THE JULIA AND MANDELBROT SETS FOR THE HURWITZ ZETA ...

THE JULIA AND MANDELBROT SETS FOR THE HURWITZ ZETAFUNCTIONLarry L. Tingen, Jr.A Thesis Submitted to theUniversity of North Carolina Wilmington in Partial Fulfillmentof the Requirements for the Degree ofMaster of ArtsDepartment of Mathematics and StatisticsUniversity of North Carolina Wilmington2009Approved byAdvisory CommitteeGabriel LugoMichael FreezeRussell HermanChairAccepted byDean, Graduate School

This thesis has been prepared in the style and formatConsistent with the journalAmerican Mathematical Monthly.

TABLE OF CONTENTSABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vDEDICATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . viiLIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .viiiLIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiLIST OF SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .xii1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 JULIA AND MANDELBROT SETS . . . . . . . . . . . . . . . . . . 42.1 Cayley’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Julia Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Mandelbrot Sets . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Previous Research . . . . . . . . . . . . . . . . . . . . . . . . 313 INTRODUCTION TO ZETA FUNCTIONS . . . . . . . . . . . . . . 333.1 History of the Riemann Zeta Function . . . . . . . . . . . . . 333.2 Adolf Hurwitz and His Work . . . . . . . . . . . . . . . . . . . 343.3 Riemann Zeta Function and Dirichlet L−Series . . . . . . . . 354 THE HURWITZ ZETA FUNCTION . . . . . . . . . . . . . . . . . . 414.1 The Gamma and Hurwitz Zeta Functions . . . . . . . . . . . . 414.2 Contour Integral Representation . . . . . . . . . . . . . . . . . 474.3 An Analytic Continuation of the Hurwitz Zeta Function . . . 514.4 A Functional Equation for the Hurwitz Zeta Function . . . . . 545 CALCULATING THE HURWITZ ZETA FUNCTION . . . . . . . . 645.1 Euler-MacLaurin Summation Formula . . . . . . . . . . . . . 645.2 Using Euler-Maclaurin Summation Formula on the HurwitzZeta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 74iii

6 JULIA AND MANDELBROT SETS OF ZETA FUNCTIONS . . . . 766.1 Images of the Hurwitz and Riemann Zeta Functions . . . . . . 776.2 Julia and Mandelbrot Sets of ζ H (s, a) + c . . . . . . . . . . . . 796.3 Values of the Hurwitz and Riemann Zeta Functions . . . . . . 967 RESULTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007.2 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1017.3 Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . 1017.4 Closing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1038 APPENDIX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048.1 Julia Sets of f c (z) . . . . . . . . . . . . . . . . . . . . . . . . . 1048.2 Maple c○ Program for the Evaluation of the Functional Equationof the Hurwitz Zeta Function . . . . . . . . . . . . . . . . 1088.3 FractInt 20.4 c○ Formula File for Hurwitz Zeta function withdifferent a−values . . . . . . . . . . . . . . . . . . . . . . . . . 1098.4 FractInt 20.4 c○ Parameter File for Hurwitz Zeta function withdifferent a−values . . . . . . . . . . . . . . . . . . . . . . . . . 114REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116iv

ABSTRACTThis work explores the Julia and Mandelbrot sets of the Hurwitz zeta function byextending the Hurwitz zeta to the entire complex plane through analytic continuationand functional equations. Values for the Hurwitz zeta function are calculated usingthe Euler-Maclaurin summation formula.Computations are made using Maple c○and images are created using FractInt 20.4 c○ . Using a functional equation of theHurwitz zeta function and the Euler-Maclaurin summation formula, we created over1,500 images of the Julia and Mandelbrot sets of the Hurwitz zeta function.v

DEDICATIONTo Mandee: It has been a long road, but we made it!vi

ACKNOWLEDGMENTSI first and foremost acknowledge my tireless, whip-cracking advisor Dr. RussellHerman. It was because of him and his infinite patience with me that I was able tocomplete this thesis, and for that I thank him from the bottom of my heart. Also,I want to thank the other members of my Committee, Dr. Gabriel Lugo and Dr.Michael Freeze. They showed me that just when you think you are done, you arejust beginning. Their advice and suggestions forced me to take my work to the nextlevel, and made sure that my final product was of the highest-quality.I also want to acknowledge all those people who asked me what my thesis wasabout and how it was going. Thank you for driving me to finish so I could stopanswering the question, “So what’s your thesis about?”vii

LIST OF TABLES1 Comparison of computed ζ H (s, 1) versus Maple c○ ζ R (s) . . . . . . . . 972 Comparison of computed ζ H (s, 1) versus Maple c○ ζ R (s) . . . . . . . . 983 Comparison of computed ζ H (s, 1/8) versus Maple c○ ζ H (s, 1/8) . . . . 99viii

LIST OF FIGURES1 Newton’s method for z 3 − 1 = 0 . . . . . . . . . . . . . . . . . . . . . 72 Julia set of z 3 − 1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . 83 The first drawing by Cremer in 1925 visualizing a Julia set . . . . . . 94 Julia Set of f c (z) = z 2 + c, c = 0 . . . . . . . . . . . . . . . . . . . . . 135 Stereographic projection of S onto C . . . . . . . . . . . . . . . . . . 146 Riemann Sphere illustrating basins of attraction at 0 and ˆ∞ . . . . . 157 Julia Set of f c (z) = z 2 + c for c = −0.70176 − 0.3842i . . . . . . . . . 158 Constructing the the Cantor set . . . . . . . . . . . . . . . . . . . . . 169 The Koch curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1610 Julia Set of f(z) = z 2 + c for c = 1.0 + 0.0i . . . . . . . . . . . . . . . 1811 Julia Set of f c (z) = z 2 + c for c = 0.45 + 0.1428i. . . . . . . . . . . . . 2012 Julia Set of f c (z) = z 2 + c for c = 0.285 + 0.01i. . . . . . . . . . . . . 2013 Julia Set of f c (z) = z 2 + c for c = 0.835 − 0.2321i. . . . . . . . . . . . 2114 Mandelbrot Set of f c (z) = z 2 + c . . . . . . . . . . . . . . . . . . . . 2215 Zoomed in top of Mandelbrot set in Figure 14 . . . . . . . . . . . . . 2316 Zoomed in on border of Mandelbrot set in Figure 14 . . . . . . . . . . 2417 A Mandelbrot set and the corresponding Julia sets . . . . . . . . . . 3018 Julia sets of f c (z), c ∈ [−2, 2] × [−2, 2] . . . . . . . . . . . . . . . . . . 3119 Contour integral used for representing ζ H (s, a) . . . . . . . . . . . . 4720 The punctured rectangle . . . . . . . . . . . . . . . . . . . . . . . . . 5521 Punctured rectangles making the infinite strip Q n (r) . . . . . . . . . 5622 Contour C(N) referred to in Equation (4.4.7) . . . . . . . . . . . . . 5823 Graph of the function [x] − x, [x] is the greatest integer function . . 6724 Graph of the function [x] − x + 1/2 . . . . . . . . . . . . . . . . . . . 6725 Graph of the equation x − [x] − 1/2 . . . . . . . . . . . . . . . . . . . 68ix

26 Julia set of ζ R (s), s ∈ [0.9048, 1.8314] × [−0.347446, 0.347446] . . . . 7727 Julia set of ζ H (s, 1), s ∈ [0.9048, 1.8314] × [−0.347446, 0.347446] . . . 7728 Julia set of ζ R (s), s ∈ [−23.53, 11.33] × [−13.07, 13.07] . . . . . . . . 7829 Julia set ζ H (s, 1), s ∈ [−23.53, 11.33] × [−13.07, 13.07] . . . . . . . . 7830 Julia set of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . . 8031 Julia sets of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . 8132 Julia sets of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . 8233 Julia sets of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . 8334 Julia sets of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . 8435 Julia sets of ζ H (s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . 8536 Julia sets of ζ H (s, 1/8) + c, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . . . . 8637 Julia sets of ζ H (s, 1/7) + c, s ∈ [−2, 2] × [−1.5i, 1.5i] . . . . . . . . . . 8738 M of ζ H (s, 1/8) + c, c ∈ [−18, 6] × [−8i, 8i] . . . . . . . . . . . . . . . 8839 M of ζ H (s, 1/7) + c, c ∈ [−18, 6] × [−8i, 8i] . . . . . . . . . . . . . . . 8840 M of ζ H (s, 1/7) + c, c ∈ [−17.22, −13.22] × [−3.87i, 0.87i] . . . . . . 8941 M of ζ H (s, 1/7) + c, c ∈ [−10.27, −5.58] × [−3.83i, 6.96i] . . . . . . . 8942 Mandelbrot set of ζ H (s, 3/7) + c, c ∈ [−24, 12] × [−12i, 12i] . . . . . . 9043 Mandelbrot set of ζ H (s, 5/6) + c, c ∈ [−24, 24] × [−18i, 18i] . . . . . . 9044 Julia set of ζ H (s, 3/5) + c, c = −1.16i, s ∈ [−24, 24] × [−18i, 18i] . . . 9145 Julia set of ζ H (s, 3/5) + c, c = −1.15i, s ∈ [−24, 24] × [−18i, 18i] . . . 9146 Julia set of ζ H (s, 3/5) + c, c = −1.14i, s ∈ [−24, 24] × [−18i, 18i] . . . 9247 Julia set of ζ H (s, 3/5) + c, c = −1.13i, s ∈ [−24, 24] × [−18i, 18i] . . . 9248 Julia set of ζ H (s, 2/5) + c, c = −0.71i, s ∈ [−24, 24] × [−18i, 18i] . . . 9349 Julia set of ζ H (s, 2/5) + c, c = −0.70i, s ∈ [−24, 24] × [−18i, 18i] . . . 9350 Julia set of ζ H (s, 2/5) + c, c = −0.69i, s ∈ [−24, 24] × [−18i, 18i] . . . 9451 Julia set of ζ H (s, 2/5) + c, c = −0.68i, s ∈ [−24, 24] × [−18i, 18i] . . . 9452 Grid of Julia sets of ζ H (s, 5/8) + c, s ∈ [−32, 32] × [−24i, 24i] . . . . 95x

53 Grid of Julia sets of ζ H (s, 3/8) + c, c = 0 + i, s ∈ [−32, 32] × [−24i, 24i] 9554 Julia sets of ζ H (s, 5/8) + c, c = 1, s ∈ [−32, 32] × [−24i, 24i] . . . . . 10255 Julia set of f c (z) = z 2 + c, c = 0.835 − 0.2321i . . . . . . . . . . . . . 10456 Julia set of f c (z) = z 2 + c, c = 0.45 + 0.1428i. . . . . . . . . . . . . . 10457 Julia set of f c (z) = z 2 + c, c = 0.285 + 0.01i. . . . . . . . . . . . . . . 10558 Julia sets of f c (z), z ∈ [−3, 2] × [−3, 2] . . . . . . . . . . . . . . . . . 10659 Julia sets of f c (z), z ∈ [−2, 2] × [−2, 2] . . . . . . . . . . . . . . . . . 107xi

LIST OF SYMBOLSC The complex numbers, {z|z = σ + iτ and σ, τ ∈ R}R(s)I(s)στĈˆ∞J (f)F(f)Mf c (z)The real part of the complex number sThe imaginary part of the complex number sThe real part of some complex numberThe imaginary part of some complex numberThe Riemann sphere, or the extended complex planeComplex infinityJulia setFilled Julia setMandelbrot setThe quadratic map, f c (z) = z 2 + c, where f c : C → Cf n (z) The n th iterate of the function f(z)ζ R (s) The Riemann zeta function, ζ R (s) =ζ H (s, a) The Hurwitz zeta function, ζ H (s, a) =∞∑n=0n −s∞∑(n + a) −sn=0xii

1 INTRODUCTIONThis thesis is a study of the Julia Sets of the Hurwitz zeta function and theirbehavior.Being the “older, less attractive” sibling, the Hurwitz zeta function isalso known as the generalized Riemann zeta function, its beautiful, younger sister.This work is an extension of research done about the Julia sets of the Riemann zetafunction by Sara Ives [18] and S. C. Woon [35].Chapter Two introduces the reader to the history of the Julia and Mandelbrotsets. This story began around 130 years ago with Arthur Cayley (1821-1895) andmaking a seemingly obvious “hop of faith” (it’s so simple it is not even a leap) indealing with the iterative equation z 3 −1 = 0. It turns out he misjudged the distance,plunging him and those to come after into a world of chaos and confusion.Thirty years later we meet up with Gaston Julia (1893-1978) and Pierre JosephLouis Fatou (1878-1929), two men whose work was two generations before its time.Dealing with both real and complex valued rational functions, Julia and Fatou wonderedwhat would happen if the value of a function were inserted back into thefunction, then compute the function, reinsert the result, and repeat ad infinitum.We had to wait for the innovation of computers and the work of Benoît Mandelbrotto begin appreciating the work of Julia and Fatou.Benoît Mandelbrot (1924 - ) describes his work in the PBS Documentary Huntingthe Hidden Dimension by saying “I don’t play with formulas [sic], I play withpictures...and that is what I’ve been doing all my life.” Although he met ridiculefrom others in the field of mathematics for his research, it was pictures that lead himto a relationship between the Julia sets and what we now know as the Mandelbrotset.In Chapter Three we introduce the Riemann zeta function, the Hurwitz zetafunction, and some background about the men who discovered them. Introduced

in an eight page paper written by the German mathematician Bernhard Riemannin 1859 entitled, “Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse,”the Riemann Hypothesis, which claims all the non-trivial zeros of the Riemann zetafunction lie on the critical line s = 1/2, was introduced to an unsuspecting world.The validity of this hypothesis has remained an unsolved problem for 140 years.Over the course of that time, researchers have found that the patterns and problemswhich have come about as a result of this hypothesis are not limited to numbertheory, but extend to physics, quantum mechanics, chaos theory, and many otherfields.The Hurwitz zeta function was discovered by Adolf Hurwitz in a search for afunctional equation for Dirichlet’s L−functions. In doing this, Hurwitz found relationshipsbetween the L−functions and the Riemann zeta function, as well as therelationship between the L−functions and the Hurwitz zeta function. For this reasonit is commonly referred to as the generalized Riemann zeta function. ChapterThree goes into detail about the three-way relationship between these functions andhow the properties of each influences the behavior of the others.In Chapter Four we focus on the Hurwitz zeta function and our method of extendingit beyond the half-plane σ > 1, in which it naturally resides, to the completecomplex plane.Extending the Hurwitz zeta function beyond its natural domaintakes a considerable effort. The first tool we use is the integral representation of theHurwitz zeta function, ζ H (s, a) and its relationship to the Gamma function, Γ(s).Through analytic continuation, we are able to prove the integral representation ofΓ(s)ζ H (s, a) for R(s) > 1.This allows us to consider a new representation of the Hurwitz zeta functioninvolving contour integrals. We show the contour integral representation I(s, a) isan entire function for any complex number s. This leads to a new representationof the Hurwitz zeta function in terms of Γ(s) and I(s, a). This representation of2

ζ H (s, a) is analytic for the entire s−plane. It gives us insight into the behavior ofthe Hurwitz zeta function in the s−plane because we know the behavior of Γ(s) andI(s, a) in the s−plane.Due to computational limitations, we are not able to explore the desired detailof the Hurwitz zeta function using the analytic continuation. We set out in search ofa functional equation for the Hurwitz zeta function. Such a representation is foundin terms of the Hurwitz’s formula and the Periodic zeta functions. It turns out thePeriodic zeta function, F (x, s) is a linear combination of Hurwitz zeta functions solong as x is a rational number.In Chapter Five we introduce the Euler-Maclaurin summation formula and howto use this approximation to calculate the iterative values of the Hurwitz zeta functionfor values of s where R(s) > 0. Based on the relationship between integrationand summation, we are able to rewrite the series representation in terms of integralsand Bernoulli polynomials.Lacking prior images of the Julia and Mandelbrot sets of the Hurwitz zeta function,Chapter Six compares images and values, obtained by Sara Ives, of the Juliaand Mandelbrot sets of the Riemann zeta function against images and values of theHurwitz zeta function, calculated by the functional equation and Euler-Maclaurinsummation method, where a = 1; i.e., ζ H (s, 1) = ζ R (s) ] . This is followed by a comparisonof values of the Hurwitz zeta function, calculated by the functional equationand Euler-Maclaurin summation method, versus values computed by the Maple c○Hurwitz zeta function. The agreement of these results allows us to say, with someconfidence, that the values and images of the Julia and Mandelbrot sets of the Hurwitzzeta function shown in this work are correct.In Chapter Seven we discuss the conclusions of this work and analyze the imagesof the Julia and Mandelbrot sets and their significance. We finish with ideas forfuture research of the Julia and Mandelbrot sets of the Hurwitz zeta function.3

