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Beginning and Intermediate Algebra
9780073229713
0073229717
Summary: Intended for schools that want a single text covering the standard topics from Beginning and Intermediate Algebra. Topics are organized by using the principles of the AMATYC standards as a guide, giving strong support to teachers using the text. The book's organization and pedagogy are designed to work "for students" with a variety of learning styles and for teachers with varied experiences and backgrounds. The inclu...sion of multiple perspectives -- verbal, numerical, algebraic, and graphical -- has proven popular with a broad cross section of students. Use of a graphing calculator is assumed. BEGINNING AND INTERMEDIATE ALGEBRA: THE LANGUAGE AND SYMBOLISM OF MATHEMATICS is a reform-oriented book.
James Hall is the author of Beginning and Intermediate Algebra, published 2007 under ISBN 9780073229713 and 0073229717. Fifteen Beginning and Intermediate Algebra textbooks are available for sale on ValoreBooks.com, thirteen used from the cheapest price of $1.87, or buy new starting at $94.38 | 677.169 | 1 |
...I have also used concepts of discrete math during my PhD program at Princeton University. Also, I am a professional software developer at Microsoft Research where I use concepts of discrete math. I have used C++ during my PhD program at Princeton University.
...Persuasive: I show the student how to make arguments in written form, methods of persuasion, evidence, factual and emotional argument, figures of speech and historical forms of arguments. Process Analysis: The student learns how to describe how something works, and presents detailed steps in det... | 677.169 | 1 |
Abstract
Concerns have been expressed by employers that many young people coming into the workforce
do not have the mathematical skills that are desired. Concerns have also been expressed about
the increasing departure of school mathematics from the demands of daily life. These concerns
are not confined to Australia and it is suggested that something may be amiss with the secondary
mathematics curriculum. This study is concerned with identifying and closing any gaps that may
exist between mathematics that is taught in secondary school and that practiced in daily life.
This study opens with an investigation of the development of mathematics and
mathematics education in order that the reasons for the mathematics that is currently taught in
secondary schools can be best understood. Previous research regarding the mathematics that is
used in everyday life is extended and compared with the mainstream syllabus of today as
exhibited in text books used in the classroom. The lessons that have been learned from
observations of past and present practices are finally drawn together to propose a new secondary
mathematics course. In addition to improving the numeracy of students, it is expected that
additional benefits will accrue in the reduction of the common anxiety in students that is
associated with mathematics, and an easing of the ever-increasing shortage of specialist
mathematics teachers.
The solution that is offered is to approach secondary mathematics from a lateral point of
view with the design of a bipartite course. The core or compulsory syllabus of the proposed
course concentrates on the mathematics found in daily life so that all students leave school
familiar and confident in their ability to perform simple mathematical tasks with a high degree of
accuracy time after time. Those students who have a deeper interest in mathematics would be
able to choos e an elective mathematics syllabus where they can study with others of a like mind. | 677.169 | 1 |
...We can review arithmetical laws, the use of variables, functional relationships, and the use of equations to model word problems. We can also study linear, quadratic, exponential, and inverse functions. Methods for graphing, analyzing, and solving systems of equations and inequalities can also be covered | 677.169 | 1 |
Our goal in this set of lecture notes is to provide students with a strong foundation in mathematical analysis. Such a...
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Our goal in this set of lecture notes is to provide students with a strong foundation in mathematical analysis. Such a foundation is crucial for future study of deeper topics of analysis. Students should be familiar with most of the concepts presented here after completing the calculus sequence. However, these concepts will be reinforced through rigorous proofs. The lecture notes contain topics of real analysis usually covered in a 10-week course: the completeness axiom, sequences and convergence, continuity, and differentiation. The lecture notes also contain many well-selected exercises of various levels. Although these topics are written in a more abstract way compared with those available in some textbooks, teachers can choose to simplify them depending on the background of the students. For instance, rather than introducing the topology of the real line to students, related topological concepts can be replaced by more familiar concepts such as open and closed intervals. Some other topics such as lower and upper semicontinuity, differentiation of convex functions, and generalized differentiation of non-differentiable convex functions can be used as optional mathematical projects. In this way, the lecture notes are suitable for teaching students of different backgrounds mathematical analysis to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Introduction to mathematical analysis
Select this link to open drop down to add material Introduction to mathematical analysis to your Bookmark Collection or Course ePortfolio
״This is a text for a two-term course in introductory real analysis for junior or senior mathematics majors and science...
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״This is a text for a two-term course in introductory real analysis for junior or senior mathematics majors and science students with a serious interest in mathematics. Prospective educators or mathematically gifted high school students can also benefit from the mathematical maturity that can be gained from an introductory real analysis course. The book is designed to fill the gaps left in the development of calculus as it is usually presented in an elementary course, and to provide the background required for insight into more advanced courses in pure and applied mathematics. The standard elementary calculus sequence is the only specific prerequisite for Chapters 1–5, which deal with real-valued functions. (However, other analysis oriented courses, such as elementary differential equation, also provide useful preparatory experience.) Chapters 6 and 7 require a working knowledge of determinants, matrices and linear transformations, typically available from a first course in linear algebra. Chapter 8 is accessible after completion of Chapters 1–5.״NOTE: This book meets the evaluation criteria set by the Editorial Board of the American Institute of Mathematics in connection with the Institute's Open Textbook Initiative Real Analysis to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Introduction To Real Analysis
Select this link to open drop down to add material Introduction To Real It's all Greek to me! Sigma notation to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material It's all Greek to me! Sigma notation
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From the directions: "Click new motif to start drawing a new point set. Then select one of the colors, which correspond to...
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From the directions: "Click new motif to start drawing a new point set. Then select one of the colors, which correspond to different scattering strengths (1-9) and click on the left canvas to position your points. Then click Fourier transform to calculate the discrete Fourier transform of that point set1 to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material JFourier1
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This applet calculates Fourier transforms (FT) of a given pointset. The pointset is generated by the plane space groups (see...
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This applet calculates Fourier transforms (FT) of a given pointset. The pointset is generated by the plane space groups (see Project 22). You may transform all points or only one unit cell. Similar to JFourier3 by the same author, except for periodic (not quasiperiodic) point sets2 to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material JFourier2
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Pick a Bookmark Collection or Course ePortfolio to put this material in or scroll to the bottom to create a new3 to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material JFourier3
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This is a free, online textbook. According to the author, "This text carefully leads the student through the basic topics of...
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This is a free, online textbook. According to the author, "This text carefully leads the student through the basic topics of Real Analysis. Topics include metric spaces, open and closed sets, convergent sequences, function limits and continuity, compact sets, sequences and series of functions, power series, differentiation and integration, Taylor's theorem, total variation, rectifiable arcs, and sufficient conditions of integrability. Well over 500 exercises (many with extensive hints) assist students through the material. For students who need a review of basic mathematical concepts before beginning "epsilon-delta״-style proofs, the text begins with material on set theory (sets, quantifiers, relations and mappings, countable sets), the real numbers (axioms, natural numbers, induction, consequences of the completeness axiom), and Euclidean and vector spaces; this material is condensed from the author's Basic Concepts of Mathematics, the complete version of which can be used as supplementary background material for the present Analysis I to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Mathematical Analysis I
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'This is a book about mathematics appreciation via discovery, rather than about practical mathematics. It considers several...
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'This is a book about mathematics appreciation via discovery, rather than about practical mathematics. It considers several problems that don't appear to be amenable to ordinary arithmetic, algebraic or geometric techniques. It then guides the reader through the process of discovering the solution to each problem, using creative methods and simple techniques that arise naturally. It also indicates how each solution leads to new questions, provides a bit of history of the problem, and discusses a few related problems of current interest that have not yet been solved.''The book contains numerous examples, illustrations, problems and projects that advance the discovery process. There is an extensive section at the end of each chapter that gives answers to the problems posed in the chapter, as well as an additional section that provides supplementary material. Each topic is divided into many short sections. The Table of Contents shows the topics covered, and lists the titles of the sections and subsections. More about the goals and use of the book can be found in the Preface Discovery to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Mathematical Discovery
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Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Mathematical Induction (Part II)
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Course Wizard
MATH 1030 Intermediate Algebra
This course is designed for students who have had some algebra. Topics include: algebraic expression and real numbers; linear equations and inequalities in one and two variables; quadratic equations; polynomials and factoring; graphs of basic functions; systems of linear equations; and applications. This course has been exempted from the requirements of the Writing Across the Curriculum policy. Prerequisite: Challenge examination or MATH 1000. (Description Last Updated: Winter 2008 (200830)) | 677.169 | 1 |
1. Once you are in the class, use a phone to dial the number
for your country mentioned above followed by the access code.
2. Keep your microphone and speakers muted in the classroom to
avoid hearing an echo during the call.
Note: The call dial-in information can also be accessed in the
classroom by clicking.
About the Class
The series of Quicker Mathematics by M.Tyra aims to help the aspirants in learning the 'backdoors' (sets of tricks and methods) to solve the intricate mathematical problems in just a few seconds. The first session would expand on the quick methods for:
i am preparing for bank po, i practise a lot of maths questions papers, but still unable to get through in the exams,drg examination solve problems in maths,getting stuck in every problem.So please kindly suggest me how solve mathematics problems in quick time or any short cut methods thank you
You a really a wonderful man It was 4-5 years ago when I accidentlly get a book (for banking from a friend).I was amazed by the clearity of topic and the magical works I also studied Vedic math andcome to know the roootcause of your magic .of your hat off to your hardwork.
Respected Sir, I did my MBA in 2005,then i have been working in different sector like private banking and educational sector.In my graduation i secured only 54%in honours not in agreegate,I am 25 years old,can i prepare for bank po exam.plz guide me. prasanti,orissa | 677.169 | 1 |
edition includes the most recent Geometry Regents tests through August 2014. These ever popular guides contain study tips, test-taking strategies, score analysis charts, and other valuable features. They are an ideal source of practice and test preparation. The detailed answer explanations make each exam a practical learning experience. Topics reviewed include the language of geometry; parallel lines and quadrilaterals and coordinates; similarity; right triangles and trigonometry; circles and angle measurement; transformation geometry; locus and coordinates; and an introduction to solid geometry.
I used this book as a study aid for the Geometry Regents I took in 9th grade. It contains exams and answers, just as its title states. This book has explanations for problems as well, which is helpful to check your answers. This book helped me achieve a decent grade on the Regents, and I think it can help out any high school student who is going to be taking the Geometry Regents.
My only problem with this was that I had thought that this was the newest version, which had all the most recent tests. But I was wrong and when I got the package I wanted to return it but it would have taken too long to send this one back and then get another. By then Regents would have been over.
At first I thought I was going to buy the last years book, I would have been fine if it was because I was out of options. However, the book came and it was the new, more updated book which made me happy! | 677.169 | 1 |
Foundations of Mathematics
9780470085011
ISBN:
0470085010
Edition: 1 Pub Date: 2008 Publisher: Wiley, John & Sons, Incorporated
Summary: Finally there s an easy-to-follow book that will help readers succeed in the art of proving theorems. Sibley not only conveys the spirit of mathematics but also uncovers the skills required to succeed. Key definitions are introduced while readers are encouraged to develop an intuition about these concepts and practice using them in problems. With this approach, they ll gain a strong understanding of the mathematical ...language as they discover how to apply it in order to find proofs.
Sibley, Thomas Q. is the author of Foundations of Mathematics, published 2008 under ISBN 9780470085011 and 0470085010. Two hundred nine Foundations of Mathematics textbooks are available for sale on ValoreBooks.com, fifty five used from the cheapest price of $74.03, or buy new starting at $127 | 677.169 | 1 |
Elementary Linear Algebra
9780534951900
ISBN:
0534951902
Edition: 4 Pub Date: 1995 Publisher: Brooks Cole
Summary: This outstanding text starts off using vectors and the geometric approach, featuring a computational emphasis. The authors provide students with easy-to-read explanations, examples, proofs, and procedures. Elementary Linear Algebra can be used in both a matrix-oriented course, or a more traditionally structured course.
Stewart Venit is the author of Elementary Linear Algebra, published 1995 under ISBN 978053...4951900 and 0534951902. One hundred twenty seven Elementary Linear Algebra textbooks are available for sale on ValoreBooks.com, twenty five used from the cheapest price of $0.77, or buy new starting at $18 | 677.169 | 1 |
Comment
After spending a good 10-20 minutes looking over the material, I realize this would be quite useful for students to understand a real-life application for math and making use of the equations for it. I looked through it briefly and followed the directions that is describing how to write the equation for that particular situation when comparing different plans. It's very well organized, much like a college syllabus, so it is easy to follow and navigate. The material accurately present concepts of graphing and how to apply a simple equation using the two factors to a graph, which can help us incorporate our understanding of it to harder and much more complicated problems.
Although it is very short and intended more for the teacher as a teaching tool, it is still very easy to follow as it is broken down into clear sections, step by step, showing the methods and ways to approach the math.
Technical Remarks:
For a lesson, it's very well organized and easy to follow.
Time spent reviewing site:
10-20 minutes | 677.169 | 1 |
Posts Tagged with "Linear Algebra"
Problem 102 of Project Euler asks us the following questionfind the number of triangles in the text file for which the interior contains the origin.This problem can be solved in a plethora of ways as far as I have seen. I like one particular solution though, which involves calculating the area of triangle and compare it to triangles containing the point P which we are testing if it is in the interior of the triangle.
I have been in contact with Frederick Koh from Whitegroupmaths.com who kindly agreed that he would write a guest post for the blog to promote what he has to offer - Tutoring in A level maths in Singapore. So without further ado let me present you with the real content.I have been asked on numerous occasions by students to provide a short effective mathematical proof verifying the fact that obtaining the vector product of the normals characterising two separate non parallel planes in 3 dimensional Cartesian space produces the direction vector of the line arising from the intersection of the above mentioned planes.
This post is shameless promotion of what I think is a great site - Khan Academy. If you don't like shameless promotions, you should stop reading now. The site features a whole set of instructional videos along with a good amount of exercises. All topics are what I consider to be some basic skills of mathematics. Not because they are simple or easy, but because they are fundamental for you to build upon.It brings you all the way from addition (yep, I am not kidding) through calculus, differential equations and probability. A total of more than 2400 videos, including videos on other topics such as chemistry and finance.
Within the field of mathematics I handle every day linear algebra plays a vital role. Linear algebra is a field of mathematics that studies vectors and vector spaces. On common use of linear algebra is to solve a set of linear equations.The reason why I bring up the topic, is that I rediscovered a video version a MIT course in linear algebra taught by Gilbert Strang. I found the videos when I first studied to my exam in linear algebra.I have embedded the first video for your pleasure, and all the videos are linked in the post. | 677.169 | 1 |
...
Show More financial decisions. Despite the availability of automated tools to perform financial calculations, the author demonstrates that a basic grasp of the underlying mathematical formulas and tables is essential to truly understand finance.The book begins with an introduction to the most fundamental mathematical concepts, including numbers, exponents, and logarithms; mathematical progressions; and statistical measures. Next, the author explores the mathematics of the time value of money through a discussion of simple interest, bank discount, compound interest, and annuities. Subsequent chapters explore the mathematical aspects of various financial scenarios, including:Mortgage debt, leasing, and credit and loansCapital budgeting, depreciation, and depletionBreak-even analysis and leverageInvesting, with coverage of stocks, bonds, mutual funds, options, cost of capital, and ratio analysisReturn and risk, along with a discussion of the Capital Asset Pricing Model (CAPM)Life annuities as well as life, property, and casualty insuranceThroughout the book, numerous examples and exercises present realistic financial scenarios that aid readers in applying their newfound mathematical skills to devise solutions. The author does not promote the use of financial calculators and computers, but rather guides readers through problem solving using formulas and tables with little emphasis on derivations and proofs.Extensively class-tested to ensure an easy-to-follow presentation, Mathematical Finance is an excellent book for courses in business, economics, and mathematics of finance at the upper-undergraduate and graduate levels. The book is also appropriate for consumers and entrepreneurs who need to build their mathematical skills in order to better understand financial problems and make better financial | 677.169 | 1 |
Introductory Algebra
Description
Need more than videos to learn math? YourTeacher's Introductory Algebra app is like having a personal math tutor in your pocket. ** "It's like a private school math classroom, but you are the only student." John "I just love YourTeacher ...Read more
Description
Need more than videos to learn math? YourTeacher's Introductory Algebra app is like having a personal math tutor in your pocket.
**
"It's like a private school math classroom, but you are the only student." John
"I just love YourTeacher and the way you explain things. I felt like I was in a classroom instead of just looking at examples." Diane
"I was desperately seeking Algebra help. I am working on an Associates Degree at a local community college. Pre-Algebra and Introductory Algebra were required classes for me. I was completely lost because I have absolutely no Algebra background. I was really struggling in my class. Then I found YourTeacher. The lessons were a tremendous help. I ended up making an A in the class and passed the exit test. Thank you for your help." Catherine
"I can't thank you enough for your patient, easy way of teaching. I am taking a college introductory algebra class this semester and my math teacher runs through the material so fast it's been very hard to keep up with her. By listening to you explain what she ran through during class, I've been able to learn what she was trying to teach. Thanks again for all your help!" Janice | 677.169 | 1 |
Description:
The primary purpose of the book is to provide a convenient source of reference to those people who are appearing for engineering entrance examinations One has to refer a good number of books to understand the basics of graphs but this book will surely reduce the number of books that each of them needs to perform his job. This book has been designed for convenience.
In JEE Maths syllabus, graphs are not explicitly mentioned but For last 15-20 years questions based on graphs are being asked in IIT JEE question papers
We have tried our level best to present the graphs in easier way and good number of problems based on basic graphs and inequalities are discussed and solved in the book. All types of transformations are also covered | 677.169 | 1 |
Discrete Mathematics Using a Computer
9781846282416
ISBN:
1846282411
Pub Date: 2006 Publisher: Springer
Summary: 'Discrete Mathematics Using A Computer' offers a new, 'hands-on' approach to teaching discrete mathematics. A simple functional language is used to allow students to experiment with mathematical notations which are traditionally difficult to pick up.
Odonnell, John is the author of Discrete Mathematics Using a Computer, published 2006 under ISBN 9781846282416 and 1846282411. Three hundred eighteen Discrete M...athematics Using a Computer textbooks are available for sale on ValoreBooks.com, six used from the cheapest price of $49.75, or buy new starting at $61.54 | 677.169 | 1 |
GCSE Mathematics in a key requirement for many job roles as it demonstrates an understanding of numbers and an ability to work with them. And yet many people find numbers intimidating. With this in mind, this home study course has been designed to guide you through the subject in logical, easy-to- follow steps, building on your understanding as your confidence grows.
The GCSE places great emphasis on "doing mathematics" and relating this wherever possible, to everyday life. Certain techniques and formulae need to be learnt, but the emphasis on "doing" means that you should work carefully through all the examples and exercises in order to be able to solve problems effectively.
Mathematics is essentially a holistic subject and, as such, should be taught in this way with connections being made between the sections on Number and algebra, Shape, space and measures and Handling data as demanded in the National Curriculum. For example, Number underpins the whole of mathematics.
The Maths GCSE distance learning course gently guides the student through basic mathematical skills, progressing onto more advanced material as the student's skills and abilities develop. A reasonable level of proficiency in arithmetical skills is assumed.
Each lesson begins with a set of clearly stated objectives and an explanation of its place in the overall programme of study. Effective learning is encouraged through frequent activities and self-assessment questions. There are thirteen tutor-marked assignments and a practice exam paper.
The Syllabus:
Our Maths GCSE course prepares students for AQA GCSE Mathematics Specification 4365 for exams in May/June 2016. The course covers an intermediate and higher tier level. The course covers the intermediate level with an optional unit included to also cover the additional higher tier so you have a choice of which tier to complete.
We have chosen this syllabus as the most suited to distance learning. Assessment is by two examination papers.
Paper 1 is 40% of the total marks. The paper is a Written Paper (Non-calculator). The exam is 1 Hour, 15 Minutes for the Intermediate Tier & 1 Hour 30 Minutes for the Higher Tier.
Paper 2 is 60% of the total marks. The paper is a Written Paper (With calculator). The exam is 1 Hour, 45 Minutes for the Foundations Tier & 2 Hours for the Higher Tier.
Please Note: This Syllabus is being retired after the May/June 2016 exams which gives no option for deferral. There will be an exam in November, although this is only for students who have failed the May/June 2016 exam. A new course covering the AQA GCSE Mathematics 8300 specification is available for students if they prefer to aim for the May/June 2017 examinations. All candidates
taking this specification would sit three 90-minute written papers, the first being without a calculator.
Prerequisites:
There are no prerequisites for starting this course, however a reasonable level of proficiency in arithmetical skills is assumed.
Coursework:
No coursework is required.
Additional Reading:
The course is fully self-contained and is designed to prepare you for the examination without the use of a textbook. However, you are strongly advised to supplement your studies by reference to one of the many specially written GCSE textbooks.
Student Support:
Every student receives a personal tutor with whom they should keep in regular contact. All tutors are fully qualified experienced teachers holding either a Postgraduate Certificate in Education or a degree in education. Most work from home and are able to offer flexible contact times. The tutor will contact students with an introductory letter and a telephone call to help decide on a study plan. They will also let the student know the best ways and times for contact.
There is no limit on how often you can contact your tutor. The advice and encouragement our tutors provide are an important part of your learning experience. You can contact your tutor by Phone, Post or Email. There are tutor-marked assignments (TMA's) in every course. These help the student to consolidate their learning and prepare for examinations.
All tutors have a Freephone 0800 telephone number and most will have Skype capabilities. Whilst you are still actively submitting assignments, you should expect to be in contact with your tutor regularly. Remember, to get the most out of your tutor, you will need to tell them when you need assistance.
Examination Centres:
All distance learning students will sit their exams as a private candidate in a registered school or college. The finding of an examination centre and the booking of exams is the responsibility of the student. In December/January students are contacted and advised to start arranging their examinations. AQA have over 300 exam centres in the UK. Examination officers are available for registered students for advice on finding suitable centres or you can use AQA's Private Candidate Page which details information about being a private candidate and has a search facility showing a small selection of centres set out geographically.
Examination fees need to be paid direct to your chosen exam centre. Prices for exams will vary depending on the exam centre's administration costs. You can find out further information about the Private Candidate process and locate suitable exam centres from the AQA Private Candidate Page.
Studying from Abroad:
It is possible to study GCSE Courses from anywhere in the world. However some home study courses are easier to study from abroad than others and some restrictions to the type of tuition available apply. If you wish to study a GCSE then you will need to sit the examinations in the UK. Please remember that most GCSE's have several examinations which may well be spread over several weeks.
Sarah Potts, Manchester (Maths GCSE)
I always regretted not getting my Maths GCSE at school and had found it detailed as a requirement for a lot of jobs that I was interested in. As I work part-time and have 2 young boys, distance learning seemed like the best option for me. The course materials were easy to follow and my tutor was a great help with marking my work quickly. I have now passed my exams with a B Grade and I am even thinking of doing my A Level.... Scary! | 677.169 | 1 |
Flummoxed by formulas? Queasy about equations? Perturbed by pi? Now you can stop cursing over calculus and start cackling over Math, the newest volume in Bill Robertson's accurate but amusing Stop Faking It! best sellers. As Robertson sees it, too many people view mathematics as a set of rules to be followed, procedures to memorize, and theorems... more...
A no-nonsense practical guide to geometry, providing concise summaries, clear model examples, and plenty of practice, making this workbook the ideal complement to class study or self-study, preparation for exams or a brush-up on rusty skills. About the Book Established as a successful practical workbook series with more than 20 titles in the language... more...
This photocopy master book, which has proven extremely popular over the years, provides a range of 30+ problem solving activities using strategies such as: Developing logical thinking; Using number concepts to develop logical thinking; Logical reasoning; Developing visual imagery; and Pattern perception using number. more...
An ideal course text or supplement for the many underprepared students enrolled in the required freshman college math course, this revision of the highly successful outline (more than 348,000 copies sold to date) has been updated to reflect the many recent changes in the curriculum. Based on Schaum's critically acclaimed pedagogy of concise theory... more...
Learn how to easily do quick mental math calculations Speed Math for Kids is your guide to becoming a math genius--even if you have struggled with math in the past. Believe it or not, you have the ability to perform lightning quick calculations that will astonish your friends, family, and teachers. You'll be able to master your multiplication... more...
Any way you slice it, fractions are foundational Many students struggle with fractions and must understand them before learning higher-level math. Veteran educator David B. Spangler describes powerful diagnostic methods for error analysis that pinpoint specific student misconceptions and supplies specific intervention strategies and activities... more...
This volume supports the belief that a revised and advanced science education can emerge from the convergence and synthesis of several current scientific and technological activities including examples of research from cognitive science, social science, and other discipline-based educational studies. The anticipated result: the formation of science... more... | 677.169 | 1 |
Just for the record. Mathematica is much powerful than it appears to be. (Anyway, I hope J. Harrison is working to make notation manipulation easier.)
Some years ago I wrote a package (for Mathematica 5.0) that I called playEquation.
I used it to teach algebraic manipulation, first and second order equations and more. It worked extremely well for teaching. Extremely well. Believe-me: extremely well!!
The general idea is very simple: just select and press a button or simply press a button. Here is a simple example from a tutorial that I wrote (in portuguese).
(1) You start with the equation 3x-7==14 on input cell in your notebook.
(2) To solve the equation, you can add 7 to both members, or subtract -7 from both members. Suppose you wants the later. Then you just select -7 and press the button *-* buttom on the control palette.
(3) This is what happens in sucession:
* the pressed button is highlighted;
* a nice short "beep" is emited;
* the user selection (-7) goes to an input field and stays selected;
* the notebooks sizes (control palette and user notebook) are automatically adjusted to fit on the vertical according to the height of the selection (-7);
* a cell "Subtracting -7" is pasted below your input cell;
* and for a brief moment (1.5 seconds) the program shows another cell with a freezed intermediate step before evaluation):(3x-7)-(-7)==14-(-7)
* the freezed cell is then evaluated to 3x==21.
Finally you must divide by 3. So you select 3 and press the divide button. As before, the program beeps, the pressed button is marked, the user selection (3) goes to the input field, etc., and for a brief moment (1.5 seconds) the program shows 3x/3==21/3 (in TraditionalForm) and then this: x==7. Instead of selecting, you could also just type 1/3 in the input field and press the times button.
In order to solve a second degree equation/inequality, the user:
selects the variable to solve for, unless there is only one, in which case the program automatically identifies the variable. (In any case, the variable goes to the input field.) However, this is not a mere Solve-like command. Instead, as in a normal class, the user will have to put his equation (by using the other buttons) in one of the following traditional forms:
x^2==k
a x^2+b x+c==0.
Only then the solve button will print the result carefully formatted as the user would see it in a traditional book.
You could also solve equations with radicals, inequalities, do some algebraic replacements of variables, customize your interface, create you own operations, etc.
Unfortunately, I stopped there. The package was not updated to Mathematica 5.2 or 6.
Cordialmente,
Carlos César de Araújo
Gregos & Troianos Educacional
Belo Horizonte, MG, Brasil
(31) 3283-1122 | 677.169 | 1 |
Numbers, Groups and Codes
9780521540506
ISBN:
052154050X
Edition: 2 Pub Date: 2004 Publisher: Cambridge University Press
Summary: A thoroughly revised and updated version of the popular textbook on abstract algebra. The material is introduced with clarity and reference to problems and concepts that students will easily understand. With many examples and exercises, it will serve as the ideal introduction to this important and ubiquitous subject.
J. F. Humphreys is the author of Numbers, Groups and Codes, published 2004 under ISBN 978052...1540506 and 052154050X. Three hundred seventy two Numbers, Groups and Codes textbooks are available for sale on ValoreBooks.com, fifty six used from the cheapest price of $62.54, or buy new starting at $62.54 | 677.169 | 1 |
Math Survival Guide Tips and Tricks for Science Students
9780471270546
ISBN:
0471270547
Edition: 2 Pub Date: 2003 Publisher: Wiley & Sons, Incorporated, John
Summary: This second edition of 'Math Survival Guide' provides tips for science students in the form of a quick reference/update guide. It uses an approachable tone and appropriate level and includes good problem sets.
Appling, Jeffrey R. is the author of Math Survival Guide Tips and Tricks for Science Students, published 2003 under ISBN 9780471270546 and 0471270547. Three hundred forty Math Survival Guide Tips and T...ricks for Science Students textbooks are available for sale on ValoreBooks.com, ninety | 677.169 | 1 |
Math cheat sheet - by Dr Seiden
10 pages full of math formulas and theorems! Cheating is bad but they will be helpful when studying for your math exams. Including help on trigonometry, functional analysis, integral and derivatives, plus much more...
10 pages full of math formulas and theorems! Cheating is bad but they will be helpful when studying for your math exams. Including help on trigonometry, functional analysis, integral and derivatives, plus much more... | 677.169 | 1 |
An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
Tuesday, September 13, 2005
janet's scribe
Calculators are not that smart, sometimes! When graphing a calculator everything isn't always what they seem. That was one of the things I learned last class, I bet you didn't know that either. First exercise we did were the problem solving questions using our calculator, which really helped me refresh my memory. We wrote equations and sketched graphs to help us solve them. Also, we refreshed our memory on f(x) functions by doing reviews. Next we talked about most of the basic functions, Cubic, square root, reciprocal, absolute value, squaring and identity functions. In that lesson I learned to flip graphs over the y-axis instead of sliding them on the x-axis. It's always good learning different ways to solve problems as Mr. K's famous phrase goes, "There is more than one way to skin a cat, but don't really do that because it's cruel!" So overall it was yet another productive class, for me that is, and I do apologize for not posting this sooner. I did try! I swear ... and I guess everyone knows who the scribe is for today's class, Sarah!! Have fun! | 677.169 | 1 |
Intermediate Algebra
9780073309309
ISBN:
0073309303
Pub Date: 2006 Publisher: McGraw-Hill College
Summary: Intermediate Algebraby Baratto/Kohlmetz/Bergman is part of the latest offerings in the successful Streeter-Hutchison Series in Mathematics. By popular demand, we are now offering an Intermediate Algebra book in the series again. This book combines the best of earlier versions of Intermediate Algebra, along with new material requested by a cross-section of Intermediate Algebra instructors across the country. This firs...t edition maintains the hallmark approach of encouraging the learning of mathematics by focusing its coverage onmastering math through practice. This worktext seeks to provide carefully detailed explanations and accessible pedagogy to introduce intermediate algebra concepts and put thecontent in context. The authors use a three-pronged approach (I. Communication, II. Pattern Recognition, and III. Problem Solving) to present the material and stimulate critical thinking skills. Items such asMath Anxietyboxes,Check Yourselfexercises, andActivitiesrepresent this approach and the underlying philosophy of mastering math through practice. The exercise sets are well-organized, and clearly labeled. Vocational and professional-technical exercises have been included throughout. Repeated exposure to this consistent structure should help advance the student's skills in relating to mathematics. The book is designed for a one-semester intermediate algebra course and is appropriate for lecture, learning center, laboratory, or self-paced courses. It is accompanied by numerous useful supplements, including McGraw-Hill's online homework management system, MathZone.
Hutchison, Donald is the author of Intermediate Algebra, published 2006 under ISBN 9780073309309 and 0073309303. Eight Intermediate Algebra textbooks are available for sale on ValoreBooks.com, six used from the cheapest price of $1.18, or buy new starting at $205 | 677.169 | 1 |
9780471707080
ISBN:
0471707082
Edition: 1 Pub Date: 2009 Publisher: Wiley
Summary: This text offers a fresh approach to algebra that focuses on teaching readers how to truly understand the principles, rather than viewing them merely as tools for other forms of mathematics. It relies on a storyline to form the backbone of the chapters and make the material more engaging.
William G. McCallum is the author of Algebra: Form and Function, published 2009 under ISBN 9780471707080 and 0471707082. ...Three hundred ten Algebra: Form and Function textbooks are available for sale on ValoreBooks.com, seventy seven used from the cheapest price of $11.54, or buy new starting at $187 | 677.169 | 1 |
Crystal Lake, IL StatisticsMichael WEdward P.
...Finite math is an introductory course in discrete math. A typical finite math course is a survey course consisting of: linear functions, matrices, linear inequalities, linear programming, the Simplex Method, counting (combinatorics), and probability. I have taught finite math within the university and community college setting for the past nine years. | 677.169 | 1 |
Details about Basic Complex Analysis:
Basic Complex Analysis skillfully combines a clear exposition of core theory with a rich variety of applications. Designed for undergraduates in mathematics, the physical sciences, and engineering who have completed two years of calculus and are taking complex analysis for the first time..
Back to top
Rent Basic Complex Analysis 3rd edition today, or search our site for other textbooks by Jerrold E. Marsden. Every textbook comes with a 21-day "Any Reason" guarantee. Published by W. H. Freeman. | 677.169 | 1 |
Monday, April 27, 2009
Classifying Polynomials by Term and Degree
We learned how to classify polynomials by number of terms. Vocabulary includes monomial, binomial, trinomial and polynomial. We then learned how to classify polynomials by degree (0 to n, and by name, including constant, linear, quadratic, cubic, and quartic). Students were introduced to the standard form for writing polynomials (in order of degree).
Students received expanded grade slip information (handwritten slips attached to the printed grade. For one day only, students turning in late work will receive reduced credit, which will only help buoy up grades. This offer ends at 3:30 p.m. on Tuesday, April 28th.
Tonight's Homework: Lesson 10.1 ( 7 -12, 49 - 60, FCO ). Try a couple of problems using the vertical format and then a couple more using the horizontal format. | 677.169 | 1 |
Learning Mathematica
Test Your Knowledge of Mathematica
Volume 2, Issue 4
Fall 1992
Nancy Blachman
How well do you know Mathematica? This quiz will test your knowledge
and hopefully stimulate your thinking. Select the best answer or answers
to each question, without using a computer or referring to any books, except
the Mathematica Quick Reference, Version 2 (Variable Symbols, Inc.,
and Addison-Wesley, 1992) or the Mathematica manual. Some questions
have more than one correct answer. | 677.169 | 1 |
9780201726343
ISBN:
0201726343
Edition: 5 Pub Date: 2003 Publisher: Pearson
Summary: This text is organised into 4 main parts - discrete mathematics, graph theory, modern algebra and combinatorics (flexible modular structuring). It includes a large variety of elementary problems allowing students to establish skills as they practice.
Ralph P. Grimaldi is the author of Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, published 2003 under ISBN 9780201726343 and 0...201726343. Five hundred fifty seven Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition textbooks are available for sale on ValoreBooks.com, sixty six used from the cheapest price of $38.85, or buy new starting at $183 | 677.169 | 1 |
Are you trying to prepare for the GED® Math Test? Are you afraid of math? Then you've come to the right place. "I'm Not Afraid of Math Anymore!" is helping students learn challenging math in an easier way. Color-coding helps you to understand what is happening in the math problems.
But that's not all. Every problem is worked out for you in the answers after each lesson. If you miss a question, you will be able to see the correct way to work it.
This fourth edition is a quick guide to the math you need to be successful on the GED® math test.
You can view Sample Pages at the store below or in the top menu.
"¡No le Temeré Nunca Más a las Matemáticas!"
Spanish Math for the GED® Test
Is Spanish your native language and you need to prepare for the GED® Math Test? Then this book is for you!
"No le Temeré Nunca Más a las Matemáticas!" is a quick guide that teaches the most important lessons for the GED® Math Test. This book also contains easy-to-understand language and color-coding to make preparing for the GED® Math Test as easy as possible.
You can view Sample Pages at the store below or in the top menu.
"I'm Not Afraid of Math Anymore!"
A Guide to High School Equivalency Math Tests
Do you want to earn a high school equivalency diploma? Are you thinking about taking the HiSET Test? Do you need help with math? Then this version of "I'm Not Afraid of Math Anymore!" is here to help. Lots of color keeps the math interesting, makes it more fun to learn, and helps you to understanding the math problems better. Practice problems in this book are also completely worked out in the answers to help you understand how to work each problem.
You can view Sample Pages at the store below or in the top menu.
"Verb Time in "I'm Not Afraid of English Anymore!"
Writing English as a Second Language
Is English your second language? Do you need help to write English correctly? Are verbs difficult for you? The Verb Time Chart from "I'm Not Afraid of English Anymore!" is designed for English language learners who want to learn how to use English verbs correctly. The Verb Time Chart booklet shows how to use verbs in the present, past, and future and progresses from easy verbs to trickier verbs.
Color is used in this book, too, to help you understand the correct words to write. Verb Time practice pages are also available electronically for you to print yourself.
"Ms. Bernard's books are great! They helped me a lot. I suggest that anyone who wants to study for the GED® Test should purchase them." -- Robbie Curry-Martin, GED® Graduate
Testimonies from GED Math Teachers
"I have been using Susan's materials for over four years with my GED® students. The success rate of my students passing the GED® test has significantly increased each year, thanks to the information in the "I'm Not Afraid of Math Anymore!" series. Her lessons are directly and accurately aligned with the questions that will be accessed on the GED Math Test. The depth of knowledge that a student is required to demonstrate and perform is effectively presented in her books." -- Pam Watts, GED® Teacher at Oklahoma City Community College
"My full-time job is in the medical field, but I wanted to "give back" to my community by teaching an evening GED® class. As a part-time teacher, Susan's books have kept me going. The lessons are easy to understand and easy to teach. My students love them and often buy their own copies of the books to study at home. It's very rewarding to see so many of my students pass the GED test." -- Lawrence Beasley, GED® Teacher at Oklahoma City Community College
"I felt lost with the math on the new GED® practice test, but now with your GED® Algebra book, I'm having fun teaching again. Thanks, Susan." -- Janie Davis, GED Tutor for the Pioneer Library System, Shawnee, Oklahoma
Student Friendly Features of "I'm Not Afraid of Math Anymore!"
Common, every-day words are used to explain complex concepts.
Difficult problems are made easier through much color-coding. Students can more easily see where that x went in algebra and can more easily remember how to work the problem.
Answers showing complete color-coded solutions are at the end of each lesson. This helps students to see how to work a question that was missed.
TI-30XS calculator tips are included in the GED® math prep book so that students know how to operate this official GED® calculator. Calculator information is also included in the HSE® math prep book.
Questions are patterned after the questions found on the GED ReadyTM or HiSET® practice tests which helps students to know what to expect on the real tests.
The lessons aren't too short to understand, and they're not too long to get through quickly. Each lesson is taught with understandable explanations, but with the goal in mind of earning the high school equivalency credential as quickly as possible. Adults have busy lives and many responsibilities. "I'm Not Afraid of Math Anymore!" can help students reach their dream.
Student Friendly Features of Verb Time Chart from "I'm Not Afraid of English Anymore!"
Teaches the top 125 verbs in the English language
Verbs progress from Easy → to Tricky → to Trickier instead of an alphabetical list. | 677.169 | 1 |
Banning CalculusJesse R.
Alex C.
...Algebra 2 is a complex subject because it requires that the student hold knowledge from previous years. Not only do they have to hold the knowledge but they are expected to take it to an extreme. Conics, trigonometry, unit circles, graphing, rational expression evaluations, binomial theorems, matrices and summations, I have done them all. | 677.169 | 1 |
GeoGebra is a free and multi-platform dynamic mathematics software for education in secondary schools that joins geometry,...
see more
GeoGebra is a free and multi-platform dynamic mathematics software for education in secondary schools that joins geometry, algebra and calculus. On the one hand, GeoGebra is a dynamic geometry system--you can do constructions with points, vectors, segments, lines, conic sections as well as functions and change them dynamically afterwards; on window corresponds to an object in the geometry window and vice versagebra to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Geogebra
Select this link to open drop down to add material Geogebra History of Pi to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material A History of Pi
Select this link to open drop down to add material A History of Pi Create a Graph to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Create a Graph
Select this link to open drop down to add material Create a Graph Math--Secondary Interactives to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Seeing Math--Secondary Interactives
Select this link to open drop down to add material Seeing Math--Secondary Interactives to your Bookmark Collection or Course ePortfolio
Symmetry is used to analyze the patterns in Oriental carpets. Educational activities are provided along with other...
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Symmetry is used to analyze the patterns in Oriental carpets. Educational activities are provided along with other educational resources. This web site is a collaborative project of The Textile Museum and The Math Forum Symmetry and Pattern: The Art of Oriental Carpets to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Symmetry and Pattern: The Art of Oriental Carpets to your Bookmark Collection or Course ePortfolio
These online notes are intended for students who are working through the textbook Abstract Algebra by Beachy and Blair. The...
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These online notes are intended for students who are working through the textbook Abstract Algebra by Beachy and Blair. The notes are focused on solved problems, and will help students learn how to do proofs as well as computations. There are also some "lab" questions on groups, based on a Java applet Groups15 written by John Wavrik of UCSD Abstract Algebra Online Study Guide to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Abstract Algebra Online Study Guide
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Collectionlets: Parametric Curves and Surfaces to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Mathlets: Parametric Curves and Surfaces
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This site provides descriptions of how pi is evaluated, Gregory's series formula, connections between Fibonacci numbers...
see more
This site provides descriptions of how pi is evaluated, Gregory's series formula, connections between Fibonacci numbers and approximations for pi, and more. It is a subpage of Fibonacci Numbers and the Golden Section that is listed separately in MERLOT and the Fibonacci Numbers to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Pi and the Fibonacci Numbers
Select this link to open drop down to add material Pi and the Fibonacci to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Pythagorean Theorem
Select this link to open drop down to add material Pythagorean Theorem to your Bookmark Collection or Course ePortfolio | 677.169 | 1 |
For over thirty-five years average mathematics scores of students in the United States have been below the international average, indicating problems with performance when compared with students from other countries. The United States NationalMany secondary pre-service and in-service mathematics educators require additional knowledge and resources pertaining to the efficient and successful use of technology in the classroom. This study examined a specific form of calculator-based...
The debate over the effectiveness of a "reform" mathematics curriculum verses a "traditional" mathematics curriculum in student achievement has been widely discussed and studied. Standards-based, or reform, programs emphasize applications to solveThe purpose of this record is both to explore and enlighten our understanding of the relationship between the capabilities of highly gifted youth and the current recommendations for teaching such students, and to describe the process of one suchThis study examined whether the use of Geometer's Sketchpad had an impact on understanding of geometry and academic achievement. The study was conducted over the course of 12 weeks and took place in a public high school in Connecticut. TheNumerous studies can be found regarding the development of mathematics anxiety with secondary and college level students. However, little can be found providing evidence of mathematics anxiety in elementary school students. In an attempt to offer... | 677.169 | 1 |
Student Solutions Manual/Study Guide for Goodman/Hirsch's Understanding Elementary Algebra with Geometry: A Course for College Students / Edition 4
Contains worked-out solutions to many odd-numbered exercises, the answers to the "Questions for Thought," two additional practice tests for each chapter, and four additional cumulative practice tests.
See more details below | 677.169 | 1 |
An Introduction to the History of Mathematics (The Saunders series)
9780030620645
ISBN:
0030620643
Edition: 5th Publisher: Thomson Learning
Summary: This classic best-seller by a well-known author introduces mathematics history to math and math education majors. Suggested essay topics and problem studies challenge students. CULTURAL CONNECTIONS sections explain the time and culture in which mathematics developed and evolved. Portraits of mathematicians and material on women in mathematics are of special interest.
Howard Eves is the author of An Introduct...ion to the History of Mathematics (The Saunders series), published under ISBN 9780030620645 and 0030620643. Four An Introduction to the History of Mathematics (The Saunders series) textbooks are available for sale on ValoreBooks.com, three used from the cheapest price of $0.96, or buy new starting at $80.905th Edn. Spivak CALCULUS, 1967, p521: Most textbooks on the history of mathematics are both superficial and dull. An admirable exception is... Evesall 593pp unmarked; baby blu [more]
5th Edn. Spivak CALCULUS, 1967, p521: Most textbooks on the history of mathematics are both superficial and dull. An admirable exception is... Evesall 593pp unmarked; baby blue covers with silver letters; Used Book in Good Condition [less]
ISBN-13:9780030620645
ISBN:0030620643
Edition:5thth
Publisher:Thomson Learning
ValoreBooks.com has some of the lowest prices for cheap An Introduction to the History of Mathematics (The Saunders series) rentals, or used and new condition books that can be mailed to you in no time. | 677.169 | 1 |
- Calculate properties of solids, including stress, pressure, deformation, and more - Calculate properties of fluids, including drag, buoyancy, hydrostatic pressure, and more - Solve a variety of electricity and magnetism calculations - Perform ray optics calculations, including Snell's law, lensmaker's equation, thin lens equation, and more - Compute wave effects, including properties of diffraction and thin film interference - Do common thermodynamics calculations, including the ideal gas law, Joule's first law, and more - Look up laws of physics and common physics constants
The Wolfram Physics II Course Assistant is powered by the Wolfram|Alpha computational knowledge engine and is created by Wolfram Research, makers of Mathematica—the world's leading software system for mathematical research and Wolfram Physics II Course Assistant App for iOS to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Wolfram Physics II Course Assistant App for iOS
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If you've ever taken a graduate statistics course and discovered that you've forgotten how to divide a fraction or turn a fraction into a percentage, then this handy guide to mathematics is for you. Each topic is provided with a definition, explanation, and examples of how to solve a particular problem using the topic's technique. With ample cross-referencing,... more... | 677.169 | 1 |
Questions About This Book?
The eBook copy of this book is not guaranteed to include any supplemental materials. Typically only the book itself is included.
Summary
A Concise Introduction to Pure Mathematics, Second Edition provides a robust bridge between high school and university mathematics, expanding upon basic topics in ways that will interest first-year students in mathematics and related fields and stimulate further study. Divided into 22 short chapters, this textbook offers a selection of exercises ranging from routine calculations to quite challenging problems.The author discusses real and complex numbers and explains how these concepts are applied in solving natural problems. He introduces topics in analysis, geometry, number theory, and combinatorics.What's New in the Second Edition:· Contains extra material concerning prime numbers, forming the basis for data encryption· Explores "Secret Codes" - one of today's most spectacular applications of pure mathematics· Discusses Permutations and their importance in many topics in discrete mathematicsThe textbook allows for the design of courses with various points of emphasis, because it can be divided into four fairly independent sections related to: an introduction to number systems and analysis; theory of the integers; an introduction to discrete mathematics; and functions, relations, and countability. | 677.169 | 1 |
College Algebra
Browse related Subjects ...
Read More do the math. This program will provide a better teaching and learning experience-for you and your students. Here's how: *Strategies for success are strategically placed throughout the text giving students guidance right when they needit. *Milestones along the way help students practice the skills and check their progress. *Tying it all together chapter review exercises help students review and synthesize the material as they prepare for the road ahead MyMathLab is an online homework, tutorial, and assessment product designed to personalize learning and improve results. With a wide range of interactive, engaging, and assignable activities, students are encouraged to actively learn and retain tough course concepts.
Read Less
Good. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780321867575-4Fine. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. Almost new condition. SKU: 9780321867575Hardcover. Instructor Edition: Same as student edition with additional notes or answers. New Condition. SKU: 9780321867575Fair. 0321916603Very Good. 0321916603 ANNOTATED INSTRUCTOR'S EDITION contains the COMPLETE STUDENT TEXT with some instructor comments or answers. May not include student CD or access code | 677.169 | 1 |
Advisory: Advisory: Demonstrated proficiency in English by placement as determined by score on the English placement test OR through an equivalent placement process OR completion of ESLL 25 & ESLL 249; UC credit for MATH 48A, B & C is limited to a maximum of 7.5 units for the combination or any portion of the series completed.
Students will develop conceptual understanding of rational, exponential, logarithmic, absolute value, composite, and piecewise-defined functions. They will demonstrate and communicate this understanding by graphing, analyzing, and transforming these functions and connecting their multiple representations.
Students will solve rational, exponential, and logarithmic equations.
Description -
This course is a continuation of topics from MATH 48A. Topics include rational, exponential and logarithmic functions, piecewise functions, combination and composition of functions and an introduction to trigonometry.
Homework Problems: Homework problems covering subject matter from text and related material ranging from 30 - 60 problems per week. Students will need to employ critical thinking in order to complete assignments.
Lecture: Five hours per week of lecture covering subject matter from text and related material. Reading and study of the textbook, related materials and notes.
Projects: Student projects covering subject matter from textbook and related materials. Projects will require students to discuss mathematical problems,write solutions in accurate mathematical language and notation and interpret mathematical solutions. Projects may require the use of a computer algebra system such as Mathematica or MATLAB.
Worksheets: Problems and activities covering the subject matter.
Such problems and activities will require students to think critically. Such worksheets may be completed both inside and/or outside of class. | 677.169 | 1 |
Haha, I wasn't having a dig (well, not a serious one ;-)). For anyone else who's curious, I wrote this on - I see a lot of titles copied directly from there that I wrote but short titles can't be copyrighted so it's OK (and it's a good sign people are reading!)
The TAs maintain an archive of solutions that's pretty complete. If you're interested in the solution to a particular problem from past revisions, I might be able to help you, or put you in contact with someone who can.
Bummer. :( I love learning by comparing my solution to THE solution (especially knowing it does exist), learn from my mistakes and also confirm I understood everything correctly. Can it be bought somewhere at least? Thanks for the info!
I don't know if they intend to make them available in the future. But when I took the class, many of the in class problems and some of the problems on the assignments were taken from this book and they always had very nice solutions available for the students afterwards.
I'm an undergraduate from the US, so I know absolutely nothing about international PhD admissions. I'd probably refer you to the admissions page for more information on that.
The average amount of math knowledge for ENTERING students actually seems to have a pretty high variance - there are people who come in as undergraduates with olympiad experience, and there are people who come in with very limited calculus experience.
However, students are brought up to speed very quickly; part of the reputation MIT has is that it filters out people who aren't willing to work hard to improve. I would imagine that even if your friend claims their math isn't "strong enough," they should still give it a shot.
I'm honestly probably not the right person to talk to, though - try looking through the resources here:
Well, I am wondering about the calculus for the most part. I just finished my undergraduate degree in CS, so my math isn't terrible, I feel confident in probability and statistics along with discrete structures and linear algebra. However, calculus, not so much.
If you have any recommendations for resources it would be much appreciated. For probability and statistics I enjoyed (and by enjoyed I really mean I freaking loved it) studying from Statistics for Engineering and the Sciences.
You typically deal with more discrete topics such as Combinatorics and Number Theory rather than Calculus in most areas of CS. Spivak's book for calculus is excellent and if you're looking for more of an analysis feel, look at Baby Rudin or Big Rudin.
Well, there's only a tiny bit of simple calculus in the OP's link – mostly discrete stuff in chapter 13.
Serious calculus (excluding basic summing and rate determination) tends to be limited to special fields within computer science: graphics, video encoding, modelling of real-world phenomena, etc. You can do a lot of computer science with zero calculus.
As for your question about textbooks though: I'm not sure I ever read a text that worked well for me. I don't think I'm a person that ever learned abstract subjects well from textbooks.
I took a double undergraduate degree (comp sci and control systems engineering). I did a lot of maths as part of the control systems so it's often pretty hard to remember what I actually studied for computer science – particularly when it comes to calculus (which I remember being entirely on the engineering side).
Frankly I don't think I learned much from my actual maths courses except how to bluff during exams. I ended up needing to learn everything when I reached my last two years of engineering and they expected a huge amount of maths. I cobbled it together by leaning on lab tutors, other students and banging my head on the table a lot.
I ended up in computer vision (which is a maths heavy field) but I don't think I'm "good" at maths [for my field – I'm sure that still puts me better than 99% of the population]. I can do what I need to do and I can read the research papers.
I learned that if you are good at mental calculations you often don't get credit. Twice I had to retake a test (without writing anything down) to demonstrate that's how I worked. I'm ADHD (rather severly evidently though this escaped everyone until my wife and I went for marriage counseling and the counselor asked us during the first session about it). Back from the tangent- so if I wrote anything down during calculations I'd completely lose my train of thought and collapse. Same thing with papers- write the whole thing in my head before putting anything on paper.
Like yourself (I was CS/EE which is close) I have never felt I was in any way "good" at maths and never enjoyed them in the slightest- it was just something you had to do to get through the courses. My son on the other hand loves math, excels at it and is completely uninterested in engineering because he doesn't like "science". I blame his high school.
Please don't say "CC licensed", tell people which CC license they use. "CC licensed" can mean anything from do-whatever-you-want, public domain (CC-0) to you-can-look-but-you-can't-touch, no derivatives and no commercial use (CC-nc-nd)
Keep things simple. The average reader doesn't want to feel they need to understand the alphabet soup behind every license. They just want to know whether they should feel guilty for not paying someone to read or have a copy of the thing.
I don't think the explicit license is confusing for the average reader of r/compsci, but maybe I'm wrong.
My point is, dumbing down the legal aspects of knowledge distribution has bad effects. For example, many universities distribute learning materials under nc-nd clauses, although they don't commercially exploit them in any way (I'm not saying that's the case of this particular document). In doing so, they harm the community as a whole, including themselves, by keeping these materials within a walled garden. These materials can't be re-used in any practical way, other than limited literal redistribution.
If "users" were more aware of what these clauses mean, they would probably ask their own organisations why they're using a restrictive license if there's really no need to. With enough pressure these licenses can be changed and we all win.
The only thing I've read before college was induction and proofs in the optional philosophy class but other than that there's pretty much nothing in there that was taught in high school. If everyone should know it by 11th there would be no need to the entire first year of college dedicated to it | 677.169 | 1 |
Welcome to Mathematics Stack Exchange
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's built and run by you as part of the Stack Exchange network of Q&A sites. With your help, we're working together to build a library of detailed answers to every question about mathA good book for beginning Group theory
up vote14down votefavorite2 Answers
up vote4down voteaccept
answered 20 hours ago
skayf 1113
up vote3down vote
Perhaps A Book of Abstract Algebra by Charles Pinter? It's very cheap, and if I remember correctly, provides motivation for the theorems/corollaries/etc. It's a small book, and there's a lot of stuff it doesn't cover, but I think it would prepare the reader well for more advanced treatments of algebra.
answered 21 hours ago
manthanomen 1,660218
Get answers to practical, detailed questions
Focus on questions about an actual problem you have faced. Include details about what you have tried and exactly what you are trying to do.A good book for beginning Group theory
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CALCULUS
OF
FINITE DIFFERENCES
BY
INTRODUCTION
BY
SECOND
EDITION
CHELSEA PUBLISHING COMPANY NEW YORK, N.Y. 1950
i / 4:
PHINTED IN THE IlNlTED STATES OF AMEHItX
INTRODUCTION
There is more than mere coincidence in the fact that the recent rapid growth in the theory and application of mathematical statistics has been accompanied by a revival in interest in the Calculus of Finite Differences. The reason for this phenomena is clear: the student of mathematical statistics must now regard the finite calculus as just as important a tool and prerequisite as the infinitesimal calculus. To my mind, the progress that has been made to date in the development of the finite calculus has been marked and stimulated by the appearence of four outstanding texts. The first of these was the treatise by George Boole that appeared in 1860. I do not me& by this to underestimate the valuable contributions of earlier writers on this subject or to overlook the elaborate work of Lacroix.' I merely wish to state that Boole was the first to present' this subject in a form best suited to the needs of student and teacher. The second milestone was the remarkable work of Narlund that appeared in 1924. This book presented the first rigorous treatment of the subject, and was written from the point of view of the mathematician rather .than the statistician. It was most oportune. ' Steffensen's Interpolation, the third of the four texts to which 'I have referred, presents an excellent treatment of one section of the Calculus of Finite Differences, namely interpolation and summation formulae, and merits the commendation of both mathematicians and statisticians. I do not hesitate to predict that the fourth of the texts that
1 Volume 3 of Trait6 du Calcul Diffdrentiel et du Calcul Int&pl, entitled Trait6 des diffbences et des shies. S. F, Lacroix, 1819.
vi
I have in mind, Professor Jordan's Calculus of Finite Differences, is destined to remain the classic treatment of this subject - especially for statisticians - for many years to come. Although an inspection of the table of contents reveals a coverage so extensive that the work of more than 600 pages might lead one at first to regard this book as an encyclopedia on the subject, yet a reading of any chapter of the text will impress the reader as a friendly lecture revealing an ununsual appreciation of both rigor and the computing technique so important to the statistician. The author has made a most thorough study of the literature that has appeared during the last two centuries on the calculus of finite differences and has not hesitated in resurrecting forgotten journal contributions and giving them the emphasis that his long experience indicates they deserve in this day of mathematical statistics. In a word, Professor Jordan's work is a most readable and detailed record of lectures on the Calculus of Finite Differences which will certainly appeal tremendously to the statistician and which could have been written only by one possessing a deep appreciation of mathematical statistics. Harry C. Carver.
'
THE AUTHOR'S PREFACE
This book, a result of nineteen years' lectures on the Calculus of Finite Differences, Probability, and Mathematical Statistics in the Budapest University of Technical and Economical Sciences, and based on the venerable works of Stirling, Euler and Boole, has been written especially for practical use, with the object of shortening and facilitating the labours of the Computer. With this aim in view, some of the old and neglected, though useful, methods have been utilized and further developed: as for instance Stirling's methods of summation, Boole's symbolical methods, and Laplace's method of Generating Functions, which last is especially helpful for the resolution of equations of partial differences, The great practical value of Newton's formula is shown; this is in general little appreciated by the Computer and the Statistician, who as a rule develop their functions in power series, although they are primarily concerned with the differences and sums of their functions, which in this case are hard to compute, but easy with the- use of Newton's formula. Even for interpolation -itis more advisable to employ Newton's expansion than to expand the function into a power series. The importance of Stirling's numbers in Mathematical Calculus has not yet been fully recognised, and they are seldom used. This is especially due to the fact that different authors have reintroduced them under different definitions and notations, often not knowing, or not mentioning, that they deal with the same numbers. Since Stirling's numbers are as important as Bernoulli's, or even more so, they should occupy a central position in the Calculus of Finite Differences, The demonstration of
I
"I.l
a great number of formulae is considerably shortened by using these numbers, and many new formulae are obtained by their aid; for instance, those which express differences by derivatives or vice versl; formulae for the operations x$X, and x& and many others; formulae for the inversion of certain sums, for changing the length of the interval of the differences, for summation of reciprocal powers, etc. In this book the functions especially useful in the Calculus of Finite Differences, such as the Factorial, the Binomial Coefficient, the Digamma and Trigamma Functions, and the Bernoulli and Euler Polynomials are fully treated. Moreover two species of polynomials, even more useful, analogous to those of Bernoulli and Euler, have been introduced; these are the Bernoulli polynomiali of the second kind (Q 89), and the Booze polynomials 6 113). Some new methods which permit great simplifications, will also be found, such as the method of interpolation without printed differences (Q 133), which reduces the cost and size of tables to a minimum. Though this formula has been especially deduced for Computers working with a calculating machine, it demands no more work of computation, even without this aid, than Everett's formula, which involves the use of the even differences. Of course, if a table contains both the odd and the even differences, then interpolation by Newton's formula is the shortest way, But there are very few tables which contain the first three differences, and hardly any with more than three, which would make the table too large and too expensive; moreover, the advantage of having the differences is not very . great, provided one works with a machine, as has been shown, even in the case of linear interpolation (3 133). So the printing of the differences may be considered as superfluous. The construction of Tables has been thoroughly treated (§(j 126 and 133). This was by no means superfluous, since nearly all the existing tables are much too large in comparison with the precision they afford. A table ought to be constructed from the point of view of the interpolation formula which is to be employed. Indeed, if the degree of the interpolation, and the
number of the decimals in the table, are given, then this determines the range or the interval of the table. But generally, as is shown, the range chosen is ten or twenty times too large, or the interval as much too small; and the table is therefore unnecessarily bulky. If the table were reduced to the proper dimensions, it would be easy and very useful to add another table for the inverse function. A method of approximation by aid of orthogonal polynomials, which greatly simplifies the operations, is given. Indeed, the orthogonal polynomials are used only temporarily, and the result obtained is expressed by Newton's formula (8 142), so that no tables are necessary for giving the numerical values of the orthogonal polynomials. In 3 143 an exceedingly simple method of graduation according to the principle of least squares is given, in which it is only necessary to compute certain "orthogonal" moments corresponding to the data. In the Chapter dealing with the numerical resolution of equations, stress has been laid on the rule of False Posifion, which, with the slight modification given (§$j 127 and 149, and Example 1, in 8 134), enables us to attain the required precision in a very few steps, so that it is preferable, for the Computer, to every other method, The Chapters on the Equations of Differences give only those methods which really lead to practical results. The Equations of Partial Differences have been especially considered. The method shown for the determination of the necessary initial conditions will be found very useful (8 181). The very seldom used, but advantageous, way of solving Equations of Partial Differences by Laplace's method of Generating -Functions has been dealt with and somewhat further developed (Q 183), and examples given. The neglected method of Fourier, Lagrange and Ellis (Q 184) has been treated in the same way. Some formulae of Mathematical Analysis are briefly mentioned, with the object of giving as far as possible everything necessary for the Computer. Unfamiliar notations, which make the reading of mathematical texts difficult and disagreeable, have been as far as
5(
possible avoided. The principal notations used are given on pp. xix-xxii. To obviate another difficulty of reading the works on Finite Differences, in which nearly every author uses other definitions and notations, these are given, for all the principal authors, in the respective paragraphs in the Bibliographical Notes. Though this book has been written as has been said above, especially for the use of the computer, nevertheless it may be considered as an introductory volume to Mathematical Statistics and to the Calculus of Probability. I owe a debt of gratitude to my friend and colleague Mr. A. &tics, Professor of Mathematics in the University of Budapest, who read the proofs and made many valuable suggestions; moreover to Mr. Philip Redmond, who kindly revised the text from the point of view of English.
CHAPTER I.
ON OPERATIONS. 8 1. Historical and Bibliographical Notes. The most important conception of Mathematical Analysis is that of the function. If to a given value of x a certain value of y correspond, we say that y is a function of the independent variable x. Two sorts of functions are to be distinguished. First, functions in which the variable x may take every possible value in a given interval; that is, the variable is continuous. These functions belong to the domain of Infinitesimal Calculus. Secondly, functions in which the variable x takes only the given values x09 x,, x2, - . . x,; then the variable is discontinuous. To such functions the methods of Infinitesimal Calculus are not applicable. The Calculus of Finite Differences deals especially with such functions, but it may be applied to both categories. The origin of this Calculus may be ascribed to Brook Taylor's Methodus ,Incrementorum (London , 1717), but the real founder of the theory was Jacob Stirling, who in his Methodus Differentialis (London , 1730) solved very advanced questions, and gave useful methods, introducing the famous Stirling numbers; these, though hitherto neglected, will form the backbone of the Calculus of Finite Differences. The first treatise on this Calculus is contained in Leonhardo EuZero, Institutiones Calculi Differentialis (Academiae Imperialis Scientiarum Petropolitanae, 1755. See also Opera Omnia, Series I. Vol. X. 1913) in which he was the first to introduce the symbol A for the differences, which is universally used now. From the early works on this subject the interesting article "Difference" in the Encyclopedic Mtthodique (Paris, 17841,
2 written by l'Abb6 Charles Bossut, should be mentioned, also, F. S. Lacroix's "Trait6 des differences et series" Paris, 1800. 9 2. Definition of the differences, A function f(x) is given for x f x,,, x,, x,, . . . , x,,. In the general case these values are not equidistant. To deal with such functions, the "Divided Differences" have been introduced, We shall see them later (8 9). Newton's general interpolation formula is based on these differences. The Calculus, when working with divided differences, is always complicated, The real advantages of the theory of Finite Differences are shown only if the values of the variable x are equidistant; that is if #i+l -xi = h where h is independent of i. In this case, the first difference of f(x) will be defined by the increment of f(x) corresponding to a given increment h of the variable x. Therefore, denoting the first difference by A we shall have Af(x) = f ( x + h ) - f ( x ) . The symbol A is not complete; in fact the independent variable and its increment should also be indicated. For instance -. thus:
a -
A 1. h
This must be done every time if there is any danger of a misunderstanding, and therefore A must be considered as an abbreviation of the symbol above. -----.
1 The most important treatises on the Calculus of Finite Differences are the following: George Boofe, A treatise on the Calculus of Finite Differences, Cambridge, 1860. A. A. Morkoff, Differenzenrechnung, Leipzig, 18%. D. Sefiuonoff, Lehrbuch der Differenzenrcchnung, Leipzig, 1904. E. T. Whitfaker and G. Robinson, Calculus of Observations, London, 1924. 'N. E. Niirlund. Diffcrenzenrechnung, Berlin, 1924. J. F. Sfeffensen, Interpolation, London, 1927. J. B. Scarborough, Numerical Mathematical Analysis, Baltimore, 1930. G. Kotuafewski, Interpolation und gen~herte Quadratur, Leipzig. 1932. L. M. Milne-Thomson. The Calculus of Finite Differences, London, 1933.
3 Often the independent variable is obvious, but not the increment; then we shall write A, omitting h only in the case h of h = 2. If the increment of x is equal to one, then the formulae of the Calculus are much simplified. Since it is always possible to introduce into the function f(x) a new variable whose increment is equal to one, we shall generally do so. For instance if y=f(x) and the increment of x is h, then we put x=a+hE; from this it follows that AE=l; that is, 6 will increase by one if x increases by h. Therefore, starting from f(x) we find f(x) = f(a+Zh) = F(i) and operate on F(S); putting finally into the results obtained (x-+)/h instead of E. We shall call second difference of f(x) the difference of its first difference. Denoting it by Azf (x) we have li)W = +[$f(x)l = ;f;~+hl -$f(x) =
Remark. In Infinitesimal Calculus the first derivative of a function f(x) generally denoted by df (4 or more briefly by 7, Df (x) (if there can be no misunderstanding), is given by y(x) Df(x) = lim f (x+hLv f(x) = lim - . /a=0 IsO h Moreover it is shown that the R -th derivative of f(x) is 9"'(x) Dnf(x) = l i m ___ h=O h" * If a function of continuous variable is given, we may determine the derivatives and the integral of the function by using the methods of Infinitesimal Calculus. From the point of
I
4
view of the Calculus of Finite Differences these functions are treated exactly in the same manner as those of a discontinuous variable; WC may determine the differences, and the sum of the function; but the increment h and the beginning of the intervals must be given, For instance, log x may be given by a table from x=1000 to x=10000, where h=Z. Generally we write the values of the function in the first column of a table, the first differences in the second column, the third, and so on. the second differences in If we begin the first column with f(u), then we shall write the first difference tf (a) in the line between f(a) and f (a+h) ; the second difference A'f(a)
h
will be put into the row between
N(a) a n d ?f(a+h) and so on. We have.
f(a) p'(a)
f (a+4
?" (al
I
I
+!(a-t2h) f(a+3h)
$f(a+h)
y'bl
f (a-t2h)
b'f(afh) +"f la-i- 4
(W4 $lf (4 pf(a-1 ;"f (a-I-2hJ j 11)
ef(a-tW
b'f (a-t 2h)
fb+W
+f(a+W f b+W
+'f (at-3h)
It should be noted that proceeding in this way, the expressions with the same argument are put in a descending line; and that the arguments in each horizontal line are decreasing. The reason is that the notation used above for the differences is nof symmetric with respect to the argument.
then according to our definition Qf(-x) From this it follows that A f ( - x ) = ---(-x-h) = f(-x-h) -f(-x). a:;; -*
wi .*,I r f E?&'!,).j G .i "'tl.,
qi~y~.Li"'t.-":
=(-',-yg(-Y) F1;-{1-+ k')"Lp'(-r..,,ni\) that is, the argumeit -x of the function is diminished by h. In the same manner we should obtain 4"' f-x) = (-l)mg/(-x-mhl;
y'(x)
=
p f ( x ) ,
This formula will be very useful in the following. Difference of a sum. It is easy to show that
A[f, (4 + f, (4 + ' - ' +f,WI = Af,(x) +Af,W +.-. +AfnM
moreover if C is a constant that = CAf(x). . According to these rules the difference of a polynomial
iS
ACf(x)
-
A[a,+a,x+a,x'+ . . . +anxn] = a,Ax + a&x' + . . , + anAx". $3 3, Operation of displacement.2 An important operation
2 Boole denoted this operation by D; but since D is now universally used as a symbol for derivation, it had to be changed. De la Valfhe Poussm in his Cours d'Analyse Infinitesimale [1922, Tome II, p. 3291 denoted this operation by Pseudodelta v. We have adopted here E since this is generally used in England. So for instance in W. F. Sheppard, Encyclopaedia Britannica, the ll-th edition, 1910, in VOC. Differences (Calculus of., .) Vol. VIII p. 223. E. k. Whitfder and G. Robinson, lot. cit. 1, p. 4. J. E. Steffken, lot. cit. 1. p. 4. L. M. Milne-Thomson. lot. cit. 1, p. 31. This operation has already been considered by L. F. A. Arbogd [Du calcul des derivations, Strasbourg, MOO]; he called it an operation of "6tat varie". F. Crrsoroti proposed for this operation the symbol @ which was also used b y Pjncherle [lot. c i t . 41. C. Jordan, in his Cours d'Analyse [Second edition tom. I, p, 1151 has also introduced this operation and deduced several formulae by aid of notation was symbolical methods. His f n = f(x + nh).
6
D =
Finite Differences by was introduced into the Calculus of Boo/e [lot. cit. 1. p, 161, the operation of displacement. This consists, f(x) being given, in increasing the variable x by h. Denoting the operation by E we have Ef(x) = f(x+h). This symbol must also be considered as 5~ abbreviation of ,E;. The operation E" will be defined by E"f(x) = E [ Ef(x)l = E f ( x + h ) = f (x+z?h) and in the same way E"f(x) = E [ En-l f ( x ) 1 = f(x+nh). It is easy to extend this operation to negative indices of E so that we have + f ( x ) = E-If(x) = f ( x - h )
0 5
& f ( x ) = E-nf(x) = f ( x - n h ) . $ 4. Operation of the Mean. The operation ofl the mean introduced by Sheppard (lot. cit. 2) corresponds to the system of Central Differences which we shall see later. We shall denote the operation corresponding to the system of differences considered in § 2, by M." Its definition is Mf(xl = ?hV(4 + f(x+h) I. \. The operation M will be defined in the same manner by Mn f(x) = M[Mn-' f(x) ] = 1/2 [Mn-l f(x) + Mn-l f(x+h) 1. Of course the notation M is an abbreviation of M. --Returning to our table of 5 2, we may write in?: the first column between f(a) and f(a+h) the number Mf (a); between
correspond to p. Thiele introduced for the central mean the symbol Cl, which has also been adopted by Sfeffensen [lot. cit. 1, p, lo]. Niirlund [lot. cit. 1. p. 311 denoted our mean by the symbol Pseudodelta V. This also h a s b e e n a d o p t e d b y Milne-Thomson [lot. c i t . 1, p. 311. W e h a v e s e e n that other authors have already used the symbol V for the operation of displacement, therefore it is not practical to use it for another operation.
1) 7
central mean by ,u. Since A corresponds to d it is logical that M should
3 Sheppard denoted the central difference by J and the corresponding
7
f(a+h and f(a+2h) the number Mf(a+h) and so on. In the second column we put MAf (a) between nf(a) and Af (a+h); continuing in this manner we shall obtain for instance the following lines of our table . . . , . .
The constant R may be considered as the symbol of multiplication by k; this symbol will obviously share the properties above mentioned. For instance we have: Ah f(x) = k A f(x) s Therefore we conclude that with respect to addition, subtraction and multiplication these symbols behave as if they were algebraic quantities. A polynomial formed of them represents united by an operation, Several such polynomials may be addition, subtraction, or multiplication, For instance
4 Symbolical methods were first applied by Boofe in his Treatise on Differential Equations, London, 1859 (third edition 1872, pp. 381-461 and in lot. cit. 1. p. 16). On the Calculus of Symbols there is a remarkable chapter in Steffensen's lot. cit. 1. p. 178-202., and in S. Pincherle, Equations et Operations Fonctionnelles, Encyclopedic des Sciences Mathematiques (French edition) 1912, Tome II, Vol. 5, pp. 1-81.
Therefore in the case of the displacement operation E, division, or multiplication with negative powers of E will be permitted, exactly as with positive powers. Moreover
EWp,WiWl = f(x+h)dx+hlHx+N = Ef(4 Ed4 Eq(xl
These are not true for the other symbols introduced. Q 6. Symbolical Methods, Starting from the definitions of the operations it is easy to see that their Symbols are connected, for instance by the following relations
Knowing the successive values of the function, this formula gives its m -th difference. We may write it as follows
' Our notation of the sums is somewhat different from the ordinary notation. We denote Ix i: f ( x ) .rd.l that is, the value I(0) corresponding to the lower !imit is included in the sum. but not the value f(n) corresponding to the upper limit. The reason for tbis notation will be fiiven in the paragraph dealing with sums. f(O) + f(1) -+ f ( 2 ) + if(n-1)
This formula becomes especially useful if f(x) is such a function that we have M*f (x) = F(x+r); since then we have Em-rMvf(x) = F(x+m) that is, the argument of F(x) is independent of V. This is often important. Example. Let f(x) = Ii I. As we shall see later, we have
and therefore
10 The function expressed by differences. From the first re. lation it follows that Em = (l+A)" 1n+1 m = 1 ), Av u=o ( I
Ex~(l+A)~=s I y ) Av v=o since the operation p performed on f(z) gives f(x) for z=O if h=l Tberefore this is the symbolical expression of Newton's formula (1) f(x) =
f(O) +(;I Af(o) + [;) A'fW +-a..
I " I
Hitherto we have only defined operations combined by polynomial relations (except in the case of E-n): therefore the above demonstration assumes that x is a positive integer. But the significance of the operation Ex is obvious for any value of x; indeed if h=l we always have Pf(0) = f ( x ) . To prove that formula (1) holds too for any values of x it is necessary to show that the operation E is identical with that corresponding to the series E: = l+(;jA+[;)Y+ . . . . +{;r]6'+ I..... This is obviously impossible if the series
f(O) + (~]AW + . . . . + 1;) A"fIo) +-a
/ I is divergent. On the other hand if f(x) is a polynomial of degree n then A*+I f(x) = 0 and the corresponding series is finite. Steffensen 110~. cit. 1. p. 1841 has. shown that in such cases the expansions are justified. If we limit the use of the symbolical expansions to these cases, their application becomes somewhat restricted; but as Steffensen remarked, these expansions are nevertheless of considerable use, since the form of an
1
2
interpolation or summation formula does not depend on whether the function is a polynomial or not (except the remainder term) If certain conditions are satisfied, the expansion of operation symbols into infinite series may be permitted even if the function to which the operations are applied, is not a polynomial when the corresponding series is convergent. [Pincherle lot. cit. 4.1 To obtain Newton's backward formula let us remark that E-A = 1 and start from
Expanding the denominator into an infinite series we get
According to what has been said, this formula is applicable, if f(x) is a polynomial, and first if x is a positive integer, in the general case if the series
Ex = +.!, ("t') &i
In the same manner we have, if f(x) is a polynomial or if it satisfies certain conditions:
+ = (1&n =
that is
4, (-')'(n+;-l)
Ay
f(x-nh) = so (-l)y (n+;-l J 4' f(x)
(1-44)x = 2; I--l)> (32'M'
Expansion of an alternate function. A function (-l)*f(x) defined for x = 0, 1, 2, 3, . . . is called an alternate function. Starting from the second formula we deduce
(-1)" p =
This formula may only he applied if x is a positive integer, since otherwise even in the case of polynomials it is divergent. Itzhas been mentioned that the Calculus of Finite Differences deals also with functions of a continuous variable. In this case it is possible to determine both the derivatives and the differences and to establish relations between these quantities. Relation between differences and derivatives. If we write Taylor's series in the following manner:
This formula was found by Lagrange; it gives the first difference in terms of the derivatives If we expand the second
14
member into a series of powers of hD, and multiply this series Cauchy's rule of multiplication we obtain by itself, and apply A'; multiplying again we get AS, and so on. We could express in this way the m -th difference by derivatives, but we will obtain this later in a shorter way by aid of Stirling numbers (5 67). Starting from ehD = 1 + A we could write formally h (4)
hD = log(l++)
In this form the second member has no meaning, but expanding it into a power series it will acquire one: hD = A - I/A2 + 'I3 Q:' - I/,++ + . . . . h h This formula gives the first derivative expressed by differences. It holds if the function is a polynomial, or if the series is convergent and satisfies certain conditions. Again applying Cauchy's rule we obtain D2, then D:', and so on. We could thus get the expression of the m -th derivative, but we shall determine it in another way (5 56). $j 7. Receding Differences. Some authors have introduced besides the differences considered in the preceding paragraph, called also advancing differences, others, the receding differences. defined by A'f (x) = f(x) - f (x-h). The symbol ~1' is that used by Sheppard&; in our notation this will be
A' = -&
a n d
(A')" = g
Since the formulae containing symbols of receding differences are very easily expressed by formulae of advancing differences, and since there is no advantage whatever in they are only introducing the receding differences, mentioned here.
" Sheppard Iloc. cit. 31; different notations have been proposed for the receding differences. Steffensen [lot. cit. 1, pp. 2241 puts v f(x) = f(x) - f(x-1).
15
8 8. Central Differences. If we accept the notation of the advancing or that of the receding differences, there will always
be a want of symmetry in the formulae obtained. Indeed, to have symmetrical formulae the argument of the difference f(x+hl -f(x) should be x+Mh; and therefore the difference corresponding to the argument x would be
f (x+ sh) - f (x-'Ah)
This has been adopted in the system called Central Differences. Different notations have been proposed; we will adopt Sheppard's notation, which is the following for the first central difference'
The connection between the formulae of the advancing and those of the central differences may be established easily by the calculus of symbols. Starting from
7 An excellent monograph on Central Differences is found in Sheppard ;Ioc. cit. 3. p, 2241. Of the other notations let us mention Joffe's [see Whittaker and Robinson lot. cit. 1. p. 36; A f(x) = f(x+ W) -1(x-%). The notation of the Central Mean introduced by Thiele and adopted by Sfeffensen [lot. cit. 1. p, 101 is the following Oflxl = %[f(x+%) + f(x-%)I*
16 we obtain
82n
= g ;
82n+l
= $2;
cL2"
= Bg
and so on. Working with central differences there is but one difficulty, since generally the function f(x) is given only for values such as f(x+nh) where n is a positive or negative integer. In such cases f (x+1/$) or f (x--l/zzh) has no meaning, nor have Sf (x) and pf (x). But there is no difficulty in determining ij2"f (x), b2*f (x) and pij2"+lf (x). Returning to our table of differences (Q 4) written in the advancing system, we find for instance the following line f(a+%h) MAf(a+h) A2f(a+h) MA3f ( a ) A"f(a) , . , I
If in this table we had used the central difference notation the numbers of the above line would have been denoted by f (a+2h) v&f (a+%) ljaf (a+2h) @ (a+2h)
6*f (a+2h) . . e
1
/
That is, the argument would have remained unchanged along the line. This is a simplification. Since in the notation of central differences the odd ones are meaningless, the difference
c\2n+1
= E11+'/,62n+l
,_*I g* /f'> asaL~Mh'~
must be expressed by aid of the even differences azm and by Pa 2m+l To obtain an expression for A we will start from the above symbolical expressions, writing
E6"
+
2Ep6
= A2 + 2 M A = A(A+E+l) = 2 A E
From this we get the important relations
(1) A=cLS +%8" and
A2n+1 = E"[v.cS"""
+ g&+2], 1
Symbolical methods. We have seen in Q 6 that E = ehD and A = ehD - 1. Putting h=o, to avoid mistakes in the hyperbolic formulae, we deduce from the definitions
i
17 WJ _ 1 h = &, = e e',aDD = e"W-e-'@'o and = 2 sinh r -wD
M l+E 1 +eoJD c1= E'"=m=-----= cash ',&ND. 2e'lPD
Starting from these formulae we may determine any symbolical central difference expression in terms of the derivatives, for instance i'jzn, t$jznt* and others. To obtain tjzn we use the expansion of (sinh x)')" into a power series. The simplest way of obtaining this series is to express first (sinh x)~~ by a sum of cash YX. Since 2 sinh x = e" - e-' we have (2 sinh x)2" = z r=o (-1)" I "," I e?ln-VI".
In the second member the terms corresponding to v and 2n-P combined give e'(n-l'l* + e-'(n-l,)~ = 2 cosh2(n--e)x so that we have (2sinhx)"n=2 ViU (-l)P[2F) cosh2(n-v)x+ ( - 1 ) " (2nn).
From this it follows that )DZm+l(2 sinh x)~")+,O = 0. For m>O (2) we find
Q 9. Divided differences. So far we have supposed that the function f(x) is given for x=x,, x,, x,, . . , x,, and that the interval xi+1 - xi = h deal with the general problem is independent of i. Now we where the system of x,,, x,, . . . . , x, may be anything whatever. By the first divided difference of f(xi), denoted by Ff (xi) the following quantity is understood: Zf (xi) = f (xi:! zf,'xJ I The second divided difference of f(xJ is F'f (x;) X 2f (Xi+,) - Bf (Xi)
Xi-2 - Xi
NOW we shall deduce an expression forf(x,) by aid of the divided differences. First multiplying sf (x,) by w. (x,,J, then B2f (x0) by o1 (x,) and so on: jDvf (x,) by wIW1 (x,,) and finally Bmf (x,,) by CO,,,-, (x,) we obtain
m+1 + = i=l
$1 v=i
cd"-1
D
(x,,,)
wy (Xi) f(xi)W
20 In consequence of the relation it follows that the coefficient of f(x,) in the preceding equation is equal to one, since it can be shown that 'y IL)*-1 (x,,,) l.=~ D toy (x,) = -' ' so that the coefficient of f(x,,) is equal to -1. Moreover it can be demonstrated that
VI+1 WY-l (&) ,zl D (tiv (Xi) = O if m ' i ' O
therefore for these values the term f(xi) will vanish from the equation and we have f M = f (x0) + h---x,) w (4 + brr-x0) (x,,,--xl) 6'f(X") + (4
We shall see in Q 123 that the remainder R,,,, of the series is equal to R mil = (x-x,,) (x-x,) ' * ' (x-x,n) D,,,+lf (6) (m+ip----where 5 is included in the interval of x, x0, x,, , . , ,x,,, . Formula (3) may be written f(x) = r(x,,) + !. o;(x) zi+'f (x,,) + ,mwr;,,', Dm+'fC1,
Q 10. Generating functions. One of the most useful methods of the Calculus of Finite Differences and of the Calculus of
21 Probability is that of the Generating Functions, found by Laplace and first published in his "Theorie Analytique des Probabilites" [Courtier, Paris 1812). Given a function f(x) for x=a, a+l, a+2,. . . ., b-l u(t) will be defined as the generating function of f(x) if the coefficient of tX in the expansion of u(t) is equal to f(x) in the interval a, b. and if moreover this coefficient is equal to zero for every other integer value of X. Therefore we have
u(f) = Ii
.c=a
f(x) f".
To denote that u(t) is the generating function of f(x) we shall write
Gf(x)
= u(f)
If b is infinite, the generating function of f(x) will be considered only for values of t for which the series is convergent. Remark. Another function q(f) may be determined so that the coefficient of fX shall be equal , in certain intervals, to the function f(x) given above. For instance, denoting by f,,,(x) the number of partitions of order m, of the number x (with repetition and permutation of 1, 2, .., n), it is easy to see that the generating function cf f,(x) is the following:
u(f) = (f+f'+f3 + , . , + f")"
The numbers B,, are Bernoulli numbers (5 78) e Second method. Starting from the generating functions obtained by the first method we may deduce, by derivation, integration, or other operations, new generating functions. Of course the conditions of convergence needed for these operations must be fulfilled. For instance starting from formula (1) we remark that the series X tx is uniformly convergent for every value of t such that 1t ) < r < 1. Therefore all derivative of the series, may be in the domain determined by term by term differentiation I t I < 1. The first derivation gives after multiplication by t 13.
(Iffy =,$ x t"
Secondly, the summation may be executed directly if it is possible to express the function f(x) by a definite integral in such a way that under the integral sign there figures the x -th power of some quantity independent of x. For instance if we have
then we shall have
In this manner we obtain u(t) in the form of a definite v integral whose value may be known. Example. Let f(x) = [cl. We shall see later that the binomial coefficients are given by Cauchy's formula (5), 8 22.
from this we conclude
This is already the required solution; but since in the
25 tables of Bierens de Haan we find that this integral is equal to 1 /y l-4t simpler form ; we may give the generating function this
1 u(f) = y li
l-4t
=
G(2,2).
Remark 1. Since ("z") = 2 ( $ 7 ; ' ) = 2 (2x2), we have (19) 2x-l
G( x 1=
2]L& / -
-
Remark II. Starting from the above result we may obtain the generating functions of many other expressions. For instance according to Cauchy's rule of multiplication of series we obtain from (18)
moreover [ u ( f ) ] ' = i&y = .r=o
2 4" f".
Hence noting that the coefficients of fx are the same in the two expressions we get
This is a very useful formula. $j 11. General rules to determine generating functions. 1. The generating function of the sum of functions is equal to the sum of the generating functions
The last formula anablcs us to dctcrmine tions of .\nlf(x) or of Mlprf(x) and so on, Example. We haw seen that
Let us d~tcrminc
G A
27
Determination of the generating function of t(x) starting from a difference equation. If the function f(x) is not directly given, but we know that it satisfies the following equation for x 2 0 14) a,f(x+n) + a ,,-, f(x+n-1) +. . .+ a,f(x+l) + a,,f(x) = V(x) where the ai are independent of x and V(x) is a given function of x then we shall call the expression (4) "a complete linear difference equation of order n with constant coefficients" If we denote the generating function of f(x) by u(t), the preceding rules (3) permit us to express the generating function of the first member of (4) by aid of u(t); if we know moreover the generating function R(t) of V(x), then, equaling the generating functions of both members we obtain an ordinary equation of the first degree in u(t) ; solving this we have finally the generating function of f(x). 15) u ( t ) = {PR(t) + z=i a,$-m/f(O) + f(l)t+. . .+
G(x+l)f(x+l) = Du(f) G&+2) Ix+lVIx+2) = D'W
and so on. By aid of these formulae we may determine the generating function of f(x), if f(x) is given by a linear difference equation whose coefficients are polynomials of x. We shall call the expression x(x-1) (x-2) . . , (x-m+l) 'factorial of degree m'and denote it by (x),. If we expand the coefficients of the difference equation into factorials, the equation may be written n+1 2, f(x+m) izo hi Ix)i = V(X). (8) ,,,=Cl By aid of the above relations we deduce from this equation a linear differential equation of n(f) which will determine the required generating function. Example. Given (2x+2)f (x+1) - (2xfl)f (x) = 0. This we may write 2(x+l)f(x+l) -2xf(x) - f ( x ) = 0 hence
2Du(f) - 2fDu(f) -u(f) = 0
that is 2(I-f)Du(f) = u(f). The solution of this differential equation is easily obtained by integration. We find u [t] = c (1 - 2)-M
29 To determine the constant c let us remark that u(0) = f(0) therefore u (t> =f(O) [l - t]-h § 12. Expansion of functions into power series. If the function f(x) is unknown but we have determined its generating function u(t), then to obtain f(x) we have to expand u(f) into a power series. In the Calculus of Finite Differences this occurs for instance when solving difference equations by the method of generating functions. In the Calculus of Probability very often it is much easier to determine the generating function of a quantity than the quantity itself. In these cases also we have to expand the obtained generating function. The methods for expanding functions into power series are found in the treatises for Infinitesimal Calculus. Here only the most useful will be given. Firsf method. Expansion by division. Example: - 1 l-f 1=+ f + f' -+ . . . + f" + . . .
