Theory of Functions 9780231898300

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THEORY OF FUNCTIONS J. F. RUT DAVIES PROFESSOR OF MATHEMATICS, COLUMBIA UNIVERSITY

REVISED EDITION KING'S CROWN PRESS • NEW YORK

Copyright 1947 by Joseph F. Ritt FIRST EDITION 1946 First

printing

1946

REVISED EDITION

1947

First

1947

Second

KING'S

printing printing

CROWN

1949

PRESS

is a subsidiary imprint of Columbia University Press established for the purpose of making certain scholarly material available at minimum cost. Toward that end, the p u b l i s h e r s have adopted every reasonable economy except such as would interfere with a legible format. The work is presented substantially as submitted by the author, without the usual editorial and typographical attention of Columbia University Press.

Composed on the Vari-Typer by Marie Russell Diagrams prepared by Angela

Pelliciari

Published in Great Britain and India by Geoffrey Cumberlege, Oxford University Press London and Bombay Manufactured

in the United States of America

Preface There is presented herewith the basic material of a course in the theory of functions which I have given several times at Columbia University during the past twenty years. This course, which is given annually, is a two-semester course with twenty eight periods of seventy five minutes in each semester. The emphasis is on the complex variable. From this standpoint, about twenty periods are given to the real variable and the remaining time to the complex. By the end of the first semester, the student finds himself equipped for higher courses in the real variable, or for a course on existence theorems for differential equations. After the second semester, he can study almost any topic of mathematical analysis. Above all, he has learned what it means to understand mathematics and can deal with any mathematical book. In treating the real number system, I have used the method of infinite decimals, rather than the Dedekind or Cantor theories. The decimals have the advantage of lacking profundity and of not putting the student through a mathematical revolution. The numbers stay quite what they always were, instead of becoming new and bizarre objects. The objection that the decimals employ the special radix ten does not strike me as important. In any case, after the student has seen how simple the matter really is, he can read the Dedekind and Cantor theories with very little effort. There is hardly time in the course for a full treatment of topological questions. However, all geometrical questions are formulated in arithmetic terms and every topological assumption made is explicijjy stated. Of course, the student quickly sees that topological considerations are important only for securing a rounded theory and may be disregarded as far as cases arising in the applications are concerned. I wish to thank Mr. Milton Sobel for reading the proofs for the present edition. New York, N. Y.

J . F. RITT

July, 1947

v

Contents I. The Real Number System .

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Introduction — Counting — Sets — Sequences — Digits — Terminating Decimals— Addition and Multiplication — Infinite Decimals — Non-Negative Numbers — Bounds — Upper Bounds — 4 Fundamental Theorem .— Least Upper Bounds — Round Numbers — Addition of Non-Negative Numbers --Multiplication of Non-Negative Numbers — Subtraction — Division — Negative Numbers — Rational Numbers — Irrational Numbers — Bounds — The Least Upper Bound — The Greatest Lower Bound.

II. Theory of Limits

11

Absolute Value — Convergent Sequences of Numbers — Uniqueness of Limit — The Fundamental Convergence Theorem (Cauchy) — Limits of Sums, Products, Quotients.

III. Linear Point Sets .

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Intervals — Neighborhoods — Limit Points — The Theorem on Nested Intervals — The Bolzano-Meiersirass Theorem — Closed Sets — Borel's Theorem — Countability.

IV. Functions and Continuity

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Variable — Function — Continuity — Dense Sets — A Function Continuous at a Dense Set of Points and Discontinuous at a Similar Set — Continuity of Sum and Difference of Two Functions — Continuity of Product of Two Functions — Continuity of Quotient — Bounded Functions — Boundedness of a Function Continuous on a Closed Interval — Attainment of Bounds by a Function Continuous on a Closed Interval — Attainment of All Values Intermediate between Two Values — Uniform Continuity — Monotonie Functions.

V. The Derivative

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Derivative — Differentiability and Continuity — The Derivative as a Function— Derivatives at Extremities of Intervals — Right-Band and Left-Band Derivatives— Maxima and Minima — Necessary Condition for a Maximum or a Minimum at an Interior Point of an Interval — The Mean Value Theorem — Functions with Zero Derivatives — Increasing and Decreasing Functions.

VI. Riemann Integration

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The Integral — Condition for Integrabllity — Integrabillty In Subintervals—' Bounds for an Integral — Continuity of Integral with Respect to Upper Limit — Integrabillty of Sum of Two Functions — Integrabllity of Continuous Functions — Differentiability of the Integral of a Continuous Function — Evaluation of Definite Integrals — Composition of Intervals — Discontinuous Functions.

VII. Infinite Series of Numbers

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Infinite Series — Convergent Infinite Series — A Necessary and Sufficient Condition for Convergence — Addition of Convergent Series — The nth Term — The Remainder — A Class of Convergent Series — Series of Absolute Values — Absolute Convergence — Rearrangement of Terms of an Absolutely Convergent Series. vii

37

VIII. Sequences of Functions .

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Convergent Sequences of Functions — Uniform Convergence — Example of a Uniformly Convergent Sequence Defined on an Interval — Example of Non-Uniform Convergence — necessary and Sufficient Condition for the Uniform Convergence of a Sequence — Continuity of Limit of a Sequence — Example of a Sequence of Continuous Functions Whose Limit Is Sot Continuous — Integrability of Limits of Sequences — Differentiability of Sequences of Functions. IX. Infinite Series of Functions.

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Infinite Series of Functions — Uniform Convergence of Infinite Jfeierstrass M-Test — Continuity, Integrability, Differentiability tion Discontinuous at a Dense Set of Points, but Integrable. X. Functions of Two Variables

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Points — Functions of Two Variables — Rectangles Derivatives — The Complete Differential. XI. Complex and Hypercomplex Numbers .

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48

Series — The — A Func-

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— Continuity

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51

Partial

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Complex Numbers — Addition and Multiplication — Subtraction and Division — Change of Notation — The Modulus — Geometric Representation — Modulus of Sum and Difference — Hypercomplex Numbers and Linear Algebras. XII. Limits and Point Sets (Complex Domain).

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Limits — Operations with Sequences — Point Sets Theorem on Nested Rectangles — Boret's Theorem. XIII. Curves and Regions.

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58

in the Complex Domain — The

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Interior», Exterior and Boundary Points — Continuous Curves — Open' Regions — Functions and Continuity: Definition of Function — Continuity — Uniform Continuity. XIV. Derivatives

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— The Cauchy-Riemann Equations — The Laplace Derivative — Monogenicity Equation — Definition of Analytic Function — Conforml Mapping. XV. Continuous Curves . Inverse Functions Equivalent Curves

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(Real Variable) — On the Points of a Continuous Curve — — Simple Curves — Representation of a Simple Curve. N

XVI. Rectifiable Curves

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Segment and Length — Inscribed Polygon — Rectifiable Curves — Condition for Red if lability — Functions of Bounded Variation — Rectiflability and Functions of Bounded Variation. XVII. Curvilinear Integrals Functions grability

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Continuous on a Curve — Curvilinear — Existence Theorem — An Example.

Integral

— Condition

for

Inte-

XVIII. Jordan Curves Definition

83 — Equivalent

Jordan Curves — Application

XIX. Analysis Situs of the Triangle

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to .

Integration. .

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Statement of the Jordan Separation Theorem — Straight Lines — Triangles — Interior and Exterior — Decomposition into Four Triangles — Integration. viii

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XX. The Cauchy I n t e g r a l Theorem f o r Triangles

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An Inequality for Curvilinear Integrals — Statement of the Cauchy Integral Theorem for the Case of a Triangle — Two Curvilinear Integrals — On Differentiable Functions — A Theorem on Sequences of Triangles — Proof of the Cauchy Integral Theorem for Triangles. XXI. Extension of the Cauchy I n t e g r a l Theorem to Polygons.

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Simply-Connected Open Regions — Polygons — Statement of the Cauchy Integral Theorem for Polygons — Replacement of the Polygon C by a Simpler Polygon — Completion of Proof. XXII. The Cauchy I n t e g r a l Theorem f o r a R e c t i f i a b l e Curve .

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Curves in Open Regions — Open Regions, Curves and Approximating Polygons proximation to Integrals — The Cauchy Integral Theorem for a Rectifiable XXIII. The Cauchy I n t e g r a l Theorem f o r Several Contours

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— ApCurve.

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Sensed Circles — Sense of a Jordan Curve — Integrals along Sensed Curves —Regions Bounded by Several Contours—The Cauchy Integral Theorem for Several Contours. XXIY. Preliminaries f o r Cauchy I n t e g r a l Formula . ' Analyticity at a Point—Combinations Integral of l / ( z - a ) .

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of Analytic

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Functions

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— The number n —

XXV. The Cauchy I n t e g r a l Formula and the Derivatives of an Analytic Function . Cauchy Integral Formula — The First Derivative — The Higher an Analytic Function — Analytic Functions Defined by Contour XXVI. I n f i n i t e Sequences and I n f i n i t e Series of Analytic Functions

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Derivatives Integrals. .

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108

of .

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Uniform Convergence —Continuity and Integrabillty—Sequences of Analytic Functions — The Higher Derivatives — Series of Complex Numbers — Series of Analytic Functions. XXVII. Power S e r i e s . Greatest

Limit

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of a Sequence of Real

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Ilumbers —Power

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Series

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—Circle

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of

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Convergence.

XXVIII. Taylor's Expansion

119

Statement of Results of Expansion.

— Proof

of Uniqueness

— A Preliminary

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XXIX. L i o u v i l l e ' s Theorem and the Fundamental Theorem of Algebra Integral

Functions

and Liouville's

Theorem—The

XXX. On the Zeros of Analytic Functions The Chief

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Theorem — Some Consequences

XXXI. Laurent Series

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of .

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Fundamental

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the Preceding .

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Derivation .

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Theorem of .

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Algebra. .

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Theorem.

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Series of Negative Powers — Statement of the Expansion Theorem of Laurent — Proof of Uniqueness, Determination of the Coefficients — Derivation of Expansion. XXXII. S i n g u l a r i t i e s of Analytic Functions

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Functions Bounded in the Neighborhood of a Point — Isolated Singularities — — Reciprocals of Functions — Isolated Essential Poles — The Number Infinity Singularities — Behavior of a Function at XXXIII. Products and Quotients of Analytic Functions Multiplicity of a Value tions — Principal Part

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— Product of Two Functions of a Laurent Development. ix

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— Quotient

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Two Func-

134

XXXIV. Rational Functions

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Decomposition of a Rational Function into Partial Fractions — Functions Whose Only Singularities Are Poles — Frequency of Attainment of Values — Polynomials — Degree of a Rational Function. XXXV. The Functions e z , sin z , cos z

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140 ez

The Definitions — First Relations -- The Multiplication Formula for ~ The Relation sin 2z + cos 2 z = 1 — The Addition Theorems for sin z and cos z — Periodicity of sin z, cos z, e z — Other Relations — Graphs — Parametric Representation of Circle — A Formula for the Length of a Rectiflable Curve — Proof that p = n — Amplitudes of a Complex Number. 147

XXXVI. Periodic Functions Period Strip -- Fundamental sin z and cos z

ez

Domain — The Function

XXXVII. Indefinite I n t e g r a l s , Logarithms.

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Functions

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Independence of an Integral of Path of Integration — The Indefinite Integral Logarithm of an Analytic Function — Integral Functions Vhich Are Nowhere Zero — The Development of Log (1 + z ) .

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XXXVIII. I n f i n i t e Products Convergent Infinite vergence —Absolute

153 Products .— Necessary and Sufficient Condition Convergence — Infinite Products, of Analytic

XXXIX. The Weierstrass Factorization Theorem. Integral Functions with a Finite with an Infinite Number of Zeros Zeros — Convergence Proof.

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Number of Zeros — Construction

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— On Integral of a Function

XL. Meromorphic Functions and M i t t a g - L e f f l e r ' s Theorem

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for ConFunctions. .

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Functions with Assigned

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Definition of Meromorphic Function —Representation of a Meromorphic tion as a Quotient of Integral Functions — Mittag-Leffler 's Theorem.

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Func-

XLI. Theory of Residues

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Residue at a Finite Point — Residue at Infinity — Formula in a Curve — The Residues of a Rational Function — On the tive of an Analytic Function — The Fundamental Theorem of Residues — Special Cases — Application to the Fundamental XLII. Certain Important Theorems .

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for Residues withLogarithmic Derivathe Calculus of Theorem of Algebra. •

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Rouchö's Theorem — Application to the Fundamental Theorem of Algebra— Preservation of Neighborhoods — On the Maximum of the Modulus of an Analytic Function — On the Minimum of the Modulus of an Analytic Function — Fundamental Theorem of Algebra — Sequence of Analytic Functions X L I I I . Variation of the Amplitude of a Continuous Function along a Continuous Curve . Variation

of Amplitude

— On the Zeros

of an Analytic

Function.

XLIV. The Functions OJz, log z The Function

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s/z — The Function

XLV. Analytic Continuation

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log z. .

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Elements and Their Continuations — Monogenic Analytic Function — tion along a Curve — Singular Points — Singular Points on a Circle Convergence — Natural Boundaries.

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. Continuaof

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I The Real Number System INTRODUCTION 1. The theory of functions can be approached only on the basis of accurate notions concerning the real number system. In any treatment of the number system, there will be ideas which have to be taken for granted. The choice of the ideas which are to be considered as primitive depends always on questions of expediency. In what follows, we shall explain what we understand to be known in advance and we shall add a set of definitions and proofs which will furnish us with an adequate theory of the real number system. COUNTING We accept as a primitive idea the idea of counting. That is, we assume familiarity with the cardinal numbers 1, 2, 3, •••. n, ... to which we attribute intuitive meaning and which we shall use to measure the multiplicities of collections of objects. SETS 3- A set of objects will be called a finite that the set contains exactly n objects.

set if there exists a cardinal number n such

A set of objects will be called an infinite set if, given any cardinal number n, the set contains more than n objects. For instance, the set of cardinal numbers is an infinite set. SEQUENCES 4. Let n be any cardinal number. By a sequence of n objects, we shall mean a set of n objects . Oi, 0 2 , ..., o n , arranged in a one-to-one correspondence with the first n cardinal numbers. Such a sequence will be said to be finite. By an infinite

sequence

of objects, we shall mean an infinite set 0 o , 0 2 , ..., 0 n , ..., which can be arranged, and is arranged, in a one-to-one correspondence with the complete set of cardinal numbers. DIGITS 5. Independently of their appearance in the symbols for the cardinal numbers, we shall employ the digits 0, 1, 2, 3. 4, 5 . 6 , 7, 8, 9 as pure symbols, with which we shall play a iame which will be called arithmetic. The game will consist in performing certain operations with entities, called numbers, which will be constructed with the digits. TERMINATING DECIMALS 6. By a terminating decimal we shall mean a finite sequence of digits with a decimal point written after some one of them, the first digit of the sequence being distinct from zero if the decimal point does not follow it immediately.

1

Examples: 56.47; 0.3420; 0.00. If no digit follows the decimal point, the decimal point may be omitted with no danger of confusion. Thus, we may write 65 instead of 65. • For the present, we shall say "decimal" rather than "terminating decimal." We shall regard the decimals as pure symbols, possessing no quantitative significance. We shall now explain the meaning of the word equal as applied to two decimals. Let a and b be two decimals. If b is either identical with a or obtained from a by adjoining; a set of zero digits to St f 8L and b will be called equal. Thus 5«4 and 5*400 are equal. It is to be emphasized that the adjective equal is merely a convenient word for describing the relationship just set forth. No meaning other than that given is to be read into our language. The word "equal," whose meaning is thus purely technical, will justify itself in later developments. Two decimals which are not equal will be called unequal. The terms "greater than" and "less than" as applied to two unequal decimals, will be employed in the manner which one would expect and we shall not use the space or time to give formal definitions for them. Thus, 64 > 53.1

0.32 < 0.323.

The student already knows how to compare two unequal decimals and all that is important is to consider the comparison as a type of game playing rather than as an operation with concrete quantitative significance. ADDITIOH AND MULTIPLICATION

7- We perform the operations of addition and multiplication upon decimals exactly as in popular arithmetic, with the distinction that we regard the operations as moves in a game. Addition has the properties: 1.a+b=b+a 2. a + ( b + c ) = ( a + b ) + c

(Commutativity) (Ass«ciativity)

Multiplication has the properties: 1. ab = ba (Commutativity) 2. a(bc) = (ab)c (Associativity) 3. a(b + c) = ab + ac (Distributivity) We note that, if a = b, c - d, then a + c = b + d , ac» bd. INFINITE DECIMALS

8. Consider a symbol composed of an infinite sequence of digits with a decimal point written after some one digit, the first digit being distinct from zero if the decimal point does not follow it immediately. If there is no digit in the sequence such that every digit which follows is the digit 9> the symbol will be called an infinite decimal. Examples: 2.333... 0.1212... 0.0000... The symbol 2.43999-.-9-•• is not an infinite decimal. In a decimal such as 64.24000..., which concludes with zero digits exclusively, the zero digits with which the decimal concludes may be omitted with no danger of confusion. In particular, we shall write 0 for 0.00.... When we say that two infinite decimals are "equal," we mean that they are identical. Two (infinite) decimals which are not identical will be called unequal. For two unequal decimals, we use the terms "greater than" and "less than" in the manner which the reader would propose himself if asked to establish meanings. If a > b and b >c, then a>c.

2

NON-NEGATIVE NUMBERS 9. An infinite decimal will be called a non-negative from 0 will be called positive.

number. A non-negative number distinct

BOUNDS 10. Let E be any finite or infinite set of non-negative numbers. If there exists a positive number G such that every number of E is less than G, we shall call E bounded. Thus, the set of non-negative numbers which do not exceed 3 (that is, 3•00...) is bounded. The sequence

2, 2.1, 2.11, 2.111,

...

is bounded. The sequence 1» 2, 3> • • • > n,... is not bounded. Every finite set of non-negative numbers is bounded. UPPER BOUNDS 11. Let E be any bounded set of non-negative numbers. There exist non-negative numbers G such that, for every number a in E, one has a s G. Every such number G will be called an upper bound of E. If G is an upper bound of E and if G' > G, then G 1 is also an upper bound of E. A FUNDAMENTAL THEOREM 12. Theorem: Let E be a bounded set of non-negative numbers. bound of E which is less than every other upper bound of E.

Then there

exists

an

upper

Remark: This is noteworthy because there exist sets of non-negative numbers which contain no least number. For instance, the sequence

1, 0.1, 0.01, 0.001, ... has no least number. Proof: By the integral part of a non-negative number a, we shall mean the number secured by replacing every digit which follows the decimal point by zero. Considering the bounded set E, we examine the integral parts of the numbers of E. The boundedness of E certainly implies that there are only a finite number of distinct integral parts. We consider those numbers of E which have the greatest integral parts of all of the numbers of E. These numbers constitute a set which we shall E 0 . The numbers of E 0 all have the same integral part. Let a represent the finite sequence of digits which precedes the decimal in that unique integral part. From the numbers of E 0 , let us select those for which the digit in the first decimal place is as great as it can be. Let be that greatest digit. The numbers of E 0 in which appears in the first decimal place constitute a set Ei. From the numbers in Ei, we select those whose digit in the second decimal place is a maximum, say (3a. We continue in this fashion, selecting an infinite sequence of digits: 0ij Pst • ••> 0 n j • • • Consider now the symbol (1)

a . p^ps...

in which a decimal point follows a. Suppose first that there is no digit Pj in the symbol

3

such that = 9 for j > i. Then the symbol (1) is a non-negative number. We designate this number by n. Very plainly, no number in E exceeds M. Thus, M is an upper bound for E. We have to show that M is less than any other upper bound. This is certainly true if M is zero. Suppose that M is positive. We shall show that no number less than M is an upper bound. Let Mi be some number less than M. Then there must be some number a . 0! 3, ... pr 000... which exceeds Mi. On the other hand, the set E r is made up of numbers no less than a . gi |3r 000 Thus Mi is not an upper bound for E.