2 JULIA AND MANDELBROT SETSThis chapter reviews the history of Julia and Mandelbrot sets and defines therelationship between the two.Both sets arise from dynamical systems and mayexhibit fractal behavior. In order to discuss the Julia and Mandelbrot sets of theHurwitz zeta function, we define Julia and Mandelbrot sets and explore how theycame about.2.1 Cayley’s ProblemIn 1879, Arthur Cayley spotted the chaotic behavior of a dynamical system andwas the first to describe what we now known as its Julia set [27]. Cayley was workingwith iterations of Newton’s method,z n = z n−1 − f(z n−1)f ′ (z n−1 ) ,to find the appropriate roots of f(z). For the equation g(z) = z 2 − 1, where z ∈ C,we havethusg(z n−1 ) = z2 n−1 + 1= 1 (z n−1 + 1 ),2z n−1 2 z n−1g ′ (z n−1 ) = 1 2(1 − 1 ). (2.1.1)zn−12Set Equation (2.1.1) equal to zero and we find it has two solutions, z = 1 andz = −1. Starting with the point z 0 , where R(z 0 ) > 0 [R(z) represents the real partof the complex number z], Newton’s method returns approximations that convergeto the solution z = 1. The closer the starting point z 0 is to the solution z = 1, thefaster the results of Newton’s method converge on said solution. Similarly, start ata point z 0 , R(z 0 ) < 0, and Newton’s method will converge on the solution z = −1.Just as with the other solution, the rate of convergence to the solution is directly4

elated to the proximity of z 0 to z = −1. However, if we begin with z 0 , R(z 0 ) = 0,the results are all points on the imaginary axis.At this point we introduce some notation and vocabulary that will help us alongthe way. First, iterates of a function f will use the notation f 1 = f, f 2 = f ◦ f,and, f n = f n−1 ◦ f.Definition 2.1.1. A dynamical system is a collection of states and a rule,f : C → C, which determines the present state of our system in terms of past states.The actual dynamics of the system are found in the behavior of points z n = f n (z 0 )under iteration of f, where n = 0, 1, 2, . . . .Definition 2.1.2. Given an iterative function f, the sequence of iterates{z 0 , f(z 0 ), f 2 (z 0 ), . . . , f n (z 0 ), f n+1 (z 0 ), . . .}, is called the orbit of z 0 under f.Definition 2.1.3. A point of f(z) for which f(z) = z is called a fixed point of f.Definition 2.1.4. The basin of attraction A(z 0 ) of a point z 0 is the set of allz ∈ C such that f n (z) converges to z 0 .Cayley had discovered that for the equation z 2 = 1, the basin of attraction ofthe point z = 1 is the half-planeA(1) = {s ∣ ∣ s = σ + iτ, σ > 0},and the basis of attraction of the point z = −1 is the half-planeA(−1) = {s ∣ ∣ s = σ + iτ, σ < 0}.Note. The notation σ = R(s) and τ = I(s) will be used when describing anycomplex number, commonly represented by the variables c, s, and z.5

Cayley then moved on to the iterations of Newton’s method for the equationz 3 = 1. The three solutions are z = 1, z = ω, and z = ω 2 , where ω = e 2πi/3 . Dueto the results of his work on the equation z 2 = 1, it seems plausible that he wouldfind three basins of attraction, each corresponding to one of the three solutions, andeach sweeping out one-third of the complex plane C. Nothing could be further fromthe truth. In the words of Arthur Cayley, “the next succeeding case of the cubicequation appears to present considerable difficulty” [27]. Instead of a crisply dividedcomplex plane similar to the “peace sign”, we end up with what you see in Figures 1and 2. To give an idea of the complexity, suppose we have a point z a such that itsiterates converge on the solution z = 1, and suppose there is another point z b withina closed neighborhood N(z a ) centered at z a , with radius r = ε, such that the iteratesof z b converge to the solution z = ω. Then there exists a point z c that converges tothe solution z = ω 2 , which is within a smaller neighborhood M(z a ), centered at z a ,with radius r < ε [27].In other words, if two friends were at the park and were moving toward oneanother, no matter how close the two were to get, there would always be a thirdfriend between them! The visual result of this is a “knot-like” pattern as seen inFigures 1 and 2 [32].2.2 Julia SetsWhile Julia sets were originally discovered over 90 years ago, they have been rebornin the modern computer age. Named after the mathematician Gaston Julia whoannounced their discovery in his 1918 work “Mémorie sur l’itération des fonctionsrationnelles,” Julia sets, or Julia fractals [19], thrust Gaston Julia to the forefrontof the French mathematical world. The then-25-year-old war survivor would laterreceive the Grand Prix of the Académie des Sciences for his work which dealt withthe iteration of a rational function mapped from the complex plane to the complex6

Figure 1: Application of Newton’s method for equation z 3 − 1 = 0, z ∈ [−2.2, 2.2] ×[−2.2i, 2.2i]plane. Julia’s rise to prominence would be followed by a descent from fame of equalif not greater speed. One of the earliest attempts at illustrating the work of Juliawas done by Cremer in 1925 and can be seen in Figure 3 [12]. Due to the intensivework involved in creating a single image, years passed until Benoît Mandelbrot’swork with computers brought new focus to Julia sets.Gaston Julia worked with Pierre Joseph Louis Fatou (1878-1929) [31] on multipleiterations of the complex quadratic map f c (z) = z 2 + c where z and c are complexnumbers. Julia denoted the set of points whose n th iterate, fc n , does not diverge toinfinity as n approaches infinity. Today this is known as the Julia set.To determine if points belong to the Julia set we need to be able to classifythese points. Consider the general iterative equation z n+1 = f(z n ) in the followingdefinitions.7

Figure 2: Julia set of z 3 − 1 = 0Definition 2.2.1. A point z 0 is called a periodic point of f if f n (z 0 ) = z 0 forsome positive integer n. Thus, z 0 is a fixed point of f n .Definition 2.2.2. For any periodic point z 0 , the smallest p ∈ Z + for which f p (z 0 ) =z 0 is called the period of z 0 .Definition 2.2.3. The orbit of a fixed point z 0 , {z 0 , f(z 0 ), f 2 (z 0 ), . . . , f p−1 (z 0 )} isknown a cycle.Fixed points have different behaviors. These behaviors are defined below.Definition 2.2.4. A fixed point z ∗ of f is called a stable or an attracting fixedpoint if, for any neighborhood U of z ∗ , ∃V ⊂ U containing z ∗ where, for someinteger N, f k (V ) ⊂ U for k > N.In simple terms, if the orbit of points near the fixed point z ∗ tends toward the8

Figure 3: The first drawing by Cremer in 1925 visualizing a Julia setpoint z ∗ , then it is considered an attracting fixed point. If orbits near a fixed pointz ∗ do not converge on z ∗ , then z ∗ is considered an unstable or repelling fixed point.To determine the stability of fixed points we apply the next theorem from analysis.First assume f is an analytic function with a fixed point z 0 . We define themultiplier of f at z 0 as the value λ = f ′ (z 0 ).Theorem 2.2.5. Let z ∗ be a fixed point of an analytic function f. If |f ′ (z ∗ )| < 1,then z ∗ is a stable fixed point. If |f ′ (z ∗ )| > 1, then z ∗ is an unstable fixed point.Proof: Consider the orbit z n+1 = f(z n ), where n = 0, 1, . . . . Letz n = z ∗ + ε n , where ε 0 ≠ 0,be a point near the fixed point z ∗ . This means ε n = z n − z ∗ is the displacementfrom the fixed point at the n th iteration. Using the analyticity of f, we look at9

the Taylor series expansion of f(z) about z ∗ :z n+1 = f(z n )z ∗ + ε n+1 = f(z n )≈ f(z ∗ ) + f ′ (z ∗ )ε n + 1 2 f ′′ (z ∗ )ε 2 n + . . . .Now for small values of ε n , the terms 1 2 f ′′ (z ∗ )ε 2 n + · · · can be neglected, leavingus withz ∗ + ε n+1 ≈ f(z ∗ ) + f ′ (z ∗ )ε n . (2.2.1)We know z ∗ = f(z ∗ ) by the fixed point property. Substituting this into Equation(2.2.1) allows us to rewrite the right side asz ∗ + ε n+1 ≈ z ∗ + f ′ (z ∗ )ε n .Simplifying, we haveε n+1 ≈ f ′ (z ∗ )ε n , where n = 0, 1, . . . . (2.2.2)Now we have a linear dynamical system for the ε’s which can be solved. Fromthis linear system, we can determine the stability of the fixed point z ∗ by thebehavior of the orbits beginning nearby. Using Equation (2.2.2) with λ = f ′ (z ∗ ),we have10

ε 1 = λ ε 0ε 2 = λ ε 1 = λ 2 ε 0ε 3 = λ ε 2 = λ 3 ε 0.ε k = λ ε k−1 = λ k ε 0 .We note thatlimk→∞ |λ|k =⎧⎪⎨ 0, if |λ| < 1,⎪⎩ ∞, if |λ| > 1,Therefore, fixed points where |λ| = |f ′ (z ∗ )| < 1 will cause |ε k | → 0, indicating thepoints surrounding z ∗ have stable orbits. On the other hand, if |λ| = |f ′ (z ∗ )| > 1,|ε k | → ∞, meaning the points around z ∗ have unstable orbits.Theorem 2.2.5 allows us to classify fixed points as follows [24]:Superattractive ⇔ |λ| = 0,Attractive ⇔ 0 ≤ |λ| < 1,Indifferent ⇔ |λ| = 1,Repelling ⇔ |λ| > 1.Now we are ready to explore some of the complexities of the Julia set.Definition 2.2.6. The filled Julia set F(f) consists of the points z ∈ C for whichf n (z) does not diverge to infinity as n → ∞.Definition 2.2.7. The set of boundary points of the filled Julia set F(f) is known11

as the Julia set, denoted J (f) for some map f. This set contains all z ∈ C forwhich f n (z) remains bounded as n → ∞.This means the Julia set of some iterative map F is the collection of the boundariesof the basins of attraction of F . If some point z is a boundary of any givenbasin of attraction of F , then it is a boundary of all basins of attraction for saidmap F . For example, consider the complex function f : C → C where f(z) = z 2 .After a handful of iterations, we obtain:f(z) = z 2 ,f 2 (z) = ( z 2) 2= z 4 ,f 3 (z) = ( z 4) 2= z 8 , . . . .We can see the value of f n (z) can speed off to infinity very quickly if |z| > 1, whereasf n (z) will retreat toward zero if |z| < 1.These two cases account for all complex numbers z except those for which |z| = 1;the unit circle of the complex plane. Looking at Figure 4, know that any point alongthe unit circle has an orbit of iterate values with magnitude one. This means onceyou find yourself on the unit circle, you will remain there forever.We have two regions with the function f(z) = z 2 , {z ∣ |f n (z)| → 0 as n → ∞}and {z ∣ |f n (z)| → ∞ as n → ∞}. The unit circle, {z ∣ |f n (z)| = 1 as n → ∞},is actually the “boundary” between falling toward zero and iterating off to infinity,the Julia set of f(z) = z 2 .It may help to think of the Julia set as an “eternalmathematical purgatory,” trapped between basins of attraction, unable to reacheither.As we can see in Figure 4, there are a lot of orbits that diverge to infinity. Wewant to say more those orbits, and to do this, we need to introduce the concept ofcomplex infinity. We extend the complex plane C by the stereographic projection12

Figure 4: Julia Set of f c (z) = z 2 + c, c = 0of a unit sphere S onto C. The projection map associates to each point A ∈ C acorresponding point α = P (A) ∈ S by the intersection of a line segment from theNorth pole P to A. In this model (seen in Figure 5), we identify the North polewith complex infinity, noted ˆ∞. The unit sphere S in this model is known as theRiemann Sphere, also called the extended complex plane. With this model, we nolonger have orbits that diverge. Instead these orbits can be described as convergingto complex infinity. The extended complex plane is represented with the notationĈ = C ∪ { ˆ∞}.The two large regions of the equation f(z) = z 2 are {z ∣ |f n (z)| → 0 as n → ∞}and {z ∣ |f n (z)| → ∞ as n → ∞} are the basins of attraction A(0) and A( ˆ∞),respectively. The unit circle, {z ∣ |f n (z)| = 1 as n → ∞}, is actually the “boundary”of each basin, made up of all values of z for which f n (z) does not fall into A(0) norA( ˆ∞) as n → ∞. This idea is illustrated in Figure 6.13

Figure 5: Stereographic projection of S onto CWe now consider a more complex Julia set.To do this, we alter the simplefunction f(z) = z 2 by adding a complex constant c. This results in the new functionf c (z) = z 2 + c, where c = σ + iτ. (2.2.3)Figure 7 illustrates how a small change in the value of c can extremely modifythe result. When we look at this shape it is difficult to describe. One major aspectis a repetitive pattern we see. The two large “spoked” pattern seen just off-centerare repeated off the sides of each spoke, then are repeated again. If we continuedto zoom in on this image, that pattern would appear again and again, a property isknown as self-similarity. The question is how to how do we classify these shapes?Answering this requires a look back into the history of dimensions and these mostinteresting figures.Fractal shapes are incredibly difficult to describe in terms of Euclidean geometry.In 1911, Luitzen Brouwer (1881-1966) proved that dimension is a topological invari-14

Figure 6: Riemann Sphere illustrating basins of attraction at 0 and ˆ∞Figure 7: Julia Set of f c (z) = z 2 + c for c = −0.70176 − 0.3842i15

ant [13]. If we start with a line with the topological dimension one, no matter howmuch we bend, twist, and turn it, so long as we do not cut or glue it, the line will alwayshave the topological dimension of one. Similarly, if you have a two-dimensionalplane, so long as you do not cut the plane or glue the edges together, the plane willalways have a topological dimension of two.Figure 8: Constructing the the Cantor setFigure 9: The Koch curveAt the same time, Felix Hausdorff (1869-1942) was working with shapes andhow they filled the space around them. His work lead to shapes that could have anon-integer topological dimension. For instance, the Cantor set, seen in Figure 816

[33], is created by starting with a line and removing the middle one-third. Then,from the remaining two pieces we take the middle one-third from each, continuingthis pattern. Thus the Cantor set contains two one-third-sized copies of itself. It isnot a line, but is certainly more than a point. Using box-counting [2], the fractaldimension of the Cantor set must be less than one, but greater than zero.A Koch curve, seen in Figure 9 [34], is created by removing the middle one-third(just like the Cantor set), then replacing it with two lines of the length one-thirdwhich meet over the newly created gap.This means the Koch curve consists offour one-third-sized copies of itself. Koch curve lies somewhere between a line and asquare, so using box-counting, its topological dimension must lie somewhere betweenone and two. These shapes led to the discovery of the similarly dimension. In 1919,Hausdorff extended this similarly dimension to cover all shapes[24], setting the tablefor the most famous of problems: How long is the coastline of Britain? This newfractal dimension describes the fractal complexity of an object. For example, boththe Koch curve and the British coastline have a fractal dimension of around 1.26[25]. On this scale, 1 is completely smooth, and the closer to 2 the dimension gets,the more complex the fractal.With fractals on the mind (the brain itself has a fractal structure), take a lookat Figure 7 and admire the complexity and intricacy of these images.In termsof fractals with the Julia and Mandelbrot sets, the fractal itself is the border orcoastline between the basins of attraction, what we call the Julia set. Figure 6illustrates how basins of attraction look and work on the Riemann sphere and howthe border between the basins, the Julia set J (f), can take on interesting shapes.While the addition of c may seem simplistic, the results of the new functionhave revealed the underlying chaos our function f(z) = z 2 . The system is chaoticbecause the small adjustment of the initial condition leads to a vast transformationof the final result. This concept is difficult for people to understand. “We are, as17