Putting h=O every term will vanish except that in which s+i=n. Therefore we find
yns = $ ,
1 $i (pT]r=o = $ ( *Ss) (2t)2s-".
Example 3. Let y = l/t. We have
Proceeding in the same manner as in the preceding example we get
Deiivation of a function of function by Faa Bruno's formula. Given u=u(y) and y=y(x), let us write
Y
= ~0 + (x--x,) DY, + (x-x,)' q + b-%I3 D3u, + -jr
and
+....
*
=
,~o~[(x-x,,)Dy+~x-xo)~~
f.,. '-dl'u dg" y=#O
The coefficient of (x-xo)n in the expansion of u is equal first to
II 1
l
34 and secondly in consequence of the above equation to
I:
,y! , (Dy,,)"' ( q)"'. , , . p, , (f2, . . . a,.
.,,,
so that finally we may write
'I =.vo
. .
where and
ff,
x a,.(R!. cl,,', (DY,,)"I ,.
D"y,, an n! 1
c
4
+UL'$-".+ff,,
II .
. a, + 2~2, + . , . + ncf,, = R.
Remark. Comparing formulae (8) and (9) we find
(IO) -& I& [Ay(~)jj=~ = x ,I,! "! , . , . 'f,,.
(DY,,)
'I (F )"" '
D . , , Rn "vo
( n! 1 -
0 13. Expansion of functions by aid of decomposition into partial fractions. Let the following function be given
where u(t) and t/l(t) are polynomials; the degree of (r)(t) being less than that of ?/p(t). We may always suppose that ,, (t) and (,1(f) have no roots in common; since if they had, it would always be possible to simplify the fraction, dividing by f-r,,, , if r,,, i s . the root. A. Let us suppose that the roots t,, r3, , , , , r,, of t/*(f) are all real and unequal. We have u= tt+l ai z a=, f--c
35 Reducing the fractions to a common denominator we obtain u if for every value of t we have
*+1 44 Y(t) = iz, Qi-ri
therefore this is an identity: so that the coefficients of P', in both members, must be equal, This gives R equations of the first degree which determine the coefficients oi. But these may be determined in a shorter way. Indeed y(t) = c(t-r,) (t-r,) . , . . (t-r*); therefore putting f=ri into the above equation every term will vanish except that of oi and we shall have
v,(t) Y (4 = ai lim t--r = ai D+(fi): I
so that
and therefore
Finally the coefficient f(x) of tX in the expansion of u will be equal to
We could have obtained the same result by solving the difference equation deduced in Sfirling's method (Method II, 6 12). On the other hand the third method giving the expansion of the trinomial would lead to Putting (2) 2 $ 2: (;) t"+"+'
B. The rootsr,, t,, ,.,., r, of y(t) are all single, but there The coefficients of y(t) are complex roots among them.g are real: therefore if rl = a+@ is a root of y(t) = 0, rs = a-pi will be a root too. The preceding method is still applicable but the corresponding values of a, 'and a, will be complex conjugate:
+ - 2a(Ga+Hb) X-P. , s # , .. az+p2 -( P )I
C. The roots rr , r2, . . . , r, of the polynomial q(f) are all real, but there are multiple roots. Let us suppose that the multiplicity of the root T,, is m,..; where m,. may be equal td 1, 2, 3, * * . , and so on. The general formula of decomposition into partial fractions is the following:
If we multiply this expression by (f-ry)m~ we get (4) 233 (f---ry)Nlt, = t&l (f--fTy%-l + . . . a,em,, + R(f) , (f-r,)"'v WI We have l)(f) = C(f-rp (f--r2)m~*, . , (f-r")% Hence if we denote the first member of (4) by A,.(t), A,(r,) will be different from zero: A"(f") = am, In the same manner the first m,,--s derivatives give (5) D"~-~dv(rJ = (m,-s)! (I,*.
This operation repeated for every value of v will give the required coefficients ova , Finally we shall have
n+: Q (4 * = ,s,
This may be written
A!L=
WV) n+l
x
"i+' (-l)s
s=l
D'"u-~d~(r,,) ! TVS
r=l
Im 7-s)
39 Therefore the coefficient f(x) of tX in the expansion of this expression will be f(x) = z w=l
n+l m,+1
s Dm, -'A, (rd ,zl (-l) (m 4) ! f yz+s ("+:-')
D. Multiple complex roots. Let us suppose that the roots u+/?i and a-pi are of the multiplicity m. We may write the fraction in the following manner
The last two equations-permit US to determine om and h. But the second one is. superfluous; indeed knowing that a, and b, are real, by equating first the real parts in both members of (7), and then the imaginary parts we get the two necessary equations. The first derivative of A(t) for t=a+/% JS
To obtain c we multiply A(f) by (f-l) and put f=l, getting 1~16~. If the coefficients as and b, are calculated we may expand each term a,f + bs
[v-d2+B21s
using Method IIIfor the expansion of a trinomial. Here we have to put in formula (6) of Q 12 R= -s, c,=l, c, = -2a and co = a2+p2* 8 14. Expansion of functions by aid of complex integrals. If the function u(f), of the complex variable f, has no poles or
41 other singularities in the interior or on the boundary of the circle of. radius p, then f(x), the coefficient of t" in the expansion of u(f), is given, according to a known theorem of Cauchy, by the following integral taken round this circle:
Remarking that the coefficient of tx in the expansion of u(t) is equal, according to Newton's rule, to 2 , and moreover 0 that the above integral taken between the limits 0 and l/in is equal to the half of that taken between the limits 0, z, we conclude that (i) = Gk""-snacos(n-2x)ada. This is Cauchy's formula expressing the binomial coefficient by a definite integral. Q 15. Expansion of a function by aid of difference equations. If'a function of the following form is given, where R(t) is a function expansible into a power series,
then u(f) may be considered as being the generating function of a function f(x) determined by the difference equation f2) U,,f(X+n) + a,,-,f(X+n--1) + , , . f a,f(x+l) + aof = V(x).
e +
Indeed, we have seen (formula 5. Q 11) that the generating function of f(x) deduced by aid of the difference equation (2) is P&f) (3)
u(f) =
+ 5 tlk=l w=o
n-i-1 z
ix0
5
a, f(i) fnemti
a, fn-m
I
where R(f) is the generating function of V(X); that is, V(x) is the coefficient of fX in the expansion of R(f). Knowing V(x), let us suppose that we are able to solve the difference equation (2) by a method which does not require the expansion of the generating function (3). If we find f(x) = = @(x7 Cl, cy * * *, c,,) where the c are arbitrary constants, then we shall determine them by identification of (1) and (3), and thus obtain the n equations:
nt1
(4)
a
,
= z I>'=1
a,f.(x+m-rz).
I
Putting in the corresponding values of f(x) we may determine the n arbitrary constants a,,, , and obtain in this way the required coefficient f(x) of fx in the expansion of u(f). Example 1. The function u(f) = f / (8f2-6f+l) is to be expanded into a series of powers of f. According to what has been said, if we denote by f(x) the coefficient of fX in this expansion then f(x) will be the solution of the difference equation f (x+2) - 6f(x+l) + 8f(x) = 0. We shall see in 3 165 that the solution of this equation is f ( x ) = ~~4~+c~2*.
By the aid of these values we obtain from (6) step by step f(3) = 6, f(4) ~36; f(5) = 216---8 = 208, f(6) = 1248-48=1200 and so on. R e m a r k . This method is identical with that of Stirhg (Method II, Q 12). Given the function
Q 16. The Factorial. In Infinitesimal Calculus the simplest function.is the power. Its derivative is very simple. Inded we have D x" = nx"'. On the other hand the difference of a power is complicated; we have 2x"=
Therefore powers and power-series will be less useful in the Calculus of Finite Differences than in Infinitesimal Calculus. But there are other functions whose differences are as simple as the derivatives of x". These are the products of equidistant factors, for instance the function 'x(x-l) (x-2) . . . . (x-n+l) = (x), called factorial of degree n, which we denoted by (x), , or the function (1)
tx)n,h *
0
n 1
x"-l h + (;) x"-~ h2 + . . . . + hn,
x(x-h) (x-2h) , . ,
(x--nh+h)
=
(x), h
called the generalised jactotial of degree n and denoted by Of course the above definitions are valid only for positive integer values of n. It was Stirling who first recognised the importance of the factorial, but he did not use any special
46 notation for it. The first notation is due to Vandermondel" who was the first to extend the definition of the factorial to any value of n. To do this he proceeded in the following way: If n and m are positive integers such that n > m then it follows from the definition (1) that (2) (Xl ", h = (x)rn, h (x-mh)n-m, h . This formula will be considered valid for any value of n and m. Putting m=O, gives that is (4
n,h = (X)o,h(&,h
Remark. Several authors have introduced special notations for x(x-h) (x-2h) , , , (x-nh+h) and for x(x+h) (x+2h) . . , (x+nh-h) but this is needless, since both factorials can be expressed by the same notation, or&y the argument being different; for instance, the above quantities would be, in our notation, and (x+-nh--h) ,,. I~ . (XL II One should always use the fewest possible notations, since too many of them, especially new ones, make the reading of mathematical texts difficult and disagreeable. Particular cases. If h=l and x=0 we have (0)-n, = 1. 2 .,.. . 31 m
In this manner the definition is extended to negative integer values of the index and to zero.
48 It is easy to see that (x+l)n = Eom this it follows that (x+2)" = (4, + 24x),-, + (n)*(xL and so on;
Ix+m)n = *:I (~)(m)i(x)n-i-
(x), +
ww*
This formula, due to Vandermonde, may be deduced from formula (14) Q 22. Putting into it x+n, instead of x and multiplying both members by (x), we find
CQZZChy'S
(x+m+n).W., = x [?)(m)iIxl-n(x+nl.j
but in consequence of (2) we have
(Xl-*(X+n)nj = (X)-i ;
therefore (x+m+n).(x)-, = 'zi I:] (mli(4-i.
This formula is symmetrical with respect to n and m, and may be written in the following way: * (4
(YL(xL
= :$I (r 1 M-i b--x----n)i. considered valid even for fracby an infinite product in the into formula (2) instead of JL get
The above formula can be tional values of n. We may express (y) "(xl-,, following manner. Putting Y h=l and m=r, we and
This is W&s's well-known formula. Since from formula (2) it follows that hence ( ?4? 1'18 = P//2) 16--%) -I/¶
(--'/2) A,* = p* We shall see in 8 17 that (x), = r(x+l) so that the above quantity is equal to r(l/,), moreover that I'(l/i) = 1 'T Application 2. Putting y-l instead of y and v=l into formula (y) we have according to (1) 6 24 J = X
j*fx-I 0
Differences of the factorial with negative arguments. It has been shown in Q 2 that the differences of a function with negative arguments are not of the same form as the differences of the same function with positive arguments, Let us determine for instance the differences of the following generalised factorial: (-x)~, ,, , The simplest way is, first to transform this factorial into another with positive argument and then apply the above rule. Since
(-x)n.h = (--l)n(x-l-nh--h),.h.
we have
~(-x)n,h = (--l)"hm(n),(x+nh--hl.-,,h.
transNow the factorial in the second member may be formed into one with a negative argument, so that finally we have ~(---x)~.I, = (-l)mhm(n),(--x-rnh),,.h.
'&at is,in the m -th difference the argument is diminished by mh. In the particular case of h=l, this formula gives (5)
Am(-x).
= (-l)m(n),(-x-m)n-~.
52 Differences of the reciprocal factorial. The differences of (xh, = l/(X$nh)n.h are determined in the usual manner;
M(X),, = '/?I (x+1);, + W,,) = '/2(4,,-,[2~+2--nJ.
The higher means are complicated. Computation of factorials. If in an expansion x, (x) 2, (x):, , (x),, are needed, to calculate them it is best to multiply first x' by ( x - l ) , then to multiply the result (x):. by (x-2) and so on. If x, x1, . . . , x" were wanted we should proceed' also by multiplication. The amount of work is in both cases nearly the same. If only (x),, is required, and x is an integer, as it generally will be, then we may obtain log(x),, = logx! -log(x--n)! using Dude's Tables; I1 and the computation will not be longer than that of logx".
n=l 'I F . J. Dunrfe, Nou\elles A n=3000. C;cntw. 1927. T a b l e s d e l o g n ! i 3 3 dkimales d e p u i s
53 Q 17. The Gamma-Function. The factorial may be considered as a function of the< index. Then in (n), the independent variable is x. The differences of this function with respect to x are not simple; since
A(n), = (&[n--x--11Hence the higher differences are complicated. This function is seldom considered in the generdl case; on the other hand in the particular case, if R=X and x is an integer, we get a very important function, (x), = 1. 2. 3. , . . x. In Kramp's notation [lot. cit. 101 this function would be lx". Considering its importance, shorter notations were introduced. For instance Camp used later (in 1808) 1 ,2 ,3 . . . . . x = x! We will adopt this notation, which is the most in use to-day. In England in the past the notation h was often used for 1.2.3, , , X. The first difference of x! is simple: Ax! = x.x! but the higher differences are complicated. The gamma function denoted by I'(x) is given for ;r> 0 by the following definite integral: (1) r(x) = ie-f
0 P-1 d t
It may be considered, we shall see, as a generalisation of (x--lLl. From formula (1) it follows that r(l) ~1, and by integration by parts we get r(x) = ( x - l ) Jrn em'
0
54 may be considered as a characteristic property of the gamma function. If x is an integer, then the solution of (21. is given by T(x) = ( x - l ) ! Therefore I'(x+l) may be considered as ageneralisation of x! Another definition of the gamma function may be had starting from the definite integral below (beta-function) in which n is a positive integer. By successive integration by parts we get
1
n-
i
u"-' (1--u)"du
=
(xJ;,
II+ I
let us put now u = t/n, then
If n increases indefinitely, we have
n, some caution must be used" but here the proceeding is
To determine the limit of an integral containing a parameter
justified, and we shall have (4)
I'(x) = lim n! ny
,I-.'2-
(x-i-n),,-,
valid for every value of x. If x is a integer, then formula (4) may be deduced from formula (2), § 16, indeed from the latter it follows that
n! nX (x+n),,, , ZIZ nX (n),T,.-.x Ix-l).,-, (xl-n), (n),,- ,--I
since
- - = I a n d !~~ (x;;1,z
(~---l)~...,
= f ( x ) .
Hence formula (4) is demonstrated. A third definition of the gamma-function is given by' '
1" See Hobson, Functions of a Real Variable, Cambridge. 1907, p. 599 ':I The demonstration of this formula may be found. for instance, in E . Artin. Einfiihrung i n d i e Theorie d e r Gammafunhtir,n, 1931. Herl!n p 1 4
55
where C is Euler's constant. The formula holds for every real or complex value of x. Gauss has given a multiplication formula for the gamma function: (6)
.
I.(*)+++] r(z ++) . ..r(z+q] = = (2n)%C-11 r(nz). n""-'IZ
A demonstration is given in Arfin 110~. cit. 13. p. 181. Owing to Euler's formula, the values of the gamma function corresponding to negative arguments may be expressed by those corresponding to positive arguments: (7) r(-x) =% sinn x r(x+l) *
Computation of r(z). Equation (2) gives r(z) = (z-l)F(z-1) = (z-l),F(z-2) = (z-l)" T(z-n). Therefore if z-l > n > z-2, the argument I-n=x of the gamma function will be such that 1 < x < 2. Consequently it is sufficient to have tables for these values. The best tables are those of Legendre giving logr (x) from x=1*000 to x=2'000 and the corresponding first three differences, to twelve figures.'* Particular values of the gamma function. If x is a positive integer, then we have seen that r(x) = (x-l)! ; moreover from (8) it follows that
14 A. M. Legendre, Tables of the logarithms of the Complete r-function to twelve figures. Tracts for computers. Cambridge University Press. 1921,
For n = 0 (9) will give (x), = 1; if n is a negative integer n=-m, from (9) we obtain
wn~ = (x+1)
rx+2; . . ,
(x+mj
= (xim),:!
conformably to our definitions of $j 16. Moreover, the extension of the definition of the factorial by gamma functions conforms to the extension by formula (2) of 8 16; indeed, putting the values (9) into this formula, we find the following identity:
Pearson published a double-entry table giving the values of the function Z(u,p), to seven decimals, from p = -1 to p = 50
57 (where Ap=O'2), for the necessary values of u, (where Az~=o'l).~~ The table also contains several formulae enabling us to calculate Z(u,p) corresponding to values of u and p outside the range of the table. Equation (1) gives by integration by parts
From this we conclude that
putting (4) we may write
The function $(m,x) has also been tabulated. There are for instance tables of this function in Pearson's Tables for Statisticians and Biometricians, [vol. I, pp. 115-121.1 from m=O'l to m=15 (where Am=O'l) for all the necessary values of x (if Ax=l). It can be shown by the tables of the function Z(u,p) that for U= vp o r m=Vp(p+l) w e h a v e ~(v$,p) < 1/2 a n d f o r - u= 1J p+l or m=p+l we have I( \,m,p) > 1/2. If p is an integer, by repeated integration by parts we obtain from (1) (6) Z(u,p) = 1 - 2: 5 e-m = 1 - Lzo + (m,x).
,
Remark. From this we may deduce the sum of a first section of the series em; indeed
P+l ms
8 19. The Diamma Function. Pairman called the derivative of log r(x+l) digamma function and denoted it by F(x). (Digamma is an obsolete letter of the Greek alphabet.)'O To obtain the derivative of log r (x+1) we shall start from formula (4) of Q .17. It will give
logr(x+l) = lim [log n! .f
"Em (X+1) log R - I: v=l log (X+v)].
It can be shown that the derivative of the second member may be obtained by derivation term by term, moreover that after derivation the second member tends to a limit if R increases indefinitely; therefore we shall have n+z 1 F ( x ) = Dlogr (x+1) = l i m [logn- Z1 x+v I. (2) ?t=OJ From this we deduce (3) F(0)=l&[lognE] = -c
where C is Euler's constant. It may be computed by formula (3) as exactly as required. We find C = 0'57721 56649 01532 86060 65120 90082, A function denoted by t/v(x), which differs but little from the digamma function, has already been considered, first by Legendre in 1809 and later by Poisson and Gauss. (See N. Nielsen, Handbuch der Gammafunktion, p. 15). We have q(x) = F(x-1). First difference of the digamma function: A F (x) = DAlogr(x+l) = D log ;;;$# = Dlog (x+1) iz 1 x+1 *
1s E. Pairman, Tables of the Digamma- and Trigamma-Functions. Tracts for Computers, Cambridge University Press, 1919.
i
59
Therefore we have F (1) = F (0) + 1 and if n is an integer F(n) = f (0)+1+'/z+%+~..+~. Hence F (x-l) is a function whose difference is equal to 1,/x; it is analogous to logx, whose derivative is l/x. The higher differences of the digamma function are easily obtained by aid of formula (7) Q 16; we find
60 x+1 instead of x, Putting in term is justified, we get
F(x) =-&- c+
since derivation term by
5 1, l r=, I y x+l+Y
This may be written more simply (7)
1
-
F (x) =
Computation of the digamma function. Pairman's Tables quoted above give the values of the digamma function and its central differences a2 and a4, to eight decimals, from x=0%0 to x=20%0 where Ax=o'O2. If F (I) is needed, and 2>20, then we will use formula (6) and put z=nx so as to have 2 -= x < 20. It may be useful to remark that the logarithm f?guring in this formula is a Napier's logarithm. 8 20. The Trigamma Function. Pairman denoted the second derivative of log F (x+1) by F (x), calling it trigamma function.
61 The higher differences are complicated. From this we obtain F (1) = F (0) - 1 and if R is an integer F(n) = F(o)++-+ . . . -$. The values of negafive arguments arguments. Starting tion with respect to the trigamma function corresponding to may be expressed by those of positive from formula (5) f$ 19, we obtain by derivax .
Compufafion of the frigamma function. In Pairman's Tables we find F (I) and its central differences b' and V to eight decimals from r=O'OO to 2=2000 (where Az=o'O2). If z > 20 we shall use formula (3) putting z=nx so as to have+=x<20.
$ 21. Expansion of 10g r (x+1) into a power series.
If 1x I c 1 then we shall have l o g I' (x+1) = 5 qlDmlogf' (x+1) ],ti.
d m.
We have seen in 8 20 that
If 1 > I xl this series is uniformly convergent, and the derivative of the second member may be obtained term by term, so that we get for m>l Dm l o g r (x+1) = % (-';;E;l) 1 I.=1 and therefore
The series is convergent, but the convergence is very slow. 8 22. The Binomial Coefficient. In Infinitesimal Calculus sequences of functions which satisfy the condition (al D fn(x) = fn-1 (x) are very important. Such sequences are for instance
fn-1 I-4 = c,, (nz;) , + c, &!) I + . . . . . + c ,L,
from this we conclude that the coefficients ci of the polynomial (p) are independent of the degree R of the polynomial, so that they may be determined once for all. This is a great simplification. If a function F(x) is expanded into a series of f,(x) funct i o n s (u) F(x) = 2, C"f"W then we may easily determine the integral and the derivatives of this function. Indeed we have on the hypothesis that these operations are permited term by term s F(x) dx = and h + 9, C"f",l(Xl
DmF(x)
= !i
c,,f,-m(x).
64 In the same manner, in the Calculus of Finite Differences the sequences satisfying
Afnb) = fn-, (4 (Y) are important. Such sequences are, for instance, the binomial coefficients x(x-1) (x-2) . . . (x-n+l) = x f"(X) = 1.2.3..:..n n (I and a n d f,(x) = 5',(x) f,(x) = 9"(X)
where q,,(x) is the Bernoulli polynomial of the second kind of degree n, and 5,(x) the Booze polynomial. Equally important are the sequences which satisfy the condi tion
Af,(x) = In+, (4.
Such a sequence is for instance -
..,..(n-I) ffr(x) = (~!+:~~~:2~(x3+3). , . (x+n) -
If f,,(x) is a polynomial of degree n then, expanding it in the following manner,
Therefore we conclude that the coefficients ci of the polynomials (8) are independent of the degree R of the polynomial; they may be determined once for all, which is a great simplification. If a function F(x) is expanded into a series of such functions FIX) = m; c,f,,Ix)
and moreover as we shall see ($32) also the indefinite sum o f F(X]. The binomial coefficient is without doubt the most important function of the Calculus of Finite Differences, hence it is necessary to adopt some brief notation for this function. We accepted above the notation of J. L. Raabe [Journal fiir reine und angewandte Mathematik 1851, Vol. 42, p. 350.] which is most in use now, putting
X = x(x-1) (x-2) . . . (x-n+l) 1.2.3....n 0n The binomial coefficients corresponding to integer values of x and n have been considered long ago. Pascal's "Triangle Arithmetique", printed in 1654, is formed by these numbers; but they had been published already a century earlier in Nicolo Tartaglia's "General Trattato di Numeri i Misure" (Vinegia 1556, Parte II, p, 70, 72). In Chu Shih-chieh's treatise ,,Szu-yuen Yii-chien" (The Precious Mirror of the Four Elements), published in 1303, they are indicated as an old invention, Omar Khayyam of Nishapur (d. 1123) knew them already in the eleventh century; and this is our earliest reference for the subject. Omar Khayyam's fame as a poet and philosopher seems to have thrown his eminence in mathematics and astronomy into the shade; nevertheless it must be recorded that these sciences owe much to him. [ Woepcke, L'Algebre d'Omar Alkhylmi, Paris 1851.1 In those early times no mathematical notation was used for these numbers. It seems that the first notation used was that of
n = (xd.. This notation, though seldom used, would be the best; indeed, if x and n are complicated expressions, for instance fractions, then in all other notations the printing is very difficult; moreover, the formulae lose their clearness.
l
0 0
X
=
X(n)
*
was considered first as 0 the coefficient of f" in the expatsion of (1 +f)x, for instance by
The binomial coefficient
X
67 Tartaglia, and later, as the number of combinations of x elements taken n by n by Pascal. The definition (1) of the binomial coefficient given above may be extended to every value of x, provided that n be a positive integer. From (1) we get, putting x=n,
For x=0 we have
Moreover, if x is a positive integer smaller than n, from (1) it follows that
On the other hand, if x is a positive integer larger than n, we deduce (2)
II
X
n
=
0.
Putting into (1) x = --z we obtain (y) = ( - 1 ) " (r+;-1) -2 is equal to the number of combinations with repetition of z IR1 elements taken n by R. We may extend the definition of the binomial coeffigient x In ) to every real or complex value of x and n, by writing (4) Remark. If z is a positive integer, then the absolute value of
II
X
n
-
- n!(n-x).! =
-
I X.
x
xt - n
1
It is easy,to show that, in the cases considered before, the definition (4) leads to the same results as definition (1). Moreover, putting n = 0 into formula (4) we have
0
X
--J&=
I'bfl)
n
Wrt
r(n+l) r(x-n+l)
'
68
This is true even if x is also equal to zero If n is a negative integer and if x is not a negative integer, putting n = -m we get
since then r (l-m) = 00 , If x and R are negative integers, then putting into (4) x = --I and n = --m we find
From this we conclude, z and m being positive- integers, that if z>m or if z=O, then
and if z=m
I I 0,
-m =
-2
I - m1
= 1.
0
-
-
0
I 1
- m - m
Besfdes from (4) it follows that n = 1 for every value of n. I 1 The definition of the binomial Zefficient may be extended also by Ccwchy's formula. (See Q 14, and N. Nielsen, Gammafunction p, 158.) If x > -1 then
co+ p cos(x-2n)q dp, =
F(n+l) r(x-n+=
Ux+ll
69 Differences of the binomial coefficient with respect to the upper index. If n is a positive integer, we have (6)
A[;] = I";') -(;) = (Ll] [ x+l-;+n-l] =
and therefore
Formula (6) may be written (7) This gives a very good rule for computing step by step a table of the numbers x , II Equation (7) shozs that the numbers
The general solution of this equation is given later (8 182 and Q 183) where it is shown that starting from .the initial conditions f (0,O) = 1 and f (n,O) = 0 if n>O or 60, we find that f(n,x) is equal to the coefficient of t" in the expansion of (1 +t1=* Differences of the function with negative argument. we have seen that
(ix) = (-1)" (x+n"-'1;
therefore we have
and moreover
70 Let us remark that, if the argument of the binomial coefficient is negative, it is diminished by one in the difference. This will be useful later. Differences of the reciprocal binomial coefficient. Since this latter may be written
1 -= I% = n! (x-n)-, n 0
X
we may apply formula (6) $j 16, giving the differences of a reciprocal factorial. We find
Ad- = nl Am(x-n), = n!(-n),,, (--1l*n-~ n = II
X
(x-n)+m =
General&ad binomial coefficients. We will denote the generalised binomial coefficient of degree n by is = x(x-h) (x-2h) . . , (x-nh+h) , 1.2.3.....n To determine the differences of this function in a system in which the increment of x is equal to h, rhat is Ax=h, we will write
and applying our formula giving the difference of the generalised factorial (4) 8 16, we find
and in the same manner
lnx * nsA(1 - hm I n-xm I .
1
71 We have
therefore the higher means will be complicated.
to X. We have seen [formula 14, 8 lo] that this generating function is:
Generating function of the binomial coefficient with respect
Starting from this expression we may deduce a forniula analogous to that of Cauchy given below (14). Obviously we have
-Fy)
GI 1
X
n =
(&;"+I =
x p. t n II
5 g x g p+Y = I II I
n m
tn+m (1~f)n+m+**
Putting x+y=z and noting that the coefficient of t' in both members is the same, we have
In the sum of the first member every combination of the numbers x,, x2 and x3, with repetition and permutation, must be taken so that x, + x, + x3 = 2.
12 Continuing in the same way, we should obtain a still more general formula:
where
n,+n,+ , . . +n, = n
and
x1+x2+ , . . +xm = 1.
In the sum of the first member every combination of the numbers x1, x2, , . . , xm should be taken, with permutation and repetition, satisfying the above equation. The derivatives and the integral of the binomial coefficient will be deduced later by aid of the Stirling numbers of the first kind. Here only the results are mentioned.
Dm (;] = -$ z (Y),
JI
Xv-':s:
; dx
1
We shall see later that this integral may be expressed by
the Bernoulli polynomial of the second kind of degree n+l,
that is by %+l(x)
The binomial coefficient may be considered as a function
J( ; 1 d x = %+,(x1
+ k.
.
07 the lower index. Let us write it thus:
The difference of this function with rtspect to x is (12)
A[;)=[x;l)-(:],=(;)[~-l]
The mean of l (1 is
=
The higher differences are complicated.
73 and therefore (13) Hence in this case the higher means are very simple. The generating function of z with respect to x is I 1
From this we may deduce a useful formula found by Cauchy. Putting (l+t)"(l+t)" = (l+f)"+m and noting that in both members of this equation the coefficients of tz are the same, we get (14) This is Cauchy's formula. It may be extended for any number of factors. From (14) we deduce
In the sum of the second member every combination with repetition and permutation of the numbers x,, x2, . . ,, x,,, should be taken so that x1+x2+x3+ . . . +x, = z. Computation of binomial coefficienfs. If in an expansion
in paragraph 16, and then divide respectively by 11, 2!, . , . , nl. If only one term c is required, and if x and n are positive ( 1
74 integers, less than 3001, then we may use Dude's Tables [Ioc. cit. 111, and the formula log(+logx!-log*!-log (x-n)!
There are also small tables giving E ls (1 To compute a table of the binomial coefficients corresponding to positive integer values of x and n, it is best to start from the difference equation (7'), which will be written in the following manner: f(n,x) = f(n,x--1) + f(n-1,x-l), taking account of the initial conditions f(n,O) = 0, if n is different from zero and f(O,O) = 1. These conditions are necessary and sufficient. Indeed, starting from them we may obtain by the aid of the above equation every value of f(n,x) Putting x=1 we find f(n,l) = 0 if n>l or n<O
polynomial Pi(x) of degree i is given for every value of i then , it is easy to show that it may be expanded into a series of polynomials Pi(x) in one way only. Indeed, putting
Is S e e Ch. Jordan, Approximation by orthogonal polynomials. Annals of Mathematical Statistics Vol. 3, p, 354. Ann Arbor, Mich, 1932. The table gives the values for x<lll and n<ll.
Q 23. Expansion of a function into a series of binomial coefficients. If the function is a polynomial of degree n, and the
~-
f(x) = c".+c,P,(x) + *. . + c,P,(x)
1 75 and noting that the coefficients of x* are in both members identical, we obtain for Y=O, l,, , . n, in all, n+l equations determining the n+l unknoti coefficients cO, C, , . . . c,, . If the polynomial f(x) is given by (1) and f ( x ) = a,+a,x+a,x2+....+a,x"
Pi(x) = ( xT )h
I I
then the required expression is the following: (2)
f(x) = b,+b,(XTa)*
+b,(x7),+...+b"(xy3-
The shortest way to determine the coefficients bi is to use formula (10) of 8 22; to do this, let us write the m -th difference of the expression (2) of f(x); we have (3) yf(x) = hm [ brn + b,,, ( "T" jh + . - ' + bn [ ;fm)i ] b, = pf (a) / hm and the expansion will be (41 f(x) = f(a) + ( xya)h &f (4 +[""],,+7 This is the general form of Newton's. formula for a polynomial; the function f(x) and its differences for x=a must be known. From formula (4) we may easily deduce the differences of f(x) for any value whatever of x:
therefore
I,f f(x) is a polynomial of degree R, then from (3) it follows that the m -th difference of f(x) is of degree n-m; the n -th difference is a constant, and the higher differences are equal to zero.
76 Therefore in the calculus of finite differences it is always advantageous to express polynomials by Newton's formuia. It has been shown elsewhere [Ch. Jordan lot. cit. 19. p. 257~-3571 what importance Newton's formula has for the statistician, This is not yet sufficiently recognised, since nearly always the statistician expands his polynomials in power series in spite of the fact that he is generally concerned with the differences and sums of his functions, so that he is obliged to calculate these quantities laboriously. In Newton's expansion they would be given immediately. If f(x) corresponding to a given value of x is needed; the computation is not much longer in the case of a Newton series. than in that of a power series. In the latter it is necessary to compute x, x2, x3, . . . , x" and these are obtained most readily by multiplication. In the case of a Newton series it is necessary to compute (x-a), (x-a) (x-e--h), (x-u) (x-a-/r) (x-a-2h) and so on; these should also be obtained by multiplication; the only additional work is division by 1, 2, 3, and so on. But if a table of the values f(a), f (a+h), f (a+2h), . . . , and so on, is required, this is obtained by Newton's formula with much less work than by a power series20 The method used is that of the addition of differences. Since f(x) is a polynomial of degree n, kf(x) = & f ( u ) i s constant. Putting x=a+[h the differences Arlf(x) are then h given step by step by A"-lf (a+&) = A"-If (a) + &Vf (a) for 6~123 I". The differences Ar2f(x) are given by 1 , Ar2f (a+Eh+h) = Ar2f (o+Eh) + A,'f (a+G) and so on: k'-,f (a+[h+h) = ,&'-If (a+lh) -f A"-, "f (a+th). Finally f(a+5h+N = f(a+th) + Af(o+W In this way we obtain by simple additions not only a table of f(a+th) but also that of the corresponding differences.
20 The method is shown in lot. cit. 19, p. XI, and an example is given on p. 301.
78 x we have multiplying the preceding equation by y , ( I (7) Example. Expansion of
The expansion of (x) vf (x) would be made in the same manner. Formula (7) is not only useful for determining the differences but as we shall see later also the sum of I :I f(x)* Newton's backward formula. In Newton's formula treated above, the argument of each difference is the same. Sometimes it is useful to have decreasing arguments in the expansion, In 8 6 we found by symbolical methods E I= l/(1- $ ), hence
f(F) being a polynomial of degree n, if we apply to f(a) we have (8) f (a+x) = ,z [ x+:-l ]Amf (a-m).
This formula may serve to determine the ,D -th difference of the first member; we obtain
Ap I(:) (;)I = 2 (':") (1:;') (y;;!!p),
A function f(x) which is not a polynomial may nevertheless in certain cases be expanded nto a Newton series (6 124); but then the series will be infinite. Applying formula (10) we get for instance 2"=% x;m -&. ) w=o ( An arithmetical progression is a polynomial f(x) corresponding to ~=a, a+h, a+Zh. . . . whose first difference is constant; therefore it can be expressed by Newton's formula in the following way: y (4 f(x) = f(Q) + (x-cQh, Example 3. Given the series 1,3,5, 7, , . . , Here we must put h=l, a=l, f(a)=1 and Af(a) ~2. Therefore we have f(x) = 1 + 2(x-l)
80
The general term of an arithmetical progression of order n is a polynomial whose R -th difference is constant. It may be given by formula (4).
Example 4. Arithmetical progression of the third order.
Given: 1, 8, 27, 64, . . . In this case n=3, h=I, a=1 and f(a)=l. To determine the differences for x=1 let us write the table of the successive values of f(x):
1
81 This formula may serve to compute B(x,y) if x is a positive integer. On the other hand, in the general case, the binomial coefficient may be expressed by a Beta function; writing in formula (3) n+l instead of x and x-n+1 instead of y we get
(1
X
n = (x+1) B(n:l , X-n+l) '
From the expression (2) we may deduce by aid of formula (2) 8 17 B(x+Ly) =
Ux+l) F(Y)
r(x+y+l~ = & B(JGY)-
This gives the difference of B(x,y) with respect to x
$B(x,y) = --+W,Y),
In the same manner we should obtain !Bhu) = - -&B(x~Y). Therefore we conclude that B (x,y) satisfies the following Dart ial difference equation (4)
Instead of B,(p, q) the function Z,(p, Q) is generally introduced. This is obtained by dividing the incomplete Beta-function by the corresponding complete function:
The numerical values of Z,(p, Q) may be evaluated by integration by parts or taken from Pearson's tables, if p 5 50 and q s 50.=
*I H. E. Sopet, Numerical Evaluation of the Incomplete B-function. Tracts for Computers, Cambridge University Press, 1921. K. Pearson, Tables of the incomplete Beta-function, Cambridge University Press, 1934.
84 In the tables, I,@, 9) is given only for the values p L q, This is sufficient, since we have I.Y(P, q) = 1 - 1,-.x(9, PI. (3) Let us remark moreover that Zx(Pv PI = 1 - f 1% (Pt ?a w h e r e x1 = I - (4x--1/2)". The tables give the function for p and q from 0'5 to 50. The intervals Ap and Aq are equal to 0'5 from 0'5 to 11, and to 1'0 from 11 to 50. Moreover x is given from 0 to I; and Ax=O'Ol. Some mafhemafical properties of the function. The function ZJp, q) satisfies the following difference equation: ( 4 ) L (p+l, 9+11 = xl, (Ps 9+1) + (l-4 L(P+L 9 1
+ L(p+m q-m).
We have seen that Z,(p+m, q-m) diminishes with increasing m, hence it is often possible to continue the proceeding till this quantity becomes negligible. To perform the computation Sopet transformed formula (11) by factoring the first term of the sum, in this way: a, (1 + X , . . then again factor ing the first term of the new sum thus: 4 + 41 + 2. , , , and so on. In the end he obtained (12) UP9 d =
t-q-1 -
I"(p+q) qp+l) r(g)xp (l--x)q-l I: 1 +
x - [l+q-2 - - [1+.*..+ x
p + l ( l - xI
p-t-2 l-x1 t
1 1 The computation is continued till become small enough. Is-m+l)x/ (P+m--1) (1-x) If xq > (1-x)p then it is better to determine 1 -ZIA(q, p) instead of I, (p, q). It may happen that the desired precision is not attained
+ q---m+1 p+m-1
ex Ill*-.. 1 + Lb+m, 4-m) -
86 before q+l-m becomes negative. Should we still continue the proceeding then the absolute value of the terms would begin to increase and the series to diverge. To obviate this inconvenience Soper advises us then to use the method of raising p. Let us suppose that by repeated integration by parts Z,(p, q) has been reduced to Z,(p', Q') so that 1 > Q' > 0. Now repeating operation (7) n times, we shall have (13) Z,(p', q') = J$$f$ (l-x)'r' 2; ;;;;;, Z,(p'+i, q'). +
To shorten the work of computation Soper transformed this formula in the following way: (14) -1+ L T(p'+q') ' G(P', 4') = r(p'+l)jT(q') xp (l--x)*' I ,+p'+q' x P'fl *
p'+q'+l x 1+*,,.+ p'+q'+n--2 x + L(P'+n* q'). p'+n-1 _ p'+2 I 1s '*'* I This can be continued till the required precision is obtained. In case p and q are integers we shall have from (11)
Remark 2. If in p+q-2 repeated trials p-l favourable events are obtained, then, according to Bayed theorem the probability that the probability of the favourable event does not exceed x is equal to I, (p, q). Remark 2. The quantity I,, (x, n+l-x) may be considered as the probability that in R trials the number of the favourable events should not be less than x, provided that the probability of the favourable event is equal to p. Indeed, from (15) it follows that (16) 1, (x,n+l-x) = x (;) p'(l-p)"-'. From this we may deduce ,$ (y") a" - 1 -w,n+1-g (1-P)" where we have to put p = a/ (l+a). (17)
87 Moreover if p=1/ we get WI 4 (: ) = 2"[1 -ZII,,(x, n + l - x ) ] .
Particular case. If np is an integer, then v=np is the most probable number of the favourable events in n trials. Putting into (16) x=np+l we get the probability that in n trials the number of the favourable events should be more than np. This will be equal to 1, (v+l, w). Moreover the probability that v is less than np is 1 --I, (np, nq+l). According to Simmons' theorem this second probability is greater than the first if p<q. g 26, Exponential functions. The exponential function is as important in the Calculus of Finite Differences as in Infinitesimal Calculus; the differences and the means of this function are easy to express. Differences of ax. We have (1) and therefore (2) AaX z ax'"- ax = aX(ah-1)
h
Amax = ax(ah-I)". h Particular cases. If h= 1 and a=2, then 02" = 2".
If h=I and a=1/2
AM1" = - Md""
A"(1/2)" = (-l)m(~)x+m. Meahs of the function ax. According to what we have seen we have , (3) and (4) Max = l/zax(d+l)
h
In the same manner we could obtain p sin (ax+b), but we may deduce this difference directly from (1) by putting into it b-+&z instead of b; we find (2)
p sin (ax+b) = (2 sin l/zah)m sin [ax+b+J&m(ah+n)].
From formulae (1) and (2) it follows that if the period of the trigonometric function is equal to h then its difference in a system of increment h will be equal to zero. T h e p e r i o d o f c o s (axfb) o r o f s i n ( a x + b ) w i l l b e 2% equal to h if a = h but then sin$$ah = 0 and both differences will vanish. Therefore we conclude that
= cos l/zah cos (ax+b+l/zah). Mm cos (ax+b) = (cos $'zah)m cos (ax+b+$mah), y sin (ax fb) := (cos l/ah)m sin (ax+b+l/zmah),
From (3) and (4) it follows that if a = t then the corresponding means in the system of increment h will be equal to zero. Hence
In the a:me manner we should get (4)
hjlcos
(cx+b) = 0
and
lfj sin (tx+b) = 0.
I
90
Differences and Weans of cosn x and sin" x. To determine these quantities, if n is a positive integer, the best way is to transform these powers into sines and cosines of multiples of x by Euler's formula. Two cases must be distinguished, n odd or even. First if n is odd: n=lm+l then Euler's formula will give:
pn+lCos2m+lX =
(eix +
Particular case. If in the forgoing example b ~0, h=l and for every integer value of x we have q(x) >O, then f(x) = (--lPP(X) is an alternate function in the system of increment h=l. The difference of this function will be A [Wl"~(xll and therefore = (-1)x+1 qJ(x+l) - (-l)Xp(x) = = - 2 (-l)"Mp(x)
93 (2) Am[(-l)xq(x)] = (-2)'"(-l)"Mm~(~).
before we obtain: (3)
Means of the alternate function (1). In the same manner as
~m[cos(~+b)q+)]= (--1/2)m~~~(y+ b)yq'(x).
It is obvious that if f(x) is equal to a constant, this equation is satisfied. But if w(x) is any periodic function whatever with period equal to h, that is if w(x+h) = w(x) for every value of x, then f(x) = W(X) equation (l), Example: is also a solution of
f(x) = ..,r$T + b) , The Mean of a function will be equal to zero if this function satisfies the difference equation (2) f(x+h) + f(x) = 0. If we put f(x) = cos( f + 6) then this equation will obviously be satisfied. But if w(x) is any periodic function whatever with period equal to h, then
cos(F + b) w(x)
will also be a solution of equation (2). 8 30. Product of two functions. Differences. The operation of displacement performed on a product gives
The first sum is to be extended to every combination of the first order.of the m elements u,, u2, , . . . , u,; the second sum to every combination of the second order of the same elements; and so on. The total number of terms will be 2" - 1. Higher Differences. Starting from formula (3) we may obtain by repeating the operation A2(uu) = M"uA% + 2MAuMAu and in the same manner (6) + M2uA2u
Am(uu) = 'x (F) Mm-vA'uMvAm--u.
This formula is not often used, since it presupposes the knowledge both of the differences and means of the functions
96 u and v; and we have seen that, except for the exponential and. circular functions, whose differences and means are simple, genera!ly, if the differences of a function are easy to obtain, means are complicated, orcbrversely. Therefore formula the (6) is to be transformed so as to contain differences only or means only. Example for formula (6). Given uv = ax sinx. Let us put u=ax and vzsinx. Then moreover
Mm-VA k k
We will deduce an analogous formula for differences. Using the symbolical method we shall write
This operation performed on uv gives (7) I = 2 (-1)~ (;) En-r u En-v.
97
If we put in the place of En-* its value expressed by differences
(7')
we shall have
Remarking that
and
(9)
21 (-1) v [ "Ti ] p-* 0 = An-iv;
we conclude that
(10)
b(w) xz ;z [ 1) aizz b-i E'v.
This is the formula that corresponds to that of Leibnitt. Since it needs only the knowledge of differences, it is more advantageous than formula (6), is required if Example 1. The n -th difference of 2" i I 1 h=l. Let us put 21~2~ and vz , formula (1Oj will give
We have seen that there are functions whose difference is a function of the same kind, but the argument is diminished by one:
ff we apply formula (10) to the product of such a function, for instance to [:I ( Gx) putting u= (z) and II= (ix) then the argument of v will be diminished in consequence of An+ by ,n-i and moreover it will be increased in consequence of E' by i, SO that it will always remain equal to -x-n. Often this is a great simplification, as we shall see later.
Example 2.
A./ [ ;] [ ;']I = :$; Wi[ s:i ) [ ;;;i) Differences of a product expressed by means. Since there are functions whose means are more easily determined than their differences, it may be useful to determine the differences of a product by aid of means. Putting into formula (7) En-l expressed by means
(11)
En-lu
z (2M--I)"-tu
= X0 (-1)" v--i
M
"fl
I
nfY (2M)k 1
we obtain
A"(uv) = le:o
PI+ 1
(--1)~1-i(2Mj iun;z'[ ;] ["i') En- vq;
applying again formula (8) and remarking that (12) it :follows that
WI-1
This formula is analogous to (10) but means figure in ii instead of differences. Example 3. The n -th difference of 2~ (z) is to be determined if h=l. Let us put ~1% and o = (t). Then we find
2iMiu = 2.~ 3i
and
p
-
i
Mn-iu = ( :$:Kii)
therefore
A" 1 2" (;] 1 = ;.i; (+3'(7] 2" [ n;;;i] ,
Q 31. Product of two functions. Means. To obtain the mean of a product, we will apply again the method of displacement used in the preceding paragraph. We have and therefore
Q 32, Indefinite sums. The operation of differences was defined by A&4 = f ( x + h ) - f ( X ) . From the point of view of addition, subtraction, and multiplication, the symbol A behaved like an algebraic quantity; but division by A or multiplication by A-l has not yet been introduced. Let us put Af(x) = f ( x + h ) - f ( x ) = (p(x). (11 By symbolical multiplication with A--' we should get A-lq~(x) = f ( x ) . The significationof theoperation A-l is therefore the following: a function f(x) is to be determined, whose difference is equal to a given quantity q(x). It must be remarked in the first place, that this operation is not univocal. Indeed, if o(x) is an arbitrary function whose difference is equal to zero, then and therefore (2)
A[fW +w(x)l = v(x) A--'pC4 = f(x) + wIx)e
We have seen in 8 29 that w(x) may be any arbitrary periodic function with period h. If the variable x is a discontinuous one, then o(x) is equal to a constant. From (2) we conclude that the operations A-1 and A are not commutative. Indeed we have
t
101 AA-1 = 1 and /j-IA =/= 1.
On the other hand we have seen that the operations A, M , D and E were commutative. The operation A-' being analogous to the inverse operation of derivation which is called indefinite integration, therefore the operation A--' has been called indefinite summation, and instead of the symbol A-1 the symbol Z is generally used. It may be useful to remark that the two symbols must be considered as identical. Determination of the indefinite sum. q(x) being given, the problem of deducing A-lp,(x) = f(x) is identical with the resolution of equation (1). The variable being considered as a discontinuous one, it is always possible to obtain a system of differences in which the increment is equal to one, and the variable takes only integer values. For this it is sufficient to put x=a+th. Then if x=a+h, the new variable will be 5~1 and so on. Therefore we may suppose without restriction, that h=l and x is an integer. Starting from this supposition equation (1) will be t3) f(x+ll - f ( x ) = q(x). This is a linear difference equation of the first order. AndrP considered equation (3) as equivalent to the system of x equations: f(x) - f ( x - l ) = 9(x-1) f ( x - l ) - f (x-2) = 94x-2) , . . , s f(a+ll -WI =%(a) where a and x are integers and f(a) is arbitrary. From these x-a equations we may determine the x-a unknowns f(x), f (x-l), . . ,( f(a+l) but it is sufficient to determine f(x)* For this, let us add together the above equations; we get:
102 It is easy to verify that this solution satisfies equation (3). In this manner the operation A-l is expressed by a sum and an arbitrary constant f(a). This is one of the reasons why this operation is called indefinite summation. Formula (4) presents more theoretical than practical interest. Indeed, it is the expression of A-' by a sum; and we shall see that generally to evaluate sums we make use of the operation A-'. In other cases formula (4) may be used. For instance it may be shown directly that A-lx = f(o) + i i = kl Therefore according to (4) we have A-lx= I ; 1 +k. General rules. The symbol A-r is distributive, hence A-t(~+~+w+...J = A-l~+A-lv+A-lwjand, if c is a constant, . . . .
A-=CY,(X) = CA-~ q(x).
Therefore we may determine the indefinite sum of a function F(x) by expanding it first into a Newton series, and then applying the above rules: i-1 F(x) = ; 4% h" CO We have seen that &Y]* and therefore 4-l ["J")* = + [ :T;jb +k. Finally we obtain = qxTy*
I
In this chapter k will signify a quantity whose difference is equal to zero.
+ ln;l) bF(O)*
I f w e h a v e A-l f ( x ) = y(x) + K t h e n A-'f(-x) = - y(-x+h) + k that is, if in the first case the argument does not change, then in the second the argument -x is increased by h. This will be useful later on. Remark. The inverse differences may be determined by symbolical methods. We have
1 a =
-+E =-(l+ E+E'+....).
Taking account of what has been said in 5 6 concerning the necessary precautions in the case of infinite series of symbolical expressions we may perform the operations figuring in equation (6), starting from f(x). We find
(7)
A - ' f ( x ) =-
5 f(x+i) = - ii f ( i ) .
i=O i=x
This may be easily checked; indeed, the difference of the quantity in the second member is equal to f(x). Finally we conclude that the inverse difference may be expressed by aid of a sum, if certain conditions of convergence are fulfilled. 8 33. Indefinite sum obtained by inversion. There are different methods of determining the indefinite sums. The first is that of the inversion of the formulae obtained by calculating the differences. For instance, if we have found Af(x) = cq~(x--a) then we conclude that
104 A-~,(X) =+ f(x+a) + k . In this way we get the following formulae:
105 Remark. In the formulae 10, 12, and 13, the argument x is diminished by h. Q 34. Indefinite sum obtained by summation by parts, Starting from the formula of the difference of a product obtained in Q 30 (formula 2),
AWN V,Hl = u(x) AV,(4 + V,(x+h) A~W.
Writing AV,(x) = V,(x), and performing the operation A-l on both members of this equation, we have (1)
A-l [U(x) V,(x)-I = U(x) V, (4 -A-l [V,(x+h) AU(x)],
This formula, being analogous to that of integration by parts, is called formula of summation by parts. It becomes useful if the indefinite sum of the first member is unknown, while that of the second member may be determined. Examples. In the following examples we will denote the first factor by U(x) and the second by V,(x). We find (2) (3) ii-1 XQ~ = & - A-1 g ELI XQX -
106 Repeated summation by parts. If we introduce the following notation
A--' Vi(x) = Vi+,(x)
and start from formula (l), repeating the summation by parts on the last term of the second member, we obtain
A-'[V,(x)W)] = V, O)U(4 - V,(x+h)AU(x) + A-l [V,(x+2h)A'U(x)].
+
Again performing the summation by parts on the last term, we get
A--'[V,(x)U(x)] = V,(x)U(x)-VVP(x+h)AU(x) -i- V,(x+2h)A2U(~) -A-' [V,(x+3h)A3U(x)]
and so on; finally we have
+
IlO)
A-'[V,(x)U(x)l
= V,
(x)U(x)
- V,(x+h)AU(x)
+
+
+ I',(x+2h)A*U(x) -. . , . +
(-l)"-'V,,(xfnh-h)&-'U(x)
+ (--l)nA-l[V,,(x+nh)A"U(x)].
This formula will be especially useful if U(x) is a polynomial of degree n-1 ; in this case AflU ~0, the last term of (10) vanishes, and the problem is solved. If U(x) is not a polynomial, the last term will be considered as the remainder of the series. Example 1.
We have seen that there are functions whose indefinite sum is a function of the same kind but in which x is diminished by h. For instance if h= I we have A-' 2-x = - 2-x+1 + k, A-' '( Gx] = - ( ;:-' ) + k. Therefore if V, is such a function, the argument will remain constant throughout the operations of repeated summation by
Formula (10) may also be applied in some cases in which U(x) is not a polynomial, provided that the corresponding series (10) be convergent. Then we have (14)
A-'
[V,(x)U(x)] = ,+)m-' V,(x+mh-h) Am-W(xj.
Example 2. Given ~12~; if we put V,=x and U=1/2" then formula (14) will give
A-$&=(;] 2-x+
(xt;') 2-"-i+...+ (x,+t;)2-'-n+SS.S
It is easily seen that this is equal to 2-(l+x);2"' which result could have been obtained directly by putting into (14) V, = l/2" a n d U=x. Example 3. Given 2-"/x. Putting V,Z~-~ and U=ljx by aid of formula (14) we find according to formulae 7, $j 16 and 6,
§ 33
A-1 2-" X-
2-x+1
+
k
ti (---I)"+~ B(n+l ,x) + k. d Remark. Condorcet, in his "Essai sur 1'Application de 1'Analyse a la Probabilite des Decisions" (Paris, 1785, p, 163) has found a formula of repeated summation by parts, somewhat different from (lo), which written in our notation is the following:
= 2-H'
A-'
[V,U) = V,U - (V, + V,)AU + (V, + 2V, + V,)AW -, , .
+ k--l)"'Am~ x (7) Vm+li,
the argument x in every term being the same.
I
108
9 35. Summation by parts of alternate functions. We have seen that A(-l)X = 2(-l)"+l. From this we conclude A-' (-l)r = 1/2 (-l)X+1 . Putting into formula (10) 5 3 4 V,(X) = (-1)" and U(x) = f(x) we find (2) A-' (-l)X f (x) = 1/ (-l)X+1 [f(x) - &V(x) + ++f(x)-.... + ( - 1 ) " & all{(x) 1 + K. It must be noted that this formula holds only if f(x) is a polynomial of degree n; moreover if in this alternate function x is an integer. Example 1. Let f(x) = x; then (3) A-'(-1)'~ = ~~(-1)X+1(x--1/2) + k. If f(x) is not a polynomial, we apply formula (14) of Q 34, but the condition of convergence must be satisfied. Example 2. Given f(x) = l/(x+1). We have (4) A-'(-1)s -!-. _. x+1
It is easy to see that in this case the condition mentioned is satisfied, Remark. It would be possible to calculate in the same way A-l(--l)'aS but it is shorter to put A-' (-a)X and then apply the known formula of exponential functions: A-1(-)x = _ (--alx. a+1 Summation of alternate functions by aid of inverse means. From formula (3) of 5 30 we deduce A-' [Mu Avl = uv - A-l [Mv Au 1. Putting into this equation v=(-1)x and u=f(x) we find Mv=O, hence we have
109 (5) A-' [ (-1)x f(x) / = g(-l)*'+l M-' f(x).
',%at is, the indefinite sum of the alternate function figuring in the first member, is expressed by aid of the inverse mean of f(x) (formulae 3, 8 38 and 10, Q 39).
8 36. Indefinite sums determined by aid of difference equations, The indefinite sum is according to our definition the
where q(x) is given. There are several methods for the resolution of these equations, but for our purpose only the methods may serve in which the solution is not obtained by inversion of differences. One of these methods is that of the generating functions, due to Laplace, which is applicable if x is an integer. According to formula (3) Q 11, if u is the generating function of f(x), then that of f(x+l) will be Gf(x+1) z u-tf(o) Let us denote by R(f) the generating function of q(x) and not that the corresponding generating functions satisfy the difference equation (1). We have u-f(O) -a = R(t); t this gives u= fR(f) + f(o) -. 1 --t where f(O) is an arbitrary constant. Finally the coefficient of t" in the expansion of zz will be equal to the required indefinite sum f(x), Example. Let q.(x) = x2. To determine the generating function R(f) of x2 let us remark that x2 = 2(z) + (F) and the generating function of the second member is, according to formula (14) 6 10,
110
Gx= R ( f ) = flL;,*3 ;
therefore
The expansion of u gives (t'+f3) (l-f)+ = (f'+f") z Ii"] (-l)i fi = = (f'+f") x ( i;3] f'. Putting into the first term i=x-2 and into the second i=x-3 we have f(x) = d-1 x" = (z-t; J-i [xA3] -j- K =; x(x-1) (2x-l) + k. @ 37. Differences, sums and means of an infinite series. Let us suppose that the function f(x) is expanded into an infinite series which is convergent in the interval a, b: f(x) = U"(X) + u,(x) + , , . . +u"(x) +..*. (1) The function f(x+l) = u,(x+l) + 2$(x+1) +. . . . + u,(x+l) +. . . . will be convergent in the interval a-l, b-1. It is known that the difference of two convergent series is also a convergent series, whose sum is equal to the difference of the sums of the given series. Therefore I4 Af(x) I i(x+l) - f ( x ) =
AU, (4 + AU, (4 + . a . + Au, Ix) + , . .
Consequently if f(x) is given by its expansion into a convergent series, the difference of f(x) may be obtained, by taking the difference of the series term by term. 'We have Mf(x) = lhlf(x+l) + f(x) 1; therefore for the same reasons, if f(x) is given by a convergent series, then to obtain ttle mean it is sufficient to determine the mean of the series term by term. To obtain the indefinite sum of f(x) given by formula (1) we determine it term by term.
111 (3) A-If(x) = A-'+(x) + A-'u, (x) + . . < + A-k(x) + . ,
If this series is convergent, then in consequence of what has been said before, the difference of the second member is equal to f(x) and therefore the required indefinite sum is given by (3). Q 38. Inverse operation of the mean. The mean has been defined by the following operation ryx! = ?LJ [f(x+h) + f(x) I* We have seen that from the point of view of addition, subtraction, and multiplication the symbol M behaved like an algebraic quantity; but division by M or multiplication by M-l has not yet been introduced here. Let us put 0) y(x) = 'h (f(x+h)
t
f(x) I = e(x)*
Multiplying both members of this quantity by M-' we shall have
f ( x ) = M-'p,(x).
The significance of the operation M-l is therefore the following: a function f(x) is to be determined so that its mean shall be equal to a given quantity y(x). Here the same difficulty presents itself as in the case of the operation A-l. In 5 29 we have seen that .there are functions whose mean is equal to zero. We may write these functions in the form cos (; x + b) . w(x) where w.(x) is an arbitrary periodical function with period h. Therefore in consequence of (1) we shall have also fy [f(x) + cos (+ x$-b) w(x) I = (AX) and moreover,
M-' cp(x) = f ( x )
h
+ cos (- x+b) w(x).
;:
From the preceding it follows that MM-' = 1
112 but we may have Remark. If the required function f(x) is necessarily a polynomial, then the operation M-l is univocal and there is only one solution. Determination of M+J(x). In the case of a discontinuous and equidistant variable we may always suppose without restriction that x is an integer and h= 1, since by introducing a new variable this can always be obtained. Supposing h=l and x integer, M+(x) = f(x) is the solution of the equation f(x+l) + f(x) = &4x) (2) where p(x) is given. This is a linear difference equation of the first order: it can be solved by AndrB's method [kc. cit. 221, in which equation (2) is considered as being equivalent to the following system of equations + f ( x - 1 ) z 2g,(x-1) f(x) f (x-l) + f (x-2) = 259 ( x - 2 ) ,.,. f(a+l) + f(a) = 2d4 where x and a are integers and f(a) is arbitrary. Multiplying the first equation by (-l)s, the second by I-1)" and so on, the n th by (-l)"+i; then adding them together we obtain (3) f ( x ) = (--i)fiaf(a) +Z 5 (-l)x+'+lq(i) = M-'Q?(X).
i==a
113 Formula (3) is more important from the theoretical than from the practical point of view, but nevertheless it may be useful in some cases. It is an expression of the inverse mean by alternate sums. If the alternate sum of V(X) is known, formula (3) may be applied directly. Example. Let v(x) = cX, and a=O. It can be shown that
5 (-l)ici = i==o i (-)i = 1-(--c)x c+l
therefore formula (3) will give M-lcx = (-l)xK+ s This may be verified by inversion. Q 39. Other methods of obtaining inverse means, A. Znversion. In the preceding chapter we have determined the mean of several functions t(x); now inverting the results we may obtain formulae for M-l. For instance, having found Mf(X) we deduce M-1 cp(x) 1 -l-f(x-4 + x* c In this paragraph x signifies an arbitrary function, whose mean is equal to zero. By this method we obtain the following formulae: (1) 12) (3) (4) (5) (6) (7)
H
This formula applied to polynomials presents no difficulty; indeed in this case the series of the second member is finite. Formula (12) would lead in the case of the preceding example directly to the result obtained above. Formula (12) may be applied to other functions, if certain conditions are satisfied.
115 C. By Euler's Polynomials. If a function f(x) is expanded into a convergent Maclautin series
If this series is convergent, then according to what has been said above, the mean of the second member will be equal to f(x) and therefore the inverse mean of f(x) is given by (13). From this we may deduce a formula giving the indefinite sum of an alternate function (-1)" f(x). By aid of formula (10) we get (14) A-' [(-l)Xf(x) 1 = '/z(-l)X+1,n; Particular case. Given f(x) = x" (15) A-l I(-1)" x"J = $5(-l)X+1 nl E,(x) + K. D. By aid of p, functions. A function f(x) being expanded into a convergent reciprocal power series; (16) f(x) = ,,;os. E , ( x ) Dmf(o) + k.
116 Hence, performing the operation term-by-term we get M-1 f(x) = co + nio (-1)" c"+l ppJp + x. If this series is convergent it gives the inverse mean of f (3c). By aid of (10) we may obtain A-l (-1)" f(x). $j 40. Sums. It has been mentioned before (Q 2) that in the case of a discontinuous variable with equal intervals, we may always suppose that x is an integer and that h=l. Let us con- ' sider in this case the indefinite sum: A-ly((x) = f ( x ) +k. If a and n are integers we may write A-' y(a+n) -A-'&I) = f(a+n) -f(a). In Q 32 we have seen that the operation A-r may be expressed by a sum, so that
Hence to calculate the sum of q(x) from x=a to X=Q+R it is sufficient to determine f(x), the indefinite sum of v(x), and then obtain f (a+n) by putting x equal to the upper limit, and f(u) by putting x equal to the lower limit. The required sum is equal to the difference of these quantities. The process is exactly the same as that used to determine a definite integral. Moreover we have, as in the case of these integrals, f Y(X) + and i* Y(X) = xi,x Y,(X) = f(c) --r(a)
117
This harmony and the above simple rule are due to the definition of the sum used in the Calculus of Finite Differences, which is somewhat different from the ordinary mathematical definition. Indeed, the definition used here is the following: IT+* da) + da+11 + ""da+--1) = ~au;(4 that is, the term v(a) corresponding to the lower limit is included in the sum, but n?t the term rp(a+n) corresponding to the upper limit. In the ordinary notation the sum above would be denoted by a+#-1 s x==u v(x) with this notation the concord between the two calculi but would cease, and the rules of summation would be complicated. However, it should be mentioned that our notation is* not symmetrical with respect to the limits; this inconvenience is due to the want of symlhetry of the notation of forward differences. Remark. In 8 32 formula (7) the inverse difference has been expressed by aid of a sum. Starting from this we obtain the sum of f(x) from X=Q to x=z, indeed
3. Sum of an arithmetical progression of order n. We have seen in 8 2 that the general term of this progression is a polynomial f(x) of degree n. To obtain the sum of f(x) it is best to expand it into a Newton series, We have f ( x ) = f ( o ) +(;]Af(O) +I;) A'f(0) +....+(;I A"f(Ol. The corresponding indefinite sum is ($ 33)
i
119
n+1 Y A-' f(x) = Z n=O I m';l I
Amf (0, + k.
Finally the sum of first u terms is 5 f ( x ) = I$[ ml;1] A*f(O). x=0 Hence, to obtain the sum required, it is sufficient to know the differences of f(x) for x=0. Example 2. Given f(x) = (2x+1)*. To obtain the differences the simplest way is to write a table of the first values of the function: 1 9 25 8 16 8 Therefore f (0) = 1, Af(0) = 8, A2f(0) = 8. Since the polynomial is of the second degree, the higher differences will be equal to zero. P In this manner it would be possible to determine Z x" x=0 but later on we shall see shorter methods, by aid of Stirling numbers, or by Bernoulli polynomials. If the sum of f(x) is required for x=a, a+h,, , ., Q+ (j~-l)h, that is, if the increment is equal to h, this should be indicated in the symbol of the sum; for instance (5) To determine this sum we generally introduce a new variable 5 = (x-Q)/~; then 5 will be an integer, and At= 1, and the preceding methods may be applied, We expand f(a+lh) into a Newton series, determine the indefinite sum, and put into it the limits. We shall have, if f(x) is a polynomial of degree n (f-4
fi t=o f(Q+lh) = zi[ m~l]Amf(a).
The significance of Amf(a) in this formula is jA"f(a+Sh) 1~~0. Example 2. Given f(x) = x2. Introducing the new variable, we get f(a+lh) = (a+lh)"; determining the differences with respect to ,$ we have
120 [Af(~+[h)]~~ = 2ah + h' and [A2f(a+th)]i=o
= 2h2.
According to formula (6) we find (7)
'!$ f(x) = 5 f (a+th) = ~2 +[ 5) (2ah+h2) +[ :]2h2. w
On the other hand, it is often possible to obtain the sum (5) directly , by determining b-1 f(x) and putting the values of the limits into the results. For this we expand f(x) into a series of generalised binomial coefficients whose increment h is the same as that of x in f(x) ; if this function is a polynomial of degree n, then we shall have
and the sum of f(x) is required for x=a, x=a+h, x=a+2h, .,.. x=a+(~bl)h, then according to formula (10) of (j 33 we have, if n>l 1
A-l (X),,
h
=
- n-l) h(
tX)-n+z,h
+
K=
.
= - - h(n-1) (x;nh-h)GI<
and therefore M-~h
X5, tx)-", h =
+ '
1
= h (n-l)
a n d i f u=m
(a+nh~h).-I, h
(.flLh+~h-&,-.l, h
1
xE (x)-n, h
=
h(n-1) (a+nh-h),,h
l
Example 1. Let n=2, h-2, and a=l. Then
(x1-2.2 =
_ %:
~x+2Ax+~
1 1 -= 2(l) (1+4-2) = -a e (x+2j1(x+4)
Remark. If the increment h of the factorial is not equal to that of the variable, it is advisable to introduce a new variable whose increment is equal to h. There is another method: to introduce a new variable whose increment is equal to one and expand the function into a series of reciprocal factorials whose increment is also equal to one.
I
I
122 This should be done also in cases when the sum of a fraction is required whose numerator f(x) is a polynomial and the denominator a factorial, the degree of the numerator being less than that of the denominator. For instance, the sum of
'
is to be determined, where u, p, . . . , 1. are positive integers in order of increasing magnitude. This problem has been solved by Stirling in his memorable treatise "Methodus Differentialis" (Lo&n. 1730). Let us multiply the numerator and the denominator of this expression by the quantity x(x+1) , , , necessary to make the denominator equal to (xfi) ),+I . The numerator F(x), whose degree is necessarily less than 1+1, expanded into a series of factorials gives:
From this the required sum is obtained without difficulty; only the determination of the numbers A,,, is needed. Example 2 (Stirling, Meth. Diff. p. 25). If h=l the following sum is required: OD 1 2, x ( x + 3 ) I
123
Here we h a v e 3,=3, and F(x) = (x+1) (x+2). Therefore according to (3) we have
Q 43. Sums of exponekial and of trigonometric functions (same method). The indefinite sum of the exponential function qX is according to Q 33 equal to
(1)
A-1 qx = & + k. Q
124 Applications. 1. Sum of a geometrical progression. Given f(x) = c ox, the sum of f(x) is to be determined for x=a, a+h, a+2h, . , , , a+ (n-1)h; From formula (1) it follows that
LZ+?Zh
Zh X==,'
cqx
= +$ [a""--11, 4
2. We may obtain by formula (1) the sums of trigonometric or hyperbolic functions. For instance if hpl, we have
A- l
and therefore
9,
eirpX
-
2'"" .e's@- 1
+
k
z
x=0
ei(xt14~j
=
efhj
- =
Q*'l'- 1
Qhwp- 1
sin 1bznq Q /a+ _ 1+2wp 1 sin l/229,
.
Since cos (x+b)p, is the real part of e'@+*)g) sin $$mp 5 c o s (x+b)+ = sin 1,2q ~0s lh W-1 + 2b) w r=O This method becomes especially useful in the case of several variables; for instance if vve have to determine
s = 3 . . . 5
XpJ
xm=O
cos (x,+x,+x:;+. , . +xm+b)q.
S is the real part of
w=l x,=:0 VZl
and therefore S = c o s $L(Xn,--m+2b)~;; "2' %ne
I'== 1 /2
In the particular case of n, =n for every value, of v, we have S = cos I,- (mn-m +26) ~1: \sii'fnJJJ The sum of a product of a polynomial and an exponential function is determined by repeated summation by parts (Q 34). Example. The sum of xaS is required from x=0 to x=n+l if Ax=l. The indefinite sum is as we have seen (formula 2,s 34),
125
A-1 xd = QX (ax-x--a) + k
(a-l)2 a"+1 (an--n-l) + a (a-l)* '
therefore
a+1
z x&z x=0
In the particular case of Q = +$ we have x=0
y-z- = 2"
2--
n+2
and
r=l
i+2,
The sum of trigonometric functions may be obtained directly. According to 6, Q 33 the indefinite sum of cos ax is, if h=l
and that of sin ax A--' sin ax = 2 sin l/za Therefore (3) and (4) We will now examine several particular cases of trigonometric functions which play an important part in the trigonometric expansions of functions and in the resolution of difference equations. 2. First form. Trigonometric functions occurring in expansions : 2nk sin 2* x cos -. P ' P If k and p are integers, and if k is not divisible by p, then in consequence of formulae (3) and (4) it follows that E cos * x = 0 ; P 2nk f sin -X = 0. P
40s (ax-l/~,a)
P Therefore if v and p are different and if v+,u and v-p are not divisible by p, then, the sum of the first member will be equal to zero. If v is equal to ,u but different from l/zp, then we shall have
+ 1/ e cos 2nb-A x.
E [czos~x]2 = l/zp. x=0
If we have 2v=2p=p, it is obvious that this sum is equal to p. In the same way we could obtain, if v is equal to ,u but different from l/zp, (6) x=0
E [sin p"1 2 2nv
- l/zp.
If 2v=2p=p then this sum is equal to zero. Moreover, if v and p are different, and if v+,u and v-p are not divisible by p, then . 2TLv sm- x sin23 x = 0. (7) P P If v and ,U are integers we always have
E
(8)
3 x=0 sin? xcos~ x = 0.
Divisibilify. If k is divisible by p, that is if k=rZp (A being an integer) then 2nk 2nk% x = p ; E [cos-x]2 T p; 5 [ s i n p"1 = = 0. x=o .%=a P The other formulae will change according to these. For instance, if v+p=lp and v=p then we shall have P and so on. E
COs*X
cos*x P
= yip;
E sin&x sin2-x = -&p x=0 P P
127 If 2R=p then x=0
5 cos- x = 0;
2.7k P
2.7k ., s Icosp xl- = p. xzo
The corresponding sine values are equal to zero. The above formulae will be used later (5 146). 2. Second form. Trigonometric expressions occurring in certain difference equations. nk .7k sin - x cos-xx', P P where k and p are integers. If k is not divisible by p, or if k= (21+l)p, 3, being an integer, then e ..+ = -$$).l-(-l)k). x=0 On the other hand, if k=21p, then ;rk E cos-x = p . x=0 P If Y and p differ and are integers, moreover, if v-p and v+,u are not divisible by p, it follows that (9)
= l,,:! $ a==u P S = I: sin"' x sin !I!.2 X = x=0
COS
("--,").7
P
*p .--l/z
X;rO
~pcos~v~~xr
=
0
and
W-Y
= I,!,
c = x=0 cos~xcoq~x = 5
g COS x=0 +---rc)n
P
x +
l/2
t=o
5
COS
(Y+I'L)n
P
x
=
0.
On the other hand, if r=,u but v is different from p and from '/zp, then and c = */$p then S=O and C=p If v=,u=p a n d C=O. If v=p=lAzp then s=p s I= lhp
v, ,u and p being integers, the following expression will always be equal to zero: 5 sinyxcosyx x=0 = 0 .
Therefore if v is not divisible by p we have e co, y (2x+1) = 0 x===o and if it is divisible, so that v=lp, then 5 cos 5 (2x+1) x=0 b) We have 2nvx cos A-' sin $ (2x+1) = ' +k 2 sin Z P therefore whether Y is divisible by p or not, E sin F (2x+1) = 0. r=o c) If 'V is different from /L and if moreover V+,LL V-P are not divisible by 2p, we have S = So sin: (2x+1) sin y (2x+1) = 0 and and = (-1)"p.
129 (13) c = 2 cos F (2x+1) cos 72x+1) = 0.
If v+p and if 2v is not divisible by p then Sz1.p and C=$/,p. If v=p and if 2v is divisible by p then S=p and C=O. d) In consequence of formula (12) we always have E sin: (2x+1) c o s 7! (2x+1) = 0. .x=0 whether 2v is divisible by p or not. The sum of a product of a polynomial and a trigonometric function is obtained by repeated summation by parts. Example. The sum of x sinx, for x=0, h, 2h, . . . , (n-1)h is to be determined. The indefinite sum will be, according to formula (3) Q 34, A-'xsinx
h
3. Let us mention here two formulae, which we will deduce later. The first is (1, () 83) :
9
130
" x x=0 X"' = !B+n)*+' en+1
mfl
'
This is a symbolical formula, in which Bi must be substituted for Bi. The numbers Bi are the Bernoulli numbers given by the symbolical equation (B+l)m+'-Bm+l = 0. The second is given by the symbolical formula (31 of 8 108: xF, ( - 1 ) " x"' = & [%-(-l)n (@+2n)"] . The numbers Gi are the tangent-coefficients of 8 104, given by the symbolical relation (G+2)m + Q, = 0. 4. To determine the sum of logx fromx=a to x=a+m if Ax=1 we make use of formula (11) fj 33: A - ' l o g x = logr(x) + k from which it follows that
n+n,
(5)
x l o g x
1 '(a+m) = log--,,(*) .
x=-t,
5. To obtain the sum of the digamma function from x=0 to x=n if Ax=1 let us remark that we have found formula (6) Q 34: A-IF(x) Hence (6) ii F ( x ) = d(n)-n. .a- 0 = xF(x)-x+R.
131 Adding the numbers F(x) from x=0 to x=20 in Pairman's Table we find 4'57315614. The error is equal to 9 units of the eighth decimal. 7. Sum of alternate functions. Let f(x) be a polynomial of degree n; then the indefinite sum of (-1)" f(x) is given by formula (2) of Q 35: @I
A-' (-l)xf(x) = 1,4(-l),+
nL==o
"$' k$
Amf(x).
From this formula we deduce, for instance, 2 (-1)x x = 1/2(-l)r+ (/k--1/+-~& .T=l If the function f(x) is not a polynomial, then the series in formula (8) is infinite; but the formula holds if the series is convergent. For instance if (91
then Amf(x) = -i;x-;+. The series
m+r
1/2m(m+l) [ ximm,tl-l f is convergent, therefore we shall have
(10)
20 x+l=
I'
(-1)"
l/$(-l)r+l
-rn ! g ", 0 2m(~~+m+llm+l
+ .f, &*
If ,u=~, in consequence of what has been said before, the first sum of the second member will vanish, and we get
(11)
OD (-1)x _ 5 1 - = - log (l---1/2) = log 2. $ZO x+l - ,,I=~ rn2"
Q 45. Determination of sums by symbolical formulae, In Q 6 we have expressed the n -th difference of a function by its successive values: we had h" = %; (---I)~+~ from this it follows immediately that ( ", 1 Ex;
132 (II *g (-I)$] f ( x ) = (-l)"Ff(O).
Therefore to have this sum it is sufficient to know the R -th difference of the function f(x), for x=0. Example 1. Given f(x) = 1/(2x+1). To determine the n-th difference if Ax=1 this may be written as a reciprocal factorial of the first degree; then we get
1 1 %b x+1/2= -5 -
This operation performed on f(x), if it is a polynomial or if it satisfies certain conditions, gives 5 f(x+ih) = I=0 4, (-l)'+' [ n+;-l ] $,-1 f (x+nh).
Indeed, since for n=O the first member vanishes, we have also C=O. 8 46. Determination of sums by generating functions. It is often possible to determine sums by the method of generating functions. This will be shown in a few examples. 2. If the following sum is required
This has already been obtained in 8 43 by summation by parts. 4. We could easily determine by summation by parts n+t s = x0 x(n-x) but it may also serve as an example for the use of generating functions of two variables, Let
u(zJ) -p+1 = -=J *a+1 = z 2% x=0 tn-"
from this we conclude that s = [3% I fzz=l that is s = [(n+l)zn + (n+l)P](z---f) + 2(t"+l--z"+l) (x---t)3 I
x=1=1
If we put z=t this giv& O/O; therefore we must apply HBpifal's rule. The derivative with respect to z of the numerator divided by the derivative of the denominator gives again O/O for z=f; and so do the second derivatives also. Finally the third derivatives give s=( .,l)Remark. According to § 34, summation by parts would give A-l x(n-x) =
(;] k----xl +A-'( "f') =
I
tie required sum is therefore
++1 x .Z=O x(n-3) =- (n~l)+(n+=(n~l).
9.47. Determination of sums by geometrical considerations. It occurs sometimes that the result of a geometrical problem is expressed by a sum. If it is possible to solve the problem in
139 another way the result will be equal to the sum obtained previously. As a simple example let us consider the sum
n-t1 (1) s zz z I*=1 v(n-v)
which we have already determined by other methods. This sum is equal to the number of points whose coordinates are x = 0, 1, 2, . , . , (n-2); y=O, 1, 2, . . . . (n-2); zzz 0, 1, 2, .,., (n-2) provided that x+y+zln-1. The points above are situated in a tetrahedron whose summits are the points of coordinates: x=0, y=o, z=o, x=0, y=O, z=n-2, x=n-2, y=O, z=O. The number of the given points contained in the plane x+y = V - l is equal to v(n-v) ; and the number of points contained in the planes parallel to this one, and corresponding to v=l, 2, . , ,, (n-l) is equal to the required sum S. On the other hand, if we consider the plane x+y+z = s the number of the points in this plane will be, as is easily seen, equal to 1+2+3+... + (s+l) = ( s;2 )-
x=0, y=n--2,2x0,
The number of the points contained in the planes parallel to this one, corresponding to s=O, 1,2, . . . , (h-2) will be
this is the total number of the points, therefore equal to S. SO that
140
Sometimes the result of a problem of probability is obtained in the form of a sum; if it is possible to solve the problem in another manner, the result will be equal to the sum above.23 Example. Given an urn in which there are m balls marked 1,2,3 ,,,,, m. Successively R balls are drawn, putting the ball taken, back into the urn before the next drawing. The probability that the sum of the R numbers obtained shall be less than x is equal, according to Monmorf's solution, to
g 48. Determination of sums by the Calculus of Probability.
Since the probability that the obtained sum shall be less than x=nm+l is equal to unity, we have
Finally putting pu=n--~ we get
sums. Often if we know a sum we may deduce the correspond-
Q 49, Determination of alternate sums starting from usual
ing alternate sum, Example. We have, as we shall see in formula (6) Q 82,
(1)
s= x=o (x+&" = 5
?4icw2"
IB,,I (24 !
where B,, is a Bernoulli number. This multiplied by l/2'" gives
*a Let us remark that Crofton determined by the same method the values of several very interesting definite integrals, which verified were later, by analytical methods, by Serref. Morgan W. Crofton, On the Theory of Local Robability, the method used being also extended to the proof of new Theorems';; the Integral Calculus. Phil. Trans. Royal Society, London, 1868. Vol. 158, pp. 181-199. J. A. Serret, Sur un probleme du calcul integral,Comptes Rendus de 1'Academie des Sciences, Paris, 1869. Vol. 2, p. 1132. Sur un problhme de calcul integral, Annales Scientifiques de 1'Ecole tiormale sup. Paris, 1869, Vol. 6, p, 177.
Q 50. Expansion of factorials into power series. Stirling's numbers of the first kind. The expansidn of the factorial into a Maclaurin series gives
n+1
(1)
(x)n
=
,zl
x"
[$! D"
(x)n j x=o
Denoting the number in brackets by Sr we have (2)
S: = [A Dm(x),lz=,
(X)" = six + SfX2 + six3 + , . . + S;x".
so that equation (1) may be written (3) The numbers 3: defined by equation (2) are called Stirling's Numbers of the first kind. They were introduced in his Methodus Differentialis, in which there is a table of these numbers (p, 11) up to n=9.24
?a 0. Sch&niZch, Compendium der Hijheren Analysis, Braunschweig. 1895, II, p. 31, gives a table of these numbers, which he calls Facult%tencoefficienten. His notation is different; his , p-m corresponds to our c: / rr I A larger table is to be found in: Ch. Jordan, On Stirling's Numbers, TBhoku Mathematical Journal, Vol. 37, 1933, p. 255. J. F. Steffensen, Interpolation, London, 1927, p. 57, introduces numbers which he calls differential coefficients of nothing, denoted by DmOt-nl, and gives a table of these numbers divided by m!; his DmObl corresponds to our 1x1 ml Niels Nielsen, Gammafunktionen, Leipzig, 1906, p. 67, introduces the "Stirlingschen Zahlen Erster Art"; his / St:;-"' / c; corresponds to our [nearly the same notation as Schlomilch's).
1 143 Stirling's Numbers are of the greatest utility in Mathematical Calculus. This however has not been fully recognised; the numbers have been neglected, and are seldom used. This is especially due to the fact that different authors have reintroduced them under different names and notations, not mentioning that they dealt with the same numbers. Stirling's numbers are as important or even more so than Bernoulli's numbers; they should 'occupy a central posiltion in the Calculus of Finite Differences. From (2) we deduce immediately that Sjl=O; S:=l a n d S: =O i f m>n. The other numbers could also be calculated by (2), but we will give a shorter way. Let us write the identity (xl,,, = (x--n) (~1~~. (4) If we put u = x-n and u = (x),, then Leibnifr's formula giving the higher derivatives of a product (S 30) is
Dm(uu)
that is-
= uDmu + ( 71 Du 0'9~
Dm(~)n+l = (x-4 DmW. + morn-l (4,.
Dividing both members by m! and putting x=0 we get, in consequence of (2), s;+, = St-l - n S:. (51 This could have been obtained from (4) by expanding the factorials (x),+1 and (x),, into power series (3), and writing that ihe coefficients of xm in both members are the same. of the first Equation (5) is a partial difference equation order. It is true that the general solution of this equation is unknown; but starting from the initial conditions we may compute by aid of this equation every number Sr. According to 6 181 one initial condition is sufficient for computing the numbers Sz; for instance S;l' given for every positive or negative integer value of m. But from definition (2) it follows directly that S; = 0 if m =t= 0 and Si = 1. Therefore, putting into the equation n=O we get S;; =Sr--l and this gives
From equation (5) we conclude that .the Stirling numbers of the first kind are integers.
145 Putting x=1 into equation (3) we get if n>l (6) s,t +S,Z+S~+#..+S~=0.
That is, the sum of the numbers in each line of the preceding table is equal to zero. This can serve as a check of the table. To obtain a second check, let us put x = -1 into equation (3) ; we get n+1 (-l)nn! = z (-1)m s; . St--l By aid of (5) it can be shown step by that the sign of S:: is identical with that of I-l)"-"; therefore we have the second check :
viz., the sum of the absolute values of the numbers in the line n is equal to n! . Generating function of the Stirling numbers of the first kind with respect to m. From equation (3) we conclude that the generating function of SF with respect to m is (81 u(f,n) = VI,,
that is, in the expansion of u(t, n) in powers of t the coefficient of t" is equal to S;'. In fact, the Stirling numbers, in their definition (2), are given by their generating function. Q 51, Determination of the Stirling numbers starting from their definition. The expression x(x-1) (x-2) . . . (x-n+l) = 0 may be considered as an equation whose roots are Ui=i for i=O, 1, 2, . . . , n. Hence in this equation the coefficient of xm is equal to (1) St = (-l)n-m z u,u,. . . u,-m where the sum in the second member is extended to every combination of order n-m of the numbers 1, 2, . , , (n-l), without repetition and without permutation.
10
146 Equation (1) may be transformed by multiplying the first factor of the second member by (n-l)! and by dividing the quantity under the sign X by (n-l) 1 . This gives sz = (-l)n-m ( n - l ) !
u1ug.. . u,-1
x - -
1
.
The sum in the second member is extended to every combination of order m-l of the numbers 1,2,3,. . . (n-l), without repetition and without permutation. Formulae (1) and (2) show also that the sign of S;; is the same as that of (-I)"-". Later on we shall see that it is possible to express the sum (2) by others having no restriction concerning repetition. We may obtain a third expression analogous to (1) and (2)) starting from the following equation, which will be demonstrated later (Q 71). We have (3) On the other hand, from the known series of log (l+t), applying m times Cauchy's rule of multiplication of infinite series, we get 44) s l [log (l+t) lm = ( - 1 ) " n;m (-1)"f" x u,u,_...
Qm
It would be possible to continue in this manner and determine S: and so on, but the formulae would be more and more complicated. 4. If we put m=n+l into equation (1) we have S %+I = s:: = . . . . = s: = 1.
a-51
150 of the 6 -th degree; indeed, the degree is increased by one by the multiplication, and by one again by the summation. Continuing in this manner, we obtain St-"'which will be a polynomial of the variable R of degree 2m. Let us write it in the following way (3) s;-'" =C~,~(2"m)+Cm.~(2mn_~)f.~.+C~,,~. After multiplication by --n the coefficient of C,,,. will be -n I 2nLv 1 - -(2m--v+l)[ therefore we have 2mTy+1) -Pm--4 (2m5v)
Sy--l = z cm+,,,. [ 2tn;2-v I v=o From the last two equations we deduce (4) G+1,* = - (2m-V+ I) [C,. a, + C,,,. J and k = Cm+,. 2m+2 . equation The general solution of this partial difference is unknown, but in some particular cases the solution is easily obtained. p-' = - ;I From 9' I 21 we conclude that C,, o = -1 and C,,,. = 0, if v > 0. Putting m=I into equation (4), it fohows that C';,, = 0' if Y > 1. From this we conclude, putting m=2 into (4), that
2nzJ-3
151 C,,,. = 0 i f v>2 and so on; finally we find Cm,,. = 0 if v > m-l, Therefore we have also K = C,, 2m = 0; and equation (3) will be
Remark. From formula (5) we may deduce the differences of SZ-~' with respect to n: m I1 4" S,:--"' = ,z, cm.r I 2m--,,-jJ I and in the particular case of ,u=2m A-m S y = C ,,,, ,, . #I From equation (5) we deduce the important formula:
The difference of the quantity in the brackets is understood tobe with lrespect to n. From (9) we conclude, r+l instead of m z ?I-" --CT @ iw =
ISpI
(p-1) I '
Remark 2. Formula (5) is advantageous for the determination of Si-" if R is large and m small. For instance we have
Q 53, Transformation of a multiple sum "without repetition" into sums without restriction. Let us consider the following sum
(1)
cTrn=
8' f 1 ,a,=1
I:
n+l
z
,,.
*+1
r,
f(u,)f(u,)...f(u,)
"F 1
'a,,,- '
from which repetitions such as ui = nj are excluded. Putting n+l ,,x, [f (Ui)-Jk = a/c (2) i it is easy to show that 8; may be expressed by sums without restriction, We have &I = z f (u,) x f (u,) - Bf (u,)f (u,) = ff12-(J2 * Indeed, to obtain 4 it is necessary to subtract from the first expression the terms in which u1 =u2. In the case of 05 we should proceed in the same manner: from ui3 we must subtract the terms in which uI=u2 or u,=u, or uz=u:,; we obtain o~~--~o~o~. But in this manner we have subtracted the terms in which u1 =u2=zzzl three times, therefore we must add a, twice. Finally eq zz a, 3-3a,n, + 20,. Continuing in this manner we could show that & may be expressed by a sum of terms of the form a0,j.l 02L ,..arnj.m
154 where 1.1 may be equal to 0, 1,2,3,. . . , but are such that
rl,+2f.2+3?.3+...mi.,=m,
and o is a numerical constant. To distinguish these constants, we will introduce first the index Eli which is equal to the degree of the term in 0. Since there may be several terms of the same degree, we introduce in these cases a second index pi ,u= 1 will correspond to the term in which i., is the greatest (Z%i being the same), ~=2 to the following, and so on. If there are several terms in which r7,i is the same, then these are ranged in order of magnitude of 1, and so on. The number of terms in which Zl.i is the same, is equal to the number of partitions of the number m, with repetition but without permutation, into Xl.i parts (i. e. of order 23.i). Adopting Neffo's notations (Combinatorik, p. 119) this number will be written:
It is difficult to determine these numbers in the general case, but there are formulae by aid of which a table containing them may be calculated step by step. For instance
where Zi&=m and the sum is extended to Xlli = 1,2,3,. , . m, and ~1 varies from 1 to P( fm). Example. Let m=4, then the number of the terms will be, according to the second table, equal to five. Moreover two terms will coriespond to Cli = 2 (First table). Indeed the partitions are 1+1+1+1, therefore e will be <=a4u14 + a3 o1202 + a2,1 0~0:~ + a2.2 022 1+1+2, 1+3, 2+2, 4,
+ alo,,
Determination of the coefficient a, ,, , First we wiil remark that they are independent of n and of the function f(u) chosen. Therefore we may deduce these numbers by choosing the most convenient function. Let us put first
156 f(q) = 1 ; according to formula (2) we have cr = R, for every value of k. Hence @m becomes oYA = a, nm + a,-, nm-' + , . . , + (az+l + aI+2 + , . , .) n'"i + + .,. + a1 n but the first member is equal to the number of combinations of order m, with permutation but without repitition, of the numbers 1,2,3 ,.,., n, 1 and 1 therefore equal to (n), . From this we conclude, taking account of equation (3) of 8 50, that the coefficients may be expressed by Stirling's numbers of the first kind: (4) and a , = S;; a,-, = S;-* ; a, = S!,,
aa,. + azli,2 + . , . . = S,f;"i This gives m equations, since the number of the unknown coefficients is equal to .I'(J m ) an daccording to the second table we generally have r(! m) > m; therefore we need more equations. To obtain them we will put f(q) = a"i where ui=1,2,3 ,..., n. Since the coefficients a are independent of n we may put n=w therefore if a < 1 it follows that: c-2 0, = - * l-afi Hence, according to (3), we have (5) Pm = X axj.i.!, c~"'(l-(il)-~~(l--~'))-~~ . . . (1-a"*)+n where, as we have seen, Xi& = m. On the other hand, in the case considered &I Gxl = 2 . . . , x a%+%+, . . +t4tj, . The coefficients of a"' in (5) and (6) are equated, for il) = m+l, m+2, m+3,. , . . and so on, till the necessary number of equations for determining the unknown number a.,,, is obtained.
1
157 The coefficient of aw in (6) is equal to the number of partitions of w into m parts (ui = 1,2,3, , . . ) with permutation but without repetition. Netto denoted this number in his Combinatorik (p. 119) by v$w; 1,2,3,, . . ) = v++ The general expression for this number is complicated, but we know that (6') and Vm(Jo) = m ! i f ~=(~$l]. Indeed the smallest sum, since there cannot be repetition, is 1+2+. . . * +m=("$'). The coefficient of am in (5) becomes, after the expansions, equal to Vm( Jw) = 0 i f w < jm$i)
b-1) x,+x*+-.
.+xtn axl+2q+.
.+NIx,+m,
In the sums above, xi takes every value of 0, 1,2, . . a with repetition and permutation, but so as to have 03) x1 + 2x, +, , , . + mx, = o-m.
Moreover the first sum in (7) is extended to every value of Xl.i and ,u such that Zili = m. In this way we get from (6') with the m equations obtained previously, in allm+1 1 equations, which will generally be t 2 sufficient to determine the a,,+,, since according to our table
But if necessary it would be easy to get even more equations. Example. Let us determine ~5";. Here Xiii = m = 4; we have seen already that
Since there are five unknowns, we need one equation more. To obtain it, let us remark that since 0=5is less than m+i =lO 2 ( 1 the coefficient of cs in (6) according to what precedes, is for m=4 equal to zero. We get an expression for this coefficient starting from (8): x,+~x,+~x,,+~x, z 1. 'fhe only solution of this equation is x, = 1 and x2=x2= =x,=0; therefore from (7) we obtain the required coefficient
159 can only be l/l .2.3.4, and this in every permutation, that is 41 times. Q 54. Stirling's numbers expressed by sums, Limits. From formula (2) of Q 51 we have
where the sum in the second member is to be extended to every combination of order m of the numbers 1,2,3,. . . , R, without repetition and without permutation. If there is no repetition the last restriction may he suppressed when dividing the second member by m! . In order to suppress the first restriction concerning repetition, let us put
According to formula (3) of Q 53 the sum in expression (1) may be transformed into ordinary sums, writing
543839'8. In this result the error is equal to permits us to determine the approximate n is large. Since according to (14) 0 87 for we have approximately
a, = F(n) + C - log (n+l) + C St+, - (-I)"+1 n! (log (n+l) + C].
The logarithm in this formula is a Napier's logarithm. If for instance n=ll then log (n+l) + C = 3'062123. Multiplying it by ll! we obtain the result with an error of 1'4% in excess, Knowing the first values of the coefficients a12z,P we may deduce from (2) the limits of certain expressions containing Stirling numbers, if n increases indefinitely. For instance ,%J ( n + l ) ! Since
I s:,::: I
=lim-+[s-(!J) s +,.,,]. ,,=cs Tn.
and moreover
is Uk I - 6
if
k>l
lim a= m log;+l) = lf I
it 'follows that (4) and (5)
Ilog (n+W = o ;A: ___n
I s:$j I 1 Et n! [log (n+l)]m =m!*
161 From this formula asymptotic values of S'$: can be obtained, for instance
IW
- f=Jf [log n + Cl"'.
Variation of S; in the lines of ihe table; that is if n remains constant and m increases. Formula (2) permits us to determine approximately the value of m corresponding to the greatest term. We have, writing n instead of n+l
01 m-
If n>2 then for m=l this ratio will be greater than one, that is the numbers will increase at the beginning of the rows. This can be verified in the table from r&3 upward, S,X will be the greatest term of the row, if
The first two items can be checked by our table. Change of S; in the columns of the table; that is if m remains constant and n increases. From formula (2) we may obtain
11
162 (7) This will lead, for instance, to the following asymptotic formula (8) Sm n+1 nSz
which gives acceptable values if m is not too large. If m=l, the values given are exact. We have seen that, m being given, the limit of is equal to zero; now we will determine the differences of this expression with respect to n. Formula (9) 8 52 gives (9) Writing s:: In ! I and applying to the second member the formula (1) of 8 30, which gives the difference of a product, we get
(10)
( S"-- 1 IS9 f-T =.-( n - l ) ! + nil ' --ii?- = ,, n! (n-kiln Ai--/
Therefore we conclude that if A /SF-~ I > 0 ,I# then we have
Therefore the quantity S$ ' will decrease with increasing I I n if m is smaller than the index corresponding to the maximum of ISt-ll in the row n. We have seen that this maximum is obtained if m-l is equal to the greatest integer contained in lognfl. H e n c e i f m-l>logn+l, 2 will increase I n! I
with increasing n.
163 i f n=B, t h e sy: quantity ] S;-11 will be maximum if m-l=3 and 2. will I I decrease with increasing n if m < 4. This can be checked by the table. Remark. From formula (9) it follows that For instance, since log 8 = 2'079,
/ p+1 1
A-lIsr'
n!
-
2
(n-ijT + k'
§ 55. Some applications of the Stirling numbers of the first kind.
1. Expansion of factorials and binomial-coefficients into power series. We have seen that
(1)
a n d (2)
0
X
n
+$ s:x*. . m-l
The expansion of the generalised factorial was *+1 (X),J = x1 s:l Xrn Pm. (3) The above formulae enable us to determine the derivatives or the integral of factorials and binomials. We have (4) and
Moreover we may determine by aid of the preceding formulae the factorial, or the binomial moments of a function f(x) expressed by power-moments, Let us recall the definition of these moments. Denoting the power moment of degree n by dn, the factorial moment of degree n by 9X, and the binomial moment of degree n by a,, , we have d/Y, = 5 x"{(x); .X=0 mm, = E x=0 (X)"f(X)
164
&I = 2 (;] f(x).
Therefore if we expand (x), into a power series (1) we have
Expanding z into a power series (2) we obtain I I *+t 1 Particuht cases. Expansion of the factorial (x+11-1), into a power series. Since we have (x;tn-1)" = ( - 1 ) " (-X)" it follows that *+1 (x+n-l)n = XI s+l (--l)*+m s; xm = z, I sr I xm.
Therefore it is sufficient to express by differences the derivatives of f(x) for X=Q. To obtain them we will expand f(x) into a generalised Newton series
pf(Q)
(2)
f(x) = .y""1,, h"'
According to the formulae of the preceding paragraph we have
( X - Q ) " h"-p s:.
From this we deduce
165
For x=u this expression gives
and finally (3) The value thus obtained put into (1) gives the required formula, If t(x) is a polynomial, there is no difficultyinapplying formulae (1) and (3). If f( x ) is not a polynomial, the Newton series (2) must be convergent and the Taylor series too. Example 2. Given f(x) = l/x; if h = 1 then we shall have Dmf (a) = %A a n d moreover from (3) it follows that -(-1)" n! , Awd = (a+n),, - f
DmW = .z, (~+n) Sit.
Therefore
m
(--i)nm!
If a=1 we get O" ICI ' = 2 (n+l)! * Let us remark, that the last sum is independent of m. Using formula (10) of Q 54 it can be shown that its difference with respect to m is equal to zero for every value of m. Example 2. Let f(x)=2". Putting a=0 and h=l we obtain A"f (x) = 2" and Dmf(x) = 23 (log2)"; therefore according to (3) it follows that
Remark. Formula (3) becomes especially useful if we deal with functions whose derivatives are complicated, and the differences simple. Q 57. Stirling's numbers of the first kind obtained by aid of probability. Let us consider the following particular case of Poisson's problem of repeated trials: n trials are made, and the probability that the i -th trial is favourable is pi = i/(n+l). According to the general theorem of repeated trials, the probability that among n trials x shall be favourable is:
The second sum in formula (1) is to be extended to the sum of the products of the combinations of order s (without repetition and without permutation) of the numbers 1,2, . . . n. But we have seen, in Q 51 (formula l), that this sum is equal to the absolute value of the Stirling number of the first kind qz:-'. Therefore p(x) = "$' (-1)X'" ("x) 's&L.-;.-. s:.x
167 'j a n d I 1 summing from x=i to x=n+l we make use of formula (3), 5 65: Multiplying both members of this equation by
that is, the sum is equal to zero if s+-i and equal to one if s=i. Consequently we shall have
If we denote by B(t) the generating function of the probability P(x), then 'LI, the binomialmoment of order s of the function P(x) is given by
-
It is easy to show that in the case considered above generating function is the following: *+I W) = g (Qi+Pif) (4) where pi = l--pi From this we conclude that
.:
(5)
IS*+l--rl= @y[.g @(f)lt& *+I
This formula may serve for the determination of the Stirling numbers. From (4) it follows that D@(t) = o(f) Determining the s-l -th derivative of this quantity by Leibnifr's theorem, and putting f=l into the result obtained, we get: B e(l) = 2 ( ~']D- 8(l) 2: ( - 1 ) " ~1 pi"+l and in the particular cases:
168
D" e(l) = (Xpi)2-Xp,2 D" 8(l) = (&i) '-3 Xp,
Xpi2 + 2 I+&'
and so on. For instance, if n=6 and 9~3, we have
s:
984.1 x " = 2,
441 +2K3
13 58. Stirling's numbers of the second kiid.25 Expansion of u power into a factorial series. Let us first expand xn into a Newton series:
( x ) , [ $ &. I
1
= 735.
We will call the number in the brackets, a Stirling number of the second kind, and denote it by a;. (11 Hence we have (21 x"=B;x+q(x), q=$ . [ I .X=3 +...+a~(X)m+..,+q(X)..
Starting from the definition (1) we conclude immediately t h a t 8,0= 0 and e$+' = 0 if m>O. Moreover, putting into equation (2) x=1 we get GL= 1.
ss The first table of these numbers has been published in: Jacob0 Stirling, Methodus Differentialis., . , Londini, 1730, p, 8, up to n=9; but the author did not use any notation for them. There is a table in: George Boole, Calculus of Finite Differences, London, 1860, p. 20, of the "differences of zero", which Booze denotes by A"om; this corresponds to A@m = [Anxmlxd = m! 6: the range of the table is up to n=lO. A smaller table of the Stirling numbers of the second kind is given by Odor Schliimilch, Compendium der Hiiheren Analysis, Braunschweig, 1895, p. 31; his notation C," corresponds to our GzTz. Niels Nielsen, Gammafunctionen, Leipzig, 1906, p. 68, calls these numbers, for the first time Stirling numbers of the second kind; his 6:: corresponds to our (--l)"+m ez<A+, In c. Jordan,
10~.
.
cit. 24, p, 263, there is a table of 6," up to n=12.
i
1
169 From the definition we may deduce a general expression for the Stirling numbers of the second kind. Indeed, dealing with symbolical methods in 8 6 we found that the operation of the m -th difference is equivalent to the following: A " = (-l)m 2,' (-1)' (7 ] E'. If f(x) = xn/m! the operation gives for x=0
If a few of the Stirling numbers are wanted, this formula is a very convenient one to determine them; but if we want to compute a table of these numbers, then there is a better way. The formula giving the higher differences of a product is the following (Q 30):
Am(m) = v(x+m) Amu + ("If ]Av(x+m-I) A*-lu +. . uAmu(x).
Putting u=x and u=xn, gives Am xn+' - (x+m) Amxn + mAmsl xx"; dividing both members of this equation by ml and putting x=0 we obtain according to (1) (4) =:+I = a;--' + me;. We may also obtain this equation starting from x"+l = x . x" and expanding xn+l and x" into series of factorials, then writing x . (x), = (x)i+l + i . (x)~ and finally equating the coefficients of (x), in both members. We shall see in Q 181, dealing with partial difference equations , that the solution of equation (4) leads to formula (3). But if we want a table of the numbers 6," there is no need to solve this equation. Starting from the initial conditions G,O=l a n d Ga=O i f m=/=O which follow directly from the definition (l), we may obtain these numbers step by step, by aid of the equation (4). From this equation it follows that the Stirling numbers of the second kind are positive integers.
170 Stirling numbers of the second kind 6",
4"
2 3 4 5 6 7 8 9 10 11 1 2
1 1 1 1 1 1 1 1 1 1 1 1 1 7
2 1 3 7 15 31 63 127 255 511 1023 2047
3
4
5
1 6 25 90 301 966 3025 9330 28501 86526 8
1 10 65 350 1701 7770 34105 145750 611501 9 10
1 15 140 1050 6951 42525 246730 1379400 11
1 21 266 2646 22827 179487 1323652 12
4" 1
2 3 4 5 6 7 a 9 10 11 12
1 28 462 5880 63987 627396
1 36 750 11880 159027
1 45 1155 22275
1 55 1705
1 66
1
If we put x=-l into formula (2) we get
!5)
( - 1 ) " = mz, (-1)" m ! a;.
n+1
This equation may be used for checking the r?umbers in the table. In some particular cases the resolution of the difference equation (4) is simple. The results obtained may shorten the computation of the table. For instance, putting into (4) m=l we have ~:I-1 =q=q =l. Putting m=n+l it follows that
Hence to have 6iw2we multiply a:'== ( ; 1 by n-l and perform the operation A-': consequently s,:-~ will be of the G;--3 of the sixth, and so on. et-'" fourth degree in n, will be a polynomial of n of degree 2m. Let us write it as follows: (7)
172 s+ 0. Starting from this, we obtain by aid of (8), as in 8 52, step by step, that C,,,-- if s > p-1; hence we have also k=O, -0 Therefore, putting s=m into (8) we get: cn+,,nI = cm. m-1 = Cl9 o = 1 . If s=O, from (8) it follows that L,o = I2m+lL, o. The solution of this difference equation is
From (8) let us deduce an equation which may serve for checking the numbers of the table. Multiplying both members of the equation by (-1)" and summing from s=O to s=m+l we get m+l z (--llS G,,, = ,,i I(-11s (m+l---4 C+, + Gzo + (-1)" (2m+l-s) C,,S]. Putting into the first part of the sum in the second member s instead of s-l and simplifying we obtain
nI + 1
.s=o
z
I-11"cn+,,s
=
( m + l ) ,E I-~)'CW.
Denoting the sum in the second member by f(m), we have f(m+l) = (m+l)f(m).
173 The solution of this difference equation f(l)=C,.,=l &=I and finally we have (9) is f(m) = ml k. Since
r=o
5 (-l)sC,,s = ml
This relation may be used for checking the table. From (7) we conclude that
Remark Formula (7) is advantageous for the determination of GZ; if n is large and m small, For instance df R= 12 and m ~3, we have
a72 =15/~j+10(tj2]+[ '42]=22275.
The computation can be made very short by aid of a table of binomial coefficients. 5 59. Liits of expressions containing Stirling numbers of the second kind. From equation (3) Q 58 it follows that
; ..
If n increases indefinitely, every term of the second member will vanish, except that in which km; therefore we have (2) This gives an asymptotic value for large n
The values thus obtained are acceptable if m is small. For m=l the value is exact; for m=2 it gives P1 instead of 2"-l-1. If m>2 then the error increases indefinitely with n. From (1) we may deduce
I
174 Neglecting the terms which are not written in this formula we may determine the greatest term of the row n in the table, E: will be the greatest term if for m=X the ratio (3) is smaller than one, and for m=X-1 greater. According to (3) the conditions fo;. a imaximum are (x+.1)"-' + Xl%--l)"< 22" y-1 + (a-1) (x-2)"-> 2(%-l)". By aid of these inequalities we find, for instance, the following greatest terms: C$ @i, a:,. Our table shows that this is exact. From formula (3) we may deduce the following limits:
and from (2)
If m is small in comparison with R, then the asymptotic formula G31 y me,:: gives a relatively good approximation. For instance from S:, we should get E:, - 85503 instead of 86526. Starting from the polynomial expression of @;-"I, formula (7) 8 58, we may deduce: (6) 1.3.5.. . (2m-1) 1 lim 11= A!&.~ =m12"* (I=rn .2m (2m) !
~?l-wI
We have seen, Q 52, formula (S), that the limit of the corresponding Stirling number of the first kind is the same. 8 60. Generating function of the Stirling numbers of the second kind, with respect to the lower index, Starting from the difference equation of these numbers (4, 0 58) (1) Emil - - #,I s1 H+l - ( m + l ) G;+l = 0
we denote the generating function of Zr with respect to n by u(m, t) and determine it by the method of Q 11. We have
175 Ga; = u(m,t) = 5 GEt". I-1)
0bviolrsly it follows that G@,m+l Since 6,"+' = 0,
= u(m+l, t),
Finally writing that the corresponding generating functions satisfy equation (l), we obtain (l-f-mf) u(m+l, f) - f zz(m, f) = 0. This is a homogeneous linear equation of the first order, with variable coefficients; its solution is, as we shall see later,
11 u(m, fl = w(f) ,fl, (l--tt_if)
where co(f) is an arbitrary function of f, which may be determined by the initial conditions. Since $ 1 = 1 and therefore ~(1, f) =f / (l-f), we conclude that w(f) ~1: so that the generating function with respect to n of the numbers ,Gc will be (2) u f ii qf", (l-f) (1-L) (:;S , , , (1-mf) = n=a,
Remark. The expansion of this function in a series of powers of f gives the solution of the difference equation (1). This is Laplace's method for solving partial difference equations. To expand the function u let us put f =1/z; then we may write
zz = z/2(2-1) (z-2) . . . (z-m). This decomposed into partial fractions (Q 13, Example 2) gives
u = c---1)"' z "$1 C-lV[~)
m! I=0 r-i
Putting again z=l/f we find
'
176
u&!.L x ml r=O
m m+l
Therefore the coefficient of fn is the following (-l)m $ y$ (-1)' f 71 i" . . I This formula has already been obtained in 5 58. The generating function (2) may be written in the following form (31 (4) u= (l+f+f2+, . .) (1+2f+2?fz+. , .), .(l+mf+m2f2+. , .) f". From this we conclude that Gt , the coefficient of f" in the expansion of this function is equal to the sum of the products of the combinations, with repetition, but without permutation, of order n-m of the numbers 1, 2, 3, . . . , m. Example. S; = 1.1+1.2+1.3+2.2+2.3+3.3 = 2 5 az, = 1.1.1.1+1.1.1.2+1.1.2.2+1.2.2.2+ t-2.2.2.2 = 31. We may deduce another rule to obtain these numbers. f2 t3 a'---1 = f+2!+3!+..,; ,
a; =
therefore using Cauchy's rule of multiplication of series wz get 1)m
(e'-
=
5
n=,,,
f"
z
r ,t (l
1' 2. . . .r,l
where the second sum is extended to every value of ri>O (with repetition and permutation) such that t, + t2 + . . . + t, = n. We shall see later that (ef - 1)m = ,,Z,"$Q f"; therefore we conclude that ..rm.
(51
where the sum is formedjas above.
e=mq z rlrll. 1 1 2
177
Examples.
q = 5.4 lt1f3,+ 3 112121 3 65; = 6 . 5 . 4 . 3
1
= 25
+3131. 1 1 - 31. -
Q 61. The Stirling numbers of the second kind obtained by probability." Ja Let an urn be given, which contains the numbers 1,2,3, , , . ( m. Let us draw successively n numbers, putting back every time the number drawn, before the next drawing. The probability that in R drawings the number one should occur rl times, the number two r2 times, and so on, finally the number m should occur r,,, times, is, according to the generalised Bernoulli theorem of repeated trials, equal to
(1)
n! 1 - * rl ! r2 ! , . . r," ! I m I
Let us now determine the probability that in n drawings every number out of 1,2,3,. , . , m, should be drawn at least once. This is given by the theorem of total probabilities (21
P= ($"
z -nf t, ! t2 I,, . fm !
where the above 'sum is extended to every value of ri > 0 with repetition (and permutation), but satisfying the condition
m! S; , so that tb.e required probability is
(3) On the other hand this probabilrty can be determined by using the generalised Poincard theorem, which gives the probability that in n trials every one of the m events should occur at least once. According to the theorem, this probability is ob'5. Ch. Jordan, Theo&me de la probabilitk de Poincare #nCralis& Acta Scientiarum Mathematicarum, Tome VII, p. 103. Szeged, 1934,
178 tamed by subtracting from unity the probability that one of the events should not occur; this is
then adding the probability that two of the events should not occur I:] [?I" then again subtracting the probability that three of the events should not occur 1:) pg]" and so on. Finally we have p = A;; 2 *0 (-1)" [y ] (m-s)".
Equating this value to that obtained above (3), and putting m-s=i, we get the expression forthe Stirling number obtained before (55 58 and 60):
Remarks. 2. From formula (1) we may deduce in the same manner the probability that in n trials, out of the m numbers there should be drawn any ,IL numbers, no more and no less; this is
where the sum is extended to every value of ti > 0 (with repetition and permutation) such as r, + r2 + . , . + r,", = n. Then according to (5) Q 60 we shall have
l
and finally the required probability will be (4)
179 2. From formula (3) it follows that the number of such combinations with repetition, of order n, of m elements in which each occurs at least once, is equal to m!E+,," . 3. Moreover from (3) we may obtain the probability that in R trials every one of the m numbers occurs, but the last of them only at the n -th trial. This is:
where every ai is a different prime number, Let us denote by f (n, V) the number of ways in which w, may be decomposed into Y factors (without permutation). For instance, if w3 = a,azan we shall have f (3,l) = 1 ; f (3,2) = 3 ; f(3,3) = 1 ;
(ala2agl
(alazlas t kwh I (a3alla2 (4 (4 (4 From these we may easily deduce the number of ways in which o4 = a,a2aJa4 may be decomposed into factors. For instance, the decompositions of wI into three factors will be obtained: First by adding a, to each decomposition f(3,2) : (ala21 (a:,1 (aA q (a24 b-4 (4 I b,a,l Ia,) (a,1 -
Thus we have obtained every decomposition of o, into three factors, and each only once; therefore we have f(4,3) = f(3,2) + 3f(3,3). Proceeding exactly in the same manner we should obtain f (n, V) starting from f (n-l, y-1) and f (n-l, Y); we will obtain them first by adding the factor a, to the decompositions
180 f (n-1, r-l), secondly, by multiplying successively every factor of the decompositions f (n-l, V) by a,, ; therefore each of these decompositions will give Y new ones, so that we shall have f(np) = f ( n - 1 , V - l ) + vf(n-+). But this is the equation of differences which the Stirling numbers of the second kind satisfy; moreover the initial conditions are the same, indeed i(n,l) =l if GO and f(n,l) =O if n 4 0. Therefore we conclude that f(n,v) = e;. That is: a product of n different prime numbers may be decomposed into v factors in (5): different ways (no permutation). Starting from the number of the decompositions of
w,-* = ala*. . . . ans2
into v factors, denoted by f(n-2,v) we may deduce the number of decompositions of
0, = a"ala2 . . . . a,-*
into v factors, which we will denote by F(n,v). Let us remark that into w, the prime number a enters twice. The decompositions of w, into v factors are obtained: First, from those of w,-~ into v-2 factors by adding to each of them the two factors a. a ; the number of decompositions obtained in this way will be equal to f (n-2, v-2), Secondly, from those of w,-~ into v-l factors by adding the factor a and then multiplying each of the Y factors by a; the number obtained in this manner will be equal to v . f (n-2, r-1). Thirdly, from those of w,,-* into v factors by multiplying two of these factors by a; the number of ways in which this may be done is equal to the number of combinations with repetition but without permutation of v numbers taken two by two, that is, r+1 2 ' Hence the number of decompositions thus obtained will be equal to I 'J1 1 f(n-2,v),
Expanding the factorials figuring in the second member of (2), into a series of powers of x we find, if h=l, (4) A-'x" x+1 v+z = I: &:zs B xl" . v+1 "== 1 v+1 g=l
182 Equating the coefficients of x1 in the expressions (3) and (4) we get
(5)
This is an expression for the Bernoulli numbers by Stirling numbers of the second kind. 3. Expression of the power moments A'&, by factorial moments ?m, and by binomial moments B,,,. In $ 55 we denoted by JM, = 2 x"f(x) m, = Xi x ( x - l ) . , , . b--n+11 f(x)
From (1) it follows that (6) and (7) A!,= *+1 I: cf:rm, PIi=1
A,=
X ?/#=I
*+1
m!G,"C%,.
Q 64. Some formulae containing Stirling numbers of both kinds. Let us expand (x), into a power series, and then expand the powers into a factorial series again. We get
a+1
(x), = z1 St' xi =
r+1
z i=l
83-l z s; 6; ( rnn=l x ) , .
Since the coefficients of (x) y in the first and in the last member must be identical, we conclude that
(1)
i=tn
n+1 x q
q= n -Qm
I
1
that is, this sum is equal to zero if n is different from m and equal to one if n=m; the binomial coefficient z being equal I 1 to zero for every positive or negative integer value of m ac0 cording to formula (4) of Q 22. Moreover 0 = 1. 1 1
183 In the same manner expanding x" into a factorial series, and then expanding the factorials into a power series again, we find
X" 1 n+l zl 6: (X)i = n+l X id i+l I: n=l a: S;lXm*
Since the coefficients of xp in the first and in the last member must be the same, we have
This is also equal to zero if n is different from m and equal to one if n=m. The limits of the sums (1) and (2) can be made independent of n and m, since for i<m or ih the expressions under the sign B are equal to zero. We may sum from i=O to i==; this is often very useful. We shall see in the following paragraph that the above formulae may serve for the inversion of series. Q 65. Inversion of sums and of series. Sum equations. In certain cases it is possible to perform this inversion, by which the following is understood. Given (1) f ( x ) = 4 fp(i) q(x,i) kU
where f(x) and y (x,i) are known and qt (n) is to be determined: Let us suppose that equation (1) holds for every integer value of x in the interval d h x 2 y. If it is possible to find a function w(x,n) such that
; x=y 9(x4 w(w) = [ .ti]
that, is if the first member is equal to zero for every value of i different from n, and equal to one for i=n (supposing P 2 n L a), then multiplying both members of (1) by w(x,n) and summing from x=y to x=8 we obtain i f ( x ) co(x,n) = y ( n ) . I-Y Indeed in the second member of (1) each term will vanish except that in which i=n.
184 The only difficulty is to find the function O(X,R) corresponding to q&i), We have deduced already a few formulae of the form (2). For instance we had in 5 45 (3) (4) and in Q 64 (5) (6) In these formulae, the limits may he made independent of n and m since for m>xkO, the first members of (3), (5) and (6) are equal to zero and also for x>n. Therefore we may sum from x=0 to x=00. Example 1. Given the Newton series
Indeed, in consequence of formula (2) Q 64, the second member summed from n= 1 to m+l would give m; but since the first member is to be summed from n=2 to m+l and therefore the second member too, we must subtract from m in the second member its value corresponding to n=l, that is
so that it will become equal to m-l. 3. Formula (2) may be written
I&rItiplying both members by G; and summing from n=v to n=p+l will give
multiplied again by Er and summed from y=rn to v=p+l we obtain (9) 5. Starting from x(x), and expanding the factorial into a power series, and then expanding the powers again into a factorial series, we get I.+1 ni2 n+2 X(X), = Z Sk-' X' = Z ,", Si-'G;, (X)i
t-2 ,.=:2
but we have also xIx)n = (x)n+1 + n (X)" therefore we conclude, writing vfl instead of v, that
That is, the first member is equal to zero if i is different from R and n+l; moreover, it is equal to one if i=n+l and equal to r~
if i=n.
6. Starting from x. x" and expanding x" into factorials and then the factorials into a power series again, we find
188
Eom this we conclude (11)
n+1
V=l
x s;+,
@" + z1 VS: q
n+l
=
I&]~
~
Bat is, the first member is equal to zero if i is different from n-tl, and equal to one if i=n+l. Ih consequence of (6) it follows that WI if ra is different from i-l. This formula multiplied by 67 and summed from i=m to i=n+l gives (13) Equation (12) multiplied by S; and summed from n=i to n=m+l gives "2' (-l).-ci ( i21) Sk = m Si, .
n+2 is equal to zero, since the quantity in the second member is
at both limits equal to zero. That is n+l z (-1)m ( m - l ) ! G; = 0. (17) m=l Moreover if m varies from p+l to n+2 formula (16) gives
n+2
Q 67. Differences expressed by derivatives, Given t(a), DW, D"f(4,. . . and so on; it is required to determine &f(x). Starting, from Newton's series,
if this series is convergent, we may determine the differences of f(x) term-by-term
Therefore it is sufficient to express, by derivatives, the differences of the function f(x) for x=u. To obtain them we will expand f(x) into a Taylor series:
f ( x ) = 5 (x-fq n==o
p (4
n.I
*
But according to formula (1) Q 58 we have
[I;\"(x-u)~]~ = ml fP Q;
190 therefore (3) yf(o) = Ii $ h"D"f(a) n=m . 6;.
The value thus obtained put into (2) gives the required formula. Formula (3) may be deduced by inversion of the series (3) § 56. Indeed, multiplying this series by hm65;,lm! and summing from m=R to m=m, we get
We have seen (5 5 65) that the second sum of the second member is equal to zero if n is different from K and equal to one if n=k. Accordingly we have 9" f ( u ) = E, -$ h" D"f(a) ali,. This formula is identical with (3). Example 1. Putting into formula (3) h=l and f(x) = log x we find: (4) Amlogx = l o g e 2 (-l)"+lm' t$ n x" "=,,I
From this we conclude according to 8 6, that the last sum of the second member is equal to
But in consequence of formula (1) , 8 59 we have
191
therefore
We remark that if p is odd, the terms of this sum ~wrrespending to i and 2v-i are equal and of opposed sign, moreover the term corresponding to i=v is equal to zero. Therefore if p is odd the sum is equal to zero, so that we may write Z insted of p in formula (5). Determining the first two terms of (5) we have r(2v+l) I (2-l) ! A2r logx = loge - (x+V)2r 12 (x+rP+ C
Particutar
cases. 1 ' 2(x+1)4 - 3(x+1)6
A210gx = loge [- (x:1)" -
A'
log x = log e
- (xt2)J -
-..,. .
Two terms of the first expression and one of the second are sufficient for an exactitude to twenty decimal8 if x h lw. In the notation of central differences this is 6210gx=loge and a410gx=loge I
Q 68. Expansion- of a reciprocal factorial into a series of reciprocal powers and vice versa, We found in 9 60, that the
193 generating function of the Stirling numbers of the second kind a:, with respect to n, is the following
(l-t)(l-2;. . . (f-nzf) = & =z t"*
Putting t=l/z (1) we get CP = (&, j& + * If z>m this series is convergent. To show it, let us write 22" = qyz"" and remark. that according to formula (2) 8 59 we have
therefore if z>m this limit is equal to zero. Moreover from formula (5) Q 59 it follows that
Hence if z>m then this limit is smaller than one and the series is convergpnt. Formula (1) may be transformed by writing I=-,; then we have (2) ( - l ) " ( x ) , = (;;l$ In = L (--i)"$.
In consequence of what has been said, this series is convergent if x>m. The expansion of a reciprocal power into a series of reciprocal factorials may be obtained by the inversion of the series (2). Let us multiply each member of (2) by (-1)' S; and sum from m=O to m= =. According to (5) Q 66 every term of the second member will vanish except that in which n=~. Hence we have
Multiplying both members by l/x and writing r+l=n, we obtain a second formula for the expansion of a reciprocal power:
13
194 (4) f = I ft-l I mL (x+dn+, 5
-1
=
I q-,-it (x-l),-, .
This formula was first deduced by Stirling (Methodus Differentialis, p. 11). Application of the above expansions. Formula (1) may serve for the determination of the derivatives of a reciprocal factorial. We find: (5) The'integral of the reciprocal factorial is also obtained by aid of (1) g G,:: - +k. ('3 s &=- (L=m nl Equations (3) and (4) may serve for the determination of the differences, or of the sum, of a reciprocal power. From (3) we deduce (7) and
As f = 5. t-d I S;l (XL,,
A-'$ =-
(8)
(9) and (10)
Starting from formula (4) we get in the same way As $ = ,j-, (-m-l), I S:-ll (~-l)-,,,-~~ 5 r '-l f = -nc=n-l In 1sy I (x-l)-#Jl.
By aid of (8) or (10) we may obtain the sum of the reciprocal powers, x varying from x=1 to x=p. For instance from (10) it follows that
and if ,~=a, (11)
195 This was the formula that Stirling used for the determination of sums of reciprocal power series (p. 28). In the same manner we should have obtained from (8) (12)
.I==1
where the numbers Ci above are independent of the function on which the operation isperformed.Por their determination we can choose the most convenient function. Let f(x) =x1, then we have @xl =axi ; 8"X 1 =a"%% ; Dv x1 = (a), x1-. Putting these values into (1) and dividing by x2 we get =+1 a" = I: c,(n),.. v==l Talcing account of formula (2) of 8 58 we conclude that C, = @; and therefore *+1 8"~ 2, a;xvDv. M This is the expression of the 8 operation in terms of derivatives. Particular cases. @k=o @logx= 1
@X=X.
197 On the other hand from (2) it follows that na+i 0" f ( x ) = W!t q x* ( n ) (x+1)"-*. (5) . Equating the second members of (4)and (5) and putting x=1 we get (6) Determination of the power moments of a function. If u(t) is the generating function of f(x), that is if u ( t ) = j. f ( x ) tx then 0" u(t) = E X" f ( x ) t". x=0 Therefore, if the 0" operation is performed on the generating function of f(x) and then we put t=l , we obtain the n -th power mcment of the function f(x): (7) Example 3. .of repeated trials (8) is equal to u(t) = (q+Pt)"* 'fherefore the m -th power moment of (8) is in consequence of (7) and of (2) equal to I : 1 P' q"' Jz; = [O" u(t)]k,. The generating function of the probability (where Q = l-p)
Jq,=
n+i z v" ", pp q- = "I!, G:, (n), p'. I l-=1
Inversion of the sum (3). Multiplying it by Sz and summing from n=l to n=m+l it results that (9) This formula gives the m -th derivative in terms of the 8 operations.
Sometimes it is possible to determine both members of the equation (9) directly, and they lead to useful formulae. Remark A formula analogous to that of Faa de Bruno given in Q 12 is applicable for the @ operation performed on a function of function. Given u=u(y) and y=y(t), we have
du dy du dt= -- - 9 dy dt du du &l=tz =dy
d2u d2u dy 2 du d2y dtz=dyzz 1 +&a(
and so on; on the other hand
8Y, wu=
and so on. It is easy to see that the two sets of formulae will remain similar, so that to obtain 0"~ starting from a formula
d"u dSy we dt' it is sufficient to put in it @y instead of z . For instance in the case of Faa de Bruno's formula (9, Q 12) we obtain:
The second sum is extended to every value of ai=O, 1,2, . . . such that al + a2 + . . . + as = v
Let us remark that for i= 1 we have [@y]1=, = 0 and therefore pI=O; from this we conclude that in the case in hand, in formula (lo), aI must be always equal to zero. So that a2 + , , . + a, = Y and 2a, + . . . + sa, = s; that is, the partitions of s, formed by 2,3,4, . . . only are to be considered. Formulae (10) and (11) solve the problem.
('J+J)Q . . . . , [q"f(-py* .
Q 70. The operation \v, In the Calculus of Finite Differences there occurs an operation analogous to that of 8, which we will denote by Y and define by y = %A. The operation repeated gives, according to the formula for the difference of a product $ 30,
The numbers C, being independent of the function on which to calculate them we may the operation is performed, f(x) = (x+A-l),, choose the most convenient function. This is %en Awt(x) = (A) (x+;L--l)i-, and Yf(x) = xAf(x) = I(x+&l)~ , Y"f(x) = F(x+A--1)~ . Putting these values into equation (1) and dividing by (x+&1)2 we get
202 We have seen in Q 58 that putting x=0, the difference in the second member will be equal to a>,,, so that (3) This formula is useful for the determination of the multiple integral
Sometimes the operation A"D-" maybe performed directly on the function f(x); then equating the result to the second member of (3) we may obtain interesting relations. Example 2. Let f(x) = ext. We have
Dnf(0)
= t",
D-mf(X) ._ ir,
Am
g = eX'(e'-l)m ; f"
'
therefore from (3) it follows that
(5)
Or putting e'-1 =z we have (6)
zm= L, + 6: Clog (z+l)Y. x
From this formula we may obtain an important expansion by inversion of the series, Let us multiply each member of equation (6) by SE/~! and sum from m=n to m=m. According to the formulae of (j 63 we get (7) [log (1 +z)]n = m1 $:I-. z
This formula has been applied already in $ 51. Operation D"A-". This operation is the inverse operation of the preceding. Indeed, from it follows that
D"A--" f Ix) = q(x)
f(x) = AmD-" y(x).
t
203 The second operation is t&vocal, that is, to a given function q(x) there corresponds only one function f(x) ; on the other hand the first operation is not univocal; to a given f(x) there correspond several functions q(x). Indeed, we have seen that if A-'f (x) =f,(x) then the following equation is also true:
A-W = f, (4 + 44
where o(x) is a periodic function with a period equal to one; the If m = 1 then q(x) = Df,(x) + Dw(x). But derivative of a periodic function is a periodic function with the same period, hence this may be written
Since for t=O the first member is equal to zero, I k=O, and formula (8) btzomes identical with our formula (7) obtained above. $72. Expansion of a function of function by afd of Stirlfng numbers. Semi-invariants of Thiele. Ia 8 12 we found SchZ6miZch's formula (8) giving the n- th derivative of a function of function. If u=u(y) and y=y(x), then (1) y=logx we have
1. Expansion of u= u (logx). Putting into formula (1)
Lyogxl~ =
log(1+ 511' [ and according to formula (7) 8 71
i
1 205
the n -th derivative of this quantity with respect to h is
5 Y I (m),h'"-" sy
In=0
ml xm
(II-
Putting into it h=O every term will vanish except the term in which m=n, that is Y! S;/x"; therefore we shall have (2) -=
d"u dx"
"2 s:, d'u
,.=I x " dy"'
or y=O Taylor's
Finally writing in this equation x=1, theorem will give
(3)
u(logx)
=
II=0
5
(x-l)" y 7 v=o
This is the required expansion.
2. Expansion of u=u(eX). Let us put into (1) y=eX. We have
[$exly =epx (en-l)* . In 8 71 we found (formula 5)
The n -th derivative of this quantity, with respect to h, is
Putting every term will vanish except that corresh=O ponding to m=n, and this will be Y! @);. Therefore we shall have
(5)
We have seen (formula 9 8 12) that if u=u(y) and y;.y(x) are given, the derivatives needed for this expansion are supplied by
Expansion of a function of function by Faa de Bruno's formula.
206 = Where ng d'u ,.=, dy' Gvo and
[1
x
~..__ (Dye)9,,,
n a,!...a,!
a, + a2 + . . . + aR = v
a, + 2a, + . , . . + na, = n.
This formula is useful even in cases where u=f(x) is given, but the direct determination of d"u/dx" is difficult, provided it is possible to find a function y=p,(u) such that the calculation of dny/dxn and of d"b/dy"' is easy. (See examples 4 and 5). Example 1. Derivatives o f u=eOy a n d y=y(x), T h e derivatives for x=0 are obtained by aid of formula (6). Remarking that
U"U
we find
F-1.2/=y, W'
= a*'@.
The numerical coefficients in this formula are independent of the function y; to determine them we may choose the most convenient function, that is y=eX. Then we shall have yo=l and [D'yIXd = 1; therefore from (7) we get
ll! = ea I: a* Z _ dx" a , ! . , . , a,! (21)" . . , , (n!)"" ' *=l I-J xr=o d"eay n+1
.
Comparing this result with that obtained in formula (4), we get an expression of the Stirling numbers of the second kind: n! e:=z J-. . a,1 (21)"s ,. 1 . . . . (?a!)"" '
where the sum is extended to every value such that a,+a,+ + . . . + a,=v and a,+Za,+. . . na,=n. From this we conclude that in the derivative (7) the sum of the numerical coefficients of the terms multiplying a" is equal to B;. If v=l, the coefficient of a(py) is Gf, = 1; and if v=n then the coefficient of a"(Dy)" is a;= 1; moreover if v=n-1 the
207 is equal to @z-l= i . Finally rl the sum of all the numerical coefficients in the derivative n+l dneay/dxn is equal to Z G; ; moreover the number of terms in V==l this expression is equal to the number of partitions of R with repetition, but without permutation; this number, I'(Jn) in Netto's notation. is given in a table in Q 53. For instance in D6eay this will be r(j n) = 11. Moreover we have D6eay = cay . aD6y + coefficient of a"' (Dy)"-'
where a,+a*+ . . . +a, = v and al+2a,+. . .+ na,, = n. The sum is extended to every partition of n, of order v with repetition but without permutation. This formula is different from (5) Q 51, in which partitions with permutation are considered. Example 4. The generating function of the probabilities in Poisson's problem of repeated trials is
To obtain the factorial moments of the probability function the derivatives of u with respect to t are needed for t=l. Indeed
92, = [D"u]r=r..
Since the derivatives of log u are less complicated than those of u we shall use formula (6) putting u=ev and y=logu; the derivatives of y with respect to t are
210 therefore the second member of (10) will be equal to this definite integral. Example 6. Semi-invariants of Thiele.* Let xi and f (xi) be given for i = 0, 1,2, . . . , N-l. Denoting the power-moment of order R of f(xJ by
dn = i Xl"f(X,)
and by A, the semi-invariant of degree n of f(xJ with respect to x0, Xl, ****,XN-, the definition of 1, is
The expansion of the second member gives (12) To expand the first member of (11) by aid of formula (6) % we denote it by u and put y = I,o +1,&o* +yo3 +. , ,; remarking moreover that [Y]~=~ = o , [Dsyl we = & and I $f y=. = l; 1 '
therefore the expansion of the first member of (11) into powers of o will give according to (6)
Since the expansions (12) and (13) must be identical, the coefficients of w" in both expressions must be the same; therefore
giving the power moments in terms
l
of the semi-invariants.
tistiques. Oslo, 1926.
Ragnar Frisch, Semi-invariants et Moments des Distributions Sta-
T. N. Thiele, Theory of Observations, London, 1903.
211
Particular cases:
=$ = CAa + 3&l, + a,q 244 gd4 = [A, + 42,1, + 3122 + 61121, + A,4].i4j
The sum of the numerical coefficients in ?g" is equal to 33; from y=l to ~=n+l. To deduce the formula which gives the semi-invariants in terms of the moments, let us now denote by y the second member of (ll), whose expansion is given by formula (12) : = e-J& lo" y= .zo~on! and moreover let us write u = logy; then dyy=
dw
(-1) *--l (V-l) I Y"
Since for w=O we have y=l, it follows that
On the otherhand the first member of (1 l), equal to y gives u = logy = Zt, W/Y! Therefore, by aid of (6) we find
r-1
d'U
dy' w=o
= ( - 1 ) "-l (v-l)! a n d [Dy],,,=o = 2 .
Particular cases
I
212 Remurk. The sum of the numerical coefficients in the derivative (15) is, according to what has been said concerning formula (6), equal to
n+l
(16)
2, ( - 1 ) w-l ( V - l ) ! e,:
but this is, in consequence of formula (17) § 66, equal to zero. The semi-invariants are useful in ihe expansion of functions, since if the origin of the variable x is changed, the semiinvariants, except the first, do not change. If the unit is chosen c times greater, the, semi-invariant A, becomes equal to @A,, . Therefore unity and the origin may be chosen so as to have A,=0 and 1,=1; this is a great simplification. The invariance with respect to the origin is a consequence of the fact that the sum of the numerical coefficients in 1, is equal to zero, according to formula (16).
$ 73, Expansion of a function into reciprocal factorial series, and into reciprocal power series. Given the function
F(x) for x>O if the solution q(f) of the integral equation
I I
(1)
F(x) = j' y (f) tx-1 dt
0
is known, then we may expand F(x) into a series of reciprocal factorials. For this purpose v(t) is expanded into a series of powers of (l--t) : (21 Since
1 s
q(t) = : (--I)* (l--t)" D"p,U)
n==o n! ( l - f ) " f"-' df = B(n+l,x)
.
0
where the second member is a Beta-function (9 24), we have (3) F(x) =
fi 9 B(n+l,x) D"q(1). . r=o nl (x-111
Bbtw = (n+xlI ;
If x and R are integers, then
213 therefore (4)
=
F(x) = "; &-- D"p,(l) =
"4 b--l)" (x--lL, D%(l) Dwl).
This is the required expansion. If the series is convergent (Q 37) then it may serve for the determination of A-'F(x) and A"F(x). From !4) we get 2. A - ' F ( x ) = ?Z (-l)n+l $ ( x - l ) - , D"p,(l) + q(l)F(x-1) +k a=1 and AmJ'(x) = .:i (m-1)"
(-n-l)m
This problem has been solved by Stirling in another way for a=+&$ (10~. cit. 25. p. 27). Remark Formula (6) is useful for the summation of l/x(x+a). For instance
215 ,g 1
$ (---l)"(a-& ='
*=1
=I
++a) =
n. n!
' (l--t)"-1 s t a0
& .
For a---1/z the integral gives 4 log 2. Remark. There is a simple method for the expansion of ~,,,(x)/IJ.J,,(x) into a series of inverse factorials, if q,,,(x) and w"(x) are polynomials respectively of degree ti and n (where n > m), Let us write
Fm (xl YLGj= i!lJ (X+T'l)i '
It is easy to show, multiplying both members by (x+1"-l),. that u,, =O if v < n-m. Multiplying by (x+n-m-l),,-,,, we
have
.b+n--m--l)nqvm(x) = =n-m + ,jn,+, (x+i21) W" (4 -
&n+m
therefore a,, is equal to the constant obtained by the division figuring in the first member, Denoting the rest by o,-,(x) which is of degree n-l, we have
multiplying both members by (x+n-m)
(x+--m)
W" (4
we find
ai (x+i---l)cn+m-1 '
W.-AX) = Q-+1 +
5
k4+m+2
an-R+1 is the constant obtained by division. Denoting the rest (of
degree n-l) by o,,-,,,+, (x) we continue in the same manner to obtain step by step any coefficient oi whatever. Applying this method to the preceding example, the first division gives x+1 1+ l--a X+a= x+a the second ( l - a ) I=$ =1-a+ u-4 w-4~ x+a the third (l-a) (2-a) x$f = (l-a) (2-4 + (l-4 P-4 v-4 and so on. uz=l, u,=(l-a), o,=(l-a)(2-+).
x+=
I
I
I
I
216
# 74. Expansion of the function l/y" into a series of powers of x. Suppoeir? that for x=0 we have y=l, Maclaurin's
theorem gives
(1)
Since
(2) we have
1 yt(=
To determine this value corresponding to x=0, let us remark that if ib then we have [DY (y-l)'l,1 = 0; , therefore it will be sufficient to make i vary in the first sum from zero to ~-j-l, Moreover we may put in the second sum also
r+l for the upper limit, since for m>i we haveI h ~0, so that 'I
the additional terms are equal to zero. But if tbe upper limit of the second sum is independent of i then the summa tion with respect to i can be performed. Let us write therefore
The part of the second member containing i may be written
Finally we obtain (3)
[&]- = n In?) 2, s [;) [D'u"l-.
I
Therefore, knowing the derivatives of ym for x=0, that is for y=l, this formula gives the derivatives of l/y" figuring in the expansion (1) so that the problem is solved.
217 In each particular case [DY y*JI-&. is to be determined, which is generally much easier than 0" WfIJ I,. Example 1. If the expansion of x"/(Qx--I)" is required, then we put y=(e* -1)/x* and the derivatives of y" are to be determined. It is easy to show that for x=0 we have y=l. Moreover to determine DYym let us write . xmym = (CC-l)rn. (4) The v+m -th derivative of the first member is given by Leibnifz's theorem (Q 30) :
mg [v+m] Du+m-iym Dip. i
ml
For x=0 each term of this sum will vanish except that of
km so that we shall have I 'irn 1 [Drym]x=o.
The second member of (4) may be written according to (5) 6 71: (e"-l)" = f=0 5 a& ml (i+m)! a!+m.
From this expression it follows that for x=0 its v+m -th derivative is equal to ml Q+,. Equating this to the preceding result we get (5) Therefore from (3) it follows that
This can also written in the following form:
218 Finally by aid of (6) we obtain the required expansion: (8) (A)"= i 3 n [":'I zl ;+z Sometimes it is possible to determine in another way the coefficient of x" in the expansion of l/y"; fhen equating the result to that given by (3), we may thus obtain interesting relations. For instance in the case of Example 1 we may write:
(91
x I8 - A(w) ex- 1 I -1 = 20 ( n - l ) * x* *
To determine the coefficients A(n,v), we will equate the first derivatives of both members of this equation. We shall find -n- [(-&p-x, - [ --xy'+l] = jl. s, x-1
X
this gives after simplification A(n+l,v) = A(n,v) - nA(np-1). The difference equation giving the Stirling's Numbers of the first kind (5, 8 50) may be written S;$:-v = S;-v -n S;'V+l therefore these numbers satisfy equation (9') and moreover, since the initial values A(n,O) 11 and S; = 1 are the same, we conclude that A(w) = SE-~ so that
Finally, equating the second members of (7) and (10) we obtain after simplification
219
which gives the Stirling number of the first kind in terms of those of the second kind. This is the simplest known expression of the Stirling numbers of the first kind, It has been found in another way by Schliimilch [Compendium 1895, Band II. p, 281. Remark. Since x/(ex- 1) is the generating function of B./v! where B, is the v -th Bernoulli number, therefore putting into (7) n=l the first will be equal to B, ; after simplification we get
Y+l Bv = mTl c--l)m
In Q 63, we found already a much simpler expression (5) giving the Bernoulli numbers by aid of the Stirling numbers of the second kind, Q 75. Changing the origii of the intervals. If the values of the function f(x) are given for x=a, u+h, a+2h,, , . and so on, then the differences
$w) = "2 (-I)- [;] f(o+vhl
are known, and also the following Newton expansion of the function
(1)
f(x) = ; (x;u]*F.
Sometimes it is needed to compute the values of f(x) corresponding to x=c, c+h, c+2h,. . . . This problem often occurs in mathematical statistics. To solve it, the best way is to determine the Newton expansion of the function f(x) in the form given below: (2)
f(x) = 2 (x7],, T.
I
I
220 Therefore it is necessary to compute the differences yf (c). They are deduced from formula (1)
Particular case. If c-armh, where m is an integer, then pf(c) = g f(a+mh).
Therefore if the table of f(a), f (a+h), f (a+2h), . . . contains also the differences of f(x), then the numbers above will already figure in the table. Q 76, Changing the length of the interval, Sometimes when a function f(x) is given by its differences in a system where the increment is equal to h, it is necessary to express this function in another system of differences in which the increment is k; that is, the differences of f(x) are required in this system for x=0, i. e. A"f(O), for m=l, 2,3, . . . This pioblem is identical with the following: Given a table of the function f(x) corresponding to equidistant values of x, the interval of x being equal to h; another table is to be computed in which the increment of x is equal to k. To solve the problem we first expand f(x) into a Newton series with increment h (formula 4 5 23)
Hence it is sufficient to expand I n I , into a series of (x) ,,Q, For this we will write, according to formula (3) Q 55:
X
Formula (1) Q 63 gives *+I x* = z e,m (X),J Pm. %I=1
221 Therefore we have
Now we may write the differences of E R in a system of 1 1 increment k:
The difference figuring in the first member is deduced from (3) ; to have that in the second member we apply the rule which gives the higher differences of a product; we shall find, after putting x=0 into the result obtained, ;y;;; p-@+llm) = w-4 ml
(n+l)
n!
P(nsm)
+
- (m-n1/ ! '" P(n,m-1); (TJl) .
simplified this gives (5) P(n+l,m) = (mw-n) P(n,m) + wP(n,m-1). From (2) it follows that P(n,m)=O if mh; moreover that P(n,O)=O. Therefore, putting m=n+l into (5) we get P(n+l,n+l) = wP(n,n) &cording to (2) we have P(l,l)=w, hence the solution of this equation is P(n,n) =a". This value has been found directly. In the same manner we could deduce from (5) the values of P(n,n- 1) and P(n,l) obtained above. This equation is especially useful for the computation of a table of P(n,m). Application. 1. Determination of the differences or of the indefinite sum of n ,, in a system of differences where the I 1 increment' is equal to one. Putting k=l into formula (3) we get
X
and from this moreover
1
223
0 a F
N
Let us remark that in this case o=l/h . 2. Expansion of the binomial a; into a Newton series with I 1 increments equal to one. I n 1
ax = ax(ux-1) . . . , , , (ax-nfl)
ll!
- = a"
n 06
X
where h=l/a. Putting A=1 and w=a into formula (3) we have
Example. Given 3 . Since w=3, by aid of the table of I 1 P(n,m) we get
3x
(33x) =I;) +l+] ++).
3. We shall see later that Cotes' numbers may be determined by aid of the P(n,m) and the coefficients of the Bernoulli polynomials of the second kind too. 9 77. Stirling polynomials. We have seen (Q 52) that the Stirting number of the first kind SZ-~ is a polynomial of x of degree 2m.
If x > m then in consequence of formula (1) the numerator is a polynomial of x of degree 2m+2 divisible by (~+l)~+? ; therefore 1ym(x) is a polynomial of degree m. Indeed we have (3)
m+l c Yh(X)= k--1)"-' i20 2m ";"' i) ! k--m---l),-i *
Particular cases:
1
225 1 1 Y*(x)= g+ij f x(x+1).
2(x) =
write down a SfirZing polynomial of any degree whatever. From
c m0 it follows that the coefficient of x" in qm(x) is equal to l/(m f1)2'"+'. Notable particular u&es. Putting x=m+l into formula (2) we obtain 1 (-l)m+l Ym(m+l) = (m+2)I S' a+2 =m+2* If we put x=m+2 into formula (2) we get (p. 148)
Ym(m+4 =
but this is, according to (3), equal to (X+2) YmIX+l) C"Li (x-m)
+ (m-4
The first term of the second member of (4) is
id (2m-i)
!
md = (x-2m+l+i)
Wm-l(X).
The second term may be transformed by putting i+l instead of i and we shall have
"S' (-1 I m G,f
i=O (2ll?.--l--i)!
(X-m),-,-i =
(2m-i) ymel (x) a
We may write m+l for the upper limit, since for i=m this expression is equal to zero. Therefore the second member of (4) will be equal to (X+llYm-1(X)Equating this to the value (5) of the first member obtained before we get the difference equation of the Stirling polynomials: (6)
(x+1) Y m - 1 (4 = ( m-4
Ymlx)
+
(X+2) Ym(x+l).
Starting from y,,(x) =1/, we may determine by aid of this equation the Stirling polynomials step by step, For instance, putting y1 (x) =a,,+a,x we have ?h (x+ 1 I - u-4 b-',+q4 - (x+2) (a,+a,+a,x) = 0. Since this is an identity, the coefficients of x' must be equal to zero for every i. This gives 6~7,,+4a, ~1 and 8o, =l so that
Yl(Xl
1 = $+fi.
Continuing in the same manner we could obtain the polynomial of any degree.
We will show that this definition leads to the same polynomial ym(x) as our definition (2). Indeed, multiplying both members of equation (7) by (x+l)t-x it becomes: 1= (x+1) m; ym(x) P-x. Ix+' The derivative of this expression with respect to t divided by x+1 will be (8) (1,@)x+1
1
-
(9)
-e-t (l-e-t)X+2 + & = $ ( m - x ) 7pm(x) tmmrl,
Writing into (8) x+1 instead of x and adding the result to (9) we find
(lo) (~~~t)m =
i
[(x+2)Ym(x+l)
+ ~~-4~~(~)ltm-p1~
Finally equating the coefficients of t'"-~l in the expressions (8) and (10) we find the difference equation (6) of the Stirling polynomials obtained before. Moreover from (7) it follows directly that Y&4 = % Therefore the two definitions lead to the same results. Putting x=0 into the generating function (7) we have
1 1 ---= s Gm(0) 1-e-' t m=O tm.
Multiplying this equation by t, and then writing in it -t instead of t, it will become t e'--l = 1 + 2 (-l)m+l ym(O) tV+'.
We shall see later that the first member of this equation is the generating function of B,lmI, where B, stands for the m -th Bernoulli number; therefore we have
Bernoulli polynomial of tbe first kind of degree R by Q)"(X), let
its definition bew2'
se) Different authors have given different definitions of the Bernoulli olynomials. Our definition is that of Nielsen [Trait6 des Nombree de bl rnoulli, Paris, 1923, p, 401. He denoted the polynomial by B,(x). Seliwanoff [Differenzenrechnung, Leipzig 1904, p, 49 and Encycl. des Sciences Mathem. Vol. I. 20, p 1111 g ave the following definition for the case of integer values of x:
This differs from our definition only by a constant. In SuaZschiitz's Bernoulli'sche Zahlen [Berlin, 1893, p. 911 the definition is for integer values of x the following
v (x,n) = i rn-'.
m=Q
This definition differs from ours by a factor equal to (n-l)! and by a constant, N6rkund [Differenzrechnung, Berlin, 1924, p 191 defined these polynomials by A B,, Ix) = nx*-' so as to have
v&4 = A-l &; I + 4.
Hence the definition (1) is not a complete one; there still remains an arbitrary constant to dispose of. Let us write the Bernoulli polynomial of degree R in the following way: (4 n+1 %(x1 = m!. x"'
Urn (n-m) I *
By derivation we obtain from (1)
but the second member is, in consequence of (11, equal to
A~,,,_,(xl so that
AD P.(xI = An-,(4 .
Performing the operation A-' on both members of this equation, we may dispose of the arbitrary constant, which enters by this operation, so as to have (3)
D vnpn[4
=
%-1(4;
then the Bernoulli polynomial cpPn (x) is, by equations (1) and (3), completely defined. ,In Q 22 a class of functions important in Mathematical Analysis has been mentioned in which
D Fn (4 =
Frt-,b4
for n=l,2,3,... According to (3) the Bernoulli polynomials belong to this class of functions. Moreover if Fn(x) is a polynomial of degree n .belonging to this class
% have seen that the coefficients ai are independent of the degree n of the polynomial, so that they may be calculated once for all.
and so on. In the Table below we see that the Bernoulli numbers increase rapidly with n whereas the coefficients a,,, decrease rapidly. The Bernoulli numbers are very important in Mathematical Analysis, especially when dealing with expansiona2'
27 The Bernoulli numbers were first introduced by Jacob Bernoulli [Ars Conjectandi, Basileae, 1713, p. 971. They have been denoted differently by later authors. Our notation is that used by Niirlund [Differenzenrechnun , p. 181, Steffensen [Interpolation, p. 120], Pascal [Repertorium, I. p. 1217 I !i There is another notation in use, in which the Bernoulli numbers are considered as being positive, and in which B, corresponds to our 1 B,, 1. This notation is found in Saalschiifs [Bernoullische Zahlen, p. 43, in Hagen [Synopsis I, p. 911, in Seliwanoff [Differenzenrechnun p. 45 in and Encycl. de? Sciences Mathematiques I. 20. p. 1111 in Nfeleen l% ombres de Bernoulli, 421 in E. Lindelof [Calcul des Residue, Paris, 1985, p. 331 and in G. teano [Formulaire Mathbmatique, Paris, 1901; p. 1981. .In the last work we find an extensive table of these numbers,up to & in our notation,
Expansion of the Bernoulli polynomial into a series of factorials. Replacing in formula (I), the power x"+ by factorials,
using formula (2) Q 58 we obtain
Since
A-'(& = (x)m+l T&r+&
we get, performing the operation
A-'
on both members,
I
236 The constant k, which entered by this operation, has been disposed of, so as to have v" (0) =a, . This is the expansion of the Bernoulli polynomial into a factorial series. It shows immediately that
A~'vn(O) = (P---1) ! @p-l (n-l)l n--i
moreover that f?-%(O) = P"(l) = an and so on. Formula (8) permits us to determine. the coefficients a,,, and therefore also the numbers B, in termis of the Stirling numbers. For this, let us replace in the second member of (8) the factorial (x),+1 by its expansion into a power series (3, 8 50); we get
Since the coefficient of x' in the first member is equal to
ati/i! and ati = Bd/ (n-i)!,
This may be simplified by putting i=l, and writing n+l instead of R. The required expression will be
(10)
n!a,=B,=
"+I (-l)mm! Z #II=1 m+l
em " *
Example. Let us determine B, by aid of this formula: B, =- +++$.6+$+,.
We may transform formula (10) by putting into it the value of Q," given by formula (3) $ 58:
Q 7% Particular cases of the Bernoulli polynomials. From the table of the coefficients a,,, we deduce
Q 80. Symmetry of the Bernoulli polynomfals. definition of these polynomials it follows that A9)2nb) =
~)zn(x+i)-~)2nbd = f2;:;jI'
From the
Putting into this equation -x instead of x we get
but the first member is equal to -Aqzn( l-x) ; hence we conclude that
ArppzU-4 = Avwbd.
Let us sum this expression from x=0 to x=2; since the indefinite sum of the first member is q2"(l-X) and that of the second member q2,,(x) we find after putting into them the values of the limits: wzn (l-4 - Q)2n (1) = %n (4 - Evl(O) * We have seen that ~,,,(O)=Q)~(~), (1)
cp2"W-r) =
if m>l; hence
f?J,z"bL
Formula (1) has been demonstrated only for integer values of x, but this expression m a polynomial of degree 2n-1, and it is satisfied for more than 2n-1 values of x, hence it is an identity, and is therefore satisfied by any value whatever of X. This is the expression of the symmetry of the polynomials of even degree. This can be written in another way; putting r=1/2+x we get
Hence the polynomials of odd degree are symmetrical to the point 'of coordinates x=1/, y=O. Putting x=0 into (3) we have 972%lM2) = - pp.; (%!I = 0. Putting xii/; into (3) we get q2"-r (1) = - q2"-r (0) , but we have seen that if m>l then q,,,(l) =vm(0) ; therefore we conclude that q2G1 (0) = 0 or a,,, = 0 if n > 1. (4)
Roots of the polynomials. We have already seen that the equation q2"+r (x) =O has thr ee roots if n>l, namely x=0, x=1/2 and x= 1; hence the polynomial of odd degree is divisible by x ( x - 1 ) ( 2 x - l ) i f f&l. We shall show now that it cannot have more than three roots in the interval 0 6 x 5 1; that is, more than one in the interval O< x < 1. If it had at least four in the first interval, then its derivative should have at least three in the interval 0 <x < 1 and its second derivative at least two. But
D"
vzn+1(4
=
v2n-164.
Hence we conclude that if q2,,+r (x) =O had four roots in the interval 06 x 5 1, then q2,,-r (x) =O would have also at least four in this interval, and so on; q3 (x) =O too, which is impossible, since ~)s(x) is only of the, third degree. Summing up, we state that q2n+l(~) =O has three roots in the interval 0 ZZ x s 1 and only three (if GO).
D
~)nn (4
= en-1 (xl
and the polynomial of the second member has no roots in the interval 0 < x < l/z; therefore Quip cannot have more than one root in the interval 0 5 x 2 Y2. But
Dea+,W =
~)2n(x)
and we have seen that vtn+r (x) is equal to zero for x=0 and x=42; therefore q2"(x) =O must have at least one root in the
240
interval O<x<l/z, therefore q,,(O) and 0)~~(4/2) are both different from zero, so that ~~"(0) =aZn $- 0. Finally qzn(x) =0 has one root and only one in the interval 06&/$,, In consequence of the symmetry, qsn (x) =O has another, root in the interval l/z<x<l. Exttema of fhe poZynomiaZs. If n>l, then vlkl(x) is equal to zero for x=0, y', 1; therefore Q)~"(x) will have extrema at these three places. qzR(x) has only one at x=4/2, If GO then qpz,,+i (x) has two extrema, one in the interval O<x<4/2 and one in ?/2<x<l; vi (x) has none. Sign of the numbers a,,, and B,. Let us suppose that a,,>O; then in consequence of qzn(0) =a2, and of the fact that q+"(x) =O has a root in the interval O<x<1/2 without having an extremum in it, we conclude that qzn(x) must be decreasing in the vicinity of x=0. Therefore we must have
D v&d = ~)m--1(4 < 0
but if x is small, the sign of qzndl(x) is equal to that of aZn-2x therefore we have azb2 < 0. Had we supposed that a,,<O, then we should have found that qnn(x) must be increasing in the vicinity of x=0 and consequently we should have
a2n-t > 0.
Finally summing up, we always have a2n azh2 < 0 that is, the aumbers a2n and also B,,= (2n) I a2" are alternately positive and negative with increasing n. Since a, is positive, we have or
242
where, as we shall see, in 8 100, E,,-,(x) is the Euler polynomial of degree u-l.
Q 82, Expansion of the Bernoulli polynomial into a Fourier series. Limits. Sum of reciprocal power series,
If y is a periodic function of period equal to one, with limited total fluctuation (or,a fortiori, if it has a limited derivative), then it may be expanded into a Fourier series (5 145)) such as 00 y = IJ2a, + Z a, cos2nmx + 2 pm sin2nmr (1) III=1 ml where (2)
am
=2 fy cos2nmx dx i
0
Pm
= zoJ' y sin2nmx dx.
Hence we conclude that q*"(x), may be expanded into a Fourier series between zero and one. By aid of formulae (2) we obtain 1
a0 = 2 o s
w&4
dx = van+rU)
-v2n+1K')
= 0
if n > 0. Moreover we find by integration by parts that
%% = s
0
1
y2,,(x) cos2nmx dx = sm22nzx ['
1
'0
sin2nmx
2nm v2n-1(4
3%
Y2"W
1
1
0
-
s
dx.
The quantity in the brackets is equal to zero at both limits, and the integral in the second member becomes, after a second integration by parts:
cos2nmx
%m = [ (2nm12 V2n-1(4
The quantity in the brackets has the same value for x=0 and x=1, if n>l, therefore the corresponding difference is equal to zero. We conclude that the operation of integration by parts performed twice, gives an expression similar to the initial; only the sign has changed, the degree of the polynomial is diminished
1J
1 o-o
1
-
cos2nmx (2nm)-2
v2n-2(4
dx.
243 by two, and it has been divided by the quantity (2nm)s. Therefore in 2i operations we should get
Writing in it i=n-I we get by integration by parts
The quantity in the brackets is equal to zero; further integration by parts gives
The integral in the second member is equal to zero. Since 4)1(1)=--9)1(0)=1/2 we find a m = 2W)"' (2nm)2n '
f
Determination of the coefficients /3,,, . It would be possible to determine them by integration by parts in the same way as the coefficients a,,, , but it is easy to show by aid of formula (1) Q 80 that pm=O. Indeed, putting l-x into (1) instead of x we should obtain ~)~"(l-x) =v2"(x) for every value of x. and cos27cm ( l - x ) = cos2.7mx sin2nm (l-x) = - sin2nmx;
hence the coefficients of sin2nmx in the expansion (1) must all be equal to zero, so that we have B,,,=O. Finally, the expansion of q2,,(x) will be (3) V)2n(X) = 2(-l)"' %, ~~~.
To obtain the expansion of ~~,,-~(x) we determine the derivatives of both members of (3), and find (4) Vrn-I(X) - 2 ( - l ) " 2, $g.
244
write x=0 in equation (3), we have (5) qJ2"(0) = a,, = 2 a ,
Ill=1
Applications of the trigonometrical expansion. 1. If we
= 2(--1J--m:, & (2n)2:1
This shows again the rule of the signs of u2,, obtained in 8 80. From formula (5) the sum of the reciprocal power series may be deduced. (6) ii & =1/2(2z)2" X=1 cases: la,,1 = g$B*"I. .
If the central value qzn(4/2) of the polynomial is known then this expression gives a formula for the sum of the alternating reciprocal power series. We will determine ~)~~(l/z) in 8 86; but in (j 49 we have seen that if we denote the sum (6) by szn then the sum of the alternatingeries will be
therefore (7) 5 (-1)x+1
X2n
x=1
cases:
$
x=1
= (22n-1-l) z2n I a,, I. (-1)x+1 7324 =-Es* 2, x4
Particular
(-1)x+1
X2
_ x2 12
'
From this we may obtain the value of q2"(4/2) ; since the sign of azn is that of (-l)"', we get from (7) (8)
92&I = & ( - i) 42n *
Starting from f (0) = x+C Lacroix determined by aid of this equation step by step A-lx, A-lx2 and so on. But we shall solve this difference equation by aid of AndrB's method in 8 178. Example 3. The result obtained there is the following (3) n-t1 = ' n1 /J+J (n$YYbL)! f(n) I A-lx0 = P+l
2, (kl+l)!(k,:if)lf #.. (ki+l) 1 *
In the second sum k,, k,, . , ,, ki take every integer value with repetition and permutation such that k, + k, + . . . + ki = ,U and k, >O. that is, every partition of ,u with repetition and permutation of order i. But on the other hand from formula (1) it follows that
where k,+k,+ , . , +ki=,u and k,>O. This is an interesting expressionfor the Bernoulli numbers, which may be obtained directly, starting from their generating function (p, 251). Example. Let ,u=4. For i=l we have k,=4 and the corresponding term is -l/120. For i=2 we have kl=l, k,=3 or k,=3, k,=l or k, =2, k,=2. The corresponding terms are 2148 a n d l/36. F o r i=3 w e h a v e k,=l, k,=l, k,=2 o r k,=l, k,=2, k,=l or k,=2, k,=l, k,=l the corresponding term . IS -3/24. For i=4 we have k,=k,=k,=k,=l corresponding to l/16. Finally we find
248 B4 ==a,=- '+$+&-g++ 16&-J = - &.
2. One of the most important applications of the Bernoulli numbers is the expansion of certain functions by aid of these numbers, as we shall see later. Q 84, Expansion of a polynomial f(x) into a series of Bernoulli polymdals. It is known, that if q,,(x), am, . . . , Q'"(X) are polynomials of degree 0, 1, . . . , n, then it is possible, and only in one manner, to expand any polynomial whatever of degree n, in the following form:
n+1 11) 04 = 2 QW(X).
If vi(x) is the Bernoutti polynomial of degree i, then the coefficients cl may be determined in the following way. Integrating both members from zero to one, every term of the second member will vanish in consequence of formula (1) Q 81 except that of c,, ; since Q,,(X) =l, we obtain (2)
Consequently,if we know the integral of a polynomial f(x) from zero to one, and its derivatives for x=0 and x=1, then the expansion (1) is known. Formula (1) may serve for the determination of the indefinite integral of f(x), of its derivatives and also of its indefinite sum. This last is obtained by aid of (2) 5 81. Example 1. Given f(x) = xn/n!
s
f(x) dx ,= GU + R a n d
D*-lf (x) = (nTrtl) ,;
therefore
249 c0 = l/(n+l)l a n d and finally c, = l/(n-m+l)l
Example 2. Given t(x) = c , We shall see later that
I 1
1 JO
;
dx=b,,
0
where b,, is the general coefficient of the Bernoulli polynomial of the second kind. From (2) it follows that co=bn . Moreover since
Putting into this expression x=0 and writing qpnr (0) =Bm/A! we have (5) This is an expression of the coefficient b, by aid of
Bernoulli numbers B, .
According to § 65 we may obtain by inversion an expression of the Bernoulli number B,. Multiplying both members of (5) by -(n-l) I q-l and summing from 12~2 to n=Y+2,every term of the second member will vanish except that in which m=r+l; therefore we find
250 (6) _ F b,, ( n - l ) ! a;-' = 3 .
nl
Q 85, Jhpadon of functions into Bernoulli polynomials. Generating functions. If the function to be expanded is not a polynomial, the series will be infinite, and considerations of convergence must be made; but the coefficients c,,, will be determined in the same manner as in the preceding paragraph. Example 1. Let us determine the generating function f(t) of the polynomials v"(x). In the case of the generating function we must have c,,, = AD+(O) = fm and 1 co = .I f(f) tit = 1. Since Dm-1 ,xr = tm-1 eat and Aexr = ext [et-l] it is easy to see that
ADm-lg]
therefore
_ = tmi .x 0
fi Vm(x) fm
(1)
f(f)
= -g =
m=O
is-the genetafing funcfion of the polynomial vm(x). From this we may obtain the generating function of the number am by putting x=0. t - = 2 a, P. et-l m-* From equation (2) we conclude that Z a, value of the function (3) et-l -= 5
t i=l tm
is the reciprocal
q. .
On the other hand, by aid of formula (4) 5 78, we could show that the product of the series (2) and (3) is equal to one,
251 and deduce in this way the generating function of the numbers We have seen that the series 2 Q, is convergent; therefore, putting into (2) f =l we obtain the sum of the coefficients a,,,
m=O
This result has been obtained in (8) 8 82. We may obtain another expression of (pn(l/z) starting from v"(x) = (x+B)*/n! . We get
(4)
2. Putting into (2) n=2m and p=4, taking account of the preceding result and remarking that in consequence of the symmetry v2rn ($1 =pzrn (:) we find that
3. Putting again n=2m and p=3, we obtain in the same way
92m !- 1 t!~
c----Q 2
1
2m
I32~1
1
11 .
§ 87, The Bernoulli series. If certain conditions are satisfied we may expand a function into a series of Bernoulli polynomials in the same way as has been done in the case of polynomials (8 85) ; but the series will be infinite and its convergence must be examined. Let us start from f(x+a) ; its expansion will be, in consequence of formulae (2) and (3) 8 83, (1)
f(x+u) = f+' f(t)& + 5 qm(x) ADm-lf(u). mzl "
Putting into this formula u=O we obtain the expansion of f(x) into a series of Bernoulli polynomials fo,und before. On the other hand, putting x=0 we have
or
f(u) = s"' f (fjdt + ,., a, AD"-' f(u) "
254 flu) = f+' f(t)dt + 5 * AD-~(U). m=il m. * This expansion might be termed a Bernoulli series, owing to the Bernoulli numbers figuring in it, ;It often leads to very useful formulae. For instance, starting from cos z we may obtain the expansions of cot 2, tanz, etc., into power series. Sometimes these formulae may serve for the computation of the values of the function f(u) as for instance in the cases of F(U) and F(u). Remainder of the Bernoulli series. If the series is infinite, its remainder should be determined. For this we will start from f(x+u) integrating by parts, u varying from zero to unity; proceeding in the following way: I 1 @I (3)
0
+ (Ix,,, AD2r3f (xl + agn D2"f (x+8) s
If Wf(x) does not change its sign in the interval 0, 1 moreover if D2nf(x) D 2ti2f (x)>O, it is possible to obtain a still more simple form of the remainder. Indeed, since a2na2n+2<0 in this case we have R,, R,,,, < 0. From (6) it follows that R,, = azn AD21r1f (x) + R,,, . This equation may be written in the following way; if A=B+C and AC < 0; then, from A2=AB+AC, it follows that AB > A2 > 0; dividing by AB we get l> $>o
SO
that we may put A = 5B where 0 < 6 < 1. Finally we have R,, = 6 a2" AD2n-1f (x).
(7)
Formula (6) may be transformed by putting DF(x) 'f(x); we find
(8)
DF(x) = AF(r) + 2;' j+ AD'F(x) + R,, .
This formula may be useful first for the determination of DF(x) if the derivative of F(x) is complicated but the derivatives of NJ(x) are simple. Example 1. If F(x) = logr(x+l), then DF(x) = F(x) and AF(x) = log(x+1) moreover
ApF(x)
= (-~~-+i~~*.
Since from formula (1) $ 21 we have D2n+lF(x) = D""" logr(x+l) = - (2n) I s, (x+A12n+l therefore D2"+'P(x) D2"+V(x)>0 so that the remainder of the form (7) may be used, that is
(9)
R 2n = - AD2"F(x) :z"n;;
where 0 C 6 < 1. Finally from (8) we get: (10) f(x) = log(x+1) + 2; (4)"' 4 ($-J i- t* (&)l: Although the infinite series is divergent, the formula may be used to compute the values of F(x) in P&man's Tables (lot. cit. 16). Indeed at the beginning of the series R,, is diminishing with increasing R. In Q 82 we have seen that @n+l) @+a4, (W2 neglecting 5, this gives approximately
I
B -z!!e
I<
;
R 2!!tz< I I
R2n
@n+l) (24 (242 (1+x)" .
So that R,, will diminish till the second member becomes equal to one: or,approximatelyt till n reaches the value of 3x+3. Therefore if x is large it will be eas% generally to obtain the prescribed precision. The most unfavourable case is that of x=0; this should give Euler's constant, Writing x=0 we get:
257 For n=l this would give C=O'5+ & ; the best value of C is obtained for n=3, that is C=5'575 + A.
therefore the remainder will be given by (9) so that we have according to (8):
Though the infinite series is again divergent, the formula is still useful for the computation of F(X); since at the beginning of the series R,, decreases with increasing n. (Approximately till n becomes equal to 3x+3.) The less favourable case is that of x=0; then we obtain F(o) =; =1+1/+
i=l
i Bzi+5'B,,.
For n=l this gives F(O) = 1'5 + $. The best value of F(O) is obtained for n=3, that is F(O) =1'633 + f Secondly formula (8) may 'be useful to determine AF(x) if the derivatives of
I
258 [AD'F(~,t)lx=o = ci f'w (f) where ci is a numerical constant. From this it follows that [AF(x,t)]x=o = co 0 (4. Writing in (8) F(x,t) instead of F(x), putting x=0 and dividing both members of the equation by co w(f), we find (12) I DF(x,f)]+=o -=
co 41
2z;y++
0 *
R 2n. co 0 (4
This equation will give the expansion of the first member into a power series. Example 4. If F(x,f) =eXt then DF(0) =f and A.F(O) =ef-1 moreover [A~F(x,f)]r=o =fi(ef-1) hence ci=l and o(f)=e'-1. Since D2"+lF(x,f) D*"+V(x,f) > 0 it follows that the remainder is given by (9) and from (12) 'we get zn-J & t (13) = z i, f' + ;;;; t"". e'-1 .
i=O
In Q 82 we have seen that lim uzn k*" = 0 if 0 S k < 2~; there*=fore if I t I< 2n we shall have lim R,, = 0. Moreover, according to
formula (10) (j 82 we have "=Q2n+2 t*n+* Q2" t*"
hence we conclude that if I t I < 2n the first member of this inequality is less than one and the series (15) is convergent. Writing in (15) B,i/ (2i) ! = Q2i and subtracting it from 2 +4/d we find 2 + I/f(l-cot'&) = 1 + 1/f + "5' (-l)i+l
Q2i
t*' +
- R,, =
i=O
x
m
i=l
1Qi 1t' .
Hence the first member is the generating function of the absolute values of Q,. From this formula we may deduce the sum of the absolute values of the a,, by putting t=l:
m=O
ii
1Q, 1= 2'5 - 4/2cotl/ -e 1'5847561
(Let us remark that arcl/z = 28O 38' 52" 40). The numbers a,,, in Q 85 would give ii IQ, 1 = 1'58475 61391. m=o Writing in formula (15) t=2z and dividing by I we get the well known formula
+ (-1)"2*"+1 We have
&
n .
12"
ca32t!?z --. sin22
In=,
(-l)m-l2*m
(2*m-1) B2m z*~-I + R,, . __
(24 1
260 From (16) and (17) we may obtain several other formulae. For instance by integration , - tan xdx = log cos 2 1
0.
or by derivation Dtanz= - , -'D cot z = & . cos"z Since l/~[tanl/~x Hence we have ( 1 8 ) L=$ + nil (-1)" (2--22m) $&$ x2m-I + R,, . sin x 8 88. The Maclaurin-Euler Summation Formula. This is a formula by aid of which the sum of a function may be expressed by its integral and its derivatives, or the integral by the sum and the derivatives. Apart from the remainder, the formula may be easily deduced by symbolical methods. In Q 6, we had formula (3) A I eh'J- 1 from this it follows that + cotlAx] = &.
1
But according to formula (2) 3 85 the expression in the brackets is equal to the generating function of the numbers ai, the coefficients figuring in the Bernoulli polynomial; since ai = Bt/i! where Bi is the i -th Bernoulli number; hence we may write
A-1 = izo $ (hD)i-1 =+ D-1 + $j . f!!
i=t .
(hD)"'.
This operation performed on f(x) gives, if the sum from x=o to x=z is calculated, i %U
f(X)
= + ),-f(x) d x + ,p, $ hi-1 jD'-'f(z) -DD'-'f(u)], a*
261 This is the required summation formula. To obtain the summation formula with its remainder we start from the Bernoulli series (6) of Q 87. Let us write it in the following way:
x+1
and moreover if D2"f(x) does not change its sign in the interval (a, I) then, according to 8 87, the remainder may be written: (4) R,, = [ a2"[D2r*f (2) - Dznwlf (a)] where O<t<l. 2. If f(x) is a function such that for every value of m=l, 2,3,. . , we have Dmf (-) ~0, then we will use form (3) of the remainder; and equation (2) may be written i f ( x ) = J" f(t)& + >i: a,Dm-'f(z) x=a a _
- a2" % D2+(x+CT) Z=I
where C, denotes the part of of I, that is,
+ CT
i f(x) which is independent x=a
262
zn-t c, =- x m=t
am
Dmlf(Q) + azn *=a 5
D'nf(X+c). (5)
We may obtain another expression for C, writing z=* in (61
C, = Ti f(x)- jmf(t)dt-a,f(w). .lZ=LI a The value of C, must be determined for each particular function. From (6) we conclude that C, is independent of n. Formula (5) will be really advantageous in cases in which one of the expressions Z f(x) or *f (x)dx is unknown, J Example 1. Given f(x)=l/(x+l) and CEO. Here we have Dmf(~)=O for m=l, 2,3,. . . , therefore formula (5) may be used. Since moreover we have f (m)=O, it follows that
but according to formula (3) 8 19 the second member is equal to Euler's constant C; therefore Cf =C:=o'57721 56649 01532.. . . D2nf (x) D"n+'f (x) > 0 and Dznf (x) does not change its sign in the interval (0,~): therefore according to Particular case 1, the part of the remainder dependent on I may be written: agn 6 D*-*f(z). For instance, we shall have, stopping at n=2 5 x=0 or f(x) = Cf + j'f (f)dt + a,f (4 +
0
Comparing this with the exact result, we find, neglecting the remainder, that the error is only -3,lV.
1
263 In some cases the Euler-Maclaurin formula performed on f(x) will lead to the same result as the expansion of ml = xzo f(x) into a Bernoulli series (6) Q 01. For instance the expansion of r(x)+C by formula (10) Q 87 is identical with the expansion above.
B&X + $ 2m(2m-1) zZm-l + 2n(2n-1) Zzrl ' ' "'
This is Stirling's celebrated series for log zl . According to what we have seen in Q 82, the general term increases indefinitely with n and therefore the series obtained by putting R== will
264 be divergent; nevertheless formula (8) is useful for the computation of log Z! , The best value of R would be approximately n(z+l). Remark. Of course the logarithms figuring in these formulae are Napier's logarithms, To have Briggs' logarithms, the result must still be multiplied by log 10 = 2'30258 50929 9, E x a m p l e 3 . G i v e n f ( x ) = I/(x+I)~ a n d a=O. S i n c e D"f(w)=O for m=l, 2,3,. . . , moreover D2*f(x) D"""f(x) > 0 and D2"f (x) does not change its sign in the interval (0,~) therefore we may use formula (5) and the remainder may be written
The best value of R would approximately be n(c+I). The series obtained by putting n=m would be divergent, but nevertheless the above formula is useful for the computation of I: I/(x+~)~. For instance, let 2~20: ;r2/6 --l/(21) -l/2(21)2 -l/6(21) 3 and R,~t/30(21)~ 1'64493 4067 -0'04761 9048 -0'00113 3787 -0'00001 7997 1'59616 3235 0'ooo00 0008 6.
265 The expansion (9) is the same as that obtained by expanding T'l a - ,io &)2 = J(4
into a Bernoulli series in Q 87 (Ex. 2). Q 89. The Bernoulli polynomials of the second kind. We will denote the polynomial of degree n by W"(X), Its definition is the following:**
(1)
D~n(4 = (n:l)
from this we obtain by integration
Hence by the definition (1) the polynomial is not wholly determined, there still remains an arbitrary constant to dispose of. This will be done as follows: The expansion of the polynomial into a Newton series may be written (2) YntX) =bO(~)+bl (n.Ll] +***+h*
The operation A performed on both members of (1) gives
but this is also equal, in consequence of (l), to D yndl Integrating both members of
(x).
ADynW = Dyne,(x1
and disposing conveniently of the arbitrary constant figuring in ~~~(4 we get (3) AY, (4 = ynn-1 (4. The Bernoulli polynomial of the second kind is completely determined by equations (1) and (3). In 5 22 it has been mentioned that the sequences of functions F,,(x) satisfying
sa CII. Jordan, Polynomes analogues aux polynomea de Bernoulli.. . Acta Scientiarum Mathematicarum, Vol. 4., p. 130, 1929, Szeged.
266
A~nb) = Fn-, (4
f o r x=1,2,3,..., play an important role in the Calculus of finite differences. Moreover, if F,(x) is a polynomial of degree n:
then we have seen that the coefficients ci are independent of the degree n of the polynomial, so that they may be determined once for all. From (1) we deduce Dyr (x) =1 and therefore vr (x)=x-& so that b,=l. Moreover if n>l then formula (1) gives &(O)=O. On the other hand we have
Since this is an odd function of Y 1 be an even function*of y; so that (1) or
YA b-1 +Yi ,? yan (n-1-y) y*" (X) = 'yen (2n-2-x).
-.'
.-
Hence the polynomial y*,,(x) is symmetrical. The symmetry axis is x=n-1. . B. Potynomi& of an odd degree. The operation A performed on both members of ._ Yzn+* (n+u) = Y2n+2b--Y) will give, according to what we.have seen (p, 5) (2) y2n+l (n+v) = - Y2n+l b--3-11. Writing in this equation y=n-l-x, we get Y~~+~(x) =~-~&+~(2n--I---x) or putting y=x+/z into equation (2):.
(3)
~2n+li+-?h+X)~ = ~~~~+~b---l/--x).
This is the expression of the symmetry of the polynomials of odd degree. Putting into it x=0 we obtain ~~~+~+--1/2) = 0. The polynomials of odd degree are symmetrical with respect . to the point x=n--1/2.
'General expression of the symmetry of the polynomials:
ynP@-l+x) = (-1)" Yn(%n-l-x).
Remark 2. If 0 I; x < m then there are -x+l"positive and
m-x-l negative factors in so that its sign and that of y,,,(x) will be the same as the sign of (-1)'"WX-1. x+ll is negative Remark 3. If x < 0 then every factor of ,m 1 1 and the sign of vm(x) will be that of-(-l)".. From this we conclude that (2)
Y2n+1Gw ' 0
if
~20.
-
270 and (3)
%n+1 @y--1 1 < 6
i f
O<rSn.
The inequalities (2) and (3) demonstrate the theorem. Moreover in consequence of Remark 3 we get Y,"+~ (-xl < 0 if x > 0.
B. Polynomials of an even degree. Theorem 2. The maxima of the polynomial yzn(x) correspond to x= 1,3,5, . . . , 2n-3 and they are all positive. The minima correspond to x=0,2,4, . , . ' 2n-2 and they are all negative. From Remark 2. we deduce that
(4) Yzn @r--1 1 ' 0
hence the maxima are all positive. Moreover from 2. it follows that (51 apzn(2r)<0 i f OSySn-1
hence the minima are all negative. Finally in consequence of 3. we have yzn(--x) >O i f x>O.
Theorem 3. If x is an integer, then the absolute value of y,,,(x) decreases from x=-l up to the point or axis of symmetry, and then it increases again. So that we have
(6) I ~~(-1) I > I y,(O) I >. . . B I ym(l/m-1) I <. . . < I ym(m-1) I. To prove it we shall show that (7) I Y"(X) I ' I Ym(x+l) 1
if -1 S x 6 l/$m--2. For this we shall determine the mean of y,,,(x). We find MYnIb)
= l/2
EYm(4
+
Ym(X+l)
1
and in consequence of formula (8) Q 89 we have:
271
(8)
MwmIx) = $4 j'[[ ,+A+') + 1 xmf"]] du =
The sign of the mean will be equal to that of (x+u+ l--Z/zm) , (~+&-l~ In the interval considered (9) -lz+-2
the first factor is always negative. The number of the negative factors in (x+u)m-r is equal to m-2-x; therefore if m and x are both odd or both even then the mean is negative; that is W&4
Y2n+1
+
Yzn(2~+1)
<0
< 0 *
w--i)
+ Yzn+1W)
On the other hand if one of them is odd, the other even, then the mean is positive and we have
YJznw4) Wzn+1cw -t -l- Y2nW ' 0
YZn+l @y+l)
'0.
1
Hence from (2), (3) and (4) we conclude immediately that in the interyal (9) we have for every integer value of x: I YmW I ' I Ym(x+l) therefore Theorem 3 is demonstrated. Since the maxima of ym(x) decrease from x=0 to the symmetry point or axis and then increase again, hence the maximum nearest to this point or axis is the minimum maximorum. The minima of y,,,(x) increase from x=0 to the symmetry point or axis, and then they decrease again, so that the minimum nearest to this point is the maximum minimorum. If m=2n+l and n is odd, the minimum maximorum corresponds to x=n-1 and the maximum minimorum to x=n. If n is even, then x=n corresponds to the first and x=n-1 to the second. In both cases we have
yzml b-1) = - lyzwl (4,
Let us remark that the highest maximum corresponds to x=0 and is equal to
272
Wzn+1(O) = &",I
moreover, that the lowest minimum corresponds to x=2n-1 and is equal to
Wzn+1(2~--i) = - Yzn+1uI = -h!"+, *
If m=2n and if R is even, the minimum maximorum is on the symmetry axis and is equal to w,,(n-1) ; the maximum minimorum is then at x=n-2 and we have ynn (n-2) = y,.(n). If n is odd, then yzn(n-1) is the maximum minimorum and yLnn (n-2) =ysn (n) the minimum maximorum. Roots of the polynomials. The roots of yn(x) =O are all situated in the interval -1, n-l ; indeed in this interval the function changes its sign n times; since y,,(x) is of degree n, hence all roots are real and single. Q 92. Particular cases of the polynomials. From formula (2) Q 89 it follows that YlW =.++-
then the absolute value of each factor of the second member is less than one if 0 < x < 1 and we have
II II
X
n
1 <n
273 and therefore
lb.1 c$
so that (1)
limlb,l = 0. "COO Moreover we may write
b,+,= s'[ &) cfx= $'[;I 5 dx.
0
Y
Figure 2.
Y
2/ = VI (2) I
z/ = 'P*(X)
does not change its sign in the interval (0,l); hence we may apply to thi's integral the obtain mean value theorem. we 1.
bn,, = -$ b,
from this we deduce
i f 0<6<
gl I b, I < I b,,, I < -& I b, I
and 18
274
(2)
bn,l< nI bn I n+l'
that is, by increasing n, the absolute values of the numbers b, diminish. Siice the series
b. + b, + b, + s . . . . + b, + 0 0,.
is alternate, from (1) and (2) it follows that it is convergent. In the preceding paragraph we have seen that in the interval O-=x~n-2 (3)
Iyn(x)I C Ibnl.
From this we conclude by aid of (1) that limvn(x) = 0 *=oI
(4)
if 0 < x C n-2. But this interval may be somewhat enlarged. Indeed, according to formula (1) of Q 89 we have
Y"(X) = j("t;"] du
therefore yn(-1) = .s'[ "';;I J du. Starting from this formula A. Sziics has shown that the limit of 1yn(-1), for n= 00 is equal to zero. Let us write the quantity under the integral sign in the following way: (-l)+-+) From 1 + x I ex, remarking that 0 < u < 1, we deduce [l-+]. . . . (1-g.
and then
275
Since for n=m the limit of the last member is equal to zero, we conclude lim ~~(-1) = 0. m=oD But according to formula (3) Q 89 we have (5) yn(-1) = (-l)"[bo4,+b,4J, or +, . . . + (-l)"b,] %+I
Iy,(-1)l
= 2- x lb,1 * m=o
and if n increases indefinitely :0
x lb,1 = 2. m=o
Hence the series 2 I b, I is convergent. Moreover, in consequence of (3) the series Z I y,,,(x) I is convergent too, if n-2 > x 2 0.
Q 94, Operations on the Bernoulli polynomials of the second kfnd. Differences of the polynomials. According to formula (3)
5 89 we have
A"wM = wn-.mW. (1) Hence the indefinite sum of the polynomial is
(2)
A-l
yn
(4 =
wn+l
(4 + R.
From this we deduce the sum of the polynomial; for instance, taking account of (3) 6 90 we find
211-l
x%
Y,,(X)
=
y2n+l (2n-1)
- Yzn+l (0)
=
- 2 b2,,
or in consequence of (1) Q 90 ?f: ylkl (x) = yzn(2n--2) - wn(Ol = 0.
Mean of the polynomial: Since M = 1+1/2A, therefore (3)
MY"(X)
= Y"(X) + ?h
Yn-1 (4.
276
Inverse operation of the mean: According to formula (12) Q 39
we have
M-1 W"(X) = m=o v yn-m(X)s "il
Derivatives of the polynomial:
I
Zntegral of the polynomial. By integration by part9 we obtain
I y"(x)
(6)
s
dx = (x-n+11 yn(x) - J (x--+1) [ 21) dx.
X
Since the quantity under the integral sign of the second member may be written n n we have I 1
y.(x)dx = ( x - n + l ) y"(x) - n y,,+I(x) + k';
from this we obtain
Y
[' yn(x)dx = (2-n)& i- (l-n)&,
Q 95. E&mion of the Bernoulfi polpomials of the second kind into a series of Bernoulli polynomials of the first kind. According to § 84 the coefficients of this expansion are given by 1 co = J yn (x)dx = (2-n)b,-, + (l--n)b,
0
cl = by,,(O) =
in-,
= b-1
c,,, = [Dm-'Ay,(x):lx=o
= [Dm-'y,,(x)].=o = (;I;; 'I S:zi '
The values of the above integral and of the derivatives have been obtained in the preceding paragraph. The required expansion will be
Putting x=0 we get another expressson of the Bernoulli numbers in terms of the coefficients b,: (5) % = (n-l) I a, = - z2 (m-l) I b, Grzi
From this formula we may obtain the number br by inversion (5 65). Multiplying both members by S;': and summing from n=2 to n=i+l we get WI (i-l) I b, = i+l n=2
X
(n-l) I a, S't_:
279 Remark. To obtain the expansion of the Bernoulli polynomial of the first kind into a series of the second kind, it is sufficient to sum formula (4) from x=0 to x=z; we find (7)
%+1(4 = %+1 + %Z +
If the function to be expanded is not a polyndmial, then the series (1) will be infinite and questions of convergence will arise. Example 2. Given f(x) = 2X. We find, using the preceding method c,,, = log 2 for m=O, 1,2, . , .; therefore the expansion will be
2x=1082[1 +Ylw +y,w +,.**I;
putting x=0, we have & = 1 + b, + b, + . . . . We have seen in $93 that both series are convergent. Example 3. By aid of the expansion (1) it is possible to determine the generating 'function of the Bernoulli polynomial of the second kind. For this it is sufficient to dispose of f(x) so a s t o havec,,, = tm. SinceAm-l(l+f)*=tm-l(l+t)x and D(l+f)x= =(l+t)Xlog(l+f) it is easy to see that
r LT.
f(x) =
t(i+tjx log(l+fl
indeed, the coefficients of the expansion of this function will be
c,,, = [DA*-'f(x)&,, = t";
therefore if -1 < t ~5 1 then
fu+~)x log(l+f)
5 = m=o Yln(4 tm
is the generating function of the polynomials y,,,(x). Putting x=0 we obtain the generating function of the numbe.rs b, ,
t log(lS-t)
ii b, tm. = VI,=0
280 After what we have seen, if -1 < t S 1 then both series are convergent. Putting into (8) and (9) t=l we obtain the results of Example 2. Putting into (9) t = - 1/2 we have 5 (-l)"b, 1 2m m=o --=TG& This may )e written in the following manner: s m=o &L-2l 2log2 2m DA-' logx = F(X)
We have seen that X I y,,,(x) I is convergent if m-2 2 x 2 - I, therefore the series (10) is absolutely convergent. Putting x=0 *we find
c= 5 Ib,l d m
This is an expression of Euler's constant in terms of the numbers b,,, , but the convergence is very slow, Q 98, The Bernoulli series of the second kind, If we put A-' 1 = - y1 (u-x) then the operation of summation by parts executed on X f(x) may be written
(1)
A-l f(x) = - y1 (u-x) f (4 + A-l[yl (u--x-l) Af (xl].
Putting again
281 A-l y1 (u--x---l) = -y&--x) the operation of summation by partsperformedon the second member of (1) will give
-y2(u-x) Af(x) + A-'
[vv,(u-x-l) A'f(x)]
continuing in the same manner we shall obtain, after having applied the operation of summation by parts n-l times, A-' f ( x ) = - 2
wm(~--x) Am-l f (xl +
+ A-l [ '~'n-~ (u-x-l) A"-l f (x)-J.
As we may add to an indefinite sum any arbitrary constant, the next summation by parts may be written
(3Lz$-lf(x) = - ~l~m(~-~)Am-lf(z) + bnAr'f(z) + J(U)+K
where n(u) is a function independent of I, obtained by putting into every term of the indefinite sum (2), except the last, x=u-1. To abbreviate we have written R, for the remainder (4) R, = .=t-,
will never be negative if n=2m; and it will never be positive if n=2m+l and u-x 5 2m. Therefore, supposing that u-&n the expression (6) will never change its sign in the interval x=rz-1, x=z so that we may apply the mean value theorem We shall find, starting from (4)
4, = AW .=t-, [in b--x--i) - b.1
where u- 1 < 5 < z. We have seen above that the indefinite sum of yn(u-x-l) is --y*,,(u--JC); therefore the summation in R, will give Rn = AWl[----Yn+lb-4 --b,z + ~n+l(l) + bnb+-l)]. Now the derivation with respect to z gives
D R, = [-~n+lIu--l) - bnz + yn+l(l) + bn(u--iI1 Dhnf15) +
I
I
If we put into this equation z=u, the first term of the second member will vanish, and we shall obtain [D RJ.=, = -b, A"f (q) where u-l < q < u.
Putting this value into equation (5) and writing x instead of u we obtain the Bernoulli series of the second kind: (7) f(x)
= [D 2-, f(f)lz=x + ,f, b,, Darn-'f(x) + bnA"fb~l 1
where x-l < r] < x. Remark. If A"f (x) and An+'f(x) do not change their sign in the interval (0,l) and if and therefore &f(x) P'f (x) > 0
whose period is equal to one, which function vanishes in F(z)-F(x-1) ; so that then the derivative of this quantity with respect to z is not necessarily equal to that of F(r). Formula (9) may be useful, first if the derivative of the function is to be determined and the differences of the derivative are ,simple. Example 2. If F(x) = log T(x+l) then AF(x) = log(x+l) and P [log r(z+l)-logr(x)].=, = f(x) and moreover
Remark. The first term of the second member is generally equal to W(x), except if F(x) contains a periodic. . function
DA"F(x) =
therefore
(-l)"-l ( m - l ) ! (x+m)m
This expansion is similar to that of F(x) obtained by aid of the Bernoulli series of the first kind (formula 10, Q 87), but that series was divergent and the present series (10) is convergent. Indeed the absolute value of the general term is smaller than I b,/m I and we have seen that this series is convergent.
i b,,, I <Ib,l m x4-m m I m 1
if x > 0,
If x is large enough, the convergence is rapid, as we shall see in a numerical example (0 118). Putting x=0 into (10) we should obtain the expression of the Euler's constant found before (1 I, Q 97), Secondly, formula (9) may be useful for the determination of AF(x) if the derivative and its differences are simple.
Thirdly, formula (9) may be useful still in other cases; for instance if [DA"F(x, t)]=d = cm fm o(t) where c,,, is a numerical constant. From this it follows that [DF(x, f)lX=o= c0 o(f). Dividing both members of equation (9) by c0 o(f), we find
The formula above has already been obtained (in 9, 8 97), but not the remainder. 8 99. Gregory's Summation Formula. This is a formula by aid of which the integral of a function may be expressed by its sum and its differences, or the sum by the integral and the differences. Apart from the remainder, the formula may be easily deduced by aid of the symbolical method. In 5 6 we had formula (4) :
hD =
From this it follows that
log(l+A). *
285
& = +
log(l+A)
A
Since according to formula (9) (j 97 the expression in the brackets is equal to the generating function of the coefficients b, of the Bernoulli polynomial of the second kind, we have
1
Performing this operation on f(x), and calculating the sum corresponding from x=a to X=Z, we get + J'f(X) dX = 5 f ( X )
and a=0 This is Gregory's formula. Putting h=l, z= 1 we get formula (1) of 8 96. To obtain the summation formula with its remainder, we start from equation (7) § 98, integrating both members from x=a to x=z; the first term of the second member will give i: f ( x ) . X=G In consequence of the mean value theorem the remainder will be r'bd.f (7) dx = (z-a) b, CInf (C) a* where a < 5 < z. Finally it follows that (1) a' ['f (x)dx = 5 f ( x ) + L, b,[Am-if(z) X=8 -Amlf ( a ) ] +
a
X==(l
+
j, bi
[pl-'f(z)-~-'f(a)l'
+ b, (z-a) AV (0.
This formula is more advantageous than Euler's summation formula, if we deal with functions whose differences are less complicated than their derivatives. Moreover there are functions which lead to convergent series if we use formula (l), while the corresponding Euler formula is divergent. For instance, this may to the fact that D" 1;) increases indefinitely with n and tends to zero if n increases,
286 There are some particular cases in which the remainder may be given in another form. First, for instance, if p{(x) and A*+lf(x) do not change their sign in the interval, a, z, and moreover if Ff (x) Flf (x) > 0 then we may obtain the remainder by integrating the expression (8) of $j 98 from ~=a to x=z; we find (2) 5 b,[A*lf (z) - A"-lf (a)] where 0 < 5 C 1.
Secondly, let us suppose that Amf (00) = 0 for m=l, 2,3, , . . and write the remainder of (1) in the following form: R, = b, i A"f(C)dx-b, f A"f(l)dx. a * If we denote by C, the part of the second member of (1) which is independent of t, then C, = b,, j- llnf (t)dx - j, b, Am-If (a) (I and consequently (3) ( 4 ) j&dx = Zi f ( x ) + jl b,A"-'f(r) -b. fb"f(C)dx+C,. X==C? D B Putting ZZW we obtain an expression which permits us to compute the number C>. Indeed, in consequence of Am!(a) = 0 we obtain from (1)
(51
C,= - =:a f(x) - liizb,f(r) + jmf (x)dx. 0
Remark. In some cases, when the operation DA-' gives the same result as A-'D; then the expansion (1) becomes identical with that of F(z) = f f(x) dx a expanded by aid of formula (7) 0 98. Example 2. Given f(x) = l/(x-l-l). We find
- i cz~m,, -- F+-] + bb-4 L-1)" n1 * (5'+n+l)n+1
From this we conclude that the remainder is smaller than h(~--av(n+l)~ This formula may serve for the determination of f(z), if a is not too small; otherwise the convergence is too slow. If a=0 it is much better to use formula (4). The constant Cf will be determined by (5): Cf = l i m log(z+l) - Xi --& =- c *=oE L 1 where C is Euler's constant. Since s - = F(2) + c l x=0 x-+1 formula (4) will be
log(z+l) = F(Z) + p, b,Am-'f(4 --n j$hc)d= '
This is the same formula as that we obtained in Example 1, fj 98, except for the remainder. Remark. Formula (1) but without the remainder, has been discovered by Gregory in 1670 [Whittaker and Robinson, Calculus of Observations, p, 1441; this was the earliest formula of numerical integration,
CHAPTER VI. EULER'S SUMS
AND
BOOLE'S POLYNOMIALS.
OF
RECIPROCAL
POWERS.
Q lo;), Euler's polynomials, We shall define the Euler polynomial E,,(x) of degree R by the following equation (1)
ME,(x) = $:
,
that is, the mean of this polynomial is equal to x"/n! . From (1) it follows that
but according to (1) this is also equal to ME,-,(x) therefore (2)
DM&(x) = ML (4
Performingthe operation M-1 on both members of this equation we obtain (3)
D&W = En-, (de
Since we are dealing with polynomials only, this solution is, according to 5 38, univocal. From (3) it follows that the Euler polynomials belong to the important class of polynomials in which the derivative of the polynomial of degree n is equal to the polynomial of degree n-l. From (1) we deduce E,,(x) z M-1 s ,
289 We have seen in (12) Q 39 that M-' = Lo (-lP$ hence the above equation gives
E,(x) = ; 2; (-l)m G . . I
From x" = n+1 ,z, a ; (x),
it follows that
2 Am 9 = w=lIl (Y)m (XI v-m 6:
and finally
nt1
I
This is the Newton expansion of the Euler polynomials. Let us write the expansion of E,(x) in a power series in the following manner:
(51
E,(x) =e,,g +e 1 (n-l) ,- + . . . a + en-, 5 + en. n!
x"-'
c-2
From (3) it follows that the coefficients ei are, according to ($ 22, independent of the degree of the polynomial E,(X). TO determine these coefficients we start from (l), writing ME, (x) = 1 and conclude that E,(x) =l and eO=l. If R > 0 then we have in consequence of (1) ME,, (0) = 0; moreover if i > 0
[Mx'],,o = 1/2,
Hence, performing the operation M on both members of (5) we obtain, if we put subsequently x=0,
Starting from e,=l we may determine by aid of this equation step by step any number ei. For instance we have 19
291 If n>O then from (1) it follows that ME.(o) therefore En(o) +&(l) =o which gives En(l) =-eB.* Moreover from (1) we deduce
= 0 and
ME,(A) = l/z[E.(--1) + En(o)1 =yFigure 3.
Y
I H
t
&,/&
t-i-.-"
& ;. ; z Y = E,(4 Y = J%(x)
From this we get &(-1) = (-1)" -$ -e,. Putting r=S/2 into equation (5) (7) The coefficients e, may be computed by aid of the Stirling numbers of the second kind. Indeed, putting x=0 into equation (4) we obtain gives
(8)
292 5 101. Symmetry of the Euler polynomials. Let us start from
and put into it -z instead of x; we find
(1)
[ME,,,, (x)]x=-E = - 12"+1- = - ME,,+,(z). Pn+l)!
On the other hand we have
MG,, (4 = ?h I&n+lIx) + E,n+,(x+l)l;
writing x=-z, we obtain (2)
[ME,,,, (x)] x= --I = ?h [Ezn+l (-4 + Em,, (l-d] =
= ME,,,, F-4,
From (1) and (2) follows, if we write again x instead of I,
Putting x= '/z we obtain = &n(l), But we have seen that E,(O) = -E,,(l) if n>l; hence we shall have, if n>O (5)
Equation (5) gives by aid of (8) 8 100,
Ezn(0) = E,(l) = can z 0. From this it follows that E2,(x) is divisible by x(x-1).
293
Roots of the polynomials between zero and one. According to (5) &n(x) = 0 has roots at x=0 and x=1. Let us show that there are no roots in the interval O<x<l. For if there were we should have at least three roots in the interval 02x21, the first derivative should have at least two, and the second Ezhz(x) at least one in the interval O<x<l, and therefore at least three in the interval 02x21. Continuing in this manner, we should find that E,(x) had at least three roots in the interval 05x51, which is impossible. Finally we conclude that E2,(x) has no roots in O<x<l. We have seen that E,,+,(x) =O has a root at x=1/2. We will now show that it has no other roots in the interval 05x$1. Indeed if it had at least two in this interval, then its derivative Ezn(x) should have at least one in O<x<l; and we have seen that it has none. Therefore Ezn+l(x) = 0 has only one root in the interval 02x21 so that e,,,, + 0. Exttema of the polynomials. From what precedes we conclude that in the interval 0 S x S 1 the function y=Ezn$) has only one extremum, and this at x=l/z; moreover y=Ezntl(x) has only two extrema in this interval, and these correspond to x=0 and x=1. If E,,,(O) = er,,-i < 0 then E2,(x) will decrease at x=0 and the extrema of E*"(x) corresponding to x=1/2 will be a minimum. On the other hand if E,,, (0) = e,,-, > 0 then Et"(x) . will be a maximum. Let us suppose first that e.+i < 0. Since E,,, ('/2) =O its derivative has no roots in this interval; therefore the derivative must be positive in the whole interval, so that
Since in the vicinity of x=0 the sign of E,,-,(x) is identical with that of ezn+, we conclude that and therefore (6) e2h3 > 0 e,,-, e2"-:, < 0.
294 Starting from the supposition that ezG1 >O, we should have also reached the result (6). Consequently from e, = - yz it follows that (7) O' e4hl > 0, e4ntl < 0
(-1)" e.+i > 0.
E. Lindelli! has shown in a very interesting way [Cal& des Rksidus, p. 371, using the Theory of Residue4 discovered by Cazzchy, that the following form may be given to the Euler polynomials :
(61 (7)
Ezhl(x)
= K;F %- =;f;+;;?:x
'
4(-l)" - sin(2m+l)nx Em (xl = - p i - 20 (2m+l)*"+l
These formulae are valid if 0 ~5 x 6 1. From the first it follows that I Ezkl (x)1 is maximum in this interval for x=0. For this value we find
(8)
E2"-, (0) = e2R-1 = cys 2 (2m~l)2n .
IE2,,-11x)l 5 le2R-ll.
Prom this it follows, if 0 s x s 1, that (9)
In Q 49 we found (formula 3) that the sum figuring in the second member of (8) may be expressed by Bernoulli numbers or by the coefficient a,,, of the Bernoulli polynomials. Indeed we have 5 m=O (2m;l)2n = 1/2(22"-1) z2= I Qz* I = 1/2(22"-1) ?z2" #:
therefore 1eZhl I = 2(22"---1) I Q2"I. We have seen that the extremum of Ezn(x) is reached if x=4/2, Putting this value into (7) we obtain w
295 This formula is very useful, for it gives the value of the alternate sum, since we have already determined E,,(l/) in other ways (7, (j 100 and 4, Q 102). So that we have
(11)
- n2n+l E,, (1/),
ghis sum is necessarily smaller than
hence from formula (8) we deduce that in the interval 0 I x 2 1 W)
nlE,,(x) 16;21E,,(1/~)I<le2n--11.
% 102. Expansion of the Euler polynomial into a series of Bemoullf polynomials of the first kid. According to § 84 we have
1 co = 0 cm s
E,(x)dx=-2e,+1,
c,=[~~(x)],..o=-~~~
= [AD"' En (4&o = EGm+, (iI- E-t+1 (0) = - 2 en,+1 .
En (xl = - 2
Therefore the required expansion will he (1)
m=O
T e,+1 pm(x)s
If the Euler polynomial is of an even degree, then we shall have, putting x=0, (2)
42n =
- 2
n-Cl I: e2n-am+l m=O a2m =
0
where azm is a coefficient of the Bernoulli polynomial (8 78). The polynomial being of even degree, if we put into it x=l,$ we get the central value of E2.(x)
(3)
E2nM2) = - 2 x e2+-2m+l~2mC4.
m=O
n+l
From this, by aid of formula (3) 0 86, it follows that
n-Cl
E2nPh)
= -
2
m5
e2n-2m+l
aznl
l
P&:X
-1);
finally in consequence of (2)
296 (4)
I n+1 &"(?4 = - x Ban-zm+l
m=O
a2m & -
The value of En (4/2) has already been obtained in another form (7 s w, If the pclynomial is of odd degree, then we find nt1 E,,, (4 = - 2 mzo e2n--2m-l vzmel (4. (5) definition of the polynomials it follows tlmt the derivatives are (1)
Dm &M = Ln(4.
Q 103. Operations on the Euler polynomials, From the
Therefore the indefinite integral of the polynomial will be s En(x) dx = E,,(X) + k and the integral between zero and one
1 s
E,(x) dx =-2e,+l.
0
Hence, if the polynomial is of an odd degree this integral will be equal to zero. The mean of the polynomial follows from its definition: (2) ME,(x) = $ s
The differences of the polynomials are more complicated. From the Newton expansion of the polynomial, formula (4) Q 100, we deduce
this gives, putting x=0, (3)
Am E,(o) = -$ "ii'
. I'==,,,
(-2tF; " q
From formula (4) Q 100 we may determine in the same way the indefinite sum of En(x), But we may obtain a more simple formula in another way. Dealing with symbolical methods we saw that
This formula may be applied to the Euter polynomials; but since their differences are complicated, the formula will not have practical value. We obtain a simpler one, starting from the formula forthe mean of a product deduced in § 31, M[uv] = uv - VMU + EuMv; performing on this equation the operation M-l we get
Supposing that the polynomial is of an even degree, then, putting into it 2n instead of R, writing x=0 and equating the will lead to the result to that of the preceding equation, important relation found by Euler in another way:
(71
2(2n+l)
e2n+l
= 5 m=O
e2n-2m-1 e2m+l .
Q 104. The Tangent-coefficients. Dealing with Euler polynomials we could proceed as has been done in the case of the Bernoulli polynomials, where we have written
n!a, = B, and B, was called a Bernoulli number. If we now put n!e,= &n the number 8, would be interesting, and equation (6) Q 100 would lead to the symbolical relation 11 + cr;]n + d",, = 0 in the expansion of which grn is to be put, instead of d'm. Starting from 4: it would be possible to determine step by step the numbers 4'm. But it is much better to introduce instead of d:m the following numbers (1)
6. = 2"n! e,.
The number (5, , which as we shall see is the coefficient of An! in the expansion of tan x into a power series, has been called tangent-coefficient.30
sa Several authors have introduced the tangent-coefficients, using different definitipns and notations. For instance in Leonhurdi Euleri. Opuscula Analytica, Petropolis, 1783-85, p. 372, the number 2rl e, has been introduced, which is the coefffcient of 2%" in the
1
299 Euter in his work quoted below has given several formulae for the determination of these numbers; one of them is equivalent to our formula (7) 6 103; multiplying it by 22n (2n) I, by aid of (1) it will become M (3 2n+1
=
5 "1 =o
2n l 2m+l 1 @2m+l
@2*-2m-I
or symbolically (3) 6 2n+* = [G * fq'*'.
We may deduce other formulae for the determination of the numbers 8,. For instance, multiplying both members of equation (6) 5 100 by 2" n! we get the symbolical formula (4) [2 + qn + (3" = 0.
From formula (5) Q 100 we obtain in the same way the symbolical expression of the Euler polynomials: (5) E , ( x ) = L2;-$f1".
§ 105. Euler Numbers. If we put x=1,/2 into the polynomial E,(x) and multiply the result by 2" n! we obtain a number which will be denoted by E, , (1) E, = 2" n! E,*($G).
This number was first called a secant-coefficient and later an Euler number. E, is, as we shall see, the coefficient of x"ln! in the expansion of set x. It would have been better to call the tangent coefficients "Euler numbers", because in consequence of (5) Q 104 they figure in the coefficients of the Euler polynomials, whereas the secant-coefficients are only particular values of the polynomials.31 Putting x=1/2 into equation (5) 8 104, we obtain in consequence of (1)
From this formula it follows that the numbers E, are inte31 Several authors have dealt with Euler numbers: Leonhardi Euleri, Institutiones differentiales, 1755, p. 522, or Opera omnia Vol. X, p, 419. Euler denoted the number E,, 1by a$ the number /fi by ,% 1 El I by Y and so on. He gave a tab I e of these numbers up 1.3 . Scherk, Mathematische Abhandlungen, Berlin, 1825 p. 7, gives a table of these numbers up to E,, , This table is reproduced by Saalschiitz; Vorlesungen iiber Bernoulli'sche Zahlen, Berlin, 1893, p. 22. His (I, corresponds to our [ E,, [ . N. Nielsen, Trait6 des Nombres de Bernoulli, Paris, 1923, p. 178, also reproduces Scherk's table. His number En corresponds to our lEzn 1. Lindefiif, Calcul d e s Residus, Paris, 1905, p. 33, Same notation as Nielsen's, Niirlund, Differenzenrechnung, Berlin, 1924, p. 458. The notation WE have adopted here is the same as Niirlund's notation.
1
301 gers. The formula gives the Euler numbers in terms of the tangent-coefficients. It is easy to get an inverse formula: indeed, multiplying both members of (2) by (-l)m-n(y] and summing from n=O to n=m+,l we obtain, according to Q 65 (Inversion), (3) Em = X0 (-l)"+,
m+l
; E, = [E-llm. I I
Ez,,,l(l/) = Oitherefore E,,,, = 0. Moreover we have seen that asi = 0 if i > 0, consequently from (2) we deduce, writing 2n instead of n
Hence in consequence of (1) (4) The series Z e, is absolutely convergent. In $ 101, formulae (9) and (12), we have seen that I E2,, (4 I 5 I e2n-1 I a n d nlE,,(x)I < le2n-ll i f O4x5;l;
from (2) it follows that I
E,n-,
(XII < +
and 1 Ezn(41 < &.
7
303 Therefore we conclude that (5)
lEn( 5& i f
lim E,(x) =O.
05x41
and in the same interval
"CC3
The series X E,(x) is absolutely convergent in the interval OzSxXl. In Q 104, formula (1) the tangent-coefficients were given by E2n-1 = 22*-1 (2n-1) ! e,,-,; from this we deduce by aid of (2)
I Q,,-, I ~- < * 2 "IF" (2n-l)! 3 'i:
l 1
:3m -__ = OS (2n-I)!
e2n-*
:.
@2n-r From these relations we conclude that the series Z -(2n-1) I is absolutely convergent. In 6 105 we defined the Euler numbers by
L.
E,, = 22" (24 ! E?,(x).
From (3) it follows that
IEznMI
< &
Therefore
and
The series Z E,,/ (2n) ! is absolutely convergent. Q 107. Expansion of the Euler Polynomial into a Fourier series. The expansion of the polynomial of even degree in the interval 0 S x S 1 may be written:
The quantity in the brackets is equal to zero at both limits; a second integration by parts gives: 'hJ"l = cos2mzx E,,-, (x) dx. (2m;c) z E,n-, (x) -I:, - oj' e;;
The quantity in the brackets is equal to
moreover the integral is the same as that in (2), only the degree of the polynomial has been diminished by two, and it has been multiplied by -1/(2m,z)', Therefore repeating the above operations v times we get:
Putting the integral in the second member will v=n be equal to zero and we have (3) urn = +cJ1 I=, (-1)' 4 h+l--2f (2mn)" '
305 The expansion of E,,, (x) will be obtained, in the interval 0 S x I 1, by determining the derivative of both members of equation (l), We find
E , , - , ( x ) = - 5 2mna, sin2mnx = g /?;,, sin2mnx;
w&=1 m=l
from (3) it follows that 14) Let us remark, that the function EzGl(x) considered as a periodic function with a period equal to one, is discontinuous at x=0, being equal to + e,,, ; therefore the Fourier series will give E,,-, (0) =O. Remark. Putting x=0 into equation (1) we get n+l E , , ( O ) = 0 = - 2 ezn+l -I- 2 X e2n+l--2,. 5 -):. nr=l (2mn)"v W=l But we have seen (6), 8 82 that the second sum is equal to --az,.; therefore we have PC1 n+1 e2.+1 + X ezn+F-21. a,, = X azl. e2n+l--2,. = 0 . (5)
?=I
w=o
This equation has been obtained before (2 (j 102). To obtain the central value of E2,,(x) expressed by aid of the coefficients of the Bernoulli polynomials, let us write in equation (1) x+/i. We find J%(%) = ?hao+ %, (--l)marni in consequence of (3) this will be
and according to 8 49 we have
Therefore
n+l &(1/2) = 2 ,z, a~,, e&-zv [ I-- :,$I,- I
20
306 Since according to (5) the first sum is equal to zero, find we
Ean(-Q$) = - ;- a2$y2r.
This formula is the same as that of 4, 5 102. Multiplying first both members of (6) by (2n)f 22" and then the numerator and the denominator of the second member by (24 ! (2n+l-24 ! remarking that (2v)!az1. = B2v is the Bernoulli number, we get, in consequence of (1) Q 104 and (1) 8 105, E,, = or
_ ,B
2n+1 2' 21
This may be written symbolically: E,, = - & [@+B]'"". Equation (7) gives the Euler numbers in terms tangent coefficients and the Bernoulli numbers. Example. Ed=-+ [&+lO Q, B, + 5 & B, ( = 5. Remark By aid of formula (10) Q 101 we may eliminate azv of (6) and get an expression of the Euler numbers by tangent coefficients; or we may eliminate e2n+,...2v and obtain the Euler numbers in term s of the Bernoulli numbers. s 108. Application of the Euler polynomials, 1. Determination of the inverse mean of f(x) . Let us expand f(x) into a Maclaurin series: f(x) = of the
f[o)
+xDf(O) + $ D'f(o) + . . . . + 2 D"f(o) + . . .
The inverse operation of the mean gives
(1)
f&f(x) = f(o) + E,(x) Df(0) + E2(x)D2f(0) +, . d. +E,(x)D"f(O) +.:a.
1
307 Starting from this formula we may determine moreover the inverse difference of the alternate function (-l)"f (x), According to formula (10) 5 39, we have A-' (-l)x f(x) = 1/2(-l)X+1M-1f (x). Therefore the indefinite sum of the alternate function will be (2) A-' (-l)Xf(x) = l/i(-l)X+'[f(0) + E,(x)Df(O)+. . . +
Hence if m + n cm = and if m=n cn = 1. The required expansion will be 1 z(n--m)l'
309
(5)
If f(x) is not a polynomial the series becomes infinite and its convergence must be examined, but the coefficients are determined in the same manner. Example 3. Given f(x) = eX'. It follows that
Dm
,xr = fm ,y~
and
MD" exl = l/z P (eX'+' + ex')i cm = $5 tm (e/+1)
putting x=0 we have and the required expansion is (6)
ex' = 1/2(e'+l) 1, E,,,(x) tm.
The generating function of the polynomial E,,(x) may be obtained immediately from (6) :
GE,,(x) = g = 5 E,(x) f".
m=O
We have seen in 0 106 that, if 0 S x S 1, we have
l&,(x) I < +
consequently if I t 1 < Z, the series (7) is convergent. Putting into (7) x=0 we obtain the generating function of the numbers e, : (8)
Ge, = & = m=O
Z
e,,, tm = 1 + e,t + e,t3 + . . . .
This may be written in another form: 5 e,,f"'="i 1 or putting t=2iz we find tan z = 5, (-l),+l esm+l (22) Im+l. According to Q 104 (formula 1) this is equal to -=-fanhl/zf ef + 1
310
(10)
tans=
1
m=O
5 (-l)m+l ts2m+l (2r;;)F =
. ii m=O IcG2m+,l --zm+l (2m+l)!
where the numbers 6, are the tangent-coefficients. We have [Dmtanzl.,o = ;@,I. Let us remark that we have already expanded tan z into a power series in § 87 (formula 17). Writing that the coefficients o f 22"' in the two expansions are identical, we find the important relation e2,, = 2(1-2*") a2" (11) which has been already obtained in § 101 (formula 10). V&kiting x=1/2 in (7) we get 2d2 1 -= = sech T/z t = 5 E, (1/2) tm. 1 + et cash 1/2 t m=o This is the generating function of the numbers Em ('/2). Let us write again t=2iz; we shall find (12) secz = mS (-l)m Ezm(l/z) (2z)2m ;
according to formula (l), Q 105 this may be written secz =
m=O
5 (-l)mEpn&!
= ,I, IE2m
I -. _
z2m
Pm) 1
where the numbers Em are Euler's numbers or the secantcoefficients. We have
Putting into it t=2z we obtain formula (10) found before. If i t I < z then the series (13) is convergent for 0 5 x I 1, in consequence of (5), Q 106. Q 110. Multfplication theorem of the Euler polynomials. To deduce the theorem, let us expand the following polynomial into a series of Euler polynomials:
F(x) =
we have
DF(X)
@P+l)"
I$ (--t)"&[ sl I-
According to Q 109 the coefficient of E,(x) is ci = MD'F(O) (Z!p+I)"-' y,S!: (--1)"E"-i( Gz$
=
moreover
If in the second term of the sum we put m instead of m+l then it will become equal to the first term, but with a negative sign; the new limits will be m=l and m=2p+2. Therefore we shall have, putting x=0:
ci
= MD'F(O) = '/$(2~+1)"-~ [En-i(O)
i-E,-,(l) 1.
Hence cn=l and ct=O if i $r n. In consequence F(x) will be equal to E,(x). Finally putting x=(2p+l)r we obtain the first multiplication formula (for odd factors):
312
(1)
E"[(2p+l)zl
= (2p+l)" *s (----iP& 2 + &i' 1 t
2p+1
Writing in this formula 2x0 we obtain an expression for e,, . For instance, if p= 1 we have
To obtain the second theorem of multiplication (for even factors) we shall expand, into a series of Euler polynomials, the following polynomial:
F(x) = - 2(2p)"' mL (-lP% 1%)
where q"(5) is a BernouNi obtain polynomial of the first kind. We
D'F(X) = -2(2p)"'-' So (-l)m qn-i [ 9 1 .
The operation M gives
Putting m instead of m+l in the second term of the sum, it becomes equal to the first term, except the sign and the limits, which will be m=l and m=2p+ 1. Therefore every term will vanish except those corresponding to m=O and m=2p; it results, after having put x=0, that Ci = (2P)"'-'
[%-i(l) -
9%i(O)]*
In consequence of what we have seen in (j 79, the second member will be equal to zero for every value of i except for i=n- 1 where ci-r = 1, Therefore the expansion will be
F(x) =E,-l(x).
I f w e w r i t e x=2pz, formula (21 we get the second multiplication
E,, (2~2) = - 2 (2~) "-I
313 Particular case. Let p=l, then %I-I(24
= - 2" CP)"M - v"(z+%2)1.
Putting z=O we have e n-1 = - 2" Can-h Rut we have seen that (8, (j 82) vn(!4 = an (&i -11; therefore we find e n-l = 2a,(l-2"). This equation has already been deduced in Q 101 (formula to), and in Q 109 (formula 11). $j 111. Expansion of a function into an Euler series. Starting from the function f (x+u) and integrating by parts, we may write ,j-' f(x+u) dx = [E,(x) f(x+u)l; - [' E,(x) Df(x+u) dx. 0' The quantity in the brackets is equal to -2e,Mf(u). The integral in the second member gives by integration by parts
The integral in the second member is the remainder; let us Ezn(x) does not change its sign in the denote it by R,, . interval 0 5 x S 1 therefore; in consequence of the thetuem of mean values, we shall have
(3)
&,=-SD
zn+lf (q) oj'E,. (x)dx = eantl D2*+lf (7)
where u<q<u+l. Particular cases. 1. If D*n+lf (u+x) and D2n+3f(u+x) have the same sign and do not change their sign in the interval 0 ~5 x S 1, then we shall have R,,R,,, < 0; the same ratiocination as in the case of the Bernoulli series ($ 87 will lead to
(4)
The preceding series has been obtained before (9, $j 109), but not the remainder,
315 Q 112 :Boole's first Summation Formula, It gives the inverse mean of the function f(x) or the sum of the alternate function (-1)X f( x) ex p ressed by aid of the derivatives of f(x). Apart from the remainder, the formula may be deduced by aid of the symbolical method. In 5 6 we found M = 1 + $Qj = !h (1 -t ehD), From this it follows that
but according to formula (8) Q 109 the second member is equal to the generating function of the numbers err the coefficients figuring in the Euler polynomials. Therefore we have M-' = ii ei (hD)'.
This is the first form of the formula. To obtain the second we remark that in 6 39 formula (10) we had M-1 = 2 (-l)x+l A-' (-1)". This gives 6:' (-1)" = 1/2(-l)X+1 g eJhD)? ho The above operation performed on f(x) gives, if the sum of the alternate function is calculated from x=a to x=z; A (-1)" f(x) = - l/i i hi ei [ (-l)'D'f (z)-(-1)" D'f (a)]. To obtain the formula with its remainder in the particular case of DZntlF(u) DP"+3F(~) > 0 and if D2""F(u) does not change its sign in the interval 0 ~5 x 5 1 we consider the expansion of F(u) into an Euler series according to formulae (2) and (4) of 5 111:
(1) FIN = W(U) + i$ e2i+lMD2i+'F(U) + ~e,,+,MDzntlF(u)
where 0<9< 1.
316 Putting now F(u) =M-'f (u) we find M-'f(u) = f (u) +
i=O
5
e*i+,W+lf (u) + 8 eZn+lDZn+lf (u) +(-I)" k
where R is an arbitrary periodic function with period equal to one. Since according to formula (11) 8 39 we have A-1 = 1/2(-1)x+1 M-' (-l)x. the indefinite sum of (-1)" f(u) will be A-l (-I)" f ( u ) = 1/(-l)"+l
I- @em+1
This is Booze's first summation formula for alternate functions, which plays the same role in these cases as the MaclaurinEuler formula for ordinary functions.32 Example 2. Let f(u) =eUf. From equation (2) it follows that i u;=o (-I)" cut = 1/ [I + (-I),+, eBr] [ I+ & e2i+l t2'+l + + 6 e2n+l t2"+l].
Particular case of Boole's formula. Let us suppose that we have D"f (a) = 0 for m = 0, I, 2, 3, . . . . 9 and D"*"f(u) D""+"f(u) > 0 and that D2"+lf(u_tx) does not change its sign in the interval 0 < x < I; then we may write formula (2) in the following manner: (3) i (-l)"f(u) lb=0
J1 B.
The constant C, must be determined by aid of (4) or (5). Example 2. Given f(u)=l/(u+l). We may determine C easily, using equation (5). Indeed we have (6) Integrating both members of this equation from t=O to t=l we get O" k-1)" = logs* UzJ u+l Hence according to (5) we have Cf = log 2. Since moreover -(2i+l) ! D"'+W = (u+1)2i+2 the required alternate sum will be (7) * ( - 1 ) " --=log2+1/2(-1)~[~+ 2* u+l "
+
(2i+l)!
"C&l+ "2n+l
(2n+l) !
-1) I '
2 (1+1)2i*2
The series obtained by putting n=a is divergent, nevertbeless the formula is useful for the computation of the alternate sum if t is large enough. On the other band for t = 0 the series is practically useless, According to (3), Q 106 the best value of n is approximately Heinz. Q 113. Boole's polynomials, We shall call Boole's polynomial of degree n the polynomial 5,(x) satisfying tbe equation (11
ML(x) = [;I.
Performing the operation M-1 we obtain
318
(2)
1;,(x) = M-' I;]
one SOlUtiOn Of
Since c",(x) is a polynomial, there is Oaly equation (1 J, as we have seen in s 38. The operation A performed on (1) gives
AMLW = [Ll)Now executing the operation M-l we find, according to (2),
(3)
AM4 = Lbd.
X
Therefore Booze's polynomials belong to the important class of functions mentioned in Q 22 (p, 64). The coefficients of I i Iin the expansion of Booze's polynomials into a binomial series, could be determined in the same way as has been done in the case of the Bernoulli polynomials of the second kind; but there is a shorter way. Indeed we have seen, formula (12) () 39, that M-1 = 5 (-l)m g In=0
Applied to equation (2) this gives
This is the expansion of the Boole polynomials into a Newton series. The coefficients of f are in consequence of Q 22 inI I dependent of the degree of the polynomials; moreover they are very simple,
Particular
cases.
and so on. (See Figure 4.)
319 Remark. (-1)" 2" 5, (x ) is equal to the first part of the expansion of (l-2)x. Indeed we have
Roots of the Boole polynomials. According to formula (5') the polynomial C,(x) changes its sign between 0 and 1, between 1 and 2, and so on, between n-l and n. Hence the roots are all real, single, and situated in the interval (0, n), Application of the Boole polynomials. 1. Determination of alternate sums. Since according to (lo), $ 39 the operation I@' may be expressed by
M-1 = 2(-1)x+' A-1 (-1)" we have
321 we expand i into a power series and determine the derivative ( 1 term by term. We get
and putting x=0
DC n (0) _ (-1)"+'
2"
nt=~
"2'
2" .
m
The higher derivatives are easily obtained from (2). Integral. !I ; dx = y * + , ( x ) + k 1 where Y"+~ (x) is the Bernoulli polynomial of the second kind of 8 89; therefore from (1) we obtain: and
J C,(x) d x
j'
0
= %I .';;!i-" ~m+~(x) + k m==O d x = "gl ';t!r b,
m=O
c,(X)
Mean of the polynomial. From the definition of the polynomial it follows directly that
Znverse operation of the mean. In consequence of formula (12) Q 39 we have
M-'&(x) = 5' wi-"e.(x).
m=O
Q 115. Expansion of the Boole polynomials into a series of Bernoulli polynomials of the second kind. According to formulae (1) and (3) of Q 91 we have
(-1)" 32 2"
co = DA-l f (0) = DC,,, (0) = and
pt 1
n4 m=i
m m
ci = D&-l f(0) = D~,-j+,(0) = $:g "-i? 2 . al
322 The above values of DC",(O) are those deduced in the preceding paragraph. Finally we find
g 116. Expansion of a function into a series of Boole polynomials. If f(x) is a polynomial of degree n we may write it as follows: (1) f(x) = C" + c,c, (4 + c,L(x) + * * * * + C"M-4. It is easy to determine the coefficients cm ; indeed, since
it follows that Moreover
c,, = Mf(o).
A" f(x) = cm + cm+, l-1 (x) + * * . a + G Lrn (xl and
cm = MA"f(O).
If f(x) is not a poynomial, the series (1) will be infinite and considerations of convergence will arise; but the coefficients a.re determined in the same way as above. Example 1. Given f(x) = (z), the formulae above give c,,-~ = 111 and c,, = 1. The other coefficients c,,, are equal to zero. So that we find
The second member is equal to (1 +r&)&i(x) or to M<,,(X) therefore this equation would immediately follow from formula (2) 8 113. Example 2. Given f(x) = yrr(x), the Bernoulli polynomial of the second kind. We have
ImS21 t F(x) = ',fz - c + 5 Wlm-' --- -m Ixl ' 2m(m+l) ",:--I
(See also the expansion of fi, (x) in 5,122). The Boole series. The expansion of f(x+u) into a series of Boole polynomials gives f(x+u) = c,, + c,C1(4 + c&(x) + * , * * where as we have seen c,, = MAmf (u), Putting into this equation u=O we obtain the preceding expansion (I) of f(x) into Booze polynomials; and putting into it x=0 we obtain the expansion of f(u) into a series which we will call a Boole series, just as we have done in the cases of the expansion of f(x+u) into series of Bernoulli and Euler polynomials, Since C, (0) = (-l)"l/2"' the Boole series will be
An example is given in 8 122. S 117. Boole's second summation formula. It gives the inverse mean of the function f(x) and the sum of the alternate function (-l)x f(x) expressed by aid of the differences of f(x).
324 The formula may easily be deduced by aid of the symbolical method. In 8 6 we found
M=l+?hA
from this it follows that
this leads to the first formula. To obtain the second, we remark that, according to formula (10) 6 39. M-1 = 2 (-1) x+l A-' (-1) X so that A-' (-1)x = 1/2(-1)X+1
i=O
5 --fli L\' 2'
l
This operation performed on f(x) gives, if the sum from x=a to x= z is to be calculated, i (-l)"f(x) = i=,' 5
i=O
-$$[(--L)?Ef(a)-(--l]'SY(z)I.
We may obtain this formula by aid of the Boole's polynomials introduced in the preceding paragraph. The expansion of f(x+u) into a series of Booze polynomials may be written f(x+u) = rd$,
this may be useful for the determination of the sum if z is large. 8 118. Sum of reciprocal powers. Sum of l/x by aid of the digamma function. This function has been already treated of in $j 19; here some complementary formulae are added. The definition of the function was the following f(x) = D logr(x+l). From this we deduced AF(x) = $i and therefore A - ' ; = F(x-1) + K; moerover if n is a positive integer Z L = F(n)--F(O). x=1 A' We have seen (5, 8 19) that the digamma function with negative arguments may be expressed by a digamma function with positive arguments:
nc1
326 F ( - x ) = F(x-1) +ncotnx. From formula (2) 8 19 we deduce
therefore we conclude that the function F(x) is a monotonously increasing function. Since we have f (0) < 0 and F (1) > 0 has only one root, and this in the the equation F(x) = 0 interval zero to one. Determining it we find x = 0.46163 21. The indefinite sum of the digamma function is given by (6) Q 34:
A-' f(x) = x F(x) - x + k
from this we obtain, if n is a positive integer:
5 F ( x ) = nF(n) - n . a=0
For instance 5 f ( x ) = lOF(10) - 10 = 13.51752 592 .I-=0 This result may be checked by adding the numbers F(0) + F(1) + F(2) + . . . . , + F(9) taken from the tables of the function We have seen (7, 3 19) that the function f(x) may be expressed by the aid of sums. We found:
Let us remark, that this series is uniformly convergent in the interval 0,N where N is any finite positive quantity whatever. Indeed, if 0 s x < N, we have
From (1) we may deduce the derivatives of F(x); we obtain D~,EF(~) and = ; k-l)"+' m! (x+v)m+' 1.=;,
i
327
or if we introduce the notation S", then 1 2 -L ,.=* lfl'
DmF(0) = (-l)m-l m! s,+~ -
Hence the expansion of F(x) into a power series will be (2) F(x) = - c + 5 m-l I-l)"-' s,+r x"'.
S ",+r 5 n2/6; therefore this series is convergent if (xl < 1. The function F(x) may be expressed if x is a positive integer by the definite integral
F(x) =-C+j
0
..' 1-p -dt.
bdeed the expansion of the quantity under the integral sign will be 1+t+t2+ ,,,, -[fX+tX+'+fr+2+..~.]; integrated from t =O to t =l it gives therefore from (1) it follows that the second member of (3) is equal to F(X). Let us remark that this is true even if x is not an integer. From the above formulae it follows that
(4) s, = +-$ D"-'F(O).
In formula (6) of 8 82 we expressed spn by aid of the Bernoulli numbers. Using this formula we obtain from (4)
D2=lF (0) = q IB,,l.
Shce AF(x) =
Expansion of the digamma function into a Newton series.
l/(x+1), we have
328 (4')
Amf(x)
= (-l~~;$L1)! m
a n d
AmF(0) = (-zm-'
so that f(x) = -c+ ; (-l)m-' [ ; ) ,
m=l
m
If 0 < x < 1 the absolute value of the general term is smaller than l/m' therefore the series is uniformly convergent in this interval. The values of P(x) for x>l may be computed by other series. For instance if x is large the Maclaurin-Euler series is indicated, We found in 8 88 an expansion of the sum of l/(x+1), from which, by remarking that Ii - = F(2) + c , l x=0 x+1 -2(x:il
we deduce: (6)
F(x) = Mx+l)
- mi, y+;;z! =2m -
- 5azn
(2n-1) ! (x+1)2"
where 0 < 6 < 1. The logarithms figuring in these formulae are Nupier's logarithms. The best tables for these are Schultza's Recueil de Tables Logarithmiques (Berlin, 1778). If we do not possess such tables, we may obtain the Napier logarithms by multiplying Briggs' logarithms by the modulus log 10 = 2.30258 50929 94045 68402. If in the expansion (6) we put n=a, the series obtained will be divergent; but nevertheless formula (6) is useful for the computation of F(x). E. Pairman has used it for computing her tables.34 Since from formula (6) it follows that the remainder is I R,, I <
IB,nI
2n(x+l)2n
%' Eleanor Pairman. Tables of the Digamma and Trigamma Functions, Tracts for Computers. Cambridge, 1919. They contain F (x) and F (x) from x=0.. 00 to x=20 q 00 (Ax=0 .Ol) and the corresponding second and fourth central differences, to eight decimals.
The greater x is, the closer the approximation that may be obtained. For instance for x=0 from Table II of Q 78 it follows taking account of (7) that the smallest remainder will be obtained for n=3. Then formula (6) gives: f (0) = - 0.57517 - E ,0,0040398 the error will be less then 4 units of the third decimal. But for x=9 already a precision of twenty decimal places may be obtained for n=12. Example 2. Determination of F(99) to ten decimal places. We have F(99) =logloo-&--l~ --5 $ *
We found in Q 78 that a, = l/12 and o, = -l/720; therefore we shall have
-
log 100 -l/200 -l/12.10'
= 4,605170 1860 = - 0.005 = - 0.000008 3333 4.600161 8527 F(99) =
According to the remainder the error is less than one unit of the tenth decimal, If x is large, F(x) may be calculated also by the aid of the series obtained in (lo), $j 98: ( 7 ) F ( x ) = l o g ( x + l ) - 5 (--i(!~;~~-')(b.-b"a" ill=, log(rj+l)
where x < 11 < x+1. We have seen that this series is convergent for x > 0. This is the expansion of f(x) into a reciprocal factorial series. The formula applied to the preceding example gives b 4 + 66, L(99) =loglOO-- + 9 2!L 10100 100.101.102 100.101.102.103 -
The error is less than one unit of the tenth decimal. Finally F(x) may be determined by the aid of Pairman's tables, and though these tables contain F(x) only up to x=20 we may employ them also for x > 20 by using the multiplication formula given by (6), Q 19. F(nx) = log n + i l!!I F x- + .
i=O I 1
Example 2. Determination of F(lOO). Putting into the preceding formula x=20 and n=5 we get . F(lOO) = log5 + -; (F(20) + F(19,8) + f(19,6) + F(19,4) + F(19,2)]. By aid of Pairman's tables we find = 1.609437 91 3.000723 94 F(100) = 4,610161 85. log 5 Since F(lOO) = 0.01 + F(99), the above result is in good accordance with those obtained before. s 119, Sum of l/x* by the aid oi the trigamma function. The trigamma function has already been treated of in Q 20. Its definition; is the following therefore F(x) = DZlogr(x+l) = DF(x);
331 and finally if I is an integer S-cl z Jg = F ( 0 ) - F ( z ) . x 1 In (7), $j 34 we obtained by aid of summation by parts the indefinite sum of F(x) A-' F(x) = xF(x) + F(x) + k where k is, as tion of x with Fom this from x=0 to has been said in Q 32, an arbitrary periodic funcperiod equal to one. it follows that the sum of F(x) is, when x varies x=n,
XL0
5
f(x) = c + d(n) + F(n).
The trigamma function may be expressed by a sum. Indeed the derivative of F(x) given by formula (1) Q 118 is
(1)
. This series is uniformly convergent if 0 < x. It is easy to show that the difference of this expression is equal to -l/(~+l)~. From (1) it follows that F(l) = 0; moreover by aid of formula (6) Q 82 (1)'gives
.-
Since AF(O) =-1, F(1) = g - 1, The values of 1(x) for negative arguments are deduced by derivation from the formula giving F(x) for negative values (Q 118); we find
where we introduced the denotation s, of the preceding paragraph. Hence the expansion of the trigamma function into a power series will be (2) F(x) = ,zo (-1)"' ( m - t - l ) s,+,xm.
The series is convergent if 1x 1 < 1. There are other series useful for the computation of f(x) if r' is large. We found in the case of the Maclaurin-Euler formula an expansion into a reciprocal power series (9, 5 88) from which, by remarking that
we deduce
The series obtained by putting n=x is divergent but nevertheless formula (3) is useful for the computation of F(X) if xh. Pairman's tables were computed by aid of this formula. In consequence of formula (2) p. 302 the best value of n would approximately be z (x+ 1 ), Example 1. The value of f(20) is required to ten dximals; let us write: F(20) = & - -__ "-+&-kg+. (2)~ -+ (21)3 Determining the remainder, we see that the error is less than two units of the eleventh decimal. Moreover
Formula (4) $j 68 gives the expansion of 1/x2 into a series of reciprocal factorials:
1 q= 21 (%;!,I,
If x 2 1 then the general term of this series is less than 1 ,n2 and therefore the series is uniformly convergent; hence the sum of this expression may be calculated term by term; we obtain by the inverse operation of the differences (4) --$(x-l) = - ,,i* $-y\) II + w(x)
i
where w(x) is a periodic function with period equal to one. Therefore the sum from x=1 to x=z will be
-_
5 5 F(O) - F(z-l) = tr=, + - n=1 -W)! + w(z) n(z+n--l),,
-o(l).
We remark that, since z is an integer, o(z) = w(1) moreover that the first sum of second member is equal to F(O), so that this equation may be written f(z-1) or = 2
n=,
?&=I
(n-l) ! n(z+=--1)"
1
(5)
F(x) = E
x+n ' n2 (. n 1
From this it is easy to see that the general term of (5) is less than l/n" and that the series is convergent. If x is large, the formula is useful for the computation of F(x).
334 Remark. Formula (5) has been demonstrated above only for integer values of x. Example 2. F(lOO) is to be calculated to nine decimals.*j Tbe successive terms are: 0,009900 9901 48 5343 6282 136 4 F(100) = 0.009950 1666 If we calculate by the aid of this formula F(20), it would be an excellent example to show that a convergent series may lead less rapidly to the result than a divergent series. To obtain the same precision as in Example 1, where we calculated four terms, here it would be necessary to calculate more than ten terms; indeed we have 0.047619 0476 1082 2511 62 7392 5 8818 . 7529 1207 230 52 12 5 F(20) = 0.048770 8232. Comparing this result with that obtained in Example 1, we remark that there is here still an error of three units of the tenth decimal. Determination of F(x) by the aid of Pairman's tables. If 0 < x S 20 then F(x) may be obtained by interpolation from the tables. If x > 20 the tables still give F(x) by using the multiplication formula of the trigamma functions
tables, for instance such tables as are to be found in the "Annals of
so The binomial coefficients occurring in (5) may be taken out of
matical Statistics", Vol. III, p. 364, ff.
Mathe-
1
335
F(mx) = -$ #so F (x- f)
Example 3. Computation of k(100). We shall put m = 5 and x = 20; it follows that F(lOO) = & [F(20) + F(19,8) +F(19,6) + F(19,4) + F(19,2) 1. By aid of the tables we find F(lO0) = 0,009950 166 in good accordance with the result of Example 2. $ 120. Sum of a rational fraction. Given a fraction in which the degree of the numerator is at least two less than that of the denominator such as, (1) a,, + a,x + , . . . . + a,r+.,m-LI xn+2m-z F(x) = (x+6,) (x+-b,) . * * (x+6,,) (x-kc,)'(x+c,)" . , . (x+c,,,)Z this expression into
To begin with, we shall decompose partial fractions, so as to have
We have seen in $j 13 that if we denote the numerator of the fraction (1) by '1. (x), and the denominator by v!(x) then we shall have A, = (P Wi) ' D(/)(A)
and denoting ,
we have (formula 5, Q 13) Cj =
I?)(-Ci)
and
Bi = DP;(-ci)
Reducing the fractions (2) to a common denominator, the numerator obtained must be identical with q(x) ; since the degree of Q(X) is at most equal to n+2m-2, therefore the coefficient of xr:+ 2n,- I obtained must be equal to zero. That is, we must have (3) A , + A , + . . . . + A,, + B, + B, -j- , . , + B, = 0.
336 The operation of A-' performed on both members of (2) gives A-IF(x) = I:
n+1 kl -
Ai F(x+bi-1)
m+l
m+l
+
iz,
BiF(x+Ci--l)
-
i=l
z C#(x+c,-1)
+ K.
sum
From this we deduce without difficulty the from x=0 to x=z, if t is finite and integer: (4)
X=0
of F(X)
2 F (
X
) = 2: Ai ]F(z+~,-1) - rF(b,-1)] + n1+1
Bi[F(Z$Cj-1) - F(C,-1) 1 +
+ Z i=l m-Cl + iz,
Ci rF( q-1) - F(r+c,l)l.
This formula is not applicable if z is infinite, since then the digamma functions in the second member will become infinite too. The following transformation of formula (1) will meet the case : Let us add to the second member of (2) the quantity - & lA,+A,+ . . , +A,+B,+-B, + . . . +&I zz o. Since according to formula (7) 8 19 we have
r.
X-LO
5 -& -I
1 I x+1
= --F(bi-1) I
-c
the sum of F(x) will be, taking account of (2), equal to (5)
x-0
g F (
X
)
= - 2: AiF(bi-I) - 2: BiF
mtl
(Ci-1) +
+
i=O
Z
CjF(Ci-1).
The coefficient of C is in consequence of (3) equal to zero. The last term of the second member has been obtained by aid of the indefinite sum, Example. To determine by this method the sum of a reciprocal factorial let us decompose l/(x+n), into partial fractions. Since now q(x) =l and v(x) = (x+n), we shall have
337
(-l)i-1
Ai = D[(X+iZ)"]s-i So that ( 6 ) *" Ai me?-- = -= i=l x+i (x+4"
= (i-l)! (n-i)! '
Performing on both members the operation A--' we get
(n-l) ;;n+ = Zl' Af(x+i-l) + k
nt1 -(*-l);n--i)! - - izl Aif(i-l);
therefore the sum of (6), x varying from zero to z, will be
To determine the sum, x varying from zero to 00, we must use formula (5); we get
338 According to formula (5) of 5 118 the sum in the second member is equal to F(n) +C; therefore we get the interesting formula ( 8 ) DA"f(o) =A"F(O) = + If (n)+Cl = e i, +l. integral of the reciprocal factorial. The integral from zero to I is obtained from formula (6):
n I n+1
o J ($iz],,
=
2,
Ailog
z+i y-'
This formula is not applicable if z=w, since then the terms of the second member become infinite. To remedy this inconvenience we add to (6) the following quantity (which is equal to zero in consequence of (3):
This may he verified by aid of Stieltjes' table.36 From (3) it follows that lim s,' = 0 moreover that the series \' s,,' is convergent. y Since
from (2) we get (4)
nr2
2 s,,' 1
1 1 5 --= _--li --I, (=I i ( i + l ) i II I
We have seen moreover that
hence putting m=b+l into the first equation (2) we obtain 1 / Sfv 1 - 5 --1=.5 : SAf.,'X i ,=2 2i(i+l) 4 i.: L , i i(i+-l)! I==, On the other hand, /when L putting m=21; we must a!so consider the case of i= 1, so that we have (5) (6) Formulae (5) and (6) are given in Stieltjes' papzr. Multiplying both members of the first equation (2) by (-l)"'G;T,-, and summing from m=2 to m=m we obtain, according to formula (5) Q 65 (Inversion):
341 We have already obtained this result in Q 67 (formula 6). B. Determination of the sum of reciprocal powers by aid of the derivatives of logr(x+l). In 5 21 we expanded log I'(x+l) into a series of powers of x; but in consequence of what has been said at the beginning of this paragraph it is better to expand it into a series of powers of x-l. ~logr(x+l)~,,l = 0 and IDlogr(x+l) !X=l= f(l)=1 -C moreover (1, 8 21)
This formula is very useful for the computation of C if the quantities s,' are known, as Stieltjes has shown in his paper,ao in which s, is given,up to s;,,, at an exactitude of 33 decimals. Moreover we have seen in 5 82, dealing with the expansion of the Bernoulli polynomials into Fourier series, that
This may also be deduced by starting from formula (7) of 8 71, putting there 2=--t, dividing by f and integrating from f=O to t=l. For m=2 we obtain the trigamma function expressed by a definite integral. Derivatives of y,,,(x). From (11) we get and therefore from (13)
Sum of the alfernafe function (-1)" y,,,(x). This is obtained without difficult+ Indeed putting first into formula (12) x=0 and then x=1; subtracting the second result from the first we get (-l)m (m-l) !/l". Putting now x=2 and x=3, proceeding in the same manner as before we find (-ljm (m-l) !/3"; and so on. Finally we shall have
In the same manner as in 8 49 it can be shown that the sum in the second member is equal to (l-2-m) s,, therefore (20) Parficular case. For m=2 we find j. (-llX YI/,(x) = f ?I?,(01 =$ = 1'2334 0055.
Remark. By aid of Pairman's tables of the trigamma function we get !ii (-l)Xy2(x) = 1'2087 0132
x=0
and
21 2: (-l)X yz(x) = 1'2574 7214.
C. Sum of the reciprocal powers determined by the Maclaurin-Euler summation formula. Let us put into formula (5) Q 88 f(x)=l/(x+l)" and a=O. Since we have
1 o J (x$ = ~ m-1
.*
(m-1) ;*+I)"-'
and D' (x~*)"' =
( - 1 ) ' (m+i-1)i and Df(w) = 0 (x+l)"'+i
moreover as D'"f (x)Dzn+zf(x) > 0, and D'"f (x) does not change x=0 its sign in the interval the remainder (4) of Q 88 may be applied and omittiz the :rG independent of z,
346
so that the expansion will be
Writing z=m we obtain
. OD cf = zz, (xi*)" - &$f =*,- m!-1
so that we have
Though the series corresponding to n=w is divergent, nevertheless formula (27) may be useful for the computation of the sum, if z is large enough. The expansion of Y,,,(x) into a Bernoulli series leads to a similar result. In 8 87 we have seen that the quantities to be determined for this are
Since we have D2RY',I(x) D""+"Ym(x) > 0 and D""Y,,,(x) does not change its sign in the interval (0,x) the remainder will be (7, 8 87) where O<E<l. Finally the expansion is:
R,, = iazn AD'"-' Y,,,(x)
347 This formula may be useful if x is large enough. 9 122. Sum of alternate reciprocal powers, by aid of the p,(x) function. Besides the digamma, the trigamma and the y,,,(x) function, useful for the summation of reciprocal powers, it is advisable to introduce still another function denoted by p1 (x) serving for the summation of alternate reciprocal powers. It has been considered first by Sfirling (lot. cit. 25). Its definition is the following
(1)
p,(x) =-2DlogB I++).
Hence 8, (x) is the derivative of the logarithm of a particular Beta-function. It may also be expressed by gamma functions:
A-' B,(x) = F(x) - ViP,(x) + k. Finally
the sum of PI(x) if x varies from x=0 to X=Z is i PI(X) = F(z) + ci- 'h/J,(O) --?A/%(4 r=O where C is Euler's constant. Expansion of fi, (x) into a series of Boole polynomials. According to 8 116 the coefficients of this expansion are given by therefore Finally (3) Putting into this formula x=0 we obtain (Q 113)
cm = MA"%(O);
(-l)m ml (x+m+l)m+, (--l)m 1 z=~ =x&i-
c0 = 1 and c, =
It is easy to see that this is equal to the expansion of -2 log (l--t) into a power series, where t=l,; hence p1 (0) = 2 log 2. Expansion of /?, (x) info a Boole series. According to Q 116 we have i! (4) /L?,(x) = s J--lJi M$$l (x) = .s 2' t---o ix0 2'(X+l +i)i+, From this it follows that p,(m) ~0; moreover, putting into it x=0 we obtain the result above, Example 2. Computation of ,8, (200) to 12 decimals.
l/201 l/2(202), l/2(203),, 3/4(205), 6/4(205), = 0.004975 124378 = = = = 0.004987 12 314664 60663 446 4 500155
349 Expansion of /!I, (x) into a series of Euler polynomials. In Q 109 we have seen that the coefficient c,,, in this expansion is given by
cm
therefore so that
= MOmPI (0);
c, = 1 a n d cm = (-l)m m!
This series is divergent. Expansion of p,(x) into an Euler series. According to § 111, 0'"" ,bl (x) 0'"'" &(x) > 0 the remainder of the since series may be given by (4) Q 111; moreover, (-1)' i! MD',:, (x) = Di --!I- = .X-)-l (x+l)i+r So that the required formula will be
-1. c L
This is the expansion of b, (x) into a reciprocal power series, For n=m the series is divergent but nevertheless formula (5) is useful for the computation of p, (x) especially if x is large. From (4), Q 106 it follows that the remainder will decrease until we have approxiniaLe1 I n k + x.
Expanding l/(l+u) into a power series and integrating, we find: ,dl (x) = 2 &
0'
j" (--l)m uXtm du = 2 m;. xr--;l .
This is identical with formula (7). It is easy to show that the mean of (9) is equal to l/x+1. Formula for negative values of the argument of Bl(x). Putting into (6) first -x instead of x and then x-l instead of X, we obtain the two equations
PI (--xl + Ah--l) = -&
Particular case. Putting x=1/2 we find /!?,(--~) =7c. Derivatives of p, Ix). D"'P, (x) may be expanded into an Euler series in the way used above for /I1 (x); we find
(11) DmBl(x) = 6--l)m ( ,x~l\m+l - ,io ff$~~$,,$ ez,+l +
I where 0 5 71 S 1. The series corresponding to n=aJ is divergent, nevertheless it is useful if x is large. According to formula (4) S 106 the best value of n is approximately (12) n - '//z (x.-c-m) . We may obtain the derivatives of PI(x) expressed by an alternate sum; starting from formula (7) we find (13) Putting x=0 we get (14) and if we introduce the notation (m+2n+l) ! - 11 (x+l)m+2"tG- ew+1
353
then we find WI
Dm& (0) = 2 ml (--l)m a,,, .
Therefore the expansion of @I (x) into a power series will be
We may obtain the sum of the alternate function in (13), from zero to x in the following way: Putting x=0 into (13) we obtain D"& (0) given by (14) ; now writing in (13) v instead of x+v we have
(--llx D"B, (4 = 2 (--llm ml ,% (y:7fk .
Finally
(17)
vfo (.:if',:l
F--llm 2 m! = - [D"B,W--~--1I"D"B,
(4l.
In ($ 49 we have seen that the alternate sum a, may be expressed by the ordinary sum s, as follows: : i: (38) a, = s, 1 l t -2m-" 1
-
Since the numbers s, are given to 33 decimals up to m=7O in Stielfjes' tables [lot. cit. 361 therefore by aid of (15) we may calculate DrnB1 (0). Moreover D"&(x) may be easily computed by the aid of formula (ll), if x is large enough, so that the alternate sum may be obtained by (17). Example 4. The sum of (-l)x/(x+l) 3 is required to 12 decimals. We have x-0
,
$ _1__1," = $ [D",t?, (0) - D'B, (100) 1. 7
(X+1)3
According to (18) we have ~7~ = G S, ; and in consequence of (15), D'pl (0) = 3s,. The tables mentioned give D"& (0) = 3.606170 709479. D'&(lOO) is then computed by aid of (11). Stopping at n=2, the remainder will be less than
CHAPTER VII.
EXPANSION OF FUNCTIONS, I NTERPOLATION . CONSTRUCTION OF TABLES . fi 123. Expansion of a function into a series of polynomials. Given the infinite series of polynomials, P, , P,(x), P,(x), , , . , P"(X), * * * 1 where we denoted by P,(x) a polynomial of degree R. If certain conditions are satisfied, the function t(x) may he expanded into a series of the polynomials P,(x) f(x) =cnPo+clP,(X) +c,P,(x) +*..+c,P,(xl+...
and determine the constants ci so that the equation (1) shall he satisfied for x=x,, x, , . . . , x, , then for any other value of x the second member will only give an approximation to f(x). Let us suppose that cn is different from zero, and add to the second member of (1) a term denoted by R,,,, and called the remainder of the series. It must be equal to zero for x=x0, x1, . . . , x, so that it does not change equation (1) for these values; therefore we shall write R,,, = (x-x0) (x-x,) . . . (x-x,) 12, Now we shall dispose of the function Q so that f(x) = cop,, + c,P, (4 + , e , . + d',(x) + Rn,, shall be exact for a given value x=z; hence U will be a function of I only. To obtain it we shall write y(x) =f (x)-COP,,--c,P, (x)- . . . -c,,P,,(x)(x-x,). . .(x-x,,)!?.
356 Let us suppose now that f(x) is a continuous function whose n+l -th derivative has everywhere a determila;evalue. In consequence of what has been said before, y(x) i's equal to zero for the n+2 values x = x,, x1, X*, . . , X", 2. From this we conclude that the n+l -th derivative of q(x) must be equal to zero, at least once in the smallest interval containing these numbers. If this occurs for x=5 we shall have D"+'y((5) R
n+1
Finally writing x instead of z the required expansion will be (3) f(x) = C"P, + c,p, (4 + czp2 (4 + * * * + GP" (4 + + (x---x,) k-x,) * * ~(x-- Dn"f(& (n+ll! This is exact for any value whatever of x; E being a function of X, whose value is included in the smallest interval containing the numbers X", x,, x,, * * - Xnr x. We conclude that the remainder R,,, depends only on numbers, and on the derivative of f(x) and not on the nomials P,,,(x) chosen. If a second series of polynomials Q,, , Q, , . . . Q, is and if we expand the function f(x) into a series of polynomials, stoping the expansion at the term Q.(x) we these polygiven Qm(x) obtain
f (4 = k,Q, + R,Q, 04 + . . . + RnQ, (4 (4) and if this equation is satisfied for the same values x=x,, , x1 , . . . . x, as expansion (l), then the two expansions are identical and so also wili be the remainders of the two series. and therefore the obtained precision too. But if the coefficients ci of formula (3) are tabulated to Y decimals only, so that their error is less than ~=5/10'+~ then this will be another cause of error; therefore the absolute value
1
357 of the error of f(x) will be 15)
Idf(X)\ <
IR,+,I-tE
II+1
X i-0 IPi I*
From this it follows that, finally, the precision will nevertheless depend on the r&ire of the expansion. Remark 1. If f(x) is a polynomial whose degree does not exceed n, then the remainder is equal to zero and the second member of the expansion (1) will be equal to f(x) for every value of x. Remark 2. Should cn be equal to zero, R,,, given by (2) could nevertheless be considered as the remainder of the series (l), but then it would be possible to chose a remainder R, more simple, in which the range of 6 is smaller. # 124. The Newton series. We have seen when treating the symbolical methods that the operation E is equivalent to l+A; therefore tie have
Since E"f(a)=f(a+x) the following manner: (I)
we may write the above equation in
f(a+x) =f(a) +(:) A{(o) +(i) A'f(a) +-..+ -t (E) @f(a) + . . .
This is Newton's series. It can he transformed in different ways. For instance, putting x=z-a we have (2) f(z) = mc ( 'ia) Amf(a)
which gives the expansion of f(z) into a series of binomial coefficients. Expansion into a generalised Newton series. Putting into (1) x=(z-a)lh we find
358 Indeed we have z--a (z-a) (z-a-h) . . , (z-a-mh+h) h =h"m! m 1-l The last term is the generalised binomial coefficient of 8 22. (p. 70). If we write moreover
f( a+y)=F(I)
then from the above it follows that
f(a) = F(a),
Emf(a) = rF(a),
Amf(a) = PFM.
Finally the expansion of F(z) into a series of generalised binomial coefficients will be: (3) F(s) = F(a) + (z~]r~X-+,
AF(al
AmF (a) . e + (zg]A%;;-- + . .
Expansion by aid of Newton's backward formula. This formula has been deduced by symbolical methods in Q 6. We had for a polynomial of degree n f(x) = 2: x+m-1 ( m ) Ff(----ml
Expansion of a function into a Newton series. If the function f(x) is a polynomial of degree n, then we shall have P+"*f(x)=O i Therefore the series will be finite: (5) f(x) = f m>O.
f(o) + (;I Aft01 + . . . . + (;I A"f(o).
In the case considered f(x) is a polynomial whose degree does not exceed n; therefore according to what we have
359 seen in the preceding paragraph equation (5) is an identity true for any vaIue whatever of x. Problem. Given y,,, yl, , . . , y,, corresponding to x=0, 1, 2 I ***, n; a polynomial f(x) of degree R is to be determined so that for the given values of x we shall have f(x) = yX . The solution may be obtained from formula (5), which gives for x=0, 1,2, , . . , n the necessary n+l equations for the determination of the n+l unknowns f(O), At (0), . . . , &f (0). But it is easier to proceed as follows: From A = E-1 we deduce A"' = ( E - l ) " = ;?I (-1)' 1 T) Em-' ; this gives Amf (0) = 2: (-I)' (7) f (m-i);
--
putting into this f (m-i) = ym-i the problem is solved. Remark. The problem above is identical with that of determining a curve of degree n passing through given n+l points of coordinates x and yX . If f(x) is not u polynomial, the series will be infinite. Stopping at the term pf (0) the second member of (5) will give the exact value of f(x) if x is an integer such that (6) O<xSn.
_. L.2 -
Indeed, for these values of x the terms i Amf(0) will vanish if I 1 m > n, so that equation (5) will give the same value as the infinite series would give. If x is an integer satisfying (6), then equation (5) may be written (l+A)Xf(0) = E"f(0) = f(x). That is, the second member of (5) gives the exact value of f(x) for x=0,1,2,. . . , n; but for other values of x it will be only an approximation of f(x). To remedy this inconvenience we may add to the second member the remainder R,,,, of (2), 8 123. This can be done since, as we have seen above, the conditions enumerated in this paragraph are satisfied.
360 Finally we shall have ( 7 )
f(x)=f(o)+(;)Af(o)+...+(;]A"f(o)+
-t
(&) D""f(E)
w h e r e 5 is included in the smallest interval containing 0,1,2 I..., r&X. Remark. If f(x) is a polynomial of higher degree than n and we stop the expansion at the term & then a remainder must be added as in the general case before. Putting into this formula n=O, we obtain one of Lagrange's formulae f(x) - - r ( O ) = Of(E) x according to which in the interval 0,x there is at least one point of the curve y=f(x) at which the tangent is parallel to the chord passing through the points of coordinates 0,f (0) and x,f (x). and f 7 = F(z) ( I we obtain the expansion of the function F(z) into a series of generalised binomial coefficients, with its remainder (8) F(z) = F(a) +('T"),,q +. . . +['")$;g + Putting into formula (7) x=(z-a)/h
+ (;;) h D"+lF (a+:h)
%--a where 6 is included in the smallest interval containing O,R, h. In this formula, by F'F we understand $>y. therefore in the last term of (8) the number hnil will vanish from the denominator. Q 125. Interpolation by aid of Newton's Formula and Construction of Tables. The problem of interpolation consists in the following: Given the values of yi corresponding to zi for LO, 1, 2, . . . r~; a function f(z) of the continuous variable z is to be deiermined which satisfies the equation
361 (1) yi=?(Zi) f o r i=0,1,2,...n
and finally f(z) corresponding to z = z' is required. The solution of this problem is not univocal; indeed, if we write F(z) =: f(2) + (z-z,,) (Z-2,) , . . . (z---z,) O(r) where O(z) is an arbitrary function; then from (1) it will follow that (21 F(G) = Yi for the given values of i. But if the n+l points of coordinates yi, z; are given and a parabola y=f (z) of degree n passing through these points is required, then there is but one solution. This problem has been solved for equidistant values of x in 9 124. If we put z=a+xh and z is given for x=0,1,2,. . . , R, then equation (8) of $j 124 will become: (3) F(t) = F(a+xh) = F(a) + xAF (a) + . . . + ; A"F(a) + 0 h"" D"+'F(a+$t). A second case of interpolation is the following: A function F(z) of the continuous variable z is given in a table for z=z,+-ih where i=O, 1, 2, . , . , N. A polynomial f(z) of degree n is to be determined so as to have (4) F(z) = f(z) for ~=a, a+h, , . . . , a+nh. Finally the value of f(z) is required for z=z' different from a+ih. First we must choose the quantity a. This choice is arbitrary, but it is considered best to choose a so as to have on each side of I' the same number m of values for which (4) is satisfied. To obtain this, it is necessary that the number n be odd, that is n=2m-1, then a must be chosen so that a+mh-h < z' < a+mh. But sometimes it is not possible to choose a in such a way. This happens for instance at the beginning or at the end of the table, when we are obliged to take respectively the first,,and the last 2m numbers,
-.
-2 -
362 In some cases it may happen that we have to determine f(Y) corresponding to a value of I' outside the range of a stable (extrapolation). This is considered as unfavourable. For instance, given the table
f(O) f(l)
f(2) f (3)
Af(Oj Aft11
Af(2) . &l-l)
" Azf (0) A'(l)
. aif(n-2)
A"j (0)
: A3i(n-3)
!
and f(n+l) is to be calculated. It is best to use Newton's backward formula (8 124) which would give f ( n + l ) = f ( n ) + Af(n-1) + A?f(n-2) + A:lf(n-3). General case of parabolic interpolation of odd degree by aid of Newton's formula. Let us write 2m-1 instead of n in formula (3), then the polynomial obtained will give exactly F(z) for z=a, a+h, . . . , a+2mh-h. As has been said, it is best to choose a so as to have a + m h - h < z< a+mh or m-l<x<m since then the polynomial will give on each side of z the same number m of exact values. To determine the maximum of the binomial coefficient zxrn I I figuring in the remainder of (3), if m-l<x<m, we shall put x=i)+m- l/2 so that --'h<K l/z. We h a v e
1 it+m---lh 2m
I
=I--1)"'1 -@~)[~--o') (2m)! I 4
. . . . /(m--1/2)2-&2);
moreover 1'1' 5 l/4, hence the maximum of the absolute value of this expression will be obtained if 8~0, or x=m--l/2, So that
I I
if
m-l<x<m.
363
For instance
and so on. Finally we shall have if m-l < x C m
(5)
I R,, I < h2m I( mz) D2"F(a+5h)
/
where 6 is included in the smallest.interval containing 0,2m-1,x. Construcfion of Tables. A table of a function F(z) should always be computed from the point of view of the interpolation formula to be used for the determination of the values of F(Z). Let us suppose that an interpolation of degree n is chosen, the error of F(z), denoted by 8F(z),'will arise from two sources: First, by putting approximate values of F(a) and +'F(a) -. -. 5 5' i into formula (3) instead the exact ones; secondly, by neglecting the remainder of the formula. If in the table F(u) and yF(a) are given exactly up to Y decimals, then the absolute value of their error will not exceed 6 E=lO'qi* Hence in the case of Newton's formula the corresponding part of the absolute value of the error of F(a+xh) will be less than
Since the errors of the function and of its differences are not all of the same sign, the resulting error will generally be much less. We have
The two parts of the error should be of the same order of magnitude. If the remainder is much greater than WE, then there are superfluous decimals in the table, since with fewer decimals the same precision of F(z) would be obtained by formula (3). If
364 the remainder is much smaller than WE. there are not enough decimals in the table; indeed, increasing them, the same formula would lead to a greater precision. Therefore we conclude that if the degree of the interpolation formula, the range and the interval of the table, are chosen (the remainder depends on these quantities), they will determine the most favourable number of decimals for the table. Linear interpokation. Putting into formula (3) n=l or m=l we get (6) F(r) = F(a+xh) = F(a) + xAF(a) + R, I R, I < ; I D"F(a+th)i. Since we chose a so that a<r<a+h o r O<x<l, we shall have O<s'<l. If the table contains the numbers F(a+ih) and their differences, the error being the same for eac!l item, for instance less than F ; then, according to what we have seen, (bF(a+xh)(<
where according to (5)
.-
(I+x)F+~R,~<~E+(R~I.
But if the differences are computed by aid of the numbers F(a+ih) of the table, then their error may be much greater. Parabolic interpolation of the third degree. From formula (3) it follows that (8) F(r) = F(a+xh) = F( a ) f xAF(a) +(g) A"F(aj +
(';) A"F(a) + (i) A'F(a) + (g) A5F(a) + h6 ( G)Dt'F(o+Fh).
This formula is of practical use only in tables where the first five differences are given, but there are hardly any such tables. Using this formula it is best, as has been said, to choose a in such a way that a+2h < z < a+3h; then as the curve passes through the points corresponding to Z=Q, a+h, a+2h, a+3h, a+dh,a+Sh, there will be three points on each side of z, and we shall have 2<x<3. According to (5) the remainder will be R , ; < $$;- D"F(a+;'h) I where 0 < 5 < 5. It can be shown that interpolating by aid of formula (8) the absolute value of the error of F(z) is G(z) < 8E + IR,;). Parabolic interpolation of even degree is seldom used, since in this the curve passes through an odd number of points, and therefore Q cannot be chosen so that the position of I shall be symmetrical. Putting into formula (3) n=2m we get (91 F(z) = F(a+xh) z
F(Q)
+ xAF(a) + . . , . , + 2mx; 1 D'""'F(a+Sh), 'I
a
A""IF(~) + h"n'1
I
In this case the best way is to choose
so as to have
366 a+mh < z < ai-mh+h or u+mh-h < z < a+mh
but neither of these dispositions is symmetrical. In both cases we have 0 < E < 2n. If the differences A, AZ, A3, . . . A" are given by the table, then the interpolation by Newton's formula is the simplest. The difficulty consists in the differences being printed, which makes the tables bulky and expensive. Moreover to obtain the differences with the same precision as the values of the function, they must be calculated directly, and not by simple subtraction starting from' the values of the function given by the tables, and therefore the determination of the differences is complicated. $ 126. Inverse interpolation by Newton's formula. If a function F(z) is known for z = z,, t,, , . . . , z,, , then the *following problems may arise: a) Interpolation: To determine F(z) if z is given. b) Inverse Interpolation: To determine z if F(z) is given. The second problem is not univocal; to a given F(z) there may correspond several solutions; but if we add a condition, for instance that z should satisfy the inequality a < z < 6, the interval being small enough, then there will be only one solution. If F(r) is given by a polynomial of degree n and by a remainder, as in Newton's series, then we will have in the second problem an equation of degree n to solve; this is geherally done by a somewhat modified method of the Rule of False Position. If the polynomial is given by a Newton, a Lagrange or any other series the procedure will always be the same. Let us suppose that for the required precision, the function F(Z) is given by a Newton series of degree n, for z=a+xh (1) F(r) = F(a-l-xh) = F(Q) + xAF(Q) + . . . , + (z) A"F(a) + hrffl[ n$] D"+'F(a+fh).
where z1 is the first approximation to Z. If F(s) is considered exact and if the precision of the numbers F(a), F (a-l-h) in the table is equal to E, then the absolute value of the error of z, due to this cause, will be less than (8 133):
I62 < -&, ,
12&+ ziD'F(o+Eh)Ij =b,,
From this it follows that if 8, is smaller than the corresponding quantity in linear direct interpolation, then inverse interpolation will give better results than the direct; and if 6, is larger, then it will give inferior results. Example 2. Given a table of F(z) = log z to seven decimals, of the integers from 1000 to 10000. Here we have h=l; the number o is chosen in such a way that .:
F(a) <F(z) < F(a+l)
the absolute value of the error of z will be less than
It can be easily shown that this is less than l/103. That is, z will be exact to three decimals, which will give with the four figures of the integer part of z a precision of seven figures. . If a greater precision is wanted it is best to repeat the operation. For this it is necessary to determine F(z,) with the required precision, by putting into equation (1) x=x, . If we find F(z,) < F(t) then we start from the points z,,F(z,) and (a+h),F(a+h) and get
F(z) - F(z,) x2 = F(=+h) -FF(~,)
Or t? = '1 + X&l
where h, =a+h-z, . z2 will be the second approximation of z. The error of z2 caused by neglecting the remainder will be less than
368
If a still greater precision is wanted, then we determine F(zJ by putting into (1) x = y . Since we shall have F(z*) < F(Z) < F(a+h), we may determine t, in the same manner, and continue till the required precision is attained. If we had F(a) < F(z) < F(z,) we should have proceeded to determine z2 in a way similar to the foregoing. But we may shorten the computation by the following method: having obtained z, , and shown that the error of z1 is smaller than 6, we may put h,=d, and determine F(z,+h,). Then we shall have Starting from these we determine z2 with a much greater precision, since the interval now considered is much smaller than in the method above. An example will be given in 8 133. # 127. Interpolation by the Gauss series. In the following we will give only in a summary way the deductions of the Gauss, Bessel and Stirling interpolation formulae, as they are now very seldom used, and are merely of historical interest. If we stop one of the series at the term of degree 2n-1, and if the series gives exact values of f(x) for every integer x such that -n + 1 ZZ x s n, then the error of the approximation of f(r) for other values of x will be measured by the remainder given by formula (2) of Q 123, that is, by
(1)
R,,, = (x+;-1 ) D-f (5)
where [ is included in the smallest interval containing -n+l, x, n. If the curves pass through the same points, the remainder of every series will be the same; therefore from this point of view there is no advantage whatever in using any one of them. Gauss's first series may be obtained, starting from the symbolical expression for Newton's formula:
369
I?= ,; (;)Am. A+1 _ the term AZ up by E - 1. This will not change the value
of the series. We find Let us multiply the second member of this equation from
This may be simplified again by remarking that the quantity . Then the above operation is in the brackets is equal to repeated from AU up, and so on; finally we shall have
This is the symbolical expression for the first Gauss series. Written in the ordinary way, starting from f(O), we get (31 f(x) = f(o)+(
;jAf(O)+(;j A"&----11+ (X$l)A:'+l)+
-I
I*,, 1 A'f(-2) +[";"j A~f(-2)+(x~2)A"f(-31+~~~~ x$m~~l]
A2m-lf(-m+l)
-4
+ [ x+TL') A'"f(-m) + , , . ,
If the differences of the function f(x) are known, then its
24
370 expansion in this series presents no difficulties. If f(x) is a polynomial of degree R, the series is finite and the second member is always exactly equal to f(x), Example 1. Given the function f(x) by its Newton expansion f ( x ) =8+6[xfl)+4[*f1) +2["f'].
Starting from this we may easily construct a table of f(x) and of its differences, in the same manner as in Q 2.
X
- 1 0 1 2 3
f(x) 8 14 24 40 64
Af(x) 6 10 16 24
d'f(x) 4 6 8
A:'f(x)
A'f(x)
2 2
Formula (3) gives by aid of this table f(x) = 14+1,[;) +41;] +2[";1] From this we conclude that only the terms of the rows of f(0) and Af (0) are figuring in the formula. If the function f(x) is not a polynomial, then the series is infinite. Stopping the series (3) at the term [xzn1), we shall obtain f ( x ) = "i, [I x+*;-1) A-"lf (-m) + 1 g$Tl) Az:rn+*f (-m) 1 + (4) -t R,n. This will give exact values of f(x) for every integer x such that -n+l<xSn.
Indeed, in these cases every term of the sum in which m > n will vanish and the limited series will give the same value as the
371 infinite series. From this we conclude that the remainder R,, will be given by (1). Interpolation by the first Gauss formula. Let us put
x=(z-a)/h and f
= F(z), we shall have
F:(Z) = mi, [( x:tA1 ) A2mF(a-mh) +I
+ [ ;$;I] A2m+1FW-mh) ] +
X+~n-')h?"D'"F(a+Eh),
It has been said that it is considered advantageous when interpolating to choose a so that the points for which the equation above gives exact values, shall be symmetrical to the point s,F(r). Then
a<t<a+h or O<x<l and
-n+l<l<n.
According to (5) of 8 125 the absolute value of the remainder will be less than (5)
A+1 the terms from A up by - = 1. We get E Es=1+
The second Gauss series may also be obtained, starting from the symbolical form of Newton's expansion, by first multiplying
I)
x A 9 Am+, 1 E t,:, E
Noting that the quantity in the brackets is equal to we repeat the operation from A3 up. This gives
We note that the quantity in the brackets is equal to and repeat the operation from A5 up; and so on. Finally we find
372 (6) This is the symbolical expression of the second Gauss series: written in the ordinary way it gives, starting from f (0): (7) f(x) = f(o) + [ ~]Aff--ll
If f(x) is a polynomial, this series is finite and the second member exactly equal to f(x). Example 2. Expansion of the function given in Example 1. By aid of the table computed there, we obtain f(x) = 14 + 6 (;) + 4 ("$'I + 2 ( xtl 1. The table shows that in the second Gauss formula only the numbers of the rows f(0) and Af(-1) figure. If f(x) is not a polynomial, then the series is infinite. Stopping at the term Azn-l we get a polysomial of degree 2n-1; which is an approximate value of f(x). But for x=-n, . . . , O,l,..., n-l it gives exact values. This can be shown by remarking that if x is an integer such that -n<xIn-1 then every term of the infinite series vanishes in which m 2 n, and therefore the limited series gives the same value as the infinite series, For other values of x the error is measured by the remainder (1). Interpolation by the second Gauss formula. Let us again put z-a = F(z). We have .x= (z-a)/h and write f I II I
1
373 It is best to choose a so that the points, for which the second member of (8) gives exact values, be symmetrical to the point z,F(z) . Then a-h<z<a o r -l<x<O a n d --n<t<n-1 and the absolute value of the remainder will be less than (5). Stopping the first or the second Gauss formula at the term A"" we get an interpolation formula of even degree. The exact values correspond to x=-n, . . . . , 0, . . . , n. The results are the same as in the case of Newton's formula of even degree. $ 128. The Bessel and the Stirling series. To deduce the Bessel series, we start from the second Gauss formula (6, 8 127) writing in it x-l instead of x and multiplying it by E; in this way its value will not change. We get
NOW let US determine the mean of this series and of the first Gauss series (2, 5 127) ; we find ( putting x(E+l) = M
This may still be simplified so that we shall have
This is the symbolical form of Bessel's formula. Fully written, starting from f(0) it will be (2)
If f(x) is a polynomial, then the above series will be finite a.nd the second member will give f(x) exactly.
I
374 Example 1. Given the function figuring in Example 1 8 126, we find, using the corresponding table, that f(x) = 19 + 10(x-'*) f 5 1;) +2(;Jq From this table we conclude that the numbers in Bessel's formula are all in the same row of the table, e. i. in that of Mf (O), Mf (0) = 19, Af(0) = 10, MA'f (-1) = 5, A:jf(---1) ~2. If f(x) is not a polynomial, and if we stop the series at the term of AZ"-* then we obtain a polynomial of degree 2n--I which is an approximation of f(x) giving for x= -n+l I...,, 0 ,"'., R the exact values of f(x). Indeed, formula (1) shows that if x is such an integer, every term of the infinite series in which m 2 n will vanish; therefore the limited series will give the same value as the infinite series. Since these numbers are the same as those corresponding to the first Gauss se!ies; therefore according to 5 123 the remainder will also be the same (1, Q 127). Stirling's series. This series may be obtained by determining the mean of the two Gauss series. Of course Stirling obtained his series in another way a century before Gauss. The first Gauss series (2, 9 127) may be written
I
E"=l +
and the second (6, 8 127)
I
Hence the mean will be
After simplification this gives the symbolical expression for Stirling's formula
From this table it is readily seen that the coefficients in Stirling's formula are all in the same row, in that of f (0). If f(x) is not a polynomial, and if we stop the series (5) at the term MA'"-' we obtain a polynomial of degree 2n-1; by aid of (4) it may be shown that this polynomial gives f(x) exactly for the following 2n values x=-n+l,. . , , n. Therefore the remainder will be given by formula (I) of Q 127. Remark. Stirling's formula is in reality a formula of central differences. We have seen in Q 8 that .
Therefore we shall have Stiding's central difference notation
formula (4) expressed by
or fully written
376 ( 7 ) f(x) = f(O) +[;J $Y(O) + (;)+f(o) +~X~lJLLs"f(o) +
+[x;'); S'f(0) + . . , +[x2+mm_-J p.'-'f(0) + 2% S""f(0) 1, +... Let us remark that in this formula the argument in each tern1 of the second member is the same. @ 129. Everett's formula. If from Stirling's formula expressed by central differences (6, $j 128), we eliminate uij by aid of the following equation (found in § 8), we get FIS = E--l--+$
Let us write in the first two terms of the preceding sum m+l instead of m; then m must vary in these terms from zero to mJ . Therefore we shall have
The third and the fourth terms will give
So that the above formula may be written
From this we obtain after simplification the symbolical expression of Eoerett's formula
Or written in the usual way, if the operations are performed on f(0) we have
1
377 (2) f(x) = (r] f(1) +(Xf'JVf(1) +("f") H(l) +
+[x~3)G"f(l)+....-~x~1)f(0)-~;)8~f(0)- x31 6'f (0) - 1 x;f-2 J 6"f (0) - * . . . I 1 If f(x) is a polynomial, the second member of (2) will be finite and it will give f(x) exactly for every value of x. Example 1. Given the function of Example 1, $) 126. Formula (2) will immediately give the expansion of this function into an Everett series by aid of the table (Q 127). We find f(x) = 24 [ ;] + 6 ( x$1 ] - 14 [ "T-' ] - 4 I';]. The mentioned table shows that in Everett's formula there figure only the numbers of the rows f (0) and f (1). If f(x) is not a polynomial, the series (2) is infinite; stopping it at the term 6**-* we obtain a polynomial of degree 2n-1 which gives the exact values of f(x) if x=--n+1 ,.*., 0 I.#.( n, therefore the remainder will be given by formula (1) of Q 127. The precision of Eueretf's formula will be the same as for instance, that of Newton's formula of the same degree; but while Newton's formula requires a knowledge of the differences A, A2, A3r . . . A'"-', Everett's formula needs only the even differences A*, Aa,. . . A*"-"; this is an advantage. Inferpolafion by Everett's formula. To obtain the general formula, we put x=(2--a)/h into equation (2) and write f[*T ] = F(r). Stopping at the term iY2 we shall have (3) F(Z) = F(Q+Xh) = 2 m=o ] D'"F(a+5h).
a+m-1 2m+l ) tj'"F(a) 1 + fz*" [ x+2n;;-1
The curve passes through the points corresponding to Z= (a-nh+h), . , , .(a+ nh), To determine F(Z) it is best to
-378
choose, if possible, as has been said, the number o so as to have the same number of points on each side of z, that is
(4)
acz<a+h o r
O<x<l.
If we interpolate in a table where the given values correspond to ri=ro+ih (for i = 0, 1, 2,. . . , N) then condition (4) can only be fulfilled if &z-l < 2 <
ZNn+,
*
Putting into equation (3) x=6 and 1-x=q it will become
(5) F(a+@h) =
+ h"n ( x+.rl 1 D'"F (a+$h)
where -n+l<E<n.
A. J. Thompson has constructed a table which gives to ten decimals the coefficients figuring in this formula,::i
&m(W = k--1)" I&$....* ] for m=l, 2,3 and 0 I4.J S 1 where AAS = 0,001. Formula (5) may be written therefore
where -n+l<t<n. If 2 < ZRel then we are obliged to put into equation (3) a=z,+(n-1)h and we shall have z-u=0h < 0; the formula thus obtained is called by Pearson, a first-panel interpolation formula. To compute the value of F(z) it is necessary to calculate
37 A. J. Thompson, Table of the Coefficients of Euerett's CentralDifference Interpolation Formula. Cambridge University Press, 1921.
379 the binomial coefficients, since 6 is out of the range of Thompson's table. If z > z,~-"+* we are obliged to put a=q,+ (N-n+l) h and we shall have z-u=&h > h; the formula obtained is called an endpanel interpolation formula. Since 19 is out of the range of the tables, the coefficients must also in this case be calculated. An interesting particular case of Everett's formula (5) is that of 9=q=1/2. For this value the formula will become N6'm F(a) + R,, .
Hence 6 = 0.53426 and v = 0,46574. Since Thompson's Tables can be entered only with three decimals of 6, then, if there are more, as for instance in this example, E,, may be determined by different methods, depending on the number of figures contained in ?jzmF(a) or tjzmF (a+h), If, as in the present example, there are four figures only in a*F(a+h) then a precision of 6 decimals of E2(8) is large enough. This may generally be obtained from Thompson's Tables by a rough linear interpolation. In the present case we have in this manner E&Y) = 0,063627 The computation will give 8F(a+h) t,vF (a) -E,(ql)8'F(a) -E,(B)t?F(a+h) = 5.220 987 5 - 4.447 782 5 1 307 8 = 329 7 9.668 132 5 If a greater exacitude is required for E,,,(8) and B has more than three decimals and less than seven, it would be E, (1~1) = 0,060786.
381 possible to determine Ezm(8) by interpolation, using Everett's formula and Thompson's Tables; but the computation would be too long, It is far better in these cases to calculate first 8' and then
Thompson has given another method, which we shall see in the next paragraph, 3 130. Inverse interpolation by Everett's formula. We start from Everett's formula (3, $j 129) in which n is large enough for the required precision. (1) F(z) =F(a+xh) + ( xl1 )
b"F(a+h)
=xF(a+h) -
(X-l)F(Q)
+
-.- ($1 ij'F(a) + lxz2 1 8*F(a+h) --
d
The function F(z) is given by a table for equidistant values of z, the interval being equal to h and the necessary differences 8, 6' and so on, are given too. z is to be determined corresponding to a given F (2). We choose u among the values of the table, so as to have or F(a) <F(z) < F(a+h) F(a) > F(t) > F(a+h) the method is similar in both cases. We will suppose that the first inequality is satisfied. By aid of a linear interpolation we deduce x, the first approximation of x, keeping only three figures in (z,-al/h. (2)
FW--F(Q) X, = :Ia+h) _ JJQ) and =I = Q + x,h.
I
382 Now we determine F (2,) by aid of formula (1) with the necessary precision. Let us suppose that F(q) < F(s) ; then we determine F @,+A,) also by formula (l), where for instance h, = h/1000. If we have Fkl < F(z) < %+h,l then we may determine xt the second approximation of x by formula (2) in which we put respectively z, and h, instead of a and h. The first member will be equal to x2, in which we again keep three decimals, and then determine zz = z1 + q.h, . Should F(q+h,) b e 1ess than F(z) then we should be obliged to calculate F(r,+2h,) and so on, till we have F(z,+ih,-h) <F(t) < F(z,+ih,). Then starting from this we would determine the second approximation of 2. If we had then the determination of z, would have been similar. Now we determine by formula (1) the quantity F(z,) putting into it x, instead of x, then x, instead of a and h, instead of h. Of course the differences a2, a',, . must be given now in the system Az=hl . It would be possible to calculate these differences by aid of formula (l), but this is complicated, and generally superfluous, since we may nearly always consider the third differences of F(x) in the system hi negligible, and therefore the second We have differences constant.
The determination of F(z,) is simple, since in x2 there are only three figures, so that Thompson's table is applicable directly. If we have F (t2) < F(z) then we compute in the same manner F(z,+h,) where we have chosen h, = h,/lOOO. If
F(z2)
< 04 < F (z,+h,)
1
303 then by aid of (2) we determine x, the third approximation of x to three decimals, and z3 = zs + x,h, . Now we determine the differences t2F(z2) = ~"F(Q) [2)' in the system Az=h2 ; if they are small enough to be neglected in determining F (z&, then the linear determination of zs above will give I exact to the required number of decimals, and the problem is solved. If the differences are not negligible, we compute I;&) and F (z,+h,), where hs=h,/lOOO, in a way similar to F(r,). Then we determine I, by linear interpolation, and continue in this way till the differences may be neglected. Remark. It would be possible to determine every value such as F(q), F(z,+h,), F(z,), F(z,+h,), F(z,), , . . by aid of the same equation (1); this would make the determination of the differences PF (z,), a2F (z2), . . , superfluous. But in this case the evaluation *&ould & complicated, since then, for instance, (Z,-Q)/h would have a great number of decimals, and Thompson's tables of the coefficients would not be directly applicable. Example 2. Given F(z) = log z = 7.95717 32271 83589 39035 and z is to be determined by aid of Thompson's Logarithmetica Britannica tables to twenty decimals. There we find F(Q) =X95717 13373 70099 19928
Q Q+h
We conclude that the fourth differences multiplied by E,, having no influence on the first 20 decimals, may be neglected. From formula (1) we get zy? F(a+h) + E++.- F(a) = 1.95717 32258 25789 51451
-E,(6)~"F(cz+h)-E,(q)8"F(a)
=
6315 03894
F(q) = 1.95717 32258 32104 55345
I
Remark. In computing this value before performing the multiplication of F(a+h) by (r,-o)/h and that of F(a) by (a+&z,jlh since the sum of these factors is equal to one, therefore the first 6 figures common to F(a) and F(a+h) have been set aside and only added to the result. The new interval will be h, = l/lOB'. To determine F(r,+h,) let us remark that now (z,+h,--a)/h is equal to 6, = 0.395. Thompson's tables give E,(0.395) and E,(0.605). Finally from formula (1 j we get
F(z,+h,) z i-95717 32306 25144 88111. Since the condition F(z,) < F(z) < F(z,+h,) is fulfilled, we may proceed to determine the second approximation of z by formula (2). We find X, = $ (z,-z,) = 0.281 and z, = 0.90609 39428 1 . 1 To compute the value of F(z,j we must start from F(z, j and F(z,+h,) and form the differences
F(G) < F(z) < F(z,+h,)
is again satisfied, we may proceed to the evaluation of the third approximation of z; but this time the second differences will be equal to &!j7(a) , 10-I" = _ lj, 10-2" therefore they will not have any influence on the first 20 decimals; hence we conclude that z, may be determined by the linear formula, up to the 20 -th decimal. We have z = z, = 0.90609 39428 19681 74509. The error is equal to -3/10"'; z being equal to e/3."* 5 131. Lagrange's interpolation formula. This interpolation formula differs from those treated before chiefly by the fact that it does not need the knowledge of the differences of the function and that the abscissae xi are not necessarily equidistant. This is a great advantage, since the determination of the differences causes much work, their printing makes the tables bulky and expensive. On the other hand the interpolation by aid of this formula is more complicated. Given R points of coordinates Xi, f(xi) for i=O, 1,2, C . , , R let us put o(x) = (x-x4,) (x-x,) (x-x,) . . . . (x-x,) (1) and moreover
It is obvious that we have for m=O, 1,2,. . ., n, Lti (Xi) = L,i(x,,) = 0 if i + m and 1 .
:"l This method is due to Thompson, Logarithmetica Britannica, 1932, Cambridge University Press, Part IX, Introduction p. vii. The example is Thompson's example too.
25
If f(x) is a polynomial of degree R it may be represented exactly by formula (3). If f ( x) is not a polynomial' or if it is a polynomial of degree higher than R. then the second member of (3) will give f(x) exactly for x=x0, x1,. , . , x, but for the other values of x it will be only an approximation to f(x). The error may be measured by the remainder, which is given according to (2), 5 123 by
Finally we shall have
where f(x) may be any function of x whose n+l derivative has value; and where ;i is in the smallest interval a determi containing the numbers x,,, x, , x, , . . . x,, , x. Lagrange's formula becomes much simpler in some particular cases. Let us suppose for instance that the values of x are equidistant, say that xi = i for i = 0, 1, , . . ,
R.
Then we shall have
387
w(x)=x (x-j[x-f]...
(x-3 =(x),+,,*
where h = i that is w(x) is equal to the generalised factorial of 5 16. Moreover
finally remarking that nh= 1. we have
= (-l)"ti i! (n-i)! ; n"
(6)
L,,j(x) = (-l)"-'n" I;),, [x-$rh],, = nn[ ;), [ :I:)*
Expansion of L,i(x) into a Newton series. From (6) it follows that L,,i(O) =O (for i+O), The values of A"L,i(O) may be deduced
from (6), by aid of the formula which givis the higher differences of a product (§ 30) and of the formula giving the differences of a function with negative argument (p. 6). We find
putting x=0, and ,remembering
that h=l/n we have for n 2 m
(7)
y L"i(O) = (-l)m-' v-m (7) ('2: Jh = (-l)m+i(T). Ax=l.
Let us now express these differences in the system According to (4), Q 76 we have
The P(m;v) were given by a table in (j 76. Let us remark that in the case considered we have to put into the formulae of the table OCR. L"i(x) is a polynomial of degree n; therefore in the preceding formula, for the upper limit of m we may put n-j- 1 instead of 00. From (7) we get
388
and therefore
This is the Newton expansion of L,,i(X). From this formula we may deduce the Cofes numbers defined by c,,j = J"' L,,j(X) dx.
0
Putting these results into the preceding formula we obtain Czt2 = 3/S. Interpolation by Lagrange's formula. If the quantities f(xi) are given for i = 0, 1, 2, , n then from formula (5) we obtain f(x) corresponding to a given value of x. If the interpolation is
I
I
389 made by aid of a table, it is best to choose the numbers x0, x,, * * ' ' 1 xn symmetrical to x. Linear inferpolafion. Here we have x, < x < x, . Formula (5) will give
+ %(=--~c,) (x--x,) D"W1.
The absolute value of the remainder is
(11)
where i is in at least one of the intervals x,, x, and x0, x. Example 1. x is given, and f(x) = log x is to be determined by aid of a logarithmic table. If a < x < a+1 then we put xO=a and x,=a+l. From (10) it will follow that log x = (a+l-x) log a + (x-a) log(a+l) + loge + Vi b---a) Ia+l--4 ~2. Of course the remainder and the precision are the same as in the linear Newton interpolation formula. The formula above may be useful in logarithmic tables where there are no printed differences. The error committed is less than loge < 1 16a"' 85' Lagrange's formula is especially useful for interpolation if the values of xi are not equidistant, since in this case the only other available formula is Newton's expansion into a series of divided differences (Q 9). But since in general the divided differences of a function cannot be given in tables, hence they must be computed in each particular case. Therefore this formula is more complicated than that of Lagrange. Lagrange's formula may be useful in some particular cases too, for instance if the xi are roots of a Legendre polynomial of degree n, (85 138, 157) or the roots of a Tchebichef polynomial of degree n, ($j 158) and finally if the xi are equidistant.
390 Inverse interpolation by Lagrange's formula. This may be done in the same manner as in the cases of Everett's or Newton's formula If f(x) is given, and we have to determine x by aid of a table giving xi, f(xi) for certain values of i (the xi may be equidistant or not). If f(a) < f(x) <f(b) where f(a) and f(b) are two consecutive numbers of the table; then starting from these values we deduce from (lo), putting x,=a a n d x,=b
is positive then we have x' <x. NOW we compute f (x') by aid of (10) ; if we find f(a) < f(x') < f(x) < f(b) then we determine x" the second approximation of x, by putting into (12) x' instead of a ; the corresponding remainder is obtained by writing in (13) x' instead of a. Then we compute f(x") and continue in the same maimer till the prescribed precision is reached. # 132. Interpolation Formula without Printed Differences, Remarks on the Construction of Tables. We saw that a parabolic interpolation of degree n, by Newton's formula presupposes the knowledge of the differences Af (0), A2f (0), A3f(0), . . , , , .pf(O); so that they must be first calculated or given by the table. This last is preferable, but it makes the printing of the tables expensive.
391
Everett's formula is much better from this point of view, as it needs only the even differences; thefalculation is somewhat shorter than that necessary to compute all the differences; and if given by the tables, as the odd differences are omitted, they are smaller and more economical. Lagrange's formula does not need differences at all; but the interpolation, especially with higher parabolas, necessitates a great amount of computation. There is another interpolation formula3g which dispenses with printed differences, reducing the tables to a minimum of size and cost, and requiring no more work than Everett's formula, but it is applicable only for equidistant values of x. The reduction thus obtained will generally be more than one-third of the tables. Instead of printing the differences it would be far more useful to publish a table of the inverse function. To obtain this formula we start from Newton's formula, which gives the expansion of a function f(x) into a series of divided differences (Q 9).
The remainder has been obtained by remarking that the polynomial of degree n of the second member gives exactly f(x) for x=x0, x,, . . . . , xnel, x, ; from this it follows, according to 6 123, that the remainder is equal to the quantity above, and that 5 is included in the smallest interval containing the numbers:
392 will now give Putting 2n-1 instead of n into formula (1) the same results as Newton's ordinary formula, if the curve passes through the points corresponding to x=-(n-l), . . . . , 0 I***, n. The remainder will be, according to 8 123, equal to (2')
Rgn = [ x+;n-1 j D-f($)
where 5 is included in the smallest interval containing -n+l, x, n. Determination of the divided differences. The (2m-1) -th divided difference of f(x) for x=x,, , given in Q 9 is (3)
p?a-lf(x,,) = g _ -__ fh)
c= o D~.l,n (xi)
where opnr (x) has been written for (x-x,,) (x-x,) . . , (x--x~,,,-~). Putting into it the values (2) we get o&(X) = x(x-1) (x+1) . * . (x+m-1) (x-m) = (x+m-l),, Therefore equation (3) may be written .
To transform this expression, let us determine the sum of the terms corresponding to F and E-"+'. If Y varies from 0 to 2m+2 then k will vary in the sum from 1 to m+2. To have the term corresponding to E" we put into (7) m+l--Y=K and find
to obtain the second we put m+l-v=-k+l and get
To combine the two last expressions, let us write (8) It = &[(x+k--1)
E'+
(k-x) E-""1 f(0).
According to Lagrange's formula this quantity is the approximate value of f(x) obtained by linear interpolation between x=-k+1 and x=k. Introducing In into formula (7) we get (9) f ( x ) = ,,,i& (-l)m ( "$2-l) ;;r2 (-l)'+'
395 To simplify this formula let us introduce the numbers
The following table gives these numbers, sufficient for parabolic interpolations of the eleventh degree.
To check the table let us remark that from (10) it follows if m > 0 that m+2 m+2 X B,,,k=O a n d z IB,,,k! = 110') k=i I-=1 From the first we conclude for intance, that the sum of the numbers in each row is equal to zero. We shall put, moreover, C,(x) = (-l)"[ x+2;-']. Starting from C,(x) =l these numbers are rapidly calculated step by step by aid of the formula (12) c,(x) = (x+m-1) (m-4 ~ G&4(x). 2m(%--1)
where Ci is an abbreviation for Ci(x). Remark. In the particular case
zk = 1/2
[ f ( k ) + ?(-k+l)].
Stopping at the term I, we get a linear interpolation; stopping at the term C,(x), we have a parabolic interpolation of degree 2m+l. If the absolute value of the error of F(a+ih) in the table is less than E, then from (8) it follows that the absolute value of the error of Zr is also less than E, if 0 < x < 1 ;
moreover according to (10') and to (13) the absolute value of the error of f(x) is
ISf(x) I < & f, ('i ) I Cm (xl 1 R,, . +
Since according to 8 125 the maximum of C,(x) in the interval (0, 1) obtained for x = */2 is equal to
Particular cases. Linear intetpoZation: ldf(x) < E + IR,i.
397 Interpolation of the third degree:
ldf(x) ! < + E + )R,l.
Interpolation of the fifth degree:
From this we conclude that interpolation by aid of formula (14) is much more adventageous than by Newton's formula; indeed, it not only dispenses with the calculation and printing of the differences, but moreover the precision is greater.
-.
The table containing the numbers F(a+ih), to obtain F(r) we choose a so as to have a < z < a+h. But then an interpolation of degree 2n+l is only possible by (14) if the table contains the numbers F(a-nh+h) and F(a+nh;, in which case the formula is called a mid-panel formula. We put into (14) x=(z-a)/h and write
Interpolation in a table in which the interval is equal to h.
= F(r) = F(a+xh).
-. (15) Hence according to (8)
Ik
may be written
Zk = G-3 [(x+k--1) F(a+kh) + ( k - x ) F(a---Kh+h)l.
Ik is therefore the result of linear interpolation between the points corresponding to r=a-kh+h and z=a+kh. C, is given by (11) and the remainder will be (16)
Rqn = Ix+;;') h2" D2"F(a+[h)
where - ( n - - l ) <E<n. In the case of the mid-panel formula we have 0 < x C 1; hence according to $ 125 the remainder will be
(17)
j R,, 1< hzn ( nG') D2" F(a+th) / .
The interpolation by formula (14) needs no more work of computation than Everett's formula, even a little less; there-
I
I
398 fore the printing of the even differences in the tables is superfluous. Linear Interpolation. Stopping at the first term in (14), we put n=l, and get F(r) = I, + (; 1 h"D2F(a+:h) where 0 < 6 < 1; moreover x=(z--a)/h, and I, = xF(a+h) + (l-x) F(a). Though the computation of I, is easy to make especially if a calculating machine is used, nevertheless it may be useful to indicate the shortest way to follow. Let us denote by A the place on the calculating machine where the result appears, by B the place where the number to be multiplied is put in, and finally by C the place where the multiplicator appears when the handle is turning First F(a+h) is put into B, then it is multiplied by x, which number appears in C. Without reading the result in A, we cancel F (a+h) in B and put in its place F(a), leaving the numbers in A and C untouched. Then we turn the handle till the number x in C becomes equal to one. The result in A will be equal to I,. The remainder is the same as that in the Newton series. Since 0 C x < 1 we shall have according to (17) I R, ! < j +- ~WF (a+5h) 1. In 8 125 it has been shown that the error dF(z) of F(Z) is due to two causes; first to the inexactitude of the numbers F(a+ih) contained in the tables, and then to the neglect of the remainder. If the tables are computed to v decimals then their error will be less than &=5/10~+~ ; moreover the resulting error will be I dF(zl I< -?- + I R2 I. 1oy+, We have seen that E and R, must be of the same order of magnitude (8 125).
399 Example 1. Let F(z) = log z; given a logarithmic table to v decimals from b to c, the interval being h=l. We shall have
f-fence the most favourable number Y of decimals, in the case of linear interpolation, will be b c
V
From this we conclude that if the interval is equal to h the best value for the number of decimals v is h 0.01 0.001 0,OOOl
Y
3 5 7
Remark. The usual seven decimal logarithmic table contains 182 pages, whereas according to what precedes a table giving the same precision by aid of linear interpolation, containing the logarithms from 1000 to 10000 to seven decimals, together with an antilogarithmic table from zero to one (h=O.OOOl) would take only 38 pages that is, hardly more than one fifth. J* edfq2 df. Example 3. Probability integral, F(Z) = 1 \/2x --'x) Table beginning at x=0; interval h. We have
400 ! R = h' z- e--m:2 ( 2 8 1: '.,_#. ._. 3h". h' '- - 1 0 0 ' sip?
The interval being equal to h the best number v of decimals is h 0,Ol 0,001 ! v 5 7
The best value of h corresponding to v is given by the table of Ex. 3. Example 5. F(r) = sin z. Range of the table 0 - J+r. Interval h. We have I R, ) < is I sin(a+$h) I < G. A. Five-decimal table. Determination of the best magnitude of the interval. According to what we have seen we should have
!L
8
5 10~ t h a t i s h=0.006324
If the circumference is divided into 360 degrees, then h=0°36 and if it is divided into 400 grades then h=OY4. Steinbecher's table (Braunschweig, 1914) in which h=O"Ol is much too large; if we choose h=Og2 the table would be twenty times as short, giving the same precision. The table would be, as we shall see, too large even for a seven-decimal table. B. Seven-decimal tables, In the same manner we find h=0°036 or 0" 04. H. Brandenburg's table (Leipzig, 1923) in which h=lO seconds or OOOO277 . . , is twelve times too large. Example 6, F (2) = tan z. Range of the table 0- :, Interval h. It is sufficient to consider the above range; indeed if we have
1
401 s < z < 3 then instead of tanr we put l/tan(l/zn-2). We have 1R j < _he 2 tan(a-!-Eh) < E 2 8 cos2(a+(h) 2 * Five-decimal tables. The best magnitude for h is given by he 5 - = --. 2 10" Comparing this with the previous example, we see that now the interval must be twice smaller, that is h = 10' or Og2 and in the seven-decimal table h=l'.. o r h=OgO2.. Therefore we should obtain a rational five-figure trigonometric table by choosing the interval 10' or 0~2 both for the sine and the tangent function. This would occupy only one page and a half. The interval chosen in seven-decimal table should be 1' or 0.02 grades. This would take 15 pages Parabolic interpolation of the third degree. Putting n=2 into equation (14) we obtain, if 0 < x < 1: (19) F(z) = F(a+xh) = I, +C,(x)IZl-Z2]t(X~1)hWF(a+s'h) where -1 < 5 < 2. The number C,(x) is given by the table mentioned. Moreover I, = 3 [(x+1) F(a+2h) + ( 2 - x ) F ( a - h ) I* This is, as has been said, the approached value of F(Z) obtained by linear interpolation between F(a-h) and F(a+2h). The remainder is (17) : (20) IR, I < f$ ID'F(a+Sh)l. The error of F(r) will be, as we have seen, IbF(z) I < ;~+lR,l.
26
402
Example 7. Let F(z) = log z and h= 1. The table contains the logarithms of the integers from b to c, then
The most favourable value of v is given by
b ______
10 32 100 1000 10000 32000
c
100 100 1000 10000 100000 100000
1 v
5 7 9 13 17 19
The error of F(z) does not exceed one unit in the last decimal. If the table begins with 32, and I is less, then it is necessary to bring it within the range of the table by multiplying it, for instance by 3 and then log3 must be subtracted from the result. Example 8. Let F(z) = lOi, the range of the table being 0- 1 and the interval equal to h; then
,R 4 ,<% 'O"+" <7h4
128 (loge)+ The best value of Y corresponding to h is
'
h
0,01 0,001 0,OOOl 0,00001
-
V
-
7 11 15 19
Remark. A 13 decimal logarithmic table could be constructed from 1000 to 10000 and a 13 decimal antilogarithmic table (h=O..OOOI) taking together 76 pages and giving 13 exact decimals by the aid of an interpolation of the third degree. This would be the rational logarithmic table for high precision.
The quantity Qp, (x)/2"l is to be found at the end of Czuber's Wahrscheinlichkeitsrechnung (Vol. I) and in Jahnke's Funktionentafeln. Linear interpolation would give in this table at the beginning 5 decimals: from ~~3.5 up, 7 decimals; and from 2=5 up, nine decimals exactly. Interpolation of the third degree. We have
I R, I < -f& r ) (~~-32) 1e-*9* . li 2r Into this r=a+[h must be put. Remarking that
"-
-&
we have
I!
'2;;
(32-23)
e-"/2
= &-
v=
(3x-&3)
e -*
= ?$I-.
,R ,< h' @P,tx) 4 40 8
From the tables mentioned it follows that I Gr (x) 18 I < O,55 and therefore I R I < 2/1010. The error of F(z) will not exceed one unit of the ninth decimal. From the point of view of the third degree approximation the table should contain 9 decimals. Example 10. Let
404 to six decimals for every integer value of x; the interval being equal to Am = h = O,l.* According to formula (2) of 8 148 we have D"y hx) = G,, (mx) y (m,x) PI where D is the symbol of derivation with respect to m. In consequ&ce of formula (2) and (3) Q 148 it follows that x m Therefore according to (17) the remainder of an interpolation of degree 2n-1 will be, if 0 < x < 1,
Dy(m,x) = (-l)"Py(m,x--n).
A*" y(a+[h,x-2n) i . x Linear interpolation:
I R, I < T I A2 y(a+th,x-2) I
x
where O<&<l. R, may be simplified by remarking first that if m < 1 then 0 > Ay > -1 and therefore I A"y ! < 1. On the other hand, if m21, then the maximum of y is reached for x=m; hence according to Stirling's formula we have o<y< 1 <0,4 li 2nm
lim
and consequently I Ay I C 0,4/ lrn; moreover I A2y I < 0,8 / vz So that we always have I A2y I < 1. Of course it is possible to obtain lower limits for A"y. For instance, denoting by y (m,i) the largest of the quantities y (m,x), y(m,x-1) and y(m,x-2) it follows that
I Ay(m,x-1)l < y(m,i)
and
I A*y(m,x-2) I < 2y(m,i).
Finally the remainder may be written, in the case of linear interpolation,
I R, I < $ y(m,i).
.* Pearson. Tables for Statisticians and Biometricians.
405 The exactitude will not be much greater than three decimals. Interpolation of the third degree. We have, if 0 <x < 1 R, < g I A4y(a+&v-4)1 where -1 c t < 2, Starting from I A\'y I < 1 we get I A'w 1< 4, and if m 2 1 from IA$1 < 0.8 we find I A"u, /<z; or denoting ,m 1' vm by y (m,i) the greatest of the quantities ~&n,x), w (mJ--1) 9 . . . y (m,x-4) we have I A"y I < 8yImi). From this it follows that I R,! < = or 401/m This gives five exact decimals. ' Example 11. Let 32 r F(m) = F(a+dd = nx+l) o J
1 IR 4 ICg: o
I R,i <$ y(m,i).
.me-ftxdt = Z(a,p).
Pearson's Tables of the incomplete Gamma-Function give Z(zz,p) considered as a function of u, to six decimals; the interval _~ being equal to &z=h=O,l. dm - -\/si and du
--
S i n c e u=m/l!i,+l w e h a v e
PZ = l/x+1 FZ = I/x+1 y(m,x) therefore
V'Z = (x+1) Du, = (xl-l) GIy = -(x+1) Ay(m,x-1). n'
(The polynomials Gi will be introduced in 8 148,) In consequence of what we have seen in the preceding example, if m 2 1: ID'Ij <0.4(+)
u
or in the general case "
I'm
i D"Z I < (x+l)y?(m,i)
where y(m,i) is the larger of the numbers y(m.x) and tp(m.x-1).
The first will be used if m is large or x small, and the second if m is small or x large. Third degree interpolation. We find YZ = (x-l-1)" G,y = -(x+f)2A3 y(m,x-3). We have seen that I A3y I < 2; moreover I A3yi < 4yr(m,i) where v(m,i) is the largest of the quantities y(m,x), w(w=--11,. . . w (m,x-3) and if m 2 1 then I A3y ( < 1,6 l/m Therefore: 2 IR,l< +$ (x+1)2 o r I R, I < $ (x+l)$(m,i) moreover if m 2 1 ,R , < 3h4(x+l)'
4
80Vk
Parabolic interpolation of the third degree will not give, in the case considered, much over five exact decimals. (24). We shall have
This is the approached value of F(r) obtained by linear interpolation between F(a-2h) and F(a+3h). The remainder is equal, according to (17) :
R, = h" [ x;2 j D"F(af:h)
and therefore
I R, I < s4 1DaiF(a$-$h) 1
where 0 < x < 1 and -2 < 6 < 3.
1 407 The error in F(o+xh) will be, as has been said, I d~(a+xh)I < g e + I R, I Example 12. Let F(z) = log Z. The table contains the logarithms of the integers from 10000 to 100000, the interval being equal to one, h=l. Therefore the remainder will be IRJ<&. =F < 1%.
If the table contains the logarithms to v decimal places, then the error of F(a+xh) will be
Hence it is best to compute the table to 24 decimals; indeed, then the error will not exceed one unit of the last decimal. Example 23. Let F(m) = Z(u,p). (See Example 11.) We have p';Z = (%+I):~ G, y = - (x+1)" Ajy(m,x-5). From the preceding we deduce . !A"I/<~; moreover, IA"vl<16y(m,i) where v(m,i) is the greatest of the quantities y(m,x), , , . , w(m,x-5) and if m > 1 then 1ASy,l I < 6,4. Therefore we have vm 6 R,; < g (x-l-1)" or IR,l <$- (x+l)"y(m,i) moreover if m 2 1 l&I< h"(X+l)" 321/m '
The third degree interpolation will in most cases be sufficiently exact; that is, the fables should always be computed, ii possible either for linear or for third degree interpolafion. Using a calculating machine, the shortest way to obtain a third degree interpolation is the following. First I, , will be
408
computed, as has been described above. I, will be noted; then we compute I,. To begin with, we put F(a+;Zh) in B on the machine (p, 398) and multiply it by (1+x) ; then cancelling F(a+2h) in B we put in its place F(a-h) without touching the numbers in A and C; we turn the handle till (1 +x) in C becomes equal to 3. The result in A is then equal to 31,. We divide it mentally by three, putting the quotient into B and checking it by subtracting it thrice from the number in A where now we must hay : zero. Turning the handle backwards again once, we have -1, in A; we add to it I,. The difference I,--I, will now figure in A. We remove it and put it into B. [There are calculating machines which permit us to transfer the numbers from A to B by turning a handle. This is useful if we have to calculate a product of three or more factors.] We multiply it by C, taken from the table. The product appears in A. We cancel the number in B, put there I, obtained before, and add. In A we have the number F(z) desired. Example 10. To find the value of the probability integral corresponding to t = 0.6744898 in Sheppard's tables (hx0.01). Putting a=0,67 we have F(a-h) = 0.745 3731
WI
= 0.748 5711
F(a+h) = 0.751 7478 F(a+2h) = 0,754 9029.
Of course it is not necessary to copy these numbers out of the table, since they can be transferred directly to the machine when they are needed. Putting x=0.44898 we first determine I, by linear interpolation as has been described before. We note the result: I, = 0.749 99737, x contains more than three figures, hence C,(x) cannot be taken from the tables mentioned; it must be computed by multiplication; C, (x) = l/zx(l-x). \ The result is noted too: C,(x) = 0.123 6985. Now we determine I,--I, as has been said above; then without noting it, we multiply it by C,(x) and add I, to the product. We find F(z) = 0.750 00002. This is exact to seven decimals.
409 Interpolations by parabolas of higher order than three are performed in a similar way; by calculating first the numbers I,, I,, . * . and noting the results. Secondly, by forming ZB,,, 1, and finally multiplying them by C,,(X) and adding the products. 1A table giving the numbers C,,(X) is found in lot. cit. 38,I Let us suppose that a table contains the numbers I,, I,, . , . , zs and that F(z) is to be determined by a parabola of degree 2n+l where . zr,, < z < z,,+, * If moreover n+lIm<N-n-l that is, if there are n+l points on each side of z; then, putting z, = a and (z-o)lh=x we have 0 < x < 1. The corresponding formula will be termed, as has been said, a mid-panel formula. If m<n+l ?hen we put z,,-, =a and we shall have x < 0. The corresponding formula is a first-panel formula, which will serve for all interpolations of degree 2n+l if z < z,,,, . N-n-l < m then we put a=~~~-,,-, and we shall have x > 1; then formula is an end-panel formula; the same will serve for all interpolation of degree 2n+l if z > z-v-,,-, . Remark. In the case of the first-panel and of the end-panel formulae the maximum of the remainder (16) will be greater than that given by (17). x < 0 or x > 1; therefore the corresponding values of C,,(x) are not in the tables mentioned , and they must be calculated in each case. Conclusions concerning the computation of tables. A rational table should always be calculated taking account of the interpolation formula to be used. For instance: A five-decimal table for linear interpolation by (18) should contain the logarithms of the integers from 100 to 1000 and the antilogarithms with an interval 0.01. Four pages altogether. A seven-decimal table for linear interpolation by (18) should contain the logarithms of the integers from 1000 to 10000 and the antilogarithms with the interval 0.001 (38 pages).
;
410 For greater precision a 13 decimal table intended for interpolation of the third degree by (19) should contain the logarithms of the integers from 1000 to 10000 and the antilogarithms with an interval of 0,OOOl (76 pages). A 19 decimal table for third degree interpolation containing the logarithms of the integers from 32000 to 100000. (Thompson's Logarithmica Britannica could serve for this.) Finally for an extreme precision a table of 28 decimals could be constructed for interpolation of the fifth degree containing the numbers from 32000 to 1OOOOO. Generally, in order to obtain a given precision of Y decimals, the intervals could be chosen greater at the end of the table than at the beginning The table may be shortened in this way too. Or if the intervals are the same throughout the table, then at the end more decimals can be given than at the beginning. This has been done in Sheppard's table of the probability function. Problem. Sometimes if F(u) is determined by aid of a parabola of degree 2n-1, the tangent to the point of coordinates u,F(u) is also required. For this a knowledge of DF(u) is necessary. Putting x=(u--a)/h into formula (14) (21)
The derivative of C,,(x) is obtained by the aid of Stirling's numbers (24), Parficutar case of the third degree (formula 19). Neglecting the remainder, the derivative of F(u) will, in consequence of (221, be DF(u) = -; [(i-x) (Z,-Z1) + AF(a) - 5 C, (x)A"F(a-41.
1'
The p -th derivative of F(u) can be obtained by aid of Leibnifz's formula. Starting from (21) and remarking that D'Z, = 0, we have 1 (23) D,lF(u) = +, [ ,,,;, D~Crn(x) u x
To begin with, we determine by aid of the remainder in formula (l), the number R of terms necessary to obtain the prescribed precision. Generally n=2, corresponding to a third degree parabola, will be sufficient. If F(Z) is given, and the corresponding value of Z is to be determined by aid of a table containing the numbers F(z) corresponding to equidistant values of z (the increment of z being equal to h) then we choose F(U) in the table, so as to have (3) or F(a) C F(Z) < F(a+h) F(a) > F(Z) > F(a+h).
The method is similar in both cases. We will suppose that the first inequality is satisfied. Now we compute z1 the first approximation of Z, by linear inverse interpolation between a and a+h, using Lagrange's formula (12) of 5 131. Putting into it a+h instead of 6, we get (4) h ( a + h - Z ) ( Z - a ) D'F(a+:h) 4FW---F(a) 1 *I = a + F(a.+h)-F(a) + ___- 2IFk+hl -F(a) I
where O<t< 1. The maximum of the absolute value of the remainder 27 will be hQ2F (a+[h) ""' < 1 8JF(a-+h)--F(a)] ' Denoting by E the precision of the numbers F(z) contained in the table, for instance E = 5/IOB+* (Y exact decimals), the absolute value of the error of z, produced by the inexactitude of the numbers in the table will be: he 3he F(Z)--F(a) ---~- 1 1+2 F(a+h) ---FM I f I Fla+hl ----F(a) II < I F ( a + h ) - F ( a ) I therefore the absolute value of the total error of Z, is 3he 162, I < qqq---- /+'8'<& -F(a) , =h,.
413 Neglecting the remainder in formula (4) we determine Z, keeping only the r1 exact decimals. If Z > 0 we necessarily have
a<r,<Z<afh.
Then adding one unit of the P, -th decimal, that is h, = l/lo"1 , to z, we have zl < Z < z, + h, . (5)
F (z, +h,), putting first into this equation x= (q-)/h and then x=(z,+hl-a)lh, using the necessary n terms for the required
precision, In consequence of (3) and (5) we shall have F(q) < f'(Z) < F(z,+hl. Then we proceed to the determination of z2, the second approximation of Z, obtained again by linear inverse interpolation, but between z, and r,+h, . We find (6)
Moreover the absolute value of the error of F(z,) or by the inexactitude of the numbers F(Z) contained in the table, if the interpolation executed was of the
third degree 1, being less than + E, if follows, in consequence of what has been said above, that /(jz, ( < -...~ 3hl& + 1~8, I < A; = h, . I F(z, +h,)--Fo We keep in z, only the ~~~ exact decimals. If a greater precision is necessary, then adding one unit of the v2 -th decimal, that is h, = l/10'! to z,, we find (7)
z2 < Z < z,+h,.
Now we determine F(z.,) and F(r,+h,) by (1) in ihe same way as F(z,) before. Then starting from
414
W,) < F(Z) < F(z,+hz)
we determine zH the third approximation of Z, by writing into (6) respectively z,, z,, h, instead of z2, z, , h, ; the error 62, is given by (7) in the same manner. If Bz, is negligible compared with the prescribed precision, or if .A; is smaller than the first part of the error CJZ, due to the inexactitude of the numbers of the table, then the problem is solved; if not, we continue the proceeding as described before. Example 1. To compare this method of inverse interpolation with that of Thompson, by aid of Everett's formula, using the even differences, we will again choose Thompson's example. Given F(Z) = 1ogZ = 1.95717 32271 83589 39035 Z is to be determined by Thompson's logarithmic table to twenty decimals. In these tables we find F(a) = i.95717 13373 70099 19928 , F(a+h) = 1.95717 61304 04846 19226 , a = 0,90609 a + h = 0,906lO
therefore h=l/106; moreover, AF(a) > 4/10° and E = 5/lO"l. Starting from these values, we determine first the maximum of the remainder in formula (4). We get
)A')<
Hence
16a2AF (a)
h3
<&.
therefore the first ten decimals of z, will be exact. From (4) we obtain, neglecting the remainder: z, = 0.90609 39428, Since the last decimal is exact, hence z, < Z and adding one unit to the tenth decimal of z, we have z,+h, ~0.90609 39429; moreover, z, < Z < z,+h,. Before computing F(z,) and F(z;+h,) by aid of formula (1) we must determine the number n of terms necessary to obtain
1
415 a precision of 20 decimals. For this we determine the remainder of (1) in the case of n=2, and find IR ,< - -__- < 2 . 3h4 6loge 4 128 a* 10"" hence a third degree interpolation is sufficient; so that we have only to determine I, , I z and C, (x). For this we put first into equation (1) x = Ly = 0,39428 and k=l; we find I, = 0.39428F(a+h) + 0.60572@) =
= i;95717 32271 67839 24368.
-.
'7 I
Remark. Before performing the multiplications by F(a+h) and F(a), since the sum of the factors x and (l-x) is equal to unity, the first six figures common to F(a) and F(a+h) have been set aside and only added to the result. In the same way, putting into formula (2) x=0,39428 and k=2 we get
1 '2 = lq39;128
F@+'&) + 1'y72
F(a-h) =
z-i,95717 32271 14941 49591
F(a+2h) and F(a-k) were taken out of the logarithmic table;
and before the multiplications, again, the five common figures were set aside. From the above results we deduce I+, = 0.00000 00000 52897 74777; moreover and C, (x,) = (0.39428) (0.30286) = 0.11941 16408 C, (x,) II,--Z,] = 0.00000 00000 06316 60686 and finally
F(q) =i.95717 32271 74155 85054.
We had F(U) < F(Z) < F(a+h) and z, < Z; therefore we must have F(z,) <F(Z), and this is what we really find. Now to obtain F(r,+h,) we put into (1) x=0.39429 and get
F (q fh,) =i.95717 32272 22086 25392. As was to be expected, we have
FM <F(Z) < Fb,+h). Now we shall determine the maximum of the remainder corresponding to formula (6). We find
Therefore the error caused by neglecting the remainder will be less than one unit of the 20 -th decimal. The problem is solved. The error due to the inexactitude of the numbers of the table cannot be overcome, and we shall have in consequence of (7)
417 Remark. Taking account of the fact that 1ogZ is equal to log (e/3) the error being less than 5jlO", it follows that the difference between tr and e/3 should not exceed 4,/10""; and really this difference is equal to 3/1020. g 134. Precision of the interpolation formulae, In 9 125 we have seen that the precision of an interpolation formula of degree 2n-1 having the same remainder may be measured by the maximum of the possible error caused by the inexactitude of the numbers in the table, that is by WE, if. the error of the data in the tables is less than E = 5/10'+', and if UI is the sum of the absolute values of the coefficients figuring in the formula. Therefore, to compare the different formulae we have to determine the corresponding values of o. 2. In the case of Newfon's formula ((j 125) we find:
if 0 I x 2 1, the series below is then convergent and
.,zo w,"(;J =o
-therefore tie have w1 S 2. 2. In the first Gauss series we had (2, 8 127)
moreover in the second Gauss series (6, $j 127)
From this we immediately deduce that for every value of x and m we have ws > w2. Indeed from the preceding it follows that x + m > m - x o r x >o; if this is satisfied, the maximum of the error of the first GQUSS formula is less than that of the second of the same degree; thus the former is preferable, 2-i
418 3. From Bessel's formula (4, Q 128) we obtain
It is easy to see that if x > '/z then wI < w2, Indeed for every value of m this gives x--'/z < x + m. But even if I/,< x < 42, then we have l/z- x 5 x + m. So that if x > 1/1 then Bessel's formula is preferable to the Gauss formulae. 4. Stirling's fyrmula gives (5 128)
W>i
211
w [Ii
z
x+m-1 2m 1 m&i + llzxmf+mi]l]'
From what precedes, we may easily deduce that Stirling's formula is preferable to the Gauss' formulae, if 0 < x < 1/. 5. In the case of Everett's formula we get (§ 129) if O<x<l
From this it follows that wa > w6. 7. The interpolation formula dispensing differences of Q 132 will give
with
printed
Other conclusions may be obtained by comparing the inter: polation formulae in some particular cases. A. Linear interpolation. The Newton, the two Gauss and the Stirling formulae give 0=1+x. To Bessd's formula corresponds it will lead to results inferior to the therefore if 2. X < '14 to better ones. former formula, and if x > I/~ Everett's formula and that of 8 132 give o = 1; therefore these are the most advantageous. B. interpolation of fhe third degree. We obtain the following values:
7. Formula of ,$ 132: w7 = 1 + x(1-x). From the above we conclude that in the case of an interpolation of the third degree the error is generally the smallest when using Everett's formula. Moreover if x > l/2 then Newton's formula is preferable to the Gauss formula I. If x=l/& then both lead to the same result; and finally, if x 5 l/z, then Gauss' formula is better. Comparing Newton's formula with that of Stirling, we find exactly the above result. On the other hand, comparing Newton's formula with Bessel's, we are led to x2 -9x + 2 > 0; therefore if x = 0,278 both formulae give the same result; Newton's formula is superior or inferior to Bessel's as x is smaller or larger than 0.278. Comparing Stirling's formula with Bessel's, we find that both lead to the same result if XX I/, moreover that Stirling's formula is preferable to Bessel's if x < I/*. The formula of (j 132 gives better results than the Gauss, kewton, and Stirling formulae. Comparing it with Bessel's formula, we find that if ~~0.386 or if ~~0.613, both formulae give the same result; and that Bessel's formula is preferable if 0.386 <x < 0.613; for the other values of x the formula of () 132 is better.
420 This is another reason why the tables constructed for interpolation by aid of the formula of (j 132 may be shorter than those in which the interpolation is to be made by Newton's, Gauss' or Stirling's formulae. 9 135. General Problem of Interpolation. A. The Function and some of its Differences are given for certain values of the variable. We have seen that if a function f(x) and its differences are given for x=a then the function may be expanded into a Newton series, Moreover if the function f(x) and its even differences are given for x=a and for x=a+h, then the function may be expanded into an Everett series. Now we shall treat the general problem. The function f(x) is given for x=aO, aI, a*, . . . , a, ; moreover some of the differences Af (ai), Aaf (ai), . . , A"i-'f (ai) are given. Let us suppose that vo+v, +v,+v,+...+,v~=R. that is, there are given n .quantities in all. We may obtain the required interpolation formula of f(x) satisfying the above conditions by aid of a Newton series. Let us write f(x) = and
f(o) + [ ;j Afto, + (;] A-f(o) +. . . . + lnTl) A'7-'f(o)
Putting into these equations the n given values AUif (al) for ,ui = 0, 1, 2, . , . , (y-l), we get n equations, which determine the n unknowns: A f(0) for K=O,1,2 ,,.., n-l. The remainder of the series will be that of a Newton series stopped at the term A"-'. Hence the interpolation formula will be f(x) = f(o) + [ yj Af(0) + . . , + ( n:l j AI-'!(O) + [ ;] D"F(:) where 0 < 6 < n-l or 0 < 5 < x.
421 B. The Function and some of its Derivatives are given for certain values of the variable. If a function f(x) and its derivatives are given for x =a then the function may be expanded into a Taylor series. If the function f(x) is given for x=a,,, a,, . . . . , a,, moreover if Df(ai), DV(ai), 1 . . . , D"i-'f(q) are given too, then putting v,, + Y, + vq + , . . . + vi = n we may expand the function f(x) into a Taylor series. Writing f(x)=f(O)+xDf(o)+ ;.j D'f(o)+. , . + $I& D"-'f(o) and D:lf (x) z.z Dl,f (0) + xD!' "f (0) -t , . , + (n "T":,)! D"-'f (0). Putting into these equations the n given values D"if (a,) for ,IL~ = 0, 1, 2, . . . , vi-1
we may determine the n unknowns: D'f(O) for k = 0, 1,2, , . . , n-l,
The remainder of this expansion will be &at of the Taylor series .which has been stopped at the term D"-'f (0). Hence the required formula will be f(x) = f (0) + xDf (0) + , , . , + &- D"-'f(O) + 5 Dnf(:) where 0 < 5 < x.
CHAPTER VIII.
A PPROXIMATION
AND GRADUATION .
$ 136. Approximation according to the principle of moments. When solving the problem of interpolation, f(x) was givenforx=0,1,2,..., R, and we determined a curve of degree r! passing through the points of coordinates x,f(x). The corresponding problem of approximation is the following: The points of coordinates x,y=f (x) are given for x=0,1,2,. . ,, N-l and a function F(x) satisfying certain conditions is to be determined so that the deviations &=F(x)-y shall be, according to some principle, the smallest. . Such a principle is for instance the principle of least squares, according to which, a function F(x) containing disposable parameters being given, the parameters must be determined so that QF= g E? x=0 = X:0 IF[x)-yy
shall be a minimum. A second principle of approximation is that of the moments. Let us denote by (1) dLrn = I5
x-0
xmy.
This is the m -th power-moment of y; given a function F(X) containing n+ 1 disposable parameters, these must be determined in such a manner that the moments &G, PM;, . . . , d* of y shall be identical with the corresponding moments of F(x). There are other principles of approximation.
423 A second case of approximation is the following: a function f(x) of the continuous variable x is given, another function F(x) satisfying certain conditions is to be determined so that the deviations of the two curves shall be the smallest according to some principle. For instance, according to the principle of least squares, F(x) containing disposable parameters being given, these are to be determined so that 8= (*= If(X)-F(X)]'dX -cd shall be a minimum. On the other hand, according to the principle of moments, if (2) z-4&= --OD[I xrnf(x) dx
is given for m=0,1,2,..., n; the parameters of F(x) are to be determined so that the first n+l moments of F(x) shall be equal to the corresponding moments of t(x) given by (2). The simplest case of approximation is that in which F(x) is a polynomial of degree n containing n+l disposable coefficients. Let us suppose first that the variable is discontinuous, x = 0, 1, 2, . , . , N-l, and that F(x) = a0 + a,x + . . . . + a,,x". -4ccording to the method of least squares the equations determining the parameters ci will be
for Y = 0, 1, 2, . . , , n. But these are also the equations determining the parameters if the principle of moments is applied. From this we conclude that if F(x) is a polynomial the principles of least squares and of moments lead to the same resulf, and therefore if the polynomial F(x) is expanded into a
424 series of any polynomials whatever, the method of least squares will always lead to the same result. Consequently we shall choose the expansion which will require the least work of computation, This will happen if F(x) is expanded into a series of orthogonal polynomials. Remark. 1. The above results are also true in the case of a continuous variable. If in order to obtain an approximation the principle of moments is chosen, it is often adventageous to introduce, instead of the power-moments given by formula (l), the factorialmoments ?J& or the binomial-moments ";iS, given by the following definition:
(5)
'ms = .go (x)sf(x)
m as = z I 5 I f(x). x=0
The approximation obtained will be the same whatever th:z chosen moments are, but often the calculus needed is much simpler in the case of binomial-moments than in that of powermoments. Moreover the computation of the binomial-moments is shorter than that of the power moments, as is shown in Q 144. Remurk. 2. If u(t) the generating function of f(x) is known, then we have u(f) = 5 f(x) P x=0 so that a n d D%(t) = Xi (x)Sf(x) tXmS
On the other hand, if the binomial-moments are known we may determine the generating function of f(x) by Taylor's formula
425 Remark. 3. The principle of approximation applied is nearly always that of the moments; indeed, the other principles, for instance that of the least squares, or Fisher's principle of likelihood, are as a rule used only in cases when they lead to the same result as the principle of moments. The reason for this is not that the principle of moments is more in agreement with our idea of approximation than the others (indeed from this point of view, the first place belongs to the principle of least squares), but that the calculus is the simplest in the case of the principle of moments. If the principle of approximation is chosen, we have still to choose the approximating function F(x) containing disposable parameters. In this, the interval in which x varies, the values of the function f(x) to be approximated, at the beginning and at the end of this interval, finally the maxima and minima of f(x) play the most important parts. In the case of the principle of least squares, it is the meansquare-deviation that measures the approximationlobtained . In the case of the moments, we have not so practical a measure, but we may proceed as follows: If the function F(x) has been determined so that its first n+l power-moments shall respectively be equal to the corresponding moments of f(x), that is to CM,, OK, , . . , . A,, then to measure the obtained approximation we have to compare Xx"+'F(x) with d(+,; the less the difference is, the better the approximation may be considered. Remark. 4. 1 If the function f(x) is expanda'd into a series c,+c,(P,(x) -tc.'cr2(x) +....+c,"%z(x)+.... and if we stop at the term c,,r/l,(x) it may happen that the coefficients c,, of this expansion are ihe same as those we should have obtained by determining them with the aid of the principle of moments, putting the given moments J,, , 1, , . . . , A,, of f(x) equal to the corresponding moments of @n(x) = co + c,v, (4 + . * ' + W/h(X). This will occur: 1. If (I)",(X) is any polynomial whatever of degree m; for instance a Legendre polynomial, if the variable is continuous
426 (§ 138), or an orthogonal polynomial of §Q 139-141, if the variable is discontinuous (x = 0, 1,2, . . . , , N-1). 2. If the variable x is continuous, and if q,,,(x) = H,Q -x'J2 where H, signifies the Hermite polynomial of degree m (Q 147). 3. If the variable is discontinuous x = 0, 1,2, . . . and if cpm(x) =G, F , where G,,, is the polynomial defined in § 148. . This adds to the importance of the principle of moments. On the other hand it may happen that, stopping the expansion of f(x) at the term cnqn(x), the coefficients c, are the same as those we should have obtained by determining them according to the principle of least squares by making minimum either the sum or the integral of the quantity
If(x) -@n(412
according as the variable x is discontinuous or continuous. This will occur: 1. If v,,,(x) is any polynomial whatever of degree m. 2. If the variable x is continuous and if f(x) is expanded into a Fourier series (5 145), in the interval (0, l), and c,,ym(x) = a,cosZzmx + &sin;Znmx. 3. If the variable is discontinuous, and if f(x) is expanded into the trigonometrical series of Q 146, where cmrpm(x) = a, cos F + /j,sin?$Z-. This is in favour of the principle of least squares. $j 137. Examples of the function F(x) chosen. Example 1. Function with two disposable parameters. Continuous variable. The range of x extends from 0 to 00. If f(0) = f(m) = 0 and f(x) > 0, th& we may try to approximate the function f(x) by aid of
F(x) = c r;;x;;).
We have J'-F(x)dx = C
0 0
jmxF(x)dx = C $:;;:I k (p+l)C
427 Let us write rJi -= j"- x'f(x)dx.
0'
According to the principle of moments we must put C = Jl, and (p+l)C = ,d< ; so that the mean of x will be ~4 (x) = =p+ 1 =x,/uu, . The formula (1 J presupposes that p > 0; therefore we must have cd(x) > 1. Writing in (1) C Y d<, and p=d(x)-1 we get an approximation of f(x). To measure the obtained approximation let us determine 4, - I'- x'F (x)dx = ~4~ - (p+2) (p+l)C = ,?Y< - (p+2)@3
0'
The less this quantity is, the better the approximation, Let us determine moreover the mean-square deviation 1~ corresponding to f(x) , 2 being the mean of Ix----,~-~(x)]~; therefore
-
.. .
,~
on the other hand, starting from F(x) we get a2 .wp+l= ,-J'(X). Since the function (1) is maximum for x=p, if f(x) is maximum for x=x,, then the approximation will be useful only if approximately x,,, x .4(x) - 1. If the approximation of f(x) by F(x) is accepted, then from (1) we conclude that .X 1 f(t)df .x* 1 x F(f)df = ~+4dZ(u,p).
0' 0'
Here Z(u,p) represents the incomplete;gamma-function of 8 18, and u=x/ \/p-/-l. The median Q of x is obtained from the equation
0
0'
F
f(x)dx=Z[&,
p) = 'I,.
The tables of the incomplete-gamma-function show that p < 0 < p+l ConcZusion. and therefore x, < Q < d(x).
The approximation of f(x) by the function (1)
428 may be accepted if x varies from 0 to x1 if f(0) = f(x) =O and if f (x)20; moreover if we have approximately 02 - c&f (x) XIll - d(x) - 1 ?-/(X)-l <Q< @Y(x). Remark. 1. In certain circumstances formula (1) may be used also for the approximation of a function f(x), if f (0) = x and f(x) = 0; but then we must have -A(X) < 1 and -1 < p < 0; moreover there should be no extremum in the interval between x=0 and x=00, Example 2. Function containing three disposable parameters. If x is a continuous variable and if its range extends from 0 to 1: moreover if f(0) = f(1) = 0 and f(x) 2 0 then we may try to approximate the function f(x) by (2)
g-1 (1-x) q-1 F ( x ) = c B(PJ71
where p>l and ~>l. Since 1" F(x) dx = C,
0'
an d
0'
c xF(x)dx = c "p;;" = A!!PfQ
1' x'F(x)dx _ Cp(p-+I)
Ip+a) (P+g+l)
I
0'
we have to put
c=Gf$; -/+~-d(, P+4
and
-"//2=;
PIP-t11 (Pf4) (P+q+l) 4.
From this we conclude that the mean of x is given by d(x) = p/(p+q) i a n d -s the mean-square deviation by
Starting from the above equations we may determine the parameters p and Q ; first we get
i
429
.u,q = (G/lo---d%,)p
and finally 'Ii, (J/,-J/,) p = -J&J&--dfc,"
and
d&q = ('//,---'/l.J
(p-+-l)
q = (~"-4) vAl,-'-4 --,ll,,,ll,--,1(1'--' P----lp+T--2 1 l+ s .
To find the mode x, we have to put DF(x)=O, this gives x,,, =
then from the above equation it follows that x, > 2 , 0 dl and if q>p then z4 >x,.
p>q
If
It is easy to show that the sign of ID2ylx=xm is the same as that of !P-1) (1-q) p+q-2 -. Since p > 1 and q > 1, this is negative, so that a maximum corresponds to x=x,. If the approximation of f(x) by formula (2) is accepted, then we have also fl(t]dt = (.'F(t)dt =
0' -.. 0' Zs(p,q)
-
where Zs(p,q) is the incomplete Beta-function of $j 25. The median Q of x given by 0 r f Wx = 1, hd = ?h
0'
is determined by aid of the tables of this function. If it can be shown that Pi-Q--2 < q then P--l P -- >e> - or x,,, > g 2 d(x) P+Q
p
>
q
then
and if
p
x&psqx). If X, and Q are nearly equal to the above values, then the approximation is admissible. To check the obtained precision we
determine first ds, the third power moment of f(x) and then that of F(x); the smaller the difference
If the approximation of f(x) by formula (4) is accepted, then we have also I: f ( x ) = 1-Z(u,p) k=o where Z(u,p) is the incomplete-gamma-function of 5 18, and __. p=z. u=m/brz+l a n d The median of Q is drrermined by aid of
I+1
that is, by l- I [" ,@-1) -I IA, e]. e v e+*
V
Since Q must necessarily be an integer, hence this equation will be only approximately satisfied. From Example I it follows that m-l < g < m + 1. ConcZu~ion. The approximation by (4) is advisable if we have d ( x ) - 1 < x, < -d(x) ( and d(x) - 1 < @ < J (Jr) + 1
If the approximation by (5) is accepted, we have also i f(t) = 1 - ZP (x,n+l-x) t=o where I, (x,n+l-x) represents the incomplete-beta-function (16), 5 25. The median Q is obtained from e f(x) - fig f(x) x=0 x=p+ 1 or from 1 - 1,) (e,n+l---e) - L (e+M--4. From example 2 it follows that np-1 S Q 5 np. Conclusion. The approximation by (5) will be useful if we "R(x) < x, < d(x) + 1
28
have
434 moreover if p > l/r or $$s'(x) > n" then J(x)-1 <g< J(x) and if p < l/z or l/z ad(x) < o' then d(x) < @<d(X) + 1. Q 138. Expansion of a function into a series of Legendre's polynomials. The Legendre polynomial of degree n denoted by X,(x) is defined by (1)
In consequence of the symmetry of the polynomials, if xi is a root of X,(x) =O then -xi will also be a root. In Mathematical Analysis it is shownqO that
1 c
-i
X,(x)X,,(x) dx = 0 if n =/= m
(3)
-.
- i
([X.(x),2dx =2n+l* 2
That is, the polynomials are orthogonal in the interval (-1, 1). A function t(x) of limited total fluctuation in the interval (-1,l) may be expanded into a series of Legendre's polynomials. Let us write f(x) =co+clxl(x) +c,X,(x) +...,+c,X,(x)+ ....
Multiplying both members of this equation by X,(x) and integrating from -1 to +i in consequence of the equations (3) we find
c m =
2m+l ' f(x) X,,,(x) dx. 2 - ir
Putting into this equation the value of X,(x) obtained in (2) we have c, = $4(2m+l) 5 i=o k$![T] [2m;2i] t(x) xm-zidx
436 and denoting by A!s the s -th power moment of f(x) in the interval considered:
ah<= J ! xs f(x) dx
- 1
we finally find that (5) where ,u is the greatest integer contained in $$m+l. The determination of the coefficients c,,, is very simple in consequence of the orthogonality of the polynomials; otherwise it would be a laborious task. Approximation of a function f(x) by a series of Legendre's polynomials. Stopping at the term X,, let us write (6) The coefficients c,,, must be determined according to the principle of least squares so that (7) #= (? If(x)-f,,(x)j"dx -i
shall be a minimum. Putting into it the value of f,,(x) given by (6) the equations determining the minimum will be i)@ -=O f o r m=0,1,2 ,.,., n .
i)C,
In consequence of the orthogonality of the functions we shall find for c,, the value (4) obtained above. Moreover by aid of (3) and (4) we find
437 These polynomials were first considered by Tchebichef" and since then several authors have investigated this subject. In the general case the polynomials are complicated and of little practical use. But if the values of xi are equidistant, then simple formulae may be obtained. Determination of the orthogonal polynomials, if the given values of x are equidistant. x=a+hE and t=O, 1,2,. . . , N-l. lnstead of starting at (1) we shall employ the following formula:
438 were to expand F,-,(x) into a series of U,(x) polynomials we should return to equation (1) again. Since F,,-, (x) is of degree m-l, formula (IO) of ($ 34, giving the indefinite sum of a product, may be written, putting there U(x) = Fm-, (4 in the following way: !A-'-'rF,-, Ix) Un,Ix)'l = F,-, (4 A-'Urn (4 -AF,-, (x) A-'U,(x+h) + A'F,-,(x) ,A--"U,(x+2h)-. . . . + and V,,(x) = Urn (4
+ (-l)m-'Am-'F,,-,(x) A-mU,(x+mh-h).
Now. since we are considering polynomials only, A-W(x) an arbitrary constant, to which may be assigned such a value that [A-W,(x) IX=,, is equal to zero. But AW,(x+h) contains an additional constant, which may be chosen so that [A-"U,(x+h)]xz ,z = 0. Cont inuing after this fashion, we may dispose of all these arbitrary constants in such a way as to have
contains
A-"U,(x+nh-h) = 0 for x=a, and for every value of n satisfying to n <, m; therefore
: A-' IF,,,-, Ix) U,(x) ] >,=a = 0.
But in order that the definite sum may be equal to zero, it is necessary for the above expression to vanish also for the upper limit x=a+Nh=b. But since A' F,-,(b) is arbitrary for all values of v, it follows that each expression A-P-l U, (x+vR) obtained for v=O, 1, 2, . . . , m-l must vanish separately for x=b. From this we conclude that (x-a) and (x-b) must both be factors of A-' U,(x). Considering for the moment only the first of these factors, we may therefore write A-W,(x) = (x--@(x). Applying to this expression the formula for the indefinite sum of a product (10, Q 34), we find A-W,,(x) = ,.2 (-1) r & [ "-;;"+;'") Ari(x). (x-o) must be a factor of A-W,,,(x) too: therefore the
m+1
439 additional constant must be equal to zero; it follows that (x-a)z,,, is a factor of A-W,(x)* By successive summation we should find that (x-o),,n is factor of AbmU,,,(x) so that it may be written
A-"V, (x) = [ x;a ]I Y(X).
As (x-b). must also be a multiplying factor of A-lc,,,(x) we deduce by the same reasoning that
A-W,(x) is of degree 2m; therefore C is an arbitrary constant, and we conclude that the general formula for the orthogonal polynomials with respect to x=a+Eh, where l=O, 1,2, . . . , N-l and b=a+Nh is the following: (3)
U,,,(x) = CAm[(x;")h (x;b)h] +
Starting from this expression, there are two different ways of deducing the expansion of U,(x) into a Newton series. First utilising formula (10) Q 30 which gives the m -th difference of a product; we obtain (4)
U,(x) = Ch"l
Secondly, we can develop AsmU,
of generalised binomial coefficients i I we shall have A-W,,(x)
x-b
into a Newton series 1 ,, . According to 5 22
= C ":I: IXTblh$qz-"]. 'r,
(*;P]*].+
butAfj';p].
[";;;")nl = 2; h ' ( t ) [;zq* (;:iijY$
Putting x=b, w e g e t
A-u",(X) = c 2m+l z i
( m II %zhJ, ( xTb)h i=O
and finany putting i=m+v, and determining the m -th difference of the above expression, we find
440 (5) U,(x) = Ch" Zo' lrnfzvJ[ byTFh)* [ ";"I,, .
As U,,,(x) is symmetric with respect to a and b, we can get two other formulae from (4) and (5) changing a into b and inversely. For instance, remarking that b-a=Nh, from (5) we get (6) U , ( x ) = ChXm '2; h-' (mm+v) [ ;I;] ("s;"1,.
Remark 1. If in this formula we put m=N then every term will vanish except that in which ,v=m, and we obtain u/,,(x) = Ch2." (71 ( xT;a)h. Therefore, if x is equal to the given numbers x=a+th 6=0,1,2 I.,,( N-l we shall have Us(x) = 0. 2. If m > N then every term of (6) will vanish in which Y < N; but if x is equal to x=a+th and 6~0, 1,2,. , . , N-i, then the last factor of (6) will be if v > N [ ";;"],,= h" [ j] ~0. Conclusion. If m > N and x is one of the given values, then we have U,(x) = 0. Introducing into (6) the variable 6=(x-a)lh, become (7) U,(a+,th) = Chl'm We have seen that
,v 2 U,,(x) &O U:,(x) = 0
for
it w-ill
if ,u =j= m. It remains to determine the value of this expression if .u=m. That is
441
This may be done by determining the indefinite sum of the product U",(x)U, (x). According to formula (10) $j 34 we have 19) A-'[U,(x)U,(x)l = U,(x)A-IUrn -AU,(X)A-~U~(X+~) + . .
, . . , + (-l)mAmU,n (x)A-~-W, (x+mh). To obtain the sum (8) we have to put in this expression x=b and ~=a; but we disposed of the arbitrary constants in U,(X) SO as to have: A-W,(x+nh-h) = 0
forn=l,2,..., m at both limits x=o and x=b. Hence it remains to determine the value of the last term at the limits. For this we start from
A-mUntbd = C 1";;;" 1, [ x;;;b),,;
the above quoted formula of the sum of a product will give
Putting x+mh instead of x into the preceding expression it will become A-"-W,,,(x+mh) = $'I%: (-l)'( xzzh)L (x-~~v:++ymh)~]~
At the upper limit x=b every term in the second member is equal to zero, so that (9') [A--"-W,,, (x+mh) 1-6 = 0.
this gives for x=a . Finally by aid of (9) we obtain the required sum:
It may be useful to remark, that this quantity is independent of the origin of the variable x. 15 140. Some mathematical properties of the orthogonal polynomials." Symmetry of the polynomials. Putting into formula (4) Q 139 afb-h-x instead of x we find U,(a+b-h-x) = Ch" X ,, A I 11
nt+t m
b-h+vh-x
V
but this is equal to II,,, (a+b-h-x) = (-1) m Ch" ys [ y ] [ x-agy?-,~h)h (x;b)l* Now putting into it ,u=m-v from (4), 8 139 it follows that (11 U,,(a+b-h-x) = (-1)"' U,(x). This equation shows the symmetry of the orthogonal polyQ They have been described more fully in lot. cit. 41. Jordan. 314-322 a n d e) p p . 309.-317.
U,"~,, (=+F] = OS
Functional equation. We may easily deduce a functional equation which is satisfied by the orthogonal polynomials; this can be done by expanding xv,,,(x) into a series of orthogonal polynomials. We find (3) xUm(x) = An, m - 1 Urn-, (4 + &mUm(x) + Am. m+Jm+l (xl ; as in consequence of the orthogonality of the polynomials the other terms vanish. Indeed
i x=u xu,"(x)u:, (x) = 0
i f p> m+l or p<m-1 (2, Q 139). Hence we have only to determine the above three coefficients, Multiplying by U,,, (x) and summing, we obtain from (3) (4) i It=U xU,(x)U,+, (4 = An. m-i,
5 x-;1 [Urn+, (x)1".
We know already the sum in the second member; to determine the first member let us apply the formula giving the indefinite sum of a product (10, 8 34). A-' I xu,,, (4 - U",, , (4 I =
= ixU,,,(x) 1A-'U,,,,, (x) - AIxU,(X) 1 A-'U,,, (x+h) +
+*... + (-l)m+l Amil\xUn,(x) 1 A-nl-WIII+, (x+mh+h). When determining the polynomials U,,(x) we disposed of the arbitrary constants so as to have AW,(x+nh-h) = 0 for R= 1,2,3 ,,,., m at both limits: x=a and x=b. Therefore every term of the preceding seiies will vanish at
444 these limits except the last term, which according to formulae (9') and (10) of 5 139 is equal to zero for x=b; and for x=a it is [A-"-"U,+,(x+mh+h)].=, = (-l)"C,+, h2m+2 I '2;:-' ] <
In this formula we have written C,,, instead of C. since this constant may depend upon the degree m+l of the polynomial. Since formula (11) Q 139 gives AmU,,, (u) = C, h"m "," I 1 we shall have for x=c Am+l[xUm(x)] I Finally we find = h(m+l) AmU,, = C, h 2m+1 (m+l) 12:),
C (m-t-11" u,,, (x), +nc,,, 2(2m+l)
Application. Let us determine the central value of the func. . a+&h If we tion U,,(x), that is U,, I 2 putinto (3) m=2n+1 and x=a+l/z(h'h-h),.then in consequence of (2) it will become - A 111+1.
PI
zn+1.2n+2
U 2n+2
Nh-h
a+ 2 1 *
Writing m=2n+l we obtain A,,,,, 2n from equation (8) and A 2n+I, 2n+2 from equation (7). Therefore the preceding formula may be written (12) F(n+l) = - N~~~~l~z1)2 F(n)
where, in order to abbreviate it, there has been put
446
h4" c 211 The solution of equation (12), which is a homogeneous linear difference equation of the first order, with variable coefficients, is as, we shall see in f 173, the following: -[NZ-(2i+l)"] =,$ - 4(i+l)r [E&i) (+.] = IL0 L?J (i+l) (it-11 "
Difference equafion. It can be shown that the polynomial U,(x) satisfies the difference equation (14)
(x--a+24 (x4+24A2U,Ix)
+ (m+l)hW,(x) = 0 .
+ /2x--a--b+3h-m(m+l)hIhAU,,(x)-m
[See lot, cit41. Jordan a) p* 315; e) p, 316.1 Roofs of the polynomial. L. Fejhr has given [See lot. cit." Jordan a) p. 3191 the following theorems concerning these roots: The roots of U,(x) =O are all real and single, and they are all situated in the interval a&-h. Whatever 5 may be, in the interval a+lh, a+th+h there is at most one root of U,(x) ~0. Fe@ showed moreover that if Pm(x) is a polynomial of degree m, and if in its Newton expansion the coefficient of m I I is unity, then the polynomial which minimizes the following expression
X
447
is the orthogonal polynomial U,(x) with the constant C suitably chosen. $ 141. Expansion of a function f(x) into a series of polynomials orthogonal with respect to x=a+Eh, where 6=0,1,2, . ..*. N - l . Supposing first that fix) is a polynomial of degree R such that n < N, then we have f (4 = c, + c,u, (xl + 42 (4 + * * * * + GUI, (4 multiplying both members by U,(x), and summing fromx=a to x=b (that is from E=O to 5=N, since a+Nh=b), we find in consequence of formula (1) 8 139, that 11) Cm
z [Urn( = tz f(X)Urn(X)*
Hence the coefficients cm are easily obtained. In the general case of t(x) the series will be infinite. Stopping it at the term U,, (x), the series will nevertheless give exactly the values of f(x), if x is equal to one of the given values x=a+th. Indeed, according to what we have seen in the preceding paragraph, every term in which m h N will vanish for these values; so that the limited series will give the same value as the infinite series. To have f(x) exact for the other values of x we must add the remainder to the limited series; according 'to Q 123 this will be (2) RN = & (x-a) (x-a-h) . . , (x-a-Nh+h) DNf (a+Ch) .
where O<tSN-1 or a<a+th< b. If m 2 N then the coefficients of U,(x) in the expansion of f(x) cannot be determined by formula (1) since in consequence of formula (7) U,(x) is equal to zero for each of the values of x corresponding to E=O, 1,2, . . . , N-l ; hence both members of equation (1) will be equal to zero. In the preceding paragraph we have already determined z[Um(x)12s T heref ore to determine the coefficients cm by aid of
According to 5 136 the last sum in the second member is equal to the binomial moment of order v, denoted by &., of the function f (a+[h) ; therefore this may be written: Therefore (3) Ii U,(x)f ( x ) = Chzm *=,,
As will be shown later, there is a far better method for rapidly computing the binomial moments than is available in the case of power moments. If we operate with equidistant discontinuous variables, it is not advantageous to consider powers; it is much better to express the quantities by binomial coefficients. Indeed, if an expression were given in power series, it would still be advantageous to transform it into a binomial series. Several statisticians have remarked that it is not advisable to introduce moments of higher order into the calculations, In fact if N is large, these numbers will increase rapidly with the order of the moments, will become very large, and their coefficients in the formulae will necessarily become very small. It is difficult to operate with such numbers, the causes of errors being many. To remedy this inconvenience, the mean binomial moment has been introduced. The definition of the mean binomial moment ,& of order Y of the function f(x+Eh) is the following
therefore
The mean binomial moment will remain of the same order of magnitude as f(x), whatever N or Y may be. For instance, if
1
449 t(x) is equal to the constant k then we shall have dy = k for any value of Y or N. On the other hand the power moment of order v k Ti +I 5'"
will increase rapidly with v and N. Introducing into formula (3) dy instead of Q% we shall have
This may be written in the following form (-*)'a"- (m+l) [mN+* 1 2 (-l)l' [ mm'y] (; J -& To simplify the formula we shall write
Since these numbers are very useful they are presented in rhe following table, which gives all the numbers necessary for parabolas up to the tenth degree.
II m\V 0
The, following relation can be used for checking the numbers: Al, + Bm, + Pm+ * * * + Bmm = 0 that is, the sum of the numbers in the rows is equal to zero. Moreover let us put m+l z prn,,2c = 0,. (6) r=O If we already know the mean binomial moments, the values of 8, may readily be computed with the aid of the table above. Finally we obtain (7)
ja U,(X) f(X) = CfPm (m+l 1 lrnjJ1) Qm -
As this expression could be termed the orthogonal moment of degree m of f(x), therefore we can consider 0, as a certain mean orthogonal moment of degree m of f(x). The mean orthogonal moments are independent of the origin, of the interval, and of the constant C. Particular case:
is equal to the arithmetic mean of the quantities f (xi). By aid of equation (7) and of (l2), § 139 we deduce from (1) the coefficient cm: 03)
c m _ (2m+l)Qm Cpm I N+m ' m 1
The coefficient c, is independent of the origin. In particular we have c,, = 0,/c.
451 $ 142. Approximation of a function y given for i=O, 1, 2, . . . , N-l, by aid of a polynomial f(x) of degree n, where x=a+th, according to the principle of least squares, that is, so that the sum of the squares of the deviations y-f(x) for the given values of x 8 = go [Y--f(412 shall be a minimum. If the polynomial f(x) is expressed by orthogonal polynomials: (1) f(x) = c,, + c&J, (x) + c&J2 (4 + * . *' + CJJ" (4
then the conditions of the minimum will be
for m=O, 1,. , . , n. In consequence of the orthogonality (1, Q 139) most of the terms will vanish and we shall have
--
Since this expression is identical with equation (1) of Q 141, which gives the coefficient c,,, of the expansion of y into a series of orthogonal polynomials: from this we deduce the important result: To obtain the best approximation possible of a function y, according to the principle of least squares, by aid of a polynomial f(x) of degree n, it is sufficient to expand x into a series of orthogonal polynomials, and to stop the series at the term U,,(x). Moreover, if the approximation obtained by aid of a polynomial of degree n should not be close enough, then to obtain the best approximation possible by aid of a polynomial of degree n+l it is sufficient to determine only one additional coefficient, c,,, ; the others would not change. This is an important observation, since, if the expansion were not an orthogonal one, then, passing from the approximation of degree
452 n to that of degree n+l every coefficient would have to be computed anew. If the approximation of degree n+l is still unsatisfactory, this can be repeated till the required precision is reached. Since we have seen in the preceding pragraph that the coefficient c,,, is given by
c
m
= em+11 @,
Ch'm( Nm+m]
hence if we know the mean orthogonal moments of the y quantities, the problem is solved, It remains but to determine the obtained precision. Measure of fhe precision. In the method of least squares the precision obtained is measured by the mean square deviation o,,~ (or standard deviation), that is by ana = dVN. From this it follows that % 2 = f ,io Iv" + W)12-2Y
f(x)1
putting into it the above expression of f(x) by orthogonal polynomials, the equation is much simplified in consequence of the orthogonality of U,(x), so that we have
0, 2 = -
moreover, multiplying it by c,," taken from (3) we obtain after simplification
1 - cm2 fz [U,(x)]') = N
,g=O
m+11(
To abbreviate, let us put
N;l N+m J em2 I I
m
Ymo z (-1)"*(2m+l) [ ";;r' I/ [Nz" 1. Finally we have R n-t1 1 0"' = - ): y' - x I urn, I c&2. N ;=o III=0 Remark 2. '&,, is easily computed by formula (4) if we have a table of binomial coefficients; moreover there are tables giving this quantity up to N=lGO and m=7, that is, up to a hundred observations, and for polynomials up to the seventh degree. [Lot. cit. 41, Jordan, e) p. 336-357.1 Remark 2. All quantities figuring in formula (5) are independent of the origin, of the interval, and of the constant C; consequently this formula is valid for all systems of orthogonal polynomials. If the approximating parabola is known in its form (1) then the problem is solved; but if it is necessary to compute a table of the values f(a+ih) for 6 = 0, 1,2, . . . , N-l, then the corresponding values of U,(x) must first be computed by aid of formula (7) $j 139. This also seems easy enough, especially when using the tables of the binomial coefficients; yet if N is large, the computation is a tedious one. At all events, the calculation would not be shorter if the U,(x) were expanded into a power series. The labour will be decreased considerably, however, if tables giving the values of U,,,(a+Eh) are available.43 But tables with a rangelarge enough would be too voluminous, and we shall see that they are superfluous, as by a transformation of formula (1) into a Newton series we can get the required values by the
Essher
':I I adopted this procedur; in my paper published in 1921, and later and Lore&z did the same.
454 method of addition of differences Q 23; and if an interpolation is necessary for any value whatsoever of x, Newton's formula will give it in the shortest way. Moreover by this method we shall be independent of the value of the constant C, that is of the orthogonal polynomial chosen. Transformation of the orthogonal series (1) into a Newton's expansion. Since the approximating parabola and the mean square deviation are independent of the constant of the orthogonal polynomial used, it is natural to transform equation (1) so that it shall also be independent of this constant, This can be done by a transformation into a Newton series. For this it is sufficient to determine the differences of f(a+th) for 5~0 and &=l. Starting from (7) Q 139 we get
Knowing the differences for t=O the problem is solved. The equation of the approximating parabola is
f(a+N = 44 + (i) AW + . . . + (i) A"f(al.
The numbers '%& may be computed by aid of (6) and a table of binomial coefficients, but there is a table giving them up to p = 7 (parabolas of the seventh degree] and to N = 100 ILoc. cit. 41, Jordan e) pp, 336-3531. Remark. Having obtained, by the above method, the Newton
455 expansion corresponding to the approximating parabola of degree N say y "f"(x), it may happen that the expansion corresponding to a parabola of degree n-+1 is desired. Then only the calculation of O,,, is necessary, and the coefficients of the new expansion will be
Acsf,+, (4 = A!+,,(a) + %+I,,,, On,, .
The work previously done is therefore not lost. Summary. Given the points x,y for x=a+Eh where (f = 0, 1, 2,. . * , N-l the equation of the parabola of degree n approximating these points is, according to the principle of least squares, or that of moments, f(a+<h) = f(a) + ( f ) Af(a) 4 ($)A'f(a) -;- . . . . i- [:I A'f(aj where A:'f (a) is the ,LL -th difference of f (a+fh) with respect to & for <=O if Al=l. We have h-1 ;. (g&, being given by formula (6) and
-
Moreover Pm. is given by formula (5) or the table in 8 141 The precision obtained is measured by nz " Cl where J+' (y') is the mean value of the y", that is Ey'/N. Remark. If there are two sets of observations y,x and z,x given for x=a+ch where t=O,l,. . , , N-l; and if we denote the mean square deviation of z by (I', and the orthogonal moments of L by ($,,I, then the coefficient of correlation between y and z will be
Particular case. Approximation of the values of y,x by a
456 function of the first degree, if y is given for x=0, 1,2, , . . , N-l. We have
Finally the mean-square deviation will be 3(N-1) &" N+l Therefore to obtain the required approximation by formula (7) it is sufficient to determine q, the mean binomial moment of the first degree; so that even in this, the simplest case, the approximation by orthogonal polynomials is preferable to the usual method. But the great advantage of the orthogonal polynomials, in shortening the computations, is shown, if approximations are to be performed by parabolas of higher degrees. Q 143. Graduation. by the method of least squares. If the observation of a phenomenon has given a set of values y(xJ where i=O, 1, 2, *.#,, N-l then the y(xJ will be affected necessarily by errors of observation; if the quantities y(xi) are statistical data (frequencies), then they will show accidental irregularities so that the differences Amy will be irregular. I? Ix d (y') -
457
The statistician generally wants to "smooth", to "graduate" or to "adjust" his observations, that is to determine a new sequence with regular differences, which differs as little as possible from the observed series. The simplest way of smoothing is the graphical method; but this gives no great accuracy. A far better method is that of the means in which, writing f (xi) for the smoothed value of y (xi), we put for instance fh) = & (y(xr-kh) + y(x,-kh+hj + . . . .
+ Y (xifkh) 1
The best method of smoothing is that of the least squares. According to this method, to obtain the smoothed value of y (xi), let us consider first the points of coordinates x,y(x) where x=x,-kh, x4-kh+h, . . . .q xi+kh. Then we determine the parabola y=f(x) of degree R which is the best approximation of these points according to the principle of least squares. Finally f(xi) will be the required smoothed value of y (xi). This will be repeated for every value of i considered. Proceeding in the usual way this is complicated, but when using orthogonal polynomials then the parabola of degree R approximating the given points may be written
E
-
where according to the notation of 8 139 we have N=Xk+I; a =xi-kh and therefore xi = 1/2(a+b-h) = a + M(Nh-h). But we have seen in 3 140 (formula 2) that and b = xi + kh + h
u,,,, [+h ) = 0.
458 Moreover, according to formula (13) 5 140 we find
Since in consequence of formula (3) 8 142 the coefficient czm in the expansion (1) is equal to Czm = (4mfl) %I _ C,,,,h In' I N;tm"")
hence /(xi) the smoothed value of y(xi) will be
To abbreviate, let us write
The number ;'zm could be easily computed by aid of a table of binomial coefficients; but the following table gives yarn up to parabolas of the tenth degree and up to N=29 points (kz14). The calculation of f (Xi) is now very simple:
all we need is to compute the mean orthogonal moments O,, , and f(xi) will be the smoothed value of y(xi) obtained by aid of a parabola of degree 2n approximating N=2k+l points. Remark. It is useless to consider parabolas of odd degree, indeed in consequence of formula (2) a parabola of degree 2n will give the same smoothed value as would a parabola of degree 2n+l. In the next paragraph an example will be given.
460 preceding paragraphs we have seen that in the calculus of approximation and also in that of graduation, knowledge of the binomial moments is needed. C. F. Hardy44 gave a very useful method for the determination of the binomial moments, which dispenses with all multiplication and may be executed rapidly by aid of calculating machines. If y is given for x=a+Eh, where 5=0,1,2,. . , , N-1, then the binomial moment of y of order m is given by
a 144. Computation of the binomial moments. In the
The method consists in the following: Denoting by y(S) the value of y corresponding to t ; jn the first column in a table, the values of y(E) are written in the reverse order of magnitude o f 5, that'is
YW-l),Y(N----21,
*** ,YU),Y(O).
the first line of every column we write the same In number y(N-1). Into the Y -th line of the 1~ -th column we put the sum of the two numbers figuring in the line r-1 of column ,u, and in the line 'V of column p-1. Therefore, denoting the number written in the line Y of column ,u by ~(v,,u), the rule of computation will be (1) Y h-4 = Y k-- L,u) + Qbw--I). The solution of this equation of partial differences of the first order, by Laplace's method of generating functions (5 181, Ex. 5) is the following:
Ybw)
r+1 = z 11-2 $ iv-i 1 I p-2
i=l
y (N-i).
The initial conditions are satisfied; indeed for v=l we get ;(id;)z;(N-1). Putting into the formula obtained v=N-/.~+2 -6 we get
o6 G. F. Hardy, Theory of construction of Tables of Mortality. 1909, London; p, S9 and onwards.
461 92(N-,U+2,/() = ._ i! /c--2 Y a5 I p3+2 I
Therefore the number figuring in the line N-,~+2 of column ,M is equal to the binomial moment of degree ,u-2. Hence if we want the binomial moments d,, , aI, . . , a", then we must compute n+l columns. The results obtained give by aid of formula (4), 6 141 the corresponding mean binomial moments 3; . Remark. If we put v=N into formula (2) we get the following moments
XZtl
z
h
I
x+yc-2 ,(4+2
I
Y(X)
that is the numbers in the last line of the table arc equal to these quantities. Example. Given the following observed values:
x=a,
x=a+h, x=a+2h, x=a+3h, x=a+4h,
y = 2502 ~~2548 y=2597 y=2675 ~~2770
x=a+5h, x=a+6h, x=a+?h, x=a+8h,
~~2904 ~~3064 y=3188 y=3309
The graduated values of y corresponding to x=a+4h are to be determined corresponding to nine-point parabolas of the second and of the fourth degree. To begin with, we shall first determine, by aid of the preceding method, the binomial moments aO, a,, . , . , a,. For this purpose we write in the first column below the values of y in reverse order, and in the first line of every column WC put the last value of y; the other numbers of the table are computed by aid of rule (1).
463 The constants ;'? and 1, corresponding to N=9 (nine-point parabola) were taken from the table of Q 143; we found i'z = - 1.81818182 and y4 = 1.13286713.
0 145. Fourier series. A function f(x) of a continuous variable x, with limited total fluctuation in the interval a, b may be expanded into a Fourier series. Let us write
The determination of the coefficients a,,, and B,,, is simple in consequence of the orthogonality of the circular functions. Indeed we have, if m +,a b 2.7m ( x - a ) aJ cos - b - - a
= x- o.!
b
-
2x
a 1. cosmt cos ,u;' dZ = 0
where 2~2.7 (x-a) / (b-a) has been put, In the same manner we should have
2% L I 0' 2n
sinm< sinff$ d$ = 0
and
-
0'
1
sin rnf cosp5 d: = 0.
The last equation holds for every integer value of m and p. Moreover if m is different from zero, then
b
(1. J
I
2rrm(x--a) 2 dx _ b--a 'OS - b-a 2n o'
cos"m$ dg = l,G(b-a)
and in the same way
a*
' I
.
b
sin
2.7m(x-a) b-a
'
dx = lh (b-a).
cos
Therefore, multiplying both members of (1) by we obtain, after integration from x-a to x=b:
2.7m ( x - a ) b--a
464 2 b - - f(x) am - b - a II. 1 2rrm(x-a) cos bdx
2.7m ( x - a )
(2)
'
in the same manner, integrating, we find (3) & = & :
multiplying by sin ~b-a 1" f(x) sin 27mEa) dx.
a n d
Putting these values into (1) we obtain the expansion of f(x) into a Fourier series. Approximation of a function f(x) of a continuous variable by a Fourier series of 2n+l terms. Let
II+1
f,(x)
= 1/2a,
n+l
+ Z a ,
m=l
c o s
2.7:(za)
+
+
IT Pm mrz,
27rm ( x - a ) sin b-a -
--
The coefficients a, and Pm are to be determined so that according to the principle of least squares
shall be a minimum. Putting into it the value (4) of f,,(x), the equations determining the minimum will be
a8 - x0 aa,
and
__ IO. ap, '
a?7
This gives 2n+l equations which determine the coefficients U, and p,,, . It is easily seen that in consequence of the orthogonality we obtain the same values as before in (2) and (3). From this we conclude that, expanding f(x) into a Fourier series and stopping at the terms a,, pn we obtain the best approximation attainable by aid of these terms. Putting into (5) the value (4) of f,(x) we get in consequence of the orthogonality and of the equations (2) and (3)
465 Example. In the calculus of probability and in mathematical statistics a function f(x) of a continuous variable is often considered, in which f(x) =y; is constant in the interval a+# <xc a+Eh+h; and this is true for every ititerval h, though the yg generally differ from one interval to another. For instance, f(x) is the probability that the number of the favourable events does not exceed X. This function is a discontinuous one; but since the total fluctuation is limited, therefore it is representable by a Fourier series. Putting into the formulae above b=a+Nh, we find u+r,,' 1 3 sin 2nm&-' dx
a
n
d Bmd&
5 YE icosT - cos :=0
N yf
2rrm(5+1) N I
finally
Pm = $ s i n T X0
sin ?!!!!w. .
In the same manner we should have cl, = yi cos m-zCX+ll N '
8 146. Approximation by trigonometric functions of discontinuous variables. Let us suppose that the numbers y(x) are given ,for x=a+th and l= 0, 1,2,. . . , N-l. A function f(x) is to be determined which gives for the above values f(x) =y (x). T h is will be done in the following manner : Starting from equation (1) f(x) = l/a. + 2: [p,sin 2"mfh-) + %cos -I]
where n is the greatest integer contained in N/2, the coefficients a,,, and pm are to be determined so as to have f(a+$h) = y(a+[h) for t=O, 1,2,. . . . , N-l. If N is odd, N=2n+l, then this gives N equations of the first degree with N unknowns to solve; if N is even, N=2n, then the
30
466 number of the unknowns is also equal to 2n, since then the term &sir&$ vanishes for every value of 6. The resolution is greatly simplified owing to the orthogonality of the introduced circular functions established in 8 43. Remarking that 2m+lI N and 2,~+1 S N, the formulae found there will be, writing in order to abbreviate t=(x-c)!h if m is an integer different from ,u: L 2.-rmE . 217~5 = 0 and 2nd i $firn3N sin - sin N po N g=o
Putting (3) and (4) into equation (I) the problem is solved. Approximafion. y(x) is given for x=a+th and l=O, 1,2, . . . . , N-l. A function 2;rm (x-cz) (5) f(x) = ? + Ifi 1 /ill, sin '*;p + u,,, cos ____ Nh I where n < M(N-1), is required so that it shall be the best
467 approximation of the given values, according to the principle of the least squares, that is, which makes 36) a minimum. Therefore the coefficients a,,, and pm are determined by the following equations - =O a a%l
a#
n
d
iT=O m
this gives the necessary 2n+l equations to determine the coefficients. Putting into (6) the value of f(x) given by (5) we obtain for a,,, and &,, the same expressions as before, (3) and (4). Moreover from (6) we deduce, in consequence of the crthogonality by aid of equations (3) and (41, that
(7)
8 = 'S? [f(x)]'- $ cfo2-$ 2:
%==a
(an2+/?,2).
Formula (5) may be useful for detecting some hidden periodicity of the numbers y(x). Q 147. Hermite polynomials. In the general case this polynomial of degree m could be defined by4s i H, = eL'# Dm[e-ns] but it is better to do it in the following particular case, in which the formulae are the simplest, and which is very suitable for approximation purposes, by aid of the probability function. (1) H, = H,(x) = e*B D"[e-"'n] . This may be written (2)
Hm = Am.mx* + Am.m-lxml + * * * * + Am.0
H, = DH,-~-xH~-~
writing by aid of (2) that the coefficient of xi in both members is the same, we get the following equation of partial differences: (3) &,i = (i+l)Am-I.I+~ -&-I,I-~ * According to Q 181 the conditions A,,, = 1 and A,,i = 0 if i > m or i < 0. are sufficient for the determination of the coefficients A,,i, Indeed, starting from these values a table of the coefficients may rapidly be computed by aid of (3): m\i 0 1 2 3 4 5 1 0 - 1 0 3 0 0 1 - 1 0 3 1 0 -1 0 10 1 0 - 1 2 3 4 5
470 Remark 2. The roots of the equation H,(x) =O are all real and are included in the interval -2 vi and 2 YL Expansion of a function into a series of Hermite functions H,e+/*. If a function f(x) and its two first derivatives are continuous and finite from -m to 00, and if f(-+_s) = 0
Df(+-) = o,
D'f(+-) 10
then f(x) may he expanded into the following convergent series: f(x) = I.c,+c,H, + c,H, + . , .I e+"*. WI Multiplying both members by H, and integrating from -90 to m we find, in consequence of the relations of orthogonality (11)) (13) 1 C", = irn H,f (xjdx. m! 1/ g-m'
To obtain the integral contained in the second member, let us introduce the power-moments of the function f(x) by A,, = I'- x"f (x)dx.
We have seen (formula 14) that in the expansion of a function f(x) into a series of Hermite functions, and stopping at H,, f,,(x) = [c,+c,H,+. . .c,H,] e-"*h the coefficients c, are determined by aid of the moments A,,, 'MI,..., '/l,, of the function f(x), Let US show now that the first n+l moments of f,(x) are the same as the corresponding
472 moments of f(x) ; and therefore f,(x) is an approximation of f(x) according to the principle of moments. To obtain the moment of order m of f"(x) let us multiply f,,(x) by xm given by formula (15) ; we find X"'f,, ( x ) (-l)"m! c, = xz ~--Hm-ni HI, c--x'~' i! (m-2i) ! 2'
Integrating from -co to O", in consequence of the orthogonality (11) we get x F--1) m m! cmepi \i-Sr i! 2'
Putting into it the value of ~,,,-~i taken from (14), it gives
xx
writing i+s=p:
1
i! S! 2i+S(m-2i-2s) ! '
(-1)" m! JLL.~~ .
!ZJ
,d 2!L (m--2p) !
m! Jk2:,
T S=="
(-1)s (;I.
The second sum is equal to zero for every value of ,L' except for ,u=O; but for ,u=O the preceding expression is equal to I,. Consequently, A, the moment of order m of f(x), is also a moment of f,(x). Q . E - D. Numerical evaluation of H,,(x)B-~'~~ I v--- Since there are 2~. no good tables giving the derivatives of the probability function zz T 277 II we have to compute H,,(x) (formula 4) and multiply the obtained value by z taken from Sheppard's Tables.46a Particular values c H = 0 H ?"Z (0) = I"'! m!2m
g-x'/2
To obtain the coefficients c, in (19) let us multiply both members of this equation by G,(x), and sum from x=0 to x=x; this may be done, if we put f(x) =O for every integer value of x such as x 2 N. In consequence of the orthogonality (15) we shall have
iii f(xlG,(x) x=0
= c, X:0 ~'G.l"w;
hence according to (18) the coefficient cn will be
w
', = 5 E, G(x)f(x). .
This may be transformed by putting into it the value of G, given by equation (4). We get C" = izo (-1)' ;I z . x=0 I n?i 1 f(x) finally introducing the binomial moments ,& of f(x) we find (211
C"
#+I (D
llC1 = ,zo (-1)' $ 'gn-i .
Knowing the coefficients c, the expansion (19) is determined. But in consequence of (3) this equation may be written in another manner: I22) /lx) = .fo (-l)"c,, A"vh=-4.
x
Remark. We may dispose of m so as to have the coefficient c, in the expansion (19) equal to zero. For this we shall put c, = '28, -m '-H,, = 0.
479 Then m will be equal to the mean of the quantities x; m= .~g~/&.
Example 1. Expansion of
into a series of poly-
nomials G, =co+c,G1 +c2G2+ . . . . +c,G,.
This expansion is identical with that of [G) y(m,x) into a series (19). Therefore the coefficients ci may be determined by aid of (21); but owing to formula (17) there is a shorter way. Indeed, multiplying (23) by G,. y J ( m,x) and summing from x=0 to x=00; in consequence of the orthogonality we find:
& (E] G,.y(m,x) = tic,. . my
Equalizing this result to (17) we have
m* n c,. = n! ( v I
and finally .(24)
! n 1 = 5 "i. (:) G,. .
X
I:
From this we may deduce the binomial moments of a function f(x) expressed by (19). Indeed, multiplying (24) by f(x) and summing from x=0 to x=00 we get by aid of (20) (25) di?" = x c,, mn-r . (n-v) ! v=o
n+l
Sum of a function f(x) expanded into a series (19). Starting from (19) we may determine A--'f (x) . For this purpose let us remark first that for x=0 the quantity (8) will be equal to zero. Indeed we have
w (m,-1) = 0. This is readily seen if we write y(m,x) in its general form
y(m,x) = emrn mx/r(x+l)
the denominator is infinite for every negative integer value of x. Moreover, according to formula (3) Q 18 we have
Therefore, if m=O every term of the second member will vanish except that corresponding to y=n-l-x; so that we shall have = (-l)n-1-x ';' . ( 1 Id x > n-l this quantity is equal to zero, (29) it follows that 1 WJ) s G, (m,x) y (m,x)dm = G-, VJ) ~(44
0
if x>n-1. Approximation of the function y(x) given for x=0, 1,2, . . , , N-l, according to the principle of moments by aid of f,(x) = [co + c,G, + c,G, + . . . + c,G,lw(m,x). Putting y(x)=0 for x L N, let us show that the coefficients CI corresponding to this principle are the same as those obtained by the expansion of y(x) into a series of G ,.y (m,x) function& when stopping at the term G,. According to (25) the binomial moment of f,(x) of order s is equal to
-21 da, = r+1 x v=o =+ m"-' (s-v) ! . ;
moreover cr is given by (21) cv = is ( - 1 ) '
Vfl
-$&-i,
Uihere &-i is the binomial moment of y(x) of order ~4. From these equation we conclude that
[S s = &lx kl)',s+i-v(,,,-.
i!
(S-V)!
'
"
Writing s+i--r=p we find
The second sum is equal to zero for every value of ,u except for p=O, and then we have
31
482
-7 P9& = cd;
for every value of s=O, 1'2, . , a , n. Therefore the corresponding n+l moments of y(x) and f(x) are the same. Q. E. D. Remark 1. If we determine the coefficients ci so that .&+= i [ f " ( X ) -Y(X)
x=0
l"lwh4
shall be a minimum, we find again the same values of cl as before. Moreover the minimum of @will be equal to &= x50 lY(x)12/y(m,x) - ri -$ CF.
Remark 2. Among the polynomials P, of degree n in which the coefficient of t t I is equal to unity, the polynomial which makes
2 Pn2y (m,x)
a minimum, is equal to m" G,/n! . Computation of the numerical uaIues. The function f(x) is expanded into a series of G,,y(m,x), or into a series of ky(m,x-rz) and f(x), corresponding to a given x is to be determined. Since there are no tables of G,y (m,x) nor of the computation of f(x) by aid of the preceding A% hx) formulae would be complicated. To obviate this difficulty there are two ways: A. We may start from y hx) G,, = h-1 )"A% (ma---R) and express the differences contained in the second member by the successive values of the function (5 6). We find G,,y(m,x) = 1: (-l)nti (7) w(w+--iI. Putting this into formula (19) we get (331 f ( x ) = Ii y(m,x-i) "3, (-l)"+i[y)cn.
-
In the same way from (26) we obtain
1
483
i+1
+=0
x
f(x)
=
co[l--lb&P)1
+
+ 2: y(m,A-i)
S, (-l)n+i (7') c,.
By aid of Peurson's Tables (5 18) giving y(m,x) the values of f(x) may easily be calculated. Z(u,p) is computed by aid of the tables incomplete r function (8 18); but if 1 is small and if the value of m is contained in the table of the function y, then we may determine
2+1 l--I(sP) = xs y Iv4
by adding the corresponding values of y. B. There is still a better way to compute the function f(x) corresponding to a given value of x. Indeed we may start from (4) and write G,(x) = i% (-l)"+l then formula (19) will give
n+l
z [;] 1
WI
and (26)
f
(
X
)
=
y(m,x) "5' q (;1 5 (-1)'
m
r--O
(S)iCs
1+1
(36) +
z f(x) = co[l-z(~,Pl I + (-1 j s+i (s-l) i ,
h 1+1 0i hd.1 x m' i,
i=O
Cs
i. are rapidly calculated, and there is I! only one value of y to be determined by the table; this is important in cases when y(m,x) is to be determined by interpolation. Example 2. Let us denote by P(x) the probability that an event shall happen x times in n trials, if the probability of its occurrence is equal to p at each trial. The formula of this probability was given by Jacob Bernoulli: The coefficients
Of i
404
P(x) = ( ; 1 pxq"-*
where q= 1-p. This is to be expanded into a series (19). In order to obtain the coefficients cl we must first determine the binomial-moments of P(x). We have seen in 8 136 that if u(t) is the generating function of P(x) then its binomial moment of order i will be equal to
8.' = D'UV) I i! 1 I=1
And since the generating function of the given probability is (q+pt)", therefore we shall have Qq =
I1
n i
pi.
Finally the coefficient c,. of the expansion will be (21)
To simplify we shall dispose of m so as to have cl=Q for this we put m= ~2, /OS" or m =np. This will give in consequence of formula (4'):
where G,. (n,n) signifies the particular value of G , corresponding to m=n and x=n. Finally from formula (4") we get
Q 14% Method of False Position, or Regula Falsi. The problem of the numerical resolution of the equation f(x) =O is identical with the problem of inverse interpolation of y=f (x), when y=O. The most important method of the resolution of equations is that of the False Position. We have already used this method in the inverse interpolations of paragraphs 131, 132 and 134, with slight modifications. Given the equation f(x) ~0; if f(x) is a continuous function in the interval a, b, and if f(a) f(b) < 0, then the equation will have at least one root in the interval mentioned. If b--a is chosen small enough, there will be only one root in it. Let us suppose, moreover, that the function f(x) has a first and a second derivative which do not vanish in the interval. These last conditions serve only for the determination of the error. The curve y=f(x) will pass through the points of coordinates a, f(a) and b, f(b). Let us consider the chord passing through these points; it will cut the x axis in a point whose absciss is x1. We have necessarily a < x, < b; x1 may be considered as the first approximation of the root. To obtain x, and the corresponding error 8, , we shall write Lagrange's linear interpolation formula: f(x) = Fa f(b) + s f(a) +%(x-4 (x--b)D"f(O where a < E < b. Putting into this equation f(x) ~0 we find
neglecting the remainder we obtain x1 ; the maximum 8, , of the remainder will be equal to the maximum of the absolute value of the error. Determining f (x,), there are two cases to be considered : First fW(x,) <O then, denoting the root by r we shall have (Figure 5) a<t<x, this will occur if Df (x) D'f (x) < 0. Figure 5.
Putting into formula (1) &x1, we determine x, the second approximation of r and 8, the corresponding maximum of the error. Necessarily we have Q < t < x2, and continue in the same manner. Secondly, f(xAf(b) <o then we have (Figure 6) x,<r<b this will occur if Df (x) Dzf (x) > 0. We put into formula (1) X, instead of a and determine x2 the second approximation of r. We have x,<r<b and continue in the same way.
488 It is possible to shorten the method in the following manner: If we have obtained, for instance, in the first case the approximated value x, and 6, the maximum of error, then, since in the case considered the error is positive,* we have
Xl -dl<r<xl
and we may apply the method in this interval. Since 6, is considerably smaller than x,-a therefore the second approximation will be far better. In the second case the error being negative we shall have x1 < rl < x, + 6, and we may apply the method to this interval. Figure 6.
Example 1. Let us choose for our first example that of Newton's equation given by Wallis f(x) = xa-2x-5 = 0 which has been solved subsequently to 101 decimal places."' Since 1(2)=-l and f(3)=16 therefore between x=2 and x=3 there is a root; this must be determined- to twelve decimals. It would be possible to put a=2 and b=3 and apply the method; but remarking that f(3) is very large in comparison with f (21, therefore the root will be near a=2. We may try to p u t a=2 a n d k2,l. Since f (2,l) ~0,061 hence 2 < r < 2,l.
* If x1 is the obtained value for I then by error we understand the difference x1-r. 47 J. Wallis, Treatise of Algebra, London, 1685, p. 338. Whiftder and Robinson, Calculus of Observations, p. 106 and p. 86 where 51 decimals are given. ~~2,094 551 481 542 326 591 482, , . . , .
489 Putting ~=2 and b=2,1 we obtain by aid of formula (1): - = 2,0942.. . . x1 = 2 + 1,061 The corresponding maximum of the error is also given by formula (1); we find 6, < 6. 10e4. Therefore 2.094 < x C 2.095. We repeat the operation, putting a = 2,094 and b = 2,095. If follows that f(2,094) = -0.00615 3416 f(2,095) = 0.00500 7375. Therefore from formula (1) we get x, = 2.09455 134228. . . , Determining the corresponding maximum of the error we find S2 < 2.10-j hence we shall have 2.094551 < x < 2.094552. Repeating the operation, starting from these values we obtain -. f(2,094551) = -0.OOOOO 53747 03234 f(2,094552) = 0,OOOOO 57867 28439. Formula (1) gives x3 = 2.09455 14815 4245,. . The corresponding maximum of the error will be 6, < 2, lo-r3 ; hence the problem is solved and we have x = 2.09455 14815 42 exact to the last decimal. 8 150, The Newton-Raphson method of numerical solution of eqoations. Given f (x)=0 let us suppose that between X=Q and x=b there is a root denoted by r, and that f (a)f (b) < 0. We will suppose moreover that f(x) is a continuous function whose first two derivatives are different from zero in this interval.
(41
490
We shall distinguish two different cases. First let 11)
D&d D"f(d < 0
(figure 7) then the tangent to the curve y=f(x) at the point of coordinates a, f(a) will cut the x axis at a point x=x1 such that we have aCx,<r<b while on the contrary, the tangent at the point b,f (b) will not necessarily cut the x axis in a point belonging to the interval a, b. Since we want to consider x, as the first approximation of the root r, it is generally advisable to start from the point a,f (a) if inequality (1) is satisfied. Figure 7.
An exception to this rule is the following. If Df (x) is large, that is, if the curve y=f(x) is steep in the interval, then the tangent to the curve at the point a,f (a) will meet the x axis near x=a and therefore the approximation obtained will he not much better than that of x=a; but in this case the tangent at the point b,f(b) may cut the axis nearer to x=r, giving thus a more favourable approximation. Hence, although the inequality (1) is satisfied, we may start nevertheless from the point b,f(b) if Df(x) is large. Secondly (figure 81, if we have (2)
Df(4 D'f(4 '0
then, to obtain certainly a number x, belonging to the interval a,b, we must start from the point b,f (b). This we shall generally
491 do unless the curve is very steep, in which case we may obtain a better approximation by starting from the point a$@). Having found x, we compute f(x,) and determine the tangent at the point x,,f (xl) which will cut the x axis at a point x=x,, serving as a second approximation of the root r. Then we compute f(x,) and continue in the same manner till the required approximation is reached. We shall proceed in the following way: First we expand f(x) into a Taylor series. (3)
f(x) = f(q) + (x--x,) Df (x,) + %(x-x,) * D"Qx,+t)
where 0 < E < x-x,, . If we start from the point corresponding to x=a then we put into this formula x,,=a; in the other case we put x0=6. Figure 8.
Omitting the remainder in formula (3) we obtain the equation of the tangent at the point x,,f (x,): this will cut the x axis in
The corresponding error will be
(5)
, D'f(x,+Sl ', = -9; (x,-x,,) L Df(r
The second approximation and the corresponding error are given by (4) and (5), if we put into these equations x2 instead of xl and x, instead of x0. Having obtained x2 we continue in the same manner till the required precision is obtained.
492 Example. We will take again Newton's example. Let f(x) =x3-2x-5 = 0. There is a root between x=2 and x=3; indeed, f(2) = -1 and f(3)=16. We have Df (x) = 3x' - 2 and D'f (x) = 6x. Since in the interval considered we have Df (x)DPf(x) > 0 we should start, as has been said, from the point 3,f(3); but since Df (3) ~25, the curve is very steep; it is advisable to start from the point 2,f(2) where Df (2) =lO. Since this is the second case, we shall have a < r < x, . Starting from x0=2, we find f (x,) =-1, Df(x,) ~10 and therefore according to (4) and (5) x1 = 2,1, 6, c 7/1000. Starting from x,=2,1, we have f(x,)=O.O61, Df(x,)=11,23 and therefore x2 = 2.0945681 , . .
18,1<2/105,
Starting from x,=2,09457, we obtain f (x,) =0.000206694976 and Df (x,) = 11,16167 0455 and therefore x3 = 2.09455 14817 26, Id, I <3/ 10'0. § 151. Method of Iteration, Given the equation f(x)=O, let us suppose that it has a root x=r in the interval x,, x,+h. If we write f(x) = f,(x) -fr,(X). Then the root of f (x)=0 will be equal to the absciss of the point in which the two curves y=f,(x) and y=f,(x) meet. We will suppose that the decomposition has been made so that the second curve is steeper than the first in the interval considered; that is
I Df, (x,,) I ' I Df, Ix,) I.
The method of iteration described below is applicable if it is possible to determine the algebraic solution of y=f,(x) and obtain x=~~(y), Then we start from the equations
493 Y =f*W and x = e(y) and put x, into the first equation to obtain y,,=f (x,) ; then putting y,, into the second equation we get x, = q2(yO). This is the first approximation of r. Since generally we have b-xx,1 < It-x01. The second move is to put x1 into the first equation and get y1 , which gives by aid of the second equation x2, the second approximation of the root. Continuing in this manner we reach the prescribed precision. Figure 9. . Figure 10.
We may consider two cases; first, both curves are increasing (figure 9), or both are decreasing, and therefore
Df, (4 Df, (xl ' 0
in this case we shall have x,<x, <x,< . . . . . . cr. Secondly, one of the curves is increasing and the other decreasing (fig. lo), and therefore and we shall have
494 error committed when stopping at the term x,; in the second case since xln <r < ~s"+~, therefore the error is obviously less than 1 x2n-x2n+1I Example. The equation f(x) = x3-2x-5 = 0 has a root between 2 and 3, since f(2)f(3) < 0. Let us put f,(x) = 2x+5 and f,(x) = x3. Our condition, that the second curve shall be steeper than the first in the interval 2,3, is satisfied. Indeed (first case)
Df, (xl = 2
and therefore
a n d Df,(x) = 3x2 in (2, 3).
Df, (4 ' Df, (x)
3
The second condition was that it must be possible to solve algebraically y=f2 (x). We have
x
= Voz(Y)
= I/Y.
Starting from x,=2, the first equation will give y0=9. For this value of y, the second equation gives x1=2:08. . , . The value of x1 put into the first equation gives ~~~9.16, and this by aid of the second: x,=2.094, and so on. The operations are given in the following table: i Xl 2 2.08 2.094 2.0944
If the roots of the characteristic equation (1) are all different, say x1, x2, . . . , x,,, then we shall see (Q 165) that the solution of (2) is equal to y(f) = CIXl + $X2' + . . . . + C"X,' . If we denote by x, the root whose modulus is the greatest, tben it results that (3)
x1+ c2x2 3 I 1 = Xl ,
lim Y('+l) = f=0 Y(f)
f+....+ -p(z)'
=liln lim
1+"1~]~
+,*..+
; ($1'
Hence if we determine cp(f+l)/y(f) for f sufficiently great, we shall have the value of the root with the required precision. It is easy to show that equation (3) holds even if the roots of the characteristic equation are multiple roots. If some of the roots of (1) were complex (except the root x, with the greatest modulus), we should have the same result; but if x, were a complex root, then, as the root conjugate to x1 would have the same modulus, the ratio (3) would not tend to a limit but would oscillate. To determine q(f) by the aid of (2) we may start from any n initial values of this function, Let them be y(0) = v(l) = (p(2) = . . . . = ~(n-2) =O a n d ~(n-1) = 1 ; then the difference equation (2) will give
y(n) =---ah1 = - a"-lY (4 --a,-,dn+l)
By this method we may also obtain the root whose modulus is the smallest. For this we have to put into (1) x=1/y, and apply the method on the equation y(y) =O obtained. 8 153. The Ch'in-Vieta-Homer method for solving namerical equations, This method was known in China as early as the XIII Century. 48 It was given by Ch'in Chiushao in his book on the "Nine Sections of Mathematics" published in 1247. The method was first used in Europe by Vieta about 1600. It is probable that he had not heard of the Chinese method, but that he rediscovered it, since his procedure is more complicated than Ch'in's. Indeed it was considered as very difficult, was seldom used, and soon superseded in Europe by the Rule of False Position or by other methods. In 1819 Homer published a numerical process for carrying out the method, which was very convenient and in consequence it became widely known and is now generally considered as the
48 Y. Mikumi The Development of Mathematics in China and Japan, Leipzig, 1913, pp. %-I?.
7
497 most practical of the methods for solving numerical equations. Horner's process differs but very little from that of Win. It consists in the following: Given the equation f(x) = 0, let us suppose that by means of a graph or in another way, we know that this equation has a root x,, between r,, and r,+l ; where r, is an integer. Therefore we have f(r,lf(rO+ll < 0. The method consists in the following operations: First we deduce from f(x) =O another equation, such that the roots of the new equation shall be equal to those of f(x) =O but each diminished by rO, This equation will be f(x+r,,) ~0, which expanded into a Taylor series gives
(11
f (x+r,) = f (r,) + xDf (rO) + , . . + 5 D"f(ro) = 0.
From this, an equation is deduced, such that its roots are each ten times larger than those of (1). This will be
expanded into a Taylor series, and multiplied by 10" it gives:
_'^.
(2) 10"f [rO +G] = lPf(r,,) +lCPxDf(rJ +,.,+sDf(r,). Since equation (1) has a root in the interval 0, 1 therefore (2) will, have a root in the interval 0, 10. Let us denote this root by x1. From (2), by inverse interpolation we obtain approximately -1Of (rJ
Xl Df(ro) .
-
If rl < x1 < r, + 1, where r, is an integer, then the approached value of the root obtained by the above operations is x0 = r,, + 6 + . , . . To have another digit of x0 we start from (2) and determine an equation (3) whose roots are those of (2) diminished by rl,
32
and then an equation (4) whose roots are each ten times larger than those of (3). From this last equation we deduce Y* < x2 < t* + 1 and we shall have
and so on till the prescribed precision is attained. Let us show Homer's method for determining the coefficients of the transformed equations in the case of an equation of the fifth degree. Given (5)
f(x) = F + Ex + Dx* + Cxa + Bx' + Ax3 .
The coefficients of the equation whose roots are those of (5) diminished by r,, are
The equation, whose roots are each ten times larger than those of (6) will become, if multiplied by 106: (7) lo5 f [rO + $) = 105 F, + 10' E,x + lo3 D3x2 + lo* C4x3+ +10B,x4+Ax5=0. From this we get by linear inverse interpolation
If we have rl <x1 < rl + 1, where rl is an integer, rl will be the second digit of the root. Then the above operations are repeated till the required number of digits is obtained. The error will always be less than one unit of the last decimal. Example. Given f(x) = x372x-s =o. Since f (2)f(3) < 0 therefore this equation has a root x0 in the interval 2, 3; so that we shall put r,,=2. To obtain an equation whose roots are those of f(x) =O but diminished by 2, and multiplied by ten, we will perform Horner's computations:
500 1 0 2 2 2 4 2 6 The transformed equation will be %3+60x2+1000x-1000=0. From the last two terms we conclude that r-,=0; so that there is no need to diminish the roots by rl . We make the rooti ten times larger and get (8) x3+6OOx~+lmx-lOOOOOO=0. - 2 4 2 8 10 . - 5 4 - 1
From the last two terms we deduce rz=9. To determine the coefficients of an equation whose roots are the same as those of (8) but diminished by nine and multiplied by ten; we follow Homer's method and find 1 600 9 609 9 618 9 627
SO
The last two terms give r3=4. Now we diminish the roots of this equation by 4 and multiply by 10, using the same method as before. We have
501 1 6270 4 6274 4 6278 4 6282 Therefore the fifth equation will be x3 + 62820x2+1115450809x-6153416ooo = 0. This will give r,=5; and we could proceed in the same manner as before; but let us suppose that the root is only required to six decimals exact; then we may obtain the last three decimals by simple division. In fact we may write 6153416000 -62820x* -x3 x-1115450800 ' The term in x3 is less than 216 (the root being less than 6), hence if we neglect this term the error in x will be less than 2/105 and that of the root x,, less than 2/10". Moreover the term in x2 is less than (62820) (36) =2261520; therefore neglecting this term too, the error of x will be less than 3/103, and that of x0 less than 3/107. Accordingly we have r, = 6153416000/1115450800 and finally x0 = 2.094551 65. The equation of this example has been solved by this method, in 1850, to 101 decimals. One of the advantages of the method is the following: If for instance the root is exactly equal to two figures, then we obtain these in two operations, whereas by the other methods a great number of operations is necessary to show that the root is approximately equal to these two figures. If the root to be determined is greater than ten, it is advisable to deduce first an equation whose roots are the same as those of the given equation, but divided by 10k; where k is = 5.5165 11104309 25096 11129396 , 25112 11154508 -50671000 44517584 -6153416
502 chosen so that the root under discussion should be in the interval 0, 10. For instance, equation a& + e * e . + qx + 00 = 0 will be transformed into
a cl&" + -fg a?-' + . . . + lo;;-k x + -=o 12"
and then we proceed as before. Ch'in Chiushao's method differs but little from the preceding. Indeed the only difference is that if the first digit r, of the root of f(x) =O is found, then Ch'in first deduces an equation whose roots are ten times larger than those of f(x) ~0, and secondly deduces from this another equation whose roots are diminished by iOr,. The result will be the same as in the method described above, moreover, Ch'in's computation is exactly the same as that of Homer. Indeed, starting from the equation -
Ax4+BXR+CxZ+Dx+E=o
we shall have
A
1OB
1O'C
lOAt, Bl iOAr,
4
lOB,r, Cl lOB,r,
c,
1WD lOC,r,, Dl
lOC,r, 4
104E lOD,r, El
lOAr, 47
lOAr, B4
1 O&r,, c::
and the new equation will be AxJ + B,x" + C,,x" + D,x + E, = 0.
Example. Ch'in solves the following equation
X4
- 763200x" + 40642560000 = 0.
503
It is easily seen that 100 <x,, < 1000 therefore we deduce another equation whose roots are those of (1) but each divided by 100; so that the root will be between 0 and ten. This equation is x4 - 76,32 x2 + 406,4256 = 0 (8') since !(8)f(9) < 0 therefore there is a root between 8 and 9. Ch'in's computation for multiplying the roots by ten and diminishing them afterwards by 80, will give 1 0
80 -7632 6400 0 -98560 -98560 925440 826880 -7884800 -3820544
Since the last term vanishes, hence 4 will be the last digit of the root, so that the root of equation (8') will be 8'4 and that of the given equation x, E&U).*) Q 154. Root-squaring Method of Dandelfn, Lobetchevsky and Graeffe. Given the equation: (1) a&" + U"-#-' + . . . + alx + 0, = 0.
l ) dlihmi's interpretations of Ci'ln's processes, in the book quoted above, are not quite correct, and in consequence of tkis the coefficients of tke equation correspondiip to (9). given on page 77, are erroneous. Indeed the coefficients of x and x' are ten times smaller and the absolute term is 10000 larger than they should be.
it is obvious that multiplied by %(x+x1) (x+x,) * * ' (x+x") = 0 it will give the required equation, if we put into the product obtained x instead of x2. But the last equation is identical with (2) 4,x" - U"-1X n-l + . , . . + (-l)"Uo = 0
therefore if we write the new equation in the following way: b,xn + b,-lxn-l + , e . a + b, = 0 (3) then we shall have 6, = (-l)"-" Uv2 + ,z, 2(-l)"+y-i
U*e+i
a&n-i.
The upper limit of the sum in the second member is equal to the smaller of the numbers v+l and n-r+l. Starting from (3) we may obtain, repeating the same operation, an equation whose roots are x,~, x,~, . , . , x,,~; and repeating it again several times we get an equation whose roots are
2' Xl x2 2m I...., 2m x, .
Let this equation be
u&xn + w,-lx"' + . . . . + w,x + W" = 0. (4) Moreover we will suppose now that the moduli of the roots are unequal, so that we have in descending order of magnitude
Let us suppose first that the roots of equation (1) are all real; then we must have w,Jw~ < 0. In the same way we have lim
In- (D
X
X~""Xj*"
x1 2'
x2zm
= lim
m=-
on-2
w,xl 2" x2 2m-
-1
and therefore
2" Xl X2 % 2m %-2 r-.d -
or
a.nd finally
X V2'" N Qb8-w %-V-l
starting from these equations the roots are easily calculated by aid of logarithms. In this way we get in the same computation every root of equation (1) without needing any preliminary approximate determination. This is one of the ad&ages of the method. There is but one small difficulty; the method gives only the absolute values of the roots. The signs must he found out by putting the obtained values into the equation. Often Descartes' rule giving the number of the positive and negative roots is useful, and sometimes the relation I: xi = - u,~/cJ,, gives aho some indications, It is advantageous to make the computations in the following tabular form:
1
507 If the coefficient corresponding to X x12" is only the square of the coefficient corresponding to X $'"+' then we stop the computation, since by continuing we should get the same value for x,. In the present case this happens at the 64 -th power. For the determination of X xi2'" x/" the computation could have been stopped at the 8 th power; therefore to minimise the error we will choose this value for the determination of x2. To have x, we start, for the same reason, from x,x,x,=14. We have log x1 z d log 8..8971.1043 = 43.94924/64 = 0.68671
Since the numbers of the first column oscillate between positive and negative values, therefore we conclude that the root with the greatest modulus is a complex one. So that we have according to Case I, - 2 = 2.0, lo cos 16$P, = - 1.96497.10" 2 = e13" = 2.32829.10"' @I2 x3 = 50. From the second we obtain ,oi = 5,OOOO. The third equation divided by ei2 gives I x3 I = 2. Finally from the first we deduce cos 1*, = 1,96497.10'0 515 o r ; ql=--- + 8O 7'48"
where K is equal to one of the values 0, 1,2, . . . ,7. T O determine k we choose one of these values, determine the corresponding root, and put it into the given equation: this we repeat till the equation is verified. But generally there is a shorter way. For instance, in the present example we know that the only real root x, must be negative, since for x=---o0 the first member of the equation is negative and for x=0 it is positive. Therefore x,=-2. Moreover the sum of the roots must be equal to 6, that is, x3 + ~Q,COS~, = -2 + lOcos~,, = 6 this gives rp,=36O 52' 12" corresponding to &=l. Finally the roots are xl =4+3i, x2 = 4--i, x, z-2.
510 roots whose moduli are the same, the treatment of the problem is somewhat different. Let us suppose for instance that we have
Multiple roots. If the given equation has multiple roots or
I x1 I = I x2 I > i x, i .
From our computation table we conclude that
-- t!?kL
wl
=2x12m;
!!kg2
=X12n+q;
-?!A!
= X32mX12m+q,
On the other hand if we have
Ix11>Ix21=IxsI then
0,1 % =x 2". 1,
*n-2 __
%
=zx 1
2" x2
2mi
%-a --=x wl
2m
1 x2
p+t
*
Therefore if in a column of the table of computation the number corresponding to the 2m -th powers does not tend to the square of the number corresponding to 2'"+' but to half the square of this number, then we conclude that there are two roots corresponding to this column whose moduli are the same. There is no difficulty in deducing, in the same way, the rule ' corresponding to any other case of multiple roots. Example 3. Given the equation
Since the number of the first column corresponding to the 64 th powers is approximately equal to half the square of the number corresponding to the 32 nd powers, we may stop the computation, concluding that there are two roots having the greatest modulus. Ix,1 = Ix,J > Ix,I. Hence we have
2X,64
-
= 6,870:1 OS*.
This gives I x1 I = 3$00. In the second column we should get the same number from x13* = 3,4335:103o. Finally x,x,x,=18 gives I x3 I = 2. From x1-l-x2+x, = 2 we conclude that x1=3, x2=-3 and x3=2. We have seen that in the Win-Vieta-Hornet method the fewer figures there are in the root, the less is the work of computation. The above example shows that the computation in Graeffe's method is independent of the number of figures by which the root is expressed, and this is a drawback; but the method is nevertheless very useful, especially for the determination of complex roots.
is to be determined from x=u to x=b. The problem becomes univocal if the condition is added that the function f(x) shall be a polynomial of degree R. Then by aid of Langrange's interpolation formula (Q 132) we obtain this function n+t f(x) = z WlYi(Xl (1)
i=O
In this expression the terms: x-xi in the numerator and x,-Xi in the denominator are suppressed. The required integral will be I= Jb f(x)dx _. ;; Y(Xj) JLJx)dx. a a To determine this integral we will introduce a new variable f that the limits of the integral shall be zero and & ,We put (3)
SO
(b-4 x=a+- t n
The transformation gives
and
Xi=U+
P----4 f.L. ___ n
(f--to) (f-f,) , . . (f-f,) Ux) = Ir,f,) (f,-f,) ' . * K--f"l This is of the same form as the corresponding expression of X; we shall denote it by 9$(f). As we have dx=(b-a)df/n therefore Z = (&a) yg y(xi) $ oj".9,;i{f)df. If the same values of f,,, f,, . . . . , f, occur often, it is useful to construct a table of the integrals
1
513 +,J 2$(t)dt f o r i=O,l,...,n
which are independent of the function f(x). Having these values, the computation of the integral I is reduced to a few multiplications and additions. The above integral has been determined for several systems of t, , as we shall see. Formula in the case of equidistant values of x. Putting b-a=nh and x,=a+ih we have ti= & = i. The integrals corresponding to these values of t have been denoted by
514 Putting into this Q=O, b=l, f(x)=1 and y(xJ=l we obtain an interesting relation between the Cafes numbers: I c no i- C", + * * ._. + C", = 1. It can be shown moreover that C n.nf
= CM.,
Determination of the Cotes numbers by uid of Stirling numbers. If ti=i, then Lagrange's polynomial Z&(t) will become
z,(t) = (i] [ :zg
i+l
Replacing the binomials figuring in this formula by their expansion into a power series (Q 55) we obtain if, n > i:
$
L$i(t)
= R"--n(i) I (n-i) ! W5
n-i+1 XI Sl bS;-i t'(n-t)".
The integration, if t varies from zero to n, gives, according to Q 24,
In the case of i=n we have
Integration from t=O to t=n gives:
c,,= + 2' n!!. vi1 v+1 If the Stirling numbers of the first kind S: are known then the Cotes number C,, may be computed by aid of this formula; but we may obtain them. still in another way. By aid of formula (5) of Q 89 from (6) we get
C"" = $ [Yn+l(n) - v~.+1(0)1
where v,,+~(x) is the Bernoutli (Q 89) of degree n+l. polynomial of the second kind
7
515 In consequence of the formulae of p. 269 we may write (9) c,,= g1- lz' IW-&+,[l+(-l)"]).
On the other hand if the Cotes numbers C,, are known the sum of the coefficients bi in the Bernoulli polynomial of the second kind may be obtained by (10)
i=O
s
I bl I = 2 - (2n-1) c,,,, 21-* .
Examples. If C,, , from (7) it follows that
then n=3 and i=2, and every term will vanish except those in which ,u= 1; since Si = 1, hence we shall have c32 = 1
p1 s; 3 3 -z *?I (v+2) (v+li = 8'
If c,,,
and from (9) by aid of the table of p. 266 we find
In Q 131 we have seen another method for obtaining these numbers (Formula 9). Application of the Cotes numbers. Trapezoidal rule. If n=l then C,,=C,, = '/z and the integral (4) will be
By the aid of the above five equations we may determine the coefficients Cl. For instance we find l ' c,= ggj\8 1 [y(c+2)+y( 4-2) I-16ry(a+3)+y(a-3)]-130y(a)}
c4=
&j :4~~@+3)+~@-3)~I
--~TY(~+~)+Y(~-~)]+~OY(U):
Now we may determine the integral of f(a+x) from a-3 to a+3. It will give I= -j J.&+x) dx = 6c, + 18C, + 7 C, .
Putting into it the obtained values of C, , C, and C, we get
1
517 (2)
1 = f
{110~(4 +8lTvb+2)
+~@--2)1+14[~(a+31+
+ Yb+3)1~~
This is Hardy's formula of numerical integration. Weddle's formula is obtained from (2) by adding to it a quantity equal to zero, therefore it will give the same values as Hardy's formula. The quantity added is the sixth difference of f (a+x) divided by 50; since f(a+x) is of the fourth degree, obviously the sixth difference is equal to zero, According to 8 6 this quantity is equal to
Now f (a+l) and f (a-l) should be determined by aid of equation (l), but if we do this, then the obtained formula becomes identical with Hardy's formula. Weddle's formula is obtained from (3) by putting simply fb+l) = y(a+l) and f(a-1) = y(a-1) which is inexact. 8 157. The Gauss-Legendre method of numerical integration. This method differs from the preceding methods especially in the following: The given n+l points through which the parabola of degree n shall pass do not correspond to equidistant abscissae but to the roots of Legendte's polynomial of degree n+ 1. The roots of the polynomials of the first five degrees are given in 5 138. The equation of the parabola is given by Lagrange's interpolation formula 8 132: n-t1 f(x) = iz Li(X)f (4 (1) where
then it will represent any curve passing through the given n+l points. If the integral of (1) is required from x=-l to x=1, then we must determine the integral of L,,(x). We shall have
These numbers may be determined once for all and given by a table. They may be considered as Cafes numbers corresponding to the Gauss-Legendre method of numerical integration. The numbers Ani are always positive; this is an advantage over the numbers corresponding to equidistant abscissae, which as has been mentioned may be negative. Table of the numbers Ani. 0 1 2 3 4 1 0.5555556 033478558 0.2369268 1 1 0.8888889 0'6521452 0.4786286 2 0,5555556 0.6521452 0.5688889 3 . 4
0.3478558 0,4786286 and
0.2369268
Remark. It can be shown that Ani=An,n-i @@+1 t: Ani = 2. i=o Finally the integral of f(x) will be 1 (4)
- 1
s
f(x)
dx
=
4,~
(4
+&Y
(~11
+
l
0
.
+
&Y
(x.1.
Gauss introduced this method because the value of the integral (4) is the same as that corresponding to a parabola of
519
degree 2n+l passing through the given n+l points and through any other n+l chosen points. Indeed, if we add to (1) the following remainder
4+, = ?;I;; n .
Qnb)
where Q,,(X) is a polynomial of degree n, then, first the curve will still pass through the given n+l points; as 2"+, is equal to zero at these values; secondly, we may dispose of the coefficients in Q"(x) so that the curve shall pass also through the other n+l chosen points. Moreover, in consequence of the orthogonality of the polynomial X,(x) we have j1 &+, (%I -1
QnIx)
dX
= 0.
From this we conclude that the integral corresponding to the new curve will be equal to (4). L. Fejhr's step parabola. So Fejdt disposes of the polynomial Qn(x) so that the tangents at the points of coordinates xi, y (xi) are parallel to the x axis. He found
where the xi are the roots of X,,, (x)=0; and y(xi) the corresponding given values. The advantage of this formula over the other is the following. If a given continuous function F(x), is approximated by the step-parabola (5) then if n is increasing indefinitely the curve (5) tends to F(x). This is not so in the general case. Q 158. Tchebichef's formula for numerical integration. Given xi, y (XI) for i = 0, 1,2, , . . , n, if a polynomial f(x) is determined so as to have f (Xi1 = Y (Xi1
:a L . Fe@: AZ interpolbci&bl. M a t h e m a t i k a i 6s Term&szettudom&nyi l?rtesltB. Vol. 34. 1916. pp. 209 and 229, Uber Interpolation, Nachrichten der Gcsellschaft der Wissenschaften. Giittingen, 1916.
520 for the given values, then according to Lagrange's formula (5 132) we have n+1 f(x) = is &ix) Y (Xi) and the integral of f(x) from x=-l to x=1 will be jf(x)dx = "i y(q) / L"l(X)dX -1 - 1 or if we denote the integral in the second member by &I (the Cotes numbers corresponding to this formula of numerical integration), then we have
1 (11
-1
s
f(x)dx
=
n+1
i=l
I:
B"iy(xl).
Since the numbers B,i are independent of y(xi) therefore putting y(xJ=l and f(x)=1 we find *+1 x B,, = 2 kl whatever the system of values x, , , , . . , xn may be; but the B,i themselves depend on these values. If the quantities y(xJ are results of observation, then owing to the errors of observation, which necessarily occur, they will not be absolutely exact, and consequently the integral (1) will be affected by the errors. Denoting the error of y (xi) by &it the error E of the integral will be E = &A + Bnl~~ + . e . + &A,, Supposing that the probability of the error Ei is independent of i, and denoting by 0 the standard error of y (xi), then the probability P(A) of having I E I < 1, as shown in the Calculus of Probability is given approximately by
where we have
521
Now Tchebichef wanted to dispose of the numbers B,i so that the probability P(1) should be maximum. This problem is identical with the following. Given I: B"i = 2, to dispose of the B nl so that Z Bg$' shall b e minimum. This will be obtained when the equations
a aB [X B,i'-tl(I: B,i-2)] = 0 RI
are satisfied for every value of i. we get
2B,i--1 = 0
or B,i = l/&J. The numbers B"i are all equal and therefore B,i = 2/(n+l). Finally the integral will be
by expanding the second factor into a series of inverse powers; neglecting, after multiplication by x", the fractional parts. Let us denote the polynomial of degree n by T,. Tchebichef has shown that the roots of the polynomial T,+l satisfy the conditions (2) .61 If y(x) is a function of x such that y (O)=O then the expansion of ey will be ey = 1 + $ [Dey] H -+ x$ [D'ey]y=O
+ $ [D"ey],=o + . . . .
The expressions Dmey are given by formulae (6) of $ 72. If we put
61 Tchebichef, Oeuvres, Vol. II, pp, 165-180.
522 y=n z2r 5 F, 2r(2v+l)
*
Writing in the exponential of (3) 1/x=1 we shall have
where r is the greatest integer in (m+l)/2. From this we conclude that [P+ly],=o = 0 and [D2my],=o = - 'ammo)'
Therefore from the above-mentioned formulae it follows that and ID 2m+1ey]2=.0 = 0
This method is only useful if the roots of the polynomial used are all real; it has been shown that in certain cases some of the roots may be complex; for instance if n=8 or if n=lO. Let us mention here the numerical integration corresponding to the "Tchebichef abscissae" given by
Fe@* has shown that in both cases the correspondingcotes numbers are always positive, like those in the Gauss-Legendre
* L. Fejdr, Mechanische Quadraturen mit positiven Cotesschen Zahlen. Mathematische Zeitschrift, Berlin, 1933.
524 method (g 157) ; moreover that if the abscissae are chosen so that the corresponding Cotes numbers are positive, then the integral obtained by numerical integration will converge to 1 I F(x) dx: 9 159. N&Cal f&ration of functions expanded into a series of their differences. In the preceding paragraphs we have integrated Lagrange's formula in different particular cases; now we shall proceed to the integration of series expressed by differences. 1. Newton's formula. Given f(a), Af (a), . , , kf (a), the integral of the corresponding polynomial of degree n is to be determined. We have
7.' determine ('f(x)dx we need the integrals of the generalized binomial coefiicients. These we may obtain by expanding the coefficients into a series of powers of (X-U) by aid of the Stirling numbers of the first kind. (Q 55).
therefore
I
x - a = -$ "i' srn (x-a)*, h"'-" m h y=O
1
Particular cases. The most important particular case is that in which the integral is taken between a and a+nh. Then we obtain a+*t/, (3)
B
J'
fix)dx=
h 'sl gf(a) "g sw w+l m! *' y+1 nt=O ,=o
If we put in this formula n=2, it becomes identical with Simpson's formula; and for n=4 with Boole's formula (5 155). We get a more simple expression if the limits are a and a+h. Since according to formula (7) $j 89 we have . 1 tn+1 - x d?" = b,, m ! vzl v+l
525 where b,,, is a coefficient of the Bernoulli polynomial of the second kind, hence from (2) it follows that a+/# a+1 f(x)dx = hmzo b,,Pf(a). (4) 1 aThis is a particular case of Gregory's formula. 2. Everett's formula. Given f(O), f(l) and the corresponding central differences ij"f(O), h"f(l), , . . , , h2"f(0), a"'f(l), the integral of f(x) is to be calculated. In Q 130 we had (1) (5) f ( x ) = ;,g (;T1) a"+(l)- y$ ["ztr;') Pf(0).
In 8 89 we have seen that
where Y~+~(x) is the Bernoulli polynomial of the second kind of degree m+l, Therefore from (5) we deduce (6) 1 * f(x)dx = :g y,v+,(x+")tP f(l)-*+I x &CO
'/.'21.+2(X+V--1)8~1~f (0) + k.
Let us consider first the particular case of this integral if the limits are 0 and 1. Since Ayi(x) =vi-, (x), we have (.'f (x)dx = "2' y2,.+, (v) 6"vf (1) - %' yzr+1 h-1) a"#, f (0); ,,=o B---O
0'
remarking moreover that
Y2*.+1 b) = - Y2v+1 b---l)
finally we get (7) (.'f(x)dx = 2; v2,+l(x)16"lf(l)
0'
+ i-jzP f ( O ) ] .
3. Numerical integration by the Euler-Maclaurin formula. According to 5 88 this formula may be written (the interval being h):
526
(8)
:
jbf(u)du = h $ f(x) - j, yf [Dn-lf(b)-D"-lf(a)]
where Ax=h, and b=a+zh, z being an integer, and 0 < 6 < 1. The numbers B, are the Bernoulli numbers (Q 78). Given are the values of f (a+ih) and the derivatives of f(x) corresponding to x=a and to x=b. Since B,,, = 0 if n > 0, therefore formula (8) may be written:
where Au=h and 0 < gh < z or 0 < 5 < n. The numbers bi are the coefficients of the Bernoulli polynomial of the second kind (8 WExample 1. [Whittaker and Robinson, Calculus of Observations, p, 1451. We have to determine
A" + = (x+m),,, *
(-1)" m!
527 Therefore from formula (10) we deduce
The terms in the three brackets will give
-105
100'
I$= O.OS?!XJ
95-OBOOOOO 76-O.OOOOOO = 0,048790 18.
01 =
There is an error of one unit of the last decimal, since the integral is equal to log 105-log 100 = 0.048790 169432. . . Q 160. Numerical resolution of differential equations. The best method is that of 1. C. Adams. Let us apply it to a differential equation of the first order. Given the quantity ya corresponding to x, and the equation (1)
x2*
ai1 by) + af, (XSY) D'y 3 --i&Dy = +$ +f, % =f,(x,y). aY From this we obtain in the same manner Day = f:, (x,y), and SO on: D"y. First, putting x0 and y0 into the obtained formulae we compute the values of Dy, , D"y,, D*y, and D4yo. Then we may obtain y(x) by the aid of Taylor's series: (2)
Y(X) =
m=O
m (xiT)m 2
Dmy,,
.
But it must be remarked that x-x,, should be chosen small enough, so that stopping formula (2) at the term m=4, this latter should be negligible; hence putting x =x,+5/1 the quantity h must be small, and moreover let us say: 6 < 5.
528 Now we may compute y1 , y2, ys , y4 by aid of formula (2), and then DyI , Dy, , Dy, , Dy, by aid of (1). Finally let us form by simple subtractions the following table of the differences of hDy,: ~DY,
hDr, hDy, hDu,
~DY,
!hDn AhDr, a $hDy, $hDys
;"hDYo +QDyl
i3hDYo e3hDyl
G4hDy,
+2hDYa
Consequently h should be small enough, SO that A'hDyi shall also be negligible. The expansion of Dy(a+Eh) by Newfon's backward formula
(81, § 23, &es Dy(a+th) = $ [ '+I-' ) FDy (a--nh).
Multiplying it by dx=hd[, the integration of this expression will give, if x varies from a to a+h, or 5 from 0 to 1:
+ ;$ +'hDy,, .
T h i r d l y , b y a i d o f (5) w e c o m p u t e y,=y,+Ay4 a n d Dy,=f, (xstu,lThen we add the number hDys to the table above, and compute the differences ahoy,. +'hDy,, FhDy, and $'hDy,. After this we may put a=x,,+5h=q into equation (4) and proceed to the determination of y. in the same way; and SO on. In this way we may obtain yy generally with sufficient exactitude for any value of Y.
34
CHAPTER X.
FUNCTIONS OF SEVERAL INDEPENDENT VARIABLES.
Q 161. Functions of two independent variables. Given a function s=f (x,y) we may apply the methods already established for determinin g the differences of z with respect to the variable x, considering y constant. The first difference will be denoted as follows: L?'(X,Y) = f (x+h,y) - f (x,y). Moreover we may determine the difference of z with respect to y, considering x constant:
py) = f(x,y+k) --r(x,y).
The above differences are called partial differences of the function t. The second partial differences are obtained in the same way:
A*f (xty) = f (x+%y) - 2f (x+W + f (xty) yxi =
f(x+hy+R)
- f(x,y+k) - f (x+hy) + f (xty)
i" (X,Y) = f (x,y+ak) - 2f (x,y+R) + f (x,y). The other operations will be defined in the same manner. For instance the operation of displacement with respect to .x is
,E"'(X,Y) = f (x+nh,y)
and that of the displacement with respect to y rf (x,y) = f (x,y+mk) . An equation of partial differences will be written symbolically
531
@(Et El x Y
WGY)
=
V(X,Y).
where @ is a polynomial. The operation of the mean with respect to x will be
ff fby) = vmx+w + fwi
and with respect to y
y fw) = 1/2rfb,Y+4 + fhy)i.
Strictly in the symbols A, A, . , , the increments of x and of x I y should also be indicated, for instance thus:
A, 0
A li.k
but if in the following formulae we always have Ax=h and &CR then, to simplify, the increments. may be omitted in the notations. The operations will be executed exactly as has been described in the foregoing paragraphs. For instance: Expansion of a function of two variables by Newton's theorem. Given the function z=f (x,y), first we expand it with respect to x considering y constant; we shall have
Now considering x constant we will give y an increment equal to k and obtain
If f (x,y) is a polynomial the expansion (1) is alwayS legitimate, since then the series is finite. If not, formula (1) is applicable only if certain conditions are satisfied. Putting into equation (1) x=a+Eh and y=b+qk we find f(a+G,b+qk) = .; "2 (i 1 (A) $"$"'f@.b).
Let us suppose that such a table is given, where the values f(ui,v.,) are equidistant, that is, where Ui = U, + ih and vj = U, + jk. The following problem occurs very often: the value of f (u,u) is to be determined corresponding to given values of u and v which do not figure in the table. We choose from the numbers ug, q, u,, . . * and from v,, u,, v,, . . . the numbers a and b so as to have respectively a<u<a+h and b<v<b+k.
If linear interpolation is considered as sufficiently exact, and if the table contains, besides the values of f (uiquj) also the first differences of this function with respect to u and to u then, using Newton's formula, we have
(1)
f(u,o) = f (a,b) + (u-a) %-- + ( u - b ) %-- .
Af WI
Ai (4)
In this way the first approximation of the point U,U is obtained by the plane (1) passing through the three points correspmding to a,b; a+h,b and a,b+k.; but these are not symmetrical with respect to u,u (Figure 11). If the differences are given, the determination of f(u,u) by aid of this formula presents no difficulty. But if greater precision is wanted, then the application of Newton's formula requires the knowledge of the following five differences:
Here w=f (u,v) is an equation of a second degree hyperboloid which passes through the six points corresponding to 44; e,b+k; a.b+2k; a+h,b; a+h,b+k; a+2h,k. Since these points are not symmetrical with respect to the point corresponding to u,v, to be determined, therefore the interpolation formula is not advantageous (Figure 11) . There are hardly any tables containing the five differences mentioned above. A third degree approximation by Newton's formula would require the knowledge of ten differences; these it would be nearly impossible to give in a table, and it would be superfluous, since by the aid of Everett's formula for two variables, the same approximation may be obtained by the two central differences c12f (a,b) " and Pf (a,b). c
In fact, in Pearson's celebrated Tables of the Incomplete Gamma-Function [§ 18, lot. cit. 15) the first four are given; to apply Everett's formula only the last is to be calculated. We will not describe the proceedings of interpolation of a function of two variables by Everett's formula, since there is another formula which permits us to interpolate as rapidly with the same amount of work, and has the advantage that no printed differences at all are necessary, and therefore the table may be simplified by suppressing the differences. The reduction thus obtained will generally be more than one-third of the tab1e.O': Let us suppose that a table contains the values of f(u,v) corresponding to u = u,,, u,,+h, u,+2h, . , . and to v = v,,, v,+k, v,+2k, . . . If f(u,v) is to be determined for a given value of u and v, then we will choose from the table the numbers a=u,+nh and b=v,+mk so as to have a<u<a+h Putting moreover we shall have x = ( u - a ) / h a n d y = (v-b)/& O<x<l a n d O<y<l. and b<v<b+k.
To begin with we start from the formula (16) of the third degree of Q 133 (3) F(z) = I, + C,[Z,--l,I + lx,, 1 lYF(a+:h)
If we apply formula (3) to F(u,o), considering u variable and v constant, we get -
* C/I. Jordan, Interpolation without printed differences in the case of two or three variables. London Mathematical Society, Journal, Vo. 8, 1933.
535 (5)
F (wl = 1, (4 + c, (4 [I, 04 - bW1
I, (II) = xF(a+h,v) + (1-x)F(a,v)
where according to (4) and Z2(v) = + [(x+l)F(a+2h,v) + (2-x)F(o--h,v)]. Let us again apply formula (3) to the above quantities, but considering now v variable and u constant. We find
4 = &I + Cl(Y) [Sl, -Ll 1, = $2, + c, (Y) &I - S,*l
where S,, is the result points corresponding to interpolated between the and v=b,b+k (Fig. 12). $1 = + + + of interpolation of Z,(v), between the v=b and v=b+k; or that of F(u,v) four points corresponding to u=a,a+h Therefore S,, is equal to + + +
536 And finally S,, is obtained by interpolation of Z,(u) between the two points corresponding to v=Lk and v=b+2k; or that of F(u,u) between the four points corresponding to u=a-h,a+2h and v=b-k,b=2k. Therefore
The numbers C,(x) and C,(y) may be taken from the table mentioned in 3 133. If we had first interpolated with respect to u and after obtaining I, (u) and Z,(u) interpolated with respect to u, we should have found the same formula (6). This is an advantage over some of the other methods. The term S,, may be considered as the first approximation of f(u,u); in which the hyperboloid z=S1 of the second degree passing through the mentioned four points is substituted for the surface z=f (u,u). Formula (6) may be considered as the second approximation of f (u,u) in which we substitute for the surface z=F(u,~) an hyperboloid of the sixth degree which passes through the sixteen points corresponding to u = a - h,a,a + h, a + 2h, and v=b-k,b,b+k,b+ak. Remark. Equation (6) of the hyperboloid does not contain terms with x4, x6, x", y4, y", y 6, or terms of the first degree or a constant. Formula (6) is identical with the Newton expansion of a function of two variables, in which the partial differences of higher order than three are neglected and in which the first term is F (a-h&-k), The sixteen points through which the hyperboloid passes
538 In these cases C,(x) and C,(y) must be calculated by C,(x) = 1/2(1--x)x and C,(Y) = %(I-Y)Y as the corresponding values of x and y are outside the range of the mentioned tables. Exumpte. Let us take Pearson's example of the Incomplete Gamma Function [Lot. cit. 15 p. XXXIII]. The table contains . . and the numbers F(u,u) corresponding to u=zr,,,u,+h,u,+2h,. U=U,,U,+k,u,+2b, . . . where h=O,l and k=O.2. The value of F(4.025;7.05) is to be computed by interpolation. We shall have a = 4.0 ; b = 7.0 and
This result is exactly the same as that obtained by Pearson using Everett's formula and requiring the second central differences. Our computation by formula (6) is in no way longer than that by Everett's formula, perhaps even somewhat shorter. It is not necessary to copy out of the table the values contained in the second columns; they can be transferred directly to the machine. Should the precision of the result obtained by formula (6) be insufficient, it would be possible to make a further step and obtain a third approximation (of the tenth degree) by starting from
F WI = 1, (ul + c, (4 [I, (4--l, (4 1 + + c, (4 [2Z, (u)-3Zs (0) +I, (4 1
obtained by interpolation of the fifth degree with respect to U; this followed by an interpolation of the fifth degree with respect to v will give an hyperboloid of the tenth degree passing through the 36 points corresponding to ' and We find
In formula (6) we had four such values to determine, therefore the work of computation would be now somewhat more than the double of the preceeding example. If a machine is used, this may be undertaken in important cases. Nevertheless, constructing a table of a function f (u,v], the intervals should be chosen so that the second approximation by formula (6) should be sufficiently exact. This may generally be done. Remark. If the tangent plane is wanted at the point corresponding to u,v we have to determine aF(u,u)/a
u
and
aF(u,u)/av.
In the case of the first approximation we shall have
aF(u,u)
au =
1 aS,!.
'7; ax
i as. aF(u,o) and ~.. = _-e-U.
aV
k
ay
aS,Jax and i)S,,/ay may be expressed by the differences of the function F (u,v). Indeed we have
OS 2; = and
ax
Y
e+F(a,b) + OF(G)
u
a!3 --G = xAAF(a,b) ay u1
+ AF(a,b). I
541 . In the case of the interpolation of the sixth degree (6) we should find
- b-?4Kl(Y) P~~+s22-&--%1~ Therefore it will be necessary to compute the numbers aS,,,/ax. To obtain them we have only to change in formula (8) of S, the first factors. For instance in the case of a&,/ax they will be y, (l-y), -y, -(l-y), The term aF(u,u)/au is obtained in the same manner. $j 163. Functions of three independent variables, It is easy to extend the methods of the preceding paragraphs to functions of three variables. Interpolating between the 8 points corresponding to u=a,a+h, v=b,b+k and w=c,c+j we obtain a hyper-surface of the third degree t = Q,,, passing through these points and giving the first approximation of f (u,u,w) ; Q,, L is given by formula (9) below. Its computation is easy enough but generally the precision obtained will be insufficient. To remedy this inconvenience we shall determine a hypersurface of the ninth degree passing through the 64 points corresponding to u=a-h,a,a+h,a+2h ; v=b-k,b,b+k,b+2k; w=c+,c,c+j,c+2j.
To obtain it we shall start from formula (6) considering the quantities S,qc as functions of w. For instance S , , ( w ) = xyF(a+h,b+k,w) + x ( 1 - y ) F(a+h,b,w) + (1-x)y F(a,b+k,w) + ( l - x ) ( l - y ) F(a,b,w) and so on; then we interpolate with respect to w by formula (3). Denoting by Qyf,l the result of interpolation of S, (w) between the two points c-Aj+j and c+Aj, or that of F(u,u,w) between the 8 points corresponding to u=a-vh+h,a+vh ; v=b-pk+k, b+,uk ; w=c-q-j-j, c+i.j.
Finally the interpolation formula giving the second approximation of f(zz,u,w) will be
F(w4 = Qm + C, (4 :Qn,--Qml + C,(Y)[Q,,,-Qml + + C,(4 [Qm-Qml + C,(xlc,(~)IQm+Qm- Qm-Qml + C, (xlC, (4 IQm-Qm-Qm+ -t- Qml + C,(Y)G@) [Q,,,-Q,,,-Q,,,+Q,,,l + + C, WC, (~1C, (4 [Q,,,- Qm+Q,,,-Q,,,--- Qn,+Qm--Qm +Qml
where z= (w-c)/j. Since we have to compute nine values of Q,i and each is twice as long as that of S, hence the interpolation will give a little more than four times as much work as the example in $ 162.
CHAPTER XI.
DIFFERENCE
EQUATIONS.
Q 164. Genesis of the difference equatfw Let y be a function of the variable x, which is considered as a discontinuous one, taking only integer values. If we had a function of the variable z taking the values of z,+ih (where i is an integer) then we should introduce a new variable x=(2--z&h, which would take only integer values, and c\x=l. Given (1) w(x,y,a) = 0 where a is a constant parameter. If x is increased by one, the increment of y will be Ay. As y+L\u=Ey, we get (2) w(~+l,Ey,~l = 0. Eliminating a from (1) and (2) we obtain
Wc,y,Ey) = 0.
As in this expression one is the highest exponent of E, this is called a difference equation of the first order. If we start from a function containing two parameters u and b, we have y (x,y,asb) = 1) and deduce in the same manner y(x+l,Ey,ab) = 0 and repeating the operation,
y(r+2,E2yd) = 0
then eliminating from the last three equations a and b, we find
then we should be led in the same way to the difference equation of the n-th order: (4)
F(x,y,Ey,E"y, . . . . , E"y) = 0.
I
If the quantities ci instead of being constants should be periodic functions with period equal to one, then we should. be led to the same equation. Example 1. Given y=c,+c2x; it follows that and
EY = c, + cz(x+l)
E"Y = c, +
c,(x+2)
I
after elimination of c, and c, from the three equations we obtain a difference equation of the second order
E2y--2Ey +
Y
= 0.
Generally, instead of deducing the difference equation of order n, starting from a function containing n constants, the inverse problem is to be solved; that is, the difference equation of order n being given, a function y containing n arbitrary periodic functions with period equal to one is to be determined, which satisfies the given difference equation. If starting from a difference equation of the R- th order the function obtained contains n arbitrary periodic functions, then it is called a general solution; if it contains fewer, it is a particular solution. Remark. If the highest power of E in the difference equation (4) is equal to n and the lowest power to m (where y is considered as being equal to E"y), then the equation is of order n-m only, and there will be only n-m arbitrary constants in the solution, In the particular case of n=m equation, (4) becomes an ordinary equation and its solution py=f(x) or y=f(x--rz) does not contain arbitrary constants.
545 Equations of the form:
(5)
(6)
F(x,y,Ay,A*y, . . . A"y) = 0
F(x,y,My,M*y, . . . t M"y) = 0
are also difference equations according to the above definition, indeed if we eliminate from them Amy or Mmy by aid of
Am = (E-1)" o
r
Mm =(l+E)"$.
we obtain an equation of the form (4). There are some cases in which it is easier to solve equation (5) or (6) than (4). In the foregoing Chapters we have already solved several such difference equations. For instance, when hy=f(x) was given, and we determined the indefinite sum y=A-'f(x); moreover when solving My=cp(x) by aid of y=M-$(x); and also in other cases. Difference equations are classed in the same manner as differential equations. The equation
(7)
Q,,E"Y + anJ?y + , . , . + KEY + a,y = V(x)
is called a linear difference equation, provided that the coefficients oi are independent of y and Emy . If V(x)=0 then the equation is termed homogeneous; if not it is a complete equation. The coefficients oi may be constants or functions of X; we shall consider the two cases separately, Remark. Sometimes, though the increment of x is equal to cne, the difference equation may concern a function of a continuous variable. Then for instance
f(x+l) -f(x) = V(x)
is true for every value of x. These equations will be solved by the same methods as those of a discontinuous variable; hut as we shall see some precautions will be necessary.
but according to (2) this equation is satisfied by '% (d Y(X) = ylr) therefore
In this way the operation F(E) has been performed on P if F(E) is a rational fraction. (3) F(E) [tX] = r" F(r). Remark. In the particular case when F(E) = l/(E-1) =A-' we find the formula A-' rv=rV/ (r-l) and if F(E) =2/(E+l) =
547 = M-1 then we have M-l rX = 2r"/(r+l) as it has been obtained before. Now if r, is equal to one of the roots (real or complex) of the equation (4) y(r) = l7$" + a,,r"-' + . , . . + ult + a, = 0 which. is called the characteristic equation of (1) then it follows from (2) that crlX is a particular solution of (1). If it has R unequal roots, we Shall have R SOlUtiOllS Of the form yj=Citix . It is easy to see that if clyl , c2ya, . . . , cnyn are particular solutions of the difference equation (1); then their sum. y = CltlX + c,r2s + . . . . + c"m"x will also be a solution; since this function contains R arbitrary priOdiC functions Ci, i t is the general solution. In the Calculus of Finite Differences the variable is discontinuous and takes only equidistant values; the particular solutions yi, yr, . . .) y,,, will be called independent if starting from any n initial values such as y,Ey, . . . . Ply, the corresponding constants ci may be determined by aid of the equations I
Y = ClYl + %i, + - * ' - + CJIYn
is different from zero. On the other hand if it is equal to zero for every value of a then the constants ci corresponding to given initial values E'y cannot be determined; and the particular solutions are not independent. It can be shown that in this case one of them may 'be expressed by aid of the others. For instance if yz, y3,. . . yn are given, then y1 may be determined by aid of the linear equa-
548 tion of differences of order n-l, obtained by writing that the determinant is equal to zero. From the above it follows that an equation of differences of order n has only n independent particular solutions; moreover that if the n initial values y(a), y(a+l), . . . y(a+n-1) are given, then every value of y(a+t) may be computed step by step by aid of the equation of differences (1). Example 1. Given the difference equation (5) f(x+2)-3f(x+l) +2f(x) =o
the corresponding characteristic equation is rr-33r+2=0 therefore r, =l and r2=2. So that the general solution is f(x) = c, + c, 2X, Putting this value into (5) it is easy to see that it verifies this equation. Example 2. Fibonacci's numbers are the following 0,1,1235813,21,.., 1 1 9 1 I each of which is the sum of the two numbers immediately preceding it. If we denote the general term of this series by f(x), we shall have f(x) + f(x+l) = f(x+2). The corresponding characteristic equation is rs-f-1 =o and therefore rl =
SO
1/2(1+V3), t:! =
lh(l-VS)
that the general solution will be
U-4 To obtain the general term of the Fibonacci numbers, the arbitrary constants c, and cs should be determined by the inital conditions. Let us put f (0) = 0 and f (1) = 1; we find -
549 Cl + c* = 0 therefore
cl=-cc,=-. ;-
(6, -c*) y s = 2
5
Finally expanding the powers in (6) by Newton's formula we have
$j 166. Characteristic equation with multiple roots, If the characteristic equation y(r) = 0 has multiple roots, then, proceeding as before, the solution will contain less than n arbitrary constants or periodic functions, and therefore it will be only a particular solution. To obviate this we must find new solutions of the difference equation y(E)y = 0. Let us show the method, first in the case of double roots. If rl is a double root of the characteristic equation, then we shall suppose that rl and tl+& are roots of this equation, and later on we will put E=O. If t, and tl+& are roots of the characteristic equation, I~J (t) = 0, then obviously
(1)
y = (rl+~)x-r,X _ A,.'. & E
will also be a solution of y(r) ~0. The expression (1) will really be a solution, if putting E=O; but
Where D denotes the deriva.tive with respect to t, . Therefore a2xrlxe1 is a new solution, putting a2=c2r this solution will be c,xr,". If r, is the only double root of the characteristic equation, then the general solution will be y = (c1+c2x)rlx + c3tzx + , * , + cnrnX . If there were other double roots we should proceed in the same way.
550 It is easy to verify that xfIx is a solution of the differenue equatidn y(E)y=O if rr is at least a double root of the characteristic equation y(r) ~0. Indeed, putting y=x tls the operation Es gives
aspxflX = a* (x+-s)r,X+* = asxf,"(flq + a*f,X+'(sf,'q
from this we conclude, summing from s=O to s=n+l, that
v(E)m, X==lX~(f,) + rlx+l Dw(rJ.
Since fl is a root of y(f) =O therefore the first term will vanish and y=xrlX will be a solution if we have DY(r,) =O that is, if rr is at least a double root of the characteristic equation. Example 1. Given the difference equation f (x+2) -6f(x+l) +9&r) = 0. The corresponding characteqstic equation is
t=
-6r+9=0
so that tr =tr = 3. According to what we have seen the complete solution of the difference equation is f ( x ) = (c,+c,x)3~. Example 2. Calculus of Probabilify. The first player has Q shillings, the second b shillings. The probability of winning one shilling in each game is 1/ for each player; play is finished if one of them has won all the money of his adversary. The probability is required that the first player shall win, Let us denote this probability by f(x) if he possesses x shillings. He may win in two ways. First, by winning the next game; the probability of this event is $,; then his fortune will be x,1 and the probability of winning will become f(x+l). Secondly, by losing the next game, the probability of this event is also '/z ; then his fortune will be x-l, and the probability of winning will become f (x-l), Hence applying the theorem of total probabilities we have w = %zf(x+l) + WC-1) that is f(x) will satisfy the difference equation f(x+2)-2f(x+l) +f(x) =o.
551
The corresponding characteristic equation is r2 -2r+l=O h e n c e rl=tP=l. Therefore the general solution is f(x) =c,+c,x. The arbitrary constants are determined by the initial conditions, If x=0 the first player has lost and f(O)=O; if x=u+~ he has won, f(a+b)=l. From this we deduce c, = 0 and c, = l/(a+b). So that the required probability is f(x) =x/(a+b). This gives at the beginning when x=a, f(a) =a/(a+b). Multiple roots. Let us suppose now that rr is a root of multiplicity m of the equation y(r) =O. To obtain the general solution of the difference equation y(E)y=O we may proceed as before considering first the m roots of the characteristic equation as being different, and equal to
rl,rI+E, r,+2&, . . . . , rl+(m-l)E.
Supposing i < m, then a solution of the difference equation is given by
,& [(r,+i&)X-( :)(r,+ia--E]'+.
Putting E=O this solution will become
, . + (-l)'riX] = 4$
y = a1 k Dir," =
or
Y
i<m.
Therefore the solution corresponding to rI containing m arbitrary constants will be
This formula holds for all real or complex values of rl . Verification. r, being a root of v(r) = 0 let us determine the conditions so that We have r," shall be a solution of y(E)y=O.
552
*
In consequence of Cozzchy's rule (14, 8 22) this may be written
I I
Hence, summing from s=O to s=n+l we must have
Y(E) [[;J rlx] = 2 (& ] 5 D'Y(~,) ~0.
Therefore 7 rlx is a solution of the difference equation if 11 D'*y(t,) = 0 for Y = 0, 1,2, . , . , i. That is, r, must be at least a root of multiplicity i+l of y(t)=O; hence, if rl is of multiplicity m, then a particular solution with m arbitrary constants is given by (2). g 167. Negative roots. We have seen that the obtained solutions of the difference equation y (E)y=O were applicable also in cases when the roots of the characteristic equation y(r) =0 were negative or complex. But in these cases the solutions may appear in a complex form, and generally the real solutions are required; therefore they must be isolated. If the function considered is one of a discontinuous variable the increment of x being equal to one, and the root r1 is negative, then the preceding methods may be applied without modification. Example 1. Let us consider the series O,l, a, ;, ;, ;, ; I'.., obtained in the following manner: each number is equal to the arithmetical mean of the two numbers immediately preceding it. If we denote the general term of the series by f(x), we have
The arbitrary constants are determined by the initial conditions. We have y(0) = 0 a n d y(1) = 1 therefore c,+c2=0 a n d c,--1/2c2=1. This gives 2 c,=-cc:!= -. 3 (--1P Finally the general term of the series will be y = -a- 3. Remark: 2 limy= -. 3 X==oD Secondly, if the function considered is one of a continuous variable and if T, is negative, then the solution y = rI+ is complex. Let us write the general expression of a number t, (1) rl = a1 + i/l, z ,oo(cosgi, + isintpl) = eleiFl From (1) it follows that (2) tlx = ~~~(cosxq~~ + i sinq,) = eIxeixW If rI is a negative number, then from (1) we deduce qI=z and eI =-r1 . Therefore
t1
w h e r e i = I'=; el > 0 ; ply = aI2 + B,' a n d tanv, = PI/al.
x = ~,~(cosnx + i sinzx) = ~,Xei~X. y = celx cosnx .
The real part of the solution is i According to what we have seen, if rl is a double root of i;*(r)=0 then we have the particular solution corresponding to r,: y = (c, + c2x)@,x cos7c.x.
554 If the coefficients in the difference equation are all real, then the real and the complex part of T,~ must each satisfy separately the difference equation; so that the solution may be written y = elx Ic, cosnx + c2 sinnx]. But since two particular corresponding takes only the have the difference equation is of the first order, the solutions cannot be independent; and indeed the determinant is equal to zero. Moreover since x values ~=a+5 (where 5 is an integer) hence we cosicx = (-1) kosna
8 168. Complex roots. Given a homogeneous difference equation with real, constant coefficients y(E)y=O. If the root r, of the characteristic equation y(r) =O is a complex one, fl = el(cosy, + i sinq,)
then, since the coefficients of the difference equation are real, it follows that rz = e1 (cowl - i sin& will also be a root of the equation. Therefore if the characteristic equation has no multiple roots, the general solution may be written
Example 2. Given the difference equation f (x+4) - f(x) = 0. The corresponding characteristic equation is rJ - 1 = 0. It is necessary therefore to determine the fourth roots of unity. Let us write the general expression of a number a in the following way a = gJcos(++2kT) + i sin(rp+ak.-r)-1. Its R th roots are given by
hence r = cosp, Ifr i sinq and therefore (2) f(x) = c, cosqx + c, sinpx. The constants must be determined by the aid of the initial conditions. Let for instance f (0) =O and f (n)=O, then equation (2) will give - 0 and c, sinpn = 0. Cl From this we conclude that if Q, is not equal to .-w/n, where Y = 1,2,3,. . . , n- 1 then cz=O and the only solution is f(x) =O. On the other hand if q=nv/n then we shall have f(x) =c2sin y. .
Since this equation still contains an arbitrary constant, therefore we may impose upon f(x) a further condition. Recapitulation. Given the homogeneous linear difference equation, with real, constant coefficients: a,f(x+n) + a ,,-, f(x+n-1) +. . . . + a,f(x+l) + aOf =.O if r, = Q,, (cosyl + i sin,, V) is a real or complex root of the characteristic equation, of multiplicity m, then the solution of the difference equation is
w(E)Y = 0.
From this we conclude that f(x) =u+y is the solution of equation (1). Therefore first we have to determine the general solution of the homogeneous equation; then we must find one particular solution of the complete equation. This last may be attained in different ways; sometimes a particular solution may be obtained by simple reasoning. Often the symbolical methods lead to it easily enough, or even the direct method of determination. Symbolical methods. They may be useful in some particular cases, A, First let V(x)=Ca*. If the difference equation (1)
w(E)f(x) = CaX
is given, where a is not a root of the characteristic equation y(r) =O that is y(a) + 0; (the equation ,v(t) may have multiple root8 or not). Dividing formula (1) by p(E) and using formula (3) of 8 165 we have 1 (2) f(x) = y(E) which is a particular solution of equation (1). Denoting again by y the general solution of the homogeneous equation corresponding to (l), the solution of the complete equation will be (3)
f(x) = yz
+y.
Particular case. a=l, V(x)=C. If r=l is not a root of y'(r)=0 then according to (3) the general solution will be (4)
f(x) = &+y*
Example 1. Given the equation
f(x+2)
- 5f(x+l) + 6f(x) = 2.
559 The corresponding characteristic equation is y(r) =r2-5r+6=0 so that r,=2 and r2=3; since moreover a=1 and ~(1)=2, the general solution of the complete equation will be given by (4) f(x) = 1 + c,2" + c,3.r .
Example 2. Given the difference equation
f(x+3) - 7f(x+2) + 16f(x+l) - 12f(x) = CaX. The corresponding characteristic equation is y(r) = r3 - 7t2 + 16r -12 = 0. Hence r, ~2, r2=2 and r,=3. Therefore if a is different from 2 and from 3, then the general solution of the complete equation is given by (3)
f(x) =
~Q&5jaM-12
+ (c, +c,x) 2"+c, 3X.
Secondly, given the difference equation t+f(E)f(x) = Ca*'
where a is a simple root of the equation q!(r) ~0. To solve the difference equation we put
t/r(E)f(x) = C(a+E)X
then the particular solution will be as we have seen:
u= C(a+&)r V?(a+t) '
moreover we have D~+j(r) = 2r-4. The solution will be, according to (6): f ( x ) = - x + c, + c23x. B. Let V(x) =&p,(x) where v(x) is a polynomial of degree n. Before solving the equation of differences (7)
vIEIf = a%+4
561 we will establish the following auxiliary theorem. It is easy to see that
'
EmI:axq(4 I = [Emax][Em~(41
= aX[aElmq(x),
This holds for every positive or negative integer value of m. Therefore if y(E) is a polynomial of E and q(x) is any function whatever, we have (8)
w(E)la%Wl
Let us show now that, fraction, t(E)=U(E)lV(E) where U(E) and V(E) are polynomials, we have (9) lndeed, multiplying both members by V(E) we find:
= a+v(aEIdx). if y(E) is a rational
U(E) [a"y(x)] = V(E) Iax B Y(X)1
but according to (8) the second member will be equal to asV(aE) w v(x) a that is, it will be identical with the first member. Therefore formula (9) is demonstrated; so that if F(E) is a rational fraction we have (10)
F(E)a%(x) = aWaE)y(x).
Let us execute the operation l/y(aE) performed on a polynomial q(x) of degree S; if y(aE) is a polynomial of E of degree n; and if a is not a root of the characteristic equation y?(r)=O. (The roots of this equation may be multiple or not.) From the equation of differences (7) it follows by aid of (10) that
(11)
f(x) = ax J--
v(aE)
Y(X).
Eliminating E from y(aE) by aid of E=A+l we get a polynomial @(A) of degree R. Let us suppose first that the 36
562 coefficient of A" in @(A) is different from zero. Then l/Q(l) may be expanded into a power series of A:
co + c,A + c,A' + . . . + csA=
v(x) being of degree s the series may be stopped at the term c,k, since As+$ (x) =O (if i > O), Finally we have
and the required particular solution will be given by (11). Since the higher derivatives of l/y are necessarily complicated, it is better, if n 2 2, to determine the coefficients in another way. If we have
y(a+aAl = h, +&A + , . . + b&
then the coefficients ci of the expansion (12) are given by the following equations:
' &(a) = 2a-4=-2
from (13) it follows that II = (1+2&x=x+2. The second method would give
y(a+aA) = u2(l+A)"--4a(l+A)
Therefore and co = 1 and
+ 4 = 1 - 28 + A".
c,-2c, = 0
u= (1 +,2A)x = x+2.
Example 7. Given r(x+2) + f(x) = xa= .
We have y(r) =r2+1 ~0; the roots of the equation are complex. The particular solution will be, according to (11) :
I
564 . I Particular case of the difference equation of the first order. In the second case (B) the difference equation of the first order may he written (14) f (x+1) -r,f (x) = a"y(x) where q(x) is a polynomicil of degree s. Let us suppose that a + r, , then we shall have u=Q" - X a"+1 2a" Ia'+l)'
u=-l;]-1;) - (;I-1. Formulae could be deduced for cases in which a is a single or a multiple root of y(r) ~0; but it is simpler then to use the method given in the next paragraph. Summing up we conclude that this method is independent of the order n of the difference equation, and therefore especially useful for equations of high order. It does not matter if the roots of v(r) =0 are single. or multiple, real or complex; to obtain the particular solution it is not even necessary to determine the roots of the characteristic equation; but the formulae are different if a is a root of y(r) ~0. Since the number of terms in the solution given by (11) increases with the degree s of the polynomial yj (x) , the method is not advantageous if it is of a high degree. Q 170. Determination of the particular solution in the general case. Given the difference equation
v~(E)f(xl = V(x).
565
The particular solution obtained by the symbolical method is
Q = J- V(x).
v(E)
To determine the second member we shall decompose l/y(E) into partial fractions. Let us denote the roots of the characteristic equation y(r) =0 by r, , ti , . , . , r,, ; first we suppose that these roots are all real and single. If we write
then according to (j 13 we have (2)
b,,, = l!ID&)]r=,., .
Let us apply now to each term & V(x) the following
transformation. It is easy to see that (E-a)
[ax-IF(x)]
= a+AF(x).
Putting the second member equal to V(x), it gives
A-'IQ-"V(X)] = F(x)
moreover from the preceding equation we get
- v(X)
EL
= ax-IF(X).
Eliminating F(x) between the last two equations we obtain the important formula
(31
& v(x) x mv-'A-' 1Q-v(X)]
this gives the required particular solution in the form of indefinite sums:""
(4)
u = blrlX-' A-'If,-W(x)] + . . . + b,,t,"' A-l[r,-W(X)].
The formula supposes that t=O is not a root of v(r) =0 but this may always be attained, For instance, starting from
Guldborg,
33 This formula is identical with that given by G. Theorie der linearen Differenzengleichungen,
Wahlenberg
und A. Berlin 1911, p, 175.
566 f(x+3) -2f (x+2) + f (x+1) = V(x) we get by putting x-l instead of x f(x+2) -2f(X+l) + f(x) = V(lr-I). I f &)=O h a s multiple roots, for instance if m is the multiplicity of the root rv, then the decbmposition of l/y(E) into partial fractions will contains terms of the form:
(E!?&)P '
Denoting
for ,u = 1,2,3, . . . , m.
then the numbers b,,,, are given according to (4) 8 13 by
Now we have to perform the operations
For this let us remark that, according to (3), the operation l/(E-u) performed on
u = 2: b,!, t,?-f' A-:' [r,.-" V(x)] + . . ,
This method may also be applied in the particular cases of
567 5 169. Sometimes it is even more advantageous than the prev,&us methods. For instance, given the equation
w(EVW = +4x)
where p(x) is a polynomial, then in this method it does not matter if u is a root of v(r) =O or not. The number of terms in formula (5) is equal to the order of the difference equation, therefore the method is especially useful for equations of low order. For instance, in the case of an equation of the first order we have only one term. The degree of q(x) does not increase the number of terms. If the roots of v(r)=0 are multiple or complex, if the equation is of a high order, then the method is more complicated than that of Q 169. The following example will serve as a comparison: Example 1. Given the equation of differences f (x+3) - 7f (x+2) + 16f (x+1) - 12f (x) = CaX . We have y(r) = r3 - 7t2 + 16r-12 = 0, hence rl=3, r2=r3=2. Moreover Dy(t) =3r"-14r + 16 and A(r) = l/+3). By aid of the preceding formulae we find b,=l, b2,=-1 a n d b2,=.-1 therefore the required particular solution will be u = 3"' A-' 13-x &xl - 2x-1 A-1 12-x &x] _ 2x-9 A-2 [2-x Cax] determining the indefinite sums in the second member we obtain u = Cax j (u-2) * (a-3). This result would have been given directly by formula (3) of Q 169. Exampte 2. Given the equation
f(x+l)
-2f(x) =2x@.
Since we have y(r) =r-2 and r1=2, hence the particular solution is given by (4):
u =
2x-l
A-'
[2-" 24 ;)I
=
2x-l
(; ) .
568 The method of Q 169 would have been much more complicated. Indeed, since a=2 is a root of y(r)=O, hence after executing the derivations in the second member of formula (9), putting a+& instead of a and determining the limits for E=O we should have had
(:1 * Negatiue roofs. If the root r of v(r)=0 is negative, then we may put t I..v = Q,, + and formula (4) will give u=.... + b,.~v~-' cos(x-1)~ A-' ]gr-" cos-cx V(x)] + . , . . Complex roofs. If the root r, of the characteristics equation is complex t, = g(cosy, + i sinq) then r2 = ~(cosp - i sinv) is also a root of this equation, since the coefficients of the given equation are supposed to be real. The decomposition of l/y(E) into partial fractions will give
cos;rx
8 171. Method of the arbitrary constants. Given the equation of differences (1)
Y(E) f(x) = V(x).
We will suppose that the coefficient CT" of f(x+n) in this equation is equal to one, and that V(x) is of the following form (2)
V(x) = ax
[aD+al(if)+a2[~]+~~,+am[~)].
We shall consider a as a root of multiplicity R of the characteristic equation y(r) ~0. If (I is not a root of this equation then k=O. According to what we have seen, the particular solution of this equation may be obtained also by the method of Q 169, or by that of Q 170. No& we will try to write a particular solution in the following way
and dispose of the coefficients ,!Ir in such a manner as to make this expression satisfy the difference equation (l), Putting there 1 (x) =u we shall have
In consequence of formula (8) Q 169 we get
on the other hand we have
t,.~(uE) = tp(u+uA) = p=k D~e,u(u)~~ "g
Pl
-
I '
I
I
570 Indeed, y(r) being of degree n, hence D""y(a) ~0; moreover a is a root of multiplicity K of y(r)=O, therefore D.?p(a)=O if p < K. Finally we shall have
V(x)
writing k+v-p=i we find
m+l a+-i Bv
D+-iv(a)
= v(xl
I
,.5,
(k+v-i) !
.
NOW we shall dispose of the coefficients /?,, so as to make the first member identical with V(x) given by (2); we obtain
VI+1 ak+v-i Dk+v-i v[a)
(4)
ai
= ,zi
br
(k+v-i) !
= .zo B,,+t-k$ Dh4 ,*
OD
fori=O,1,2 ,..,, m. If D&y(a) and a are different from zero then the last of these equations will give pm, the last but one Brnml, and so on; since o is a root of the equation y(x)=0 of multiplicity k, therefore the first condition is satisfied. The second is satisfied too; indeed, if we had a=0 then equation (1) would be homogeneous. The 8,. being determined, the problem is solved. This method is applicable whatever the roots of y(r)=0 may be, and it is not even necessary to determine these.
Purticular case of the equation of the first order.
(5)
f(x+l) -rr,f(x) ==~[a.+al[~)+,...+~~(31.
From (4) we deduce (6)
ai=bi-ky(a)
+ @i+l-kaDy(a)-
Here we have y(r) =r-r1 , &(a)=1 and D2y(a)=0. Let us suppose first that o+r, and therefore k=O, we find
ai = BiY (4 + /La
8 172. Resolution of linear equations of differences by aid of generating functims,u=u(t) is the generating function of f(x) :if in the expansion of u(t) into a series of powers of f, ths coefficient of fX is equal to f(x); that is if (1)
Gf(x), u=f(O) +f(l)f+f(2)f"+ ,.,, ff(x)P+ .,.,
The generating function of f ( x ) was denoted in Q 1 0 b y Starting from this generating function we deduced in 8 11 the generating function of f (x+m) :
Remark 1. The characteristic equation of the difference equation (3) is the following a# + . , , , + a,r + a, = 0 hence the denominator of u(f) is obtained from this equation by multiplying it by f" and putting into it r=l/f. Remark 2. Knowing u(f) the generating function of f(x), it is easy to deduce w(f) the generating function of the indefinite sum of f(x), that is of A-If(x) =F (x). Indeed we have we get that is F(x+l) - F ( x ) = f ( x )
Wb-04 t
--w(f) = u(f)
w(f) = fu(f) +F(o) -. l - f
Having determined u(f) by aid of (4), the expansion of this function into a series of powers of f will give the required function, In the general solution of a difference equation of order R we must have n arbitrary constants; in fact the generating func-
I
I
574 tion (4) contains the n constants f(O), f(l), . . , , , f(n-1). Hence it will lead to the general solution. Finally the constants will be disposed of so that the initial conditions shall be satisfied. Formula (4) gives the generating function of the difference equation of the firef order if we put into it R= 1: u - - elf +fwl al+%t * This expression expanded into a series of powers of t will become
This formula is identical with that corresponding to (4) (j 170. Example 1. Given the equation f(x+l) -2f(x) = xd. In consequence of (6) we have f(x) = 2Xf (0) + 2x-1 ,x0 p(l/$J) p.
.s
I
According to formula (2) Q 34 the second member is equal to
f(x) = 2xf(()) + ti+l(x--~ + u2x .
From formula (4) we may obtain the generating function of the difference equation of the second order. Putting into it n=2 we find (7) ll=
d(O) + Calf(o) + %fU)lf + t2w
a2
+ ad + =ot2
.
Example 2. Given the equation
f(x+4 -3f(x+l)
+2f(x) =o.
Since t1 = 1 and t2=2, hence the denominator of u is equal to (l-f) (1-2t) and the expansion of u will give
575 u= ( f ( 0 ) Therefore and + [f(l)--sf(O)]f} % 5 2"fY+f'. v=Q p=o
x+1 f(x) = f(O) ,zo zr+
Cf(lF--3f(OIl
Vi0 2"
f(x) = 2f(O) -f(l)+[f(l)-f(0)]2X.
Example 3. Calculus of Probability. A player possessing x shillings plays a game repeatedly; he stakes each time one shilling, if he wins he gets two. The probability of winning is equal to p. It is required to find the probability f(x) that the player will lose his money before winning m-x shillings. This problem is a little more general than that of Example 2, (j 166. Applying the theorem of total probability as in this example, we shall get f(x) = pf(x+l) + gf(x-1) where q=l-p. Hence we have to solve the difference equation pf(x+2) -f(x+l) + qf(x) = 0. Applying formula (7) we get Pf u= - (0) + [PfU) -fOlP P --f+qtz *
Since the roots of the characteristic equation are rr=l and r,=q/p, hence the denominator will be ~(14) (1- % t) and the expansion of u will give
To determine the arbitrary constants let us remark that if x=0 the player has lost so that f (0) ~1; and if x=m he cannot
576 lose before winning m-x, as he has already done SO; therefore f(m) 10. Putting into (8): f(0) 11, x=m and f(m) =O we find f(1) = and finally f(x) = [ q"'-p"I [$'I/ (,"I-P"'). Remark. In the method considered above we supposed that x is a positive integer, but putting the obtained result, in order of verification, into the given equation of differences, it is easy to see that the result holds for any value whatever of x, the operation E being always the same. Since the method does not presuppose the resolution of the characteristic equation, it may be useful if this equation cannot be solved and if we are able to expand the generating function without the knowledge of the roots of the characteristic equation. Generally this method leads to the same formulae as the methods applied previously, so that its real advantage will show only when applied to functions of several independent variables, when the other methods fail. Q 173. Homogeneous linear equations of differences of the first order with variable coefficients. We may write these equations in the following way (1) f(x+l) -p(x)f(x) = 0. Let US suppose that x takes only integer values such as x 2 a. Then we shall introduce a function y(x) so as to have y(a)=1 and for x>a:
Y(X) = P(Q) PtQi-1) P(Q+2) . .'. Pk-1).
Q (pm-l-q~~-l)
Pm-v
Now dividing both members of equation (1) by y(x+l) and putting Q(X) = f(x) /Y(X) we obtain 22(x+1) therefore --u(x) = 0 or Au(x) = 0
577
a(x) = c
(2)
f(x)
a n d f(x) = Cy(x)
moreover f(a)=C; hence from x > a it follows that
= f(a) ii p(i). i=R
Example 1. The general term f(x) of the series given by the following equation of differences is to he determined if the initial value is f (0) = 1. [Stirling, Methodus Differentialis, p. 108.1 2x+1 f(x+l) - - f(x) = 0. 2x+2 According to (2) we have f ( x ) = f ( 0 ) ii g$ = [Y] &. Example 2. Let us denote by f(x) the number of permutations of x elements. Starting from the permutations of x elements we may obtain those of x-i-l elements, if we insert the x+1 th element into every permutation, successively in every place. For instance, if x=2 the permutations are ab and ba. This gives for x=3 cab, acb, abc, therefore we have
f(x+l) - (x+l)f(x) = 0.
and cbtr, bca, bat
The solution of this equation of differences is, according to formula (2), f ( x ) 1 f ( 1 ) ii1 (i+l) = f ( l ) x ! . Since f (1) = 1 hence the solution is f(x) =x! . Example 3. Let us denote by f(x) the number of the possible combinations with permutation of y elements taken x by x, i . e. of order x. Starting from the combinations of order x we may obtain those of order x+1, if we add at the end of each combination successively one of the still disposable y-x elements. In this way we obtain from each combination of order x a number of y-x different combinations of order x+1.
37
578 For instance, if y=4 and x=1 we have the combinations a, b, c, d; this gives for x=2: ab, ac, ad, ba, bc, bd, and so on. Hence we shall have f(x+l) - (y-x)f (x) = 0. The solution is according to (2), f(x) = f(l) ii
(Y--i) = f(l) (Y--lL, .
Since f (1) =y therefore f(x) = (y)* . Example 4. Let us denote by f(x) the number of combinations without permutation of y elements of order x. Starting from the combinations of order x we may obtain those of order x+1 by adding to each combination one of the still disposable y-x elements. But proceeding in this way we obtain every combination x+1 times. For instance if y=4, x= 1, we have the combinations a,b,c,d. From this we obtain the combinations of order x=2: ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc
every combination has been obtained twice. Therefore we have the difference equation f(xf1) = s f(x) whose solution is according to (2): f(x) =f(l) ,i, s = f 11) (Y--l)x-, Ix!. Since f (1) =y, hence f(x) = ; . ( ) Example 5. Let us denote by f(x) the number of combinations with repetition but without permutation of y elements of order x. Starting from the combinations of order x we obtain those of order x+1, by adding first to every combination successively every element of y. In this way we should obtain every combination of order x+1 but not each the same number of times; to obviate this, we add secondly to every combination of
The function u(t) is still to be disposed of, and so are the limits of the integral. From (2) it follows that f (x+n) = j&+"v(f)dt a
580 therefore we get f (x+n) starting from f(x), if we write in (2) f%(t) instead of u(t). Multiplying equation (2) by x and integrating by parts we find xf(x) = (.hxf.~-'u(f)df : = jpo(t)]: - /f.KD"(f)df. a
Multiplying this by x+1 and repeating the integration by parts we have (x+l),f(x) = 1(x+l)f+u(f) -f"+' Do(f)]; + & D"v(f)df. a Multiplying this by (x+2) and repeating the operation, and on, in the end we obtain (x+m-l),f (x) = ["ii:
SO
(-l)i+l (x+m-l),jfrlti D'-%(t) ]I+
+ (-l)m )**tX+"'-'D"'u(f)df. us Now writing into it f%(f) instead of v(f) we get the general formula (3) x m-l)m f (x+n) = + = {x1 (-l)it'(r+fm-l)~,-ifx-l+i~-l~f.u(t)lr
b
+ (--I)"' 1' P+*-1 Dm[f"u(t)]df. (1 Since we may suppose that the coefficients a,. of (1) are expanded into a series of factorials (x+m-1), therefore formula (3) permits us to write the difference equation (1) in the following form: h [fp(x,o,f)]; + I' f" I,~(U,f) df = 0. a* First we dispose of u=v(f) so as to have y(u,f) ~0. Generally this gives a homogeneous linear differential equation whose solution is equal to u(f) . Having obtained this, we dispose secondly of the limits of the integral, so that q(x,u,f) shall be equal to zero
581 at both limits. The limits must therefore be roots of the equation in t: ql(x,u,t) = 0. Let us suppose that they are, in order of magnitude, t, , t, , , . . ; each combination of these roots, two by two, will give by aid of (2) a particular solution of the difference equation. The sum of n independent particular solutions each multiplied by an arbitrary constant will give the general solution. Remark. Having determfned f(x) by equation (2) it is easy to deduce A"V(x) and A-If(x) ; we find Amf(x) = (.'F(t-l)"%(t) d t
A-, f(x) = tz (t.X-1
.
Example 1. Given
f(x+ll
From (3) it follows that
- xf (x) = 0.
I-- t"u(t)l; + J" t"[u(t) + Du(t)]dt = 0. a Putting first
u(t)
+ Do(f) = 0
v = Ce-' . Ye+ = 0
we obtain
Secondly, the limits of the integral are the roots of therefore t,=O and t,=~ . Finally we shall have f(x) = c I'- t"-' e-'dt = Cr(x).
0'
Starting from the difference equation of I'(x), we have got its expression by a definite integral. Pctrficufar case of formula (1). The coefficients oi are all of
582 the first degree in x; so that we may write equation (1) in the following manner: nil E (a,x+/L.)f(x+r) = 0 d where a,. and By are numerical coefficients. From (2) we obtain
and from (3)
n+l s b fx D[t 'o] df.
Hence u will be the solution of the equation n+1 n+1 vz (By--yaw) frl+u~- 2 avPYDu =O which can be written (4)
Do _ E W%-+ay) .
V
Xa,W
The limits of the integral are chosen from the roots of the equation n+1 Pu 2 a,f' = 0 . First we have the root LO; moreover, if the roots of a# + . ...+ a,f + a0 ~0 are all real, different from zero and unequal, say t, , t, , . . . . , t,, then the general solution will be?' 4 f(x) = Cl s t=-ludt+.... + c, oj+ P-l u dt.
0
= ( x - l ) ! [f(l) + r"$')--11 Remark 2. In the particular case, if a=0 and p(x) is constant, say p(x) =ri , then we shall have y(x)=rlx and formula (2) will be the same as that corresponding to (4) of 8 170. Remark 2. As has been said at the end of $ 172, though we supposed x to be an integer, nevertheless, if the result may be written in such a way that it has a meaning for every value of x, for instance in the case of (3), writing
f(x) = r(X) (t(i) +[ "2+l] -11
then this will necessarily satisfy the given equation, Indeed the operation E is the same whatever the value of x may be. 0 176. Reducible linear equations of differences with variable coefficients. In some cases the equations with variable coefficients can be reduced to equations with constant coefficients. . A. If the following equation is given a,f(x+n) + k,p(xl f(x+n-11 + an-2p(4p(---ll f(x+n--2)f (11 +.... + a,, Pb)P(X-1)' . . p(x-nfl) f(x) = V(x) where the ai are numerical coefficients, then the equation may be reduced by putting
f(x) = p(x-n)p(x-n-l) . . . . p(a) p,(x).
The solution of this is the same' as that in 5 170. Example 1. Given the difference equation. f (x+2) - 3xf (x+1) + 2x(x-l) f(x) = 0 let us put f(x) = (x-i) (x-3) . , .2.1 .y(x) = (x-2)!q4x) where xlz and a = 1. Dividing both members of the given equation by X! we find v(xt2) -%,(x+1)
+ &4x) = 0
the equation is reduced to one of constant coefficients. The corresponding characteristic equation is r"--33r+2=0 therefore r,=l and r,=2, so that y(x) = c, + c,2? and finally f(x) = (x-2)! (c, +c,2q.
Q 177. Linear equations of differences whose coefficients are polynomials of x, solved by the method of generating fllnctions.5" We have seen in Q 172 that if the generating function of f(x) is u then the generating function of f(x+m) is G.f(x+m) = i; [u-f(O) --f(l) -. . . .-tm-lf(m-l)]. The first derivative of this function with respect to f multiplied by t gives the generating function of xf (x+m). The second derivative multiplied by P/2! will give that of ($]f(x+m) and so on.
(1) $1;) f(x+m)]=$
~lU-f~o)-ff(l)~~~~-~f(m-l)
Let us consider now the equation of differences (4 anf(x+n) + . . . . + a,f (x+1) + a,f (x) = V(x) where the coefficients ci are polynomials of x. If the generating function of V(x) is known, we may write that the generating functions corresponding to both members of the equation (2) are equal. This will give a linear differential equation:
@(n.. . , D*u,. . . . . , Du,u,f) = 0
whose solution gives the generating function u. Developing it into a series of powers of f we get the required function f(x). If the coefficients in equation (2) are of the first degree, then the differential equation will be of the first order. Example 1. We have seen that if we denote by f(x) the numbers of the combinations of R elements of order x, then this number is given by the equation of differences (x+l)f(x+l) - (n-x)f(x) = 0. Let us solve this equation by the method of generating functions. Noting that
j5 Laplace, Thkorie analytique des ProbabilitCs, p, 80. Seliuanof-Andoyer, Calcul des Differences et Interpolation, Encycl. des Sciences Mathimatiques, I. 21,. p.'76.
587
G [(x+lV(x+l)l = Du
from the given equation it follows that and therefore
Du- nu + tDu = 0
du ndt -=-* u 1+t u = cp+q*.
Hence
Developing this into a series of powers of t this gives f(x) = c (=I. As f(l)=n, theref ore it follows that c=l and finally f(x) = F I I Example 2. Given the equation (2x+2)f (x+1) - (2x+l)f (x) = 0 according to formula (2) we have
If instead of the f(l), f (2), . . . , f(n), the quantities y, , y, , . . . , y,, , are considered as known, then from the above system of x equations, the x unknowns, f(l), f (2), . , , , f(x) may be determined; but it is sufficient'to determine only one of them, viz. f(x). It is easy to see that the determinant figuring in the denominator of f(x) is equal to unity. Indeed it is equal to 1 0 0
0
Let us denote the general term of the expansion of this determinant by Uk, (xl)arp (x2) . . . . an, (Xi).
590 We shall show that these expressions satisfy the following conditions : 1. The terms are homogeneous with respect to the indices of a, and we have k, + k, + . . , , + kj = m. 2. The first argument is always equal to x, so that x1=x; the others are: x*=x--RI, x::=x--R1--K2, . . . .
Xi
=
X-kl-kk,--.
. . .-ki-, = x-m+ki.
Therefore the coefficient of V(x-m) will be equal to
I: ak,(X,)as,(x,) . . . ski (Xi).
This sum is extended to every partition of the number m with repetition and permutation. The number of the terms being equal to
r(J'm) =
2"'-' .
(See Netto's Combinatorih, p. 120.) This may be verified in the cases considered above of V(x- I), V(x-2) and V(x-3). For instance in the last the terms correspond to 1 +l +I, 1+2, 2-k1, 3. We shall show now that if the sum above is equal to the coefficient of V(x-m), then it follows a similar formula for V(x-m-1) and therefore the formula is true for every m such that x > m. The coefficient of V(x-m- 1) in the numerator of f(x) being equal to the determinant
-q (x) -2 (xl ' * ' * ---%,(X1 -m+i txl
We shall examine each term of the expansion of this determinant to see whether our conditions are satisfied.
!
591 The coefficient of the term a,+,(x) is equal to one, hence it satisfies both conditions. The coefficient of a,(~-1) is equal to a,(x) therefore the term aI (x)u,(x-1) satisfies the conditions. The coefficient of a,-,(x-2) is equal to that of V(x-2), Since the sum of the indices in this coefficient is equal to 2, therefore 2+m-l=m-i-1 satisfies condition (1). Moreover the arguments in the last factors of the terms in this coefficient being equal to (x-2--RJ so that the terms . . . . . Qk, (--2+&a ,,,- 1 (x-2) also satisfy condition (2). The coefficient of a,-,(x-3) is equal to that of V(x-3). For the same reasons this coefficient satisfies both conditions, and so on; finally, the coefficient of a,@-m) is equal to that of V(x-m). The sum of the indices which has been equal to m, has become now equal to mfl; moreover the argument in the last factor has been (x-m+&, therefore ......
aki(x-m+kj)a,
We have already used Andrh's method in simple cases, SO, for instance, in Q 32 determining the indefinite sum of V(x), that is, solving the difference equation /(x+1) --r(X) = V(x) in fj 38 by the inverse operation of the mean, solving the equation 1(x+1) -+- f(x) = ZV(x).
or
Now we shall apply it to more complicated problems. Particular case. 1. Difference equation of the first order with constant coefficients. We have
Multiplying .the second equation by a, , the third by a,z the fourth by a,3, and so on; after addition we get r-l f(x) = X a," V ( x - m ) + a,"-' y, . (4)
m=O
This result could have been obtained directly by formula (3). Indeed the sum of the indices must be equal to m, therefore we have a,"l; moreover, since a, is constant, there is no need to consider the arguments. Finally we shall have Z a,"V(x-m) and a,"-ly, . Particular case 2. If the given equation is homogeneous, then we have V(x) =O and the solution (3) will become
(5)
593 times, where a,+az+ . . . +a,,=1 has been put. Indeed this expression is equal to the number of permutations of rl elements among which there are a1 elements a, , moreover a, elements a2 and so on, u,, elements being equal to a,. Therefore formula (6) may be written
i,! ','(X,rn) = z a,la,l..,a,i a,"la,"*....a,"*
where the sum is extended first to every combination with repetition of the numbers a, = 0, 1,2, . , , , satisfying the equations: ~,+u~+...+~,,=i a n d u1+2a,+ ,.., +na,,=m and secondly extended to every value of 1. from the smallest integer not less than m/n to m (inclusive). Finally f(x) is given by (5). In this way we obtain the solution of the difference equation without determining the roots of the characteristic equation. This is an advantage of the method; but the result is obtained in the form of a sum generally very complicated. If we solve the problem in both ways, then, equalizing the results, we may obtain interesting formulae. Example 1. 1 Waring, Meditationes Algebraicae, Cantabrigiae, 17821 Newton has deduced a differen& equation, one of whose particular solutions f(x) is equal to the sum of the x th powers of the roots of a given equation. Writing this equation in the following way: @I he found r~-aa,rn~'-aa,r"~2-. , , .-arrmlr-a, = 0 f(l) = a, f(2) - a,f(l) = 2a, f(3) -a,f(2) -aa,f(l) = 3a,, ;(i; -'aif (n-l) -. . , , - a,,-,f (1) = nail and if x > n, then (9) f(x) - a,f (x-l) - a,f (x-2) - . . . - a,f (x-n) = 0. The last equation is a linear equation of differences of order I? whose characteristic equation is given by (8). Therefore,
3u
that is, f(x) obtained from Newton's difference equation (9) will be really equal to the sum of the x th powers of the roots of equation (8). Waring has determined the generating function of f(x); this we will obtain by An&d's method of solving equation (9), Remarking that in the problem considered the n given values figuring in this method are y* = 20,) . . . . , yn = nun. Yl = a1 9 Equation (9) being homogeneous therefore, according to formula (S), its solution will be f (4 = 2 y (x,m) (x-m)a,-, . m==x--n
Putting into this equation the value of t+l(x,m) obtained in the case of equations with constant coefficients (7) we find
Now the second sum is independent of i, therefore we may execute the summation with respect to i. Since Z: iai=x hence finally we shall have
f(x) = x x +I+
a,! . . . . a,!
* * - + ac-1)
! Qla'.
. . a =I,
"
This sum is to be extended to every combination with permutation and repetition of the numbers a,. satisfying u,-+2a,+nu,=x. This is Waring's formula. It is easy to see that f(x) is the coefficient of tX in the expansion of the following expression x X - (a,t+a,t'+ . . .+a,P)p r.=cJ v that is in log(l--a,f---Q,t'- . . . . -a,f")-" therefore this is the generating function of f(x). Example 2. Equation with variable coefficients. Problem of Coincidences. (Rencontre.) From an urn containing the numbers 1, 2, 3, . . . , x these are all drawn one after another. If at the m th drawing the number m is found, then it is said that there is coincidence at the m th drawing. The number of the possible cases is equal to the number of permutations of x elements, that is to xl. If we denote by f(x) the number of permutations in which there are no coincidences at all; then the number of permutations in which there is one f f (x-l); the number of those in which 0 there are two coincidences is equal to G f (x-2) and so on; 1 1 finally, there is only one permutation in which there are x coincidences. coincidence will be
x+1 1
596
Therefore we may write the following difference equation f(x)+(;]f(x-l)+(;jf(x-2)+...+( *Q(l) = x ! - 1 . From (2) we may deduce
Since x, =x, x,=x-k1 , xzI=x--RI-k2 putting
and so on, therefore
I kz 1 =
x2
(x-k,) ! k,! (x-k,-k,) ! ' . ' * ' '
we obtain
In this sum we have
112)
k, + k, + . . . ki = m.
Therefore if i is given and the k, > 0 then according to (5) 5 60 we have '
1 i! -q k,!k,!...ki!=m!
where the sum is extended to every value of k, different from zero and satisfying equation (12). The number EL is a Stirfing's number of the second kind. In consequence of (5) Q 58 we have mt1 iIZl (-l)'i! Ei, = (-l)m the expression of tp(x,m) will be equal to (-l)NJ(i) and finally according to (3) f(x) = & ( - 1 ) " ' [i] ) ( x - m ) !-1) = x ! 2: (-1)" -$ Example 3. In ($ 83 we deduced Lacroix's difference equation giving f(x)=A-'u"-'; we found: t(
597 ( 1 3 ) f ( x ) +(x~l)ff(x-l)+[x~')$ +-( f(x-2) +***.+
X-l y ]& f(X-Y)f....+;f(l)+
The initial condition being f(1) = u + c. Let us solve this equation by aid of the preceding method. We have x-l 1 a , . = - y (14) v+1 and
I
599 The sum being extended to every combination of order i with repetition and permutation of the numbers R, = 1,2,3, . . . such that k,+k,+...+ki=m, (k,.>O). To obtain f(x) we have still to sum the expression (18) from i=l to i=m+l. In 5 83 we found the symbolical formula f(x) = $ lu+B]*' + k. Equating the coefficient of u"-"' in this expression and in (16) we have (k 1 fl)'*2 'Ilf'; . . . (ki+l) 1 (k where k,+k,+ . . . + ki=m and k, > 0 and the sum is extended for every value of i from one to m. Moreover in consequence of (18) (20) B,,, = :
,- I k-iii Cm.",-i *
(19)
B,"=m! X
I m I Q 179. Sum equations which are reducible to equations of differences. Besides the forms of difference equations (4), (5) and (6) considered in Q 164 there is another form in which sums and differences figure. The simplest case is:
(1)
@[x,f(x), Af(x), A?f(x), . . . , 5 f ( i ) ] = 0 .
i=a
In certain cases this equation may be solved; for instance if it may be written in the following way: c4 5 1=11 f(i)=QP,Ix,f(x),Af(x),...]
then performing on both members the operation A we find
f(x) = A@,Ix,f(x),Af(xj,.
. - .I
an ordinary equation of differences. Example 1. Probability of repeated trials. In a game the player gets n times the amount of his stake, if he wins. If he
600 loses he plays again, staking each time anew, and continues till he wins in the end. Let us denote by f(i) his stake in the i th game. This quantity is to be determined in such a manner that if the player wins, say the x th game, he gets back not only the stakes he lost previously but moreover a certain sum s fixed in advance. Therefore we shall have
x+1 nf(x) = s + ,zl f(i)
where the unknown functions u and v are to be determined, Starting from these equations we may deduce a difference equation containing only one function. Indeed, executing the operation q,(E) on both members of the first equation, and the operation v,(E) on both members of the second, we get, after subtracting the second result from the first: (4
LY~E)Y(E) --~,(E)Y,IE~I~=Y~(E)F(~) -dEV,(x).
This is a complete linear equation of differences with constant coefficients whose order n is equal to the highest power of E in the first member of (2); n will also be equal to the number of the arbitrary constants in the solution of U. Denoting by V(x) a particular solution of (2), then if the characteristic equation has no multiple roots, the general solution of (2) will be
(3) u = c,r, x+.... + C"r"X + v (4,
In the same manner we could get a difference equation of order n determining the function u. Denoting by U(x) a particular solution of
(4)
It may be shown that if V(x) is a iarticular solution of (2) and U(x) a particular solution of (A), then it follows that V(X) and U(x) are also solutions of equation (1). u contains also n arbitrary constants, but they are not independent of those figuring in v. Indeed, from the first of the equations (1) it follows, in consequence of formula (2) 5 165, that n+1 I: [Ci Y(fj) + Ci' y((tj)] rix = 0.
bl
Since this is to be satisfied for all values of x, therefore
602 the coefficients of each exponential must be separately equal to zero, so that we have
Remark. If the equations (1) are homogeneous, the functions u and u will differ only by the values of the arbitrary constants. Example 1. Given (4E- 17)~ + (E--4)~ = 0 (2E- 1)~ + (E--2)u=b , from these it follows that and therefore moreover IE"-8E+15]o = 0 v = c, 3" + c* 5x 1 c,' = --cl 5 so that 1 u=- FC, 3.+,5-Y If the characteristic equation of (2) has double roots, the calculus is a little more complicated. We have seen that in this case the solution is of the form 0 = (c, + c2x)r,x + . . . . u = (c,' + c,'x)t,S + , . . . Before putting these values into the first equation (l), let us deduce the following formula. In (j 165 we had p(E)ry = r"q(r). Derivation with respect to r followed by multiplication by t gives (6) a n d c2' = -J-C., 3 -
8 181. Introduction. If z is a function of the two independent variables x and y, so that z=f(x,y), then the equation
(1) F(x,y,z, AZ, AZ, A:z, . . . . , A*z, Amz) = 0
x
Y
x
x
I
is called an equation of partial differences, or a difference equation with two independent variables. Eliminating from equation (1) the symbols A and A by aid of E=l+A and E=i+A we obtain a seconds formYof x x Y if equations (4 these
F,(x,y,z,Ez,Ez,E"z,E"z,. . . . , E"z,E"z) = 0. x !lx Y x B
The function z may be determined by aid of (2) if certain initial conditions are given. But in these cases a few particular values of z are not sufficient, here particular functions must be given. For instance, if equation (2) is a linear equation of the first order with respect to x and also with respect to y, and if the equation contains each of the four possible terms, so that we have (3) ~11 E F z + =ol if z + a,, E x z +a,,,~ = V(X,Y)
then, to enable us to compute z=f (x,y) for every integer value of x and y, two functions must be given as initial conditions. For instance: f(x,O) = q(x) and f(W) = Y(Y)
for every positive or negative integer value of x and y. If the two functions are given, then putting x=0 and y=O
605 into (3) the only unknown quantity in this equation will be f (1,l). This being calculated we put into (3) x=1 a n d y=O and get f (2,1), and so on. Moreover putting into (3) x=-l and y=O we obtain f (-1,l) ; and so on step by step we may find any value f (x,1 ) whatever. [Figure 13,) In the same way, starting from x=0 and y= 1 we now obtain r(x,2) and starting from x=0 and y=2 we get f[x,3). Continuing we may find finally any value of f (x,y) whatever. In the general case of a difference equation of order R with respect to x and of order m with respect to y, the number of the possible terms . Ei'& for r=O,1,2 ,.,. n a n d ,~uO,l,... m will be eiull to (n+l) (m+l). Figure 13.
Indeed, after having put x=0 and y=O into (2), the above equations will give mn+m+n of the quantities figuring in (Z), so that there will remain in this equation only one unknown, f (n,m). Having determined this we may proceed to the determination of f(n+l,m), and so on. But if from the (n+l) (m+l) possible terms in equation (2) some are missing and if x and y are positive, then the number of the necessary conditions may often be reduced. For instance,
I
606 if the equation is of order n with respect to x and of the first order with respect to y; moreover if there is but one term E in the equation, and if y 2 0, then instead of the nf 1 neceszary conditions, one condition will be sufficient. For instance if we have %,f(x+zY) + al~f(X+l,Y+l)+a,,f(x+l,Y) + %mY)=v(x~Y) then f (x,0) =q(x) given for every positive and negative value of x will be sufficient for computing step by step every number f(x,y). Indeed, putting y=O and x=-l into this equation, the only unknown will be f (0,l) ; this being calculated we put into the equation y=O, x=0 and obtain f(l,l),and so on. (Figure 14.) Figure 14.
We may determine in every particular case the number of the necessary and sufficient conditions for the computation of f(x,y). We have to dispose of them so, that putting the corresponding values of f (xi,yj) into equation (2), there shall remain only one unknown in it, for instance f (nm). But this must be done in such a way that having determined f(n,m) we may proceed in the same manner to the determination of f(n+l,m) and PO on. The conditions must be independent; that is, no condition shall be obtainable starting from the other conditions, by aid of the equation of differences. Let us suppose that the necessary initial conditions cocresponding to an equation of partial differences are the following functions: kY,)t f(X,Y,), * * * * 9 f (X.Yj)i f(X",Y), f(X,vY) - * * * f(Xi,Y) given for every value of x and y; if the arbitrary functions contained in the solution are such that they may be disposed of so as to satisfy the above initial conditions, then this solution may be considered as the general solution.
607
g 182. Resolution of linear equations of partial differences with constant coefficients, by Laplace's method of generating functions. If the given equation is of order n in x and of order m in
y then the complete difference equation may be written in the following manner: n+l mtl x ,,zo a,.,t f(x+y,y+p) = V(x,y). (11 v=o Let us call u(t,t,) the "generating function of f(x,y) with respect to x and y" if in the expansion of u(t,tJ into a double series of powers of t and t, the coefficient of fXf,Y is equal to f (x,y) ; if x and y vary from zero to OQ , then (2)
u(f,f,) = *;. $2
f(X,Y) fXf,Y
[should y vary for instance from one to 0~ , we should simply have to put f(x,O) =0-l. This generating function is also denoted by G . f(x,y). If first we expand zz(f,f,) into a series of powers of f %d have (3) u(f,fJ = z. w(x,fJ fX
then we say that u(f,fl) is the generating function of w(x,fJ with respect to x, that is p w (XJJ = u Wll
1, is in this formula only a parameter.
Expanding w(x,t,) into a series of powers of t, we get (4) w(x,fJ = #z. f(w) f,y = 1
G.f(x,y).
Here w(x,fl) is the generating function of f (x,y) with respect Y; in this formula x is a parameter. In this manner, instead of a generating function of two variables we have obtained two functions of one variable. From (4) we deduce directly
to
Let us denote by R(x,t,) the gknerating function of V(x,y) with respect to y, that is 9. V(x,y) = R(x,f,). Now we may write that the generating functions with respect to y corresponding to the terms of equation (1) satisfy this difference equation. Therefore we have n+1 m+l (5) .zo ,z, QWfl m-@ ~~W(XfV,f,) -f (x+v,o) - f,f (x+v,l) -. . , --q-if (x+v,p-1)] = f,mR(x,t,). This is a linear difference equation with constant coefficients of the variable x, of order n; it contains already m arbitrary functions of x f(x+v,,u-1) where !c = 1,2,3, . . . , m. Into the solution w(x,f,) the3.c will enter moreover R arbitrary functions of f, , The expansion of w(x,f,) into a series of powers of f, will give f (x,y); the arbitrary functions of f, expanded will give n arbitrary functions of y. Finally the m-/-n arbitrary functions of x or of y are determined by the aid of the initial conditions. Example 1. Problem of poinfs. In a game the first player needs x points to win the stakes; the second needs y points. The probability that the first player shall win the stakes, is denoted by f (x,y). Let us suppose that the probability of winning a point is p for the first player and 1-p=q for the second. To determine f(x,y) let us remark, that the first player may win the stakes in two different manners; first by winning the next point, the probability of which event is pi then, as he needs now only X-I points to win and his adversary y points, the probability of his winning the stakes will be f (x-l ,y) ; and the compound event, viz. of winning the next point and afterwards the stakes too is pf (x-1,~) .
609 Secondly, the first player may win the stakes by losing the next point, the probability of which is q; then for winning the stakes he needs still x points and his adversary y-l Faints, therefore the probability of the compound event will be qf (x,y-I). According to the theorem of total probabilities, the probability t(x,y) that the first player shall win in one of the two ways is equal to the sum of the two probabilities obtained above, that is, to f(w) = pf(x-1,y) + qf(x,y-I). We shall write this equation of differences in the following manner: (6)
f(x+l,y+l) - Pf(x,Y+l) - qf(x+l,y) = 0.
To solve this equation we have to put into (5) n=m=l; we find (7) (I-q~,)w(x+l,t,) -puJ(x,f,) = f(x+l,O) -pf(x,O). According to 8 181, it is easy to see that, knowing the values of f(x,O) and f (0,y) we may compute f (x,y) by aid of the equation of differences, step by step for every integer value of x. But f(x,O) is the probability that the first player wins the stakes if he needs x points and his adversary none. This probability is obviously equal to zero, Therefore f (x,0) =O for every positive value of x; f(O,O) cannot occur. f (0,~) is the probability that the first player wins the stakes if he needs no points at all; if y > 0 this probability is equal to one, since he has won already. Therefore f (0,y) =I. Since f (x,0)=0 if x > 0, therefore from (7) it follows that w(X+lrfl) - --E-I-+, If x=0 then we have Since f (0,~) = I, hence W(O,f,) = & . I The equation above is a homogeneous linear equation of
39
w(x,fl) = 0.
w(O,f,) = f(O,l)f, + f(0,2)f," + . . . . .
610 differences of the first order, with constant coefficients; hence its soiution will be
y(tJ must be determined by aid of the initial condition corresponding to x=0, therefore Hence
Example 2, The problem of coincidence considered in f$ 178 (Ex. 2), somewhat generalised, will lead to a partial equation of differences. From an urn containing the numbers 1,2,3, . . , - , y these latter are all drawn one after another. The probability is required of not having coincidence in x given drawings. The number of the different ways of drawing y numbers is equal to the number of permutations of y elements, that is, to y! If we denote by f(x,y) the number of the favourable cases, that is, those in which there is no coincidence in x given drawings; then the required probability will be f (x,y)/y! But f (x,y) may be considered as equal to the number of ways of drawing y numbers so as to have no coincidences at x-l given places, that is to f(x-1,y) less the number of ways of drawing the y numbers; so that there is coincidence at the place x, and no coincidences at the other given x-l places, This last number is obviously equal to f (x-l,y-1). Consequently we have f (x,y) = f (x-1,y) - f (x-l,y-1).
611 Let us write this equation of partial differences in the following manner : (8) f(x+l,y+l) - f (x,y+l) + f (xy) = 0. Since this equation contains only one termof x+1, moreover, since y 2 xl 0, therefore one initial condition will be sufficient for the computation of f (x,y). Such a condition is for instance f (0,y) = q(y), But f (0,y) is equal to the number of ways of drawing y numbers without any restriction at all; therefore f(O,y)=yl for y=O, 1, 2,. . . . Moreover, since there is only one way of not drawing numbers, we have f(x,O) ~1. Starting from the conditions f(x,O) ~1 and f (0,~) =y! we may compute step by step every value of f (x,y), In consequence of (5) the generating function of f(x,y) with respect to y, denoted by w(x,fl), will be given by w(x+l,t,) - (1-f,)w(x,t,) = f(x+l,O) -f(x,O) = 0 therefore, if x > 0, the solution is W(XJll = P'(h) U---t,)" moreover if x=0, then
w(O,f,) = f (0,O) + f (OJ) f, + f (0,2)f,2 + . # .
and in consequence of f (0,~) = y I we have: w(O,f,) = ; Y! f," this is equal to the arbitrary function q(f,), and we get w(x,f,) = (l-fr)" I: V!f,Y The expansion of this function will give w(x,f,) = z Y! f,v x (--1P (;)v Finally, putting Y+,u=Y, the coefficient of fly will be equal to f(JGY) = 2 w(;) b-Al
In the particular case of y=x we obtain a formula already deduced in $j 178: x+1 (-1)" f(x,x) =x! x 7 nr=o From this the probability of not having coicidences at all in x drawings is obtained by dividing f (x,x) by x! , If x increases indefinitely this probability tends to e-l . Example 3. The rule of computation of the numbers in Pascal's arithmetical triangle is given by the following equation of partial differences:
f(x+l,y+l) = f(x+l,Y) + f(XvY).
Since this equation contains only one term of ~4-1, and since y 2 x 2 0 therefore the condition, which follows from the definition of these numbers: f(x,O) = 0 i f x=0 a n d f(O,O) = 1 will be sufficient for the computation of f (x,y). According to (5) Gf(x,y) will be given by u(y+lJ) - (;+t) u(u,f) = f(O,y+l) - f (O,y), Starting from the initial conditions it is easy to show that f (0,y) ~1, therefore we have
u(y+l,t) = u+q 4YA).
The resolution of this equation gives
u(u,fl = q$) (l+w
for y=O we get u(O,f) 'v(f); but in consequence of the initial condition it follows that u(O,f) = f(O,O) + f(l,O)f + f(2,0)f2 + . . *. = 1 so that finally and
u(u,f) = u+qy f(XtY) =(g.
Example 4. We have seen in 5 58 that, denoting the Stirling numbers of the second kind a; by f (x,y), they satisfy the following equation of partial differences:
613 (9) f(X+l*y+l) - (x+1) f(x+l,y) - f (x,y) = 0.
Since this equation contains only one term in y+l, and y 2 x 2 0 hence the condition f(x,O) =O if x=/=0 and f(O,O) = 1 which follows from the definition of the Stirling numbers, is sufficient for the computation of f (x,y). According to (5) w(x,tl) the generating function of f(x,y) with respect to y is given by the difference equation: w(x+W - f(x+l,O) - (l+x)t, w(x+l,t,) - t,w(x,t,)= 0. Since f (x+1,0) ~0 therefore we have
(10)
w(x+lA) - &$-- w(x,t,) = 0.
1
This is a homogeneous equation of the first order with variable coefficients, whose solution is (5 173):
w(x,fJ = w(O,f,) ii
t0
4
1-(l+i)t, *
To determine w(O,t,) let us remark that starting from the initial condition it is easy to show that f(O,y) =O if y > 0; therefore w(O,t,) =f(O,O) +f(0,1)t, +f(0,2)t,2+... = 1 so that finally w(x9tJ = (1-fl) (l-2;;, . . , (1-xf,) * To of t, it expand. cient of expand this generating function into a series of powers is best to decompose it into partial fractions and then This has been done in (j 60, where we found the coeffit,Y equal to f(x,y) = 9 2: (-l)i(;)iY = E,". Example 5. In 8 144 we denoted by F(E) the frequency of E which was given for 5 = 0, 1,2, . , , . , N-l. There in a table, we put into the first line of every column the same number
614 F(JV-1) ; moreover into the first column we put F (N-1). F(N-2), F(N-3), . . ., F(l), F(0). Denoting by f (x,y) the number figuring in the line x and in the column y, we have f(l,y) = F(N-1) f (x,1) = F (N-x) where x 2 1 and y 2 1. These are the initial conditions. The other numbers f(x,y) of the table are computed by aid of the equation (12) f (X+l,Y+l) - f(x,y+l) - f (x+l,y) = 0, Since in this equation there are two terms of y+l and two of x+1, hence, according to Q 181, two equations of condition are necessary to compute f (x,y), The two equations given above are such. To solve equation (12) we shall denote by w (y,t) the generating function of f (x,y) with respect to x; since in our problem x > 0 therefore w(y,f) = f(l,y)i + f(2,y)f2 + . *, +f(x,y)fX+ *. . Starting from equation (12) we obtain w(u+lJ) - +y w(y,f) = 0. The solution of this linear equation with constant coefficients is
This will give, in the particular case, the number of the line x=N-y+2 and of the column y: f (N-y+2,y) = 3;' (;--;I F(N-lg. Therefore this number is equal to the binomial moment of degree y-2, of the function F(e): WV--y+2,y) = ,&.+ Computing the table mentioned above the required binomial moments are obtained all at the same time by simple additions: this is the shortest way, since no multiplications are necessary. Example 6. Bernoulli's formula of the probability of repeated trials, Let the probability of an event be equal to p at each trial. The probability is required that the event shall occur x times in n trials. Denoting this-probability by P(n,x) we obtain by aid of the theorems of compound and total probabilities (1) P(n+l,x+ll = pP(n,x) + gP(n,x+l). Indeed the probability that the event shall occur x+1 times in n+l trials is eqtial to the sum of the two probabilities: first, the probability that it shall happen x times in n trials, and moreover also at the n+l th trial; and secondly, the probability that the event shall occur x+1 times in n trials and fail at the last trial. Equation (1) is a linear homogeneous equation of partial differences of the first order with respect to both variables. To solve it, since nT XT 0, according to $j 181 one equation of initial conditions is sufficient, For instance, if P(O,x) is given for every value of x; but P(O,x) =O unless x=0 and then P(O,O) =l; indeed, if the number of trials is equal to zero, then the event can only occur zero times; hence the probability of it is equal to one.
Indeed, from the theorem of compound probability it follows immediately that P(n+l,O) - gP(n,O) = 0 hence the solution of equation (2) will be u(d) = C(pf+q)". In consequence of the initial condition u(O,f)=l, we have C= 1, and P (n,x) will be equal to the coefficient of fX in the expansion of u(n,f)=(pf+g)", that is P(n,x) = (4) px qpx.
Q 183. Boole's symbolical method for solving partial dffference equations. This method is applicable *to partial difference
equations in which one of the variables (e. g. the variable y) does not figure in an explicit manner: (1)
y(x,E,E)f(x,y) = 0. x Y
We write k instead of the symbol E, and considering it
as a constant solve the equation (1) and f&d f(w) = Y,(JG~ V(Y) = wlC$' ~4~1 where q(y) is an arbitrary function of y. Finally we obtain f (x,y) by performing the operation y1 (x$) on the function q(y).
617 As in every symbolical method, the result obtained must be verified by putting it into equation (1). The function v(y) will be determined in particular cases by aid of the initial conditions. Example 2. Given the difference equation y(x.ul - y(x.y) = 0 let us put E=k and have I
Ef (x,y) = kf (x4 x
the solution of this homogeneous equation with constant coefficients is f(XtY) = kXdY) = yY) = v(x+Y)* Where q(y) is an arbitrary function. Since the equation contains only one term of x+1, so that to compute f(x,y) for x20 only one equation of condition is necessary, and this is for instance f(O,y) given for every integer value of y. Putting x=0 into the above equation, we get f (0,y) =9(y) and therefore f(XPY) = f(O,Y+x)* It is easy to verify that this result satisfies the given equation. Example 2. Given the equation ~~f(x~Yl-~frx~Y~-f(x,Y)=O putting E=K we have ?f
k E f(x,y) - (h+l)f (x,y) = 0 *
the solution of the equation with constant coefficients is
Finally the expansion gives
This result satisfies the given equation. In this case too, one initial condition is sufficient for the computation of any value
618 of f (x,y) s For instance if x 2 0 and if f (0,~) is given for every integer value of y, we find f (0,y) =q(y) and fIx,yl = 2; (;) f(O,y--il. Example 3. Given the equation E F f(%Y) - xf (x,y) = 0 the transformation gives
Since the given equation contains only one term of x+1, hence if x 2 0 one initial condition, for instance f (1,y) given for every integer value of y, is sufficient for the computation of f(x,y). We find f(l,y) =p(y) and fkY1 = ( x - l ) ! f(l,y-x+1). Example 4. The difference equation giving the numbers in Pascal's arithmetical triangle is f(x+l,y+l) - f(x+l,y) - f(x,y) = 0 or written in the symbolical way
The function y(y) is determined by aid of the initial values. Since the given equation contains only one term of y-l-1 and y 2 0 hence one initial condition f (x,0) =O if x =j=O and f (0,O) = 1 is sufficient for the computation of f (x,y).
619 Putting into the above result x=0 we get Y(Y) = f(OtY)* From the initial condition it follows that f (0,~) = 1. Therefore e(y)=1 and f(w) = +-"l= (,y) +F(y) where F(y) is a polynomial of y; but putting x=0 into this equation we find F(y) ~0. Finally
fhY) = (;]
that is, the numbers in the arithmetical triangle are the binomial coefficients. 0 184. Method of Fourier, Lagrange and Ellis, for solving equations of partial differences, Given the linear homogeneous equation of partial differences (1) v'(~*~)f(x,Y) = 0
and likewise the necessary initial conditions corresponding to a problem, the method consists in determining a certain number of particular solutions of equation (l), multiplying them by arbitrary constants, forming the sum of the products obtained, and finally disposing of the arbitrary constants in such a manner that the initial conditions may be satisfied.57 The method was first applied in the case of partial differential equations; but there the difficulties were much greater; indeed, to satisfy the initial conditions, the number of arbitrary constants and therefore that of the particular solutions to be determined is infinite. On the other hand difference equations are generally valid only for a finite number of values of the variables, and therefore it will suffice to dispose of a finite number of arbitrary constants, in order to satisfy the initial conditions. Of course, the number of these constants will be very great, especially in the case of three, four, or more variables; so that it -
620 would be impossible to carry out the calculations in the usual way; this is only possible, as we shall see, by means of the orthogonal properties of certain trigonometric functions given in 9 43. Example 1. "Third Problem of Play." TWO players have between them a number a of counters; they play a game in which the first player has a chance p of winning one counter from the second in each game. At the beginning, the first player has x counters. Required the probability that after a number y of games the first player shall have z counters, neither of the players having previously lost all his counters (ruin). Let us denote this probability by f(x,y,z). If the first player wins the next game, of which the probability is p, then the required probability becomes equal to f(x+l,y-1,~). If he loses it, the probability of which is l-p=q, then it becomes equal to i(x-l,y-1,~). Therefore, applying the theorem of total probabilities, we get (2) 1(x,y,z) = pf(x+l,y--lJ) + qf(x-l,y-1,z). This is an equation of partial differences of two variables, which may be written (3) pf(x+2,y,zl - f(x+l,y+l,zl + qf(x,y,z) = 0 where z is merely a parameter, Since there is only one term of y+l in this equation, and moreover since y > 0, hence, according to 9 181, one initial condition will be sufficient for computing the values of f(x,y,z) by aid of equation (3). Such a function is, for instance, f (x,0,2), given for every integer value of x from -00 to +", We shall suppose that 0 < z < a. Obviously we have (4) f(x,O,z) = 0 if x $I z and f (z,O,r) = 1 this will give for x = 1,2, . . . , a-l in all a-l conditions. Starting from (4) we may compute step by step any value of f (x,y,z). There is but one difficulty; in our problem there are two supplementary conditions, viz. : (51 f (O,y,r) = 0 and f (a,y,r) = 0
621 that is, the probability of having I counters after y games is equal to zero if one of the players has already lost all his counters, since then the play is over. But the values (5) are necessarily incompatible with those corresponding to the difference equation (2). To obviate this inconvenience, we will restrict the validity of (2) to the interval 0 < x < a; then there can be no contradiction. In the end we shall have a-l condition to satisfy, therefore the number of the arbitrary constants in the particular solution must also be equal to that number. It is easy to find a particular solution of (3), by putting f (x,y,Z) = ay F(x) and disposing of a in such a way that equation (3) shall be satisfied. From (3) it follows that: (6) pF(x+2) -aF(x+l) + qF(x) = 0.
If we write a=2 cosy ],'Fq then the roots of the characteristic equation corresponding to (6) will be r = (cosfp f i sinq) (:r. In consequence of the sign + this gives, according to 8 165, two different solutions of (6), so that we obtain the general solution of (6) F(x) = [%r" [A,cosqx + A,sinvx]
To satisfy the condition f(0,y.z) =q we have to put in (7) A,=O. Moreover, to satisfy f(a,y,z) =O we put ~=wz/u, where v is an integer such that 0 <Y < a. To each value of Y there corresponds one particular solution of (3) ; multiplying it by the constant C,. and summing from r=O to ~=a we get a particular solution containing a-l arbitrary constants.
622 (8) f(x,y,z) = (4~g)"~ [:jd2 ,jio C,. s i n y ( c o s F]".
(There is no objection to beginning the summation with Y=O instead of r=l, since the term corresponding to 1~x0 is equal to zero.) The number of the conditions (4) still to be satisfied is also equal to a-l, therefore we may attain this result by disposing of the constants C,. in the following manner: Putting y=O into equation (8); after division by get for x= 1,2, , . . , a-l the equations
xl2
we
I PI 4
*/2
f(x,O,Z) = 5 C,. s i n y .
IV-=0
Multiplying the first of these equations by sin -?: , the 2p jl.TIX second by sin -;- , . , . the x th by sin ~ and adding up the a results, we find
The first member of this equation is equal, in consequence o f (4), t o 11.72 P 3iz sin '1.4 a 9 The sum in the second member, according to formula (9) 8 43, is equal to zero if JJ=[ ,U and to MaC,, if v=-!(. Therefore we shall have
Finally the required probability obtained from (8) will be (9)
This formula has been found by Ellis (lot. cit. 57, p. 210), it may be transformed in the following way; the product of sines
623 is expressed by the corresponding difference of cosines, and the power of cosine by cosines of multiples. Let us suppose first that y is odd; y=2n-1; then the sum in the second member may be written:
The products of cosines occurring in this formula are again expressed by sums of cosines. We find
(10)
_ cos 2i-1 +x+f y7 _ cos Zzkl-x-z a a
V.-l
I
In Q 43 we have seen that 0 ): cosy =o (11) v=o if a is an integer not divisible by a. Therefore the first term of (10) will be equal to zero unless we have 2i+x-z-l= l a o r i = l/z(ia-x+2+1). Since the number y of the games is odd, hence z-x is also odd; moreover, since i must be an integer, therefore ?.a must be even. If 1 is odd, we have (12) 5 cosl.vz = 1 - (-1p. r=o
But if 1 is odd, then a is necessarily even, and the above expression is equal to zero. Hence it is sufficient to consider the even values of R, If rZ is even, then (13) i u=o codvn = a.
If we put rZ=2y, then in the first term of (10) we shall have i=a~+l/~(z--x+1), and the sum of this term will be 2n-1 aZ I n-ay- fGz+ y~x--y~ . 1
624 Since we must have i 2 1, hence if z > x then 7 may be equal to 0,1,2,... and if z < x, then y = 1,2,3, , . , In the second term we have i=ay+1/ (x--1+1), therefore it gives aZ 2n-1 [ n-ay--'/2x+ 1/~--1/~ I .
If x > z, then in this sum we have y=O, 1,2, . . , , if not, then y=l, 2,3,. . . Since
(pq)" ff)e-z"2 .
Moreover, the two first terms under the Z signs corresponding to x-l and x-l-l, are 2n-1 2n-1 I n-l-ay--'/x+I/zr I + ( n-oy--Mx+yzr We have seen in 5 22, formula (13) : I *
therefore the sum of the two terms above will be equal to 2n I n-up-*/~x+l/~z I * Combining in the same way the other terms two by two, we find
From (14) and (15) it follows that for y even or odd we have
4
+
40
y=o
z
Y
I( 1h~-%x+
+ $5 z--v 1 - I %Y--?4 x--+&-ay Jl
Y
Y yz, [l%Y+Mx%+aY 1 - I %Y + ?4 x+ l/zz-q 11) -
I
I
I
626 The probability above is necessarily equal to zero if y-z+ x is odd; therefore, to simplify, we may put 2w=y-2+x. We find (16)
.
1
y=o
r,
[I
co-+-r-x-q
Y
I
-
I
Y w-x-ay II +
This formula is advantageous for the computation of the required probability. Particular Cases. 1. If the two players have the same chance for winning, then p=q= 1/2, and the term preceding &he sign Z in formula (9) reduces to 2/a; moreover the term preceding the brackets in (16) will be equal to 112~. In this case the formulae will be symmetrical with respect to x and z. 2. In the particular case of x=z, formulae (9) and (16) give the probability that the players have the same number of counters at the beginning and at the end of the play. The formulae are much simplified by this substitution. 3. An important particular case is that in which the number of counters the second player has, may be considered infinite; this occurs if he has more counters than the number y of games he will play, that is if a-x>y. But in this cast we have to put into formula (16) y=O; it will become (17) f (X&Z) = (pq)"'2 [ f]'"i" [[ w+:-x) - ( ,41_, ]] *
This formula was found by D. Arany.5s We may obtain another formula for the required probability, by starting from (9) and putting into it p)=vz/o and therefore &p=;r/a. Now if a increases indefinitely, then the sum in (9) will become a definite integral:
5x Considerations sur le Probleme de la Dur&e du .leu. TBboku Mathematical Journal, 1926, Vol. 30, p. 160. What in our notation is f(x,y,z), is in Arany's notation, in the case of a = 00: Y:-~,~-~; and in the general case: .r.a--z YI-z.z-*
In the case of z=x this may be simplified as has been mentioned before. Problem of ruin. Let us suppose now that play continues until one of the players has won all the counters. What is the probability that the first player shall lose his last counter at the y th game? This would be f (x,y,O) ; but this number cannot be obtained by putting z=O into formula (9) or into (16). Indeed, in establishing these formulae we supposed z to be different from zero. Nevertheless we may derive this probability from these formulae. First, by determining the probability that the first player has only one counter left after y-l games; this is equal to f (x,y-1,l) and may be obtained by the formulae (9) or (16) ; and secondly by writing that he lost the y th game; the probability of this is equal to 9, so that the required probability will be: gf (x,y-1,l). Putting y-l instead of y and r=l into formula (9), we find after multiplication by 9
This formula has also been obtained by Ellis, but starting anew from the difference equation (2). We may obtain qf (x,y-1,l) also by aid of equation (16), putting into it y-l instead of y and z=l ; after multiplication by 9 we find
628
where 2w=y+x-2. If moreover we put a=m, then the above formula will be
This may be simplified; indeed in Q 22 we found formula (12) :
therefore the difference in the brackets is equal to - 2 x- y - l y-x I 0-x 1 so that the required probability will be (21)
f(x,y,O) = (pq)"'" [g','$ [;zf, 1.
This formula is identical with that found by Ampere.59 We may obtain a corresponding formula by starting from (9) and putting q-an/a and therefore Ae,=n/a. Now if a increases indefinitely, then the sum in (9) will become a definite integral (22) f(x,y,O) = (4pg) V!2 [ $)"'2+ 1" sin91 sinxy (cosv)rl dp, .
0'
From this it follows the probability v(x) that the first player shall be ruined at all, by putting w = 00. We find (23)
This probability can be determined directly by aid of the difference equation v(x) = P cL (X+1) + Qvb-1) the initial values being equal to q(O)=1 and ~(a) =O. According to Q 165, the solution will be
-
y(x) = c, + c, I % r
and taking account of the initial conditions we find (24) y,(x) = a" p-" - v-1 pa--Q" ' Now from (23) and (24) we deduce the value of the definite
SUIll
where we put 2 II&= k. To have the definite sum, we must still
630 determine p and Q by aid of K. Starting from 2 / p(l-p) =K, we 1 find p=1/2(1-yl-k?) andq=?+(l+1/1-K*) or vice versa; the formula being symmetrical with respect to p and q. In the particular case, if a may be considered as infinite, we shall put into the first member of (25) ~=vn/a, then q will become a continuous variable whose range is 0,~. In the second member, if q has been chosen for the larger root, we have
Finally we shall have
axis; starting from x, it may advance one step to x+1 or it may go back one step to x-1, the probabilities of both.events being the same, that is, equal to '/2. Having moved, it can again take one step in one of the two directions, under the same conditions. The probability is required that in n steps the point shall be at x, without having touched in its movement the points x-0 and If at the first move the point has advanced one step (probability 1/), then it has still to cover a distance of x,-x--1 in n-1 steps. If we denote the required probability by y(x,n), then the probability of both the above events is l/sq(x+l,n-1). If at the first move the point has gone back to x-l (probability '/), then it has to cover a distance of x,-x+1 in n-l steps; the probability of both events is '/2cp(x-l,n-1). Finally, according to the theorem of total probabilities y (x,n) will be equal to the sum of the two probabilities mentioned.
in Bietens de Haan's book quoted above. Example 2. Problem of Parcours. A point is moving on the x
The formula may be verified by aid of formula 12, Table 64,
x=2a.
q(x,n) = '/zcp(x+l,n-1) + 1$&(x-l,n-1).
This equation of partial differences is the same as that of Example 1.; moreover, the initial conditions are also the same,
631
except that in (9) 2a is to be written instead of a, moreover R instead of y and p=q=yz. We find
Let us consider the particular case in which x=a. Since sin$+vm is equal to zero for the even values of v, hence we may put v=2i+l. Then sinQ(2i+l);r=(-1)'. Introducing moreover the new variable $=X-Q (so that E,=x,-u), the above formula will give: 2i+l 2iJ-1 " c o s -.x&!l co3 -----z F@,,R) =z ; 5 (28) 2a * 2a I i=O
1
This is the probability that the point starting from l=O shall reach in R steps the point f =[,, without having touched in its movement the point [= f a. The formula has been found by Courant and h-any"" in different ways, but in the first paper there are certain mistakes. From formula (16) we obtain by putting into it 2a instead of a, moreover x=a and p-=q=$& y=R and t-n=ll
632 n steps is 2". To obtain the number of ways starting from 6~0 and ending in tl, let us remark that to reach the point tl, in n steps, starting from 6x0 the number of steps in positive direction must be equal to w+tl and in negative direction equal to o. Since the number of steps is n, therefore Hence the number of ways ending in lr in n steps is equal to the number of permutations of R elements, among which there are w+E, elements equal to +1 and w elements equal to -1. That is
and the required probability will be P = (y&-)" [It).
Q 185. Homogeneous linear equations of mixed differences. If such an equation F (A, D, x, I) = 0 is given; where I is an
unknown function of the;wt variables x and y, and F is a polynomial with respect to the symbols A and D, moreover the variable y does not figure explicitly in F, then Booze's symbolical method is the following: Instead of the symbol D we have to write k, and considering it as a constant, to solvl the ordinary equation of differences F (A,k,x,r) = 0. Supposing that its solution is: z=~~(x,k)C,+y,(x,k)C,+...... Now we put k = D and Ci = vi(y) v where the vi(y) are arbitrary functions of y. The equation
If for instance the initial conditions are for x=0 to have z=y"/n! , then q~(y)=y~jn! and
x-t1 I= I:
x izo 0v
(n-v)!
Y"--"
'
As has been said; when solving difference equations by symbolical methods, it is always necessary to verify the results by putting the obtained functions into the given equation. In the case considered, it is easily seen that the general solution (1) satisfies the given equation.
Q 186. Difference equations of three independent variables.
Sometimes it is possible to solve a linear homogeneous equation of differences of three independent variables by using the method of $j 184 due to Fourier, Lagrange and Ellis. Example. Problem of parcours in two dimensions. A mobile starting from the point of coordinates x,y may advance one step to the point x+l,y, or to the point x,y+l; it may go back to the point x-1,y or to the point x,y-1. The probability of either of the four events is equal to l/4. Having moved, it may again take one step in one of the four directions, and so on. The probability is required that the mobile reaches in n steps the point of coordinates xI,y,, without touching the four lines, x = 0, y = 0, x = 2a, y=2b.
The same ratiocination as that employed in the case of the problem in one dimension (5 184) shows that this probability satisfies the following difference equation:
The values of the function f (x,y,n) may d computed step by step by aid of (1) starting from f (x,y,O). In the problem considered this latter is known, Indeed, for every value of X,Y except for x,,y, we have (21 f (X,Y,Ol = 0 a n d f(x,,y,,O) = 1 . Putting nzl into (1) we obtain f (x,y,l); and having determined in the same manner f (x-l,y,l), f(x-l-l,y,l), f(x,y-l,l) and f(x,y+l,l) we may proceed to the computation of f(x,y,2) and so on; finally we should obtain any value of f (x,y,n). In this problem, as in that of 8 185, there are also supplementary conditions to satisfy: f (O,y,nl = 0,
(3) f (2aJm) = 0, f (x,O,n) = 0
f (x,2b,n) = 0.
They are also necessarily incompatible with the results given by (1); to obviate this contradiction we are again obliged to restrict the validity of (1) to 0 < x < 2a and 0 < y < 2b. To determine a particular solution of (l), let us write:
(4) f (x&n) = u" 8" yy .
Now we dispose of a, 8, y so as to satisfy equation (1) ; putting the expression (4) into equation (l), we get after simplification:
(5) yp + (+--4ay+l)~ + y = 0.
If we put (61 cosy = y"-4ay+ 1
then from (5) it follows that (i= v-1) and from (6) Y" - 2(2a--cosqj)
21'
b = cosy k i sing7
y + 1 = 0;
635 Futting into it cosy = 2a - cosp, we have and moreover a = ~~(cosy+cosy), Finally, by aid of these expressions, from (4) we get f [x,y,n) = + (cosp+cosy-f)" (cowpx * i singx) (cosyy rt i sinyy). In consequence of the signs f , this gives four different particular solutions with two arbitrary parameters q and ly, Multiplying them by arbitrary constants and summing, we shall have the following particular solution: f (x,y,n) = $ (cos~+cosp)" (A, sinrpx sinvy+Az cosq~x cosvy + + A, sinvx cos yy + A, cosqx sinyy) I To satisfy the first two of the conditions (3) that is f(O,y,n)=O and f(x,O,n)=O we have to put A,=A,=A,=O. TO satisfy the other two conditions (3) viz. f(2a,y,n) =O and f(x,2b,n)=O, we put y = cosy 2 i siny,
-
where v and /t are arbitrary positive integers. Finally, writing CP,,, instead of A, and summing from v=O to ~=2a, and also from ,u=O to ,u=2b, we obtain a solution of (l), which is sufficiently general for our problem, as will be seen. (7) f (x,y,n] = zO ,4 Cc!, ($17 cos z + cosErsin z sin !!$?!
637 putting these values into (8), the terms sin 7 and sin F in which Y or ,u is even, are equal to zero, so that we may write v=Zi+l a n d p = 2j+1. Moreover, introducing the new variables 6=x--a and 7jzy-b we obtain the probability that the mobile starting from x=4, y =b that is from t=O, q=O reaches the point F1=xl-u, ql=yl-b in n steps: (9) P = i e: g&
iz0 j=O
2i+1 [ cos,=-
2j+l
n+cos~z I
*
-
2i+l 2j+l . c o s __ ;rElCOS 2b nq1. 2a If a and b may be considered infinite, then the expression of the probability (9) is transformed into a double integral; putting u=(2i+l)n/2a and ~=(2j+l)n/2b and therefore Au=n/a and Av=nlb, we get (10) P = - [[ (cosu+cos")~ COSUE, coslq, du du. 2:n' 00
But if a=- and b== then the probability P may be obtained directly by aid of combinatorial analysis. The total number of ways possible in n steps is, under the conditions of the problem, equal to 4". To determine the number of ways starting from the point t=O,q=O and ending in E1,r], let us remark, that the number of steps taken in the positive direction of the 6 axis is equal to i+[, and that in the negative direction to i; moreover, the number of steps in the positive direction of the q axis is equal to to-i+q, and in the negative direction to m-i. Since the number of steps is equal to n, hence we have 2w = n--5,-rj1. Therefore the required number of ways ending in tl,ql is given by the number of permutations of n elements, among which there are i+E, elements equal to a and i elements equal to -a, moreover (o--i+t],) elements equal to b and (o-i) elements to
638 -8, But this number of permutations must still be summed for every value of i from i=O to i=o+l. Hence we have
co+ 1 2 (i+t,ll il fl! ( w-i+v,) ! (o-i) !
(11)
The formulae found by Courant and Arany (lot. cit. 60) are particular cases of formula (9). Equating the quantities (10) and (ll), we obtain an evaluation of the double integral (10). I) 187. Difference equatfons of four independent variables. The method of Q 184 is applicable in the case of four or more independent variables. mobile starts from the point x,y,r and may take one step parallel to the axes fn one of the six directions, the probability of either direction being equal to l/6. Having moved, it can take again one step in one of the six directions, and so on. The probability is required that the mobile reaches in n steps the point of coordinates x1, yr, tI without touching the six planes x=0,
Example: Problem of Parcours in three dimensions. The
y=O, z=O, x=2a,
y=2b,
z=2c.
It may be shown, in the same way as in the case of one or two dimensions, that this probability satisfies the following equation:
639
(1)
f (x,y,r,n) = f f (x+l,y,z,n-1) + f f (x-l,y,*,n-1)
+ +
$f(x,Y+l,r,n-l) +f(x,y,r+l,n-1)
+ +f(x,y-l.lJ+l) + $f(x,y,z-l,n-1).
To compute f (x,y,z,n) by aid of this equation, it is sufficient to know the values of the function f (x,y,z,O). In the problem considered we know that this function is equal to zero for every value of x,y,z except x,,yl,z,, so that we have: (2) f(x,y,z,O) = 0 a n d f(x,,y,,z,,O)
= 1.
Particular case of x=a, y=b, and z=c. Since in this case sinl,zvn, sinl/z,un and sin$@n are equal to zero if v,~ or rZ are even, therefore we shall consider only the odd values, and put v=2i+l, ,uu=2jfl and r7=2k+l. Moreover, introducing the new variables 5=.x-o, v=y-b and [=z-c we find
643 2i+l 2k+l * 2i+l 7 cos - 2a - + cos 2b Jl+ cosypz ' = <:o0 jz kzo 3" abc 6 c 1 1 (11) 2i-L1 a.+1 2k+l . c o s - 35, c o s 2a 2b .ql cos zc n51 where ~l=~l-Q, ql=y,-b and <,=zr-c. If a, b, and c increase indefinitely, this becomes a triple integral. But on the other hand we may then determine this probability by combinatorial analysis. The total number of ways possible in n steps is 6". To determine the number of ways starting from E=O, q=O and 5~0 and ending in E,, tj,, C, let us remark that the number of steps taken in the positive direction of the 6 axis must be equal to i+tl and in the negative direction to i, where i may vary from zero to every possible value, Moreover the number of steps in the positive direction of the q axis must be equal to s+ql and in the negative direction to S, where s varies also from zero to every compatible value. Finally the number of steps in the positive direction of the 5 axis is o-i-s+~l and in the negative direction w-i-s. Since the sum of the steps is equal to n we have 2w = n-t, --,-Cl. Hence the required number of ways is given by the number of permutations of n elements, among which there are i+5, elements equal to c, and i elements equal to --a then s+ql elements equal to /? and s elements equal to -/3 moreover w-i-s+E, elements equal to y and w-i-s elements equal to -7. But this number of permutations must still be summed for every possible value of i and s. Therefore we have .
-I I
1
This may be written in the form of a sum of a product of binomial coefficients
644 The second sum is, in consequence of Couchy's formula (14) fj 22, equal to n-t,-2i I n-~l-ql-i 1 = I Therefore the required probability will be
Remark. Starting from formula (11) we may easily derive the formula corresponding to the problem in two dimensions and also that of the problem in one dimension, First we have to put into (11) [,=O and then c=l; thus we obtain the probability that a mobile starting from [=O, 9~0, c=O reaches the point Er, ql, 0 without touching the planes l= +a, *I= + b and C= f 1. But then the corresponding ways will all be in the plane c=O. This is therefore the solution of the problem in two dimensions. Only the total number of ways will now be 4" instead of 6*, hence we have still to multiply the probability (11) by ($r- Then it will become identical with formula (9) of Q 186. In the same manner, starting from this formula we solve the problem in one dimension, putting first ql=O and b=l. The corresponding ways will all be on the 8 axis. The number of ways will now be 2" instead of 4", hence we have may then total still
to multiply the probability (9) 8 186 by [-$r. The formula obtained will be identical with formula (25) Q 184.
* Courant's formula (lot. cit. 60) is erroneous. The formula given by Army. without demonstration, is another form of formula (12). | 677.169 | 1 |
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In a sense, trigonometry sits at the center of high school mathematics. It originates in the study of geometry when we investigate the ratios of sides in similar right triangles, or when we look at the relationship between a chord of a circle and its arc. It leads to a much deeper study of periodic functions, and of the so-called transcendental functions, which cannot be described using finite algebraic processes. It also has many applications to physics, astronomy, and other branches of science. It is a very old subject. Many of the geometric results that we now state in trigonometric terms were given a purely geometric exposition by Euclid. Ptolemy, an early astronomer, began to go beyond Euclid, using the geometry of the time to construct what we now call tables of values of trigonometric functions. Trigonometry is an important introduction to calculus, where one stud ies what mathematicians call analytic properties of functions. One of the goals of this book is to prepare you for a course in calculus by directing your attention away from particular values of a function to a study of the function as an object in itself. This way of thinking is useful not just in calculus, but in many mathematical situations. So trigonometry is a part of pre-calculus, and is related to other pre-calculus topics, such as exponential and logarithmic functions, and complex numbers.
Editorial Reviews
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"Cover[s] all of the basic topics that a high school or beginning university student should be expected to know.... There are...some nice touches; for example, a nice informal discussion showing that the sine of an angle in a right triangle does not depend on whether the sides are measured in inches or centimeters..."
—Choice
"Covers all the basics of the subject through beautiful illustrations and examples…. Throughout, the treatment stimulates the reader to think of mathematics as a unified subject."
— L'enseignement Mathématique
"As a teacher I enjoyed this book enormously and I will doubtless borrow many of the plums to spice up my lessons…. [For] that ideal student who is to be prepared to be challenged to think what the subject is really about, and has the patience to excavate the basic ideas for all they are worth before jumping on to the next chapter, it should prove to be a godsend."
—The Mathematical Gazette
"The authors tried to explain the results of trigonometry as simply as possible…. The exercises include a few problems of each routine type. Most of the problems exhibit a new aspect of the technique or object under discussion. One of the goals of this book is to prepare students for a course in calculus. We recommend it for teachers and students."
Top Customer Reviews
This text, which is designed as a supplement to a trigonometry course, is noteworthy for the clarity of its explanations, the connections it draws between trigonometry and other mathematical topics, its many challenging problems, and its numerous worked and illustrated examples. Of particular interest are the appendices to the later chapters in which the authors relate trigonometry to Pythagorean triples, use sequences of trigonometric functions to approximate pi, and introduce Fourier series.
After reviewing the geometry of the triangle, the authors cover right triangle trigonometry, the relationship between trigonometry and the geometry of the triangle, unit circle trigonometry, trigonometric formulas and identities, graphs of trigonometric functions, and inverse trigonometric functions. The authors make every effort to explain why the results hold and how to use them. Rather than presenting a self-contained treatment, the authors make every effort to connect trigonometry with other branches of mathematics, thereby providing the reader with many fascinating insights.
The problems are designed to be challenging. The reader who diligently studies the numerous worked examples in the text and works through the problems will acquire considerable knowledge of the subject. Solutions to the problems are not provided in the text.
Finally - a trig book that doesn't talk down to students. Gelfand treats his readers as intellegent, curious, and competent. This goes far ... especially with kids.
Most other trig books are written by educational consultants who view the subject as a odorous swamp that you have to slog through. They distract the reader with glitzy graphics and useless photos. No such chartjunk here. It's from someone who loves the subject, and places the mathematics first.
I feel like an avuncular mathematician is showing me the delights of trig ... indeed, he seems to revel in sines, cosines, and tangents. Several of the problems have tickled my 10 year old son: "Dad! Did you know that the area under the first half of the sine curve is exactly 2?"
Aaah. Now *that's* a great trig book!
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This book is a very nice change away from the overly analytical approach that most trigonometry texts take nowadays. In my view something is lost when the geometric aspects are not emphasized. There is some fantastic geometry in this book, and I especially liked the connection between the sine addition formula and Ptolemy's theorem for quadrilaterals inscribed in a circle. Those kinds of connections are not only neat in themselves, but I think they help to understand the material better. I think they can spark students' interest more than the dull plug-in-the-numbers types of exercises in most books.
Speaking of the exercises, many of them are excellent. As some other reviewers have mentioned, it's nice to see a book at this level not insult the intelligence of its readers. I wish more elementary math textbooks were like this.
The only criticism I have is that there are lots of errors and typos in the book, enough to make me give it only 4 stars instead of 5. For instance, in Exercise 8 on p.147 it asks the reader to prove that (cos a)^2 <= cos(2a). Of course, that inequality should be reversed. And there are many more annoying little errors like that, which makes me believe that no one actually proofread the book.
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I rarely compose reviews even if I happen to really appreciate a book. But THIS is one of those rare exceptions!Gelfand (who also happens to rank among the 20th Century's most renowned and prolific Mathematicians ... I noticed that no reviewer has mentioned this little fact) has written an exceptionally clear and concise, easy to read and to digest text concerning Trigonometry with advancement to higher education in mind. The perfectly logical format and development is novel, very inspiring, coaxing the reader to read on and on and one truly begins to learn right from the first sentence. The reader requires little in the way of prior knowledge and yet, I am convinced, will come away with an enthusiasm for this subject (and hopefully for Mathematics in general).
I am a member of a Mathematical Honor Society and of the Mathematical Association of America but I must confess that I was a poor math student, had some very poor teachers and, thanks to some inspiring individuals along the way, managed to discover the real magic and beauty of Math before it was too late (although I have spent the better portion of my professional life compensating for the failures of my former teachers!).
This text represents the way Math should be taught in schools. I believe that any frustrated math student will benefit from this text (and probably the other texts also by Gelfand).
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I enjoyed this textbook, especially the way some subjects were so well explained. Not only does this text cover a good bit of material, but it also reveals the way in which the author thinks about this subject. I have noticed, both with this text and the previous two which I have commented on, that certain aspects of the subject become much more transparent or understandable when reexamined with a keener mathematical ability than, at least I possessed, when I was first exposed to these subjects in high school. I had no special interest in math at that time. The limited reexposure one has to trigonomety and geometry as one learns new areas of mathematics in college doesn't seem to do justice to these foundation areas of math.
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104 - Women in Mathematics
This course is designed to introduce a variety of mathematical topics stemming from the research of women mathematicians both past and present, from Hypatia to current professors. In discussing the work of these women, we will also discuss the gender issues that are associated with being a female mathematician. Course material will be covered in lecture, research, in-class visitors and activities. Course work will include research papers, a course project and problem sets related to the mathematician of discussion. CORE REQUIREMENT MET: MATH/SCI
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Calculus 1: Scientific Modeling and Differential Calculus Many mathematical models in the natural and social sciences take the form of systems of differential equations. This introduction to the calculus is organized around the construction and analysis of these models, focusing on the mathematical questions they raise. Models are drawn from biology, economics, and physics. The important elementary functions of analysis arise as solutions of these models in special cases.
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A metamathematical investigation of the main formal language used to symbolize ordinary mathematics: first order logic. The focus is on the two fundamental theorems of logic: completeness and compactness. Gödel's completeness theorem says that every intuitively valid consequence is formally provable from the hypotheses, while compactness says that every intuitively valid consequence of an infinite premise set really depends on only finitely many premises. Prerequisite: Mathematics 210 or permission of instructor. Given in alternate years. CORE REQUIREMENT MET: MATH/SCI
352 - Computability and Complexity
The logical foundation of the notion of a computable function underlying the workings of modern computers. Representation of the informal mathematical idea of calculability by canonical proxies: "general recursive functions," "Turing computable functions." Discussion of Church's Thesis, which asserts the adequacy of these representations. Survey of decidable and undecidable problems. Prerequisites: Mathematics 210 or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
354 - Set Theory and Foundations of Mathematics
Cantor's naïve theory of sets and equinumerosity. Paradoxes and axiomatic set theory. Finite and infinite cardinal numbers, fixed point theory, applications to computer science. Well orderings, transfinite induction and recursion, the Axiom of Choice and its consequences, ordinal numbers and the cumulative hierarchy of sets. Discussion of the Continuum Hypothesis and its relation to models of set theory. Prerequisites: Mathematics 210 or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
360 - Axiomatic Geometry
Axiomatic development of Euclidian and non-Euclidian geometries, including neutral and hyperbolic geometries, and, possibly, brief introductions to elliptic and projective geometries. The course will emphasize a rigorous and axiomatic approach to geometry and consequences of Euclid's Parallel Postulate and its negations. Prerequisite: Mathematics 210 or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
362 - Topology
General topology studies those properties (such as connectedness and compactness) which are preserved by continuous mappings. A disk and a solid square are topologically equivalent; so are a doughnut and a coffee cup; but a disk is different from a doughnut. This course enables you to construct your own proofs and counterexamples while getting to know the basic concepts behind modern mathematics. Prerequisites: Mathematics 210, and Mathematics 212 or 214, or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
370 - Numerical Analysis
This is a first course in studying numerical techniques in solving a variety of problems in mathematics using computer algorithms. Topics can include methods for approximating solutions to algebraic equations, iterative methods for linear and nonlinear systems, interpolation and approximation theory, numerical integration, explicit and implicit methods for solving initial value problems and boundary value problems. Error estimation, stability and performance are themes throughout. Students will be expected to be able to execute standard algorithms using a computer programming language by the end of the course. Prerequisite: Mathematics 212 and 214 or permission of instructor. Prerequisite: Mathematics 212 and 214 or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
372 - Operations Research
Optimal decision-making and modeling of deterministic and stochastic systems. Different systems of constraints lead to different methods. Linear, integer, dynamic programming, and combinatorial algorithms. Practical problems from economics and game theory. Inventory strategies and stochastic models are analyzed by queuing theory. Prerequisites: Mathematics 210 and 214. CORE REQUIREMENT MET: MATH/SCI
382 - Graph Theory
Graph Theory is a beautiful area of mathematics with many applications. It is used in computer science, biology, urban planning, and many other contexts. Like other areas of discrete mathematics, Graph Theory has the property that the problems are often quite approachable and understandable. Sometimes the solutions to Graph Theory problems can be complex and often require clever arguments, thus the subject is quite pleasing to study. This class will build a solid foundation in Graph Theory for the students. Possible topics are graph isomorphisms, coverings, and colorings; independence number, clique number, connectivity, network flows, and matching theory. Prerequisite: Mathematics 210. Suggested co-requisite: Mathematics 380. CORE REQUIREMENT MET: MATH/SCI
392 - Mathematical Models in Biology
This course is intended to introduce students to common models used in biology. A variety of models in terms of both biology and mathematics will be covered. Biological topics include action potential generation, genetic spread, cell motion and pattern formation, and circulation. These topics span a range of mathematical models as well, including finite difference equations and differential equations, both linear and non-linear. The focus will be on model analysis and the translation between the mathematical language and the biological meaning. Such analysis will be done both quantitatively and qualitatively. Towards this end, topics seen in previous mathematical courses, such as eigenvalues, phase portraits, and stability, will be revisited. Relevant biology will be presented with each model. The course will be project based. Prerequisite: Mathematics 212 and 214, or permission of instructor. CORE REQUIREMENT MET: MATH/SCI
395 - Special Topics in Advanced Mathematics
Special topics in advanced mathematics, selected largely by student interest and faculty agreement. May be repeated for credit.
Advanced Differential Equations. The course will consists of advanced topics in differential equations not usually seen in either ordinary differential equations or partial differential equations such as delay differential equations, stochastic differential equations. boundary value problems, numerical methods, and infinite series solutions. Prerequisite: Math 340 or Math 342 or permission of instructor.
Computational Differential Geometry. Differential geometry combines calculus, algebra and geometry. It is an important tool in the simulation and analysis of nonlinear objects. The focus of this course is extrinsic and intrinsic analysis of curves and surfaces. Topics include differential forms, frame fields, the shape operator, Gaussian and mean curvature, the Gauss-Bonnet Theorem, and geodesics. Applications to computer graphics, image processing, statistics, differential equations and physics. The computational component uses Mathematica, but no prior knowledge of Mathematica is assumed. Prerequisite: Math 212 and Math 214, or permission of the instructor. CORE REQUIREMENT MET: MATH/SCI
396 - Mathematical Modeling
A project-oriented introduction to mathematical modeling. Techniques from calculus, linear algebra and other areas of mathematics will be used to solve problems from the life, physical and social sciences. Familiarity with a programming language is desirable but not required. This course may be taken up to two times for credit. Prerequisites: Mathematics 212 and 214. 2 units | 677.169 | 1 |
This book reflects more than 25 years of author involvement with business math education and the business community. The focus of this edition is on linking mathematics with real business practices in real/b>…
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This book reflects more than 25 years of author involvement with business math education and the business community. The focus of this edition is on linking mathematics with real business practices in real businesses—giving readers a better appreciation for and understanding of the concepts that are vital in the business world. The book is filled with chapter-opening scenarios, chapter-ending case studies, cases, boxed features, and exercise sets. It is organized in workbook format and opens with chapters on basic math. Coverage includes banking, business statistics, trade and cash discounts, markups and markdowns, payroll, consumer credit, mortgages, insurance, taxes, and stocks and bonds. For entrepreneurs and anyone interested in personal finance.
Identify types of fractions. Convert an improper fraction to a whole or mixed number. Convert a whole or mixed number to an improper fraction. Reduce a fraction to lowest terms. Raise a fraction to higher terms.
Adding and Subtracting Fractions.
Add fractions with like (common) denominators. Find the least common denominator for two or more fractions. Add fractions and mixed numbers. Subtract fractions and mixed numbers.
Find the cost, markup, or selling price when any two of the three are known. Find the percent of markup based on the cost when the cost and selling price are known. Find the selling price when the cost and the percent of markup based on the cost are known. Find the cost when the amount of markup and the percent of markup based on cost are known. Find the cost when the selling price and the percent of markup based on the cost are known.
Markup Based on Selling Price and Markup Comparisons.
Find the amount of markup and the percent of markup based on the selling price when the cost and selling price are known. Find the selling price when the amount of markup and the percent of markup based on the selling price are known. Find the selling price when the cost and the percent of markup based on the selling price are known. Find the cost when the selling price and the percent of markup based on the selling price are known. Compare the markup based on the cost with the markup based on the selling price.
Markdown, Series of Markdowns, and Perishables.
Find the amount of markdown, the reduced (new) price, and the percent of markdown. Find the final selling price for a series of markdowns. Find the selling price for a desired profit on perishable and seasonal goods.
10. Payroll.
Gross Pay.
Find the gross pay per paycheck based on salary. Find the gross pay per weekly paycheck based on hourly wage. Find the gross pay per paycheck based on piecework wage. Find the gross pay per paycheck based on commission.
Find an employers total deposit for withholding tax, Social Security tax, and Medicare tax per pay period. Find an employers SUTA tax and FUTA tax due for a quarter.
11. Simple Interest and Simple Discount.
The Simple Interest Formula.
Find simple interest using the simple interest formula. Find the maturity value of a loan. Convert months to a fractional or decimal part of a year. Find the principal, rate, or time using the simple interest formula.
Ordinary and Exact Time and Interest.
Find ordinary and exact time. Find the due date. Find the ordinary interest rate per day and the exact interest rates per day. Find simple interest using a table.
Promissory Notes.
Find the bank discount and proceeds for a simple discount note. Find the third-party discount and proceeds for a third-party discount note.
12. Consumer Credit.
Installment Loans and Closed-End Credit.
Find the amount financed, the installment price, and the finance charge of an installment loan. Find the installment payment of an installment loan. Find the estimated annual percentage rate (APR) using a table.
Paying a Loan Before It Is Due: The Rule of 78.
Find the interest refund using the rule of 78.
Open-End Credit.
Find the finance charge and new balance using the average daily balance method.
13. Compound Interest, Future Value, and Present Value.
Compound Interest and Future Value.
Find the future value by compounding manually. Find the future value using a $1.00 future value table. Find the effective interest rate. Find the interest compounded daily using a table.
Present Value.
Find the present value based on annual compounding for one year. Find the present value using a $1.00 present value trade.
14. Annuities and Sinking Funds.
Future Value of an Annuity.
Find the future value of an ordinary annuity using the simple interest formula method. Find the future value of an ordinary annuity using a $1.00 ordinary annuity future value table. Find the future value of an annuity due using the simple interest formula method. Find the future value of an annuity due using a $1.00 ordinary annuity future value table. | 677.169 | 1 |
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Widely acclaimed algebra text. This book is designed to give the reader insight into the power and beauty that accrues from a rich interplay between different areas of mathematics. The book carefully develops the theory of different algebraic structures, beginning from basic definitions to some in-depth results, using numerous examples and exercises to aid the reader's understanding. In this way, readers gain an appreciation for how mathematical structures and their interplay lead to powerful results and insights in a number of different settings. * The emphasis throughout has been to motivate the introduction and development of important algebraic concepts using as many examples as possible.
Top Customer Reviews
I'm a graduate student in math. We used this book for the basic year-long abstract algebra sequence: group theory, chapters 1-4 and some of chapter 5; ring/field/galois theory chapters 7-9, 13-14. Some of my fellow students took a module theory course which was at least partially based off chapters 10 and (I think) 11. I'm sure more advanced courses could easily be based off chapters 15-end. Considering the cost of university books, I consider it very nice to buy one book for essentially 3+ courses. The exercises in some sections are very diverse. My group theory professor made us do a huge number of them, and now I am amazed at how often I see questions similar to those from Dummit-Foote show up on past qualifier exams from many different universities. Regarding lack of answers in the back...well, you shouldn't need too many, and if you get really stuck, that's what the professor is for. And if you're learning it on your own then I'm thinking you should be brainy enough not to need answers! The text itself is very readable and complete. I don't think I'd recommend this as an undergrad textbook, although I've no doubt that there are some clever undergrads who could learn from it. I used Herstein's "Topics in Algebra" for my intro-to-abstract course as an undergrad. Herstein is designed to be introductory in nature, though still a wonderful book, while DF is more encyclopedic. I do have one complaint though: the binding in DF started to crack and pages started to fall out near the end of its first semester of use. It did see some moderate backpack use but not too much, certainly less than many other books I've had in the past. So take care of it!
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Excellent service from the seller. Very good binding. Book recommended for one of my course in my pure&applied math major program and looking in it, we can see why. I found the exposition of the material to be very inviting. Highly recommended.
I think I would only recommend this book to someone who has already had some exposure to algebra (or one especially gifted in mathematics). The beginning of the book is not too bad, but towards the end of Part I the pace quickens quite a bit. If you are willing to read over the text many times, and do all of the non-trivial exercises (there is an impressive olla podrida of algebra in them, most of which are the beginnings of some very deep ideas), then it should be a very rewarding experience. Namely because this is one of the most readable textbooks which covers everything from groups, rings, and fields to homological algebra and algebraic geometry. It is very rare to see this much material covered in one book, and for it to remain so structured (Rotman is an example of a book that covers a lot of material, but loses its structure somewhere).
(When I wrote this review D+F was almost one hundred and fifty dollars. It seems to have dropped in price significantly since then)
D+F tries to straddle the line between being a book for advanced undergraduates and a book for graduate students and does a decent job. It is fairly readable, with many excellent exercises and lots of examples. The book also covers all the material in the standard graduate algebra sequence. The section on group theory is particularly good.
I think the biggest problem with D+F is that it is bland. The exposition isn't a joy to read and full of motivation like that of Halmos, Stillwell, or Eisenbud and it isn't full of deep insights like that of MacLane, Lang, or Artin. In addition Category Theory is pushed off to an appendix at the end of the book rather than integrated through the text. Finally the book is expensive and the binding is terrible.
If you want to learn algebra I would recommend purchasing some of these cheaper more focused texts since almost everything in D+F is treated better elsewhere:
If on the other hand you are already fairly comfortable with algebra and are looking for a one volume reference I would just buy Lang. It is less than half the price, more advanced, and has more material.
I am surprised that this book has not got the 5 stars. It is very suitable for advanced undergraduates/first-year graduates. The book is full of examples; and the proofs are amazingly clear and succinct. The book introduces new concepts in the excercises long before the student encounters them in the sections.
This is a beautiful way to teach mathemtatics,--and indeed to learn it. The book is replete with examples that connect concepts from toplogy and real analysis with Algebra.
Dummit and Foote contains just about everything an undergraduate ought to know about abstract algebra. In addition, it is written in a more user-friendly, down-to-earth fashion than, say, Lang's Algebra is.
The pro's have been discussed in other reviews and include: clear development of group, ring, and field theory; tons of exercises at the end of every chapter; numerous examples scattered around the text; sylow theorems (for group theory, imo, it's important, and not every algebra book does sylow stuff!); great introduction to exact sequences (useful if the reader is going into algebraic topology anytime soon. ugh!); galois theory is pretty clearly laid out; and, the third section of the book has some neat topics the reader can check out (which are, I think, commutative algebra, homological algebra, and representation theory introductions, as well as a small section on category theory at the very end).
The con's of D+F are the price (it's very expensive!), the binding (it's horrible!), and some of the sections are much harder than others and D+F doesn't do as well a job at explaining them as in many of the other sections (the tensors section sticks out in my head, and they wait something like 100 pages to explain "tricks" for figuring out the structure of finite groups after explaining some of the sylow stuff (eg., they wait to tell the reader about how to "pin small groups against one-another" and to make use of the sylow n! trick). Also, D+F introduce modules before vector spaces which I have mixed feelings about --- as a student who's already taken an algebra class, I love the "flow" of the lessons; as a student who remembers what it was like to try to imagine what modules "looked like", it makes me cringe to think that they didn't introduce vector spaces first.
Overall, wonderful book. One of my favorites of all time. DEFINITELY have it, and if you study from it, you may feel more comfortable supplimenting it with Herstein's Algebra, Artin's Algebra (which are just as hard) or Fraleigh's Abstract Algebra, Gallian's Abstract Algebra, or Rotman's Abstract Algebra (which are much, much easier).
I really like (and would recommend) Dummit and Foote for those wanting to learn abstract algebra I've used others (Lang, Jacobson, and Herstein). If I had to order them it would be: Dummit, Hersein, Jacobson, Lang. With Dummit being easily in first. The material is explained very well in this book. I found it much easier to learn from this book than the other books listed. Also, this book covers everything a first year graduate would cover in algebra (not that the others don't). Lots of good stuff explained in a way that clicks with my brain. | 677.169 | 1 |
Precalculus
9780073312637
ISBN:
0073312630
Edition: 6 Pub Date: 2007 Publisher: McGraw-Hill College
Summary: The Barnett, Ziegler, Byleen College Algebra series is designed to be user friendly and to maximize student comprehension. The goal of this series is to emphasize computational skills, ideas, and problem solving rather than mathematical theory. Precalculus introduces a unit circle approach to trigonometry and can be used in one or two semester college algebra with trig or precalculus courses. The large number of peda...gogical devices employed in this text will guide a student through the course. Integrated throughout the text, students and instructors will find Explore-Discuss boxes which encourage students to think critically about mathematical concepts. In each section, the worked examples are followed by matched problems that reinforce the concept being taught. In addition, the text contains an abundance of exercises and applications that will convince students that math is useful. A Smart CD is packaged with the seventh edition of the book. This CD reinforces important concepts, and provides students with extra practice problems.
Barnett, Raymond A. is the author of Precalculus, published 2007 under ISBN 9780073312637 and 0073312630. One hundred four Precalculus textbooks are available for sale on ValoreBooks.com, fifty three used from the cheapest price of $54.60, or buy new starting at $232.64.[read more] | 677.169 | 1 |
books.google.com - Looking closely at algebra, its historical development, and its many useful applications, Algebra examines in detail the question of why this type of math is so important that it arose in different cultures at different times. The book also discusses the relationship between algebra and geometry, shows... | 677.169 | 1 |
Geometry: Fundamental Concepts and Applications
9780321473318
ISBN:
0321473310
Edition: 1 Pub Date: 2007 Publisher: Pearson
Summary: This Geometry workbook makes the fundamental concepts of geometry accessible and interesting for college students and incorporates a variety of basic algebra skills in order to show the connection between Geometry and Algebra.
Alan Bass is the author of Geometry: Fundamental Concepts and Applications, published 2007 under ISBN 9780321473318 and 0321473310. Two hundred eighteen Geometry: Fundamental Concepts ...and Applications textbooks are available for sale on ValoreBooks.com, sixty three used from the cheapest price of $4.99, or buy new starting at $20 | 677.169 | 1 |
The Honors Class: Hilbert's Problems and Their Solvers…
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Overview natural introduction to thesis writing with examples, this collection of problems has indeed become a guiding inspiration to many mathematicians, and those who succeeded in solving or advancing their solutions form an Honors Class among research mathematicians of this century. In a remarkable labor of love and with the support of many of the major players in the field, Ben Yandell has written a fascinating account of the achievements of this Honors Class, covering mathematical substance and biographical aspects.
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Fascinating historical comments, lively portraits of mathematicians, and their times. While the narrative is about the lives of some great mathematicians, it sucessfully outlines main ideas in the subject,--the personal and scientific context. The author does a great job in sharing his fascination with the rest of us. The book covers roughly the past hundred years. It is a great service to the mathematics community,-- and especially, it is an enjoyment for everyone. It reads like a novel, fast paced, and it is hard to put down. I meant to look at it before going to sleep, but instead read it to the end, finishing in the morning. As a professional mathematician, I am often saddened by how little our work is perhaps understood and appreciated. Books like this can do a lot of good. I can now tell my children that dad does stuff like that. The author brings the events and the mathematical people to life, and he has a story to tell. This book is and will be a success for a long time to come. | 677.169 | 1 |
Contact Us
New Data Science Course
By William Schellhorn
Mar 17, 2014
The Simpson College Mathematics Department is excited to offer a new special topics course called Data Science during the Spring 2015 semester. The class will be taught by Dr. Bill Schellhorn, who recently spent his sabbatical leave studying topics in the field.
What is data science? Data science is the study of the extraction of knowledge from data. In practice, it involves learning from data in order to gain insight and make useful predictions. Knowledge in the field is increasingly important for Mathematics and Actuarial Science majors.
Why is data science important? Data is being generated faster than it can be analyzed. Many of the current challenges in science, government, industry, economics, marketing, and sports are "big data" problems. Some examples include:
the Large Hadron Collider experiments;
the Sloan Digital Sky Survey;
the human genome project;
surveillance data collected by the National Security Agency;
social network data collected by Facebook;
marketing data collected by Amazon and NetFlix;
statistics from professional sports leagues.
What topics will be covered in the Math 390 Data Science course? The course will introduce methods used in data science, including techniques in data collection, data management, exploratory data analysis, prediction, and communication of results. Real-world examples will be used to illustrate the methods presented. The analyses and methods will be implemented in a statistical software package (either R or JMP).
What are the prerequisites for the course? Math 152 Calculus II and CmSc 150 Introduction to Programming. | 677.169 | 1 |
Product Description
Instead of memorizing formulas and equations, Videotext Algebra helps students to understand math through mastery learning, encouraging them to solidify each concept before moving on the next. A copy of the print materials needed for this module is included.
Module F in Videotext Algebra, this unit covers:
Parabolas (Origins, Standard Form 1, 2, 3, & 4), Intercepts)
Circles (Standard Form, Non-standard Form)
Ellipses (Standard Form, Non-standard Form)
Hyperbolas (Standard Form, Non-standard Form)
Systems (One 1st Degree, One 2nd Degree, Two 2nd Degree)
Problem Solving (Numbers, Geometric Figures)
Exponential Functions (Functions of x & y)
Logarithmic Functions(Translating, Graphs of Solution, Sets)
Operations (Properties, Finding Logarithms, Computation)
Solving Equations (Exponential, Logarithmic)
Unit Tests IX and X included | 677.169 | 1 |
Passing a Math Aptitude Test
More and more employers are using a math aptitude test to test the basic skills of prospective employees. While you may not need to be brilliant at maths for most jobs, a computer scientist or an engineer will need good maths skills.
Maths may not be your strong point, but most organizations want to know just what your mathematical ability is before they consider hiring you. Different organizations use different types of math aptitude tests for different roles, and will include questions from some of the following areas:
• profit and loss
• statistics
• algebra
• geometry
• probability
• trigonometry
• number series
• permutations and combinations
• quadratic equations
Importance of Algebra
Algebra is the basis of almost all math aptitude tests. Employers use them to determine your overall understanding of mathematical principles and your ability to put it into practice.Aptitude test questions are meant to discover what your achievements fail to tell employers. They answer a lot of questions a prospective employer may have, such as:
• Are you capable of identifying an equation to solve written problems?
• Do you understand algebraic symbols and how to use them?
• What is your ability to interpret algebraic expressions?
• What is your capacity to learn algebra quickly and efficiently?
• Can you see relationships between sets of numbers?
Understanding Mathematical Symbols
Before you take an aptitude test you need to understand the basic symbols. Here are a few and what they mean:
• √ means the square root of something (√25 = 5)
• |x| means the absolute value
• ≤ means it is lesser than or equal to
• ≥ means it is greater than or equal to
• > means it is greater than (132 > 99)
• < means it is lesser than (99 < 132)
• ≠ means it is not equal to
• : and :: means to the ratio to
Types of Algebraic Problems
Numerical tests have questions from a cross section of mathematical disciplines such as the roman system, measurements, and geometry. The point of aptitude test questions is to measure your ability to use your mathematical skills to solve problems.
You will need to identify the underlying principles and any relationships within mathematical problems to answer them correctly. Math aptitude tests tend to test your analytical skills rather than that you can add up.These tests will include questions from some of the following mathematical skills:
• greatest common divisor
• least common divisor
• polynomial addition and subtraction
• monomial addition and subtraction
• factoring the difference of squares
• factoring out a common divisor
• factoring the squares of binomials
• evaluation of an expression
• exponents
• trinomial multiplication
• monomial multiplication
• binomial multiplication
• algebraic fraction simplification
• square root of an algebraic expression
What is the Roman System?
When using the Roman system the numbers are represented by Roman numerals such as D for 500 and ix for nine. If a smaller number is on the left it indicates you need to subtract, and if it is to the right you need to add.
Answering Geometry Questions
These questions test your basic understanding of geometry, and its principles and uses to help problem solve. You need to have basic knowledge of:
• the terminology and characteristics of angles
• circles
• ordinate numbers
• the concepts of polygon
• quadrilaterals
• perimeters
• hexagons
• the rectangular coordinator system
• pentagons
Before encountering a math aptitude test, you are wise to brush up on some sample tests to test just how well you do.
Questions Involving Measurements
Nearly every math aptitude test has questions involving measurements in some way. You need to use different techniques such as weight, space, and time to solve the problems.
Statistical and Probability Data
Many jobs require some sort of statistical analysis and most math aptitude tests will determine your ability to analyse situations and reach a positive outcome using your math reasoning skills. Before sitting an aptitude test, brush up on how to collect and analyse organizational statistical data and how to interpret what it means. | 677.169 | 1 |
Freshman Scholars Program 2014
The Graceful Labeling Conjecture: A Programming Approach
Instructor: Carlos Nicolas, Ph.D.
Project Description:
Students interested in mathematics, computer science, or computer information systems.
Graphs are simple mathematical objects consisting of a set of vertices and a set of edges joining the vertices. For example, a road map of the USA is a graph in which the
cities are the vertices and the roads joining the cities are the edges. The eight
corners and twelve edges of a cube also form a graph. The study of graphs at the introductory
level requires no technical knowledge beyond high-school algebra.
There are numerous questions about graphs that still remain open. One of them is
the famous graceful labeling conjecture for trees (a tree is a type of graph). In
this project, the student(s) - assisted by the instructor - will use a Computer Algebra
System to verify this conjecture for particular cases, i.e., for trees with simple
properties (e.g., having a small number of branches). The evidence gathered from
the computer will be used to prove the conjecture for other trees with the same properties.
By considering increasingly complex classes of trees, the student(s) will understand
the difficulty involved in proving the general conjecture and why nobody has been
able to prove it.
The project does not require previous programming experience or previous exposure
to graph theory.
Carlos Nicolas, Ph.D.
Dr. Nicolas obtained his Ph.D. in Mathematics from the University of Kentucky. He
is particularly interested in combinatorics, graph theory, and combinatorial geometry.
He has published several papers in the area of combinatorial geometry. He teaches
a wide range of mathematics courses at Ferrum College, from MTH-100 to the senior
seminar, including Graph Theory.
If you have specific questions about this project, please contact Dr. Nicolas directly
at Cnicolas@ferrum.edu. | 677.169 | 1 |
Mathematics for the Nonmathematician17
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About the Book
Erudite and entertaining overview follows development of mathematics from ancient Greeks to present. Topics include logic and mathematics, the fundamental concept, differential calculus, probability theory, much more. Exercises and problems. | 677.169 | 1 |
McGraw-Hill's GED Mathematics
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Create your own path to GED success with help from McGraw-Hill's GED test series
The newly revised McGraw-Hill's GED test series helps you develop the skills you need to pass all five areas of the GED test.
Presented in a clear, appealing format, these books offer many opportunities for test practice and explain the essential concepts of each subject so you can succeed on every portion of the GED exam. The series covers: Language Arts, Reading * Language Arts, Writing * Mathematics * Science * Social Studies
"McGraw-Hill's GED Mathematics" guides you through the GED preparation process step-by-step. A Pretest helps you find out your strengths and weaknesses so you can create a study plan to fit your needs. The following chapters introduce you to math concepts on which hundreds of GED questions are based. Then check your understanding of these ideas with the Posttest, presented in the GED format. You can then see how ready you are for the big exam by taking the full-length Practice Test.
"McGraw-Hill's GED Mathematics" includes: Clear instructions to show you how to use number grids and coordinate plane grids Instruction and frequent practice with the "Casio fx-260" calculator Problem-solving strategies to help you understand word problems Easy-to-follow lessons to develop essential math skills in whole numbers, decimals, fractions, percents, ratios, data analysis, geometry, and algebra
With "McGraw-Hill's GED Mathematics," you will sharpen your study skills for test success! | 677.169 | 1 |
Product Description
Instead of memorizing formulas and equations, Videotext Algebra helps students to understand math through mastery learning, encouraging them to solidify each concept before moving on the next. A copy of the print materials needed for this module is included.
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This Quantitative Survival Guide for Operations Management comes from a desire to help students understand and succeed when faced with the quantitative portions of a course in Operations Management. If you have struggled with math and statistics in earlier courses, this is the guide for you! This supplement gives examples of the types of problems that a student will encounter in a typical Operations Management textbook. The first section reviews some basics of algebra and pre-algebra. Each of the following sections reviews quantitative material by topic covered in an Operations Management course.
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In this work we investigate students' understanding of: 1) scalar multiplication of a vector and, 2) vector subtraction. We administered two tests to 717 students completing introductory physics courses at a private Mexican university. In the first part, we used a modified version of a problem designed by Van Deventer [1] to investigate students' difficulties with multiplication of a vector by a positive scalar and we designed a problem to study students' difficulties with multiplication of a vector by a negative scalar. We compared the frequencies of the errors in these two problems to comprehend students' conceptions in these vector operations. In the second part, we designed a vector subtraction problem and identified errors that haven´t been reported in the literature. | 677.169 | 1 |
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envictoria@socialfulcrum.comCopyright 20162016-02-11T18:00:00+00:00The Power of Subject-Specific Studying Part Four: The Problem-Solving Subjects
Power of Subject-Specific Studying Part Four: The Problem-Solving Subjects
Welcome to part four of our five-part exploration detailing subject-specific studying techniques. This series details a surgically targeted approach to studying that any student can employ to flexibly, adaptably and effectively master college material.
So far we've looked at how to prepare for memorization-heavy and reading-dense informational subjects, as well as reaction-focused interpretive subjects. These share some common ground. Today, however, we will leave the language-focused disciplines and explore a new horizon that uses a different part of your brain: problem-solving subjects.
The hallmark of problem-solving subjects are that they share a mathematical basis. This category includes many common undergraduate classes:
Algebra
Calculus
Accounting
Finance
Engineering
Physics
Perhaps more than any other subject grouping, students frequently struggle to succeed with these. Much of this academic strife stems from inadequate test preparation strategies. Below, we explore five powerful ideas designed to boost your mastery and understanding of any problem-solving class.
1) Do Now, Understand Later
With almost any other subject matter, understanding comes before doing. A law student could not parrot ideas to a client without understanding the underlying statutes that make his advice practical in a court, for example. A medical student shouldn't prescribe medicine without first understanding how the pills will affect the patient's body.
In mathematics, fittingly, this equation is reversed. Yes, an enlightened understanding of mathematical principles can take years to attain and is often only achieved through consistent repetition of problem-solving. In other words, problem-solving classes are best approached from a 'do now, understand later' perspective.
First, learn to accurately and reliably solve problems through imitation. That will allow your understanding of the techniques to come with practice, patience, and repetition. Seen from the opposite perspective, striving for a premature understanding can create a mental roadblock in the way of further progress.
2) Remove the Abstraction, See The Solution
Many people are uncomfortable with the abstract nature of mathematics. Students used to anchoring their learning in concrete examples – as you might do in science or literature classrooms – can find it difficult to accept that "X" and "Y" equation variables could refer to literally anything.
Depending on how your brain prefers to operate, working with abstractions can pose a significant impediment to success. These students stare at the numbers and letters in front of them and experience a fog of confusion as they struggle to grasp the 'real-world' significance of the work.
Does this sound like you? If so, a good strategy is to remove the abstraction. Reword the problem to create something concrete. Take any abstract equation problem and invent a context for it. Give names to the variables, and turn it into a word problem. As an example, 3(x) = 30 could become:
"John needs 30 bananas for his mother's recipe. The problem is, the store only sells these bananas in groups of three. How many groups of three does he need to purchase to get thirty bananas?"
3) Get Familiar With The Language of Mathematics
Informational and interpretive subjects use familiar language to convey information. Math, on the other hand, speaks a language all its own. Your success in problem-solving classes is reliant on your fluency with this new verbiage. Keep the following principles in mind:
The letters X and Y designate variables in an equation
The letters A and B designate constants in an equation
The placement of letters and numbers matters tremendously in how the equation functions. 3(x-1) is not the same as 3x(-1).
The particular form of math you're doing will dictate the language being used. Don't be afraid to go back to basics and study the language of the class before practicing for the test. A clear understanding here will save you time, effort, and endless frustration.
4) Study the Rules, Find the Patterns
Above all else, math classes represent the study of rules and patterns. Fans of these subjects have been known to comment, "Math is the easiest subject. Once you know the rules, you can solve any problem they throw at you!" While this is certainly partly true, the ease with which some people naturally learn the rules and pick up on patterns dramatically varies!
Are you the kind of student who struggles here? If so, pause. Take a moment to ask yourself the following questions as you work through practice exercises:
What are the steps I need to take to reach my answer?
In which order do I need to perform these steps?
Am I noticing any patterns between similar problems?
What is the essential component of this problem, and what is it measuring?
How can this problem be rephrased to ask the same thing in a different way?
Which other problems does this question remind me of? How did I solve them?
By vigorously attacking your problems with this line of questioning, you will start to pick up on the rules and patterns upon which your problem-solving techniques are calling. Being able to recognize similarities in this way will help you solve problems faster and more accurately than you ever thought possible.
5) Brush Up On the Building Blocks
Problem-solving classes, and particularly mathematics, build on one another in an additive fashion. Skipping steps is not recommended. Much like you cannot begin building the kitchen of a house before laying the foundation, you shouldn't challenge yourself with advanced math courses until you fully comprehend the building blocks.
This isn't to say you need to take remedial courses (unless you feel your skills are that rusty). It does mean that it might not hurt to stop by the bookstore and revisit some of those high school math courses you haven't thought about in years. Even a 30-minute nightly refresher starting several weeks before a class begins is more preparation work than most students will ever attempt.
Additionally, if you have never taken a physics class, consider starting there. Much of the math used in other problem-solving classes was originally derived from the study of physics. A firm understanding will make assimilating more abstract math that much easier.
Keep visiting the blog for our final edition in this series, in which we'll explore performance subjects and what the thrill of success looks like from the stage!
]]>2016-02-11T18:00:00+00:005 Quick Tips to Help You Thrive This Semester
it's your first semester or your fifth, starting off on the right foot is critical to achieving success in a new academic term. So we've compiled 5 quick tips to help you get started.
1. Spend some quality time with your syllabus
This is no exaggeration: your syllabus is instrumental to starting the semester off right. So rather than reading it and immediately setting it aside, take a few minutes to deconstruct its contents. When does your instructor hold office hours? What are your major assignments and when are they due? Write this information, along with anything else you may need later in the semester, on an index card for quick reference. Keep this card on your desk at home to remind you of the course's expectations and help you stay on track all semester long.
2. Create a schedule that you can actually stick to
Scheduling your time effectively is a critical skill for every student to develop. You won't be able to complete all of your homework, attend class, and study if you don't plan ahead. Just make sure your plan is realistic. Don't plan on reading 300 pages of your textbook in one night because you'll just end up frustrated. Instead, reflect on your own working habits, prioritize your assignments, and make a schedule around the student you are, rather than the student you'd like to be. This will help you develop a schedule that sets you up for success rather than disappointment.
3. Set aside some free time this semester
While completing your assignments and acing exams should be your focus, academic success shouldn't come at the expense of your mental health. So make sure you build a little time into your schedule to have fun. Spend time with friends, go see a movie, or take up a new hobby. Scheduling this time to recharge will help keep your mind fresh and focused, making it that much easier to study when you need to.
4. Get involved
Join a club, sign up for intramural sports, or start an online study group with your classmates. These activities will help you socialize with your peers and enhance the learning process. You may develop a better understanding of a concept you were struggling with in class or simply make some friends that make you more excited for and comfortable in class. Just make sure your involvement doesn't keep you from getting your work done!
5. Go to class and do the reading
The simplest tip is also the most important. From college freshmen to adult learners, the quickest way to lose momentum at the beginning of the semester is to fall behind in the readings or skip class. Simply put in the effort to keep up with the course, and you'll be on the road to success in no time.
It's often said that the best education happens outside the classroom. While taking classes to earn a degree is essential to getting the job you want later on, one Maryland student just proved that there's much value in real-world experience"Completing my portfolio for prior learning credits not only aided my efforts toward graduation, but it also allowed me to reminisce about my personal and professional experiences that I can share with my own children," Strader wrote in her essay, Adult Learning Experiences at Stevenson University.
Strader was able to expedite the completion of Stevenson's Paralegal Studies program using her personal experience as a caretaker as well as her professional experience as a paralegal. With this degree under her belt, she's been promoted to a senior position and has obtained the skills necessary to move up in her field.
This serves as just one example of how adults can use their existing skills and passion to earn a degree in an affordable, timely way. DSST is a proud partner of CAEL in promoting college credit by examination so that online and nontraditional students can use experience to their advantage.
One study by CAEL found that 43 percent of students with prior learning assessment (PLA) earned a bachelor's degree within seven years, compared to just 15 percent of students without PLA. Credits for prior learning can go a long way in giving adult and nontraditional students the boost they need to reach their goals. For more information about how you can get on your way to a better life, click here.
]]>2016-01-14T17:08:00+00:00Join Our Session at CCME
Credit-by-Exam will be exhibiting and holding a session at the 2016 Council of College and Military Educators (CCME). For more information, please see details below:
Session: DANTES and Credit-by-Exam: How Fully-Funded National Test Centers Help Military Students Succeed When: Wednesday, February 17, 2016 Where: San Antonio Marriott Rivercenter
]]>2015-12-16T19:39:00+00:005 Tips to Help You Retain What You Read
The life of an online learner is a busy one. Between school, work, and home life, there's a lot for you to juggle and remember.
Whether you're preparing for an online exam, studying for a midterm, or just want to remember the novel you're reading, reading retention tips can help you understand and remember any subject.
Try these five tactics to improve your memory retention and recall just when you need it.
1. Make Connections
The brain makes natural connections during the learning process. So when you read, try to connect the lessons to your established knowledge and experiences for better understanding. Once you've grasped key concepts, learning patterns will start to form naturally.
2. Take Notes
The more engaged your senses are while learning, the easier it can be to remember that information later. So instead of just reading a passage, write down key points and concepts at the same time. Jot down notes, questions or ideas as they come to you so you can refer back to them later. The key is to keep your mind active; the more critical thinking you exercise while learning, the more effective your learning retention will be.
3. Quickly Summarize
If you had to summarize what you just read to someone, what would you say to them? Are there any key details or information that stood out to you? After reading, summarize what you just read out loud or write it down for a few short minutes. This will give you the time you need to file that information into your memory bank in terms that makes sense to you.
4. Sleep On It
Sleep is essential to studying, and it plays a big factor in your memory and learning comprehension. According to a 2010 Harvard Medical School study, students who took a 90-minute nap after learning performed 50 percent better over a 24-hour period than the group that did not rest. The study suggested that, "dreaming may reactivate and reorganize recently learned material, improving memory and boosting performance." So take that nap and let your brain process the new information.
5. Sing it Out
Come up with a song for the material or sing out the information you are trying to remember. This will allow you to recreate the experience of learning and associate a simpler phrase with a more complex concept. Singing this phrase later on will help jog your memory as you prepare to recall the information for a test or writing assignment.
]]>2015-11-17T17:37:00+00:00How to Find a Mentor as an Online Learner
Here are some tips on finding a mentor as an online learner. As an added bonus, many of these suggestions can also help you get on your way to building a strong professional network, which can lead to a career later on.
Look in Your Immediate Circle
This may seem obvious, but your mentor would ideally be a person with whom you have some kind of history because they'll have insight into who you are and what you want to be. Don't exclude relatives just because they're family. Aunts, uncles, cousins, or even grandparents can be great resources if they've made achievements that you admire. Look for someone in a field that's similar to your own, but significantly ahead of you in terms of progress.
Don't neglect the professional or academic connections you've already made. Even if it's been some time since you've spoken to someone, simply reaching out and letting him or her know about the impact they've made on you can be the catalyst to a strong mentor-mentee relationship.
Stand Out in the Crowd
Networking and simply being good at what you do can attract the right kind of attention. Remember that a mentorship should be a reciprocal relationship. Ask yourself what you can do to make yourself a valuable and enjoyable mentee.
Volunteer, network, and take part in activities where you can offer your unique skills and insights to be noticed. Mentors seek people who are eager to learn more and open to new ideas, and it's always better to show than tell.
Also, scope out the big names in your industry and connect with them on social media, whether you're re-tweeting a quote or adding them on LinkedIn. These kinds of connections can come in handy later, particularly if you're entering a competitive field.
Expand Your Definition of Mentorship
Not everyone is going to have a reliable person available to mentor them, but luckily, you can seek guidance in a number of places. Look for books, blogs, and podcasts that teach tools for success. For instance, TED Talks are a great resource for subjects like technology, education, design, science, business, and more. You may also consider hiring a personal coach with whom you can schedule meetings to talk about your progress and goals. Personal coaches can be found through agencies, but many also work on an independent basis, and rates can be surprisingly affordable.
]]>2015-10-26T16:48:00+00:00How to Make Procrastination Work for You
Oftentimes the most urgent thing on our to-do list is the last thing on our want-to-do list. But while procrastination is typically viewed as the enemy of productivity, the two can go hand in hand if you're smart about it.
Some academics – namely, philosophy professor John Perry and law professor and author Frank Partnoy – have voiced support for procrastination as a way to get seemingly less important tasks finished and, more importantly, to give the mind a chance to weigh all possibilities before making a decision.
So, how can you make procrastination a part of your time management toolkit? Here are a few tips that can help you be productive while you put off that less-than-fun task – without actually blowing anything off.
Keep an Eye on Your Deadline
The key to productive procrastination is knowing your deadline and having an accurate estimate of how long it will take you to complete the task at hand. Once you know that, you can set reminders for yourself, whether it's a simple alert on your phone or a detailed calendar reminder.
Here's the thing, though: This strategy is not productive if you don't make your final deadline, so stick to your schedule and be realistic about your timeline.
Procrastinate With Worthwhile Tasks
While you're putting off that history paper until the final week before it's due, fill your time with other things you need to accomplish rather than vegging in front of the TV. If you manage to finally get your closet organized or start that side project you've been thinking about, you'll not only get the satisfaction of crossing those things off your list, you'll have less on your mind when it comes time to buckle down.
Get Up and Get Moving
Being sedentary for long periods of time can make you feel stir crazy, so what better way to spend a study break than moving around? Whether you go for a walk, jog, or bike ride, physical activity will give you a boost in blood flow that could be just what you need to focus. Who knows – you may even think of your breakthrough thesis once the fresh air hits you.
Productive procrastination has its benefits, but it's important to use it wisely and always keep an eye on your final goal. Waiting until the absolute last minute to take an exam for college credit can cause big setbacks, especially if you're close to graduation. Planning ahead and taking strategic breaks are the keys to conquering your to-do list while retaining your sanity.
]]>2015-09-22T17:52:00+00:00Visit Us at NACADA 2015
]]>2015-09-16T15:16:00+00:00Study Hacks to Get You Through an Online Exam
Getting your online degree means you have the luxury of taking exams on your schedule, but you'll still need to have regular study habits. Preparing for an online exam may feel like a time-intensive process, but if you do it right, you'll be in a good place to ace your test without having to pull an overwhelming all-nighter.
Not sure how it's done?
Try one of these simple study tips while you're prepping for your next big exam.
1. Limit your distractions
Do you consider yourself a serial texter? Are you constantly flipping between Facebook and Twitter? Do you keep your Netflix playing because you enjoy the "background noise"?
You can make excuses all day long, but in the end, it's your concentration that suffers.
Whether you're allowing yourself a quick 30-minute break to watch TV or you're running to the grocery store to grab a snack, distractions can be debilitating to a study routine. As you're immersing yourself in outlines, textbooks and class notes, try limiting your distractions by doing the following:
Turn your phone off (or face-down on silent)
Keep the TV off
Only listen to music without lyrics
Stay away from social media
While it's important that you don't let your distractions get the best of you, it's equally imperative that you allow yourself small breaks between study sessions. Avoid burnout by taking a 5- to 10-minute break after each major study milestone, such as the end of a chapter or unit, where you allow yourself to surf the Web or text on your phone.
2. Keep your study materials organized
How are you preparing for your online exam? Are you doing a basic overview of all your class materials, or are you breaking it down chapter-by-chapter?
No matter how you do it, ensure that you're keeping your study routine in check by organizing your resources. One easy way to streamline your study materials is by typing any hand-written notes you may have taken while watching online lectures. Moving these notes from physical to digital documents can make it easier to file them, and the process will allow you to further digest the material by digitally writing it down again.
3. Enlist your friends
It may seem counterintuitive to add socialization to your studying process - especially since distractions can prove to be the weakest link - but if you live with roommates or you've got close friends who are anxious to see you succeed, they can be some of your most valuable study resources.
After you've gone through your material and feel confident enough to test your abilities, have your friends quiz you over your notes, textbooks and class materials. This way, you can not only test your recall skills, but you can further absorb your testing material by repetition.
4. Have power snacks accessible
You can only ignore an empty stomach for so long before it starts to overtake your entire train of thought.
Brain foods, anyone?
Eating the right snacks while studying can improve your memory skills and increase your focus, all while keeping you energized enough to stay alert until you're done studying. Try nibbling on one of the following treats to get your brain going:
Blueberries, strawberries or raspberries: Got some fruit lying around? Eat a handful of berries to boost your memory skills. Many leading medical researchers - including those at Tufts University and Brigham and Women's Hospital - have linked berries to improved memory.
Granola bars: Easy to eat on-the-go and good for your brain? Bingo. Whole grains can keep you energized and focused, so keep a stash nearby during your next study session.
You're not just blocked, you're stonewalled. Any earlier momentum is long gone, and your cursor just sits there, blinking at you.
Blink.
Blink.
You are not alone. Writer's block can strike anyone at any time, whether you are a famous author or a college student. Even Ernest Hemingway once stated that the most frightening thing he ever encountered was a blank sheet of paper. While writer's block can be overwhelming when you are on a deadline, there is a cure: a little self-help, some creativity, and a lot of inspiration.
So next time you face a blank screen and worry about what to write, kick start your inspiration with these guaranteed tactics to overcome your latest bout of writer's block.
Get Moving
Leaving your desk or study space is a proven cure for a creative block. Take a few minutes, or even a few days, to move from the space you normally associate with work. That pause - however long - will allow you to rethink and recharge so you can begin writing again.
Multitask
Focus your energy on other tasks to reignite your momentum and power through your work. Switch to shorter, less complicated assignments that put your brain in a different mindset. By taking the pressure off of a more difficult project, you will stimulate your subconscious and easily jump right back into your work.
Free Write
Maya Angelou once declared, "What I try to do is write. I may write for two weeks 'the cat sat on the mat, that is that, not a rat.' And it might be just the most boring and awful stuff. But I try. When I'm writing, I write. And then it's as if the muse is convinced that I'm serious and says, 'Okay. Okay. I'll come.'" Your subconscious is a powerful thing. When you feel blocked, stop what you are doing and write down anything you can, whether it relates to your topic or not. Write without stopping. Disregard the urge to check your spelling or correct your grammar. Free write until you uncover an idea and pick up your pace again.
Break it Up
Your paper doesn't have to be written in order. This blog wasn't. It may be easier to start writing in the middle, branching out your ideas from there and expanding on each section as you work. Once the bulk of your writing is out of the way, you'll feel less stressed by the enormity of the project and free to refocus on your introduction.
Challenge Yourself
If you feel a roadblock coming up, challenge yourself to achieve gradual word count goals. Begin with 250 words, then increase to 500 if you feel up for the challenge. By escalating your goals as you meet each one, you'll find that you've settled into a steady, consistent flow and pace that allows your ideas to tumble out onto the paper.
Go to Sleep
Your subconscious is active even after you've fallen asleep. Sleep reinvigorates our minds to unleash its ultimate creative potential. Next time you are stumped on how to continue, think of the problem before you go to bed. Often, when you wake up, the solution and a fresh perspective will just come to you.
Write Now, Edit Later
Attempting to master any skill the first time around will always end in failure. The same can be said for writing. If you try to write a perfect paper in your first go, you may feel discouraged if it doesn't meet your expectations. Aim for ideas, then edit for grammar and quality. Most writing is rewriting anyway.
Want more writing tips? Download our free writing checklist and learn how to write a cohesive and engaging assignment sure to be your best work yet. | 677.169 | 1 |
Bristol, CT PrecalculusRayhon A.
...For students who are not visual learners and logical thinkers, this course poses great challenges. Physics is an integral part of engineering and the basis for understanding engineering principles. As such, all engineering students are required to take Physics in the early stage of their curriculumRobert S.
Paula M.
After mastering with Pre-Algebra, Algebra 1 takes you further into Properties of Real Numbers. You learn to solve Linear Equations, Linear Inequalities, and Systems of Linear Equations or Inequalities. You revisit and enhance the knowledge of Powers, Exponents and Roots (Radicals), and learn how to solve Quadratic Equations. | 677.169 | 1 |
Mathematics Instructional DVD. Intended for a young adult/teenage audience. NEW. SHIPS WORLDWIDE. DELIVERY CONFIRMATION WITHIN THE UNITED STATES.
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About the Book
These text-specific DVDs cover all sections of the text and provide explanations of key concepts, examples, exercises, and applications in a lecture-based format. Additionally, closed captioning of the videos makes them accessible for the visually impaired. | 677.169 | 1 |
Overview
Overview
PUM is a physics/physical science curriculum that strongly links middle and high school physics curricula and builds on the intrinsic mathematical reasoning to develop and strengthen students' mathematical concepts at the pre-algebra, algebra and algebra 2 levels. PUM curriculum consists of logically connected modules that allow students to build their conceptual understanding of physics concepts, develop relevant mathematical reasoning and simultaneously learn how to think like scientists.
Type of Method
Full curriculum,
Curriculum supplement
Level
Designed for:
High School, Middle School
Can be adapted for: Teacher Preparation, Teacher Professional Development, Intro College Calculus-based, Intro College Algebra-based, Intro College Conceptual | 677.169 | 1 |
gotta love how they come out with a new version every couple years too with minor re-wordings, and re-numbering of questions, so you HAVE to use the new version for homework. It's not like they make any meaningful changes. You can learn calc from a 10 year old book as well as you can learn calc from a brand new book. It's not like basic Math is changing.
But... these publishers somehow convince schools to force their classes to require the latest versions, that just happen to cost an ass-load. It's quite a racket.
Make sure on your course evaluations you give the course low points in some area to reflect the cost of books. Someone needs to collect all those statics and generally the ones that look bad get attention from high up. At the very least this attention is pressure that the profs want to avoid. | 677.169 | 1 |
Fundamentals of Mathematics, 10th Edition
Jumpstart your mathematical success with FUNDAMENTALS OF MATHEMATICS, Tenth Edition. The book's clear, easy-to-follow review of all basic mathematics functions helps you see the How and Why of concepts, providing the strong foundation you need for learning and lasting comprehension. Build your problem-solving skills, understand math concepts, see how these concepts are applied in the real world, and make the connection between mathematics and your own day-to-day experiences as a college student. If you need to improve your study habits and overcome math anxiety, this is the book for you57.49
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Algebra
Algebra is the way Islamic mathematicians figured out to connect our ideas about geometry with our ideas about numbers. You can use equations to represent geometric shapes. You can also use equations to represent the movement of numbers (or real objects like trains) through time or space.
Equations can also have two variables. These look like y = 2x + 1 or x = 4y - 12. These equations also define a straight line, but the line can be tilted in various ways, and can move up or down, to the left or to the right.
What if the line you want to talk about isn't straight, but curvy? Quadratic equations help you to define a curved line. They look like y = x2 or x = 8y3 + 16.
These equations each define one particular line, but you can also use algebra to describe a whole set of lines that are all in the same category. We call simple categories algebraic groups, and more complicated kinds of groups are called rings and fields.
Bibliography and further reading about algebra | 677.169 | 1 |
Math for Economists will help you assemble a toolkit of skills and
techniques to solve fundamental problems in both macroeconomics and
microeconomics. The material covers both precalculus and calculus
concepts and should help you identify the best approach to solving
problems. For example, an economist may be called upon to determine the
right mix allocation of capital to a production process. The tools in
this course will help you evaluate the options and select from the best
alternatives. Advanced courses in economics typically utilize
mathematical techniques beyond basic calculus; so, gaining practice in
fundamental skills can serve as a good basis for further study. Of
note, this course applies precalculus and calculus; this is different
from "applied math," which economists typically use to refer to
probability and statistics. This course begins with a survey of basic
optimization tools and then applies them to solve problems over several
periods in time. These optimization tools describe feasible choices and
then direct you to the best possible solution. In essence, they will
help you evaluate an economic environment and determine the best course
of action. The role of risk in financial decisions is explored in
relation to individual choices and macroeconomic processes. The
equitable distribution of resources is then considered. In other words,
this class will explore whether an optimal solution is indeed also fair
to the participants and society. A specific application, game theory,
is presented as one of the major recent advances in economic theory.
The final topic returns to microeconomic problems such as taxes,
elasticity, and specific types of supply and demand curves. The
materials included in the course offer a framework that you can use to
apply quantitative skills. You should liberally use Saylor course
materials (MA001,
MA003,
MA005,
ECON103/MA101,
and MA102) to refresh their
general techniques. This course fits into the major as a bridge between
quantitative theory and specific applications to problems in economics.
Completing this course can greatly help students successfully build a
toolkit for the intermediate courses,
ECON201 and
ECON202.
Course Requirements
In order to take this course you must:
√ have access to a computer;
√ have continuous broadband Internet access;
√ have the ability/permission to install plug-ins or software (e.g.,
Adobe Reader or Flash);
√ have the ability to download and save files and documents to a
computer;
√ have the ability to open Microsoft files and documents (.doc,
.ppt, .xls, etc.);
Requirements for Completion: In order to complete this course, you
will need to work through each unit and all of its assigned material.
All units build on previous units, so it will be important to progress
through the course in the order presented. You will also need to
complete:
Unit 1 Self-Test
Unit 2 Self-Test
Unit 3 Self-Test
Unit 4 Self-Test
Unit 5 Self-Test
Unit 6 Self-Test
Note that you will only receive an official grade on your Final Exam.
However, in order to adequately prepare for this exam, you will need to
work through the assessments in this course.
In order to pass this course, you will need to earn a 70% or higher
on the Final Exam. Your score on the exam will be tabulated as soon as
you complete it. If you do not pass the exam, you may take it again.
Time Commitment: This course should take you a total of
approximately 84.25 hours to complete. Each unit includes a time
advisory that lists the amount of time you are expected to spend on each
subunit and assignment. These time advisories should help you plan your
time accordingly. It may be useful to take a look at the time advisories
before beginning this course in order to determine how much time you
have over the next few weeks to complete each unit. Then, you can set
goals for yourself. For example, Unit 1 should take you approximately 29
hours to complete. Perhaps you can sit down with your calendar and
decide to complete Subunit 1.1 (a total of 10.75 hours) on Monday and
Tuesday night, Subunit 1.2 (a total of 9.5 hours) on Wednesday and
Thursday night, etc.
Table of Contents: You can find the course's units at the links below. | 677.169 | 1 |
Encourage a positive attitude toward mathematics. • Set aside a place and a ... The Glencoe Pre-Algebra Parent and Student Study ... includes: • A 1-page worksheet for every lesson in the Student ... Answers are located on pages 115– 119.
Dec 1, 2010 ... and be used solely in conjunction with GlencoeAlgebra1. Any other reproduction, ... Reading to Learn Mathematics Workbook. 0-07-861060-5. ANSWERS FOR WORKBOOKS The answers for Chapter 6 of these workbooks.
0-07-827753-1. Study Guide and Intervention Workbook (Spanish) ... ANSWERS FOR WORKBOOKS The answers for Chapter 1 of these workbooks ... and be used solely in conjunction with Glencoe's Algebra1. ... Glencoe/McGraw-Hill ...
The exercises are designed to aid your study of mathematics by reinforcing ... These worksheets are the same ones found in the Chapter Resource Masters for Glencoe Pre-Algebra. The answers to these worksheets are available at the end of each Chapter Resource Masters booklet ... 1-2 Variables and Expressions . | 677.169 | 1 |
Find a Kings ParkIncluded in the course were:
1. Methods of solving various types of First Order Differential Equations including Separable, Linear, Exact, and Homogeneous.
2. The procedure for solving higher order Homogeneous Linear Differential Equations, solving Nonhomogeneous Linear Differential Equations by the Methods of Undetermined Coefficients and Variation of Parameters.
3 | 677.169 | 1 |
Anyone who has read Underwood Dudley's Mathematical Cranks could be forgiven a bit of skepticism in the face of a book that promises to offer elementary geometric methods for solving classical calculus problems. New Horizons in Geometry is about as far from crank mathematics as possible. The book begins with "Mamikon's sweeping tangent theorem", a result first conceived in 1959, and proceeds to derive numerous formulas for arclength, area, and volume that might ordinarily be deduced with calculus. Along the way, readers will be introduced or reintroduced to such figures as the cyclogon, the autogon, and the bifocal disk. By the end, recursive formulas for volume in n-dimensional space are easily handled, with no integrals required.
This volume serves as the "collected works" of a fascinating mathematical collaboration between the two authors. This is mathematics of the highest caliber; but what makes this book even more impressive is the attention paid to high-quality full-color graphics that ably illustrate the problems under consideration. The advent of four-color printing in calculus has been at times both a blessing and a curse; since most of the pictures in this book are two-dimensional, most of the downside of full color illustrations is avoided. It is rare to see a mathematics book that takes such advantage of relatively simple illustrations.
The authors have put together an impressive body of work that is a solid addition to any library's geometry holdings. For those of us who want a corner of mathematics that will continue to use calculus, there will always be the challenge of measuring fluid pressure at the base of a dam.
Mark Bollman (mbollman@albion.edu) is associate professor of mathematics at Albion College in Michigan. His mathematical interests include number theory, probability, and geometry. Mark's claim to be the only Project NExT fellow (Forest dot, 2002) who has taught both English composition and organic chemistry to college students has not, to his knowledge, been successfully contradicted. If it ever is, he is sure that his experience teaching introductory geology will break the deadlock. | 677.169 | 1 |
Objetos Educacionais em Engenharia Elétrica is a collection of learning objects in the area of undergraduate Electrical...
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Objetos Educacionais em Engenharia Elétrica is a collection of learning objects in the area of undergraduate Electrical Engineering.Each object addresses a topic and is composed of text, mathematical expressions and still schematic images to introduce to conceptual aspects of the content. Animations and/or small videos are used to enrich the object and enhance the comprehension of the subject.Important information: this collection has many authors. I am the organizer and one of the Objetos Educacionais em Engenharia Elétrica to your Bookmark Collection or Course ePortfolio
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(Portal Resources for Indiana Science and Mathematics) PRISM is a free website that provides collections of online resources...
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(Portal Resources for Indiana Science and Mathematics) PRISM is a free website that provides collections of online resources for Indiana educators in the fields of science, technology, engineering, and mathematics (STEM PRISM to your Bookmark Collection or Course ePortfolio
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This book contains 1830 ready-to-use and easy-to-find report card comments for busy teachers. To make writing of report cards...
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This book contains 1830 ready-to-use and easy-to-find report card comments for busy teachers. To make writing of report cards even easier for you, every comment has been ranked and organized according to subject, topic, length and positive or negative nature. Save tons of time by simply inserting student's name into the comment that most accurately matches the recorded level of achievement. You will find writing, search, and sight-reading of available comments easy, as key words have been bolded and pages designed uncluttered, bright, and pleasing to the eye. All comments are "copy and paste" enabled Report Card and IEP Comments to your Bookmark Collection or Course ePortfolio
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Scratch is a new programming language that makes it easy to create your own interactive stories, animations, games, music,...
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Scratch is a new programming language that makes it easy to create your own interactive stories, animations, games, music, and art -- and share your creations on the web. Scratch is designed to help young people (ages 8 and up) develop 21st century learning skills. As they create Scratch projects, young people learn important mathematical and computational ideas, while also gaining a deeper understanding of the process of design. Scratch is available free of charge drop Society of Mathematical Biology to your Bookmark Collection or Course ePortfolio
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This Starting Point Module contains information on teaching with models geoscience education. Modeling is fundamental to...
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This Starting Point Module contains information on teaching with models geoscience education. Modeling is fundamental to science, great for an interactive student-centered environment, and ideal for the Earth systems approach. Types of models include conceptual, physical analog, mathematical, statistical, and visualization. This page offers links, information and examples for all modeling types, including examples of different models for each geoscience Starting Point: Teaching with Models to your Bookmark Collection or Course ePortfolio
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'Science, technology and engineering all involve some degree of mathematics. This is true regarding the International Space...
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'Science, technology and engineering all involve some degree of mathematics. This is true regarding the International Space Station and its microgravity environment. The orbit of the space station, the amount of storage space on the station, and the scale modeling used to design station components all engage mathematical concepts. See how mathematics applies to the science and engineering projects related to the space station.'This site provides a link to 4 videos that will interest students STEM on Station -- Mathematics to your Bookmark Collection or Course ePortfolio
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...
Show More revision, Marv Bittinger continues to focus on building success through conceptual understanding, while also supporting students with quality applications, exercises, and new review and study materials to help students apply and retain their knowledge. KEY TOPICS: Whole Numbers; Fraction Notation: Multiplication and Division; Fraction Notation and Mixed Numerals; Decimal Notation; Ratio and Proportion; Percent Notation; Data, Graphs, and Statistics; Measurement; Geometry; Real Numbers; Algebra: Solving Equations and Problems MARKET: For all readers interested in basic college mathematics | 677.169 | 1 |
Fundamental Mathematics
Description
EPMATH126 Fundamental Mathematics is a non-calculator introductory mathematics course which covers mathematical skills relevant to degrees requiring a basic level of mathematical knowledge and ability. The learning outcomes include an understanding of and competence in the skills of numeracy, arithmetic, fractions, decimals, percentages, algebra, equation solving and units of measurement. | 677.169 | 1 |
mathscard GCSE
Screenshots
Details
Description
Developed by Loughborough University, the mathscard GCSE app contains hundreds of examples of maths formulae, graphs and diagrams. The GCSE app is based on the hugely successful Loughborough A-level app and is designed to help students with their exam revision when at home or on the move. Number and Arithmetic, Algebra, Graphs, Statistics and Probability, Geometry and Measurement and much more are all covered in this handy resource | 677.169 | 1 |
San Manuel SATIt is important to understand the basic concepts of algebra before continuing to Algebra II. Students will learn to solve equations and inequalities. They will become proficient in factoring and simplifying algebraic fractions.
Kacie J.
Paul K. | 677.169 | 1 |
PrecalculusPRECALCULUS prepares learners for calculus and the rigors of that course, having been written by teachers who have taught the courses and seen where learners need help--and where other texts have come up short. The text features precise definitions and exposition, carefully crafted pedagogy, and a strong emphasis on algebraic, transcendental, and trigonometric functions. To show readers how important and relevant precalculus topics are to their future coursework, an optional Looking Ahead to Calculus feature appears in each chapter. The varied examples and exercises include many that encourage readers to use and understand graphs, as opposed to simply draw them, providing additional sound preparation for calculus. | 677.169 | 1 |
Mathematics
This course focuses on creating a simulator for the game of Blackjack that would allow for two players to play with the option of automating one of them. The simulation also takes into account the ever changing probability of the deck of cards. This is done by implementing a card counting algorithm, which will also calculate the rate of success for each given situation as well as the probability of breaking.
INTD 299: DS: CAPTCHA Security
This course focuses on the study of CAPTCHA security, particularly on techniques for creating a program capable of decoding CAPTCHAs using the MATLAB programming language.
Homma Farian
Lecturer of
Mathematics
Research Interests
My research interests include computational math and computer science applications of parallel and distributed computing.
Spring 2016 Classes
MATH 230: Programming&MathProblemSolving
This course serves as an introductory programming course for Mathematics majors. Basic programming techniques for solving problems typically encountered by mathematicians will be developed. The cour
se covers basic procedural techniques such as algorithms, variables, input/output, data types, selection, iteration, functions and graphing. Good programming and commenting practices will be emphasized. The programming language for the course will be a mathematical programming language such as Matlab. Restricted to Math majors only. Corequisite/Prequisite: MATH 222. Offered every semester | 677.169 | 1 |
A comprehensive mathematical approach to teaching binary arithmetic to students in a first course on the topic is presented. The weighted sum expression to represent signed numbers in 2's complement notation is developed and logically used to derive all the rules for arithmetic operations. 1's complement arithmetic is similarly dealt with. The superiority of the present approach over the conventional approach is pointed out | 677.169 | 1 |
Basic Medical Math
The Incredible Medical School
Alfred Ricks Jr., M.D.
This book is available for download with iBooks on your Mac or iOS device, and with iTunes on your computer. Books can be read with iBooks on your Mac or iOS device.
Description
Basic Medical Math gives you a foundation for everyday calculations in the clinical use of medications. The book begins with an introduction for fractions, decimals, percentages, and conversions. Acid base is explained as well as logarithms. Learn equivalents and milliequivalents. Osmotic pressure and calculations are explained. Learn percentage and ratio calculations. Several basic medication calculations are presented. Calcium mg and milliequivalents comparisons are explained. The differences in calcium chloride and calcium gluconate are covered. The relationship between micrograms and milligrams. The next time someone asks about potassium dosage in milligram or potassium dosage in milliequivalents you will know the difference. If you are in healthcare, this book will not replace your medical math textbook or reference, but it will give you a foundation that would take much longer to find in the those large textbooks. | 677.169 | 1 |
The Complete Idiot's Guide to Algebra, 2nd Edition/i>
Overview
Just the facts (and figures) to understanding algebra. inequalities, polynomials, exponents and logarithms, conic sections, discrete math, word problemsand more.
-Written in an easy-to-comprehend style to make math concepts approachable
-Award-winning math teacher and author of The Complete Idiot's Guide® to Calculus and the bestselling advanced placement book in ARCO's "Master" series
Download a sample chapter.
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Meet the Author
W. Michael Kelley is a former award-winning calculus teacher and the author of the bestselling The Complete Idiot's Guide® to Calculus, The Complete Idiot's Guide® to Precalculus, and The Humongous Book of Calculus Problems. Among the honors he's received throughout his career are his four-year-running title of "Most Popular Teacher" and being recognized by the Maryland Council of Teachers of Mathematics as an Outstanding High School Mathematics Teacher. He has contributed to over 50 publications including Playboy, WIRED, Sync, SPIN and The Source. | 677.169 | 1 |
This book takes a distinctly applied approach to finite mathematics at the freshman and sophomore level. It features many commercial applications, including linear programming, the theory of games, and financial mathematics. more...
Combinatorial theory is one of the fastest growing areas of modern mathematics. Focusing on a major part of this subject, Introduction to Combinatorial Designs, Second Edition provides a solid foundation in the classical areas of design theory as well as in more contemporary designs based on applications in a variety of fields. After an overview... more...
First published in 1982, this book examines anti-semitism in the Western world. The author concludes that, fringe neo-Nazi groups notwithstanding, significant anti-semitism is largely a left-wing rather than a right-wing phenomenon. He finds that Jews have reacted to this change in their situation and in attitudes towards them by making a shift to... more...
Mathematics in the Real World is a self-contained, accessible introduction to the world of mathematics for non-technical majors. With a focus on everyday applications and context, the topics in this textbook build in difficulty and are presented sequentially, starting with a brief review of sets and numbers followed by an introduction to elementary... more...
This title takes an in-depth look at the mathematics in the context of voting and electoral systems, with focus on simple ballots, complex elections, fairness, approval voting, ties, fair and unfair voting, and manipulation techniques. The exposition opens with a sketch of the mathematics behind the various methods used in conducting elections. The... more...
It has long been argued that the Allies did little or nothing to rescue Europe's Jews. Arguing that this has been consistently misinterpreted, The Myth of Rescue states that few Jews who perished could have been saved by any action of the Allies. In his new introduction to the paperback edition, Willliam Rubinstein responds to the controversy caused... more... | 677.169 | 1 |
Mathematics Of Voting And Elections A Hands-on Approach
9780821837986
ISBN:
0821837982
Pub Date: 2005 Publisher: American Mathematical Society
Summary: The results of an election depend not just upon the wishes of the electorate, but also upon the mathematics used to calculate the result. Using numerous case studies & featuring discussions of actual elections from the perspectives of both politics & popular culture, this text explores vote counting systems.
Hodge, Jonathan K. is the author of Mathematics Of Voting And Elections A Hands-on Approach, publishe...d 2005 under ISBN 9780821837986 and 0821837982. Six Mathematics Of Voting And Elections A Hands-on Approach textbooks are available for sale on ValoreBooks.com, and six used from the cheapest price of $33.98.[read more] | 677.169 | 1 |
Products in Mathematics
As technology advances, so must our education system. Cloud computing serves as an ideal method for e-learning thanks to its flexibility, affordability, and availability. Cloud-based learning is especially dynamic in STEM education, as it can significantly lower the cost of building cumbersome computer labs while fostering engaged learning and collaboration among students. The Handbook of Research on Cloud-Based STEM Education for Improved Learning Outcomes prepares current and future instructors for exciting breakthroughs in STEM education driven by the advancement of cloud technologies. From...
We all swim in a sea of Big Data, dangerously vulnerable to the unscientific thinking that now replaces the critical faculties we used to rely on. We seek simple explanations where complexity is required. But as we endeavor to solve global problems of energy, food, and water shortages, a planetary biodiversity crisis, and emerging threats to our public health, the development of scientific habits of mind becomes even more essential for our survival. We fear numbers and prefer neat and simple solutions to complex problems, but scientific reasoning plays a central role in combating misinformation...
A volume in Cognition, Equity, & Society: International Perspectives Series Editors: Bharath Sriraman, University of Montana and Lyn English, Queensland University of Technology Mathematics is traditionally seen as the most neutral of disciplines, the furthest removed from the arguments and controversy of politics and social life. However, critical mathematics challenges these assumptions and actively attacks the idea that mathematics is pure, objective, and value?neutral. It argues that history, society, and politics have shaped mathematics-not only through its applications and uses but also...
Pre-Algebra and Algebra Warm-Ups for grades 5 to 8+ provides students with daily math activities to get them warmed up for the lessons ahead and to review lessons learned. Each page features four warm-up activities that can be cut apart and used separately, making it easy to adjust each activity when needed. Mark Twain Media Publishing Company specializes in providing engaging supplemental books and decorative resources to complement middle- and upper-grade classrooms. Designed by leading educators, this product line covers a range of subjects including math, science, language arts, social studies,...Math Connections to the Real World for grades 5 to 8 increases students' ability to effectively apply math skills in real-world scenarios. Aligned to current state standards, this supplement offers students the opportunity to combine math and language arts skills to successfully solve everyday problems and communicate answers. Mark Twain Media Publishing Company specializes in providing engaging supplemental books and decorative resources to complement middle- and upper-grade classrooms. Designed by leading educators, this product line covers a range of subjects including math, science, languageGeometry Basics for grades 5 to 8 targets the basic geometry concepts students need to understand and perform operations involved in higher-level math. In this standards-based series, students are given practice with lines, angles, circles, perimeter, area, volume, two-dimensional figures, and three-dimensional figures. Mark Twain Media Publishing Company specializes in providing engaging supplemental books and decorative resources to complement middle- and upper-grade classrooms. Designed by leading educators, this product line covers a range of subjects including math, science, language arts,...The ABZZ word code system is a method that can be used to convert any given number to word code equivalents. With this method, any given number from zero to zillion can be converted to word codes. The principles of this method are explained in this book. This book also contains a thesaurus that gives two examples of easily derivable and meaningful word code equivalents of every number from 0 to 9,999. It is a forerunner to a proposed online thesaurus that would list word code alternatives of every number from zero to one billion (0-1,000,000,000). For centuries, people have been attracted to...
Magic Square is no longer magic, but the era of magic square puzzle has just begun. This outstanding application of Magic Square will be played by millions throughout the world. I appeal to educators to adopt magic square puzzles to schools' curriculums. If you are interested, e-mail me at dwkmei@aol.com.
Interactive Notebooks: Math for grade 8 is a fun way to teach and reinforce effective note taking for students. Students become a part of the learning process with activities about rational numbers, multistep equations, functions, the Pythagorean theorem, scatter plots notebooks...
Interactive Notebooks: Math for grade 7 is a fun way to teach and reinforce effective note taking for students. Students become a part of the learning process with activities about integers, proportions, expressions and inequalities, angle relationships, probability notebooks...
Interactive Notebooks: Math for grade 6 is a fun way to teach and reinforce effective note taking for students. Students become a part of the learning process with activities about absolute value, ratios, evaluating expressions, one-variable equations and inequalities, surface area...
Fully updated to reflect the new curriculum, the revised edition of Transforming Primary Mathematics sets out key theories and cutting-edge research in the field to enable teachers to take a fresh look at how they teach mathematics.
The book encourages teachers to reflect on their own beliefs and values about mathematics, and asks them to question whether their current methods meet the needs of all learners, and the challenge of having high expectations for all. It provides clear, practical approaches to help implement fundamental change in classroom environments, and offers motivational teaching...
The application of technology in classroom settings has equipped educators with innovative tools and techniques for effective teaching practice. Integrating digital technologies at the elementary and secondary levels helps to enrich the students' learning experience and maximize competency in the areas of science, technology, engineering, and mathematics. Improving K-12 STEM Education Outcomes through Technological Integration focuses on current research surrounding the effectiveness, performance, and benefits of incorporating various technological tools within science, technology, engineering,...
Professional development of educators is an complex process through which teachers strive continuously for pedagogical improvement. In that sense, professional growth benefits learners and teachers while also promoting the quality of the schools, colleges, and academic departments where it takes place. Innovative Professional Development Methods and Strategies for STEM Education is an authoritative publication featuring the latest scholarly research on a wide range of professional advancement topics in STEM education with special emphasis on content, process, implementation, and impact, as well...
The American Educational History Journal is a peer?reviewed, national research journal devoted to the examination of educational topics using perspectives from a variety of disciplines. The editors of AEHJ encourage communication between scholars from numerous disciplines, nationalities, institutions, and backgrounds. Authors come from a variety of disciplines including political science, curriculum, history, philosophy, teacher education, and educational leadership. Acceptance for publication in AEHJ requires that each author present a well?articulated argument that deals substantively with... | 677.169 | 1 |
Merchantville Statistics teach it right. Algebra 2 is a lot harder than it used to be. It's also more important than it used to be because algebra 2 concepts are included on the new SAT.Aashita B | 677.169 | 1 |
3/14Geometry
Summary
Harold Jacobs'sGeometrycreated a revolution in the approach to teaching this subject, one that gave rise to many ideas now seen in the NCTM Standards. Since its publication nearly one million students have used this legendary text. Suitable for either classroom use or self-paced study, it uses innovative discussions, cartoons, anecdotes, examples, and exercises that unfailingly capture and hold student interest. This edition is the Jacobs for a new generation. It has all the features that have kept the text in class by itself for nearly 3 decades, all in a thoroughly revised, full-color presentation that shows today's students how fun geometry can be. The text remains proof-based although the presentation is in the less formal paragraph format. The approach focuses on guided discovery to help students develop geometric intuition. | 677.169 | 1 |
Concepts inOn computational efficiency of the iterative methods for the simultaneous approximation of polynomial zerosIteration
Coefficient
In mathematics, a coefficient is a multiplicative factor in some term of an expression (or of a series); it is usually a number, but in any case does not involve any variables of the expression. For instance in the first three terms respectively have the coefficients 7, ¿3, and 1.5 (in the third term the variables are hidden, so the coefficient is the term itself; it is called the constant term or constant coefficient of this expression).
more from Wikipedia
Academic Press
Academic Press is an academic book publisher. Originally independent, it was acquired by Harcourt, Brace & World in 1969. Reed Elsevier bought Harcourt in 2000, and Academic Press is now an imprint of Elsevier.
more from Wikipedia | 677.169 | 1 |
For your Common Core and STEM curriculum.GRADES 6-8Explore and apply algebraic thinking and data analysis in the context of engineering design and adventure. The books will guide students through simulations of climbing Mt. Everest...
For your Common Core and STEM curriculum.These resources meet the 5 concepts addressed by the NCTM Standards and encourage students to learn and review the concepts in unique ways. Each task sheet is organized around a central problem taken ...
Writing about math helps students clarify their own thinking while reinforcing what they have learned. A variety of prompts encourage students to log their thoughts, activities, and understanding of math. Logs allow easy monitoring of students' ma..
For your Common Core and STEM curriculum.GRADE 6-8Explore and apply algebraic thinking and data analysis in the context of engineering design and adventure. The book will guide students through a simulation of being stranded on an...
Powerful strategies for adapting mathematical lessons and tasks to address a wide range of abilities, interests, and learning styles. This research-based book has activities tailored its certain grade span. Includes differentiated tasks that are rea..
For your Common Core and STEM curriculum.GRADE 6-8Explore and apply algebraic thinking and data analysis in the context of engineering design and adventure. The book will guide students through a simulation of navigating a mission ...
For your Common Core and STEM curriculum.GRADE 6-8Explore and apply algebraic thinking and data analysis in the context of engineering design and adventure. The book will guide students through a simulation of climbing Mt. Everest....
For your Common Core curriculum.This in-demand collection of lessons explores proportionality, proportional relationships, and proportional reasoning, acknowledging that the ability to reason proportionally is at the forefront of the mid...
Gr. 6-8. These simulations are uniquely appropriate for STEM (science, technology, engineering, and mathematics) initiatives. Each of the programs uses algebraic thinking and data analysis to solve engineering problems in real-...
Gr. 6-8. In Amazon Mission, students help a native village surmount malaria and confront the environmental hazards of gold mining. This simulation is uniquely appropriate for STEM (science, technology, engineering, and math...
Prices listed are U.S. Domestic prices only and apply to orders shipped within the United States.
Orders from outside the United States may be charged additional distributor, customs, and shipping charges. | 677.169 | 1 |
Students and research workers in mathematics, physics, engineering and other sciences will find this compilation invaluable. All the information included is practical, rarely used results are excluded. Great care has been taken to present all results concisely and clearly. Excellent to keep as a handy reference! If you don't have a lot of time... more...
An important new perspective on AFFINE AND PROJECTIVE GEOMETRY This innovative book treats math majors and math education students to a fresh look at affine and projective geometry from algebraic, synthetic, and lattice theoretic points of view. Affine and Projective Geometry comes complete with ninety illustrations, and numerous examples and... more...
Analytic Geometry covers several fundamental aspects of analytic geometry needed for advanced subjects, including calculus. This book is composed of 12 chapters that review the principles, concepts, and analytic proofs of geometric theorems, families of lines, the normal equation of the line, and related matters. Other chapters highlight the application... more...
Analytical Geometry contains various topics in analytical geometry, which are required for the advanced and scholarship levels in mathematics of the various Examining Boards. This book is organized into nine chapters and begins with an examination of the coordinates, distance, ratio, area of a triangle, and the concept of a locus. These topics are... more...
The main focus of this volume is on the problem of describing the automorphism groups of affine and projective varieties, a classical subject in algebraic geometry where, in both cases, the automorphism group is often infinite dimensional. The collection covers a wide range of topics and is intended for researchers in the fields of classical algebraic... more...
Bibliotheca Mathematica: A Series of Monographs on Pure and Applied Mathematics, Volume V: Axiomatic Projective Geometry, Second Edition focuses on the principles, operations, and theorems in axiomatic projective geometry, including set theory, incidence propositions, collineations, axioms, and coordinates. The publication first elaborates on the... more...
Get ready for a trip around the world to find the many different shapes that surround you. You can find shapes in the places you go, games you play, and even the food you eat. In fact, shapes are everywhere! Can you find them? 32pp. more... | 677.169 | 1 |
This book presents the first five Abel Prize in Mathematics winners, from 2003 to 2007. Leading experts in the field detail the work of each Abel Prize winner. The book also includes a brief history of the Abel Prize. more...
To succeed in Algebra II, start practicing now Algebra II builds on your Algebra I skills to prepare you for trigonometry, calculus, and a of myriad STEM topics. Working through practice problems helps students better ingest and retain lesson content, creating a solid foundation to build on for future success. Algebra II Workbook For Dummies,... more...
Algebraic and Classical Topology contains all the published mathematical work of J. H. C. Whitehead, written between 1952 and 1960. This volume is composed of 21 chapters, which represent two groups of papers. The first group, written between 1952 and 1957, is principally concerned with fiber spaces and the Spanier-Whitehead S-theory. In the second... more...
For the past ten years, alternative loop rings have intrigued mathematicians from a wide cross-section of modern algebra. As a consequence, the theory of alternative loop rings has grown tremendously. One of the main developments is the complete characterization of loops which have an alternative but not associative, loop ring. Furthermore, there is... more...
Aimed at both students and researchers in philosophy, mathematics and the history of science, this edited volume, authored by leading scholars, highlights foremost developments in both the philosophy and history of modern mathematics. - ;This edited volume, aimed at both students and researchers in philosophy, mathematics and history of science, highlights... more...
This work presents the most important combinatorial ideas in partition calculus and discusses ordinary partition relations for cardinals without the assumption of the generalized continuum hypothesis. A separate section of the book describes the main partition symbols scattered in the literature. A chapter on the applications of the combinatorial methods... more... | 677.169 | 1 |
Students are to work in groups at the beginning of the class (usually around 15 min) to fill up the worksheet. The worksheet is loosely based on the tutorial problems. After which, the tutor will go through with the worksheet as a class.
Tutorial 4 Worksheet
Please be reminded that there will be no classes on 9 Feb. Instead, Dr Tan will conduct a mass tutorial session on 11 Feb from 8:30am to 9:30am at LT24.
Tutorial 2 Worksheet
A student asked about setting "" when we substitute . In these cases, we may obtain extra solutions by setting .
For Q3(a), we divided by to separate the equation. So, the additional solution is .
For 3(b), we divided by to separate the equation. So, the additional solution is , or , where is an integer. The extra solutions are therefore .
While this sort of accounting seems rather abstract, there is usually a "natural" interpretation for the extra solution in most engineering problems. For example, when we look at Newton's law of cooling in Tutorial 1, Q6. The extra solution corresponds to the case where the temperature of the object is equal to the temperature of the surroundings . Clearly, there will be no cooling or warming.
I've excluded the Chapters 17 to 20 from David Lerner's notes. Does that mean that orthogonal projections, inner product and trace will not be tested?
No. I excluded them because in the notes, more "advanced" concepts are discussed in the chapters. For example, the chapter on inner product introduce the concept of positive definite and bilinear, which may be confusing. However, you will still need to know that how to compute the inner product between two vectors and what it means for two vectors to be orthogonal.
Similarly, orthogonal projection and trace are classic examples of linear transformations. Hence, you should be able to determine the kernel and range of these transformations. For example, Tut 10 Q2 asks you to determine the range of a trace map, which is within your ability. However, we will not ask you to show that trace map is invariant with respect to a change of basis.
I do not understand the concept of determining a matrix associated with a linear transformation.
It looks a lot, but the videos are typically very short. You can also select the topics to revise.
For those who prefer the "abstract" flavour, you can search for the text Linear Algebra Done Right by Sheldon Axler. (The link is the link to the official website. The book is easily found online, but I would not place it here.) This text is probably more useful for Linear Algebra 2.
There are some who attempted Q5 to Q7 (in Tutorial 6). I have also provided the solutions and feedback for these questions too.
Question 1 and 5
Solution: (Q1) Since the equation contains the term , there are no values of that make the equation linear. (Q5) Since the equation contains the term , there are no values of that makes the equation.
Feedback: Unless otherwise specified, unknowns in an equation are given by , and . For an equation to be linear, any term containing an unknown must be of the form (constant)(unknown). Terms like and are not of the required form and hence, the equation is not linear.
Some mistakes:
Writing "" and "" to mean "there are no values of ". This is not the case. "" means can be . "" means is a set which contains no element.
There is a solution that states , because is in fact two straight lines. This is not correct. Prof Putinar mentioned that is in some sense "linear". Do not interpret this as saying that the equation is linear. By definition, both and are not linear equations. | 677.169 | 1 |
Alg II 2-6 Transformations
2.
Essential Understanding: There are sets of functions called families, in which each function is a transformation of a special function called the parent.Objectives:Students will be able to analyze transformations of functions
3.
FunctionsF-IF.7.Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. | 677.169 | 1 |
Introductory & Intermediate Algebra for College Students
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Read More textbook, to the enhanced digital resources in the MyMathLab course. Blitzer pulls from topics that are relevant to college students, often from pop culture and everyday life, to ensure that students will actually use their learning resources to achieve success. With an expansion of the series to now include a Developmental Math "all-in-one" text (with content spanning prealgebra through intermediate algebra), and with an enhanced media program accompanying this revision, developmental students at all levels will see how math applies to their daily lives and culture. Also available with MyMathLab MyMathLab(R) is an online homework, tutorial, and assessment program designed to work with this text to engage students and improve results. Within its structured environment, students practice what they learn, test their understanding, and pursue a personalized study plan that helps them absorb course material and understand difficult concepts192907 / 9780134192901 * Introductory & Intermediate Algebra for College Students Access Card Package Package consists of: 0134178149 / 9780134178141 * Introductory & Intermediate Algebra for College Students 0321431308 / 9780321431301 * MyMathLab -- Glue-in Access Card 0321654064 / 9780321654069 * MyMathLab Inside Star Sticker
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New. 0134178149 Premium Publisher Direct Books are Like New or Brand New books direct from the publisher sometimes at a discount. Multiple copies are usually available. These books are not available for expedited shipping and may take up to 14 business days to receive.
Good. Hardcover. Instructor Edition: Same as student edition but has free copy markings. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780321780102Very good. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. Has minor wear and/or markings. SKU: 9780321823687-3Fair. Hardcover. Missing components. All text is legible, may contain markings, cover wear, loose/torn pages or staining and much writing. SKU: 97803217293851729385.
Fine. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. Almost new condition. SKU: 9780321823687 | 677.169 | 1 |
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