert Recorde (c. 1510-1558), the
inventor of the symbol "=" of equality, was the first to
use the term in this sense. He, however, still spelled it as algeber.
The misspellers proved to be more numerous, and the current spelling
algebra took roots.
Thus the original
meaning of algebra refers to what we today call elementary algebra
which is mostly occupied with solving simple equations. More
generally, the term algebra encompasses nowadays many other fields of
mathematics: geometric algebra, abstract algebra, boolean algebra, to
name a few.
Simply
put, Algebra is about finding the unknown or it is about putting real
life problems into equations and then solving them. Unfortunately
many textbooks go straight to the rules, procedures and formulas,
forgetting that these are real life problems being solved.
What is Algebra? Why Take Algebra?
A branch of
mathematics that substitutes letters for numbers. An algebraic
equation represents a scale, what is done on one side of the scale
with a number is also done to the other side of the scale. The
numbers are the constants. Algebra can include real numbers, complex
numbers, matrices, vectors etc.
We hear a lot about
the importance that all children master algebra before they graduate
from high school. But what exactly is algebra, and is it really as
important as everyone claims? And why do so many people find it hard
to learn?
Answering these
questions turns out to be a lot easier than, well, answering a
typical school algebra question, yet surprisingly, few people can
give good answers.
First
of all, algebra is not "arithmetic with letters." At the
most fundamental level, arithmetic and algebra are two different
forms of thinking about numerical issues.
Let's
start with arithmetic. This is essentially the use of the four
numerical operations addition, subtraction, multiplication, and
division to calculate numerical values of various things. It is the
oldest part of mathematics, having its origins in Sumeria (primarily
today's Iraq) around 5,000 years ago. Sumerian society reached
a stage of sophistication that led to the introduction of money as a
means to measure an individual's wealth and mediate the
exchange of goods and services. The monetary tokens eventually gave
way to abstract markings on clay tablets, which we recognize today as
the first numerals (symbols for numbers). Over time, those symbols
acquired an abstract meaning of their own: numbers. In other words,
numbers first arose as money, and arithmetic as a means to use money
in trade.
It
should be noticed that counting predates numbers and arithmetic.
Humans started to count things (most likely family members, animals,
seasons, possessions, etc.) at least 6,000 years ago, as evidenced by
the discovery of bones with tally marks on them, which
anthropologists conclude were notched to provide what we would today
call a numerical record. But those early humans did not have numbers,
nor is there any evidence of any kind of arithmetic. The tally
markers themselves were the record; the marks referred directly to
things in the world, not to abstract numbers.
Something else to
note is that arithmetic does not have to be done by the manipulation
of symbols, the way we are taught today. The modern approach was
developed over many centuries, starting in India in the early half of
the First Millennium, adopted by the Arabic speaking traders in the
second half of the Millennium, and then transported to Europe in the
13th Century. (Hence its present-day name "Hindu-Arabic
arithmetic.") Prior to the adoption of symbol-based,
Hindu-Arabic arithmetic, traders performed their calculations using a
sophisticated system of finger counting or a counting board (a board
with lines ruled on it on which small pebbles were moved around).
Arithmetic instruction books described how to calculate using words,
right up to the 15th Century, when symbol manipulation began to take
over.
Many
people find arithmetic hard to learn, but most of us succeed, or at
least pass the tests, provided we put in enough practice. What makes
it possible to learn arithmetic is that the basic building blocks of
the subject, numbers, arise naturally in the world around us, when we
count things, measure things, buy things, make things, use the
telephone, go to the bank, check the scores, etc. Numbers may be
abstract — you never saw, felt, heard, or smelled the number 3
— but they are tied closely to all the concrete things in the
world we live in.
With algebra,
however, you are one more step removed from the everyday world. Those
x's and y's that you have to learn to deal with in
algebra denote numbers, but usually numbers in general, not
particular numbers. And the human brain is not naturally suited to
think at that level of abstraction. Doing so requires quite a lot of
effort and training.
