DOCSLIB.ORG

Solutions to Tutorial 6 (Week 7)

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Solutions to Tutorial 6 (Week 7) The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 6 (Week 7) MATH2962: Real and Complex Analysis (Advanced) Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/ Lecturer: Florica C. C^ırstea Questions marked with * are more difficult questions. Material covered (1) power series and radius of convergence. (2) double series and Cauchy products. Outcomes This tutorial helps you to (1) know and be able to apply convergence tests for sequences and series; (2) have a working knowledge of (real and complex) power series and their properties; Summary of essential material N N Let (an) be a sequence in R or C and z 2 R or C.A power series is a series of the form 1 X n anz : (PS) n=0 Cauchy-Hadamard Theorem There exists a number % 2 [0; 1], called the radius of convergence, such that the power series (PS) • converges absolutely if jzj < % (makes only sense if % > 0); • diverges if jzj > % (makes only sense if % < 1). The radius of convergence can be found as follows: (CH1) Derived from root test: 1 (formula is always valid, but the limit superior may % = pn be impossible to compute) lim sup kank n!1 (CH2) Derived from ratio test: 1 ka k % = = lim n (only valid if the limit exists!!) ka k n!1 ka k lim n+1 n+1 n!1 kank (CH3) Find other means of getting convergence for jzj < % and divergence for jzj > %, for instance by substituting certain values of z, or by using known convergence properties of series, for instance, the geometric series. Copyright c 2017 The University of Sydney 1 Some practical considerations • Formula (CH1) does not work well if the coefficients an of the series contain factorials, n! p for instance a = because we cannot easily handle n n!. It is better to use (CH2) in n nn such cases because we can cancel many factors before computing the limit. pn • If we use (CH1), then the n in kank must match the power of z. 1 X Example. To find the radius of convergence for the series 4nz2n, we can proceed in p n=0 p two ways. To use (CH1), we need to look at 2n 4n = 2, and not n 4n = 4. Indeed, we P1 k write the series as k=0 akz , where the coefficients ak for k 2 N are given by ( 4n if k = 2n for some n 2 N, ak = 0 if k = 2n + 1 for some n 2 N: pk Here, the sequence ( jakj)k≥1 has two convergent subsequences: The one with odd indices 2n+1p being identically zero (i.e., limn!1 ja2n+1pj = 0), and the one with even indices having 2pn 2n n a positive limit: limn!1 ja2nj = limn!1 4 = 2. Then pk lim sup jakj = maxf0; 2g = 2 k!1 (the largest between the two accumulation points). Hence, the radius of convergence for 1 1 the given series is ρ = = . Alternatively, we can denote z2 = y and pk lim supk!1 jakj 2 P1 n n find that the radius of convergence for n=0 4 y is 1=4 (using either (CH1) or (CH2)), that is the series converges absolutely for jyj < 1=4 and diverges for jyj > 1=4. This shows P1 n 2n that n=0 4 z converges absolutely for jzj < 1=2 and diverges for jzj > 1=2 (use that 2 P1 n 2n jzj = jyj). Thus the series n=0 4 z has radius of convergence 1=2. Double series theorem N N Suppose that xjk 2 R or C such that m n XX sup kxjkk < 1; m;n2 N j=0 k=0 and that σ : N ! N × N is a bijection. Then the series 1 1 1 1 1 XX XX X xjk ; xjk ; xσ(i) j=0 k=0 k=0 j=0 i=0 converge absolutely, and they all have the same limit. Cauchy products 1 1 X X Suppose that aj and bk are absolutely convergent series in R or C. Then, j=0 k=0 1 1 1 n X X X X aj bk = cn; where cn := akbn−k: j=0 k=0 n=0 k=0 2 Questions to complete during the tutorial P1 n 1. It is given that the power series n=0 anz converges for z = 1 + i and diverges for z = 3 + 4i. Find bounds for its radius of convergence %. P1 n Solution: By the Cauchy{Hadamard theorem a power series n=0 anz converges for jzj < %, and diverges for jzj > %. Since, by assumption, the power series converges for z = 1 + ip, it follows that 1 + i is eitherp inside the ball B(0;%) or on its boundary. As j1 + ij = 2, we conclude that % ≥ 2. We are told that the power series diverges for z = 3 + 4i. This means that 3 + 4i cannot belong to B(0;%). Hence, % ≤ j3 + 4ij = 5. Consequently, we have p 2 ≤ % ≤ 5: 2. Determine the radius of convergence % of the following power series. 