Figure 10: Julia Set of f(z) = z 2 + c for c = 1.0 + 0.0ihumans, heavily socialized to make a kind of rough approximation between causeand effect...We are trained to think that what goes into any system must be directlyrelated, in intensity and dimension to what comes out” [15]. For instance, considerthe equation at hand, f c (z) = z 2 + c. If we start with z 0 = 0 and let c = 2, theeighth iteration of f c (z) returns an enormous numberz 8 = 2 27 > 10 38 .Iterating only eight times returns a number which exceeds the diameter of the universeas measured in atomic units [27]. Now consider the case where c = −2 and18

z 0 = 0 it turns out thatz 0 = (0) 2 − 2 = −2z 1 = (−2) 2 − 2 = 2z 2 = 2 2 − 2 = 2z 3 = 2 2 − 2 = 2.z n = 2 2 − 2 = 2, n ∈ Z + .The differences in the two results results is enormous, yet all we did is change thesign of c.from such a miniscule “tweaking” is nothing short of astounding. Due tothis complexity, we will be using FractInt 20.4 c○ , an MS-DOS based program whichemploys fine-tuned algorithms used to increase the speed of calculation. To combatany fears of error due to round-off, FractInt 20.4 c○ performs integer arithmetic, givingus error free computations. With the aid of FractInt 20.4 c○ that we journey into theworld of the complex, hoping to see the chaos that lies beneath the seemingly simpleaddition of a harmless complex constant c.In the images of Julia sets, the point (σ, τ) corresponds to the value z = σ + iτ.If a point is shaded black, we know the addition of the corresponding z−value causesthe iterative function in question to fall into a basin of attraction of a stable fixedpoint z ∗ ≠ ˆ∞. On the other hand, if the corresponding co-ordinate is any colorother than black, we know that particular z−value causes the iterative function inquestion to converge to complex infinity. The different colors represent the speed ofconvergence to complex infinity. The deepest, darkest blue begin the fastest and thebrighter the color gets, the slower the convergence to complex infinity. The followingexamples use the function f c (z) = z 2 + c.19

Figure 11: Julia Set of f c (z) = z 2 + c for c = 0.45 + 0.1428i.Figure 12: Julia Set of f c (z) = z 2 + c for c = 0.285 + 0.01i.20

Figure 13: Julia Set of f c (z) = z 2 + c for c = 0.835 − 0.2321i.While fractal shapes are difficult to describe, we are able to classify Julia sets asconnected and disconnected.Definition 2.2.8. A set is connected if it is not contained in two disjoint opensets.An example of a connected Julia set can be seen in Figure 4, whereas an unconnectedJulia set it found in Figure 10.Whether a Julia set is connected or not is the link between the Julia and Mandelbrotsets. This most interesting behavior deserves further investigation as we seein the next section.2.3 Mandelbrot SetsAs we iterate the function f c (z) from Equation (2.2.3), we can determine whichvalues of the domain of f c (z) belong to the Julia set. While it is obvious that thechoice of z 0 is a critical factor in determining how the corresponding orbit behaves,21

the selection of the constant term c is just as important. Mandelbrot focused onthese c−values and how they affected the orbit of f c (z). Mandelbrot assumed theinitial value z 0 = 0 and iterated the function with different c−values. He thenmapped these into the c−plane where the x−axis is the σ-value and the imaginaryaxis corresponds to the τ-value of the complex constant c.The Mandelbrot set is made up of all the values of c for which the correspondingorbit does not converge to complex infinity as we iterate the function. The imagein Figure 14 is quite possibly the most famous and certainly the most recognizablefractal to date. The main cardioid is the large dark region in the center. It issurrounded by similar shaped regions attached along its border, displaying its selfsimilarity.This self-similarity can be seen in more depth in Figures 15 and 16 [32],which are zoomed in images along the border of Figure 14.Figure 14: Mandelbrot Set of f c (z) = z 2 + c22

Figure 15: Zoomed top of Mandelbrot Set of f c (z) = z 2 + cFor a more formal explanation of the Mandelbrot set, we need look no furtherthan this definition.Definition 2.3.1. The Mandelbrot set M of some complex map g c (z), where c ∈ C,is the set of values of the complex parameter c which, gc 0 (z) ↛ ˆ∞ as n → ∞. Inother words,{}M = c ∣ c ∈ C and lim gc n (z 0 ) ≠ ˆ∞ .n→∞Just as with Julia sets, Mandelbrot sets require extensive computation in orderto render an image. The smaller the area we wish to render, the less computingpower we need, thus the faster the pictures can be drawn. It is helpful that for ourfunction f c (z) = z 2 + c, if |z n | > 2, then that c is not part of the Mandelbrot set.Theorem 2.3.2. The Mandelbrot set M for f c (z) = z 2 + c lies in and on a circleof radius r = 2 centered at the origin of the complex plane C.Proof:Case 1: Suppose |c| > 2 and z 0 = 0. We want to show that all sequences withthis property diverge to infinity. We know |z 1 | = |f c (z 0 )| = |0 + c| = |c|, therefore23

Figure 16: Zoomed border of Mandelbrot Set of f c (z) = z 2 + c|z 1 | = |c| > 2. Now looking to the next iterate, meaning:|z 2 | = |f c (z 1 )| = |z1 2 + c| =∣ z 1(z 1 + c z 1)∣ ∣∣∣= |z 1 |∣ z 1 + c ∣ ∣∣∣(2.3.1)z 1Given |u| ≥ |v|, we know|u + v| ≥ |u| − |v| ≥ 0. (2.3.2)Apply the inequality from Equation (2.3.2) to Equation (2.3.1), we find(|z 2 | ≥ |z 1 | |z 1 | − |c| ). (2.3.3)|z 1 |Since |z 1 | = |c|, we have|z 2 | > |c| (|c| − 1) . (2.3.4)24

Moving to the next iterate, we find|z 3 | = |f(z 2 )| = |z2 2 + c| = |z 2 |∣ z 2 + c ∣ ∣∣∣(2.3.5)z 2and applying the inequality from Equation (2.3.2) to Equation (2.3.5) leaves uswith(|z 3 | ≥ |z 2 | |z 2 | − |c| ). (2.3.6)|z 2 |The assumption that |c| > 2 implies (|c| − 1) > 1.Applying this to Equation(2.3.4) we can see that |z 2 | > |c|, thus|c||z 2 |< 1, (2.3.7)implying|z k | − |c||z k |> |c| − 1. (2.3.8)Apply the result of Equation (2.3.8) to Equation (2.3.6) and we find(|z 3 | ≥ |z 2 | |z 2 | − |c| )|z 2 |( ) (|z 3 | > |c| |c| − 1 |z 2 | − |c|( 2.|z 3 | > |c| |c| − 1)|z 2 |)This pattern continues, meaning(|z 4 | > |c| |c| − 1(|z 5 | > |c| |c| − 1) 3) 4.( n.|z n+1 | > |c| |c| − 1)25

Now we prove that |z n+1 | > |z n |(|c| − 1) n for n = 0, 1, . . . using induction.Base Case: n = 2.Equation (2.3.4) states|z 2 | > |c| (|c| − 1) .Inductive Hypothesis: Assume formula holds for n = k.So we are now given the fact that|z k | > |c| (|c| − 1) k−1 .Inductive Step: Prove the case n = k + 1. Since|z k+1 | = |f(z k )| =∣ z k(z k + c )∣ (∣ ∣∣∣ ∣∣∣= |z k | z k + c ∣)∣∣∣. (2.3.9)z k z kApplying the inequality from Equation (2.3.2), we know that(|z k+1 | ≥ |z k | |z k | − |c| ). (2.3.10)|z k |We will now show that |z k | > |c|, for k ≥ 2, from our induction hypothesis. Weknow from an earlier assumption that |c| > 2, which means |c| − 1 > 1. Wealso have k ≥ 2, thus k − 1 ≥ 1. Putting these facts together, we know that(|c| − 1) k−1 > 1.Thus starting with the induction hypothesis, we obtain|z k | > |c|(|c| − 1) k−1 > |c|. (2.3.11)26

From Equation (2.3.11) we know that|c||z k | < 1,which implies|z 2 | − |c||z 2 |> |c| − 1. (2.3.12)Applying the inequality from Equation (2.3.12) to Equation (2.3.10), we findthat|z k+1 | ≥>[ (|z k | |z k | − |c| ) ] k−1(|c| − 1)|z k |[|c| (|c| − 1) k−1] (|c| − 1) .Thus,|z k+1 | > |z k | (|c| − 1) k .We know that (|c| − 1) > 1, so as k → ∞, |c|(|c| − 1) k → ∞. Thereforeno complex parameter |c| > 2 is an element of the Mandelbrot set M off c (z) = z 2 + c.This is important for us because now we do not have toconsider |c| > 2 when calculating M.Case 2: Suppose |c| ≤ 2 and |z n | > 2, for n ∈ Z + and n ≠ 0.We still need to see what happens to those points within our circle of radius r = 2,but whose orbits still converge to complex infinity. What we wish to show is thatonce the iterate of some c−value goes outside the circle r = 2, the successiveiterates with converge to complex infinity.27

We know that,|z n+1 | = |zn 2 + c| = |z n |∣ z n + c∣ ( ∣∣∣≥ |z n |z n|z n | − |c||z n |). (2.3.13)Also |c||z n | < 1, which means ( )|z n+1 | > |z n | |z n | − 1 . (2.3.14)Recalling the assumption |z n | > 2, we have|z n+2 | = |zn+1 2 + c|= |z n+1 |∣ z n+1 +≥ |z n+1 |c ∣ ∣∣∣z n+1(|z n+1 | − |c||z n+1 |)≥ |z n+1 | (|z n | − 1) . (2.3.15)Applying Equation (2.3.14) to Equation (2.3.15), we find that|z n+2 | > |z n | (|z n | − 1) (|z n | − 1) = |z n | (|z n | − 1) 2 , (2.3.16)so |z n | gets bigger as we iterate. Furthermore, we know that|z n+k | > |z n | (|z n | − 1) k . (2.3.17)Now as k → ∞, we can see the right side of Equation (2.3.17) diverges to infinity,therefore if |z n | > 2 for any integer n, then z ∈ M.Now that we have limited the region of C that can be in M, we shift our focus tohow we go about computing M. In order to compute a “true” Mandelbrot set M,28

we must iterate infinitely many times for every c, which is simply impossible.We all know the adage about “given an infinite amount of time, a monkey witha typewriter will produce the works of Shakespeare.”This applies to our “mathproblem” because in the same spirit, if we iterate this system infinitely many times,how can we be sure of the behavior of any orbit? According to Hubbard, “if we reachz 1000 and |z 1000 | < 2, there is only a very, very small chance that the sequence willdiverge to infinity” [27]. This means we can be almost certain of the behavior of ourpoint without having to calculate a very large number of iterations. We limit thenumber of iterations we calculate by programming in a “bailout value,” the maximumnumber of iterations to run. Once we have reached the bailout value, if our iteratesare still inside the circle |r| < 2, then we say, with some confidence, that the c−valuein question is an element of the Mandelbrot set.As we alluded to earlier, there is a relationship between the Mandelbrot set ofa mapping and the corresponding Julia sets. We introduce this relationship in thefollowing Theorem:Theorem 2.3.3 (The Julia-Fatou Theorem).The Julia set J(p c ) of some map p c (z) for the given value of c ∈ C is connected ifand only if limn→ ˆ∞p n c (0) ≠ ∞; i.e. if and only if the constant c is in the Mandelbrotset M of the map p c (z) for z = 0.This theorem shows how the Mandelbrot set of some map p c (z) is itself a mappingof Julia sets. The Mandelbrot set exists in the c−plane, and each point correspondsto the Julia set of the map p c (z). If the point c ∈ M, then the Julia set of p c (z)for that c−value is connected. If the point c /∈ M, then the corresponding Julia setis not connected. This idea is illustrated in Figure 17, where the Mandelbrot setconsists of points colored black. As one can see, the points in the Mandelbrot setcorrespond to connected Julia sets whereas the point outside not in the Mandelbrot29

Figure 17: A Mandelbrot set and the corresponding Julia setsset corresponds to a disconnected Julia set.FractInt 20.4 c○ has a most helpful option called the Evolver command. This letsus create multiple images of Julia sets in a grid pattern, each image representingthe Julia set for a different value of c. Looking at Figure 18 we can see the Evolvercommand applied to the function f c (z). The center image applies to c = 0 + 0i, theimage directly to the right represents c = 1 + 0i, image directly above showing theJulia set for c = 0 + 1i, and so on. We know the center image shows a connectedJulia set, so by Theorem 2.3.3, the point c = 0 + 0i is an element of M of f c (z).The corresponding image of the Mandelbrot set of f c (z) will have a black point atthe co-ordinate (0, 0i).This work parallels the idea used in the search for the Julia set of f c (z) = z 2 + c.Using FractInt 20.4 c○ , we iterate the Hurwitz zeta function with different values forc and the program will return images which describe if a given c−value is an elementof the Mandelbrot set of the Hurwitz zeta function.For this work, we will use a relationship described in Theorem 2.3.3 to find the30

Figure 18: Julia sets of f c (z), c ∈ [−2, 2] × [−2, 2]Julia and Mandelbrot sets of the Hurwitz zeta function.2.4 Previous ResearchS.C. Woon, Sara Ives, and others have used the relationship between the Juliaand Mandelbrot sets found in the Julia-Fatou Theorem to find and illustrate theJulia and Mandelbrot sets of the Riemann zeta function [35]. The current work isan extension of their work in so much as the Hurwitz zeta function is also known asthe generalized Riemann zeta function.We believe this work to be the first known attempt to calculate the Julia andMandelbrot sets of the Hurwitz zeta function. Our original idea and formula forusing the Euler-Maclaurin summation was inspired by the work of Linas Vepštas.31

He worked on finding faster and better ways to compute values of the Hurwitz zetafunctions. These methods are shown and recalculated in Chapter Five.As for work with the Hurwitz zeta function itself, I draw the majority of myinformation from the work of Tom M. Apostol.H. M. Edwards’s seminal workentitled “Riemann’s Zeta Function” [14] is the main source of information on theRiemann zeta function.In the next chapter, we introduce the reader to the history of the Riemann zetafunction and the lesser known Hurwitz zeta function, discovered by Adolf Hurwitz.32

3 INTRODUCTION TO ZETA FUNCTIONSThe work of Bernhard Riemann (1826-1866) and Adolf Hurwitz (1859-1919) hasimpacted more than just mathematics.We present here some historical remarksbefore we delve deeper into how these functions where developed and the relationshipbetween the two.3.1 History of the Riemann Zeta FunctionIf we were to imagine math as a living being, the Riemann zeta function would bethe DNA of mathematics. The Riemann zeta function is intertwined with every fieldof mathematics. The Riemann zeta function is central to the search for a patternin prime number distribution. The proof or disproof of the Riemann Hypothesis, allnon-trivial zeros of the Riemann zeta function lie on the critical line s = 1/2, is oneof the Clay Institute’s “Million Dollar problems” and one of five remaining problemsof the 23 David Hilbert’s posed in 1900.Serendipitous connections to the Riemann zeta function are abundant. It arosein the classical derivation of the Stefan-Boltzmann Law from Planck’s Blackbodyradiation spectrum distribution. The Stefan-Boltzmann law gives the power emittedper unit area of the emitting body. In order to derive this law we must calculate thefollowing integral∫ ∞0x n−1e x − 1 dx = Γ(4)ζ R(4) = π415 . (3.1.1)This integral is solved by rewriting it as the product of the Riemann zeta functionand Gamma function, where s = 4. Another connection of the Riemann zetafunction to other fields starts with a famous Princeton tea party where FreemanDyson and Hugh Montgomery the distribution of pairs of zeros as discovered byMontgomery matched exactly the calculations done by Dyson in the theory of randomenergy levels of Hamiltonians under certain symmetry [26]. Another example33

is found in the mathematical technique called random matrices, which is used tohandle the enormous amount of data generated when quantum mechanics is appliedto a system of particles. The results of this technique appear remarkably similar tothe distribution of the Riemann zeros [26]. In yet another “theoretical experiment,”Karl Sabbagh talks about how “if someone wanted to calculate the exact air pressureupon a wall. It can be done by finding the position of all the air molecules, solvingNewton’s equations, finding out how many times they hit the wall, and computingthe pressure that way. The resulting numbers could be statistically indistinguishablefrom a random sequence.” Following this idea Professors Michael Berry and JonKeating have devised what they call “quantum chaotic systems,” which, when usedto describe situations such as the air molecules in a room, can lead to a mathematicalfunction with strong resemblance to the Riemann zeta function [26].The underlying chaotic behavior of the Riemann zeta function has been well documentednot only in number theory, but in nature itself. In this work we investigateif the chaotic behavior of the Riemann zeta function remains when we move to thegeneralized Riemann zeta function, changing s to a non-integer value.3.2 Adolf Hurwitz and His Work“Adolf Hurwitz was born 26 March 1859 in Hildesheim, Lower Saxony, Germany.Working with Hermann Schubert at Healgymnasium Andreanum in Hildesheim in1868, Hurwitz was encourged to proceed in his studies by Schubert. His familyunable to send Hurwitz to university, his father’s friend agreed to help financiallyallowing Hurwitz to continue on in academia. With glowing recommendations fromSchubert to Felix Klein, Hurwitz entered the University of Munich in 1877, beforehe was eighteen years old. After a year with Klein, Hurwitz left for the University ofBerlin where he attended classes taught by the Ernst Kummer, Karl Weierstrass, andLeopold Kronceker. He then followed Klein from Munich in 1879 to the University of34