...

Suppose now that there is a digit 3, in (1) such that Pj = 9 for j > i. We consider the terminating decimal a . f3i ... ^ and add to it 0.00...01, where the digit 1 stands in the ith decimal place. The sum is a terminating decimal, which, if an infinite sequence of digits 0 is adjoined to it, produces a non-negative number. This non-negative number, we designate by M. The number M is seen as above, to be an upper bound of E which is less than any other upper bound. LEAST

UPPER

BOUNDS

13. Let E he a bounded set of non-negative numbers. The number M whose existence was proved in §12 is called the least

upper

bound or the

upper

bound of E. Evidently, if E contains posi-

tive numbers, M is positive. In that case M has the following properties, which

characterize

It:

a. For every number a of E, a £ M. b. Given any number Mi less than M, E contains a number greater than Mi. ROUND

NUMBERS

14. By a round number, we shall mean an infinite decimal like 34.72580000..., which concludes with zero digits exclusively. We define addition for two round numbers as follows: Suppress the zero digits in both numbers from some point on. There result two terminating decimals. Add these, and complete the sum into a round number by adjoining zeros. The resulting round number will be called the sum of the two given round numbers. Multiplication is defined similarly. Evidently, addition and multiplication, for round numbers, have the properties stated in §7. ADDITION

OF NON-NEGATIVE

NUMBERS

15. Let a and b be two positive numbers, which may or may not be round numbers. Consider all round numbers less than a. They form a class which we shall call Ei. Similarly, let E 2 be the class of round numbers which are less than b. We form now a class E, putting into E every round number which is the sum of a number in E t and a number in E 2 . (See §14.) Let us prove that E is bounded. Let a be any round number which exceeds a, and P any round number which exceeds b. Then every number in E is less than a + p. (Obvious on the basis of elementary arithmetic.) Thus E is bounded. By §12, E has a least upper bound M. If a and b are both round numbers, M is seen easily to be the sum of a and b as defined in §14. When at least one of a and b is not round, M will be taken by definition as the sum of a and b. 16. By definition, if a is any non-negative number, we take a + 0 = 0 + a = a . 17. Evidently, addition as defined in §§15» 16 is commutative. We shall prove it associative. Let a, b, c be three infinite decimals. We shall prove that a + (b + c) = (a + b) + c.

4

The case in which one of the three numbers is zero is trivial. Accordingly, we assume a, b, «5 all positive. Suppose that a + (b + c) > (a + b) + c. Let d be a round number such that a + (b + c) > d > (a + b) + c. Because d < a + (b + c), there must be round numbers a and 6 with a (a + b) + c. This will follow if we can show that a + 0 < a + b. Let at and a < o t < a,

P
a.

Proof: We can c e r t a i n l y find a number d, l e s s than b, of the form 0.00...0100... such t h a t a + d > a . By Lemma 1,

a + b g a + d > a . SUBTRACTION

20. Let a and b be p o s i t i v e numbers with a > b. We s h a l l prove t h a t there exists only one positive number c such that a = b + c.

one and

Proof: I t i s easy to see t h a t there e x i s t p o s i t i v e round numbers y such t h a t b + y < a . In s h o r t , there are such numbers y of the type 0.00...0100... By Lemma 2 of §19, we have b + y > y f o r any y. Hence every round number y such t h a t b + y < a i s l e s s than a . Thus, the s e t of round numbers y such that b + y < a i s a bounded s e t . Let c be the l e a s t upper bound of t h i s s e t . Then c > 0. We s h a l l prove t h a t a = b + c. F i r s t we show t h a t a § b + c . This we prove by considering any two round numbers 3 and y which are l e s s respectively than b and c and showing t h a t a > (3 + y. Because y < c , there i s a round number y g r e a t e r than y, such t h a t b + y < a . Now b + y g f B + y by Lemma 1 of §19. Thus a > 0 + y . Again, P + y > (3 + y, (round numbers), so t h a t a > P + y. Hence a z b + c . Suppose now t h a t a > b + c. We can c e r t a i n l y f i n d a p o s i t i v e 6 such t h a t a > (b + c) + 6 There are such numbers 6 of the type 0 . 0 0 . . . 0 1 0 0 . . . . Then (1)

a > b + (c + 6)

% Lemma 2 of §19, c + 6 > c. Let y be any round number which i s g r e a t e r than c but l e s s than c + 6. By (1) and Lemma 1 of §19,- we have a > b + y. On the other hand, the d e f i n i t i o n of c shows t h a t , because y > c, we have b + y 2 a. Thus, we cannot have a > b + c . Then a = b + c . I t remains to prove t h a t c i s unique. Suppose t h a t a=b+Ci,

a = b + c2

with c 2 > Ci. By what precedes, there i s a d > 0 such that c t + d = c 2 . Then a + d = (b + Ci) + d = b + (ci + d) = b + c 2 = a . But Lenma 2 of §19 shows t h a t a + d > a . Thus c i s unique. The quantity c determined above i s called the difference between a and b and i s denoted by a - b. We have seen t h a t 0 i s the onlir number which when added to a number a produces a . We define a - a , f o r any non-negative a, as 0. Similarly, because a = 0 + a, we write a - 0 = a . 6

DIVISION

21. Let a and b be two positive numbers. We wish to show that there is one and only one positive number c such that a = be. There are positive round numbers y such that by < a. I t i s easy to show that those round numbers are bounded. Let c be the least upper bound of the numbers y. One proves, following the method of §20, that a = be. The uniqueness is also proved as in §20. We c a l l c the quotient of a by b and write

For the case of a = 0, we write, f o r any b > 0,

NEGATIVE

22. Let

NONBERS

0 fl

a . PiP 2 ....,

with a a f i n i t e sequence of digits, be any positive i n f i n i t e decimal. Introducing a dash, (that is, ) , into our category of mathematical symbols, we form the symbol a .

P2....

The symbol just written w i l l be called a negative number. We use the symbol -0 as an equivalent of 0. That i s , the meaning of "-0" is to be 0. The non-negative numbers and the negative numbers w i l l be called real numbers. The arithmetic operations w i l l be performed upon real numbers in the manner taught in f i r s t courses in algebra. For instance, i f a and b are positive numbers, we are to have, by definit Ion,

a(-b) = (-a)b = - ( a b ) ; ( - a ) ( - b ) = ab. I f a > 0, b > a, we are to have a - b = - ( b - a ) . One sees from the above examples how the arithmetic operations are to be performed in other cases. I t i s seen without d i f f i c u l t y that the arithmetic operations on real numbers, described as above, have the commutative, associative and distributive properties mentioned in §7. Let a be any negative number. Let a = -b, where b i s a positive number. By -a, we shall mean b. Thus, by definition, - ( - 2 ) = 2. The real number system is ordered as follows. We say that a > b i f a - b is positive. This is harmonious with the ordering which existed at the start among the non-negative numbers. RATIONAL

NUMBERS

23. A real number in which every digit which follows the decimal point is zero w i l l be called an integer A number which is equal to the quotient of two integers w i l l be called a rational

number. Thus,

f,

1 , 0 . 1

are rational numbers. 24. We s h a l l show that between any two numbers there

7

lies

a rational

number.

Let a and b be two numbers with a > b. First, let a and b be positive. Then, between a and b there lies a round number, and round numbers are rational. A little reflection will show that the result is true in all cases. IRRATIONAL NUMBERS

25. There exist numbers which are not rational. An example will be given below. A number which is not rational will be called Irrational. Towards showing the existence of irrational numbers, we shall prove first that there exists a positive number whose square equals 2. There are certainly positive numbers whose squares are less than 2. Let E be the class of all such numbers. Evidently E is bounded. Let a be the least upper bound of E. We shall prove that a 2 = 2. First, let a 2 > 2. Let b = ' a - b > 0. Now

~ 2 . Then b > 0. We have b = | - 1 so that b < j* < a. Hence, 2a 2 a 2

(a - b)2 = a 2 - 2ab + b2 = 2 + b 2 > 2. Thus a - b, which is a positive number less than a, has a square which exceeds 2. This contradicts the fact that a is the least upper bound of E. (Note that, since a - b < a, E contains a number c which exceeds a - b. As c is in E, c < 2. But, if q > a - b, we would have c2 > (a - b)2 > 2.) Similarly, one cannot have a 2 < 2. Thus, a 2 = 2. We denote a by -J2. 26. To prove that n/S" is irrational, we assume that J2--1

(1)

q

with p and q integers. Evidently we may assume p and q both positive. We have (2)

E^ = 2, q

so that P 2 = 2q2 .

(3)

Equation (3) asserts the existence of positive integers which have the property that their squares are double the squares of positive integers. The integer p, for instance, has this property. Of all positive integers having this property, let r be the least. Let r2 = 2s2

(4)

with s a positive integer. We see at once that r is even, for the square of an odd integer is odd. Let r = 2t with t a positive integer. Then (4) gives (5)

4t2 = 2s2

(6)

s 2 = 2t2

or This is absurd, because s < r. Thus, J2 is irrational. 27. We shall prove that between any two numbers there lies an irrational number. Let a and b be any two numbers with a > b. Let a! and bi be two rational numbers such that b < bi < aa < a. Now bt < bj + 3,1 ~ < bi + (ai - b j * . 72 (Note that>/2 > 1.) Because at - K is rational, is irrational and lies between b and a.

o

~ b l is irrational. (Why?) Then K

s/T

+

&1

Jl

BOUNDS

28. Let E be any s e t , f i n i t e or i n f i n i t e , of numbers. I f there e x i s t s a number G, p o s i t i v e , negative or zero, such that every number of E i s l e s s than G, E w i l l be said to be bounded from, above. I f there e x i s t s a number G, p o s i t i v e , negative or zero, such that every number of E exceeds

G, E w i l l be s a i d t o be bounded

from

below.

A s e t which i s bounded from above and bounded from below w i l l be called bounded. Examples' I f E consists of the numbers x such that - 1 0 s x £ 12, E i s bounded. The s e t E consisting of the two i n f i n i t e sequences 1, 1 . 9 , 1 . 9 9 , 1 . 9 9 9 , . . . -1, -1.9, -1.99, -1.999,... i s bounded. Every f i n i t e s e t of numbers i s bounded. The sequence 5, 4 , 3 , 2, 1, 0, - 1 , - 2 , . . . ,

-n,...

i s bounded from above but not from below. The s e t of a l l positive numbers i s bounded from below but not from above. The t o t a l i t y of integers i s not bounded in e i t h e r d i r e c t i o n . THE LEAST UPPER BOUND

29. Let E be a s e t of numbers which i s bounded from above. Every number G which has the property that for every number a of E we have a gG w i l l be called an upper bound of E. We shall prove the existence of a number M such that (1) No number of E exceeds M. (2) Given any number Mi l e s s than M, E contains a number which exceeds M t . Clearly, such a number M i s an upper bound for E and i s l e s s than every other upper bound. We s h a l l c a l l M, when i t s existence i s proved, the least upper bound or the upper bound of E. Existence proof: F i r s t , suppose that E contains non-negative numbers and l e t E t be the set of non-negative numbers which are contained in E. Let M be the l e a s t upper bound of Ej whose existence was proved in §12. Then M has the properties (1) and (2) above. Now, suppose that E contains only negative numbers. Let - a , where a > 0, be some number of E. Let a be added to every number of E and l e t the resulting s e t of numbers be denoted by E . Let M be the l e a s t upper bound of E , shown above to e x i s t . Then M = M - a has the propert i e s (1) and (2) r e l a t i v e to E. Q.E.D. THE GREATEST LOWER BOUND

3 0 . Let E be a set of numbers which i s bounded from below. Every number G which has the property that for every number a of E a ^G w i l l be called a lower bound of E. We s h a l l prove the existence of a number m such that:

9

(1) No number of E i s l e s s than m. (2) Given any number nu g r e a t e r than m, E contains a number which i s l e s s than m t . The number m, which i s evidently a lower bound f o r E, and g r e a t e r than any other such bound, w i l l be called the greatest lower bound of E or the lower bound of E. Existence proof: Let E1 c o n s i s t of the negatives of the numbers of E. Then E1 i s bounded from above. Let M be the l e a s t upper bound of E . Then m = -M has the p r o p e r t i e s (1) and (2) above. Q.E.D.

II Theory of Limits ABSOLUTE VALUE

1. % the absolute value of a number a, we shall mean a i f a I 0 and - a i f a < 0. The absolute value of a will be denoted by !a|. l-2l

Example:

|a + b l

Theorem:

13 + 21

Examples:

l a - bl

i

I a - bI I

Proof:

+

I s/21

.

tbl.

= i 31

+ l2l ;

*

(-2)1

< 01

+

1-3 + (-3)1

= * —3 » + I - 3 I.

la I + I b I.

Clear.

Proof: Theorem:

lal

|0| = 0 ,

Clear.

Proof:

Theorem:

i

= 2,

lal -

lb I .

a = b + (a - b ) ;

Theorem:

lab I = l a l

Theorem:

Let W

lal i

Ibl

+ la - b l ;

l a - bl

§ lal

-

Ibl.

Ibl.

0. Then |ft| = M . b I b| CONVERGENT SEQUENCES OF NUMBERS

2. Consider an i n f i n i t e sequence of numbers (1)

a 2 , . . . , an

The sequence will be called convergent i f there e x i s t s a number X such that, for every e > 0, a positive integer N e x i s t s such that, for every n > N, |X - a^l < e. I t i s to be emphasized that N w i l l as a rule have to be taken different for different numbers e. The important thing i s that when some e i s assigned, an N can be found for that e. When a number A e x i s t s as above, i t i s called the limit of the sequence ( 1 ) . The sequence i s said to have or to approach or to converge to the limit X. ..., 1-^-,...

Examples:

(Limit 1)

2, 2 , 2 , . . . , 2 , . . .

(Limit 2)

The sequences 1* 2 , 3* • • 1 - 1 1 - 1 X

;

J.,

n,... I

it

do not converge. A sequence which i s not convergent i s said to be

11

divergent.

UNIQUENESS OF

LIMIT

3. A sequence (1)

a,i, a 2 ,..., ELjj,...

cannot have two distinct limits. Suppose, in fact, that X and X' / X are both limits of (1). For n large, |X - a n l < | |X' - X|, IX' - a j < I IX' - Xl. Then lx' - Xl = I (X' - a n ) - (X - a n )I g IXf - a j + |X - a n l < 1 IX' - Xl

2

+

I |'X' - X| = IX' - Xl

2

which is absurd. THE FUNDAMENTAL CONVERGENCE THEOREM (CAUCHY) 4.

Theorem:

For

the

sequence

(1) to converge, such that,

a t , a 2 ,..., a n , it

for

is necessary and sufficient every n > N and for every

that positive

'an+p

for every e > 0 a positive integer p,

integer

N

exist

- a n l .< e.

Proof:

(a) Necessity. Let the sequence have a limit X. Let any e > 0 be given. Let N be such that for n > N, |X - a j < |. Then ^ a n+p -

a

J

= I ^ ~ a n) -

- a n + p)l

g |X - a n | + |X - a n + p | < I + | = e. (b) Sufficiency. We shall say that a number G is greater than every a n of sufficiently large n if an N exists such that G > a n for n > N. We shall prove the existence of such numbers G for the sequence (1). This we accomplish by producing a number G which exceeds every a n . Let N be such that |a n + p n > N, so that, for n > N,

- a n | < 1 for n > N and p arbitrary. Then |an - a N + 1 | < 1 for

lanl = I(an - a N + 1 ) + a N + 1 l i |an - a N + 1 l + |a N+1 l < 1 + la N + 1 l. Hence, if G is greater than the greatest of the numbers a

i, a 2 ,..., a N , 1 +

then G exceeds every a n . It is possible for a number G to exceed every a n of sufficiently large n without exceeding every a n . In an entirely similar manner, we can find numbers G which are less than every a n . (It suffices to take G less than the least of the numbers ai, a 2 ,..., a^, - (1 + la N+1 |), with N as above.) Hence there exist numbers which do not have the property of exceeding every a n of sufficiently large n. Let E be the class of numbers which have the property of exceeding every a n of large n. A number which does not have this property is less than every number in E. Since there are numbers without the property, E is bounded from below.

12

Let X be the greatest lower bound of E. We shall prove that A is a limit

for the sequence

(1).

Consider any positive number e. Because X +

> X, there are numbers in E less than X + Hence X + i s i t s e l f in E, that 2 2 * 2 i s , at most a f i n i t e number of the a n are greater than or equal to X + Let N be such that for n > N , a n T , < X + -|. 2 Let M be such that n > M, la n + p - a n |


Because X - 4 i s not in E, an i n f i n i t e number of the a^ exceed or equal X - 1. 4 4 Hence an i n f i n i t e number of the a n exceed X - ^. Let P, greater than N and M, be such that aP > X Then a p l i e s between X - !• and X + p 2 2

so that la p - Xl < f . 2

For every n, Ia n — X| = I (a n - a p ) '+ (a p - X) I £ |an - a p | + |ap - X I < |an - a p | + e Now P > M. Hence, i f n > P la._ - aD| < § 2 Thus, for n > P, an

"


greater than b w i l l a and are less than or closed), we mean

a be two numbers. The set of numbers which are neither less than a nor be called the closed Interval (a, b). The set of a l l numbers which exceed b w i l l be called the open interval (a, b ) . By the length of (a, b), (open the number b - a. NEIGHBORHOODS

2. In what follows, the word point w i l l be used as an equivalent for real number. Consider a point a. Let e be any positive number. The set of points in the open interval (a - e, a + E) is called a neighborhood of a. Thus, every point has an i n f i n i t e number of neighborhoods, one f o r each e. LIMIT

POINTS

3- Let E be any set of points. A point a, not necessarily belonging to E, is called a limit point of E i f every neighborhood of a contains a point of E distinct from a. I f a is a limit point of E, every neighborhood of a w i l l contain an i n f i n i t e number of points of E. Examples: Every point is a limit point of the set of a l l real numbers or of the set of a l l rational numbers or of the set of a l l irrational numbers. A f i n i t e point set has no limit point. The point 0 is a limit point for the set 1

1 1 2' 4'

1 8 ' " "

THE THEOREM ON NESTED INTERVALS

4. A closed interval (a, b) w i l l be said to be contained in a given closed interval (c, d) i f every point of (a, b) is a point of (c, d ) . The points a and c may coincide, and so may b and d. By a point

contained

in the closed

interval

( a , b ) , we mean any p o i n t of the c l o s e d

in-

terval. Thus, even a and b are contained in (a, b ) . Theorem:

Let

(ai, bi),

(a 2 , b j ) , . . . ,

be an infinite sequence of closed intervals, preceding one, whose lengths approach zero. tained in every interval of the sequence.

Proof:

(a n , b n ) , . . .

each one after the first There exists one and only

beini contained one point which

in is

the con-

Consider the sequence of points

(1) that is, the l e f t ends of the intervals. This is a convergent sequence, since a n and a l l succeeding points of (1) l i e in (a n , b n ) . (Canchy's convergence theorem.) Let P be the l i m i t of the sequence (1). Then P is contained in every interval (a n , b ) . For, l e t us imagine that P i s not contained in ( a j , b j ) . Then a j and the points which follow i t in (1) cannot form a sequence converging to P. This proves that P l i e s in every interval.

