# Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Size: px
Start display at page:

Download "Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1"

Transcription

1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution x = f(x). Solution. The condition is described be the following Theorem. Assume x = c is solution of the eqution x = f(x). Assume further tht there re numbers r > nd q with q < 1 such tht f (x) q for ll x with c r < x < c+r. Then strting with ny vlue x 1 (c r,c+r), the sequence { } n=1 defined by = f( 1 ) for n > converges to c. b) Define the sequence { } n=1 by x 1 = 1 nd +1 = +. Explin why converges to. Solution. Writing f(x) = x +x we hve +1 = f( ), tht is, we re deling with n exmple of fixed-point itertion. Further, we hve f (x) = x +1, = x x = 1 1 x > 4 whenever 1/x < 5/4, or else x > / 5. On the other hnd, f (x) < 1/ for ll x. So, we hve f (x) < /4 for ll x in the intervl ( ) ( ( ) 5,+, ( )) Note tht x 1 = 1 is in the intervl on the right-hnd side. Hence converges to ccording to the Theorem mentioned in the solution to Prt ) (tke c =, r = / 5 nd q = /4 in the Theorem). Note. The following more generl sttement is true: Let > be rel number, nd define the sequence { } n=1 by x 1 = 1 nd +1 = + Then converges to. To see tht this sttement is true, observe tht for every n > 1. Indeed, for n 1 we hve +1 = + =, where the inequlity holds since the rithmetic men is lwys greter thn or equl to the geometric men. Now, writing x f(x) = +x, 1 All computer processing for this mnuscript ws done under Debin Linux. AMS-TEX ws used for typesetting. 1.

2 we hve f (x) = x +1 = x x So, noting tht +1 = f( ) nd = f( ), for n > 1 we hve = 1 x < = f( ) f() = f (ξ)( ) with some ξ (, ) ccording to the Men-Vlue Theorem of differentition. As f (ξ) < 1/, this implies tht +1 (1/)( ) for ll n. Therefore (1/) n ( ) (n ) follows by induction on n. Hence it follows tht tends to. Observe tht this proof simply repets the rguments used in estblishing the theorem bout the convergence of fixed point itertion, while certin technicl detils re different (since c for every n 1 with c = in the present cse). Further, observe tht when solving the eqution g(x) = with g(x) = x by Newton s method, we hve +1 = g() g ( ) = = + I.e., the itertion described bove cn be viewed s solving the eqution x = by Newton s method... We wnt to evlute dx lnx using the composite trpezoidl rule with five deciml precision, i.e., with n error not exceeding Wht vlue of n should one use when dividing the intervl [,] into n prts? Solution. The error term in the composite Simpson formul when integrting f on the intervl [, b] nd dividing the intervl into n prts is (b ) 1n f (ξ). We wnt to use this with =, b = 1, nd f(x) = 1/lnx. We hve f (x) = 1 ( 1 x (lnx) + ) (lnx). It is cler tht f (x) is decresing on the intervl (1,+ ), so it ssumes its mximum of the intervl [,] t x =. We hve f ().17. So, noting tht = nd b =, the bsolute vlue of the error is (b ) 1n f (ξ) = 1 1n f (ξ) n n. In order to ensure tht this error is less thn 5 1 6, we need to hve /n < 5 1 6, i.e., n > = Recll tht. In cse =, we cnnot require ξ (,), since then this intervl is empty. In this cse, tke ξ =. It is esy to show tht = cn hppen only in cse = 1, but it is not worth doing so for the completion of the present proof. Equlity below cn hold only in cse =.