Leipzig in October 1880, where he received his Ph.D. in 1881, superviesed by Klein,in work on the elliptic modular functions, entitled Grundlagen einer independentenTheorie der elliptischen Modulfunktionen und Theorie der Multiplikatorgleichungen1. Stufe” [30].Hurwitz would have become a professor of mathematics at Leipzig, but he didnot have sufficient knowledge of Greek to satisfy a faculty requirement [30]. Hurwitzleft for the University of Göttingen in 1882, then the University of Königsberg in1884 where he taught the likes of David Hilbert and Hermann Minkowski. Aftera move to Bonn, Hurwitz finally settled down in 1892 when he took over chair atEidgenössische Polytechnikum Zürich where he would remain for the rest of his life.He had poor health and which was plagued with migraines and renal failure, finallypassing away 18 November 1919 in Zürich, Switzerland.It was the influence of Klein that most inspired Hurwitz. His work spans modularfunctions, topology, algebra, and Lie groups. However he is best known his work onthe Dirichlet L−functions for which the Hurwitz zeta function is named.3.3 Riemann Zeta Function and Dirichlet L−SeriesHurwitz discovered his zeta function, denoted ζ H (s, a), while conducting originalresearch on a functional equation for Dirichlet’s L−functions. Hurwitz then used hiszeta function deduce the continuations of the L−functions in the year 1882. This wasmade possible by the relationship he found between the Riemann zeta function ζ R (s)and Dirichlet’s L−functions L(s, χ). These two functions are defined as follows:ζ R (s) =∞∑n=11, s ∈ C, σ > 1, (3.3.1)ns L(s, χ) =∞∑n=1χ(n)n s , (3.3.2)35

where χ is known as a Dirichlet character.Introducing the Dirichlet character requires a some background taken directlyfrom [14]. The reduced residue system modulo k is a set of φ(n) integers {a 1 , a 2 , a 3 , . . . , a φ(k) }incongruent modulo k, each of which is relatively prime to k. For each integer a, thecorresponding residue class â is the set of all integers congruent to a modulo k:â = {x ∣ ∣ x ≡ a mod k}.Definition 3.3.1. Now consider a group G of reduced residue classes modulo k.Corresponding to each character f ∈ G we define an arithmetical function χ = χ fas follows:⎧⎪⎨ f(ˆn), if gcd(n, k) = 1,χ(n) =⎪⎩ 0, if gcd(n, k) > 1.(3.3.3)The function is called a Dirichlet character modulo k. The principle characterχ 1 has the property⎧⎪⎨ 1, if gcd(n, k) = 1,χ 1 (n) =⎪⎩ 0, if gcd(n, k) > 1.(3.3.4)The L−functions L(s, χ) can be rewritten in terms of Hurwitz zeta functions.Since χ is a character mod k, we can arrange the terms of the series for L(s, χ) bytheir residue classes mod k. Let n = qk + r, where 1 ≤ r < k and q = 0, 1, . . . .The corresponding Dirichlet character mod k would be defined as⎧⎪⎨ 1, if gcd(n, k) = 1,χ k (n) =⎪⎩ 0, if gcd(n, k) > 1.36

This leaves us withL(s, χ) =∞∑n=1χ(n)n s =k∑∞∑r=1 q=0r=1χ(qk + r)(qk + r) s= 1 k∑ ∞∑χ(r)k sq=01(q +rk) s .The series∞∑ 1(q +rkq=0) sis the Hurwitz zeta function, which means we haveL(s, χ) = 1 kk∑r=1(χ(r) ζ H s, r ). (3.3.5)kDefinition 3.3.2. The Hurwitz zeta function is defined by the seriesζ H (s, a) =∞∑n=01, s ∈ C, σ > 1, 0 < a ≤ 1. (3.3.6)(n + a)sNotice that if a = 1, the Hurwitz zeta function reduces to the Riemann zeta function,Thus,ζ H (s, 1) =∞∑n=01(n + 1) s = 1(0 + 1) s + 1(1 + 1) s + 1(2 + 1) s + · · ·= 1(1) + 1s (2) + 1s (3) + · · · s∞∑ 1=n . sn=1ζ H (s, 1) = ζ R (s). (3.3.7)The representation of the L−functions shown in Equation (3.3.5) illustrates howtheir properties are dependent on properties of the Hurwitz zeta function. One quite37

important property of the Hurwitz zeta function is its analyticity in the half-planeσ > 1.Theorem 3.3.3. The series for ζ H (s, a) converges absolutely for σ > 1. The convergenceis uniform in every half-plane σ ≥ 1 + δ, where δ > 0. Therefore, ζ H (s, a)is an analytic function of s = σ + iτ in the half-plane σ > 1.Proof: First we look at absolute convergence. We need to show∞∑∞∑|(n + a) −s | = (n + a) −σ < ∞. (3.3.8)n=1n=1Starting with the denominator on the left side, we find that(n + a) s = e s ln(n+a)(σ+it) ln(n+a)= e= e ln(n+a)σ it ln(n+a)e(n + a) s = (n + a) σ e it ln(n+a) ,meaning it is also true that(n + a) −s = (n + a) −σ e −it ln(n+a) . (3.3.9)By Euler’s equation, we know that ∣ ∣ eiθ ∣ ∣ = 1, so long as θ ∈ R. Apply this to38

Equation (3.3.9) and we find that∞∑|(n + a) −s | =n=1==∞∑∣ (n + a) −σ e it ln(n+a)∣ ∣n=1∞∑∣ ∣ (n + a)−σ ∣ ∣ eit ln(n+a) n=1∞∑∣ ∣ (n + a)−σ .n=1Since there are no negative terms in this series, we can remove the absolute valuebars, thus∞∑|(n + a) −s | =n=1By the integral test, the series∞∑(n + a) −σ .n=1∞∑(n + a) −σ ,n=1converges for σ > 1. Therefore the seriesis absolutely convergent for σ > 1.∞∑(n + a) −s ,n=1Now we need to prove uniform convergence in every half-plane σ ≥ 1 + δ, whereδ > 0. From this, we know thatThen it is true that∞∑n=1∞∑n=11(n + a) < ∑ ∞1< ∞.σ (n + a)1+δn=11(n + a) < ∑ ∞1< ∞.s (n + a)1+δn=139

We choose M n = (n + a) 1+δ , and by the Weierstraß M-Test, the series∞∑|(n + a) −s |n=1is uniformly convergent in every half-plane σ ≥ 1 + δ. Therefore, we have shownit to be true that∞∑∣ 1 ∣∣∣∣ =(n + a) −sn=1∞∑n=11(n + a) ≤ ∑ ∞1< ∞, (3.3.10)−σ (n + a)−(1+δ)proving that ζ H (s, a) is an analytic function of s in the half-plane σ > 1 becauseζ H (s, a) is uniformly convergent for σ > 1.n=140

4 THE HURWITZ ZETA FUNCTIONNow that we have been introduced to the Hurwitz zeta function, we want to explorehow this function acts when we extend it to the entire s−plane. Accomplishingthis task requires other representations of the Hurwitz zeta function.It should be said that in this chapter the development of the multiple representationsof the Hurwitz zeta function and the process of finding the functional equationare all heavily borrowed from Tom Apostol’s book “Introduction to Analytic NumberTheory” [14].4.1 The Gamma and Hurwitz Zeta FunctionsTo extend the domain of the Hurwitz zeta function we use the method of analyticcontinuation, a technique commonly used in complex analysis to define a function forwhich it is initially undefined or divergent. In order to find the analytic continuationof the Hurwitz zeta function, we need to introduce the integral representation of theGamma function, Γ(s).Γ(s) =∫ ∞0x s−1 e −x dx, σ > 0. (4.1.1)The Gamma function can be extended to the rest of the s−plane except fors = 0, −1, −2, −3, . . ., where Γ(s) has simple poles [4].Other representations of the Gamma function includen s n!Γ(s) = lim, for s ≠ 0, −1, −2, . . . ,n→∞ s(s + 1)(s + 2) . . . (s + n)and the product formula,1Γ(s) = seγs∞∏n=1(1 + s )e −sn , ∀s ∈ C, (4.1.2)n41

where γ ≈ 0.57721566 . . . is the Euler-Mascheroni constant. Since the product inEquation (4.1.2) converges for all s ∈ C, the Gamma function Γ(s) is never zero [8].The Gamma function also satisfies two functional equations,Γ(s + 1) = sΓ(s) (4.1.3)andΓ(s)Γ(1 − s) =πsin(πs) , (4.1.4)which are also valid for all s ∈ C, save the simple poles of Equation (4.1.1).With these pieces of information, we are able to introduce and prove the integralrepresentation of the product Γ(s)ζ H (s, a) [4]. Note: I have taken liberties withthe proof of this theorem, adding information not found in Apostol when deemednecessary.Theorem 4.1.1. For σ > 1 we have the integral representationΓ(s)ζ H (s, a) =∫ ∞0x s−1 e −axdx. (4.1.5)1 − e−x Proof: We start with the integral representation of the Gamma function, Equation(4.1.1) and make the change of variable x = (n + a)t, where a ∈ R + , n =1, 2, . . . . Then we haveΓ(s) ==∫ ∞0∫ ∞0e −x x s−1 dxe −[(n+a)t] [ (n + a)t ] s−1(n + a) dt∫ ∞= (n + a) s e −nt e −at t s−1 dt.042

Divide both sides by (n + a) s gives us∫Γ(s) ∞(n + a) = e −nt e −at t s−1 dt. (4.1.6)s0Summing over all n ≥ 0 in Equation (4.1.6) and we haveΓ(s)ζ H (s, a) =∞∑n=0Γ(s)(n + a) = ∑ ∞ sn=0∫ ∞0e −nt e −at t s−1 dt, (4.1.7)where the series on the right of Equation (4.1.7) converges if σ > 1 [5].Now we need to switch the order of the sum and integral to prove the integralrepresentation is true for any real number s, where s > 1. The easiest way toaccomplish this is through the properties of Lebesgue integration. In particular,if a function is Riemann integrable, then it is also Lebesgue integrable, and bothforms of integration return the same result [4]. Thus we will treat the integrationin Equation (4.1.7) as Lebesgue integration.Since the integrand e −nt e −at t s−1 from the right side of Equation (4.1.7) is nonnegative,we can apply Levi’s Convergence theorem [5], which tells us that theseries∞∑e −nt e −at t s−1n=0converges almost everywhere to a Lebesgue-integrable function on [0, +∞) [16].More importantly, we now know thatΓ(s)ζ H (s, a) =∞∑∫ ∞e −nt e −at t s−1 dt =∫ ∞n=0 00 n=0∞∑e −nt e −at t s−1 dt. (4.1.8)43

There is a geometric series inside the right-hand side of Equation (4.1.8)∫ ∞ ∞∑0 n=0e −nt e −at t s−1 dt =∫ ∞e −at t s−1∞∑( ) n ∫ 1 ∞dt = e −at t s−1e t∞∑0n=00n=0allowing us to rewrite the right side of Equation (4.1.9) ase −nt dt,(4.1.9)∫ ∞0e −at t s−1∞∑e −nt dt =∫ ∞n=00e −at t s−1e t ∫ ∞e t − 1 dt =0e −at t s−11 − e −t dt.Therefore,Γ(s)ζ H (s, a) =∫ ∞0e −at t s−1, for s ∈ R. (4.1.10)1 − e−t Now we extend the proof to the complex numbers s = σ + iτ for σ > 1. We dothis by showing∣∫ ∞0e −at t s−1dt1 − e −t ∣ < ∞.We know the modulus of the integral is less than the integral of the modulus, sowe know that∣∫ ∞0e −at t s−1 ∫ ∞∣dt1 − e −t ∣ ≤ e −at t s−1 ∣∣∣∣ dt. (4.1.11)1 − e −t0We can bound the right side of Equation (4.1.11) as follows:∫ ∞0∣ e −at t s−1 ∫ ∣∣∣ ∞(∫e −at t σ−11∣ dt ≤dt = +1 − e −t 0 1 − e −t 0∫ ∞1) e −at t σ−11 − e −t dt.Let’s break this integration into two pieces,I 1 =∫ 10e −at t σ−11 − e −t dt (4.1.12)andI 2 =∫ ∞1e −at t σ−11 − e −t dt. (4.1.13)44

First we work with Equation (4.1.12). Multiplying the numerator and denominatorby e t gives us∫ 10e −at t σ−1 ∫ 1e t e −at t σ−1 ∫ 1e (1−a)t t σ−1dt =dt =dt. (4.1.14)1 − e −t 0 e t − 10 e t − 1Let 1 + δ ≥ σ > 1, where δ > 0. Since 0 ≤ t ≤ 1, it follows that t σ−1 ≤ t δ .Similarily, e (1−a)t ≤ e 1−a for t in the interval [0, 1]. Using these relationships, wecan rewrite the right side of Equation (4.1.14) as∫ 10e (1−a)t t σ−1e t − 1dt ≤∫ 10e (1−a)t t δe t − 1 dt ≤ ∫ 10e (1−a) t δe t − 1dt. (4.1.15)In the interval 0 ≤ t ≤ 1, we know that e t ≥ 1. We know the exponential functione t is equal to the sum of its Maclaurin series [28], thusThis meanse t =∞∑n=0x n, ∀t ∈ [0, 1].n!e t = t00! + t11! + t22! + · · ·= 1 + t + 1 2 t2 + · · · ,soe t − 1 = t + 1 2 t2 + · · · .Therefore e t − 1 ≥ t for 0 ≤ t ≤ 1. Applying this fact to Equation (4.1.15) tells45

us∫ 10e (1−a)t t δ ∫ 1e t − 1 dt ≤ e (1−a)t t δdt0 t∫ 1≤ e (1−a) t δ−1 dt= e1−aδ .0Thus∫ 10e −at t σ−11 − e −t dt ≤ e1−aδ .For the integrand in Equation (4.1.13) we let c > σ > 1. This means t σ−1 ≤ t c−1 ,leading us to∫ ∞1e −at t σ−1 ∫ ∞e −at t c−1dt ≤dt = Γ(c)ζ1 − e −t 1 1 − e −t H (c, a).Recombining our integrals, we find∫ ∞0∣ (∫ e −at t s−1 ∣∣∣ 1∣ dt ≤ +1 − e −t 0∫ ∞1) e −at t σ−1dt ≤ e1−a1 − e −t δ+ Γ(c)ζ H (c, a),where δ > 0 and c > 1.This shows that the integral∫ ∞0∣ e −at t s−1 ∣∣∣∣ dt1 − e −tis bounded, therefore it converges uniformly in every strip 1 + δ ≤ σ < 1, whereδ > 0, and therefore represents an analytic function in every such strip, thus alsoin the entire half-plane σ > 1. This means Equation (4.1.5) holds for all complexnumbers s, for σ > 1, proving Theorem 4.1.1.46

4.2 Contour Integral RepresentationTo completely extend ζ H (s, a) beyond the line σ = 1, we need another representationof the Hurwitz zeta function. In this case, we create a representation interms of a contour integral. We will use the contour C shown in Figure 19 notingz = 0 is a branch point of z s−1 . The contour C is a loop around the negative realaxis and is composed of the three parts C 1 , C 2 , and C 3 . C 2 is a positively oriented(counterclockwise) circle centered at the origin and of radius c < 2π while C 1 is thelower edge and C 3 is upper edge of a “cut” in the z−plane along the negative realaxis. The contour is shown in Figure 19. We use the parameterizations z = re −πiFigure 19: Contour integral used for representing ζ H (s, a)on C 1 and z = re πi on C 3 , where r varies from c to +∞.Lemma 4.2.1. If 0 < a ≤ 1, the function defined by the contour integralI(s, a) = 12πi∫Cz s−1 e azdz (4.2.1)1 − ez is an entire function of s ∈ C, where s = σ + it.Note: This proof is adapted from [4].47