16

Furthermore, if Q is a point distinct from P, then, if the length of (an, b ) is less than |Q - P|, as it will be when n is large, Q cannot lie in (an, b n ). Hence P is the only point contained in all of the intervals. THE BOIZANO-VEIERSTRASS

THEOREM

5 . In harmony with what has been said concerning the meaning of the word point, set of points will mean a bounded set of numbers. Theorem:

Every

bounded

infinite

set

of

points

has at

least

one

limit

a bounded

point.

Proof: Let E be an infinite set of points contained in the closed interval (a, b). (Note that every bounded set is contained in some closed interval.) We represent (a, b) by I t . Let c be the midpoint of (a, b). Then either the closed interval (a, c) contains an infinite number of points of E or the closed interval (c, b) does. Pick out that one of the two intervals which contains an infinite number of points of E, or either, if both contain an infinite number of points of E. Let the interval selected be represented by I 2 . The length of I 2 is half that of I t . Similarly, there is an interval I 3 whose length is half that of I 5 which contains an infinite number of points of E. We form in this way an infinite sequence of intervals Ii> I2J•••> I n >••• each contained in, and each half the length of, its predecessor. Each I n contains an infinite number of points of E. By §4, there is a point P which is contained in each of the intervals I n . Evidently, P is a limit point of E. Remarks: In the preceding proof, it may occur an infinite number of times that both halfintervals contain an infinite number of points of E. Unless some definite procedure is given for choosing one of the two intervals which are available, one will experience a sense of dissatisfaction as regards the actual existence of a set Ij, I 2 , This vagueness can be removed, for instance, by choosing the right half-interval when each half is available. In some proofs involving an infinite number of selections, one is unable to furnish a definite procedure for the selections. In such cases one tolerates the resulting vagueness. This question is discussed in the theory of assemblages, where an assumption known as "Zermelo's axiom" is invoked to authorize the selections. Q.E.D. CLOSED SETS

6. A point set will be called closed if either (a)

it

has

limit

points

and contains

all

of

them

or (b)

it

Examples

has no of

limit

type

points. (a):

(1) A closed interval. (2) The

(3)

SPt

The set

0 1 u,

1 ^

I

^

i 8' *'

1 1 1

2' 4' 8'

1 0 1

4'

1

8'

J. 16""

17

Examples

of sets

which are

not

closed:

(1) All rational numbers. (2) An open interval. (3) The point set

J,

... BOREL'S THEOREM

7. Let (a, b) be an interval, open or closed. A point c will be said to be interior (a, b) if a < c < b.

to

Borel's Theorem: Let E be a closed and bounded point set. Let 2 be an infinite set of intervals (open or closed) such that each point of E is interior to at least one of the Intervals of 2. Then there exists a finite subset of the intervals of 2 such that each point of E is interior to at least one interval of the finite subset. Proof: We suppose that we have an E and a 2 described as in the hypothesis, and that the conclusion does not hold for this E and 2. We shall produce a contradiction. Let (a, b) be a closed interval containing E. (Note that E is bounded.) We denote (a, b) by It. Let c be the midpoint of (a, b). Consider the closed intervals (a, c) and (c, b). There must be one of these intervals such that the points of E lying in it cannot be covered by a finite number of intervals chosen from 2. (When we say that a set of intervals "covers" a set of points, we mean that each point of the point set is interior to at least one of the given intervals.) Let I 2 be that one of the two intervals (a, c) and b, c) whose points of E cannot be so covered, or either, if the covering is impossible for both intervals. We find similarly a closed interval I 3 , which is half of I 2 , such that the points of E lying in I a cannot be covered by a finite number of the intervals of 2. We build in this way an infinite sequence Ii> Is>•••> I n > • • • By §4, there is a point P which is contained in all of the intervals I R . Of course, P is a limit point of E, for there is an interval I in every neighborhood of P and every I contains an infinite number of points of. E. Since E is closed, P belongs to E. There is an interval in 2 which contains P in its interior. This interval contains every I n of sufficiently large n in its interior, since the length of I n approaches zero as n increases. We have thus a contradiction of the inference that the points of E lying in I cannot be covered by a finite number of intervals taken from 2. hence a contradiction of the assumption that E cannot be so covered. Q.E.D. COUNT ABILITY 8. An infinite set is said to be countable if it possible to establish a one-to-one correspondence between its elements and the positive integers; that is, if the elements of the set can be arranged in a sequence. A set which is not countable is called uncountable. Theorem: Proof:

The positive

rational

numbers are

countable.

We arrange the rational numbers E with p > 0, q > 0 (p and q integers), according to

the size of p + q, taking care not to repeat any rational number. We secure thus a sequence: i. n .

i

s-1

n

1' 2' 1' 3' 1' 4' 3'

i 2'

1""

It is clear that all positive rational numbers will be included in this sequence.

18

Theorem: The real

numbers are

uncountable.

It will suffice to show that the numbers in the closed interval (0, 1) are uncountable, for an infinite subset of a countable set is evidently countable. We shall prove that, given any infinite sequence of numbers (1)

ai, a4,..., a n ,...

in (0,1), there is a number in (0,1) which is not in the sequence (1). Let a n have the decimal representation 0 • (3 in P2n Pan • • • Let at be any digit which is distinct from (3^ and from 9. Let be any digit distinct from 022 and from 9. In general, let c^ be a digit distinct from (3^ and from 9- Then the number 0 . oil a2... ct^ which lies in (0,1) is distinct from every a n of (1).

19

IV Functions and Continuity VARIABLE

1. We consider any set of numbers E. We introduce a symbol x, which we use to represent a point of E when such a point is called to our attention. The symbol x will be called a variable. The set is called the domain of x. For domains, we shall, as a rule, use intervals. FUNCTION

2. Suppose that with every point x of E there is associated a number whose value is known as soon as x is known. We introduce a symbol y and use it to represent the number associated with x. We call the symbol y a function of the symbol x. Sometimes, we write y = f(x). The function y is said to be defined on the domain E. CONTINUITY

3. let y = f(x) be defined at every point of an open interval (a, b). Let x 0 be an interior point of (a, b). We shall say that f(x) is continuous at x 0 if, for every e > 0, a 6 > 0 exists such that, if |h| < 6, we have If(x0 + h) - f(x0)| < e. Roughly speaking, when x is changed slightly from x 0 , f(x) changes slightly. Suppose that f(x) is defined at a as well as in the open interval. We shall say that f(x) is continuous at a if, for every e > 0, a 6 > 0 exists such that, if 0 < h < 6, |f(a+h) - f(a)| < e. A similar definition is made for the point b. If f(x) is not continuous at x 0 , f(x) will be said to be discontinuous Examples

of continuous

Examples

of discontinuous

same is true of y = x.

functions:

at x 0 .

If f(x) is a constant, it is continuous for every x. The

functions:

(a) Let y = 0 when x is rational and y = 1 when x is irrational. Then y is discontinuous for every value of x. (b) Let y = x when x ^ 0. Let y = 1 for x = 0. Then y is continuous for every x 4 0 and is discontinuous at x = 0. DENSE SETS

4. A set of points is said to be dense in an interval contains a point of the set. A set is called every

where

(a, b) if every subinterval of (a, b) dense

more briefly, every where dense, or, still more briefly, dense, arbitrarily small, contains a point of the set.

amoni

the

real

numbers,

or,

if every interval (a, b),

A FUNCTION CONTINUOUS AT A DENSE SET OF POINTS AND DISCONTINUOUS AT A SIMILAR SET

5. If x is positive and rational and equal to E, with p an integer, q a positive integer and £ in its lowest terms, we let y = -. q 4

^

20

If x is positive and irrational, we let y > 0. Clearly y is discontinuous for every rational value of x. Let x be irrational. Let e > 0 be given. Let n be an integer which exceeds Then ! < e. Plot all rational numbers whose denominators are 1 or 2 or 3 or ... or n. There is an interval with x as midpoint which contains none of these rational numbers. Let 6 be half the length of such an interval. Suppose that Ihl < 6, then, if x + h is rational, say

we have q > n, so that q I f (x + h) I = 1 < 1 < e. n q

Hence If(x + h) - f(x) I = If(x + h) - 01 = If(x + h)| < e. If x + h is irrational, f(x + h) - f(x) = 0 - 0 = 0 .

Hence, for |h| < 6, |f(x + h) - f(x)| < e. ft.E.D.

CONTINUITY OF SDH AND DIFFERENCE OF TWO FUNCTIONS 6. Theorem: If f(x) and f(x) - g(x) are continuous

g(x) are at x 0 .

both

continuous

at

a point

x 0 , then

f(x) + g(x)

and

Proof: We limit outselves to f(x) + g(x). Let e > 0 be assigned. Let 6* > 0 be such that, for |h| < 6 l t |f(xo + h) - f(x0) I < |, and let 6, > 0 be such that, for |h | < 6 a , |g(x0 + h) g(x0) I
0 be assigned. We assume, as we may, that e < 1.

Let G be a number which exceeds |f(x0) |. Ig(x0)l and unity. Let 6 > 0 be such that, for Ih I < 6, If (x0 + h) - f (x0) I < lg(x0 + h) - g(x0) I < We have, for Ihl < 6, • If (x0 + h) g(x 0 + h) - f(x0) g(x0) I = lf(x0) [g(x0 + h) - g(x0)l + g(x0) [f(x0 + h) - f(x0)l + [f(xo + h) - f(x0)l [g(xo 4 h) - g(x0)l I 3G

3G

SG ' 3G

< £ + i + i = e.

3

Theorem:

Any

polynomial

3

a0xn + a^11-1

3

Q.E.D.

+ ... + a n

is continuous

This follows from the theorems on sums and products. 21

for

every

x.

CONTINUITY OF QUOTIENT 8. Theorem: Let f(x) and g(x) be continuous at x c . Let g(x 0 ) 4 0. Then fined on an interval containing x 0 , is continuous at x 0 •

which

is

de-

Proof left as exercise. See corresponding proof for limits. BOUNDED FUNCTIONS

9. Let a function f(x) be defined on a domain E. Let E' be the set of values which f(x) assumes on E, that is, the set of values of f(x) for the various values of x in E. If E1 is bounded from above, we call f(x) bounded from above on E. We call the least upper bound of E' the least upper bound of f(x) on E. We designate the least upper bound of f(x) on E by M. Examples:

(a) Let E be the closed interval (0, 1). Let f(x) = x. Then M - 1. (b) Let E be the set of all positive numbers. Let f(x) = x above.

Then f(x) is not bounded from

10. If E' is bounded from below, we call f(x) bounded from be lew on E. We call the greatest lower bound of E' the greatest lower bound of f(x) on E. We designate the greatest lower bound of f(x) by m. 11. If E' is bounded, we call f(x) bounded on E. Thus, f(x) is bounded on E if and only if a G > 0 exists such that If(x) | < G for every x in E. Examples

of bounded

functions:

(a) Let E be the set of all integers. Let f(0) = 0; f(n) = 1 - i for n > 0; f(n) = -1 - I n n for n < 0. Then M = 1,

m = -1.

(b) Let E be the set 0 £ x g 1. Let f(x) = x. Then M = 1, m = 0. {c) Let E be the set of all numbers. Let f(x) = 1. Then M = m = 1 . BOUNDEDNESS OF A FUNCTION CONTINUOUS ON A CLOSED INTERVAL 12. Theorem: Let f(x) be defined, bounded on (a, b).

and continuous,

on a closed

Interval

(a, b). Then f(x) is

Note: when we say that f(x) is continuous on (a, b), we mean that f(x) is continuous at every point of (a, b). Proof: Suppose that f(x) is not bounded on (a, b). Let Xj be a point of (a, b) such that |f(Xi) | > 1. Let x 2 be a point of (a, b) for which |f(x2) | exceeds |f(xt) | and 2. Let x 3 be a point of (a, b) such that If(xa) I is greater than both |f(x2) | and 3• Continuing, we form an infinite sequence of distinct points, (1)

x 4 , x«,..., xn..., with |f(xR) | > n.

By the Bolzano-Weierstrass theorem the points of the sequence (1) have at least one limit point. Let x 0 be such a limit point. Clearly, x 0 lies in (a, b). Now f(x) is continuous at x 0 . Thus, we can find a 6 > 0 such that, if x' lies in (a, b) and |x1 - x 0 | < 6, we have If(x') - f(x0)| < 1. Thus, for Ix' - x 0 | < 6, we have If(x')I = |[f(x') - f(x0)] + f(x0)| < 1 + |f(x0)|. 22

There are an infinite number of points x n in the sequence (1) for which |xn - x0l < 6. If we take a value of n greater than 1 + |f(x0)l for which |xn - x 0 | < 6, we have the contradiction that |f(xn)| > n > 1 + |f(x0) | Remarks: The preceding proof employs an infinite number of selections. One can give proofs which do not have this feature. One can, for instance, use the process of bisection used in proving the theorem on nested intervals. Q.E.D. ATTAINMENT OF BOUNDS BY A FUNCTION CONTINUOUS ON A CLOSED INTERVAL 13- Theorem: Let f(x) be continuous on the closed interval (a, b). Let M and m be respectively the least upper and greatest lower bounds of f(x) on (a, b). Then there is a value of x in (a, b) at which f(x) equals M and a value of x in (a, b) at which f(x) = m.

Proof: We discuss M. The proof for m is similar. Let us suppose that there is no value of x for which f(x) = M. % the nature of M, there is a point x 1 on (a, b) such that f(xj > M - 1. Naturally f(xj < M. Thus, there is a point x 2 on (a, b) such that f(x2) exceeds both f(xi) and M - 5. Continuing, we form an infinite sequence of distinct points (1)

Xi, x2,..., xn ,... with f(x_) n > M - i. n

Let x 0 be a limit point of the sequence (1). Of course, f(x0) < M. We know that f(x) is continuous at x 0 . Let 6 >-0 be such that, for Ix1 - x 0 I < 6, f(x') - f(x0) < M ~

f(x

2

°) .

Then, for Ix1 - x0l < 6, (2)

f(x') < f(x0) + M ~

= M - M - f < x °) .

%

2

There are an infinite number of points x n of the sequence (1) for which |xn - x0l < 6. Let n be such that n >

M - f(x0)

and such that |xn - x0l < 6 . Then f (xj > M - 1 > M - M - f < x °) . n n 2 This contradicts (2) above.

Q.E.D.

ATTAINMENT OF ALL VALUES INTERMEDIATE BETWEEN TVO VALUES 14. Theorem: Let f(x) be continuous in the closed interval (a, b). Let f(a) « A, f(b) = B. Suppose that A = B. Let C be any number between A and B. Then there is a point c in (a, b) such that f(c) = C.

Proof:

To fix our ideas, let B > A.

For x close to b, we have f(x) > C since C < B. Let E be the class of all numbers C in (a, b) such that, for C ix ib one has f(x) > C. Then E is bounded from below. Let c be the greatest lower bound of E. We say that f(c) = C.

23

Suppose first that f(c) < C. Then, for any x slightly greater than c, we have f(x) < C. It is easy to see, however, that any such x is in E, so that, for it, f(x) > C. Hence, we cannot have f(c) < C. Suppose now that f(c) > C. Because f(a) = A < C, we have c > a. There is a 6 > 0 such that, in the closed interval (c - 6, c +6), one has f(x) > C. This means that c - 6 is in E, which is absurd. Q.E.D. UNIFORM

CONTINUITY

15. Let f(x) be continuous in the closed interval (a, b). We shall show that, given any e > 0, a 6 > 0 exists such that, for any x at all in (a, b) and for |h| < 6, we have |f(x + h) - f(x)| < e. We describe this situation by saying that f(x) is uniformly continuous in (a, b). The concept of uniform continuity is useful in integration and in other problems . Proof: Let e > 0 be given. For every x in (a, b), there is a 6 X such that |f(x + h) f (x) I < | for |h| < 6 X . Put about each point x an interval with x for midpoint, of length 6 X . By Borel's Theorem (page 18), there exists a finite subset of the above intervals such that every point of (a, b) is interior to at least one interval of the finite subset. Let 6 be any positive number less than half the length of the smallest interval of the finite subset. We say that 6 answers to the assigned e.

1

^ 1

I

Let x be any point in (a, b). Consider one of the intervals of the finite set which contains x in its interior. (This interval is shown in the diagram.) Let C be the midpoint of this interval. Then

1

lx - C\ < | Let h be such that |h| < 6. Then I(x + h) -

= |(x - 0

+ hi * |

+ 6 x

u

where

and x 2 are in (a, b), we have f(x2) i f(xj,

we say "f(x) is non-decreasing in (a, b)." 24

If, whenever x 2 > x l t we have f(x 2 ) i f(x t ), we call f(x) non-increasing in (a, b). A function which is non-decreasing or non-increasing in an interval is said to be monotonic in the interval. If f(x2) > f(Xi) when x 2 > x If f(x2) < f ( x j when x 2 >

n

we call f(x)

increasing.

we call f(x)

decreasing.

Theorem: Let f(x) be continuous in the closed interval (a, b). Suppose that f(x) assumes value twice in (a, b). Then f(x) is either increasing or decreasing in (a, b).

no

Remark: The assumption that (a, b) is closed is not essential and is made only for convenience . Proof:

Suppose that f(b) > f(a). We shall prove that f(x) is increasing.

Let x 2 > x t . We cannot have f(x2) = f(xi). We shall prove that f(x2) > f(xj). Suppose that f(x2) < fix,). First, let, us imagine that f(x2) < f(a). Then f(x2) < f(b). Between x 2 and b, f(x) assumes all values between f(x2) and f(b). In parf(a) f(b) ticular, it assumes the value f(a). This contradicts the , i___J hypothesis. a *| *2 b Suppose now that f(x2) > f(a). Between a and x t , f(x) assumes all values between f(a) and f(Xi), in particular, the value f(x 2 ). This contradicts the hypothesis. Q.E.D.

25

V The Derivative DERIVATIVE

1. Let y = f(x) be a function whose domain of definition comprises intervals. Let x 0 be a point throughout a neighborhood of which f(x) is defined. Consider tha expression f(x 0 + h) - f(x0) h which, for every sufficiently small h distinct from zero, is a definite number. We call this expression a difference

quotient.

The difference quotient is a function of h. It may be that as h tends toward zero, the difference quotient tends toward a limit. That is, is, a number A may exist such that, for every e > 0, we can find a 6 > 0 such that, for 0 < |h| < 6, |f(Xp + h) - f(x0) _ X | h


0 be given and let 6 > 0 be taken so that |f(xn) + h) - f(x0). _ X | h

< e

for 0 < |h| < 6. Then )f(x0 + h) - f(x0) | < h and

|f(x0

+

U |

+ g

h) - f(x 0 )| < |h| (|\| + e).