1 X zn (a) , p > 0; np n=1 p Solution: In the given series an = 1=n for n ≥ 1. We have that ja j n p α = lim n+1 = lim = 1: n!1 janj n!1 n + 1 Hence, by formula (CH2) the radius of convergence for the given series is 1=1 = 1. 1 X p (b) n(1 + i 3)nzn; n=0 Solution: We use formula (CH1) to compute %. We have q p p p p p p n jn(1 + i 3)nj = n nj1 + i 3j = n n 1 + 3 = 2 n n ! 2 as n ! 1. Hence from (CH1) we get 1 % = : 2 If the coefficients in the series are sign changing or complex as in this example, remember to take absolute values! 1 X n3 (c) zn; (1 + i)n n=0 Solution: We use formula (CH1) to compute %. We have s p p n3 n n)3 n n)3 1 n = = p ! p (1 + i)n j1 + ij 2 2 as n ! 1. Hence from (CH1) we get 1 p % = p = 2: 1= 2 If the coefficients in the series are sign changing or complex as in this example, remember to take absolute values! 3 1 X nn (d) zn; n! n=1 nn Solution: We denote a = for n ≥ 1. Because of the n! in a we compute n n! n the radius of convergence using formula (CH2). We have n+1 n n jan+1j (n + 1) n! n + 1 1 lim = lim · n = lim = lim 1 + = e: n!1 janj n!1 (n + 1)! n n!1 n n!1 n Hence by formula (CH2), the radius of convergence is % = 1=e. 1 X (e) (log n)zn; n=2 Solution: For n ≥ 2 we have p pn 1 ≤ log n ≤ n n: p p As n n ! 1 as n ! 1, the squeeze law implies that n log n ! 1 as n ! 1. Hence formula (CH1) gives % = 1=1 = 1. 1 X (f) n1=nzn; n=1 Solution: We compute the radius of convergence for the power series using formula . For n ≥ 1 we have 1 ≤ n1=n ≤ n. Hence, for all n ≥ 1 p p 1 ≤ n n1=n ≤ n n p p Because n n ! 1 as n ! 1, the squeeze law implies that n n1=n ! 1 as n ! 1. Hence, by formula (CH1) the radius of convergence is % = 1=1 = 1. 1 X n2 (g) z3n; 2n n=0 k Solution: This is a power series where the coefficient ak of z is given by 8n2 < n if k = 3n and n 2 N; ak := 2 :0 otherwise: We find the radius of convergence using formula (CH1). We cannot directly use (CH2) because we would need to divide by zero when taking the ratio of successive terms. pk When k 2 N is not a multiple of 3, then jakj = 0. If k 2 N is a multiple of 3, that is k = 3n, then r p n2 ( n n)2=3 1 pk ja j = 3pn ja j = 3n = ! k 3n 2n 21=3 21=3 pk 1 as k = 3n ! 1. Hence, lim sup jakj = 1=3 . Then, by formula (CH1), the k!1 2 1 radius of convergence is % = = 21=3. pk lim supk!1 jakj 4 An alternative approach is to consider the series as a power series in the new variable y = z3. Then, 1 1 X n2 X n2 z3n = yn: 2n 2n n=0 n=0 For the latter series we can find the radius of convergence using (CH1) or (CH2). Using either formula we get a radius of 2, that is, the series for y converges if jyj < 2 and diverges if jyj > 2. Since jzj3 = jyj, we see that the original series converges when jzj < 21=3 and diverges when jzj > 21=3. Again we see that the 1 X n2 radius of convergence for z3n is 21=3. 2n n=0 z z2 z3 z4 z5 *(h)1 − + − + − + ··· . 2 32 23 34 25 1 X 1 Solution: This is a power series a zn, where a is given by if n 2 is n n 3n N n=0 1 even, and − if n 2 is odd. It follows that 2n N 8r 1 1 > n = if n ≥ 0 is even <> n pn ja j = 3 3 n r 1 1 > n = if n ≥ 1 is odd: : 2n 2 pn Hence, lim supn!1 janj = 1=2 since the limit superior is the largest accumulation point.
Load more

Recommended publications
  • Arxiv:1207.1472V2 [Math.CV]
    SOME SIMPLIFICATIONS IN THE PRESENTATIONS OF COMPLEX POWER SERIES AND UNORDERED SUMS OSWALDO RIO BRANCO DE OLIVEIRA Abstract. This text provides very easy and short proofs of some basic prop- erties of complex power series (addition, subtraction, multiplication, division, rearrangement, composition, differentiation, uniqueness, Taylor’s series, Prin- ciple of Identity, Principle of Isolated Zeros, and Binomial Series). This is done by simplifying the usual presentation of unordered sums of a (countable) family of complex numbers. All the proofs avoid formal power series, double series, iterated series, partial series, asymptotic arguments, complex integra- tion theory, and uniform continuity. The use of function continuity as well as epsilons and deltas is kept to a mininum. Mathematics Subject Classification: 30B10, 40B05, 40C15, 40-01, 97I30, 97I80 Key words and phrases: Power Series, Multiple Sequences, Series, Summability, Complex Analysis, Functions of a Complex Variable. Contents 1. Introduction 1 2. Preliminaries 2 3. Absolutely Convergent Series and Commutativity 3 4. Unordered Countable Sums and Commutativity 5 5. Unordered Countable Sums and Associativity. 9 6. Sum of a Double Sequence and The Cauchy Product 10 7. Power Series - Algebraic Properties 11 8. Power Series - Analytic Properties 14 References 17 arXiv:1207.1472v2 [math.CV] 27 Jul 2012 1. Introduction The objective of this work is to provide a simplification of the theory of un- ordered sums of a family of complex numbers (in particular, for a countable family of complex numbers) as well as very easy proofs of basic operations and properties concerning complex power series, such as addition, scalar multiplication, multipli- cation, division, rearrangement, composition, differentiation (see Apostol [2] and Vyborny [21]), Taylor’s formula, principle of isolated zeros, uniqueness, principle of identity, and binomial series.
    [Show full text]
  • Formal Power Series - Wikipedia, the Free Encyclopedia
    Formal power series - Wikipedia, the free encyclopedia http://en.wikipedia.org/wiki/Formal_power_series Formal power series From Wikipedia, the free encyclopedia In mathematics, formal power series are a generalization of polynomials as formal objects, where the number of terms is allowed to be infinite; this implies giving up the possibility to substitute arbitrary values for indeterminates. This perspective contrasts with that of power series, whose variables designate numerical values, and which series therefore only have a definite value if convergence can be established. Formal power series are often used merely to represent the whole collection of their coefficients. In combinatorics, they provide representations of numerical sequences and of multisets, and for instance allow giving concise expressions for recursively defined sequences regardless of whether the recursion can be explicitly solved; this is known as the method of generating functions. Contents 1 Introduction 2 The ring of formal power series 2.1 Definition of the formal power series ring 2.1.1 Ring structure 2.1.2 Topological structure 2.1.3 Alternative topologies 2.2 Universal property 3 Operations on formal power series 3.1 Multiplying series 3.2 Power series raised to powers 3.3 Inverting series 3.4 Dividing series 3.5 Extracting coefficients 3.6 Composition of series 3.6.1 Example 3.7 Composition inverse 3.8 Formal differentiation of series 4 Properties 4.1 Algebraic properties of the formal power series ring 4.2 Topological properties of the formal power series
    [Show full text]
  • Math 263A Notes: Algebraic Combinatorics and Symmetric Functions