Proof: Recall that z s is defined differently for each contour. Those definitionsare listed below.⎧r s e −πis , for z on C 1⎪⎨z s = c s e iθs , for z on C 2⎪⎩ r s e πis , for z on C 3 .Let D be an arbitrary compact disk |s| ≤ M. Now we need to prove that theintegrals along C 1 and C 3 converge uniformly on every such disk. The integrandof Equation (4.2.1) is an entire function of s ∈ C, so we will prove the integralI(s, a) is defined for all complex numbers s as well.Along contour C 1 we have|z s−1 | = r σ−1 |e −πi(σ−1+it) | = r σ−1 e πt ≤ r M−1 e πM , for r ≥ 1.Along contour C 3 we have, for r ≥ 1,|z s−1 | = r σ−1 |e πi(σ−1+it) | = r σ−1 e −πt ≤ r M−1 e πM , for r ≥ 1.Hence on either contour C 1 or C 3 , we have∣ z s−1 e az ∣∣∣∣ ≤ rM−1 e πM e −ar= rM−1 e πM e (1−a)r, for r ≥ 1.1 − e z 1 − e −r e r − 1But e r − 1 > er2 when r > ln(2), so the integrand is bounded by ArM−1 e −at whereA depends on M. Since∫ ∞cr M−1 e −at dr converges if c > 0, the integrals alongcontours C 1 and C 3 converge uniformly on every compact disk |s| ≤ M, and henceI(s, a) is an entire function of s.Note: ln(x) means “natural log,” or log e (x).48

Theorem 4.2.2.ζ H (s, a) = Γ(1 − s)I(s, a), for σ > 1. (4.2.2)Proof: To prove Theorem 4.2.2, we must deal with(∫2πiI(s, a) =C 1+∫C 2+∫C 3)z s−1 g(z) dz, (4.2.3)where g(z) =eaz1 − e . On contours C z 1 and C 3 we have g(z) = g(−r). On C 2we have z = ce iθ , making dz = ice iθ dθ, where −π ≤ θ ≤ π. This means Equation(4.2.3) can be rewritten as2πiI(s, a) =∫ c+= −r s−1 e −πis g(−r) dr +∞∫ ∞c∫ ∞r s−1 e πis g(−r) dr∫ π−πr s−1 e −πis g(−r) dr + i(ceiθ ) s−1g(ce iθ )ice iθ dθ∫ πc s e isθ g(ce iθ ) dθ +c−πc∫ ∞r s−1 e πis g(−r) dr.Now that the delimiters match, we can combine the integrands of our first andthird terms and simplify, obtaining2πiI(s, a) = 2i sin(πs)∫ ∞∫ πr s−1 g(−r)dr + ic s e isθ g(ce iθ ) dθ. (4.2.4)c−πFrom here, we divide both sides by 2i sin(πs), giving us∫πI(s, a) ∞sin(πs) = r s−1 c s ∫ πg(−r)dr +e isθ g(ce iθ ) dθ. (4.2.5)2 sin(πs) −πcFor simplification, letI 1 (s, c) =∫ ∞cr s−1 g(−r) dr (4.2.6)49

andI 2 (s, c) =∫ π−πe isθ g(ce iθ ) dθ. (4.2.7)Now we want to know the behavior of these integrals as c → 0, so we take thelimit of both sides of Equation (4.2.5):πI(s, a)limc→0 sin(πs) = lim I c s1(s, c) + limc→0 c→0 2 I 2(s, c). (4.2.8)First, consider I 1 (s, c).∫ ∞lim I r s−1 e −ar1(s, c) =c→00 1 − e dr = Γ(s)ζ H(s, a), (4.2.9)−rso long as σ > 1.Now we need to show limc→0I 2 = 0.We remember g(z) is analytic in |z| < 2πexcept for a first order pole at z = 0. Ergo, zg(z) is analytic everywhere inside|z| < 2π and is therefore bounded inside the same region. Let |g(z)| ≤ A|z| , where|z| = c < 2π and A is a constant. Therefore, we have|I 2 (s, c)| ≤ cσ 2∫ πAe −tθ−π cdθ ≤ Ae π|t| c σ−1 .This is useful because for σ > 1,limc→0 Aeπ|t| c σ−1 = 0, (4.2.10)thuslim I 2(s, c) = 0. (4.2.11)c→050

This means Equation (4.2.8) simplifies toπsin(πs) I(s, a) = Γ(s)ζ H(s, a). (4.2.12)Using the identity in Equation (4.1.4), we can rewrite the left side of Equation(4.2.12) asΓ(s)Γ(1 − s)I(s, a) = Γ(s)ζ H (s, a), Γ(s) ≠ 0 (4.2.13)from which it follows thatζ H (s, a) = Γ(1 − s)I(s, a), σ > 1,exactly the statement we set out to prove.4.3 An Analytic Continuation of the Hurwitz Zeta FunctionUp to this point, we have learned:• Γ(1 − s) is defined for s ≠ 1, 2, 3 . . . .• Lemma 4.2.1 tells us I(s, a) is an entire function for all complex numbers s ∈ C.• We also proved that ζ H (s, a) = Γ(1 − s)I(s, a) is true for σ > 1.Therefore, we know Equation (4.2.2) can be used to define the Hurwitz zeta functionfor s = σ + iτ where σ ≤ 1.Definition 4.3.1. When σ ≤ 1, we define the ζ(s, a) byζ H (s, a) = Γ(1 − s)I(s, a). (4.3.1)51

We want Equation (4.3.1) to define ζ H (s, a) for all s ∈ C. To do this we mustdeal with the poles of the Gamma function Γ(s).Theorem 4.3.2. The function ζ H (s, a) as defined by Equation (4.3.1) is analyticfor all s ∈ C except for a simple pole at s = 1 where it has residue 1.Proof: We know I(s, a) is an entire function of s by Lemma 4.2.1. Thereforesingularities of ζ H (s, a) can only occur at the poles of Γ(1 − s) found at s =1, 2, 3, . . . . We know from Theorem 4.1.5 that ζ H (s, a) is analytic for σ > 1, thuswe can throw out the poles s = 2, 3, 4, . . . , leaving us with s = 1 as the onlypossible pole of ζ H (s, a).Now all we need to do is show that ζ H (s, a) has non-zero residue at the pole s = 1.We use the contour integralI(s, a) = 1 ∫z s−1 g(z) dz, (4.3.2)2πi Cwhereg(z) =eaz1 − e zand the contour C is same contour found in Figure 19. Returning to Equation(4.2.4),2πiI(s, a) = 2i sin(πs)∫ ∞∫ πr s−1 g(−r)dr + ic s e isθ g(ce iθ ) dθ,c−πand dividing both sides by 2πi, leaves us withI(s, a) = sin(πs)π∫ ∞c∫ πr s−1 g(−r)dr + cse isθ g(ce iθ ) dθ.2π −πThe leading contour integral is the sum of the integrals along contours C 1 andC 3 . Since we are dealing with the cases where s = n, n ∈ Z, our leading contour52

integral vanishes because sin(πn) = 0, leaving us with,I(n, a) = 1 z2πi∫C n−1 e az21 − e dz. zWe are only considering s = n = 1, which simplifies matters a little:I(1, a) = 1 e2πi∫C az21 − e dz z= 1 []eaz2πi 2πi Res 1 − e ; z = 0 z[ eaz= Res1 − e z ; z = 0 ].Looking at our integrand, the numerator e az is an entire function of z ∈ C, sowe only need to consider when the denominator is zero, or e z = 1. So we knowz = 2mπi, but |z| = c < 2π, which only occurs when m = 0. This gives us a poleof order 1 at z = 0, resulting in the fact[ ]eazI(1, a) = Res1 − e ; z = 0 zze az= limz→0 1 − e = −1.zWe calculate the residue of ζ H (s, a) at s = 1 as:lim (s − 1)ζ H(s, a) = − lim [(1 − s) Γ(1 − s) I(s, a)]s→1 s→1= −I(1, a) lims→1(1 − s)Γ(1 − s).53

By Equation (4.1.3), we can rewrite this expression as−I(1, a) lims→1(1 − s)Γ(1 − s) = −I(1, a) lims→1Γ(1 − s + 1)= −(−1) lims→1Γ(2 − s)= Γ(1) = 1.Therefore, ζ H (s, a) has a simple pole at s = 1 with residue 1.Thus Equation (4.3.1) provides us with an analytic continuation of the Hurwitz zetafunction. Note: Since ζ H is analytic at s = 2, 3, 4, . . . and Γ(1 − s) has poles at theintegers, Equation (4.3.1) implies I(s, a) vanishes at these points.We are then able to see how this particular analytic continuation of the Hurwitzzeta function behaves in the entire s−plane. However, for the purposes of calculatingthe Julia and Mandelbrot sets, this analytic continuation is not very helpful. Asyou may recall from Chapter Two, the calculation of the Julia set is computationallyintense, so we seek yet another representation of the Hurwitz zeta function tofacilitate the calculation.4.4 A Functional Equation for the Hurwitz Zeta FunctionNow that we have been through the trenches of the Hurwitz zeta function, weknow ζ H (s, a) is defined asζ H (s, a) =∞∑n=01(n + a) s for σ > 1.Hurwitz himself actually found another series representation for ζ H (s, a) whichhas natural domain σ < 0 [4]. To get to this representation, we need some moreinformation.54

{}Figure 20: The punctured rectangle Q(r) = z ∣ |x| ≤ 1, |y| ≤ π, |z| ≥ r , referencedin Lemma 4.4.1Lemma 4.4.1. Let S(r) represent the region that remains when we remove all opencircular disks centered at z = 2nπi, n ∈ Z, with radius r. Then, if 0 < a ≤ 1, thefunctiong(z) =eaz1 − e zis bounded in S(r).Proof: We split up the z−plane into two pieces and show g(z) is bounded inboth. Let z = x + iy and consider the punctured rectangleQ(r) ={}z ∣ |x| ≤ 1, |y| ≤ π, |z| ≥ r .Now Q(r) is a compact set so we know g(z) is bounded on Q(r).Put thattogether with the periodicity of the function g, g(z + 2πi) = g(z), and it followsg(z) is bounded on any punctured rectangle Q n (r) with width 2 and centered atz = 2nπi, where n ∈ Z. The union all these rectangles in the z−plane is the55

Figure 21: Punctured rectangles making the infinite strip Q n (r)infinite strip⋃Q n (r) =n∈Z{z}∣∣z = x + iy, |x| ≤ 1, |z − 2nπi| ≥ r . (4.4.1)Since g(z) is bounded in this strip, we only need to show that g(z) is also boundedoutside the infinite strip.We start on the right side of the infinite strip (see Figure 21 for visual help),which means x ≥ 1. Then∣ |g(z)| =e az ∣∣∣∣ = eax1 − e z |1 − e z | ≤ eax|1 − e x | . (4.4.2)For x ≥ 1 we know |1 − e x | = e x − 1. Also, since 0 < a ≤ 1, it is true thate ax ≤ e x . Thus|g(z)| ≤eax|1 − e x | = eaxe x − 1 ≤exe x − 1 = 11 − e < 1−x 1 − e = e−1 e − 1 .56

Therefore,|g(z)|

So long as a ≠ 1, this formula is valid for σ > 0.Figure 22: Contour C(N) referred to in Equation (4.4.7)Proof: Consider the functionI N (s, a) = 1 ∫2πi C(N)z s−1 e azdz, (4.4.7)1 − ez where C(N) is the contour shown in Figure 22 for N ∈ Z.First, we need to prove thatlim I N(s, a) = I(s, a), for σ < 0. (4.4.8)N→∞We want to show the integral along the outer circle tends to 0 as R = (2N + 1)πtends to infinity. On the outer circle, we have z = Re iθ for −π ≤ θ ≤ π.Remembering s = σ + iτ and the work from the proof of Theorem 3.3.3, we find∣ zs−1 ∣ ∣ =∣ ∣R s−1 e iθ(s−1)∣ ∣ = R σ−1 e −tθ ≤ R σ−1 e π|t| .58

No matter the size of N, the outer circle lies in the set S(r) from Lemma 4.4.1,thus the integrand is bounded by AR σ−1 e π|t| , where A is the bound for |g(z)|.Hence the integral is bounded by2πA R σ e π|t| ,for σ < 0, andlim 2πAR→∞ Rσ e π|t| = 0.Shifting the region of interest from σ < 0 to σ > 1 by replacing s with 1 − s inEquation (4.4.8), we see thatlim I N(1 − s, a) = I(1 − s, a), if σ > 1. (4.4.9)N→∞Now we compute I N (1 − s, a) explicitly by Cauchy’s Residue theorem. We haveI N (1 − s, a) = −N∑n=−Nn≠0R(n) = −N∑ [ ]R(n) + R(−n) , (4.4.10)n=1where[ ]z −s e azR(n) = Res1 − e ; z = 2nπi .zCalculating the residue gives usR(n) =lim (z − 2nπi) z−s e azz→2nπi 1 − e = e2nπia z − 2nπilim = − e2nπiaz (2nπi) s z→2nπi 1 − e z (2nπi) . sTherefore,I N (s, a) =N∑n=1e 2nπia(2nπi) + ∑ Ne −2nπias (−2nπi) . sn=159

However, i −s = e −πis/2 and (−i) −s = e πis/2 , soI N (1 − s, a) = e−πis/2(2π) s∑ Ne 2nπian sn=1+ eπis/2(2π) sN∑n=1e −2nπian s .Now if we let N → ∞ on both sides, we findlim I e −πis/2N(1 − s, a) = limN→∞ N→∞ (2π) s= e−πis/2(2π) slimN→∞∑ Ne 2nπian=1N∑n=1n se 2nπian s+ eπis/2(2π) s+ eπis/2(2π) s∑ Ne −2nπian=1limN→∞n sN∑n=1e −2nπian s .Using Theorem 4.4.9, we find thatI(1 − s, a) = e−πis/2 eπis/2F (a, s) + F (−a, s).(2π)s(2π)sTherefore,ζ H (1 − s, a) = Γ(s)I(1 − s, a) = Γ(s)(2π) s [e −πis/2 F (a, s) + e πis/2 F (−a, s) ] .We are now ready to be introduced to the ingenious functional equation of theHurwitz zeta function, the keystone of this work.Theorem 4.4.3 (A Functional Equation of ζ H (s, a)). If h and k are integers, 1 ≤h ≤ k, then for all s we have(ζ H 1 − s, h )= 2Γ(s)k (2πk) sk∑r=1( πscos2 − 2πrh ( )ζ H s, r ). (4.4.11)k kProof: This functional equation results from the fact that the Periodic zetafunction, F (x, s), is a linear combination of Hurwitz zeta functions when the60

argument x is rational. Consider the Periodic zeta function when we let x = h/k,for h, k ∈ Z, and gcd(h, k) = 1. Substitute this into Equation (4.4.5) and we have( ) hFk , s =∞∑n=1e 2πinh/kn s . (4.4.12)Now, let n = qk + r, where 1 ≤ r ≤ k and q = 0, 1, 2 . . . . This allows us torearrange the terms in Equation (4.4.12) according to the residue classes mod k.For σ > 1, we find that∞∑n=1e 2πinh/kn s =k∑∞∑r=1 q=0e 2πi(qk+r)h/k(qk + r) s =k∑∞∑r=1 q=0(e 2πiq ) h e 2πirh/k(qk + r) s ,Since ( e 2πiq) h= 1 h = 1, we havek∑∞∑r=1 q=0(e 2πiq ) h e 2πirh/k(qk + r) s ===k∑∞∑r=1 q=0k∑r=1k∑r=1= 1 k s k∑e 2πirh/ke 2πirh/kr=1e 2πirh/k(qk + r) s∞∑q=0∞∑q=0e 2πirh/k1(qk + r) s1[ (k q + r )] sk∞∑q=01(q + r ) s .kSo far, we havek∑r=1 q=0∞∑ (e 2πiq ) h e 2πirh/k= 1 k∑(qk + r) s k sr=1(e 2πirh/k ζ H s, r ).k61