This last inequality proves the continuity of f(x) at x 0 . The converse of the above theorem is not true. A function may be continuous at a point without having a derivative there. TFE DERIVATIVE AS A FUNCTION

3. Let f(x) have a derivative at every point of an open interval (a, b). Then the derivative is a function of x defined on the interval. We designate the derivative by f'(x) or by The methods of the calculus for determining ^ for the simple functions are entirely rigordx ous. We shall use freely the formulas: 26

xn = nxn_1 A . (u+v) = du dx dx

(n an integer). +

dv dx

etc. The derivative of a constant is easily seen to be zero. DERIVATIVES

AT EXTREMITIES

OF

INTERVALS

4. Let f(x) be defined on the closed interval (a, b). We consider f(a + h) - f(a) h for positive values of h. If a number X exists such that, for every e > 0 a 6 > 0 can be found such that |f(a + h ) ~ f ( a ) - X| < e h for 0 < h < 6, we call X the derivative of f(x) at a. A similar definition is made for b. RIGHT-HAND AND LEFT-HAND

DERIVATIVES

5. Let f(x) be defined for a neighborhood of x 0 . It may be that f(x) has no derivative at x 0 but that if h is allowed to approach zero through only positive values, the difference quotient approaches a limit. In that case, we say that the function has a right-hand derivative

at x 0 - S i m i l a r l y , we d e f i n e

left-hand

derivative.

Example: y = x for x 2 0 and y = -x for x < 0. This function does not have a derivative for x = 0 but does have a right-hand and a left-hand derivative for x = 0. MAXIMA AND MINIMA

6. Let f(x) be defined on a closed interval (a, b). If the point x 0 in (a, b) is such that for some 6 > 0, f(x0) £ f.(x) if |x - x 0 | < 6, where it is understood that x is in (a, b), we say that f(x) has a maximum at x 0 . Examples:

(a) y = 1 - x 2 ; x „ = 0.

(b) y = x for interval (0,1); x 0 = 1. (c) y = 1; x 0 arbitrary. The value of f(x) at a point for which f(x) has a maximum need not be the greatest value of f(x) for the domain of definition. Similarly, if a 6 > 0 exists such that f(x 0 ) s f(x) if |x - x 0 | < 6, we say that f(x) has a

minimum

at x 0 . NECESSARY CONDITION FOR A MAXIMUM OR A MINIMUM AT AN INTERIOR

7 . Theorem: that, at some

POINT OF AN INTERVAL

Let f(x) have a derivative at every point of a closed interval point x 0 interior to (a, b ) , f(x) has a maximum or a minimum.

(a, b). Suppose Then f ' ( x 0 ) = 0.

Proof: To fix our ideas, let us suppose that f(x) has a maximum at x 0 . We have f'(x0) > 0 or f (x0) = 0 or f'(x0) < 0. Suppose that f'(x 0 ) > 0. Then, when h is very small, f(x 0 + h) - f(x0) being very close to f'(x 0 ), must be positive. Hence, if h is smaLl and positive, f(x 0 + h) f(x^) is positive. Thus, for h small and positive f(x0 + h) > f(x0) so that f(x) has no maximum at x 0 . 9f7

Similarly, we cannot have f'(x 0 ) < 0. Hence f'(x 0 ) = 0. THE MEAN VALUE THEOREM 8. Theorem: Let f(x) be defined, (a, b). Then a point x 1 ( interior

and possess a derivative, throughout to (a, b), exists such that

the closed

interval

f(b) - f(a) = f'(x t )(b-a). Remark: It suffices actually for f(x) to be continuous on the closed interval (a, b) and to have a derivative at all interior points. Proof:

Consider the function

1

n

=

-

We are going to introduce the notion of the approach to a limit of the sum (2) as n increases indefinitely and the lengths of the intervals (x i _ 1 , x^) tend towards zero. We shall say that the sum (2) tends toward a limit as its norm tends towards ber I exists such that, for every e > 0, a 6 > 0 exists such that, for x

we have, for any C's at all

i ~ x i-i


in (x i _ 1 , Xj_)], II -

§ ftCjiXi - x ^ x H i=l

< e,

Obviously, at most one such number I can exist. If such an I exists, we shall say that f(x) is integrable in (a, b). We shall call I the integral of f(x) from a to b. We write I = fl f(x) dx. 2. Theorem: Proof: -Let ¿ x

o I

a

If

f(x) is

integrable

in (a, b), then

f(x) is bounded

in (a,, b).

Suppose that a function f(x) is integrable in (a, b) but not bounded. f(x) dx = I. Xi I

i

x

x, I

npi I

x

i

n An e > 0 being given, let 6 > 0 be taken I so that any sum (2) of norm less than 6 k differs from I by less than e.

Take any subdivision into intervals (x0, Xi), (Xi, x 2 ),..,, (x n _ x , x n ) with every x i - x i _ 1 less than 6.

29

Then there is one (at least) of the intervals (x 1 _ 1 , x¿) in which f(x) i s not bounded. To fix our ideas, suppose that f(x) i s not bounded in (x 0 , x t ) . Let C 2 ,

be chosen, and held fixed. Then f ( £ 2 ) , . . . ; f(Cn) are definite numbers.

Keep Ci arbitrary for the present. Consider the sum (3)

fteiMXi - Xo)

+

f(C 2 )(x 2 -

Xl)

+...+ f ( ^ ) ( x n -

x^).

The terms which follow the f i r s t in (3) are fixed. Since f(x) is not bounded in (x 0 , x j , we can make f ( i i ) large at pleasure by choosing Ci appropriately. Then f ( C i ) ( x i - x 0 ) can be made a r b i t r a r i l y large. Hence the sum (3) can be made arbitrarily large in absolute value by a proper choice of Ci. I f the sum exceeds |I| + e in absolute value, i t will d i f f e r from I by more than e. Q.E.D. CONDITION FOR INTEGRABILITY 3 . Theorem: For f ( x ) to be integrable in ( a , b ) , it is necessary and sufficient that every e > Ü a 6 > 0 exist such that any two sums of norms less than 6 have a difference is less than e in absolute value.

for which

This theorem is analogous to the fundamental theorem on convergent sequences. Proof: A. Necessity - Let I e x i s t . Take 6 such that any sum of norm less than 6 differs from I by less than f-. Then any two such sums d i f f e r by less than e. B. Sufficiency

- Let

Ei, e2 < Ei, £3 < e 2 , . . . ,

< £p»•• •

be an infinite sequence of positive numbers which decrease toward zero. For every p, chose a 6p > 0 such that any two sums of norms less than d i f f e r by less than e p . Furthermore, choose the 6's so that 6 t > ó 2 > . . . > óp > . . . , ana so that the sequence of 6 ' s tends toward zero. For every p, form a sum Zp of norm l e s s than 6 p . Consider the sequence (4)

Zi, Z 2 , . . . ,

Zp,....

We say that this sequence approaches a l i m i t . Given any e > 0, take p so that e p < e. Then two elements of the sequence (4) beyond the pth d i f f e r by less than e, for t h e i r norms are less than 6 p . Hence the sequence (4) approaches a l i m i t . Let I be the limit of the sequence (4). Given any e > 0, l e t 6 > 0 be chosen so that any two sums of norms l e s s than 6 differ by less than 2 We say that any sum Z of norm less than 6 differs from I by l e s s than e. For, l e t p be taken so that II - Zp| < I and 6 p < 6, where Zp i s as in (4) and 6 p i s as above. Then II - Z| = 1(1 - 2 p )

+

This proves the sufficiency of the condition.

30

(Zp - 2)1 < I • I = e.

INTEGRABILITY 4. Theorem: Let f(x) be inteirable f(x) is inteirable in (c,d). Proof:

IN

SUBINTERVALS

in (a, b). Let (c, d) be any sublnterval

of (a, b). Then

It will suffice to consider any subinterval (a, c). Given an e > 0 let 6 > 0 be so i | , chosen that two sums of norms less than 6, constructed for c b the interval (a, b), differ by less than e. a

Consider any two sums

and 2 2 , formed for (a, c), of norms less than 6. We say that

IZJ. - S2I < s.

For, let (c, b) be divided in any way, into intervals of lengths less than 6. Let points C be chosen in these intervals. Using this division of (c, b), and the subintervals and C's which appear in we form a sum Si for (a, b). Similarly, using the division of (c, b) and the subintervals and C s which appear in 2 2 , we form a sum S 2 for (a, b). Now, as |SX - S 2 | < e and as Sj - S 2 = 2i - 2 2 , we have |Zi - 2 a | < e. 5. Theorem:

Let f(x) be inteirable

Q.E.D.

in (a, b). Let c be any point

interior

to (a, b).

Then f(x) dx =

/

a

Proof: a

i

a

f(x) dx + I

c

f(x) dx.

Choose a sequence of sums approaching /a , with norms tending toward zero and with c one of the points of subdivision for each sum. Let the sec b | ^ quence of sums be

Let ^ be that part of I which comes from intervals in (a, c) and comes from intervals in (c, b). Then 2

P

The limit of I' is /a°, that of ,.b ,.c rb Hence, J = Ja + Jc •

that part of 2 p which

= 2' + 2». P

P

is f*.

BOUNDS

FOR AN

INTEGRAL

6. Theorem: Let f(x) be integrable in (a, b). Let M and m be respectively and greatest lower bounds of f(x) in (a, b). Then m(b - a) s ¿ Proof:

the least

upper

f(x) dx á M(b - a).

We limit ourselves to the second inequality. For any sum, we have f(Ci)(x! - x 0 ) + ...+

f(Cn)(xn - x n _ x ) s M(xi - x 0 ) +...+ M(xn - x ^ ) = M(xn - x 0 ) = M(b - a).

Then the integral cannot exceed M(b - a). If it did, any sum of sufficiently small norm would exceed M(b - a). Theorem: G(b - a). Proof:

Let f(x) be inteirable

in (a, b). If |f(x)| g G in (a, b), then

Clear. 31

|jrb f(x) dx| i

CONTINUITY OF INTEGRAL VITS RESPECT TO UPPER LIMIT

7 . Let f(x) be integrable in (a, b). Let x be any point of (a, b ) . Then, i f x > a, i x f ( x ) dx e x i s t s . For a = a, we define £ f(x) dx as zero. The integral fa upper limit x. We designate this function by F ( x ) .

f(x) dx i s a function of the

We s h a l l prove that the function F(x) i s continuous throughout (a, b). Case A. Let h be small and p o s i t i v e . Then F(x + h) = F(x)

+

l X + h f ( x ) dx.

Hence |F(x Let If(x) h

+

h) - F(x) I - |/ x X+h f(x) dx|.

G in (a, b ) . Then |F(x + h) - F(x)I á Gh

and ffli i s small when h i s small. Case B. Let h be small and negative. Then F(x + h) = F(x) - £ + h

f ( x ) dx

and |F(x + h) - F(x)I = l / * + h

8.

Theorem: (a, b) and

Let

f(x)

f(x

)

dx

| ¿ G(-h).

Q.E.D.

INTEGRABILITY OF SUM OF Tk'O FUNCTIONS and g ( x ) be Integrable in ( a , b ) . Then f ( x ) + g ( x )

f*

tf(x)

+

is

integrable

In

g(x)] dx = /ab f(x) dx + 4 b g(x) dx.

Proof: Given an e > 0, l e t 6 > 0 be taken so that any sum formed for f ( x ) , of norm l e s s than 6, d i f f e r s from /a f(x) dx by l e s s than also so that any sum for g(x) of norm l e s s x than 6 d i f f e r s from 4 g ( ) dx by l e s s than |. Consider any sum o f norm l e s s than 6 for f(x) + g ( x ) . Let i t be 2 = [fícj

+

e i C J H x ! - Xo) + . . . +

[f(^)

+

g(^n)](x n - x n _ 1 ) .

Now Z = Zi + Z 2 , where 2! = f ( C j ( X l - xo) + . . . + f i ^ J t X n - x n _ i ) Z2 = e ( C i ) ( x i - xo) + . . . + g ( ^ ) ( x n - x n _ x ) . As I

f ( x ) dx - Z J

< | and |/a g(x) dx - Z2 I
0, l e t 6 > 0 be taken so t h a t , i f |x1 - x| < 6 , with x and x ' in (a, b), xO *i xn-i *n w e have 1 1 1 1 1 1

i

P

C

q

If(x') - f(x)|

b

2(b - a)

We say that any two sums Zi and Z2 of norms l e s s than 6 d i f f e r by l e s s than e. This w i l l prove our theorem. Take a l l of the points of division which figure in and a l l which figure in Z 2 . The t o t a l i t y of these points gives a new division of (a, b ] . Let Z3 be any sum based on t h i s third division. We shall study the difference between Z3 and Zj. Let (p, q) be any interval which i s used in forming Si- Then (p, q) decomposes, when the points of division used for I 2 are added, into intervals (Xo, X i ) , ( x i , X 2 ) , Let C be the point chosen for (p, q) in forming (xi-i> i = 1 , . . . , n.

In forming I 3 , l e t

be chosen in

The terms of Z9 which come out of (p, q) have a sum A = f ( C i ) ( x ! - Xo) + . . . + f(Cn)(x n - x n - 1 ) . As to S i , we get from (p, q) a single item, B = f ( 0 (q - p) = f ( 0 (x t - xo) + . . . + f ( 0 (x n - x n _ 1 ) . Then A - B = ( X l - x 0 ) [ f ( C i ) - f(C)l As

-

(x n - x ^ H f ^ )

-

f(0]..

< 6 for every i , we have

"«i» - « « ' -»nr^T for every i . Hence |A

-

B

'
0 .

OF THE INTEGRAL OF A CONTINUOUS FUNCTION

Inteirable

d x . Then F ( x )

in

( a , b) and continuous

has a derivative

at some point

at x 0 and its

derivative

x 0 of

(a,

at x 0 equals

b). f(x0).

Then

F(x 0 * h) - F(x 0 ) = J l ° 0 * h f ( x ) dx. Let M be the l e a s t upper bound of f(x) in ( x 0 , x 0 +h) and m the greatest lower bound. Because of the continuity of f ( x ) at x 0 , given any e > 0, a 6 > 0 can be found such that M - f ( x 0 ) < e,

f(x0) - m < e

for h < 6.

33

By §6, mh < F(x 0 + h) - F(x 0 ) g Mh so that m g

F(x 0 + h) - F ( x 0 ) < h

M

We see now e a s i l y fron (1) that |F(x 0 + h) - F(x 0 ) _ h

f ( x o )

,

< £

Case B. h < 0 .

Proof:

Similar to above.

Q.E.D. EVALUATION OF DEFINITE INTEGRALS

1 1 . Theorem: different table

Let f ( x ) be continuous on ( a , b) such that

on the closed

dSM dx

=

interval

(a,

b ) . Let

G(x) be any

function

f(x)

for each x on (a, b ) . Then (1)

/ab f ( x ) dx = G(b) - G(a).

Proof: ¥ e use F(x) as in §10. Then F(x) i s d i f f e r e n t i a b l e on (a, b) and f

[F(x) - G(x)] = 0

dx throughout (a, b). Thus a constant c e x i s t s such that F(x) = G(x) + c . As F(a) = 0, we have c = -G(a). Thus F(b) = G(b) - G ( a ) . As F(b) i s the d e f i n i t e integral in ( 1 ) , the theorem i s proved.

Q.E.D.

COMPOSITION OF INTERVALS 1 2 . Theorem: Let a < b < c . Let inteirable in ( a , c ) .

Proof: f

f ( x ) be inteirable

In ( a , b) and In (b, c ) .

Then f(x),

is

Let G > 0 be such that |f(x)| < G in (a, b) and in (b, c ) ; that i s , |f(x)| < G throughout (a, c ) . p q 1 1—• 1 Let e > 0 be assigned. Take 6 so that b

c

(1)

0

0 be assigned. 1 1 Choose a point p, interior to (a, b) so that b - p < 12G Q q p b P Take 6 so that any sum for (a, p) of norm less than 6 differs from /a f(x) dx by less than •§•; and also so that 6 < ^ ^ . 6 12G Take any division of (a, b) into intervals of length less than 6. Let 2 be a sum for this division. Let q be the rightmost point of division which is not to the right of p. Then b - qn < LL

12G

-

66*

We shall prove that that part of I which comes from (a, q) differs from J^ by not more than Suppose that the absolute value of the difference exceeds let the absolute value be 6 o § + 7] with n > 0. 6 p Form a sum for (q, p) with intervals so small that this sum differs from /q by less than n. (If q = p, the sum, by definition, is to be zero.) Then this sum, plus the part of Z which comes from (a, q), gives a sum for (a, p), of norm less than 6, differing from by more than This contradicts the nature of 6. 6 That part of 2 which comes from (q, b) has an absolute value no greater than G • -£- = 6G 6

35

Hence 2 differs from /aQ by no more than value of which is no more than G

Finally, £

differs from /a by f^, the absolute

— = 12G 12

Hence 2 differs from /aP by less than Thus, if /aP

and

by less than

are two sums for (a, b) of norms less than 6, they will each differ from and will differ from each other by less than E. Q.E.D.

36

VII Infinite Series of Numbers INFINITE

SERIES

1. Let Ui, u 2 ,..., u n ,... be any infinite sequence of real numbers. The symbol (1)

u t + u 2 +...+ u n +...

will be called an infinite series of real numbers. We sometimes use the symbol 1

i=l to represent the series (1). CONVERGENT

INFINITE

SERIES

2. Let (2)

Ui + u 4 +...+ u n +.

be any infinite series. Let Si = u t s 8 = Uj + u 2 s a = u t + u, + u. s n = u x + u 2 +. . . + U n . Consider the sequence (3)

Si,

S

2

, , . .

f

S

n

, . . . .

If (3) converges, that is, if it has a limit, s, we shall say that (2) converges and we shall call s the sum of (2). If (3) diverges, that is, if it has no limit, we shall say that (2) diverges. Examples:

The series 1 2

+

I 4

+

+

1 - n+ 2

converges. Its sum is 1. For the sequence 1 1 i_ 1_ 2' 4 ' " " 2n' converges and its limit is 1. The two series 1 + 1 +...+ 1 +...

37

and 1 - 1 + 1 - 1 +...+ (-1)n+1 + ... are divergent.

A NECESSARY

AND SUFFICIENT

CONDITION

FOR

CONVERGENCE

3- Theorem: For ui +...+ un +... to converge, it is -necessary and sufficient that for every e > 0. a positive integer N exist such that, for every n > N and for every positive integer p, |un+1 + ...+ Proof:

u n + p | < e.

The associated sequence Si, S2 j • • • j S^j.t.

is such that S

- s

n+P

n = un+l

u

n +p

The theorem then follows immediately from the fundamental theorem on convergent sequences.

ADDITION

OF CONVERGENT

SERIES

4. Theorem: If (4)

u t + u 2 +...+ u n +...

(5)

Vi + v 2 +...+ V n +...

and

converge,

then (ui + v j + (u2 + v 2 ) +...+ (un + v n ) +...

converges

and its sum is the sum of the sums of (A) and (5).

Proof: Clear. THE nTE TERM 5. It follows from §3 that if uA +...+ u n +... converges then u n approaches zero as n increases indefinitely. The converse is not true. For instance, the harmonic series 1

+

i

+

i+...+ i+...

diverges, even though its nth term approaches zero.

THE

REMAINDER

6. Let (6)

ut + u 2 +...+ u n +...

converge to a sum s. Clearly, for every n, the series u

n+l

+

Un+2

converges, and its sum is s - (ut +...+ u n ) = s - s n .

38

We shall call the sum of the series (7), that is, s - s n , the remainder the series (6).

after

n terns

in

Clearly, the remainder after n terms, that is, s - s n , approaches 0 as n increases indefinitely. We have s = s n + Rn where R n is the remainder after n terms.

A CLASS OF CONVERGENT

SERIES

7. Let a t , a2,..., an,... be an infinite sequence of positive numbers which is such that

(a) at > a 4 > a a >...> a n > ... (b) a n approaches zero as n increases indefinitely.

We say that the series ai - a 2 + a s - a 4 +...+ (-l) n+1 a n +...

(8) converges +

Example: Proof:

conver

Ses-

We study the sequence s

(9)

l> Si,..., sn,...