The important thing
to realize is that doing algebra is a way of thinking and that it is
a way of thinking that is different from arithmetical thinking. Those
formulas and equations, involving all those x's and y's,
are merely a way to represent that thinking on paper. They no more
are algebra than a page of musical notation is music. It is possible
to do algebra without symbols, just as you can play and instrument
without being ably to read music. In fact, traders and other people
who needed it used algebra for 3,000 years before the symbolic form
was introduced in the 16th Century. (That earlier way of doing
algebra is nowadays referred to as "rhetorical algebra,"
to distinguish it from the symbolic approach common today.)
There are several
ways to come to an understanding of the difference between arithmetic
and (school) algebra.
First, algebra
involves thinking logically rather than numerically.
In
arithmetic you reason (calculate) with numbers; in algebra you
reason (logically) about numbers.
In
arithmetic, you calculate a number by working with the numbers you
are given; in algebra, you introduce a term for an unknown number
and reason logically to determine its value.
The above
distinctions should make it clear that algebra is not doing
arithmetic with one or more letters denoting numbers, known or
unknown.
For
example, putting numerical values for a,
b, c
in the familiar formula
in order to find the
numerical solutions to the quadratic equation
is not
algebra, it is arithmetic.
In contrast,
deriving that formula in the first place is algebra. So too is
solving a quadratic equation not by the formula but by the standard
method of "completing the square" and factoring.
When students start
to learn algebra, they inevitably try to solve problems by
arithmetical thinking. That's a natural thing to do, given all
the effort they have put into mastering arithmetic, and at first,
when the algebra problems they meet are particularly simple (that's
the teacher's classification as "simple"), this
approach works.
In
fact, the stronger a student is at arithmetic, the further they can
progress in algebra using arithmetical thinking. For example, many
students can solve the quadratic equation x2 = 2x + 15 using basic
arithmetic, using no algebra at all.
Paradoxically, or so
it may seem, however, those better students may find it harder to
learn algebra. Because to do algebra, for all but the most basic
examples, you have to stop thinking arithmetically and learn to think
algebraically.
Is
mastery of algebra (i.e., algebraic thinking) worth the effort? You
bet — though you'd be hard pressed to reach that
conclusion based on what you will find in most school algebra
textbooks. In today's world, algebra, and not arithmetic,
should be the main goal of school mathematics instruction. For
example, you need to use algebraic thinking if you want to write a
macro to calculate the cells in a spreadsheet like Microsoft Excel.
With a spreadsheet, you don't need to do the arithmetic; the
computer does it, generally much faster and with greater accuracy
than any human can. What you, the person, have to do is create that
spreadsheet in the first place. The computer can't do that for
you.
It
doesn't matter whether the spreadsheet is for calculating
scores, keeping track of your finances, or running a business, you
need to think algebraically to set it up to do what you want. That
means thinking about or across numbers in general, rather than in
terms of (specific) numbers.
Algebra expressions
An
algebraic expression is one or more algebraic terms in a phrase. It
can include variables, constants, and operating symbols, such as plus
and minus signs. It's only a phrase, not the whole sentence, so it
doesn't include an equal sign.
Algebraic
expression:
3x2 + 2y + 7xy +
5
In
an algebraic expression, terms are the elements separated by the plus
or minus signs. This example has four terms, 3x2,
2y, 7xy,
and 5.
Terms may consist of variables and coefficients, or constants.
Variables
In algebraic
expressions, letters represent variables. These letters are actually
numbers in disguise. In this expression, the variables are x and y.
We call these letters "variables" because the numbers they
represent can vary—that is, we can substitute one or more
numbers for the letters in the expression.
Coefficients
Coefficients
are the number part of the terms with variables. In 3x2
+ 2y + 7xy + 5,
the coefficient of the first term is 3. The coefficient of the second
term is 2, and the coefficient of the third term is 7.
If
a term consists of only variables, its coefficient is 1.
Constants
Constants
are the terms in the algebraic expression that contain only numbers.
That is, they're the terms without variables. We call them constants
because their value never changes, since there are no variables in
the term that can change its value. In the expression 7x2
+ 3xy + 8
the constant term is "8."
Real Numbers
In algebra, we work
with the set of real numbers, which we can model using a number line.
Real
numbers describe real-world quantities such as amounts, distances,
age, temperature, and so on. A real number can be an integer, a
fraction, or a decimal. They can also be either rational or
irrational. Numbers that are not "real" are called
imaginary. Imaginary numbers are used by mathematicians to describe
numbers that cannot be found on the number line. They are a more
complex subject than we will work with here.
Rational Numbers
We
call the set of real integers and fractions "rational numbers."
Rational comes from the word "ratio" because a rational
number can always be written as the ratio, or quotient, of two
integers.
Examples of
rational numbers
The fraction ½
is the ratio of 1 to 2.
Since
three can be expressed as three over one, or the ratio of 3 to one,
it is also a rational number.
The
number "0.57" is also a rational number, as it can be
written as a fraction.
Irrational
Numbers
Some real numbers
can't be expressed as a quotient of two integers. We call these
numbers "irrational numbers". The decimal form of an
irrational number is a non-repeating and non-terminating decimal
number. For example, you are probably familiar with the number called
"pi". This irrational number is so important that we give
it a name and a special symbol!
Pi cannot be written
as a quotient of two integers, and its decimal form goes on forever
and never repeats.
Translating
Words into Algebra Language
Here
are some statements in English. Just below each statement is its
translation in algebra.
the sum of three
times a number and eight
3x + 8
The
words "the sum of" tell us we need a plus sign because
we're going to add three times a number to eight. The words "three
times" tell us the first term is a number multiplied by three.
In
this expression, we don't need a multiplication sign or parenthesis.
Phrases like "a number" or "the number" tell us
our expression has an unknown quantity, called a variable. In
algebra, we use letters to represent variables.
the product of a
number and the same number less 3
x(x – 3)
The
words "the product of" tell us we're going to multiply a
number times the number less 3. In this case, we'll use parentheses
to represent the multiplication. The words "less 3" tell us
to subtract three from the unknown number.
a number divided
by the same number less five
The words "divided
by" tell us we're going to divide a number by the difference of
the number and 5. In this case, we'll use a fraction to represent the
division. The words "less 5" tell us we need a minus sign
because we're going to subtract five.
In mathematics, an
expression is a finite combination of symbols that is well-formed
according to rules that depend on the context. Symbols can designate
numbers (constants), variables, operations, functions, and other
mathematical symbols, as well as punctuation, symbols of grouping,
and other syntactic symbols. The use of expressions can range from
the simple:
to the complex:
We
can think of algebraic expressions as generalizations of common
arithmetic operations that are formed by combining numbers,
variables, and mathematical operations. Some common examples follow:
Linear
expression:
.
Quadratic
expression:
.
Rational
expression:
.
Strings of symbols
that violate the rules of syntax are not well-formed and are not
valid mathematical expressions. For example:
would
not be considered a mathematical expression but only a meaningless
jumble.
In
algebra an expression may be used to designate a value, which might
depend on values assigned to variables occurring in the expression;
the determination of this value depends on the semantics attached to
the symbols of the expression. These semantic rules may declare that
certain expressions do not designate any value; such expressions are
said to have an undefined value, but they are well-formed expressions
nonetheless. In general the meaning of expressions is not limited to
designating values; for instance, an expression might designate a
condition, or an equation that is to be solved, or it can be viewed
as an object in its own right that can be manipulated according to
certain rules. Certain expressions that designate a value
simultaneously express a condition that is assumed to hold, for
instance those involving the operator to designate an internal direct
sum.
Being an expression
is a syntactic concept; although different mathematical fields have
different notions of valid expressions, the values associated to
variables does not play a role. See formal language for general
considerations on how expressions are constructed, and formal
semantics for questions concerning attaching meaning (values) to
expressions.
Variables
Many mathematical
expressions include letters called variables. Any variable can be
classified as being either a free variable or a bound variable.
For
a given combination of values for the free variables, an expression
may be evaluated, although for some combinations of values of the
free variables, the value of the expression may be undefined. Thus an
expression represents a function whose inputs are the value assigned
the free variables and whose output is the resulting value of the
expression.
For example, the
expression
evaluated for x =
10, y = 5, will give 2; but is undefined for y = 0.
The evaluation of an
expression is dependent on the definition of the mathematical
operators and on the system of values that is its context.
Two expressions are
said to be equivalent if, for each combination of values for the free
variables, they have the same output, i.e., they represent the same
function. Example:
The expression
has
free variable x, bound variable n, constants 1, 2, and 3, two
occurrences of an implicit multiplication operator, and a summation
operator. The expression is equivalent with the simpler expression
12x. The value for x = 3 is 36.
The '+' and '−'
(addition and subtraction) symbols have their usual meanings.
Division can be expressed either with the '/' or with a horizontal
dash. Thus
are perfectly valid.
Also, for multiplication one can use the symbols '×' or a '•'
(mid dot), or else simply omit it (multiplication is implicit); so:
are
all acceptable. However, notice in the first example above how the
"times" symbol resembles the letter 'x' and also how the
'•' symbol resembles a decimal point, so to avoid confusion it's
best to use one of the later two forms.
An expression must
be well-formed. That is, the operators must have the correct number
of inputs, in the correct places. The expression 2 + 3 is well
formed; the expression * 2 + is not, at least, not in the usual
notation of arithmetic.
Expressions
and their evaluation were formalised by Alonzo Church and Stephen
Kleene in the 1930s in their lambda calculus. The lambda calculus has
been a major influence in the development of modern mathematics and
computer programming languages.
One of the more
interesting results of the lambda calculus is that the equivalence of
two expressions in the lambda calculus is in some cases undecidable.
This is also true of any expression in any system that has power
equivalent to the lambda calculus.
History of Algebra
The
roots of algebra can be traced to the ancient Babylonians, who
developed an advanced arithmetical system with which they were able
to do calculations in an algorithmic fashion. The Babylonians
developed formulas to calculate solutions for problems typically
solved today by using linear equations, quadratic equations, and
indeterminate linear equations. By contrast, most Egyptians of this
era, as well as Greek and Chinese mathematics in the 1st millennium
BC, usually solved such equations by geometric methods, such as those
described in the Rhind Mathematical Papyrus, Euclid's Elements, and
The Nine Chapters on the Mathematical Art. The geometric work of the
Greeks, typified in the Elements, provided the framework for
generalizing formulae beyond the solution of particular problems into
more general systems of stating and solving equations, though this
would not be realized until the medieval Muslim mathematicians.
By
the time of Plato, Greek mathematics had undergone a drastic change.
The Greeks created a geometric algebra where terms were represented
by sides of geometric objects, usually lines, that had letters
associated with them. Diophantus (3rd century AD), sometimes called
"the father of algebra", was an Alexandrian Greek
mathematician and the author of a series of books called Arithmetica.
These texts deal with solving algebraic equations.
The
word algebra comes from the Arabic language (الجبر
al-jabr
"restoration") and much of its methods from Arabic/Islamic
mathematics. Earlier traditions discussed above had a direct
influence on Muhammad ibn Mūsā al-Khwārizmī (c.
780–850). He later wrote The
Compendious Book on Calculation by Completion and Balancing,
which established algebra as a mathematical discipline that is
independent of geometry and arithmetic.
The
Hellenistic mathematicians Hero of Alexandria and Diophantus as well
as Indian mathematicians such as Brahmagupta continued the traditions
of Egypt and Babylon, though Diophantus' Arithmetica and
Brahmagupta's Brahmasphutasiddhanta are on a higher level. For
example, the first complete arithmetic solution (including zero and
negative solutions) to quadratic equations was described by
Brahmagupta in his book Brahmasphutasiddhanta. Later, Arabic and
Muslim mathematicians developed algebraic methods to a much higher
degree of sophistication. Although Diophantus and the Babylonians
used mostly special ad hoc methods to solve equations, Al-Khwarizmi
contribution was fundamental. He solved linear and quadratic
equations without algebraic symbolism, negative numbers or zero, thus
he has to distinguish several types of equations.
In
1545, the Italian mathematician Girolamo Cardano published Ars
magna -The great art,
a 40-chapter masterpiece in which he gave for the first time a method
for solving the general quartic equation.
The
Greek mathematician Diophantus has traditionally been known as the
"father of algebra" but in more recent times there is much
debate over whether al-Khwarizmi, who founded the discipline of
al-jabr, deserves that title instead. Those who support Diophantus
point to the fact that the algebra found in Al-Jabr is slightly more
elementary than the algebra found in Arithmetica and that Arithmetica
is syncopated while Al-Jabr is fully rhetorical. Those who support
Al-Khwarizmi point to the fact that he introduced the methods of
"reduction" and "balancing" (the transposition of
subtracted terms to the other side of an equation, that is, the
cancellation of like terms on opposite sides of the equation) which
the term al-jabr originally referred to, and that he gave an
exhaustive explanation of solving quadratic equations, supported by
geometric proofs, while treating algebra as an independent discipline
in its own right. His algebra was also no longer concerned "with
a series of problems to be resolved, but an exposition which starts
with primitive terms in which the combinations must give all possible
prototypes for equations, which henceforward explicitly constitute
the true object of study". He also studied an equation for its
own sake and "in a generic manner, insofar as it does not simply
emerge in the course of solving a problem, but is specifically called
on to define an infinite class of problems".
The
Persian mathematician Omar Khayyam is credited with identifying the
foundations of algebraic geometry and found the general geometric
solution of the cubic equation. Another Persian mathematician, Sharaf
al-Dīn al-Tūsī, found algebraic and numerical
solutions to various cases of cubic equations. He also developed the
concept of a function. The Indian mathematicians Mahavira and
Bhaskara II, the Persian mathematician Al-Karaji, and the Chinese
mathematician Zhu Shijie, solved various cases of cubic, quartic,
quintic and higher-order polynomial equations using numerical
methods. In the 13th century, the solution of a cubic equation by
Fibonacci is representative of the beginning of a revival in European
algebra. As the Islamic world was declining, the European world was
ascending. And it is here that algebra was further developed.
François
Viète's work at the close of the 16th century marks the
start of the classical discipline of algebra. In 1637, René
Descartes published La Géométrie, inventing analytic
geometry and introducing modern algebraic notation. Another key event
in the further development of algebra was the general algebraic
solution of the cubic and quartic equations, developed in the
mid-16th century. The idea of a determinant was developed by Japanese
mathematician Kowa Seki in the 17th century, followed independently
by Gottfried Leibniz ten years later, for the purpose of solving
systems of simultaneous linear equations using matrices. Gabriel
Cramer also did some work on matrices and determinants in the 18th
century. Permutations were studied by Joseph Lagrange in his 1770
paper Réflexions
sur la résolution algébrique des équations
devoted to solutions of algebraic equations, in which he introduced
Lagrange resolvents. Paolo Ruffini was the first person to develop
the theory of permutation groups, and like his predecessors, also in
the context of solving algebraic equations.
Abstract
algebra was developed in the 19th century, deriving from the interest
in solving equations, initially focusing on what is now called Galois
theory, and on constructibility issues. The "modern algebra"
has deep nineteenth-century roots in the work, for example, of
Richard Dedekind and Leopold Kronecker and profound interconnections
with other branches of mathematics such as algebraic number theory
and algebraic geometry. George Peacock was the founder of axiomatic
thinking in arithmetic and algebra. Augustus De Morgan discovered
relation algebra in his Syllabus
of a Proposed System of Logic.
Josiah Willard Gibbs developed an algebra of vectors in
three-dimensional space, and Arthur Cayley developed an algebra of
matrices (this is a noncommutative algebra).