From this fact, we know that( ) hFk , s = 1 k∑(e 2πirh/k ζk s H s, r ). (4.4.13)kr=1We have shown that F (h/k, s) is a linear combination of Hurwitz zeta functions.Now recall Hurwitz’s formula as found in Equation (4.4.7), which statesζ H (1 − s, a) = Γ(s) []e −πis/2 F (a, s) + e πis/2 F (−a, s) .(2π) sNow let a = h/k and substitute it into Equation (4.4.13), resulting inTherefore,(ζ H 1 − s, h ) (= Γ(s) e −πis/2 · 1k (2π) s k s+ e πis/2 · 1k s= Γ(s)(2πk) s= 2Γ(s)(2πk) sr=1k∑r=1k∑r=1([e 2πirh/k ζ H s, r )]k)] )[e −2πirh/k ζ H(s, r kk∑( πs2 cos2 − 2πrh ( )ζ H s, r )k kr=1k∑( πscos2 − 2πrh ( )ζ H s, r ).k k(ζ H 1 − s, h )= 2Γ(s)k (πk) sk∑r=1( πs2 cos2 − 2πrh ( )ζ H s, r ), (4.4.14)k kis true for all complex numbers s by analytic continuation.While valid, Equation (4.4.14) is not in a useful form for computing the Hurwitzzeta function ζ H (s, a) for s < 1. Substitute 1 − s in Equation (4.4.14) for s and62

earrange as follows:(ζ H s, h )=kTherefore,(ζ H s, h )k2Γ(1 − s)(2πk) 1−sk∑( π(1 − s)cos − 2πrh ( )ζ H 1 − s, r )2 kkr=1(k∑= 2Γ(1 − s) (2πk) s−1 cosr=1= 2Γ(1 − s) (2πk) s−1 k∑r=1= 2Γ(1 − s) (2πk) s−1 k∑r=1[π πs2 − 2 + 2πrhk( πssin2 + 2πrhk] ) (ζ H 1 − s, r )k)ζ H(1 − s, r k).( πssin2 + 2πrh ( )ζ H 1 − s, r ). (4.4.15)kkWe evaluated ζ H (s, h/k) in Equation (4.4.15) for different values of s and a = h/kin Maple c○ using its enormous computational power. The results were then used inFractInt 20.4 c○ , returning images for the iterative functionF c (s, a) = ζ H (s, a) + c.Equation (4.4.15) is used to calculate the Hurwitz zeta function for values of s ∈ Cwhere σ < 0. Linas Vepštas [29] proved that by using the Euler-Maclaurin Summationformula, we can confidently compute Hurwitz zeta functions for s ∈ C whereσ > 0. For this, we need to introduce the Euler-Maclaurin Summation formula.63

5 CALCULATING THE HURWITZ ZETA FUNCTION5.1 Euler-MacLaurin Summation FormulaWe can compute the Hurwitz zeta function based on the Euler-Maclaurin summationformula.Definition 5.1.1. The Euler-Maclaurin Summation FormulaN∑f(n) =∫ NMMf(x) dx +f(M) + f(N)2+p∑k=1B 2k(2k)! f (2k−1) (x) ∣ N + R 2p , (5.1.1)Mwhere B n is the n th Bernoulli number, and R 2p is the remainder term [29].To understand this formula we start with the Trapezoidal Rule, where x i −x i−1 =1, and rearrange terms as follows:∫ NMf(x) dx ≈ N − M []f(M) + f(x 1 ) + f(x 1 ) + · · · + f(N)2n≈ 1 []f(M) + 2f(x 1 ) + · · · + 2f(x n−1 ) + f(N)2n∑≈ f(x i ) − 1 2 f(M) − 1 2 f(N),i=0where x 0 = M and x n = N. So we now have∫ NMf(x) dx ≈n∑i=0f(x i ) − M + N . (5.1.2)2This tells us that integrals can be approximated by a summation. What we needis a way to approximate an integral. If we subtract the integral from the left-handside and the sum from the right-hand side of Equation (5.1.2), then multiply bothsides by -1, we arrive at:N∑f(n) ≈∫ Nn=MMf(x) dx + 1 ]f(M) + f(N) . (5.1.3)2[64

This approximation may be inadequate for certain functions, so we introduce theerror term, E f (M, N), defined asE f (M, N) =N∑(∫ Nf(n) −n=MMf(x) dx + f(M)2+ f(N) ). (5.1.4)2Inserting the error term E f (M, N) into Equation (5.1.2) gives us the equationN∑f(n) =n=M∫ NMf(x) dx +f(M) + f(N)2+ E f (M, N).To define the error precisely, we introduce the greatest integer function [x], definedas [14]:⎧⎪⎨ x, if x ∈ Z[x] =⎪⎩ ⌊x⌋, if x /∈ Z,(5.1.5)where the function ⌊x⌋ is known as the floor function, and returns the largest integerwhich is not greater than x. In other words, we truncate any fractional part of thenumber x. Due to the construction of [x], the greatest integer function has jumps ofone at the integers. This means the Stieltjes measure d ( [x] ) assigns values as follows[14]:⎧d ( [x] ) ⎪⎨ 1, if x ∈ Z=⎪⎩ 0, if x /∈ Z.(5.1.6)This is unlike Riemann integration because when integrated it sums up values only65

at integers. Hence,∫ NMf(x) d ( [x] ) = 1 f(M) + f(M + 1) + f(M + 2) + · · ·2=+ f(N − 1) + 1 2 f(N)N∑f(n) −n=Mf(M) + f(N).2Now we can rewrite our error term from Equation (5.1.4) asE f (M, N) ===N∑(∫ Nf(n) − f(x) dx + f(M)2n=MN∑n=M∫ NMMf(n) − f(M)2f(x) d ( [x] ) −− f(N) −2∫ NMf(x) dx.∫ NM+ f(N) )2f(x) dxThis can be combined into a single integral, giving usE f (M, N) =∫ NMf(x) d ( [x] − x ) . (5.1.7)It is much easier to describe the measure d ( [x] − x ) as d ( [x] − x + 1/2 ) becausethe function [x] − x + 1/2 is positive half the time, negative half the time, and iszero when x ∈ Z. The addition of 1/2 shifts the graph of the function up verticallyone-half. By convention, discontinuities are given the average of the left and righthandlimits. The difference between the two functions can be seen by comparingFigure 24 and Figure 23.Since we are using the shifted measure d ( [x] − x + 1/2 ) , we can rewrite the errorterm of f(x) asE f (M, N) =∫ NM(f(x) d [x] − x + 1 ). (5.1.8)266

Figure 23: Graph of the function [x] − x, [x] is the greatest integer functionFigure 24: Graph of the function [x] − x + 1/267

Performing an integration by parts on the right side of Equation (5.1.8), we find ourerror term can be written asE f (M, N) =∫ NM(x − [x] − 1 )f ′ (x) dx. (5.1.9)2The result of Equation (5.1.9) validates the equationFigure 25: Graph of the equation x − [x] − 1/2N∑f(n) =n=M∫ NMf(x) dx +f(M) + f(N)2+∫ NM(x − [x] − 1 )f ′ (x) dx. (5.1.10)2The advantage of Equation (5.1.10) is minor.The real gain comes in repeatedintegration by parts of the last term.This process gives numerical results withgreat accuracy. The integration by parts requires the use of Bernoulli polynomials,discovered by Jakob Bernoulli (1654-1705), a Swiss mathematician and chair of BaselUniversity’s mathematics department from 1687 until his death [31].68

Definition 5.1.2. The n thBernoulli polynomial is the unique polynomial ofdegree n with the property that∫ x+1xB n (t) dt = x n . (5.1.11)Bernoulli polynomials are useful because they are unique, allowing us to rewriteany variable power function as the integration of a specific polynomial. From therewe solve for unknown coefficients which gives us a new way to write the variablepower function in question. For instance, the cubic B 3 (x) = ax 3 + bx 2 + cx + d canbe rewritten as:∫ x+1x 3 = (at 3 + bt 2 + ct + d) dtx[] [] [] [ ](x + 1) 4 − x 4 (x + 1) 3 − x 3 (x + 1) 2 − x 2 x + 1 − x= a+ b+ c+ d4321( )( 3a a= ax 3 +2 + b x 2 + (a + b + c)x +4 + b 3 + c 2 + d .1)Equating the left and right sides we have four equations and four unknowns. Thisyields: a = 1, b = −3/2, c = 1/2, d = 0. Thus,B 3 (x) = x 3 − 3 2 x2 + 1 2 x.Similarly,B 2 (x) = x 2 − x + 1 6 ,B 1 (x) = x − 1 2 ,B 0 (x) = 1.69

By differentiating Equation (5.1.11), we discover two relationships [14]:∫ x+1x1n B′ n(t) dt = x n−1 , (5.1.12)B n (x + 1) − B n (x) = nx n−1 . (5.1.13)The differentiation of∫ x+1xB ′ n(t) dt = x n , (5.1.14)shows that B ′ n(x)/n satisfies the definition of B n−1 (x), thusB ′ n(x) = nB n−1 (x). (5.1.15)Now we can use Bernoulli polynomials to integrate by parts on the right side ofEquation (5.1.9). Let x = n + t, making [x] = n and dx = dt.E f (M, N) ==∫ NMN−1 ∫ 1∑n=M(f(x) d x − [x] − 1 )f ′ (x) dx20(t − 1 )f ′ (n + t) dt.2By Equation (5.1.15) we rewrite the error in terms of a Bernoulli polynomialE f (M, N) =N−1∑n=M∫ 10(t − 1 )f ′ (n + t) dt = 1 N−1∑22n=M∫ 10B ′ 2(t)f ′ (n + t) dt. (5.1.16)70

Now integrate by parts on the right side of Equation (5.1.16) and we find thatThusN−1∑∫ 1E f (M, N) = 1 B22(t)f ′ ′ (n + t) dtn=M0N−1∑[B2 (t)=f ′ (n + t) ∣ 1 − 1 20 2E f (M, N) =n=MN−1∑=N−1∑n=Mn=MB 2 (t)2B 2 (t)2f ′ ∣(n + t)f ′ ∣(n + t)∣ 1 0∣ 1 0− 1 2− 1 2∫ NM∫ 10∫ NM]B 2 (t) f ′′ (n + t) dtB 2 (x − [x]) f ′′ (x) dx.)B 2(x − [x] f ′′ (x) dx. (5.1.17)Equation (5.1.17) deals with Bernoulli Polynomials where x = 0 and n = 2. Withthese values, Equation (5.1.12) tells us B 2 (1) = B 2 (0). Using this on the first termin Equation (5.1.17) gives usN−1∑n=MB 2 (t)2f ′ (n + t) ∣ 1 =0N−1∑n=MN−1∑=n=MN−1∑=n=M1[]B 2 (1)f ′ (n + 1) − B 2 (0)f ′ (n)21[]B 2 (0)f ′ (n + 1) − B 2 (0)f ′ (n)21[]2 B 2(0) f ′ (n + 1) − f ′ (n) . (5.1.18)71

Simplifying Equation (5.1.18) gives usN−1∑n=M12 B 2(0) [ f ′ (n + 1) − f ′ (n) ] = 1 2 B 2(0) [ f ′ (M + 1) − f ′ (M) ]+ 1 2 B 2(0) [ f ′ (M + 2) − f ′ (M + 1) ] + · · ·+ 1 2 B 2(0) [ f ′ (N − 1) − f ′ (N − 2) ]+ 1 2 B 2(0) [ f ′ (N) − f ′ (N − 1) ]= 1 2 B 2(0) [ f ′ (N) − f ′ (M) ] .Therefore,N−1∑n=M12 B 2(0) [ f ′ (n + 1) − f ′ (n) ] = 1 2 B 2(0) [ f ′ (N) − f ′ (M) ] .This gives us a more accurate approximation of the error term∫ N(E f (M, N) = f(x) d x − [x] − 1 )M2= B 2(0) [f ′ (N) − f ′ (M) ] + 1 ∫ N22M)B 2(x − [x] f ′′ (x) dx.Now carry out integration by parts repeatedly on the right side of Equation (5.1.10),resulting inN∑f(n) =n=M∫ NMf(x) dx + 1 [ ] B 2 (0)f(M) + f(N) + f ′ ∣(x)22f (k−1) (x) ∣ N + (−1) (k+1) 1Mk!+ · · · + (−1) k B k(0)k!∣ N M∫ NM− B 3(0)3!f ′′ ∣(x)∣ N M)B k(x − [x] f (k) (x) dx.(5.1.19)If k is odd, then B k(x − [x])is an oscillating function similar to B3(x − [x]), and thecorresponding integral can be estimated (provided f k is monotone) by an alternating72

series technique. To prove that B 2k+1 (x) has an oscillating character we draw on theidentities [14]:B k (1 − x) = (−1) k B k (x), (5.1.20)[ (B k (2x) = 2 k−1 B k (x) + B k x +2) ] 1 . (5.1.21)From these identities we can see thatB 2k+1 (1 − x) = −B 2k+1 (x),( ) 1B 2k+1 = 0,2B 2k+1 (0) = 2 2k[ B 2k+1 (0) + 0 ] = 0, for k = 1, 2, . . . ,which means( ) 1B 2k+1 = B 2k+1 (0) = 0, for k = 1, 2, . . . .2Theorem 5.1.3. There are no zeros of B 2k+1 (x) between 0 and 1/2.Proof: For sake of contradiction, let r be a zero of B 2k+1 (x) where 0 < r < 1/2.This means there are two zeros of the derivative between 0 and 1/2. Therefore,there is a zero r 1 of the 2 nd derivative of B 2k+1 . By Equation (5.1.15), we knowthe second derivative isB ′′2k+1(x) = (2k + 1)2kB 2k−1 (x), for 0 < r 1 < 1 2 . (5.1.22)Now we repeat the process. Since there is a zero of B 2k−1 (x), we know there aretwo zeros of the Bernoulli polynomial B 2k−2 (x), which means there exists a root r 2of B 2k−2 (x) on the interval [0, 1/2]. This means for every odd indexed Bernoulli73

polynomial, there exists a root between 0 and 1/2, leading to the implicationthere is such a root for the Bernoulli polynomial B 3 (x). However, we knowB 3 (x) = x(x − 1/2)(x − 1),thus we have a contradiction.This means B (2k−1) (x) has one sign on the interval (0, 1/2). By Equation (5.1.20),we know B 2k+1 (1 − x) = −B 2k+1 (x), thus B 2k+1 (x) has the opposite sign on theinterval (1/2, 1), therefore B 2k+1 (x) oscillates, as stated previously.Also, we seethat many terms of Equation (5.1.19) are zero; specifically all terms containingB 2k+1 (0). In order to calculate Equation (5.1.19), one must find the values of theconstants B 2k (0), which are known as the Bernoulli numbers and are denoted as B n ,where n = 0, 1, 2, . . . .Now we have the Euler-Maclaurin summation formulaN∑f(n) =M∫ NMf(x) dx ++ B 44! f ′′′ ∣(x)∣ N Mf(M) + f(N)+ B 22 2 f ′ (x) ∣ N M+ B 2p(2p)! f (2p−1) (x) ∣ N + R 2p ,Mwhere B n are the Bernoulli numbers, f(x) is any function with 2p + 1 continuousderivatives on the interval [N, M], and where R 2p is given byR 2p =∫1 N)B 2p+1(x − [x] f (2p+1) (x) dx. (5.1.23)(2p + 1)! M5.2 Using Euler-Maclaurin Summation Formula on the Hurwitz Zeta FunctionTo use the Euler-Maclaurin summation formula on the Hurwitz zeta function,we split up ζ H (s, a) by subtracting off the first N terms, where N is large enough74

that we can bound R 2p .∞∑(n + a) −s −n=0N−1∑n=0(n + a) −s =N−1∑ζ H (s, a) − (n + a) −s =ζ H (s, a) =N−1∑n=0n=0(n + a) −s +∞∑(n + a) −sn=N∞∑(n + a) −sn=N∞∑(n + a) −s . (5.2.1)The first term on the right can be directly computed, but the second one can not,n=Nso we use Euler-Maclaurin summation for a reliable approximation∞∑(n + a) −s ≈ f(N) +2p∑n=Nk=1∫ ∞NB 2k(2k)!(n + a) −s dsd 2k+1 [ ](n + a) −sds 2k+1x=N+ R 2p .Putting this into Equation (5.2.1) and we haveζ H (s, a) =N−1∑(n + a) −s + f(N)2∫ ∞p∑+ (n + a) −s dsn=0Nk=1B 2k(2k)!d 2k+1 [ ](n + a) −sdx 2k+1x=N+ R 2p , (5.2.2)where∣ ∣ ∣∣s(s ∣∣R 2p ≤ 2 + 1)(s + 2) · · · (s + 2p) 1(2π) 2p σ + 2p (N + a) . σ+2pThis is the method we use for calculating the Hurwitz zeta function for s = σ + iτ,σ > 0. The end of Chapter 4 gave us the way to calculate for values of s, σ < 0. Thismeans we have the ability to calculate the Julia and Mandelbrot sets of the Hurwitzzeta function. All FractInt 20.4 c○ and Maple c○ code used for these calculations canbe found in the Appendix.75