Take the subsequence of (9) (10) We show first that (10) converges, then that (9) converges to the limit to which (10) does. In accordance with the fundamental theorem on sequences, we have to consider the difference s 2(n+p)+l

" s2n+l'

Now s

2n+l = a t - a 2 +...+ a 2 n + 1

s

2(n+p)+l + at - a a +...+ a 2 n + 1 - a 2 n + 2 +...+ a 2 ( n + p ) + 1 .

Hence s 2(n+p

)+l " s 2n+l ' 0 be assigned. If N is so large that, for 2n + 1 > N we have a 2 n + 1 < | and Is ~ s2n+i' < | then, for 2n > N, we have Is - s 2 n | < e. That is, every sun s L with i > N differs from s by less than e. Q.E.D.

SERIES OF ABSOLUTE

VALUES

8. Theorem.: Let (12)

Ui + u 2 +.•.+ u n +...

be an Infinite series. If (13)

converies,

luil + lu2l +...+ Iun| +...

then (12) also

converies.

Proof: We have |u n+1 + ...+

u n + p | < |un+1l +...+ |un+p|. Q.E.D.

1

Remark: The converse is not true. For instance, 1 - 1 + - - 1 + . . . ' 2 3 4 + 1 + 1 + 1 + 2 3 4

ABSOLUTE 9. Definition:

converges, but not

CONVERGENCE

Let

(14)

Ui +...+ u n +...

be a convergent infinite series. If (15)

I +•••+ l«nl +.••

converges, we shall call (14) absolutely

semi-converient.

converient.

If (15) diveries,

we shall call (14)

(If (14) is absolutely convergent, the sum of (14) cannot exceed the sum of (15). For the difference between (15) and (14) is [|«il - u j +...+ [|un| - u n ] +... whose terms are non-negative.)

REARRANGEMENT

OF TERMS OF AN ABSOLUTELY

CONVERGENT

SERIES

10. Let (16)

Uj +...+ u n +...

be absolutely convergent. Let (17)

vt +...+ v n +...

be the series (16) with its terms rearranged in any order. That is, the series (IV) is a series

40

U.1 1 + U: i2

+...+U,

n

+...

where i t , i 2 ,... constitute a rearrangement of the positive integers 1, 2 We say that (17) Proof:

converges

absolutely,

and that

its

sum is the sum of

(16).

Consider the convergent series (18)

l u j +...+ |u n | +...

Let R n be the remainder after n terms in (18). Let e > 0 be assigned and let lRn| < e for n 2 N. Let M be so large that the sum Vi + v 2 +...+ v M contains all of the terms u 1 ; u 2 ,..., u N . Let o (17).

represent the sum of the first n terms of

For n > M, the difference between a and the sum of the first N terms of (16) is a sum of terms of (16) which follows the Nth. The sum of any finite number of terms of (16) which follow the Nth does not exceed the remainder after N terms in (18). Hence the difference between a n and the sum of the first N terms in (16) is less than e in absolute value. But the sum of the first N terms of (16) differs from the sum of (16) by no more than the remainder after N terms in (18). Hence, a n with n > M differs from the sum of (16) by less than 2e. This proves that (IV) converges to the sum of (16). The series |vt I + ...+ |vn| +... is a rearrangement of (18). Hence (17) is absolutely convergent. 11. If

Q.E.D.

ii + u 2 +...+ u n +...

is only semi-convergent, we can, by properly rearranging its terms, get a series which converges to any desired limit.

41

VIII Sequences of Functions CONVERGENT SEQUENCES OF FUNCTIONS 1. Let (1)

Si(x), S2(x),..., S n (x),...

be a sequence of functions which are defined on a domain E. Suppose that (1) converges for every value of x in E. For every x of E, let the limit of (1) be represented by s(x). We have, in s{x), a function defined on E. We shall call s(x) the limit of the sequence (1). UNIFORM CONVERGENCE 2. We shall say that the sequence (1) converges uniformly on E if, for every e > 0, we can determine an N such that, for n > N, |s(x) - s n (x)I < e for every

x of E.

Very plainly, if E consists of a finite number of points, convergence on E implies uniform convergence on E. EXAMPLE OF A UNIFORMLY CONVERGENT SEQUENCE DEFINED ON AN INTERVAL 3. Let E be the closed interval (0, 1). Let the sequence be x

2

~ X, X

^

X ^ X ® X — —, X — •''f x - —,... .

The limit, s(x), is x 2 . Now, for x in (0, 1), |s(x) - s n (x)| - X $ 1. n n If e > 0 is assigned, take N > 1. For n > N,

for every x in (.0, 1).

|s(x) - s n (x)| £ 1 < 1 < e, n N EXAMPLE OF NON-UNIFORM CONVERGENCE

4. Let E be the closed interval (0, 1). Define s n (x) as follows: For For

0

*

i


Hence the sequence x S i ( x ) , . . . , s n ( x ) , . . . converges to zero. But the convergence is not uniform, for '•«¿'-••«¿»'

• 10-11

-1

Hence, i f we take e = 1, we cannot find an N such that |s(x) - s n (x)| < e for n > N and for x.

every

NECESSARY AND SUFFICIENT 5. Theorem:

For

(2) to converge tive integer

si(x), s2(x),..., sn(x),...

uniformly N exist

on a set E, it is necessary and sufficient that such that, for every n > N and for every positive

(3) for

every

CONDITION FOR THE UNIFORM CONVERGENCE OF A SEQUENCE

for every e > 0 a integer p,

posi-

|s n+p (x) - s n ( x ) I < e

x on E.

Proof:

A. Necessity - Let s(x) be the limit of (1). I f N is taken so that |s(x) - s n (x)I < ^ throughout E for n > N, then we have (3) • Q.E.D. B. Sufficiency - Let the condition be satisfied. Of course, (2) w i l l converge throughout E. An e > 0 being given, let N be taken so that (3) holds throughout E. We say that, for n > N, (4)

Is(x) - s n (x)| s e

on E. This w i l l prove uniform convergence. Let x0 be a point for which (4) does not hold, that is for which |s(x 0 ) - s n (x 0 )| > e for some n > N. Keep n fixed. I t is clear that, for p very large, we have Is n+ p(x 0 ) - s n (x 0 )| > e because s_._(x 0 ) is an arbitrarily good approximation to s(x 0 ) when p is large. Q.E.D.

p

CONTINUITY OF LIMIT 6.

Theorem:

Let

OF A SEQUENCE

Si(x), s 2 ( x ) , . . . , s n ( x ) , . . .

converge uniformly on a closed interval ( a , b ) . Let the tinuous at some point x 0 of ( a , b ) . Then s ( x ) , the limit

functions of the sequence all be conof the sequence, is continuous at x 0 .

Remarks: I f the functions s n (x) are continuous throughout (a, b), then s(x) is continuous throughout (a, b). The assumption that (a, b) is closed is not essential. I t is made in order to take care of the case in which x 0 is an end-point of the interval. Proof: An e > 0 being assigned, take n so that |s(x) - s n (x)| < | throughout (a, b). Keep n fixed. As s n (x) is continuous at x 0 , there is a 6 > 0 such that, for |x - x 0 l < 6 , we have Is n (x) - s n ( x 0 ) I < e. Then, for |x - x 0 l

0, sn(x) = 0 for n > so that s(x) = 0. On the other hand, sn(0) = 1 for every n, so that s(0) » 1. Thus, s(x) is discontinuous at 0. To see that the sequence is not uniformly convergent, we observe that sn(-l ) = i for every n. Ln £ Hence, for every n,

and, if we are given e = i, we cannot find an N which will give |s(x) - sn(x)| < e throughout 2 (0, 1) for n > N. INTEGRABILITY g. Theorem:

Let

(a, b) be a closed

OF LIMITS OF

interval.

SEQUENCES

Let

Si(x), s4(x),..., sn(x),... be a sequence of functions each one of which is inteirable verge uniformly on (a, b) to a limit s(x). Then (1) s(x) is

on (a, b). Let the sequence

con-

inteirable;

(2) /a s i (x) dx,..., /a sn(x) dx,... is a convergent sequence; (3) /a s(x) dx > lim n — «

/ab sn(x) dx.

Proof: To prove (1), we have to show that two sums for s(x) which are of small norms have a small difference. Let Rn(x) = s(x) - sn(x). Let e > 0 be assigned. Take n so that |R (x)| < e for every x on (a, b). Keep n fixed. Take 6 so that any two sums for sn(x) of norms less than 6 differ by less than eWe have s(x) = sn(x) + Rn(x). Hence 2 = s{Ci)(\i - x 0 ) +...+ s(Cp)(xp - x ^ j ) and 2' = SntCjixt - x 0 ) + ...+ sn(Cp)(xp - x p - 1 ), we have 2 - r

= Rn(C1)(x1 - x 0 ) + ...• Rn(Cp)(Xp - xp_i)

44

and - Z'| < e(xt - x 0 ) +...+ e(Xp - x p _ 1 ) = e(b - a). As any two sums for s n (x) of norms less than 6 differ by less than e, any two such sums for s(x) differ by less than e + 2e(b - a), which is small when e is small. Hence s(x) is integrable . It remains to prove (2) and (3). We consider the difference /ab s (x) dx - /ab s n (x) dx. We know that s(x) - s n (x) is integrable and that /a s(x) dx - /ab s n (x) dx = / b [s(x) - s n (x)]dx. An e > 0 being assigned, take N so that |s(x) - s n (x)| < e for n > N throughout (a, b). Then |/a s(x) dx - /ab s n (x) dx | = |/ab [s(x) - s n (x)] dx | á e |b - a|. As e |b - a| is small when e is small, we see the truth of (2) and (3). 9. When the sequence does not converge uniformly, s(x) may not be integrable, or the sequence of integrals may diverge, or s(x) may be integrable and the sequence of integrals may not converge to s(x). We give an example to show that the sequence of integrals may converge to a limit different from the integral of s(x). Let s n (x) = 2n s x for 0 I x í i , ZI1 s_(x) = 2n a (i - x) for -A- ¿ x < i, n n 2n n 0

i/2n

i/h

i

Then s(x) = 0 for every x- As

s n (x) = 0 for i ¿ x s 1.

s(x) dx = 0 and f^ s n (x) dx = | for every n, we see that the

limit of the integrals of the functions s n (x) is not the integral of the limit. DIFFERENTIABILITY OF SEQUENCES OF FUNCTIONS 10. Theorem:

Let

(5)

s1(x),...,

s n (x),...

be a seauence of functions each of which has a derivative, val (a, b). Let (5) converge on (a, b). Let (6) converge

s{(x)

uniformly

(a) The seauence (b) The limit

Proof:

n

s;(x),...

on (a, b). Then (5) converges

of (5)

(c) The derivative

s'(x) throuihout

uniformly

is differentiable

of the

limit

of

on (a, b). throuihout

(5)

is the

limit

(a, b). of the sequence

(a) Given an e > 0, take N so that lsi+pW

" shW

throughout (a, b) for n > N and also so that 45

I
N. Let = sn+pW -

s

n

M-

Then

amp (zi - z 0 ) will approach the inclination of the tangent to the curve at z 0 . Similarly, amp (wi - w 0 ) approaches the inclination of the tangent at w 0 to the image of the first curve. Hence limit [amp (wt - w 0 ) - amp (zi - z0)] which equals limit amp

Wl

~ W(? z O — Zo

is the difference between the inclination of the tangent to the first image curve at w 0 and the inclination of the tangent at z 0 to the first curve in the z-plane. Now, if f 1 (z 0 ) / 0, limit amp w * ~ w ° = amp f'(z0). Z i - Z0 Similarly, the difference between the inclinations of the second w-curve and the second z-curve at w 0 and z 0 respectively is amp f'(z0). Let 0x and 0 S be the inclinations of the tangents to the two z-curves at z 0 and Ci and the inclinations of the tangents to the two w-curves. Then Ci

—

9i - C? — 6j

Hence Cl - is = 01 - 02 That is, the angle between the two w-curves equals the angle between the two z-curves. Q.E.D.

70

XV Continuous Curves

INVERSE FUNCTIONS (REAL VARIABLE) 1 . Let y » f ( x ) be continuous and increasing on the closed i n t e r v a l (a, b). Let f ( a ) = c , f(b) = d. Then on (a, b ) , y assumes exactly once every value in the interval (c, d ) . Hence to every value of y between c and d, there corresponds one value of x between a and b. We may thus regard x as a function of y , defined on (c, d ) . VTe write x = g ( y ) . I t i s obvious that x i s an increasing (c, d ) . We are going to prove that x i s of y on (c, d ) . Let y 0 be any value of y , and x 0 the corresponding value of t e r i o r to (a, b ) . The cases of x = a and x = b are treated in the manner in values of x 0 which are interior points of (a, b).

function of y on a continuous funcx . Let x 0 be inwhich we treat

Let a small positive number e be assigned. Let f ( x 0 - t) * y 0 - 6 1 , f ( x 0 + e) = y 0 + 6 2 . Then 6 t and 6 2 are p o s i t i v e . Let 6 be the smaller of 6 t and 6 2 . Then i f y l i e s between y 0 - 6 and y 0 + 6, x w i l l l i e between x 0 - e and x 0 + e. This proves that g(y) i s continuous at y 0 . We s h a l l c a l l g(y) the function

inverse

to f ( x ) .

A similar theorem holds f o r decreasing functions. ON TEE POINTS OF A CONTINUOUS CURVE 2. Theorem: If the points stitute a perfect set. Proof:

of a continuous

curve consist

of more than one point,

they con-

Let x =

z

2>'-->

z

n

be n p o i n t s . We c o n s i d e r the continuous curve (made by t a k i n g in order t h e segments Z j Z i + 1 ) , z = Zi + t ( z j - Z i ) ; (2)

o i t i 1

z = z 2 + (t - 1) (z a - z 2 ) ; 1 s t s 2 .

z = z n _ 1 + (t - n + 2) (z n - z n - 1 ) ;

n - 2 s t $ n - l . -

The curve ( 2 ) , or any curve e q u i v a l e n t t o i t , w i l l be c a l l e d a polygon with ..., zn. I f z n = Z j , the polygon (2) w i l l be c a l l e d closed. curve, i t w i l l be c a l l e d a simple closed polygon. STATEMENT OF THE CAOCHY INTEGRAL

the vertices

zt,

I f , in a d d i t i o n , t h e polygon i s a Jordan

THEOREM FOR POLYGONS

3- Let A be a simply connected open r e g i o n . Let f ( z ) be a f u n c t i o n a n a l y t i c throughout A. Let C be any closed polygon l y i n g in A. (That i s , the p o i n t s of C l i e in A.) We a r e t o prove t h a t fc f ( z ) dz = 0. For a closed polygon of two v e r t i c e s (n = 3 ) . the r e s u l t i s immediate because f —

f ( z ) dz = - / —

f ( z ) dz.

Accordingly, in what f o l l o w s , we l i m i t o u r s e l v e s t o t h e case of n g 4 , t h a t i s , t o polygons of a t l e a s t t h r e e s i d e s . REPLACEMENT OF THE POLYGON C BY A SIMPLER

POLYGON

4 . With r e s p e c t t o C, no g e n e r a l i t y i s l o s t in assuming t h a t no consecutive pair of points z x and z i + 1 coincide. When z i and z i + 1 c o i n c i d e , we may suppress the segment z i z i + 1 as f a r as 95

integration is concerned. The assumption just indicated will apply to everything which follows. This"understood, let C be a polygon in A of n sides. In §5> where a proof by induction is made, it will be seen that we may iimit ourselves to polygons C in which no two of the points !

(1)

are

> • • • i z n-l

i>

coincident.

For a polygon C as just described, we shall prove that there lying in A, such that

exists

a polygon C 1 of n

sides,

fc f (z) dz = Sc, f (z) dz and such that no pair

of extremities

of a side

of C' are collinear

with any other vertex

of C'.

We begin by proving the lemma: Lemma: Let A be an open region. Let zlz^ be a segment ly close to z 2 , the segment zize lies In A.

lying

In A. Then,

If zt

Is

sufficient-

Proof: For every C on ZiZ 2 , there is a 6,. such that if |z - Cl < z lies in A. About each C on z t z 2 as center, let a circle of radius 56^ be described. Since z7z7 is closed and bounded, there is a finite number of these circles such that every point on zlz7 is interior to at least one of them. Let 6 be the least of the radii of the finite number of circles. Then, if C is on Z1Z2, and if |z - Cl < 6, z is in A. For if Ci is the center of one of the circles of the finite subset which contains C, then (2) Also, if Iz -

IC - Ctl < ¿ 6 ^ < 6, we have, because 6 i

(3)

Iz - CI


0 S t i 1.

Thus z = Zt(l - t) + tz 2 + t(z9 - z a ). Let t be any given value. Then Zi(l - t) + tz2 is a definite point C on ZtZ 2 and z = C + t(z s - z 2 ). Thus

z - t = t(z» - z 2 ),

and, as It I £ 1, and |z3 - z 2 | < 6, we have Iz - Cl < 6 so that z lies in A. Q.E.D. We return now to the replacement of the polygon C by the polygon C 1 . Consider the segment ZiZ 9 . Suppose that certain vertices distinct from Zi and z 9 are collin96

ear with z t and z 2 . It is easy to prove, on a rigorous, arithmetic basis, that in every neighborhood of z 2 there is a point a such that none of the vertices z 3 , z 4 , ..., 7,n_1 is collinear with z t and a. We can, furthermore, choose a so as not to be coilinear with z 2 and z 3 or with zi and z2 . If a is sufficiently close to zs, Let C' be the polygon of vertices

the segments zTa and az3 will lie in A. So also will az 2 . z 1, a, z3, ..., z n - Zi

We wish to show that Ic, f(z) dz = fc f (z) dz. Evidently (4)

+

/c ~ 4

~ 4jz 5

_

42Z3

Now the triangle ziaz2 lies in A. As A is simply connected, the interior of z!az2 lies in A. Hence (5)

*

+

-

Similarly, considering the triangle az 2 z 3 , we have

(6) Subtracting (6) from (5), we find ÏT. = and as J,z t z 2

we see that the second member of (4) is zero and fc

=

fQ,.

It is thus legitimate to assume that no vertex distinct from z t and z 2 is collinear withzt and z 2 . Suppose now that some vertex distinct from z s and z 0 is collinear with z s and z 3 . Proceeding as above, we replace z 3 by a point close to z s so as to remove this condition. The point replacing z 3 , being very close to z 3 , will not be collinear with Zi and z 2 . Continuing in this manner, we secure a polygon C' lying in A, with no pair of extremities of a segment collinear with any other vertex. The integral of f(z) along C' equals the integral along C. COMPLETION OF PROOF

5. We now complete the proof of the theorem stated in §3« If C has three sides, we may, as was seen in §4, replace C by a triangle. The Cauchy Integral Theorem is already proved for triangles, so that the case of three sides is disposed of. Now, let the theorem be supposed proved for polygons of fewer than m sides, when m is some integer exceeding 3- We shall prove the theorem, by induction, for polygons of m sides. First, let us consider the case, mentioned in §4, in which the points Zi, z 9 , ..., z n _ x are not all distinct. (Note that n = m + 1.) Let, for instance, Zi coincide with Zj, where ,} is some one of the integers 3 . 4 , ..., n - 2, We consider the polygon C 4 whose vertices are

97

Zij Z t , ..., Zj = Zi and the polygon C 2 of vertices Zj'

We have

Z

z

j+1'

n

= Zl =

Zj'

/c f(z) dz « /Clf(z) dz + /Caf(z) dz. As Ci and C 2 have less than m sides, we have J"Ci = fCa = 0. Thus /cf(z) dz = 0. We now consider the case in which the m vertices of C are distinct and replace C by a polygon C', of m sides, as in §4* We now distinguish two cases. Case 1. C' is a simple closed polygon. Let the vertices of C' be Zi, Z 2 , Za, •.•> Z n = Zi. A segment of the type z^zj where z 1 and Zj are not the extremities of a side of C' will be called a diagonal of C . It is proved in the literature on analysis situs (see Kerekjarto, Topoloiie), that, given a polygon of more than three sides, with no two adjacent sides collinear, there is at least one diagonal of the polygon which lies, except for its extremities, within the polygon. Plainly no two adjacent sides of C' are collinear so that C' has a diagonal which lies, except for its extremities, within C 1 . As any of the vertices of C' may be taken as the initial-terminal point of C', then some diagonal zlzj where j is some integer among 3» 4, . n - 2, may be assumed to be of the type described above. Consider the polygon Ci whose vertices are Zi, Z 2 , •••» Zjj Zi and the polygon C 2 of vertices ZI,

Z J y ZJ+2,,...F

Z

s

N

ZL.