3 So one needs to mke sure tht n Thus one needs to divide the intervl [,] into (t lest) 184 prts in order to get the result with 4 deciml precision while using the trpezoidl rule..) Describe how to del with the singulrity in the integrl x / (e x 1)dx if one wnts to evlute this integrl using Simpson s rule. Solution. One cn subtrct the singulrity by tking n initil segment of the Tylor series t x = of e x. We hve e x = n! ; this Tylor series is convergent on the whole rel line. Therefore, we hve where the symbol O( ) is ment s x. Hence n= e x = 1 x + x4 +O(x6 ), ( ) x / e x 1+x x4 = O(x 6 / ) = O(x 9/ ). The fourth derivtive of this ner x = is O(x 1/ ); tht is, the fourth derivtive tends to zero when x ց. So the fourth derivtive is bounded on (, 1), the intervl of integrtion; therefore, Simpson s rule cn be used to clculte the integrl. To sum up, we hve x / (e x 1)dx = ( ) x / e x 1+x x4 dx+ ) ( x 1/ + x5/ dx. The first integrl on the right-hnd side cn be clculted by Simpson s rule (t x = tke the integrnd to be ), nd the second integrl cn be evluted directly, by clculting the integrl explicitly. b) We wnt to evlute the integrl x sin 1 x dx (tke the integrnd to be for x = ) using some numericl method, such s the trpezoidl rule, Simpson s rule, dptive integrtion, or Romberg integrtion. Which method would be best suited for evluting the integrl? Give resons for your nswer. Solution. The integrnd behves unevenly on the intervl of integrtion; it behves bdly ner zero (so the intervl of integrtion needs to be divided into mny prts ner zero), while it behves well wy from zero (so, wy from zero we do not need to divide the intervl of integrtion into very smll prts). Hence dptive integrtion needs to be used. Adptive integrtion with the trpezoidl rule works better thn dptive integrtion with Simpson s rule. This is becuse the fourth derivtive of the integrnd is x 6 sin 1 x 4x 5 cos 1 x ; this fetures in the error term of Simpson s rule, while the second derivtive, which fetures in the error term of the trpezoidl rule, is x sin 1 x x 1 cos 1 x +sin 1 x.

4 The second fourth derivtive behves much worse (is often much lrger) thn the second derivtive ner. In n ctul clcultion, one would write the integrl s x sin 1 4 x dx = The bsolute vlue of the first integrl is less thn 4 x dx = x 1 4 x= x sin 1 x dx+ 1 4 x sin 1 x dx. = < The second integrl cn be evluted s.86,59,55,596 with n error less by dptive integrtion using the trpezoidl rule. So, this result gives the vlue of the integrl on the left with n error of < ; tht is, the integrl of the left-hnd side is.86,59,55,596 with 11 deciml digit precision. Trying to evlute the integrl on the left-hnd side directly, with dptive integrtion on the whole intervl [,1], will fil becuse of the bd behvior of the integrl ner. 4.) Given certin function f, we re using the formul to pproximte its derivtive. We hve f(x,h) = f(x+h) f(x h) h f(,1/8).84,5 nd f(,1/16).871,7. Using Richrdson extrpoltion, find better pproximtion for f (). Solution. We hve f (x) = f(x,h)+c 1 h +c h 4... f (x) = f(x,h)+c 1 (h) +c (h) 4... with some c 1, c,... Multiplying the first eqution by 4 nd subtrcting the second one, we obtin Tht is, with x = nd h = 1/16 we hve f (x) = 4 f(x,h) f(x,h)+1c h f (x) 4 f(x,h) f(x,h) 4.871,7.84,5 =.88,85 The function in the exmple is f(x) = sin 1 x, nd f ().88,511. b) In estimting the error of the trpezoidl rule nd Simpson s rule, the following lemm ws used. Lemm. Let g(x) for ll x [,b]. Assume f is differentible on (,b), nd for ech x [,b] we hve ξ x (,b). Then there is n η (,b) such tht f (ξ x )g(x)dx = f (η) provided tht the integrls on both sides of this eqution exist. g(x) dx, Briefly outline the proof of this lemm. 4