6 JULIA AND MANDELBROT SETS OF ZETA FUNCTIONSAs we mentioned earlier on, we have no way of knowing if our pictures of theJulia sets of the Hurwitz zeta function “look right” because there are no prior imagesof Julia sets of the Hurwitz zeta function. To check our images for obvious mistakes,we created images of the Hurwitz zeta function for a = 1, which is the Riemann zetafunction [see Equation (3.3.7) and the work prior to show this fact]. Sara Ives andS. C. Woon created images for the Riemann zeta function, so we take the imagescreated by this work and compared it Sara’s images in her work, “Julia Sets of theRiemann Zeta Function”. We compare Figure 26 [18] against Figure 27 and Figure 28[18] against Figure 29.76

6.1 Images of the Hurwitz and Riemann Zeta FunctionsFigure 26: Julia set of ζ R (s), s ∈ [0.9048, 1.8314] × [−0.347446, 0.347446]Figure 27: Julia set of ζ H (s, 1), s ∈ [0.9048, 1.8314] × [−0.347446, 0.347446]77

Figure 28: Julia set of ζ R (s), s ∈ [−23.53, 11.33] × [−13.07, 13.07]Figure 29: Julia set ζ H (s, 1), s ∈ [−23.53, 11.33] × [−13.07, 13.07]78

6.2 Julia and Mandelbrot Sets of ζ H (s, a) + cThis section contains multiple images of the Julia sets of ζ H (s, a). We were onlyable to calculate the Hurwitz zeta function for twenty-one values of a = h/k. This isbecause in our functional equation of the Hurwitz zeta function, Equation (4.4.15),k is the upper bound of summation. For k > 8, there were too many terms, andFractInt 20.4 c○ was unable to create an image for these values of a.Figures 30 through 35 contain images of the Julia sets with the frame [−2, 2] ×[−1.5i, 1.5i]. These were created in FractInt 20.4 c○ then moved into a grid patterncreated by Keynote. Within each image you will find the value of a to which theimage corresponds. The images were put in order by a−value, smallest to largest.From these below, we can see as a gets closer and closer to one, there is a basin ofattraction that seems to grow larger and larger, save the case of a = 1/2. Rememberthe frame of each image is [−2, 2] × [−1.5i, 1.5i].Figures 36 and 37 use the Evolver command to draw multiple Julia sets for thefunctions ζ H (s, 1/8) + c and ζ H (s, 1/7) + c, respectively. Each image that makes upthe grid has the domain [−2, 2] × [−1.5i, 1.5i], and the grid spans σ ∈ [−10, 10], τ ∈[−10i, 10i].Figure 38 explores the Mandelbrot set of the ζ H (s, 1/8)+c. Figures 39 through 41look at the Mandelbrot set of ζ H (s, 1/7) + c and zooms in on this image. Figures 42and 43 show the images for ζ H (s, 3/7) + c and ζ H (s, 5/6) + c respectively.Figures 44 through 47 show the Julia set of ζ H (s, 3/5) + c as the value of ais changed from −1.16i to −1.13i.Figures 44 through 47 show the Julia set ofζ H (s, 2/5) + c as the value of a is changed from −0.71i to −0.68i.Figures 52 and 53 show off the unexplained triangles in the Julia sets of ζ H (s, 5/8)+c and ζ H (s, 3/8) + c respectively.79

80Figure 30: Julia set of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

81Figure 31: Julia sets of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

82Figure 32: Julia sets of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

83Figure 33: Julia sets of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

84Figure 34: Julia sets of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

85Figure 35: Julia sets of ζH(s, a) + c, c = 0, s ∈ [−2, 2] × [−1.5i, 1.5i]

86Figure 36: Julia sets of ζH(s, 1/8) + c, s ∈ [−2, 2] × [−1.5i, 1.5i]

87Figure 37: Julia sets of ζH(s, 1/7) + c, s ∈ [−2, 2] × [−1.5i, 1.5i]

Figure 38: M of ζ H (s, 1/8) + c, c ∈ [−18, 6] × [−8i, 8i]Figure 39: M of ζ H (s, 1/7) + c, c ∈ [−18, 6] × [−8i, 8i]88

Figure 40: M of ζ H (s, 1/7) + c, c ∈ [−17.22, −13.22] × [−3.87i, 0.87i]Figure 41: M of ζ H (s, 1/7) + c, c ∈ [−10.27, −5.58] × [−3.83i, 6.96i]89

Figure 42: Mandelbrot set of ζ H (s, 3/7) + c, c ∈ [−24, 12] × [−12i, 12i]Figure 43: Mandelbrot set of ζ H (s, 5/6) + c, c ∈ [−24, 24] × [−18i, 18i]90

Figure 44: Julia set of ζ H (s, 3/5) + c, c = −1.16i, s ∈ [−24, 24] × [−18i, 18i]Figure 45: Julia set of ζ H (s, 3/5) + c, c = −1.15i, s ∈ [−24, 24] × [−18i, 18i]91

Figure 46: Julia set of ζ H (s, 3/5) + c, c = −1.14i, s ∈ [−24, 24] × [−18i, 18i]Figure 47: Julia set of ζ H (s, 3/5) + c, c = −1.13i, s ∈ [−24, 24] × [−18i, 18i]92

Figure 48: Julia set of ζ H (s, 2/5) + c, c = −0.71i, s ∈ [−24, 24] × [−18i, 18i]Figure 49: Julia set of ζ H (s, 2/5) + c, c = −0.70i, s ∈ [−24, 24] × [−18i, 18i]93

Figure 50: Julia set of ζ H (s, 2/5) + c, c = −0.69i, s ∈ [−24, 24] × [−18i, 18i]Figure 51: Julia set of ζ H (s, 2/5) + c, c = −0.68i, s ∈ [−24, 24] × [−18i, 18i]94

Figure 52: Grid of Julia sets of ζ H (s, 5/8) + c, s ∈ [−32, 32] × [−24i, 24i]Figure 53: Grid of Julia sets of ζ H (s, 3/8) + c, c = 0 + i, s ∈ [−32, 32] × [−24i, 24i]95

6.3 Values of the Hurwitz and Riemann Zeta FunctionsAnother way we can check to see if our equation is correct is by comparing theactual computed value of the Hurwitz zeta function to the value found by Maple c○ .We use the case a = 1 to compare against Maple’s Riemann zeta function and thecase a ≠ 1 to compare against Maple’s Hurwitz zeta function. This goes to showour calculations are correct, giving more credibility to the images of the Julia andMandelbrot sets of the Hurwitz zeta function.Tables 1 and 2 compare values of ζ H (s, 1) as calculated in Chapter 5 to values ofζ R (s) as computed in Maple c○ . For Table 1, the value of s = σ + iτ is in the planes ∈ [1, 4] × [−2i, 2i], whereas for Table 2, s ∈ [−2, 2] × [−2i, 2i]Table 3 looks at values of ζ H (s, 1/8) as calculated in Chapter 5 and those valuesof ζ H (s, 1/8) as computed in Maple c○ . The value of s is in the plane s ∈ [−5, 5] ×[−5i, 5i].96

s−Value ζ H (s, 1) ζ R (s) Difference1-2i 0.5982 + 0.3519i 0.5982 + 0.3519i 2.0 × 10 −10 + 7.0i × 10 −101-i 0.5822 + 0.9268i 0.5822 + 0.9268i 1.4 × 10 −9 + 8.0i × 10 −101+i 0.5822 - 0.9268i 0.5822 - 0.9268i 7.0 × 10 −10 + 2.8i × 10 −91+2i 0.5982 - 0.3518i 0.5982 - 0.3518i 1.2 × 10 −9 + 2.0i × 10 −102-2i 0.8674 + 0.2751i 0.8674 + 0.2751i 02-i 1.1503+0.4375i 1.1503+0.4375i 1.0 × 10 −9 + 1.0i × 10 −102 1.6449 1.6449 1.0 × 10 −92+i 1.1503-0.4375i 1.1503-0.4375i 3.0 × 10 −9 + 1.0i × 10 −102+2i 0.8673-0.2751i 0.8673-0.2751i 3.0 × 10 −10 − 1.0i × 10 −93-2i 0.9730+0.1477i 0.9730+0.1477i 7.0 × 10 −103-i 1.1072+0.1483i 1.1072+0.1483i 2.0i × 10 −103 1.2020 1.2020 03+i 1.1072-0.1483i 1.1072-0.1483i 1.0 × 10 −9 + 2.0i × 10 −103+2i 0.9730-0.1477i 0.9730-0.1477i 9.0 × 10 −104-2i 0.9979+0.0714i 0.9979+0.0714i 5.0 × 10 −10 + 1.0i × 10 −114-i 1.0535+0.0580i 1.0535+0.0580i 2.0 × 10 −9 + 2.0i × 10 −114 1.0823 1.0823 04+i 1.0535-0.0580i 1.0535-0.0580i 2.0 × 10 −9 + 2.0i × 10 −114+2i 0.9979-0.0715i 0.9979-0.0715i 8.0 × 10 −10 + 4.0i × 10 −11Table 1: Comparison of computed ζ H (s, 1) versus Maple c○ ζ R (s)97

s−Value ζ(s, 1) ζ(s) Difference−0.5 − 2i 0.2281 + 0.1445i 0.2281 + 0.1445i 1.52 × 10 −6 − 1.60i × 10 −6−0.5 − i −0.0008 + 0.2231i −0.0008 + 0.2231i 2.98 × 10 −6 + 7.68i × 10 −6−0.5 −0.2079 −0.2079 1.68 × 10 −5−0.5 + i −0.0008 − 0.2231i −0.0008 − 0.2231i −2.97 × 10 −6 − 7.68 × 10 −6−0.5 + 2i 0.2281 − 0.1445i 0.2281 − 0.1445i 1.52 × 10 −6 + 1.60 × 10 −6−2i 0.3147 − 0.2317i 0.3147 + 0.2317i 2.59 × 10 −7 − 5.87 × 10 −6−i 0.0033 + 0.4182i 0.0033 + 0.4181i 2.49 × 10 −5 + 4.59i × 10 −5i 0.0033 − 0.4182i 0.0033 − 0.4182i 2.49 × 10 −5 − 4.59i × 10 −52i 0.3147 − 0.2317i 0.3147 − 0.2317i 2.60 × 10 −7 + 5.87i × 10 −6−1 − 2i 0.1689 + 0.0705i 0.1689 + 0.0705i 7.50 × 10 −7 + 2.32i × 10 −7−1 − i 0.0169 + 0.1142i 0.0169 + 0.1142i −1.03 × 10 −6 + 1.06i × 10 −6−1 −0.0833 −0.0833 −1.75 × 10 −6−1 + i 0.0169 − 0.1142i 0.0169 − 0.1142i −1.03 × 10 −6 − 1.06i × 10 −6−1 + 2i 0.1689 − 0.0705i 0.1689 − 0.0705i 7.5 × 10 −7 + 2.32i × 10 −7−2 − 2i 0.0864 − 0.0205i 0.0864 − 0.0205i 1.12 × 10 −7 + 1.87i × 10 −9−2 − i 0.0292 + 0.0178i 0.0292 + 0.0178i 4.18 × 10 −8 + 6.84i × 10 −8−2 0 0 0−2 + i 0.0292 − 0.0178i 0.0292 − 0.0178i −4.18 × 10 −8 + 6.84i × 10 −8−2 + 2i 0.0864 + 0.0205i 0.0864 + 0.0205i 1.12 × 10 −7 + 1.89i × 10 −9Table 2: Comparison of computed ζ H (s, 1) versus Maple c○ ζ R (s)98

s-Value(ζ s, 1 )8(Maple c○ c○ ζ 0, s, 1 )8Difference−5 ± 5i .50697 ± .07433i .50697 ± .07433i 1.8 × 10 −7 ± 5.5i × 10 −9−5 ± i −.0071 ± .0050i −.0071 ± .0050i 8.0 × 10 −9 ± 1.4i × 10 −10−5 −.002753272616 −.002753272889 2.73 × 10 −10−1 ± 5i .50697 ± .07433i .50697 ± .07433i 9.023 × 10 −8−1 ± i −.10734 ± .05426i −.10734 ± .05426i 1.0 × 10 −6 ± 1.4i × 10 −6−1 −.0286452 −.0286458 6.0 × 10 −7−0.75 −.006288778214 −.006289026186 2.47972 × 10 −7−0.5 .04944253890 .04944653059 3.99169 × 10 −6−0.25 .1638325460 .1638629426 3.03966 × 10 −55 ± 5i −18457 ± 27075i −18457 ± 27075i ±0.00002i5 ± i −15957 ± 28619i −15957 ± 28619i 2.2 × 10 −5 ± 2.0i × 10 −55 32768 32768 01 ± 5i −3.919 ± 6.863i −3.919 ± 6.863i 2.8 × 10 −91 ± i −3.5004 ± 5.9930i −3.5004 ± 5.9930i 1.0 × 10 −90.75 1.143331897 1.143331894 3.0 × 10 −90.5 1.212035527 1.212035221 6.0 × 10 −90.25 0.7282393496 0.7282393461 3.5 × 10 −9Table 3: Comparison of computed ζ H (s, 1/8) versus Maple c○ ζ H (s, 1/8)99

7 RESULTS7.1 SummaryIn order to create, interpret, and make sense of the Julia and Mandelbrot sets ofthe Hurwitz zeta function, we had to take an incredible journey from the mind ofAdolf Hurwitz and the desk of Arthur Cayley in the late 19th century to the researchof Gaston Julia and Pierre Joseph Louis Fatou in the 1910s, ending up on the monitorsand print outs of Mandelbrot in the late 20th century. From iterating rationalfunctions to complex dynamics, from simple linear systems to complex dynamicalsystems, from crude drawings in the 1920s to the frontiers of chaos theory and vividimages, our trek has taken us across fields of mathematics thought at the time to bedisconnected, now known to be intertwined at the most basic levels.In this thesis, we took the Hurwitz zeta function and used analytic continuation,integral representation, contour integrals, and two functional equations of theHurwitz zeta function to extend it from the half-plane to the s−plane.The computation of the Hurwitz zeta function was done, by both a functionalequation of the Hurwitz zeta function, found in Equation (4.4.15), and the Euler-Maclaurin summation formula, a method in which evaluating an integral is approximatedinstead by a similar summation, a process shown to be more simple, efficient,and accurate.Through the use of Maple c○ and FractInt 20.4 c○ , we were introduced to someincredible images and illustrations. The resulting pictures were, in some cases, oneof-a-kindimages of fractals through the dynamics of the Hurwitz zeta function. Inthese images, we are able to see the line σ = 0 and how the Hurwitz zeta functionchanges behavior depending on both the value of a and c in the equationζ H (s, a) + c =∞∑n=01001+ c. (7.1.1)(n + a)s

The pictures of the Julia sets tell us the fate of any iterative function of the Hurwitzzeta function starting at any specific c−value.7.2 ConclusionWe were able to produce images of the Julia and Mandelbrot sets of the Hurwitzzeta function ζ H (s, a) for rational values of a = h/k, for k ≤ 8, resulting in 21different values for a. From there, we have access to an infinitude of images for eacha−value because we then are able manipulate the c−value in our iterative functionz n+1 = f(z n ) = ζ H (z n , a) + c.In terms of calculations, we did get the idea of using the Euler-Maclaurin summationformula to approximate values of the Hurwitz zeta function from Linas Vepštas[29], but we recalculated these formulas to ensure their accuracy. This method wasnot only found to be more efficient than others, but the resulting equation from theEuler-Maclaurin method was compatible with FractInt 20.4 c○ allowing us to finallysee these pictures up close and personal. We created over 1,500 images of the Juliaand Mandelbrot sets of the Hurwitz zeta function. These images show the chaoticbehavior found in the Riemann zeta function is still present for the generalized Riemannzeta function, ζ H (s, a). The Riemann zeta function allows only integer valuesof the denominator in the series while the Hurwitz zeta function allows the denominatorhave non-integer values. This work also led to the founding of a functionalequation of the Hurwitz zeta function due to issues the original functional equationhad with the program Maple c○ .7.3 Future ResearchThe most obvious extension of this research would be to create images for a = h/kfor k > 8.The most puzzling results we discovered in this exercise were what I refer to as101

the “Tingen Triangles,” an example of which can be seen in Figure 54Figure 54: Julia sets of ζ H (s, 5/8) + c, c = 1, s ∈ [−32, 32] × [−24i, 24i]These triangle appear in the images of the Julia and Mandelbrot sets of theHurwitz zeta function when a has any of the following rational values:14 , 3 8 , 5 8 , 3 4 , 7 8 .These triangles are always found in quadrant II with the line σ = 0 as one side.These “Tingen Triangles” are, as of now, unexplained.The simplest answer hasto be programming error, but the code and calculations have been checked, rechecked,and checked again with no error found. For now, it seems these trianglesare actual attributes of these Julia and Mandelbrot sets. I will continue working onthese “Tingen Triangles” and encourage others to look into these out-of-place shapes,what they are, and what they mean, if anything, to the Julia and Mandelbrot setsof the Hurwitz zeta function.102

7.4 ClosingIn closing, it has been said that chaos can be found in the simplest systems - youdo not have to start with incredibly complicated systems in order to get outrageousresults. I want to take this opportunity to say the contrapositive is not true. As seenin this work, from great complexity and calamity comes the same chaos found inthe simplest of systems. This is interesting because even though we take two totallydifferent paths, in the end, both cases end up giving us equally mind-blowing resultsand amazing images, being both aesthetically and intellectually pleasing.Anyone can look at these images be captivated. Human being are programmedto find patterns in what may seem entirely random, but finding that pattern is notenough for a mathematician. A mathematician must look into the interworking ofthe problem and try to understand what is causing the results. People have askedme time and time again, “why do you study math?” I study math because I haveto know why. I was the kid who broke his watch just so I had an excuse to open itup and see what made it tick, and I am still that kid today. Why do I study math?The only answer to that question is another question - why don’t you?103

8 APPENDIX8.1 Julia Sets of f c (z)Figure 55: Julia set of f c (z) = z 2 + c, c = 0.835 − 0.2321iFigure 56: Julia set of f c (z) = z 2 + c, c = 0.45 + 0.1428i.104

105Figure 57: Julia set of fc(z) = z 2 + c, c = 0.285 + 0.01i.

106Figure 58: Julia sets of fc(z), z ∈ [−3, 2] × [−3, 2]

107Figure 59: Julia sets of fc(z), z ∈ [−2, 2] × [−2, 2]

8.2 Maple c○ Program for the Evaluation of the Functional Equation of the HurwitzZeta Functionrestart:The Hurwitz Zeta FunctionHZetaA := (N,s,q) -> sum((n+q)^(-s),n=0..N-1)+(1/2)*(N+q)^(-s)+((N+q)^(1-s))/(s-1):HZetaB := (N,s,q,p) -> sum(bernoulli(2*k)/(2*k)!*product(s+r,r=0..2*k-2)*(N+q)^(-s-2*k+1),k=1..p):HZeta := (N,s,q,p) -> HZetaA(N,s,q) + HZetaB(N,s,q,p):Hurwitz Zeta Error TermHurZetaError:=(N,s,q,p)->abs((s+2*p+1)/(Re(s)+2*p+1))*abs(bernoulli(2*p+2)/(2*p+2)!*product(s+m,m=0..2*p)*(N+q)^(-(s)-2*p-1)):Functional Hurwitz Zeta using Stirling Series Approximation of GAMMA(z)G:=(x,M)->simplify(convert(series(GAMMA(x), x=infinity, M),polynom)):expand(G(x,5)/G(x,1))*G(x,1):The functional equation for the Hurwitz Zeta Function(Apostol, 261)Func_HZeta := (N,s,h,k,M) -> 2*G(s, M)*(2*Pi*k)^(s)*(sum(cos((1/2)*Pi*(s)-2*Pi*r*h/k)*HZeta(N, s, r/k, 3), r = 1 .. k)):My_Func_HZeta := (N,s,h,k,M) -> 2*subs(z = 1-s,G(z, M))*(2*Pi*k)^(s-1)*(sum(sin((1/2)*Pi*(s) + 2*Pi*r*h/k)*HZeta(N, 1-s, r/k, 3), r = 1 .. k)):108

8.3 FractInt 20.4 c○ Formula File for Hurwitz Zeta function with different a−valuesHZeta_1_2 {z = pixel, c=p1:IF (real(z)

+(81/2)^(1-z)/(z-1)+(1/2)^(-z)+(3/2)^(-z)+(5/2)^(-z)+(7/2)^(-z)+(9/2)^(-z)+(11/2)^(-z)+(13/2)^(-z)+(15/2)^(-z)+(17/2)^(-z)+(19/2)^(-z)+(21/2)^(-z)+(23/2)^(-z)+(25/2)^(-z)+(27/2)^(-z)+(29/2)^(-z)+(31/2)^(-z)+(33/2)^(-z)+(35/2)^(-z)+(37/2)^(-z)+(39/2)^(-z)+(41/2)^(-z)+(43/2)^(-z)+(45/2)^(-z)+(47/2)^(-z)+(49/2)^(-z)+(51/2)^(-z)+(53/2)^(-z)+(55/2)^(-z)+(57/2)^(-z)+(59/2)^(-z)+(61/2)^(-z)+(63/2)^(-z)+(65/2)^(-z)+(67/2)^(-z)+(69/2)^(-z)+(71/2)^(-z)+(73/2)^(-z)+(75/2)^(-z)+(77/2)^(-z)+(79/2)^(-z)+(1/2)*(81/2)^(-z) + cENDIF|z|

+(1/30240)*(-z+5)*(-z+4)*(-z+3)*(-z+2)*(-z+1)*(122/3)^(z-6)+(44/3)^(-1+z)-(122/3)^z/z+(50/3)^(-1+z)+(53/3)^(-1+z)+(56/3)^(-1+z)+(59/3)^(-1+z)+(62/3)^(-1+z)+(65/3)^(-1+z)+(68/3)^(-1+z)+(71/3)^(-1+z)+(74/3)^(-1+z)+(77/3)^(-1+z)+(80/3)^(-1+z)+(83/3)^(-1+z)+(86/3)^(-1+z)+(89/3)^(-1+z)+(92/3)^(-1+z)+(95/3)^(-1+z)+(98/3)^(-1+z)+(101/3)^(-1+z)+(104/3)^(-1+z)+(107/3)^(-1+z)+(110/3)^(-1+z)+(113/3)^(-1+z)+(116/3)^(-1+z)+(119/3)^(-1+z)+(1/2)*(122/3)^(-1+z)+(11/3)^(-1+z)+(14/3)^(-1+z)+(8/3)^(-1+z)+(2/3)^(-1+z)+(5/3)^(-1+z)+(17/3)^(-1+z)+(20/3)^(-1+z)+(23/3)^(-1+z)+(26/3)^(-1+z)+(29/3)^(-1+z)+(32/3)^(-1+z)+(35/3)^(-1+z)+(38/3)^(-1+z)+(41/3)^(-1+z))+sin((1/2)*pi*z)*(1-41^z/z+(1/30240)*(-z+5)*(-z+4)*(-z+3)*(-z+2)*(-z+1)*41^(z-6)+(1/12)*(-z+1)*41^(z-2)-(1/720)*(-z+3)*(-z+2)*(-z+1)*41^(z-4)+2^(-1+z)+3^(-1+z)+4^(-1+z)+5^(-1+z)+6^(-1+z)+7^(-1+z)+8^(-1+z)+9^(-1+z)+10^(-1+z)+11^(-1+z)+12^(-1+z)+13^(-1+z)+14^(-1+z)+15^(-1+z)+16^(-1+z)+17^(-1+z)+19^(-1+z)+20^(-1+z)+21^(-1+z)+22^(-1+z)+23^(-1+z)+24^(-1+z)+25^(-1+z)+26^(-1+z)+33^(-1+z)+34^(-1+z)+35^(-1+z)+36^(-1+z)+37^(-1+z)+38^(-1+z)+39^(-1+z)+40^(-1+z)+27^(-1+z)+28^(-1+z)+18^(-1+z)+29^(-1+z)+31^(-1+z)+32^(-1+z)+(1/2)*41^(-1+z)+30^(-1+z)))/(1-z)^4 + cELSEz=(1/12)*(z)*(121/3)^(-z-1)-(1/720)*(z+2)*(z+1)*(z)*(121/3)^(-z-3)+(1/3)^(-z)+(4/3)^(-z)+(7/3)^(-z)+(10/3)^(-z)+(13/3)^(-z)+(16/3)^(-z)+(19/3)^(-z)+(22/3)^(-z)+(25/3)^(-z)+(28/3)^(-z)+(31/3)^(-z)+(34/3)^(-z)+(121/3)^(1-z)/(-1+z)+(37/3)^(-z)+(40/3)^(-z)+(43/3)^(-z)+(46/3)^(-z)+(49/3)^(-z)+(52/3)^(-z)+(55/3)^(-z)+(58/3)^(-z)+(61/3)^(-z)+(64/3)^(-z)+(67/3)^(-z)+(70/3)^(-z)+(73/3)^(-z)+(76/3)^(-z)+(79/3)^(-z)+(82/3)^(-z)+(85/3)^(-z)+(88/3)^(-z)+(91/3)^(-z)+(94/3)^(-z)+(97/3)^(-z)+(100/3)^(-z)+(103/3)^(-z)+(106/3)^(-z)+(109/3)^(-z)+(112/3)^(-z)+(115/3)^(-z)+(118/3)^(-z)+(1/2)*(121/3)^(-z) + cENDIF111

|z|

+(32/3)^(-1+z)+(35/3)^(-1+z)+(38/3)^(-1+z)+(41/3)^(-1+z))+sin((1/2)*pi*z)*(1-41^z/z+(1/30240)*(-z+5)*(-z+4)*(-z+3)*(-z+2)*(-z+1)*41^(z-6)+(1/12)*(-z+1)*41^(z-2)-(1/720)*(-z+3)*(-z+2)*(-z+1)*41^(z-4)+2^(-1+z)+3^(-1+z)+4^(-1+z)+5^(-1+z)+6^(-1+z)+7^(-1+z)+8^(-1+z)+9^(-1+z)+10^(-1+z)+11^(-1+z)+12^(-1+z)+13^(-1+z)+14^(-1+z)+15^(-1+z)+16^(-1+z)+17^(-1+z)+19^(-1+z)+20^(-1+z)+21^(-1+z)+22^(-1+z)+23^(-1+z)+24^(-1+z)+25^(-1+z)+26^(-1+z)+33^(-1+z)+34^(-1+z)+35^(-1+z)+36^(-1+z)+37^(-1+z)+38^(-1+z)+39^(-1+z)+40^(-1+z)+27^(-1+z)+28^(-1+z)+18^(-1+z)+29^(-1+z)+31^(-1+z)+32^(-1+z)+(1/2)*41^(-1+z)+30^(-1+z)))/(1-z)^4 + cELSEz=-(1/720)*(z+2)*(z+1)*(z)*(122/3)^(-z-3)+(113/3)^(-z)+(53/3)^(-z)+(1/12)*(z)*(122/3)^(-z-1)+(107/3)^(-z)+(8/3)^(-z)+(29/3)^(-z)+(95/3)^(-z)+(101/3)^(-z)+(86/3)^(-z)+(65/3)^(-z)+(5/3)^(-z)+(35/3)^(-z)+(83/3)^(-z)+(56/3)^(-z)+(47/3)^(-z)+(116/3)^(-z)+(38/3)^(-z)+(62/3)^(-z)+(17/3)^(-z)+(110/3)^(-z)+(32/3)^(-z)+(11/3)^(-z)+(26/3)^(-z)+(80/3)^(-z)+(104/3)^(-z)+(68/3)^(-z)+(77/3)^(-z)+(92/3)^(-z)+(23/3)^(-z)+(71/3)^(-z)+(41/3)^(-z)+(98/3)^(-z)+(122/3)^(1-z)/(-1+z)+(14/3)^(-z)+(59/3)^(-z)+(89/3)^(-z)+(1/2)*(122/3)^(-z)+(20/3)^(-z)+(44/3)^(-z)+(74/3)^(-z)+(2/3)^(-z)+(50/3)^(-z)+(119/3)^(-z) + cENDIF}|z|

8.4 FractInt 20.4 c○ Parameter File for Hurwitz Zeta function with different a−values32x24-1_2 {reset=1960type=formulaformulafile=HZeta.frmformulaname=HZeta_1_2corners=-32/32/-24/24float=yinside=0colors=@goodega.map}32x24-1_3 {reset=1960type=formulaformulafile=HZeta.frmformulaname=HZeta_1_3corners=-32/32/-24/24float=yinside=0colors=@goodega.map}32x24-1_4 {reset=1960type=formulaformulafile=HZeta.frmformulaname=HZeta_1_4114

corners=-32/32/-24/24float=yinside=0colors=@goodega.map}32x24-2_2 {reset=1960type=formulaformulafile=HZeta.frmformulaname=HZeta_2_2corners=-32/32/-24/24float=yinside=0colors=@goodega.map}32x24-2_3 {reset=1960type=formulaformulafile=HZeta.frmformulaname=HZeta_2_3corners=-32/32/-24/24float=yinside=0colors=@goodega.map}115

REFERENCES[1] Milton Abramowitz, & Irene A. Stegun, Handbook of Mathematical Functions,Dover Publications Inc., Mineola, NY, 1972.[2] Kathleen T. Alligood, Tim D. Sauer, & James A. Yorke, Chaos: An Introductionto Dynamical Systems, Springer-Verlag, New York, NY, 1996.[3] Tom M. Apostol, An Elementary View of Euler’s Summation Formula, TheAmerican Mathematical Monthly, Vol. 106 (1999), No. 5 (May), pp. 409-418.[4] Tom M. Apostol, Introduction to Analytic Number Theory, UndergraduateTexts in Mathematics, Springer-Verlag, New York, NY, 1976.[5] Tom M. Apostol, Mathematical Analysis, Second Ed., Addison-Wesley, NewYork, NY, 1981.[6] Tom. M Apostol, Remark on the Hurwitz Zeta Function, Proceedings of theMathematical Society, Vol. 2 (1951), No. 5 (October), pp. 690-693.[7] Michael F. Barnsley, Fractals Everywhere, Second Ed., Academic Press, Cambridge,MA, 1993.[8] Bruce C. Berndt, The Gamma Function and the Hurwitz Zeta-Function, TheAmerican Mathematical Monthly, Vol. 92 (1985), No. 2 (February), pp. 126-130.[9] Stephen Boul, Zeros of the Riemann Zeta on the Critical Line, Masters Thesis,University of Missouri Columbia, 2001.[10] James Ward Brown & Ruel V. Churchill, Complex Variables and Applications,Sixth Ed., McGraw-Hill, New York, NY, 1996.[11] Lennart Carleson & Theodore W. Gamelin, Complex Dynamics, Springer-Verlag, New York, NY, 1993.116

[12] H. Cremer, Über die Iteration rationaler Funktionen, Jahresbericht derDeutschen Mathematiker-Vereinigung, Vol. 33 (1925), No. 2, pp. 185-210.[13] Lokenath Debnath, Brief Historical Introduction to Fractals Geometry, InternationalJournal of Mathematical Education in Science and Technology, Vol. 37(2006), No. 1, pp. 29-50.[14] H.M. Edwards, Riemann’s Zeta Function, Academic Press Inc., New York, NY,1974.[15] Malcolm Gladwell, The Tipping Point, Little, Brown and Company, New York,NY, 2002.[16] Helmut Hasse, Ein Summierungsverfahren für die Riemannsehe ζ-Reihe, MathematischeZeitschrift, No. 32 (1930), pp. 458-464.[17] Eldon R. Hansen & Merrell L. Patrick, Some Relations and Values for the GeneralizedRiemann Zeta Function, Mathematics of Computation, Vol. 16 (1962),No. 79 (July), pp. 265-274.[18] Sara Ives, Julia Sets of the Riemann Zeta Function, Masters Thesis, Universityof North Carolina Wilmington, 2001.[19] Gaston Julia, Mémorie sur l’itération des fonctions rationnelles, Journal deMathématiques Pures et Appliquées, Vol. 81 (1918), pp. 47 - 235.[20] L. Kronecker, G. Lejeune Dirichlet’s Werke: Herausgegeben auf Veranlassungder Königlich Preussischen Akademie der Wissenschaften, Druck und Verlagvon Georg Reimer, Berlin, 1889.[21] Serge Lang, Complex Analysis, Third Ed., Springer-Verlag, New York, NY,1993.117

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[32] Wikipedia. Title: Julia sets. Last Checked: October 7, 2009, http://en.wikipedia.org/wiki/File:Julia-set_N_z3-1.png[33] Miqel.com. Title: Simple Iterative Fractals. Last Checked: November25, 2009, http://www.miqel.com/fractals_math_patterns/visual-math-iterative-fractals.html[34] Title: The Koch Curve. Last Checked: November 25, 2009, http://www.jimloy.com/fractals/koch.htm[35] S.C. Woon, “Fractals and the Julia and Mandelbrot sets of the Riemann ZetaFunction,” Trinity College, University of Cambridge, CB2 ITQ, UK, 1998. Accessedat: http://arxiv.org/abs/chao-dyn/9812031v1119