    MATH 263A NOTES: ALGEBRAIC COMBINATORICS AND SYMMETRIC FUNCTIONS AARON LANDESMAN CONTENTS 1. Introduction 4 2. 10/26/16 5 2.1. Logistics 5 2.2. Overview 5 2.3. Down to Math 5 2.4. Partitions 6 2.5. Partial Orders 7 2.6. Monomial Symmetric Functions 7 2.7. Elementary symmetric functions 8 2.8. Course Outline 8 3. 9/28/16 9 3.1. Elementary symmetric functions eλ 9 3.2. Homogeneous symmetric functions, hλ 10 3.3. Power sums pλ 12 4. 9/30/16 14 5. 10/3/16 20 5.1. Expected Number of Fixed Points 20 5.2. Random Matrix Groups 22 5.3. Schur Functions 23 6. 10/5/16 24 6.1. Review 24 6.2. Schur Basis 24 6.3. Hall Inner product 27 7. 10/7/16 29 7.1. Basic properties of the Cauchy product 29 7.2. Discussion of the Cauchy product and related formulas 30 8. 10/10/16 32 8.1. Finishing up last class 32 8.2. Skew-Schur Functions 33 8.3. Jacobi-Trudi 36 9. 10/12/16 37 1 2 AARON LANDESMAN 9.1. Eigenvalues of unitary matrices 37 9.2. Application 39 9.3. Strong Szego limit theorem 40 10. 10/14/16 41 10.1. Background on Tableau 43 10.2. KOSKA Numbers 44 11. 10/17/16 45 11.1. Relations of skew-Schur functions to other fields 45 11.2. Characters of the symmetric group 46 12. 10/19/16 49 13. 10/21/16 55 13.1.
    [Show full text]
  • Complex Function Theory
    Complex Function Theory Second Edition Donald Sarason AMERICAN MATHEMATICAL SOCIETY http://dx.doi.org/10.1090/mbk/049 Complex Function Theory Second Edition Donald Sarason AMERICAN MATHEMATICAL SOCIETY 2000 Mathematics Subject Classification. Primary 30–01. Front Cover: The figure on the front cover is courtesy of Andrew D. Hwang. The Hindustan Book Agency has the rights to distribute this book in India, Bangladesh, Bhutan, Nepal, Pakistan, Sri Lanka, and the Maldives. For additional information and updates on this book, visit www.ams.org/bookpages/mbk-49 Library of Congress Cataloging-in-Publication Data Sarason, Donald. Complex function theory / Donald Sarason. — 2nd ed. p. cm. Includes index. ISBN-13: 978-0-8218-4428-1 (alk. paper) ISBN-10: 0-8218-4428-8 (alk. paper) 1. Functions of complex variables. I. Title. QA331.7.S27 2007 515.9—dc22 2007060552 Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to [email protected]. c 2007 by the American Mathematical Society. All rights reserved.
    [Show full text]
  • A Certain Integral-Recurrence Equation with Discrete-Continuous Auto-Convolution
    ARCHIVUM MATHEMATICUM (BRNO) Tomus 47 (2011), 245–250 A CERTAIN INTEGRAL-RECURRENCE EQUATION WITH DISCRETE-CONTINUOUS AUTO-CONVOLUTION Mircea I. Cîrnu Abstract. Laplace transform and some of the author’s previous results about first order differential-recurrence equations with discrete auto-convolution are used to solve a new type of non-linear quadratic integral equation. This paper continues the author’s work from other articles in which are considered and solved new types of algebraic-differential or integral equations. 1. Introduction In the earlier paper [4], N. M. Flaisher solved by Fourier transform method a second order differential-recurrence equation. The present author used in [2] the Laplace transform to derive Newton’s formulas about the sums of powers of the roots of a polynomial. In this paper, the Laplace transform will be used to solve an integral-recurrence equation on semi-axis, with discrete-continuous auto-convolution of its unknowns. Namely, applying the Laplace transform on considered equation, we obtain for the transforms of unknowns a first order differential-recurrence equation with discrete auto-convolution of the type studied in [3] and [1]. Using for this equation the general theory given in [3], we find the transforms of unknowns in convenient assumptions, the solutions of the initial equation being obtained by inverse Laplace transform. 2. Convolution products Two convolution products have been imposed over the time. The first, in continuous-variable case, is the bilateral convolution of two integrable functions u(x) and v(x) on real axis, given by formula Z ∞ u(x) ? v(x) = u(t)v(x − t) dt , −∞ 2010 Mathematics Subject Classification: primary 45G10; secondary 44A10.
    [Show full text]
  • M 3001 Assignment 5: Solution
    M 3001 ASSIGNMENT 5: SOLUTION Due in class: Wednesday, Feb. 16, 2011 Name M.U.N. Number P∞ P∞ P∞ 1. Suppose n=1 an = ∞. Let n=1 bn be a regrouping of n=1 an formed by inserting paren- P∞ theses. Prove that n=1 bn = ∞. P∞ P∞ Solution. Since n=1 bn is a regrouping of n=1 an, there is an increasing function φ : N → N such that bn = aφ(n). Note that for any M > 0 there is an N ∈ N such that n > N implies Pn Pn Pφ(n) Pn P∞ k=1 ak > M. So k=1 bk = k=1 ak ≥ k=1 ak > M. That is to say, n=1 bn = ∞. P∞ P∞ P∞ 2. If n=1 an converges absolutely and n=1 bn is a rearrangement of n=1 an, prove that P∞ P∞ n=1 |bn| = n=1 |an|. P∞ P∞ P∞ Solution. Since n=1 |an| converges and n=1 |bn| is a rearrangement of n=1 |an|. By the P∞ P∞ theorem on series rearrangement proved in Section 2.3, n=1 |bn| equals n=1 |an|. 3. Give an example showing that a rearrangement of a divergent series may diverge or converge. Solution. Since a series is a rearrangement of itself, a divergent series is its own divergent P∞ rearrangement. Suppose n=1 an is conditionally convergent. Then there is an arrangement P∞ P∞ P∞ n=1 bn = ∞ of n=1 an. In fact, let p1, p2, ... be the nonnegative terms of n=1 an in the P∞ order in which they occur; let −q1, −q2, ..
    [Show full text]
  • Generating Functions
    MATH 324 Summer 2012 Elementary Number Theory Notes on Generating Functions Generating Functions In this note we will discuss the ordinary generating function of a sequence of real numbers, and then find the generating function for some sequences, including the sequence of Mersenne numbers and the sequence of Fibonacci numbers. Definition. Given a sequence ak k≥0 of real numbers, the ordinary generating function, or generating f g function, of the sequence ak k≥0 is defined to be f g 1 k 2 G(x) = akx = a0 + a1x + a2x + : ( ) k=0 · · · ∗ X The sum is finite if the sequence is finite, and infinite if the sequence is infinite. In the latter case, we will think of ( ) as either a power series with x chosen so that the sum converges, or as a formal power series, where we ∗are not concerned with convergence. Example 1. Suppose that n is a fixed positive integer and n ak = k for k = 0; 1; : : : ; n: From the binomial theorem we have n n n n n n (1 + x) = + x + x2 + + x ; 0 1 2 · · · n n and the generating function for the sequence ak k is f g =0 n G(x) = (1 + x) : Note: The advantage of what we have done is that we have expressed G(x) in a very simple encoded form. If we know this encoded form for G(x); it is now easy to derive ak by decoding or expanding G(x) as a k power series and finding the coefficient of x : Often we will be able to find G(x) without knowing ak; and then be able to solve for ak by expanding G(x): Example 2.