5 Solution. We my ssume tht g. Indeed, if g = then both integrls in the bove eqution re zero, so the eqution holds with ny η (,b). 4 Assume m < f (x) < M for ll x (,b), where we llow m = nd M = +, if f is not bounded. Then or else m g(x)dx < f (ξ x )g(x)dx < M f (ξ x )g(x)dx = H with some H with m < H < M. In fct, one only needs to tke H = f (ξ x )g(x)dx g(x)dx. g(x) dx, g(x) dx, Since the derivtive stisfies the intermedite vlue property, 5 it is esy to conclude tht we hve H = f (η) for some η (,b). Hence bove Lemm follows. 5.) Write third order Tylor pproximtion for the solution y(x) t x = 1 + h of the differentil eqution y = x + y with initil condition y(1) = (i.e., the error term in expressing y(1 + h) should be O(h 4 )). Solution. We hve y(1) =, y (1) = x+y = 5; the right-hnd side ws obtined by substituting x = 1 nd y =. Differentiting, then using the eqution y = x+y, nd gin substituting x = 1 nd y =, we obtin y (x) = (x+y ) = 1+yy = 1+y(x+y ) = 1+xy +y = 1. Differentiting, then using the eqution y = x+y, nd gin substituting x = 1 nd y =, we obtin y (x) = (1+xy +y ) = y +xy +6y y = y +(x+y )y = y +(x+y )(x+y ) = 14. for h ner. y(1+h) = y(1)+y (1)h+y (1) h +y (1) h 6 +O(h4 ) = +5h+1 h +14h 6 +O(h4 ) = +5h+ 1 h + 67 h +O(h 4 ) b) Consider the differentil opertors D 1 = h x +k y nd D = x +f y, where h nd k re fixed numbers, nd f is function of the vribles d y ll whose prtil derivtives (of ny order) re continuous. These opertors re to be pplied to functions g of the vribles d y ll 4 To show tht f (ξ x)g(x)dx = ssuming g(x)dx = is little messy when one works with the Riemnn integrl, but quite esy if one works with the Lebesgue integrl, the ltter being generliztion of the Riemnn integrl, in common use in modern mthemtics. We cnnot go into detils here. 5 Tht is, if for c,d (,b), we hve f (c) H f (d), then there is n η (,b) (in fct, n η between c nd d cn be found) such tht f (η) = H. 5

6 whose prtil derivtives (of ny order) re continuous. Explin why one cn use the Binomil Theorem to clculte D 1 n, where n is positive integer, while one cnnot pply the Binomil Theorem to clculte D n. Do not do detiled clcultion, just clerly describe the min reson. Solution. The proof of Binomil Theorem sying tht (A+B) n = n k= ( ) n A n k B n k fortheelementsandb ofring 6 mkesuseofthessumptionthtandb commute(i.e., thtab = BA). The differentil opertors h d k y forming D 1 commute, so the Binomil Theorem is pplicble with D 1, while the differentil opertors d f y do not commute, so the Binomil Theorem is not pplicble with D. 6 The differentil opertors described generte ring.

### Lecture 1. Functional series. Pointwise and uniform convergence.

1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

### Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

### Numerical Integration

Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the

### UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

### Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

### NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

### Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

### Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

### THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

### Chapter 0. What is the Lebesgue integral about?

Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

### MAA 4212 Improper Integrals

Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

### Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

### f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

### A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

### Improper Integrals, and Differential Equations

Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

### 63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

### The Regulated and Riemann Integrals

Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

### Chapter 1. Basic Concepts

Socrtes Dilecticl Process: The Þrst step is the seprtion of subject into its elements. After this, by deþning nd discovering more bout its prts, one better comprehends the entire subject Socrtes (469-399)

### n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

### different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

### Chapter 5. Numerical Integration

Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with

### Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

### Definite integral. Mathematics FRDIS MENDELU

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### Numerical Analysis: Trapezoidal and Simpson s Rule

nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =

### The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

### 1. Gauss-Jacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),

1. Guss-Jcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on

Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics - A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the

### Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

### Calculus II: Integrations and Series

Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

### 1.9 C 2 inner variations

46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for

### SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

### 1 The Riemann Integral

The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

### INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

### f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

### How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

### P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)

1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

### Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

### Math& 152 Section Integration by Parts

Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

### Main topics for the First Midterm

Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

### Review of Riemann Integral

1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

### III. Lecture on Numerical Integration. File faclib/dattab/lecture-notes/numerical-inter03.tex /by EC, 3/14/2008 at 15:11, version 9

III Lecture on Numericl Integrtion File fclib/dttb/lecture-notes/numerical-inter03.tex /by EC, 3/14/008 t 15:11, version 9 1 Sttement of the Numericl Integrtion Problem In this lecture we consider the

### Overview of Calculus I

Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

### Integration Techniques

Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

### AP Calculus Multiple Choice: BC Edition Solutions

AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this

### An approximation to the arithmetic-geometric mean. G.J.O. Jameson, Math. Gazette 98 (2014), 85 95

An pproximtion to the rithmetic-geometric men G.J.O. Jmeson, Mth. Gzette 98 (4), 85 95 Given positive numbers > b, consider the itertion given by =, b = b nd n+ = ( n + b n ), b n+ = ( n b n ) /. At ech