10 reasons why algebra is so important
"Numbers are
everywhere, but the value of algebra is not that obvious. If you do
something once, you will not need algebra. Algebra works best when
you need to generalize your findings from a situation" (Usiskin,
1995).
The
reasons why Algebra is important:
•One
needs algebra as a requirement for the higher education acceptance.
•Without
Algebra, you do not have the control of your own life but have to
rely on others and that increases your chance of being fooled or
decreases your chance of having a better life standard. i.e., in
finance by earning more money, in stock market by analyzing the
market.
•You
may not learn much from more advanced disciplines such as Chemistry,
Physics, etc.
•Algebra
helps you share your knowledge with other people in an efficient way,
and that includes teaching it.
•You
save time with Algebra so that you do not waste time solving the same
problem over and over again in different occasions.
•Algebra
helps you become an active producer rather than a passive consumer.
i.e. you can be the one who designs the formulae for others and make
profit from it or the one who pays for it. i.e. there are people who
sell the spreadsheet templates they produced.
•You
enjoy life by adding meaningful experiences to your life. i.e. you
can watch nature or everyday happenings and make recreational,
meaningful calculations.
• Maths helps
you become more sensitive to details, thus become a better problem
solver. Life is full of problems, and analytical thinking is best
achieved by studying maths.
•Maths
works as an entertainment and recreation, and an exercise for your
brain.
•Algebra
helps to generalize your solutions to math related problems.
10 Everyday Reasons Why Algebra is Important in your Life
Mathematics is one
of the first things you learn in life. Even as a baby you learn to
count. Starting from that tiny age you will start to learn how to use
building blocks how to count and then move on to drawing objects and
figures. All of these things are important preparation to doing
algebra.
The key to
opportunity
These
are the years of small beginnings until the day comes that you have
to be able to do something as intricate as algebra. Algebra is the
key that will unlock the door before you. Having the ability to do
algebra will help you excel into the field that you want to
specialize in. We live in a world where only the best, the daring and
the humble succeed.
Taking a detour
on not
Having the ability
and knowledge to do algebra will determine whether you will take the
short cut or the detour in the road of life. In other words, ample
opportunities or career choices to decide from or limited positions
with a low annual income.
Prerequisite for
advanced training
Most employers
expect their employees to be able to do the fundamentals of algebra.
If you want to do any advanced training you will have to be able to
be fluent in the concept of letters and symbols used to represent
quantities.
Science
When doing any form
of science, whether just a project or a lifetime career choice, you
will have to be able to do and understand how to use and apply
algebra.
Every day life
Formulas are a part
of our lives. Whether we drive a car and need to calculate the
distance, or need to work out the volume in a milk container,
algebraic formulas are used everyday without you even realizing it.
Analysis
When it comes to
analyzing anything, whether the cost, price or profit of a business
you will need to be able to do algebra. Margins need to be set and
calculations need to be made to do strategic planning and analyzing
is the way to do it.
Data entry
What
about the entering of any data. Your use of algebraic expressions and
the use of equations will be like a corner stone when working with
data entry. When working on the computer with spreadsheets you will
need algebraic skills to enter, design and plan.
Decision making
Decisions like which
cell phone provider gives the best contracts to deciding what type of
vehicle to buy, you will use algebra to decide which one is the best
one. By drawing up a graph and weighing the best option you will get
the best value for your money.
Interest Rates
How much can you
earn on an annual basis with the correct interest rate. How will you
know which company gives the best if you can't work out the graphs
and understand the percentages. In today's life a good investment is
imperative.
Writing of
assignments
When
writing any assignments the use of graphs, data and math will
validate your statements and make it appear more professional.
Professionalism is of the essence if you want to move ahead and be
taken seriously.
Can you see the
importance of algebra? Your day can be made a lot easier with
planning. In financial decisions this can save you a lot of finances
or maybe get you the best price available. It all comes down to
planning and using the knowledge and algebraic skills you have to
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Can SAP C_TFIN52_65 it be definitely well worth it?
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