Because every point on ztzj is on or within C 1 , and because A is simply connected, every point on ZiZj is in A. Thus Ci and C e are in A and we have /c, f(z) dz = /Ci f(z) dz + fCi f(z) dz. Now Ci and C 2 have fewer than m sides. Thus, /ci =Jc2 - 0, so that fc, = 0 and fc

=0.

Case 2. C' is not simple. Because C1 satisfies the condition of §4, it must be that two non-adjacent sides of C' have a point in common. Such a point, by §4, cannot be an extremity of either of the sides. To fix our with ideas, let us suppose that zIzT has a point, called it i, in common with some zizi+1 2 < i < n - 1. We consider the polygon Ci of vertices and the polygon C 2

Zl, of vertices

z i + 1 , Z i+2 > ..•» Z n = Zl, Cf z 2 , Zg, ..*> z^, C.

We have Jc'

=

-fc!

+

•

As Ci and C 2 have fewer than m sides, we have /c. = 0 and the proof is completed. 98

XXII T h e Cauchy Integral Theorem f o r a Rectifiable Curve CORVES IN OPEN REGIONS

1. Let A be an open region and C a continuous curve l y i n g in A. We say t h a t there exists

6 > 0 such that long to A.

every

z of C is the center

of a circle

of radius

6 all

points

within

a

which be-

Proof: For every z of C, we can f i n d a p o s i t i v e 6 Z such t h a t t h e c i r c l e i n t e r i o r

c o n t a i n s only p o i n t s z ' belonging t o A. Let a c i r c l e of r a d i u s ¿6 Z be t a k e n , with z as c e n t e r , f o r every z of C. By B o r e l ' s theorem, t h e r e e x i s t a f i n i t e number of t h e s e c i r c l e i n t e r i o r s which cover C. Taking such a f i n i t e s e t of c i r c l e i n t e r i o r s , Let 6 be t h e l e a s t of t h e i r r a d i i . Let z be any p o i n t of C. Let C be t h e c e n t e r of some one of t h o s e c i r c l e s of t h e f i n i t e s e t which c o n t a i n z . Then |z - Cl
t i , t n = b, be n + 1 p o i n t s on (a, b ) , which a r e a r b i t r a r y except f o r the c o n d i t i o n s j u s t placed upon them. Let (1)

1

Oy

•'

be t h e p o i n t s cp(tj) + i i p ( t j ) on the curve C, j = 0, We c o n s i d e r the polygon ( i n s c r i b e d in C), whose v e r t i c e s a r e t h e p o i n t s (1). Let t h i s polygon be r e p r e r e s e n t e d by P . We s a y t h a t there exists an r\ > 0 such that, if t j - t j _ x < n , .i = 1 , . . . . n , every point on P lies in A.

Proof: Let 6 > 0 be such t h a t i f I z ' - zl < 6, where z i s on C, then z ' i s in A. The r e p r e s e n t a t i o n of C being, as above, z = qp(t) + i y ( t ) , 99

a

S t

g b,

we see that z is a uniformly continuous function of t on (a, b). Let, then, ri > 0 be such that if It1 - t| < ri, then jz' - z| < 6, where z = 0 be chosen so that: (1) If each tj - tj_x < 6, then P lies in U. (2) If |t' - t| < 6, then |z' - z| < r\, where z = ]

dz

-

J— 1 J

S i m i l a r l y , in integrating along the segment Z j _ i Z j , we have (4)

fU)

Now, f o r any z on on z J - i *J>

dz

- fUj-i)

Z j or on

(Zj " Z j . i )

• STJzr^

dz.

Z j , we have |z - Z j _ j | < tw Hence, f o r z on ZJZT^J

|f(z) - f 0 be taken. Let P be a closed polygon, inscribed in C and lying in A, such that |/c f ( z ) dz - f p f ( z ) dz| < e. As fp f ( z ) dz » 0, we have proved. Remarks on the Preceding

f ( z ) dz| < E. Because e i s a r b i t r a r i l y small, our theorem i s

Theorem:

5- As has already been observed, one of the most important cases is that in which C is a Jordan curve. Now, in this case, i t i s actually unnecessary to assume that A is simplyconnected. A may be any open region provided that C i s a r e c t i f i a b l e curve which l i e s , with i t s i n t e r i o r , in A. Proofs of the theorem in this form, f o r restricted types of curves C, w i l l be found in Osgood, Funkttonentheorie, and in Tannery, Théorie des fonctions, Vol. I I . The general case i s treated in a paper by Kamke, Mathematische Zeltschrift, Vol. XXXV (1932), p. 539. We s h a l l ,

in certain l a t e r work, assume the truth of the theorem f o r the case just discussed.

101

XXIII T h e Cauchy Integral Theorem for Several Contours SENSED CIRCLES

1. Let a + bi be any complex number. Let r be any positive real number. Consider the Jordan curve given by x = a + (r - t), y - b + Jr* - (r - t)a, 0 $ t $ 2r x » a + (t - Sr), y - b - 7r 2 - (t - 3r)2, 2r i t s 4r. It is easy to show that the points of this curve are those points z for which |z - (a + bi)| = r. to the above curve will be called a positively sensed circle with Any curve equivalent center at (a + bi) and radius r. Any curve inversely equivalent to the above curve will be called a negatively sensed circle with center at (a + bi) and radius r. SENSE OF A JORDAN CURVE

2. Consider a Jordan curve C given by (1)

x « 0 be preassigned. Let o > 0 be such that |f(C) - f(z)I < 6 for |C - z| < 6. The existence of 6 follows from the continuity of f ( z ) . Let the radius of T be taken as less than 6 and l e t r represent that radius. Then, for any C on T, IC - z| - r . Also, the length of r i s 2nr. Hence IJr n

n

£M C- z

dC|

g

L 2nr r

=

2ne.

Then, by (2) |f(z) -

2m

1

so that, by (1), 108

C- z

,

e,

As e i s a r b i t r a r i l y small, we have U

ZTtl

£ - Z Q.E.D.

THE FIRST

DERIVATIVE

2. We s h a l l prove t h a t , for any z of B,

Proof: We deal with a fixed z in fi. Let r > 0 be such that z i s the center of a c i r c l e of radius 2r which l i e s , with i t s i n t e r i o r , in B. Then, for every £ of C,

(3)

z | > 2r

-

and, i f |Az| < r , (4)

It - (z + Az)I > r .

We have, for |Az| < r , f(z

+

Az) - f(z) =

2m

JLc

f(C) [ri —y - - 1 — ] dr C - (z + Az) Z - z

. f ~ 2rci c [C - (z

Az,f(C), dC. az)] (t - z)

+

Hence f ( z + ¿z) - f-(z) _ Az 2ni

J

c

f(Q [t - (z + Az)] (C - z)

Thus (5)

f ( z + Az) - f ( z ) Az

_1_ 2rci

jc

=¿i

/

= 2rci

Jc

f ( Q dC5 (£ - z)

cf(C)

/

í(C-z-Azt ( t - z ) "fc^zT^

f ( t ) df (t - z) 2 (t - z - Az)

As f(z) i s integrable along C, f(z) i s bounded on C. Let M > 0 be such that |f(£)l < M for every í on C. Then, by (3), (4) and (5), if X i s the sum of the lengths of the contours which make up C, |1f ( z + Az) - f ( z ) . J• f ( e ) dr I , M i J L x. Az 2rti J ° (c -z)« - 2n 4r* A As the quantity H2l_ i s fixed, we see that

Q.E.D.

109

THE HIGHER DERIVATIVES

OF AN ANALYTIC

FUNCTION

3- Let f(z) be analytic in an open region A. We know t h a t f ( z ) has a derivative, f ' ( z ) , in A. We are going to prove t h a t f ' ( z ) i s analytic in A. I t w i l l be an immediate consequence of that that f ( z ) possesses derivatives of all orders In A, which are all analytic functions. We see thus a s t r i k i n g difference between the r e a l domain and the complex. A function of a r e a l variable can have a f i r s t derivative without having a second derivative. The r e s u l t i s established in the following form. Using the notation of the preceding sect i o n s , one shows t h a t , for every positive integer n and f o r every z of B, f ( z ) has a derivat i v e of order n, given by the formula f ( » ) ( z ) , -aL 2m

C

_

(c

de

z)n+i

We s h a l l carry out the d e t a i l s of the proof for n = 2. Let r be defined as in the preceding section. We have, f o r |Az| < r , f •(, . * ) - r w

- &

fcHO

- ^

Ic f < 0

[

[

-

(:

«

- z - i l ï ) 4 CC-z) " K - z - t a f * (C - ^

dc

Hence f'(z

+

Az) - f ' ( z ) Az

2! r f(C) dC . 2fti (£ — z ) 3

_ Lf(Z) [ 2 2ni J c (C - 2 - Az)* (C - z) 5 - JL c { W 2m

^

2 (C - z ) 9

I dt (C - z - Az)4 (C - z)* J C

- g) az - 2Az2 - ( C - 8z) AZ] (Ç - z - Az) (Ç - z )

= Az_ / f ( r ) - z) - 2Az 2ni (C - z - Az)* (C - z ) a

dr

Let G > 0 be such t h a t , for |Az| < r and f o r Ç on C, - z) - 2 A Z I < G . Then, i f \ i s the sum of the lengths of the contours which make up C, and M an upper bound for |f(C)l on C, 13 (C

i f ' ( z » Az) - f'.(z) _ _2!_ f Az 2ni J c

f(g) dg I < MGX IAzI (n - z ) 3 ' = 2rcrs • 8r*

so t h a t , when Az approaches zero, f ' ( z + Az) - f ' ( z ) Az approaches a limit equal to J L / f(C) dC 3 2ni (C - z ) ' Hence f " ( z ) e x i s t s and is equal to i l

2rd

f f ( c ) dg 3

Jc

(C - z )

110

'

ANALYTIC

FUNCTIONS

DEFINED

BY CONTOUR

INTEGRALS

4. On the basis of §2, we can formulate the following theorem: Let region

F ( z ) be defined,

and

continuous, Then

B. (C and B as before.)

at

every

Jr

c

represents

a function

analytic

throughout

point

the

complete

contour

W U L c - z

B . Its n

of

nth

derivative

' * (C - z) n + 1

111

is

given

by

C of

the

open

XXVI Infinite Sequences and Infinite Series of Analytic Functions UNIFORM CONVERGENCE

1. Let s 4 (z), s 2 (z),

S n (z), ...

be a sequence of functions of z (not necessarily analytic), which converges on a domain E to a limit s(z). We shall say that the sequence converges uniformly on E if, for every e > 0 an N can be found such that, for n > N, we have |s(z) - S n (z)| < 6 for every z on E. As in the theory of the real variable, one proves easily that, for the sequence to converge uniformly on E, it is necessary and sufficient that for every e > 0 an N exist such that, for every n > N and for every p, lsn+P r. Hence, for z in r ' and n > N,

113

[The length of r i s 2n (2r).] Thus, (3) converges uniformly in r 1 . Borel's theorem does the r e s t . THE HIGHER DERIVATIVES

4. Exactly as in

we

prove that i f

(4)

Si(z), . . . , s n ( z ) ,

...

i s a sequence of functions analytic in an open region A, the sequence converging uniformly in A to a function s(z) (necessarily a n a l y t i c ) , then, i f p i s any positive integer, the sequence of the pth derivatives of the functions in (1) converges to the pth derivative of s(z) throughout A, the convergence being uniform in every closed and bounded domain contained in A. One uses, in the proof, the formula s

(z) - Jg-L L «n ( z )

dC

z)P + 1

.

SERIES OF COMPLEX NUMBERS

5- If » ®>2 > ' • • > an> • • • i s an i n f i n i t e sequence of complex numbers, we mean by the sum of the (5)

series

a t + a2 +...+ a n + . . .

the l i m i t , as n increase i n d e f i n i t e l y , of s n = a t +...+ a n , provided that such a limit e x i s t s . When the limit e x i s t s , we say that (5) i s converient. in the r e a l variable, we can prove t h a t (5) converges i f (6)

| a t | + |a2| + ...+ |an| +...

converges. If (6) converges, we say that (5) i s absolutely

convergent.

SERIES OF ANALYTIC FUNCTIONS

6. Let Ui(z), u 2 ( z ) , . . . , u n ( z ) ,

...

be a sequence of functions defined on a domain E. Suppose t h a t , f o r every (z) on E, the series (7)

Ui(z) + . . . + u n (z) + . . .

converges. We then use the expression - - The infinite throuihout

series

(7)

E.

We represent by s n ( z ) the sum \i1 (z) +...+ u n ( z ) . If (7) converges throughout E to the sum s ( z ) , we c a l l s(z) - S n (z) = u n + 1 ( z ) • u n + 2 (z) + . . . the

remainder

after

n terms

of

(7).

114

of

functions

converies

As

Let (7) converge on E and let Rn(z) denote its remainder after n terms. We call (7) uniformly convergent on E if, for every e > 0, an N exists such that, for n > N, |Rn(z)l < e throughout E. The theorems on sequences of analytic functions go over immediately into theorems on series. Thus, let u t (z) +. . .+ Un(z) + . . . ,

each un(z) being analytic in an open region A, converge uniformly throughout Then s(z) is analytic throughout A. Furthermore, for every z in A,

A to a sum s(z).

s'(z) = u^(z) +...+ u^(z) +..., and the series in the second member of the last equation closed and bounded domain contained in A.

115

is uniformly

convergent

in every

XXVII Power Series GREATEST LIMIT OF A SEQUENCE OF REAL NUMBERS

1. Consider a sequence of non-negative r e a l numbers (1)

a15 a2, . . . , an,

...

which i s bounded from above, that i s , a sequence for which a number M e x i s t s such that a n 0, there

are an infinite

(3) For every

e > 0, there

are at most a finite

I - e.

number of elements

of the sequence

number of elements

which

which exceed

exceed

I + e.

Proof: Let B represent the t o t a l i t y of numbers which have the property of being exceeded by an i n f i n i t e number of elements of the sequence. (Evidently a l l negative' numbers belong to B.) Then B i s bounded from above, since M exceeds every number in B. Let I be the least upper bound of B. Then I has the properties (a) and (|3). Evidently no number d i s t i n c t from I as j u s t determined has the properties (a) and (3). We s h a l l c a l l I the greatest

limit of the sequence.

For a sequence (1) which i s not bounded from above, we s h a l l use the expression — The greatest limit of the sequence is plus Infinity. The statement j u s t made i s to be regarded merely as a convenient substitute for the statement that the sequence i s unbounded. I t does not at a l l imply that we have created a number "plus i n f i n i t y . " POVER SERIES

2. We consider a s e r i e s (2)

c 0 + Ctiz-a) + c 2 ( z - a ) 2 +...+ c n ( z - a ) n +.-..

where the c'S and a are constants (complex numbers). We s h a l l c a l l (2) a power series. going t o study the values of z for which the series converges.

We are

Consider the sequence | c 0 | , I c J , s/|c 2 I, V l c a | , . . . , \ Z l c J ,

...

By V w , we mean the unique non-negative number whose nth power equals | c n | . Let I be the greatest limit of t h i s sequence. We s h a l l show t h a t : (a) If I i s i n f i n i t e , the power s e r i e s diverges for every z except z = a . (g) If I > 0 and i f I is not i n f i n i t e , the series converges for |z - a I < i and diverges for |z - a | > 1 . 116

(y) If I * 0, the s e r i e s converges for every value of z . The r e s u l t s j u s t stated are due to Cauchy and to Hadamard. We s h a l l work under the assumption that a » 0. The extension to the general case is immedia t e . Thus, we consider the s e r i e s c 0 + CiZ + c 2 z s + . . . + c n z n + . . . .

(3) Let us dispose of case (a).

Take any z d i s t i n c t from zero. As I i s i n f i n i t e , an i n f i n i t e number of the quantities \ / | c n | exceed yl-r. Hence, for an inf i n i t e number of values of n, yicj

•

> i

or | c n z n | > 1. Thus, the nth term of (3) does not approach 0, and (3) diverges. We take now case 0 ) . Let r = i . We shall prove f i r s t that (3) diverges positive

number less

than r ,

(3) converges

for | z | > r and then that,

absolutely

certainly establish our statement above.

and uniformly

for

if r t is any

|z| < r t .

This w i l l

Consider any z with. | z | > r . Then -pr< I, so t h a t , for an i n f i n i t e number of values of n, z VIcJ > v

n

lz|

or | c n z n | > 1. This, as above, implies divergence. Now, r j being as above, l e t r 2 be any number such t h a t Ti < r 2 < r . . Hence, there is Then — > I, so that only a f i n i t e number of the quantities l j / | c | exceed r2 r2 an N such t h a t , for n > N, v / K nI

^ r- 1 2

or I cnr2n I s 1. For | z | < Tj,, each term of the s e r i e s (3i has a modulus not greater than the corresponding term of (4)

|c 0 l + ICj| r 1 +...+ | c n | r? +

Hence, i f we can show that (4) converges, we shall know that (3) converges absolutely and uniformly for | z | g r ^ We write (4) (5)

Icol

+

|ctr9|

^

+|c 2 r 2 1

|cnr?| 117

Since, for 11 large, |cnrf| < 1, the terms of (5) are ultimately not greater than those of (6)

As (6) is a geometric series with ratio less than unity in absolute value, (6) is convergent. Hence (5) is convergent and our result is proved. Of course, (3) converges absolutely for every z such that |z| < r. We consider finally the case (y) and prove that, for every rt > 0, (3) converges absolutely and uniformly for |z| < rt. Proof: Let r 2 > ^ . As -1- > 0, only a finite number of the quantities ?/|c_| exceed -1-. The n r2 r2 proof continues precisely as for case (p). CIRCLE OF CONVERGENCE 3. When 0 < l < + , we call r = y the radius of convergence of the series (7)

a + Ci(z-a) +...+ cQ(z-a)n +....

We call the circle |z - a| » r the circle of converience of (7). When Z = 0, we use the expression "the radius of convergence is infinite." In that case,' we call the entire plane "the interior of the circle of convergence." Thus, inside of its circle of convergence, (7) converges absolutely at eveiy point. Outside the circle of convergence (7) diverges. Upon the circle of convergence, various states of affairs may exist. The series may converge all over the circle, or may diverge everywhere on it, or be convergent at some points and divergent at others. Within any circle interior to the circle of convergence, the series converges uniformly. Thus, the series represents a function which is analytic in the interior of the circle of converience. According to our results on series of analytic functions, the derivative of the function represented by (7) is obtained by differentiating the series formally. Examples: 1 + z + z a +...+ z n +...