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

### Math 1B, lecture 4: Error bounds for numerical methods

Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

### New Expansion and Infinite Series

Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University

### x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

### Analytical Methods Exam: Preparatory Exercises

Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the

### ODE: Existence and Uniqueness of a Solution

Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

### ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

### Euler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )

Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s

### Section 7.1 Integration by Substitution

Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find

### APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

### Math 554 Integration

Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

### Introduction to the Calculus of Variations

Introduction to the Clculus of Vritions Jim Fischer Mrch 20, 1999 Abstrct This is self-contined pper which introduces fundmentl problem in the clculus of vritions, the problem of finding extreme vlues

### 4181H Problem Set 11 Selected Solutions. Chapter 19. n(log x) n 1 1 x x dx,

48H Problem Set Selected Solutions Chpter 9 # () Tke f(x) = x n, g (x) = e x, nd use integrtion by prts; this gives reduction formul: x n e x dx = x n e x n x n e x dx. (b) Tke f(x) = (log x) n, g (x)

### 10. AREAS BETWEEN CURVES

. AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

### 8 Laplace s Method and Local Limit Theorems

8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

### Math 360: A primitive integral and elementary functions

Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

### MA 124 January 18, Derivatives are. Integrals are.

MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

### Sections 5.2: The Definite Integral

Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

### Numerical integration

2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter

### 1 The Lagrange interpolation formula

Notes on Qudrture 1 The Lgrnge interpoltion formul We briefly recll the Lgrnge interpoltion formul. The strting point is collection of N + 1 rel points (x 0, y 0 ), (x 1, y 1 ),..., (x N, y N ), with x

### x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

### Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

### Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

### MATH , Calculus 2, Fall 2018

MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

### Final Exam - Review MATH Spring 2017

Finl Exm - Review MATH 5 - Spring 7 Chpter, 3, nd Sections 5.-5.5, 5.7 Finl Exm: Tuesdy 5/9, :3-7:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.

### Numerical Analysis. 10th ed. R L Burden, J D Faires, and A M Burden

Numericl Anlysis 10th ed R L Burden, J D Fires, nd A M Burden Bemer Presenttion Slides Prepred by Dr. Annette M. Burden Youngstown Stte University July 9, 2015 Chpter 4.1: Numericl Differentition 1 Three-Point

### W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

### 1 Part II: Numerical Integration

Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble

### Lecture 19: Continuous Least Squares Approximation

Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for

### c n φ n (x), 0 < x < L, (1) n=1

SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry

### 1.3 The Lemma of DuBois-Reymond

28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBois-Reymond We needed extr regulrity to integrte by prts nd obtin the Euler- Lgrnge eqution. The following result shows tht, t lest sometimes, the extr

### The Riemann Integral

Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function

### NUMERICAL INTEGRATION

NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls

### F (x) dx = F (x)+c = u + C = du,

35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil

### B.Sc. in Mathematics (Ordinary)

R48/0 DUBLIN INSTITUTE OF TECHNOLOGY KEVIN STREET, DUBLIN 8 B.Sc. in Mthemtics (Ordinry) SUPPLEMENTAL EXAMINATIONS 01 Numericl Methods Dr. D. Mckey Dr. C. Hills Dr. E.A. Cox Full mrks for complete nswers

### 7.2 The Definite Integral

7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

### We know that if f is a continuous nonnegative function on the interval [a, b], then b

1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

### Taylor Polynomial Inequalities

Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil

### 1 The fundamental theorems of calculus.

The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

### f(a+h) f(a) x a h 0. This is the rate at which

M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

### Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...

Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.

### Abstract inner product spaces

WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the

### Properties of the Riemann Integral

Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2

### MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

### 38 Riemann sums and existence of the definite integral.

38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

### Math 61CM - Solutions to homework 9

Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ

### Math 131. Numerical Integration Larson Section 4.6

Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n

### Riemann is the Mann! (But Lebesgue may besgue to differ.)

Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

### APPLICATIONS OF THE DEFINITE INTEGRAL

APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through

### SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

### Chapters 4 & 5 Integrals & Applications

Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

### Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued