FINDING ZEROS OF COMPLEX FUNCTIONS
|
|
- Austin Washington
- 5 years ago
- Views:
Transcription
1 FINDING ZEROS OF COMPLEX FUNCTIONS It is well kow sice the time of Newto that the zeros of a real fuctio f(x) ca be foud b carrig out the iterative procedure- f ( x[ x[ 1] x[ ] subject to x[0] x f '( x[ 0 Here x[0] represets a value lig withi the eighborhood of the root at x[]. The s i the square brackets represet subscripts. If we have f(x)=x -1 ad x[0]=1/, oe fids the zero at x[]=1. Had we started the iteratio with x[0]=- 3/, oe would fid the secod zero at x[]=-1. Thus the iterative procedure will coverge toward that zero of f(x) lig closest to the startig value x[0]. It is possible to exted this iterative procedure to complex fuctios f(z) where z=x+i. We do this i the preset article ad also show how other methods such as makig cotour maps of the absolute value of a complex fuctio to quickl locate its zeros.. Our startig poit will be a complex fuctio f(z) whose derivative f (z) exists.to fid its zeros we start with z[0] ad write- df(z[0])/dz={f(z[0])-0}/(z[0]-z[1]) Now if z[0] lies close eough to oe of the zeros of f(z), oe ca deduce form this last expressio that the followig iterative procedure will be valid- f ( z[ z[ 1] z[ ] subject to z[0] z f '( z[ 0 To get some idea of where to place z[0] it is a good idea to first obtai a rough cotourplot of the absolute value of f(z). That is, locate the eighborhood of the expected zero b otig where the cotours make small diameter circles i the z=x+i plae. The absolute value fuctio of f(z) ca be writte as- Abs{ } u( v( with u( iv( It eeds to be remembered that iterative solutios for complex fuctios will ofte ot coverge whe the startig poit lies outside of the immediate eighborhood of a zero. The boudar betwee coverget solutios ad those where the Newto method fails ca lead to ver itricate patter associated with fractals. Let us demostrate the iterative ad cotour map procedure for the special case of the complex fuctio-
2 z 1 i( z 1) Here we have u ( x 1 ad v( x( 1) 1 The cotour map for Abs(f(z) idicates small closed circles about z=1+i ad z=-1.so ruig the iterative process usig either z[0]=1++i or z[0]=-1+ as a startig poit, produces z[]=1+i ad -1, respectivel. Epsilo is take as a small icremet such as =0.1. I this case we could also have used a much shorter approach to get the aswer b otig that f(z) is a quadratic i z ad thus ca be writte as- f(z)=(z+1)(z-1-i) From this form the aswer ca be read off directl. Wheever f(z) is a quadratic, cubic, or fourth order polomial i z, the aswer for the zeros of f(z) are expressible i closed form ad the aswer ca be read off directel.thus the four zeros of- z 4 1 ( z 1)( z 1) ( z 1)( z 1)( z i)( z i) are see to be z[]=1, -1, i, -i.. The cotour plot for the absolute value of z 4-1 looks like this-
3 For quitic ad higher power polomial fuctios oe must(with a few exceptios) resort to iterative or cotour plot procedures. If the polomial form of f(z) cotais ol the highest power of z plus a costat such as f(z)=z -1, the solutios ca be give i closed form for a positive iteger. For example, take the complex fuctio- z 1 r exp( i ) 1 whe z is expressed i polar form. Solvig we have r=1 ad k / with k 1,,3,... So the zeros of f(z) all lie o a uit circle at / itervals. Expressed i Cartesia coordiates, the roots are located at- x 1 1 ta ad ta( ) 1 ta with k Aother complex fuctio isf ( z) si( z) si( x)cosh( i cos( x)sih( I this case- Abs{ } si cosh ( x)cosh ( cos ( x) ( cos ( x)sih ( A cotour plot of this fuctio follows-
4 It looks ver much like the phase plae plot for a simple pedulum. Here the zeros clearl lie alog the x axis at x={0,1,,3, etc.}..there are o zeros possible elsewhere i the z plae. This last result brigs to mid the as et uprove cojecture b Riema that the complex Riema Zeta Fuctio Z(+i) ca have zeros i the - plae ol alog the lie =1/. It certail looks like a valid statemet as show b the followig cotour map I have geerated
5 It appears that a defiitive proof that all roots of Zeta(x+i lie alog the x=1/ lie will be based a a topological argumet i which oe otes that all cotours of Abs(Zeta) for C<<1 form small radius circles about the zeros of f(z). As a fial example of fidig the zeros of a complex fuctio cosider- exp( z ) 1 u( iv( where u exp( x )cos(x 1 ad v exp( x )si(x Oe sees at oce b ispectio that z=0 is the ol possible zero of f(z)=exp(z )- 1. To cofirm this poit oe could carr out a Newto iteratio o this complex fuctio. Eve simpler is to just look at a cotour map of the absolute value of f(z). Doig so we fid the followig-
6 This icel cofirms that the ol zero lies withi the small circle cetered o the origi i the z plae.. September 1, 015
x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,
Math Activity 9( Due with Fial Eam) Usig first ad secod Taylor polyomials with remaider, show that for, 8 Usig a secod Taylor polyomial with remaider, fid the best costat C so that for, C 9 The th Derivative
More informationMath 113, Calculus II Winter 2007 Final Exam Solutions
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this
More information1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute
Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral,
More informationMATH 10550, EXAM 3 SOLUTIONS
MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More information( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!