    [Show full text]
  • A Power Series Centered at Z0 ∈ C Is an Expansion of the Form ∞ X N An(Z − Z0) , N=0 Where An, Z ∈ C
    LECTURE-5 : POWER SERIES VED V. DATAR∗ A power series centered at z0 2 C is an expansion of the form 1 X n an(z − z0) ; n=0 where an; z 2 C. If an and z are restricted to be real numbers, this is the usual power series that you are already familiar with. A priori it is only a formal expression. But for certain values of z, lying in the so called disc of convergence, this series actually converges, and the power series represents a function of z. Before we discuss this fundamental theorem of power series, let us review some basic facts about complex series, and series of complex valued functions. Series of complex numbers and complex valued functions A series is an infinite sum of the form 1 X an; n=0 where an 2 C for all n. We say that the series converges to S, and write 1 X an = S; n=0 if the sequence of partial sums N X SN = an n=0 converges to S. We need the following basic fact. P1 P1 Proposition 0.1. If n=0 janj converges then n=0 an converges. P1 Proof. We will show that SN forms a Cauchy sequence if n=0 janj con- verges. Then the theorem will follow from the completeness of C. If we let PN TN = n=0 janj, then fTN g is a Cauchy sequence by the hypothesis. Then by triangle inequality M M X X jSN − SM j = an ≤ janj = jTM − TN j: n=N+1 n=N+1 Date: 24 August 2016.
    [Show full text]
  • Course 214 Section 2: Infinite Series Second Semester 2008
    Course 214 Section 2: Infinite Series Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins 1989–2008 Contents 2 Infinite Series 25 2.1 The Comparison Test and Ratio Test . 26 2.2 Absolute Convergence . 28 2.3 The Cauchy Product of Infinite Series . 28 2.4 Uniform Convergence for Infinite Series . 30 2.5 Power Series . 31 2.6 The Exponential Function . 32 i 2 Infinite Series An infinite series is the formal sum of the form a1 + a2 + a3 + ···, where each +∞ P number an is real or complex. Such a formal sum is also denoted by an. n=1 +∞ P Sometimes it is appropriate to consider infinite series an of the form n=m am + am+1 + am+2 + ···, where m ∈ Z. Clearly results for such sequences may be deduced immediately from corresponding results in the case m = 1. +∞ P Definition An infinite series an is said to converge to some complex n=1 number s if and only if, given any ε > 0, there exists some natural number N m X such that an − s < ε for all natural numbers m satisfying m ≥ N. If n=1 +∞ +∞ P P the infinite series an converges to s then we write an = s. An infinite n=1 n=1 series is said to be divergent if it is not convergent. For each natural number m, the mth partial sum sm of the infinite se- +∞ +∞ P P ries an is given by sm = a1 + a2 + ··· + am. Note that an converges n=1 n=1 to some complex number s if and only if sm → s as m → +∞.
    [Show full text]
  • Chapter 10: Power Series
    Chapter 10 Power Series In discussing power series it is good to recall a nursery rhyme: \There was a little girl Who had a little curl Right in the middle of her forehead When she was good She was very, very good But when she was bad She was horrid." (Robert Strichartz [14]) Power series are one of the most useful type of series in analysis. For example, we can use them to define transcendental functions such as the exponential and trigonometric functions (as well as many other less familiar functions). 10.1. Introduction A power series (centered at 0) is a series of the form 1 X n 2 n anx = a0 + a1x + a2x + ··· + anx + :::: n=0 where the constants an are some coefficients. If all but finitely many of the an are zero, then the power series is a polynomial function, but if infinitely many of the an are nonzero, then we need to consider the convergence of the power series. The basic facts are these: Every power series has a radius of convergence 0 ≤ R ≤ 1, which depends on the coefficients an. The power series converges absolutely in jxj < R and diverges in jxj > R. Moreover, the convergence is uniform on every interval jxj < ρ where 0 ≤ ρ < R. If R > 0, then the sum of the power series is infinitely differentiable in jxj < R, and its derivatives are given by differentiating the original power series term-by-term. 181 182 10. Power Series Power series work just as well for complex numbers as real numbers, and are in fact best viewed from that perspective.