(A)

r » 1. Series divergent for |z| = 1. (B)

JL+a!+...+ai+... 1 2 n

r . 1. Convergent for |z| = 1. (C)

1 • Z + ¿r + ...+ 4 + ... 2! n!

r = flo. (D)

1 + l!z + 2!z2 +...+ n!zn +

r = 0. ®

"

i

' i ' T ' i '

•••

r = 1. Divergence on circle of convergence at z = -1; convergence everywhere else on circle. The last statement is difficult to prove except for z = 1.

118

XXVIII Taylor's Expansion STATEMENT OF RESULTS

1. Let f ( z ) be analytic in an open region A. Let a be any point of A and l e t T be any c i r c l e with a as center, lying with i t s interior, in A. We shall prove that f(z) admits, throughout the interior of T, a representation. f(z) = c 0 + Cj(z-a) + c ^ z - a ) * + . . . + c n ( z - a ) n

+....

We shall prove that there i s but one power series of this type (that i s , a series of powers of z - a ) , which represents f(z) for a neighborhood of a and that i t s coefficients are given by Co = f ( a ) ,

cn = l ^ a i , n!

(n > 0).

Remark: Actually the power series will converge to f(z) upon T, as well as within T, for i t i s easy to show that a c i r c l e concentric with F and of radius only slightly greater than T, l i e s with i t s interior in A. PROOF OF UNIQUENESS

2. Suppose that the series c 0 + Ci(z-a) + c 2 ( z - a ) 2 + . . . + c n ( z - a ) n + . . . converges to f(z) for a neighborhood of a. We find, putting z = a, that f(a) = c 0 . For a suff i c i e n t l y small neighborhood of a, the power series will converge uniformly, so that i t can be differentiated term by term. Thus f ' ( z ) = Ci + 2c 2 (z-a) + . . . + nc n (z-a) n ~ 1 + Hence, putting z = a, we find f1(a) =

Cl.

Again, we have f " ( z ) = 2c 2 + 3 . 2 c 3 ( z - a ) + . . . so that f11 (a) = 2c 2 . Proceeding in this way, we see that, for n £ 1, c

n

= li^lfii n!

Hence, there can be no more than one expansion of f(z) in powers of ( z - a ) , valid for a neighborhood of a. A PRELIMINARY

3- We shall need the identity, easily proved, 119

_ 1 A-B " A

+

JS. + B_ + A2 A3 '

A57*

+

A n + 1 (A-B)

Here, i t i s understood that A and B are complex numbers with A f 0 and A - B i 0 DERIVATION

OF

EXPANSION

4. We refer to §1. We have, for z within T1, f(2)

m

^Tjr£iSUL 2711 1 £ - z

l

or f(z) _ V ' ~ 2ni

L £(£) d,E r r (C - a) - (z - a ) '

J

Now 1 (C - a) - (z - a)

-

C - a

+

,z - a (C - a) 2

+

+

(z - a ) n + 1 (C - a ) n + i ( £ -

(z (C - a ) " + i

z)

Hence f(z) - - 1 - f W d£ 1 ' " ' 2 i i i Jr ^ _ a +

+

(z

r f ( Q dr r a) 2rci * (£ - a ) 2

+

+

(z - a j n r 2ni *

f(0 (¿1

(z - a ) n + 1 r _ m i d£ . 2ni "T (C - a)n+l(H - z)

Now r

2TII

f ^ m i M 1

C -

m

a

f(a)

and _1_ / f(E) d£ _ f^,,(a) 2rei r (C - a)® + 1 ~ m! f o r m > 0. Hence, for z within T, f(z) = f(a)

+

f 1 (a) (z - a) + ...+ J & L M ( z n!

a)n

+

(z - a)"'* 1 L1 — M Q B + 1 2m (C - a ) ( C - z)

To prove the v a l i d i t y of the Taylor expansion, a l l we have to show i s that as n increases,

tends towards zero. Let r be the radius of P. Then |z - a|


Jr

so that r da _ l i da - ^ 2n-

Jrzn

Let r be the radius of T 1 . Then

Hence, for r arbitrarily large, 1^1

, Jk rn-l

As n > 1, we see that L,da _ 0. i Suppose now that we have an expansion for f ( z ) f(z) =

2

n=0

c (z - a) n + ~S c (z - a) n n n=-l n

which is valid in the ring between Ci and C s . Let T be any c i r c l e , with & as center, lying within the ring. As the two infinite series in (3) converge throughout the ring, each of them w i l l be uniformly convergent along P. Let p be any integer, positive or negative. Then, by (3), ^

1

¿'teL

(z - a)P + 1

=

I n=0

c„ (z - a ) n - P _ 1 + 1 n=-l

n

c„ (z - ajn-P" 1 n

and the two series in the second member converge uniformly along T. Now (5)

/n ,(

Jr

Z

- a)P+i

= Jr t 2. c„(z - a ) n -'P - 1 dz ++JVf v ~Z Cn(z - a) 11 "?- 1 dz. n " n 0 n=-l

=

Because of the uniform convergence mentioned above, we can integrate the series in (5) term by term along T, so that fvl 1

i(i\Jf = 2 c L (z - ^ - P - 1 dz + ~f c (z-a)P+1 n=0 n 1 n=-l n

L (z - ajn-P" 1 dz.

1

Now, i f n-p-1 * 0, (z - a ) n - P _ 1 i s analytic everywhere, so that fr (z - a ) « - ? - 1 dz = 0. I f n-p-1 < -1, then, as was seen above, /r (z - ajn-P" 1 dz = 0. 127

When n-p-1 = - 1 , that i s , when n = p, we have Jr (z - a ) n - P _ 1 dz = JT dz z - a

2ni.

Thus, Jr T ^ i l f e

-

cF

or r_ f ( z ) dz 'p " 2rd -T (z - a ) P + 1 ' which i s the result stated in §2. This shows that the Laurent expansion, i f i t e x i s t s , unique.

is

DEEIVATIOH OF EXPANSION 4. Consider any ring R bounded by c i r c l e s Dj and D 2 , each with center at a and lying in the ring bounded by Ct and C,. I t i s understood that Dj l i e s within D 2 . Then f ( z ) is analytic in an open region containing Dj, D2 and the ring R bounded by them. Hence, for z in R, f (z)

=

L m / 2ni TJ« C - z 2rci D i

C- z '

the integrations being performed in the positive sense. Now the expression _!_ j

dC

since-f(C) i s continuous along D 2 , defines an analytic function within D 2 . This analytic function has an expansion : 0 + Cj(z - a) + . . . + c R (z - a) n

(6) where 1

r

f(E) dC

f

P

f(C) dC

n

>1

This because c 0 in (6) is the value of the analytic function at a and c R , for n 2 1 i s the nth derivative at a divided by n! We examine now the expression . ! _ JT f { g dC 2tu C- z which equals We have

_ r 2ni

ffc) —f~>

so that for |z - a| small, c.J 1 2 | z - a |n' z |z - a||f(z)| i s very large when |z - a| i s very small. l| i s very small. f (z) I

As n > 0,

Q.E.D.

THE NUMBER INFINITY

4 . We are going to adjoin to our complex number system a symbol oo, dogmatically s e t t i n g down the following rules for operation with the symbol. (A) J = 0(B) I f a 4 0, | = » . (C) I f a f 0, a ® = 1, f 1 ( a ) = 0, f " ( a ) = 0 ,

. . . , f t P " 1 J ( a ) = 0; f ( P > ( a ) / 0.

I f p = 1, the understanding i s that f ' ( a ) t 0. The function f ( z ) i s then said to assume the value c 0 p times at a . I f c 0 = 0 , a i s c a l l e d a zero

of order p o f

f(z).

If a function has a pole of order p at a point a , the function i s said to assume the value

p times a t

a.

Suppose that f ( z ) i s analytic at

Let i t have at ® an expansion

if with c

j 0. Then f ( z ) i s said to assume the value c 0 p times at «

t o have a zero

of order p a t

PRODUCT

OF TWO

If c 0 = 0, f ( z ) i s said

FUNCTIONS

2. By a vicinity of a f i n i t e point a, we s h a l l mean the open region obtained by removing a from the interior of a c i r c l e with center a t a . Let f ( z ) and g ( z ) , neither identically zero, be analytic in some vicinity of a , and l e t the Laurent expansions of f ( z ) and g(z) about a contain at most a f i n i t e number of negative powers of z - a , so that f ( z ) and g(z) are either analytic at a or have poles at a . Let the expansions be f ( z ) = c 0 ( z - a) m + c t ( z - a ) m + 1 + . . . g(z) « d 0 ( z - a ) n + d t ( z - a ) n + 1 + . . . where m and n are any integers at a l l and where c 0 ^ 0, d 0 ^ 0. Then, for some vicinity of a , f'(z) = (z - a) m 0. By the mean value theorem, s i n E - s i n 0 = (cos x j ¿J

where 0 < x,
0 2 2

so t h a t s i n E = 1 . 2

Using the formulas f o r s i n (z + h) and cos (z + h) with h = s i n (z + ]|) = cos z ;

cos (z +

¿j

we f i n d t h a t

= - sin z.

Hence s i n (z + p) = cos (z +

= - sin z.

and cos (z + p) = - s i n (z + | ) = - cos z . Finally, s i n (z + 2p) = - s i n (z + p) = s i n z cos (z + 2p) = - cos (z + p) = cos z . Thus s i n z and cos z a r e p e r i o d i c , with 2p f o r p e r i o d . I t follows e a s i l y , from §2, t h a t 2pi i s a period f o r e z . OTHER

RELATIONS

7 . We have >in

- z ) = s i n £ cos (-z) + cos £ s i n (-z) = cos (-z) = cos z .

Similarly, cos (? - z) = s i n z ,

s i n (p - z) = s i n z ,

cos (p - z) = - cos z ,

sin

¿i

- z ) = - cos z , e t c .

Also e z+pi

m

p

g z +2 i

_ez_ 142

=

ie2>

Q.E.D.

GRAPHS

8. From the periodicity of sin z and cos z and from the fact that for real values of z, they admit 1 and -1 as maximum and minimum values (use relation sin 2 z + cos 2z = 1 ) , we see that the graphs of these functions for real values of z are of the types exhibited below.

The graph e z is seen to be

PARAMETRIC REPRESENTATION

OF

CIRCLE

9. We are going to prove that the curve x = cos t , y = sin t

0 s t g 2p

i s a positively sensed c i r c l e of radius 1 with center at the origin. F i r s t , because cos tioned .

2t

+ sin

2t

= 1, every point of the above curve l i e s on the c i r c l e men-

Consider now any point of the c i r c l e . To fix our ideas, l e t x £ 0, y 2 0. There is a t on the interval (0, such that cos t = x, since 0 g x £ 1. Now, for 0 i t < ¿j

id

sin t = + ,/i - cos 143

2t

= +

\/l

- x2,

so that y = s i n t . Thus, every point of the c i r c l e i s on our curve. F i n a l l y , i t i s easy to see that two d i s t i n c t values of t other than 0 and 2p give d i s t i n c t points of our curve, so that the curve is a simple closed curve. Hence, by the theorem on the equivalence or inverse equivalence of two simple closed curves with the same points, our curve i s e i t h e r a positively sensed c i r c l e or a negatively sensed c i r c l e . Now the sense must be positive, because as t increases from 0 to L x decreases and y increases. ^ A FORMULA FOR THE LENGTH OF A RECTIFIABLE CURVE 10.

Theorem:

Let

a rectiftable

curve

C be given

by

x = t 0 , t2 > t j ,

t n = b.

The corresponding inscribed polygon has a length 7 M t j

- cp(t 0 )] 2

+

[ v ( t j - y(t0)]2

+

. . . + 1/[«p(tn) - ^ V ^ ) ] 2

+ [v(tn) - V(tn_x)]2

By the mean value theorem, there are in every interval ( t i _ 1 , t ^ points t^ and c^ such that /[cp' (u)] 2

+

[V(v)]2

i s uniformly continuous with respect to both variables in the closed rectangle a £ u £ b, a s v s b. Let an e > 0 be assigned. Let a 6 > 0 be taken in such a way that I s/[ . is constant throughout A. To determine the constant, we observe that, for z = a, f (z; = e l o £ f ( a ) = f(a). Hence, the constant is unity and e e
= f(z). 151

I f h(z) i s an a n a l y t i c f u n c t i o n such t h a t e h ' z ' • f ( z ) , then e v i d e n t l y , a t any p a r t i c u l a r point z , h(z) d i f f e r s from g(z) by an i n t e g r a l multiple of 2rci. As h(z) - g(z) i s continuous, the i n t e g r a l multiple must be a constant multiple of 2ni, f o r an integer cannot vary c o n t i n u ously without staying c o n s t a n t . Q.E.D. INTEGRAL

FUNCTIONS

WHICH ARE NOWHERE ZERO

4 . The e n t i r e f i n i t e complex plane i s simply connected. Hence, i f f ( z ) i s an i n t e g r a l f u n c t i o n which i s nowhere z e r o , we have f(z) = e^(z) with g(z) an i n t e g r a l f u n c t i o n . THE DEVELOPMENT

OF LOG (1

+ z)

5. Consider a c i r c l e of radius u n i t y , with center a t the o r i g i n . Inside t h i s c i r c l e , 1 + z i s nowhere zero, f o r 1 + z = 0 only f o r z = - 1 . Hence, there are functions a n a l y t i c within the c i r c l e whose exponentials equal 1 + z . We s h a l l c a l l any such f u n c t i o n a branch of loi (1 + z ) . One of these f u n c t i o n s equals 0 f o r z = 0. We c a l l i t the principal branch of log (1 + z ) . Let us get the Taylor expansion of the p r i n c i p a l branch f o r z We have

àlog

(1

+ z) = -

z) = - 1 . + z

ff, iog a »,) - '-ijr!!;;1'

Then Log (1 + z)

Z -

152

2

3

+ a_ -

4

(1 + z ) 2 '

0.

XXXVIII Infinite Products CONVERGENT INFINITE PRODUCTS

1. Biy an infinite product of complex numbers, we shall mean a symbol (1)

a ^

. . . an . . .

where each a n is a complex nunber. The infinite product (1) w i l l be said to be convergent i f both of the following conditions are satisfied: ( a ) At most a finite ( 0 ) If

the

number of m is such

integer

the d's

that

are

zero.

a n i 0 for

n g m, then

the

sequence

am> am am+l> am am+l am+2>

converges

and has a limit

distinct

from

zero.

Evidently, i f the i n f i n i t e product is convergent in the above sense, the sequence a t , a t a 2 , a^ a 2 a®, ••« converges. The limit of this last sequence is taken as the value of the product (1). An infinite product which is not convergent is called divergent.

For instance,

1 • 2 • 3 • 4 ... 1 • 0 • 2 • 0 • 3 • 0 ... 1 1 1 1 2 ' 3 ' 4

5 " "

are divergent. The above definition is so framed as to allow a convergent product to be zero only i f one of its factors is zero. NECESSARY AND SUFFICIENT 2 . Theorem:

For

the

infinite

CONDITION FOR CONVERGENCE

product

3-1 a^SLa . . . to converge, it is necessary and sufficient for n > N and for any positive Integer p,

that

for

l a n+l an+2 ••• an+p " Proof:

every

1

'

e > 0 an N > 0 exist

< 6•

(a) Necessity - Suppose that a n j 0 for n 2 m and that am> am am+l> am am+l am+2'

converges to a limit a f 0.

153

such

that,

Let e > 0 be preassigned. Let N > m be such that for n > N, ' m m+1 • • • na1 I > 9-Lâl and l1a™ ,^ . . . a^.,, m am+1 n+p - am m m+1 . . . a„| n' < -I^J9 e for p = 1, 2, 3, Now K

••• a n+ P " am ••• a n ' = lam ••• a J

l a n+l ••• an+p " H -

Hence l a n+l

•••

an+p

Me " II < j j p = e2

Corollary:

If a t

. . . a n . . . converges

then a n approaches

unity as n increases

Indefinitely.

(B) Sufficiency:Let the condition be s a t i s f i e d . Then, because a tends toward unity as n increases, at most a f i n i t e number of a ' s are zero. Suppose that a n i 0 for n 2 m. Consider the sequence a m' am am+l> am affl+l am+2'

Let M > m be such that K+l for n > M, p = 1, 2,

••• an+p " H < I

....

Let A = |am . . . aM| Then A > 0. We know that for n > M, laM+l aM+2

••• a n " H
d i s t i n c t from one another, zeros of orders >

Pi> Ps,

n>

• • • >

a

••P

n

>

respectively, and which have no other zeros. One such function is f ( z ) = (z - a J P l

. . . (z - a n ) P n

Consider any other such function, F ( z ) . We see that EM f (z) i s analytic a l l over the plane, even at the zeros of f ( z ) . Also, F ( z ) / f ( z ) has no zeros. That i s , F ( z ) / f ( z ) i s an integral function devoid of zeros. Now any such function i s of the form with g(z) an integral function. Then F(z) = e S ( z )

. . . (z - a n ) P n .

(z - a t ) P l

Furthermore, any F(z) as j u s t written has the indicated zeros and no others. ON INTEGRAL

2. Instead of saying,

M

FUNCTIONS

WITH AN INFINITE

NUMBER OF ZEROS

f(z) has p zeros at a , " we s h a l l say,

H

f(z) has zeros at p times.

Let f ( z ) be a non-constant integral function with an i n f i n i t e number of zeros. For 0 s |z| g 1, there are at most a f i n i t e number of zeros. If there are any, we arrange them in a sequence, in the order of increasing moduli. I f several zeros have equal moduli, we arrange them in any order. We now adjoin to this sequence the zeros for which 1 < |z| g 2, arranging them in the order of increasing moduli. Continuing in this fashion, we have a l l of the zeros of f ( z ) written in an i n f i n i t e sequence a

i

>

a

2

>

•••t

a

n

>

• • •

with | a n + 1 | s | a n | and with lim | a n | =

a

2>

•'•>

157

'*"

ZEROS

be any infinite sequence of complex numbers with |an+1| s |an| and with lim |aj = n —^ oo We propose to construct an integral function which vanishes for the points of this sequence and for no other points. We assume, for the present, that no a n is zero. This assumption will be removed later. Suppose first that the series la! | + |a51

|aj

converges. Then the function f(z) = (1 - a-2-) (1 - a-2-) ... (1 • a-5-) ... i 2 n fulfills our requirements and every function of the type described is of the form (1 • -2-)... (1 - -2-) ... a ! % with g(z) an integral function. But when 2 l/|an| diverges, the product ti(1 - z/a ) will not as a rule converge. We meet this situation as follows: We choose positive integers Pi,

••, Pn>

in such a way that for every r > 0, Pl i r- ^ir . a i

^ i-s-r i r ipn ^ a n

converges. For instance, we can take p n = n. In that case, since for any r, |a | > 2r for n sufficiently large, the nth term of the series has, for n large, a value less than l/2n. Thus the series will surely converge. We now consider the product a (1)

TC (1 n=l

-2-) e a. n

n

+

t(JM 2 2 a n

— 1 — (-5-)Pn_1 Pn " 1 a n

In this expression it some p n is unity, the exponential is supposed to be absent from the nth factor, and the nth factor is simply (1 • a— ) . n We shall prove that the product converges absolutely and uniformly in every bounded domain. By this we mean, firstly, that the product converges for every z; secondly, that the convergence is uniform in every bounded domain; finally, that if the nth factor in the product is written as 1 + un(z)> then n[1 + |un(z)|] converges for every z. When the uniform convergence is established, we shall know that the product converges to an 158

integral function which vanishes only at the points a t which i t s factors vanish. Now, since the exponentials vanish nowhere, the integral function w i l l vanish at the points of our s e quence and nowhere e l s e . CONVERGENCE PROOF 4.

Lemma:

For

|u| < 1,

we

have

|eu - 1| s 2 |u |. Proof:

le u - I I - lu • ¡ ;

+

»1

= |u| • |1

i

+

+

a!

+...|

< |u|(l

+

|r

+

^

• . . . ) S 2|u|. Q.E.D.

Consider now a c i r c l e with center at the origin and of radius r , where r i s any positive number, a r b i t r a r i l y l a r g e . Let m be such that |an| > r for n z m. Then, for n 2 m and for |z| < r , we have 1 - -2- / 0an

Hence, log (1 — i s analytic for |z| < r . Taking that branch of the logarithm which i s zero an for z = 0, we have for |zI < r , Log (1 —

an

=

—~ ~ a W an an

- |

an

Then, for |z| < r , 1 _ JL an

e

3n

^

3

Thus, representing by v n the nth factor of the product (1), we have for |z| < r , vn Now, for |z|
r ,

I ± (JL)P* Pn a n

+

1—_ Pn + 1

an

+ . . . | = |-S-|Fn . \JL an Pn i l-^l n -2

(Note:

P n

1-5-1 a_

• (1

+

+

_ J L _ JL Pn + 1 a n 1

1 2

+

Pn -

"We have used the f a c t that pn 2 1 and that |-2-| < 1 . ) a_ 2

Hence, for |z|


|an| =

In this sequence, multiple poles are not to be repeated.

That is, even if there is a pole of higher order than the first at a n , we write a n only once. At the pole a n , f(z) has a Laurent expansion c

J

n'

n

n (z - a ),Pn

.

-Pn*1' n -rr /_ - a.j »P (z '

+•••+ —

+ c0

+ c1

(z - an) +...,

which is valid inside of a circle about a n on the circumference of which lies that pole of f(z), distinct from a n , which is closest to a n .

161

Let G n denote the principal part of the above expansion. That is, (2)

Gn - , C - V p + ...+ (z - a n ) n

Suppose now that given a sequence (1) and any system of functions G n (each G n arbitrary), we desire to construct a meromorphic function f(z) which is analytic except at the points a n , and which has, at each a n , a pole of order p n with principal part G n . A first trial would perhaps be to form the series (3)

Gj + G 2 + ... + G n +....

Suppose that, given a circle of any radius r > 0, where r is arbitrarily large, the series obtained from (3) by rejecting those terms for which |an| < r converges uniformly for |z| < r. For instance, suppose that for n > m we have |an| g r and that the series G

(4)

m+1 +

G

m+2

converges uniformly for |z| < r. Then the series (4) gives a function analytic for |z| g r, so that the series (3) is meromorphic for |z| < r, with a pole at each a n for which n < m, the principal part at a n being G n . As it was understood that r may be taken arbitrarily large, the series (3) gives a function such as we seek. But, as a rule, (3) will not have the convergence properties exacted above. We proceed to treat this difficulty. We assume that at f 0. It will be easy to see that this involves no loss of generality. The function G n is analytic for |z| < |an|. Hence, it has a Taylor expansion (5)

d0>n

+

d1>n z

d p n zP

valid for |z| < |an|. Let

be any sequence of positive numbers such that e4 + e2 +...+ e n +... converges. Since (5) converges uniformly to Gn(z) in any circle interior to the circle |z| = |an|, we can find a segment of the series (5), that is, a polynomial = dO,n

+ d

l,n Z

n

^

such that |Gn(z) - Pn(z)| < e n for \z\
0, Jet the positive integer m be such that |an| > 2r for n > m. Then G^ - P , with n > m, is analytic for |z| < 2r. Furthermore, for |z| < r, we will have, since r


that is, by g(z) = >/ |z | (cos 1 amp z + i sin i amp z) is continuous on C and its square is z. The function -g(z) also answers our requirements. It is easy to prove that g(z) and -g(z) are the only two functions continuous on C, whose squares are z. Suppose that C is closed, that is, that b coincides with a. If the variation of amp z along C is an even multiple of 2K, that is, speaking geometrically, if C makes an even number of turns about the origin, the value of g(z) at b (the terminal point of C) is the same as the value at a (initial point of C). For 5 amp z will increase by a multiple of 2rc, so that e i i amp z w i H n o f change. On the other hand, if the variation of amp z is an odd multiple of 2n, then 5 amp z will increase by an odd multiple of 71, so that the value of g(z) at b is the negative of the value of a. Speaking intuitively, i f v T i s followed along a closed curve, starting from some point on the curve, one returns to the starting point with the same value of s/z if the curve winds about the origin an even number of' times, and one returns with the negative of the original value if the curve makes an odd number of turns about the origin. 5. The foregoing results tell, in an arithmetic way, everything which is essential about the behavior of Vz. We shall now introduce, in an intuitive way, the idea of Riemann Surface, which will furnish us with an excellent physical picture of the behavior of \fz~. Let a plane, representing the complex plane, be cut along the positive real axis. That is, let the positive real axis be deleted from the plane. There is left an open region in which both branches of-Jz are analytic. Now suppose that we have two planes like the above cut along the positive real axis. Suppose that we label the points of one of the dissected planes with the values of one of the branches of sfz and label the points of the second dissected plane with the valiies of the other branch.

(cuf) 0

3

As to the cut itself, we can imagine any point of it as belonging either to the upper edge of the cut or to the lower edge. (This is intuitive. The arithmetic has been disposed of in §§1-4.) In each plane, we attribute to -Jz, at a point on the upper edge of the cut, that value which is the limit approached by \fz when one approaches the point on the cut from the upper half of the plane. For the lower edge of a cut, one proceeds similarly. The two values of a branch ofv/z at coincident points on opposite sides of a cut are negatives of each other. 175

Now we take, the two dissected planes and fuse the upper edge of the cut for one plane to the lower edge of the cut for the second plane. We fuse the lower edge of the first plane with the upper edge of the second. We form in this way a two-sheeted surface to each point of which one value of>Jz is attached. It is clear that if we start from any point of the surface, and follow a curve which is closed but winds about the origin, we will return to the same point on the surface if the curve winds about the origin an even number of times, but that we will reach the point on the other sheet of the surface which has the same z as the starting point if one winds about the origin an odd number of times. Thus, the two-sheeted surface, labeled with the values of \fz, gives a clear picture of the behavior of the functions/z. We call the surface the Riemann surface of \fz. 6. For Vz", where n is any positive integer greater than 1, we obtain similarly an n-sheeted surface. THE FUNCTION LOG z 7. The function log z can be treated in the same way. If we follow log z along a curve which winds n times about the origin, in the positive sense, log z increases by 2nni. We therefore take the complex plane and cut it along the positive real axis. In the simply connected open region A which results, log z is analytic.

(cut)

Consider any branch of log z in A. Its two values at points on the cut on opposite edges of the cut differ by 2tu.

Consequently, if we take a second such plane and label its points with values greater by 2rti than the values at corresponding points in the first such plane, and if we fuse the upper edge of the first plane to the lower edge of the second plane, we will have a two-sheeted surface which will give a clear picture of two branches of log z. To visualize all of the branches of log z, we can use an infinite number of planes, each of which we associate with one of the integers 0

"

..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ... Consider the sheet associated with any integer n. We attribute to each point of it, a value of log z which exceeds by 2nr.i the value at the corresponding point of the first sheet. If we fu'se the lower edge of the nth sheet to the upper edge of the (n + l)si sheet, we will have a Riemann surface of an .infinite number of sheets, to each point of which is attached one value of log z. 8. In each of the Riemann surfaces considered above, the point on the surface for which z = 0 is called a branch point of the surface. Also, we adjoin to each of the surfaces an ideal point which is also called a branch point of the surface. The reason for this is that if we move around a large circle with center at the origin, the values of the functions change. (Note that a change results in the case of every circle. The reason we refer to a large circle is that it is the large circles, in other cases, which decide whether or not ® is a branch point.)

176

XLV Analytic Continuation ELEMENTS AND THEIR CONTINUATIONS 1. By an analytic

element,

we shall mean a power series c 0 + Ci (z - a) +...+ c n (z - a ) n + ...,

with a radius of convergence greater than 0. The point a will be called the center ment .

of the ele-

We shall represent the above element by the symbol P(z; a). 2. Consider the analytic function f(z) which the above element defines within the circle of convergence of the element. Let a^ be any point within the circle. We can develop f(z) in powers of z - a', obtaining a second analytic element c'0 + cI (z - a 1 ) +...+ c^(z - a 1 ) n +... whose radius of convergence is at least as great as the distance from a' to the first circle. On the other hand, the second circle may extend outside the first one. But certainly, in the open region which is common to the interiors of the two circles, the power series converge to the same values. We shall call the second analytic element an Immediate.continuation of the first one. Every analytic element has an infinite number of immediate continuations, one for every point within its circle of convergence. let us give an example to show that the second power series may converge for points outside the first circle. Consider the function which has a simple pole for z = (J. The expansion in powers of z - a, for any a / 0, has a radius of convergence |a|. For 1 is analytic inside the circle |z - a| = |a|, on which the origin lies, but is not analytic in any larger circle with a as center. Evidently, if a 1 is any point within the above circle, not on the line joining a to 0, the expansion about a 1 will converge for certain points outside that circle. 3- Each immediate continuation of an analytic element has immediate continuations, which in turn have their immediate continuations, etc. By a chain of analytic elements, we shall mean a finite sequence of elements P(z; a), P(z; a'), P(z; a"),

P(z; a ( n > ) ,

each element after the first being an immediate continuation of its predecessor. We shall call an analytic element P(z; b) a continuation (not necessarily immediate) of an element P(z; a) if there exists a chain whose first element is P(z; a) and whose last element is P(z; b). It is not hard to prove that if P(z; b) is a continuation of P(z; a), then P(z; a) is a continuation of P(z; b).

177

It is easy to see that if P(z; b) and P(z; c) are continuations of P(z; a), they are continuations of each other. THE MONOGENIC ANALYTIC FUNCTION

4. Consider any analytic element P(z; a) and the totality of analytic elements which are continuations of it. Weierstrass called the totality of those elements a monoientc analytic function. Thus an analytic function for Weierstrass was a collection of power series. We observe that any element which is a continuation of P(z; a) determines the same monogenic analytic function which P(z; a) does. This conception of monogenic analytic function is important, because the elements which can be obtained from a given element have properties in common with that element. For instance, if the first element satisfies an algebraic differential equation, every element in the monogenic analytic function which it determines will satisfy the same equation. CONTINUATION ALONG A CURVE

5. Consider a curve C which joins a point a to a point b and an analytic element P(z; a) with center at a. Suppose that there exist n + 1 points on C h 0 = a> h 4 , h 2 , ..., h n = b, h i + 1 coming from a greater value of the parameter than hj^ for i = 0, 1, n-1 and n+1 analytic elements [including P(z; a) = P(z; h0)], P(z; h0), P(z; h j , ..., P(z; h j , each, after the first, an immediate continuation of its predecessor, such that for every i that part of C which joins h. to h, , lies within the circle of convergence of P(z; h.). This chain of elements will be said to continue along C. We are going to prove that any two chains of

which

continue

P(z; a ) aloni

C yield

the

P(z; a)

elements

same element

at

b. That is, the continuation of P(z; a) with center at b is independent of the particular chain used. Proof: Consider any point £ on C and any two chains which continue P(z; a) along C from a to £; We shall prove that if £ is very close to the two chains yield the same element at £.

Take a circle r with center at a and of a radius which is one-third the radius of convergence of P(z; a). Let £ be any point of C within f such that the arc of C which joins a to £ lies within P. Any immediate continuation P(z; a 1 ) of P(z; a), for which a 1 lies within r, will have a circle of convergence which contains the arc of C joining a to £ in its interior. P(z; a 1 ) has the same values, for the neighborhood of £ which P(z; a) does. Any immediate continuation P(z; a") of P(z; a 1 ), for which a" lies in T, will likewise have the same values for the neighborhood of £ as P(z; a') and hence the same values as P(z; a). It is thus clear that an analytic element P(z; £) obtained from any chain of elements with centers within r is merely the immediate continuation of P(z; a) for the point £. Certainly then, any two chains yield the same element at £. Now suppose that there are points £ on C for which two chains yield two distinct elements. Consider those values of the parameter t of the curve which yield such points £ and let T be the greatest lower bound of those values. Let d be the point on C which corresponds to x-

178

Then for any point £ on C between a and d, the r e s u l t of continuation i s unique, whereas, e i t h e r f o r d or for points a r b i t r a r i l y close to d, between d and b, two d i s t i n c t elements can be obtained. Of course, d may coincide with b.

Consider any two chains which terminate with an element having d for center. Suppose that the center which precedes d in the f i r s t chain i s e , while that which precedes d in the second chain i s f . I f e and f coincide, the elements corresponding t o them in the two chains coincide, because e = f is between a and d. Hence the two chains give the same element for d. Suppose that e^and f do not coincide. Let, for instance, e l i e between f and d. Then, because the whole arc id l i e s inside the c i r c l e of convergence of the element with center at f , the element with center a t e is an immediate continuation of that with center a t f . Hence, the two elements P(z; f ) , P(z; e) have the same values for the neighborhood of d. Thus, the e l e ment at d obtained from either of them i s the same. Hence, the process of continuation gives unique r e s u l t s as f a r as d i s concerned. d = b, we are through with the proof.)

(If

Now, consider any point e between d and b such that the arc eie l i e s within the c i r c l e of convergence of P(z; d ) . Consider any element P(z; e ^obtained by continuation from a . We are going to show that P(z; e) i s a continuation along de of P(z; d ) . Consider a chain which continues P(z; a) into P(z; e) and suppose that P(z; d) i s not in the chain. Let the chain be P(z; a i , P(z; k j ,

. . . , P(z; k ^ ) ,

P(z; e ) .

Let d be between and k i + 1 . The entire arc k 1 k i + 1 l i e s within the c i r c l e of convergence of P(z; k i ) . Hence, P(z; d) i s an immediate continuation of P(z; k i ) . Furthermore, the entire arc l i e s within the c i r c l e of convergence of P(z; d ) , because de does. Thus, we can get P(z; k i + 1 ) as an immediate continuation of P(z; d ) . Hence, P(z; e) i s a continuation of P(z; d ) . I t f o l l o w s , as at the beginning of the proof, that i f e i s close t o d, P(z; e) is unique. This contradiction of the assumption that two chains can lead to d i s t i n c t elements a t b proves our statement. SINGULAR

POINTS

6. Let P(z; a) be an analytic element with center at a . Let b be any point of the plane, d i s t i n c t from or coincident with a . Let C be a curve joining a t o b. I f P(z; a) can be continued along C to b, we c a l l b a regular point of the monogenic a n a l y t i c function defined by P(z; a ) , relative to C. I f P(z; a) can be continued along C to points a r b i t r a r i ly close to b, but not to b, we c a l l b a singular point re lative to C.

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Example: Let P(z; a) be the expansion of i about some point a f 0. I f we take b = 0, b i s a singular point r e l a t i v e to any curve j o i n i n g a to b .

I f b i s a singular point r e l a t i v e to C, then, as we carry out the continuation of P(z; a) along C towards b, the r a d i i of convergence of the elements obtained must approach zero. I f P(z; a) cannot be continued along C to points a r b i t r a r i l y c l o s e to b, we make no statement r e l a t i v e to b. Let b be any point of the plane to which P(z; a) can be continued. I f , for every such point b, a l l curves along which P ( z ; a ) can be continued to b y i e l d the same element a t b, we s h a l l say that the a n a l y t i c function determined by P(z; a ) i s uniform or one-valued. I f there i s a point b such that for two c e r t a i n curves which j o i n a to b, we secure d i s t i n c t elements a t b, we s h a l l say that the a n a l y t i c function determined by P ( z ; a) i s multiform or many-valued. Example: Let P(z; a) be the expansion of one of the two branches ofs/% at a point a j 0. I f we continue to b along C , , we get one element of -Jz at b. I f we continue along C 2 , we get the negative of the f i r s t element. SINGULAR

POINTS

OS A CIRCLE

-OF CONVERGENCE

7 - Consider an a n a l y t i c element P(z; a ) , with a f i n i t e radius of convergence. Let C be the c i r c l e of convergence. IVe say that the analytic function determined by P(z; a) has at least one singular point on C relative to a ray emanating from a. Suppose that t h i s i s not s o . Then, for every b of C, we can get a P(z; b) by continuation from a along the radius ab. Cons i d e r the i n t e r i o r s of the c i r c l e s of convergence of the e l e ments P(z; b ) . By B o r e l ' s theorem, i t i s possible to find a f i n i t e number of them which cover C. Evidently the. points i n t e r i o r to the f i n i t e number of c i r c l e s j u s t obtained, and the points i n t e r i o r to C, c o n s t i t u t e an open region, c a l l i t A, to which every point of C i s i n t e r i o r . Consider any point k of A. I t may be that s e v e r a l of the f i n i t e number of elements which we are considering converge a t k. Consider any two such elements. I f they have the point k in common, they have an area in common which includes part of the i n t e r i o r of C. Since the two elements have the same value at any point i n t e r i o r to C, namely, the value of P(z; a ) , they have the same value a t k. Thus, the f i n i t e number of elements P(z; b ) , and P(z; a ) , define a function which is a n a l y t i c in A. Then we can increase the radius of C s l i g h t l y , obtaining a larger c i r c l e c 1 , in which there e x i s t s an a n a l y t i c function coinciding with P(z; a) within C. This c o n t r a d i c t s the f a c t t h a t C i s the c i r c l e of convergence of P ( z ; a) and proves that there i s a singular point on C. I f b i s a singular point on C and i f c i s on the radius j o i n i n g a to b, the c i r c l e of convergence of P(z; c ) i s tangent t o that of P(z; a) a t b . NATURAL BOUNDARIES

8- We s h a l l give an example of a power s e r i e s which cannot be continued outside i t s c i r c l e of convergence. That i s every continuation of the power s e r i e s has a c i r c l e of convergence which i s tangent i n t e r n a l l y to the f i r s t c i r c l e . 180

Consider the series f(z) = 1 + z + z 2 ' + z 3 1 +...+ z n I +... whose radius of convergence is 1. As we approach the point 1 from the left along the real axis, each term z n 1 approaches 1, so that the sum of the series approach It follows that 1 is a singular point of the series. In short, if any immediate continuation of f(z) had 1 inside its circle of convergence, the continuation would be analytic, therefore bounded in the neighborhood of 1 and f(z) would have to be bounded in the neighborhood of 1. Now, let e be any pth root of unity, where p is any positive integer. We have, if O s r < 1, f(re) = 1

+

re

+

(re)21 + (re)31 + ...+ (re)nI +....

When n > p, n! is divisible by p, so that (re)nl = r n I . Thus, as r approaches unity through positive values less than 1, the distant terms of our series will all approach +1. Consequently f(z) cannot be bounded for the neighborhood of e, so that e is a singular point. Now the points e are dense all over the circle of convergence. Hence f(z) is not bounded in the neighborhood of any point on the circle of convergence. Thus, every point on the circle of convergence is a singular point. The circle of convergence of any immediate continuation of f(z) is tangent to that of f(z). When every point on the circle of convergence of the power series is a singular point (radially), we call the circle of convergence a natural boundary of the function.

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