.8,.9: Taylor ad Maclauri Series.8. Although we were able to fid power series represetatios for a limited group of fuctios i the previous sectio, it is ot immediately obvious whether ay give fuctio has
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
More informationMIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS
MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will
More informationProblem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =
Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,
More information3 Show in each case that there is a root of the given equation in the given interval. a x 3 = 12 4
C Worksheet A Show i each case that there is a root of the equatio f() = 0 i the give iterval a f() = + 7 (, ) f() = 5 cos (05, ) c f() = e + + 5 ( 6, 5) d f() = 4 5 + (, ) e f() = l (4 ) + (04, 05) f
More informationPractice Problems: Taylor and Maclaurin Series
Practice Problems: Taylor ad Maclauri Series Aswers. a) Start by takig derivatives util a patter develops that lets you to write a geeral formula for the -th derivative. Do t simplify as you go, because
More informationAlternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.
0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate
More informationChapter 2 The Solution of Numerical Algebraic and Transcendental Equations
Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
More informationAppendix F: Complex Numbers
Appedix F Complex Numbers F1 Appedix F: Complex Numbers Use the imagiary uit i to write complex umbers, ad to add, subtract, ad multiply complex umbers. Fid complex solutios of quadratic equatios. Write
More informationPLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)
Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationMTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1
MTH 42 Exam 3 Spr 20 Practice Problem Solutios No calculators will be permitted at the exam. 3. A pig-pog ball is lauched straight up, rises to a height of 5 feet, the falls back to the lauch poit ad bouces
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationMath 312 Lecture Notes One Dimensional Maps
Math 312 Lecture Notes Oe Dimesioal Maps Warre Weckesser Departmet of Mathematics Colgate Uiversity 21-23 February 25 A Example We begi with the simplest model of populatio growth. Suppose, for example,
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationSolutions to Final Exam Review Problems
. Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the
More information... and realizing that as n goes to infinity the two integrals should be equal. This yields the Wallis result-
INFINITE PRODUTS Oe defies a ifiite product as- F F F... F x [ F ] Takig the atural logarithm of each side oe has- l[ F x] l F l F l F l F... So that the iitial ifiite product will coverge oly if the sum
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationNUMERICAL METHODS FOR SOLVING EQUATIONS
Mathematics Revisio Guides Numerical Methods for Solvig Equatios Page 1 of 11 M.K. HOME TUITION Mathematics Revisio Guides Level: GCSE Higher Tier NUMERICAL METHODS FOR SOLVING EQUATIONS Versio:. Date:
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationAnalytic Continuation
Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for
More informationSolutions to Homework 1
Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.
More informationCalculus I Practice Test Problems for Chapter 5 Page 1 of 9
Calculus I Practice Test Problems for Chapter 5 Page of 9 This is a set of practice test problems for Chapter 5. This is i o way a iclusive set of problems there ca be other types of problems o the actual
More informationy = f x x 1. If f x = e 2x tan -1 x, then f 1 = e 2 2 e 2 p C e 2 D e 2 p+1 4
. If f = e ta -, the f = e e p e e p e p+ 4 f = e ta -, so f = e ta - + e, so + f = e p + e = e p + e or f = e p + 4. The slope of the lie taget to the curve - + = at the poit, - is - 5 Differetiate -
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationSOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.
SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad
More informationSeptember 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1
September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright
More informationRecurrence Relations
Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial(-)); } Let t be the umber of multiplicatios eeded to calculate factorial(). The
More informationU8L1: Sec Equations of Lines in R 2
MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationN14/5/MATHL/HP1/ENG/TZ0/XX/M MARKSCHEME. November 2014 MATHEMATICS. Higher Level. Paper pages
N4/5/MATHL/HP/ENG/TZ0/XX/M MARKSCHEME November 04 MATHEMATICS Higher Level Paper 0 pages N4/5/MATHL/HP/ENG/TZ0/XX/M This markscheme is the property of the Iteratioal Baccalaureate ad must ot be reproduced
More informationMATH 2300 review problems for Exam 2
MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the x-axis, ad the lie x =. Fid the mass of the plate. Iclude uits. (b) Fid the
More informationSigma notation. 2.1 Introduction
Sigma otatio. Itroductio We use sigma otatio to idicate the summatio process whe we have several (or ifiitely may) terms to add up. You may have see sigma otatio i earlier courses. It is used to idicate
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationMath 113 Exam 4 Practice
Math Exam 4 Practice Exam 4 will cover.-.. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n
Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More informationMa 530 Infinite Series I
Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li
More informationZ ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew
Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe
More informationTHE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.
THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of
More informationTaylor Series (BC Only)
Studet Study Sessio Taylor Series (BC Oly) Taylor series provide a way to fid a polyomial look-alike to a o-polyomial fuctio. This is doe by a specific formula show below (which should be memorized): Taylor
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More information( 1) n (4x + 1) n. n=0
Problem 1 (10.6, #). Fid the radius of covergece for the series: ( 1) (4x + 1). For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio.
More informationMATH 2300 review problems for Exam 2
MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the x-axis, ad the lie x =. (a) Fid the mass of the plate. Iclude uits. Mass =
More informationSection 1.4. Power Series
Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio
More informationMath 210A Homework 1
Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called
More informationCALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=
More informationAP Calculus BC Review Applications of Derivatives (Chapter 4) and f,
AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)
More information18.01 Calculus Jason Starr Fall 2005
Lecture 18. October 5, 005 Homework. Problem Set 5 Part I: (c). Practice Problems. Course Reader: 3G 1, 3G, 3G 4, 3G 5. 1. Approximatig Riema itegrals. Ofte, there is o simpler expressio for the atiderivative
More informationMATH Exam 1 Solutions February 24, 2016
MATH 7.57 Exam Solutios February, 6. Evaluate (A) l(6) (B) l(7) (C) l(8) (D) l(9) (E) l() 6x x 3 + dx. Solutio: D We perform a substitutio. Let u = x 3 +, so du = 3x dx. Therefore, 6x u() x 3 + dx = [
More informationMath 116 Practice for Exam 3
Math 6 Practice for Eam 3 Geerated April 4, 26 Name: SOLUTIONS Istructor: Sectio Number:. This eam has questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur
More information4755 Mark Scheme June Question Answer Marks Guidance M1* Attempt to find M or 108M -1 M 108 M1 A1 [6] M1 A1
4755 Mark Scheme Jue 05 * Attempt to fid M or 08M - M 08 8 4 * Divide by their determiat,, at some stage Correct determiat, (A0 for det M= 08 stated, all other OR 08 8 4 5 8 7 5 x, y,oe 8 7 4xy 8xy dep*
More informationMEI Conference 2009 Stretching students: A2 Core
MEI Coferece 009 Stretchig studets: A Core Preseter: Berard Murph berard.murph@mei.org.uk Workshop G How ca ou prove that these si right-agled triagles fit together eactl to make a 3-4-5 triagle? What
More information1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y
Questio (a) A square matrix A= A is called positive defiite if the quadratic form waw > 0 for every o-zero vector w [Note: Here (.) deotes the traspose of a matrix or a vector]. Let 0 A = 0 = show that:
More informationU8L1: Sec Equations of Lines in R 2
MCVU Thursda Ma, Review of Equatios of a Straight Lie (-D) U8L Sec. 8.9. Equatios of Lies i R Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio
More information10.2 Infinite Series Contemporary Calculus 1
10. Ifiite Series Cotemporary Calculus 1 10. INFINITE SERIES Our goal i this sectio is to add together the umbers i a sequece. Sice it would take a very log time to add together the ifiite umber of umbers,
More information(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is
Calculus BC Fial Review Name: Revised 7 EXAM Date: Tuesday, May 9 Remiders:. Put ew batteries i your calculator. Make sure your calculator is i RADIAN mode.. Get a good ight s sleep. Eat breakfast. Brig:
More informationMath 132, Fall 2009 Exam 2: Solutions
Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationSeries Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1
Defiitio: We say that the series S = Series Review i= a i is the sum of the first terms. i= a i coverges if lim S exists ad is fiite, where The above is the defiitio of covergece for series. order to see
More informationMini Lecture 10.1 Radical Expressions and Functions. 81x d. x 4x 4
Mii Lecture 0. Radical Expressios ad Fuctios Learig Objectives:. Evaluate square roots.. Evaluate square root fuctios.. Fid the domai of square root fuctios.. Use models that are square root fuctios. 5.
More informationa 3, a 4, ... are the terms of the sequence. The number a n is the nth term of the sequence, and the entire sequence is denoted by a n
60_090.qxd //0 : PM Page 59 59 CHAPTER 9 Ifiite Series Sectio 9. EXPLORATION Fidig Patters Describe a patter for each of the followig sequeces. The use your descriptio to write a formula for the th term
More informationChapter 2 Feedback Control Theory Continued
Chapter Feedback Cotrol Theor Cotiued. Itroductio I the previous chapter, the respose characteristic of simple first ad secod order trasfer fuctios were studied. It was show that first order trasfer fuctio,
More informationTR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT
TR/46 OCTOBER 974 THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION by A. TALBOT .. Itroductio. A problem i approximatio theory o which I have recetly worked [] required for its solutio a proof that the
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical
More informationCarleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below.
Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of true-false questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified
More informationTERMWISE DERIVATIVES OF COMPLEX FUNCTIONS
TERMWISE DERIVATIVES OF COMPLEX FUNCTIONS This writeup proves a result that has as oe cosequece that ay complex power series ca be differetiated term-by-term withi its disk of covergece The result has
More informationConvergence: nth-term Test, Comparing Non-negative Series, Ratio Test
Covergece: th-term Test, Comparig No-egative Series, Ratio Test Power Series ad Covergece We have writte statemets like: l + x = x x2 + x3 2 3 + x + But we have ot talked i depth about what values of x
More informationSNAP Centre Workshop. Basic Algebraic Manipulation
SNAP Cetre Workshop Basic Algebraic Maipulatio 8 Simplifyig Algebraic Expressios Whe a expressio is writte i the most compact maer possible, it is cosidered to be simplified. Not Simplified: x(x + 4x)
More informationPractice Test Problems for Test IV, with Solutions
Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums,
More information-ORDER CONVERGENCE FOR FINDING SIMPLE ROOT OF A POLYNOMIAL EQUATION
NEW NEWTON-TYPE METHOD WITH k -ORDER CONVERGENCE FOR FINDING SIMPLE ROOT OF A POLYNOMIAL EQUATION R. Thukral Padé Research Cetre, 39 Deaswood Hill, Leeds West Yorkshire, LS7 JS, ENGLAND ABSTRACT The objective
More informationSEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
More information6.003 Homework #3 Solutions
6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the
More informationFall 2018 Exam 3 HAND IN PART 0 10 PIN: 17 INSTRUCTIONS
MARK BOX problem poits HAND IN PART 0 10 1 10 2 5 NAME: Solutios 3 10 PIN: 17 4 16 65=13x5 % 100 INSTRUCTIONS This exam comes i two parts. (1) HAND IN PART. Had i oly this part. (2) STATEMENT OF MULTIPLE
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationMost text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t
Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said
More informationWe will conclude the chapter with the study a few methods and techniques which are useful
Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs
More informationCALCULATION OF FIBONACCI VECTORS
CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationRADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify
Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL
More informationFINALTERM EXAMINATION Fall 9 Calculus & Aalytical Geometry-I Questio No: ( Mars: ) - Please choose oe Let f ( x) is a fuctio such that as x approaches a real umber a, either from left or right-had-side,
More informationlim za n n = z lim a n n.
Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationComplex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
More informationMA131 - Analysis 1. Workbook 10 Series IV
MA131 - Aalysis 1 Workbook 10 Series IV Autum 2004 Cotets 4.19 Rearragemets of Series...................... 1 4.19 Rearragemets of Series If you take ay fiite set of umbers ad rearrage their order, their
More informationFundamental Concepts: Surfaces and Curves
UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat
More informationSolutions to quizzes Math Spring 2007
to quizzes Math 4- Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x
More informationMATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and
MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f
More information