    [Show full text]
  • Homework 4 Solutions 26.5. Let ∑ N=1 an and ∑ N=1 Bn Be Absolutely
    Homework 4 Solutions Math 171, Spring 2010 Please send corrections to [email protected] P1 P1 P1 p 26.5. Let n=1 an and n=1 bn be absolutely convergent series. Prove that the series n=1 janbnj converges. P1 P1 P1 Solution. Since n=1 janj and n=1 jbnj converge, by Theorem 23.1, n=1 janj + jbnj converges. Note that for all n, we have p p p janbnj = janjjbnj ≤ (janj + jbnj)(janj + jbnj) = janj + jbnj: P1 So by Theorem 26.3 (Comparison Test), since n=1 janj + jbnj converges absolutely, we see that P1 p n=1 janbnj converges absolutely, and hence converges as all terms are positive. 27.1. Find the radius of convergence R of each of the power series. Discuss the convergence of the power series at the points jx − tj = R. Solution. P1 x2n−1 (a) n=1 (2n−1)! Note that x2(n+1)−1 x2 lim (2(n+1)−1)! = lim = 0 n!1 x2n−1 n!1 (2n + 1)(2n) (2n−1)! so by the ratio test, the series converges absolutely for all x, and so R = 1. P1 n(x−1)n (b) n=1 2n Note that (n+1)(x−1)(n+1) (n+1) n + 1 x − 1 x − 1 lim 2 = lim = n!1 n(x−1)n n!1 n 2 2 2n so by the ratio test, the series converges absolutely if jx − 1j < 2 and diverges if jx − 1j > 2. Thus P1 P1 n R = 2. If jx − 1j = 2 then the power series diverges, since n=1 n and n=1(−1) n diverge.
    [Show full text]
  • Arxiv:1311.6710V1 [Math.CA] 25 Nov 2013 Oetpc Eae Ora,Complex, Real, to Related Topics Some N Ore Analysis Fourier and Tpe Semmes Stephen Ieuniversity Rice Preface
    Some topics related to real, complex, and Fourier analysis Stephen Semmes Rice University arXiv:1311.6710v1 [math.CA] 25 Nov 2013 Preface Here we look at some situations that are like the unit circle or the real line in some ways, but which can be more complicated or fractal in other ways. ii Contents 1 The unit circle 1 1.1 Integrablefunctions ......................... 1 1.2 AbelsumsandCauchyproducts . 3 1.3 ThePoissonkernel .......................... 5 1.4 Continuousfunctions......................... 6 1.5 Lp Functions ............................. 8 1.6 Harmonicfunctions.......................... 10 1.7 ComplexBorelmeasures. 11 1.8 Holomorphicfunctions .. .. .. .. .. .. .. .. .. .. 15 1.9 Productsoffunctions. .. .. .. .. .. .. .. .. .. .. 16 1.10 Representationtheorems. 20 2 The real line 22 2.1 Fouriertransforms .......................... 22 2.2 Holomorphicextensions . .. .. .. .. .. .. .. .. .. 24 2.3 Someexamples ............................ 25 2.4 The multiplication formula . 27 2.5 Convergence.............................. 29 2.6 Harmonicextensions ......................... 31 2.7 Boundedharmonicfunctions . 33 2.8 Cauchy’stheorem........................... 35 2.9 Cauchy’sintegralformula . 37 2.10Somevariants............................. 39 3 Commutativetopologicalgroups 42 3.1 Basicnotions ............................. 42 3.2 Haarmeasure............................. 44 3.3 Directsumsandproducts . 47 3.4 Continuoushomomorphisms. 49 3.5 Sumsandproducts,continued. 52 3.6 Spacesofcontinuousfunctions . 54 3.7 Thedualgroup...........................
    [Show full text]