UniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that


 Gerard Burns
 2 years ago
 Views:
Transcription
1 Problemen/UWC NAW 5/7 nr juni Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de Lune, en PG Kluit Problem 005/4A We hve k= = Consequently prtil sums must stisfy k(k+) k(k + ) < Show tht for every q Q stisfying 0 < q <, there eists finite subset K N so tht k(k + ) = q Solution This problem ws solved by Peter Vndendriessche, Vldislv Frnk nd Arne Smeets The solution below is bsed on tht of Vldislv Frnk First note tht k(k+) = k k+ Hence k(k+) + (k+)(k+) + + (k+n )(k+n) = k k+ + k+ + k+n k+n = k k+n Consequently it suffices to represent every rtionl number between 0 nd s + k, where 3 k If two consecutive numbers re equl, they simply cncel out, so we llow equl numbers This will be useful in finl step of proof Let b be our rtionl number There is nturl number n such tht n+ < b n Consider = n b = b n bn The numertor of this frction is nonnegtive becuse b n, but less thn, the numertor of b, becuse (n + ) < 0 We hve b = n We now pply the sme lgorithm to Let m be nturl number such tht m+ < m b n The clim is tht m n Nmely, = bn < bn, hence n m n + > n If we continue this lgorithm, we obtin b = ( ( 3 ( ( ) ))) Notice tht the lgorithm cn only be repeted finitely mny times, s the numertor decreses t ech step We now hve b = + 3 ± If is even we re done In other cse we my ssume tht > nd chnge into ( ) Here ( ) nd we re done Of course =, s otherwise b = = which is impossible As generliztion, V Frnk shows tht for ny irrtionl number in the intervl [0,] there eists n infinite sum Problem 005/4B We consider the progressive rithmetic nd geometric mens of the function sequence f n () = n, n N, > 0, = These re nd A n = A n () = n ( n ) = n n( ) G n = G n () = ( ++ +(n ) ) n = n The Mrtinsproperty reds A n+ /A n G n+ /G n In our cse this gives n n+ n + n Eindredctie: Mtthijs Coster Redctiedres: UWC/NAW Mthemtisch Instituut Postbus RA Leiden Prove, more generlly, tht + + for >, > 0, = Solution This problem ws solved by Jn vn de Lune, Peter Vndendriessche, Vldislv Frnk nd Arne Smeets The solution below is bsed on tht of Peter Vndendriessche
2 48 NAW 5/7 nr juni 006 Problemen/UWC Let f (t) be (smooth) nonnegtive function tht is conve on [, b] nd let [, y] [, b] such tht + y = + b We then hve b f (t)dt f (t)dt y To prove this, consider, for given f (t),, nd y, the function g(t) defined by g(t) = f () + ( f (y) f ())(t ) y g(t) is the line through the points (, f ()) nd (y, f (y)) Notice tht the conveity of f gives g(t)dt f (t)dt Let h(t) = g(t) g(t)dt + f (t)dt, then y h(t)dt = f (t)dt By conveity we hve f (t) g(t) h(t) for t [, ] [y, b] Since h(t) is the eqution of line nd + y = + b, we hve Combining these results we find: b h(t)dt = h(t)dt y b f (t)dt f (t)dt + b y f (t)dt = + h(t)dt + b y h(t)dt + b h(t)dt b f (t)dt = = y y f (t)dt y h(t)dt y Problem B is specil cse of this result For R + 0, =, let f (t) = t Then f (t) = t log () 0 Therefore f (t) is conve We hve to distinguish two cses: (, 0) Then 0 < + < + Apply the lemm to the intervl [ +, ] [0, + ] (0, ) Then 0 < < + < + Apply the lemm to the intervl [, + ] [0, + ] Notice tht in the first cse the sign in both numertor nd denomintor chnges on the right side of the eqution: from which we cn deduce + 0 t dt + + t dt, + + ( ) It is esy to prove the generliztion z y b b z y y, where y + z = + b nd < < y < z < b Problem 005/4C A finite geometry is geometric system tht hs only finite number of points For n ffine plne geometry, the ioms re s follows: Given ny two distinct points, there is ectly one line tht includes both points The prllel postulte: Given line L nd point P not on L, there eists ectly one line through P tht is prllel to L 3 There eists set of four points, no three colliner
3 Problemen/UWC NAW 5/7 nr juni We denote the set of points by P, nd the set of lines by L Let σ be n utomorphism of (P, L ) (mening tht three colliner points of P re mpped onto three colliner points of P nd three noncolliner points of P re mpped onto three noncolliner points of P ) Prove tht there eists point P P with σ(p) = P or line L L with σ(l) = L or σ(l) L = Solution This problem hs been solved by Leendert Bleijeng nd Peter Vndendriessche The solution below is bsed on their solutions First we will prove the following lemm: Lemm Let M, L L, then M = L Proof Suppose tht M L > then M = L Therefore we my ssume tht M L = Let M = m nd L = l By Aiom 3 we know tht there eists P P such tht P L nd p M Through P we cn construct line prllel to L nd l lines tht intersect L in its l points In the sme wy we cn construct, through P, line prllel to M nd m lines tht intersect M in its m points Let us now determine the number of lines through P; this equls l + nd m + If M L = 0, pick points L nd b M Let N be the line through nd b Then by the previous rgument L = N nd M = N We conclude tht ll lines consist of n equl number of points, sy s Lemm P < L Proof Let P = p nd L = l Every two points define line, nd there re p(p ) pirs of points Ech line hs s points nd is counted s(s ) times Therefore l = p(p ) In order to show tht p < l we hve to prove tht s(s ) < p or p > s(s ) s s + The third iom tells us tht there eist three noncolliner points, b, c P Let L be the line through nd b, M the line through nd c By the prllel postulte, through every point on L there is ectly one line prllel to M Strting with s points on L, we find s lines, ll consisting of s points Therefore p s Suppose tht σ(p) = p, for ll p P, nd tht σ(l) = L nd σ(l) L = for ll L L Consider the function µ : L P given by µ(l) = σ(l) L µ is well defined since σ(l) L is lwys unique point Now suppose tht µ(l) = µ(m) or σ(l) L = σ(m) M = p, nd σ(q) = p Then q L nd q M We know tht q = p Therefore L = M nd µ is injective However, if µ is injective, then P L, which contrdicts the previous lemm Problem 005/4* stisfy We hve k= /k = (π /6) Consequently prtil sums must k < π 6 Given ny q Q stisfying 0 < q < (π /6), does there eist finite subset K N \{} so tht k = q? Solution This problem ws solved by PG Kluit The solution below is bsed on his solution Let q = ki, where k i re different integers Let m be the lest common multiple of ll k i in the sum For ech such k i number k i eists such tht k ik i = m Then q = m (k i ), tht is, q cn be written s frction with denomintor m nd the numertor sum of squres of different divisors of m This rises the question: given m, which numbers cn be written s sums of squres of different divisors of m? We will show tht for highly composite numbers m, more specificlly m = n!, the nswer will be tht sufficiently mny integers cn be written s sums of squres to prove the problem
4 50 NAW 5/7 nr juni 006 Problemen/UWC Lemm Let n 5 be n integer nd let 3 = d < d < d m = n!/3 be ll divisors of n! between 3 nd n!/3 Then d k > d k+ for k < m Proof Let us prove this by induction For n = 5 the divisors d,, d re 3, 4, 5, 6, 8, 0,, 5, 0, 4, 30, nd 40 It is esy to verify the lemm We ssume the lemm is true for n We hve to prove tht the Lemm holds for the divisors of (n + )! The divisors of (n + )! tht re less thn n!/3 clerly stisfy the lemm Even though there my be more divisors, this cnnot influence the inequlity Suppose tht d k nd d k+ re two successive divisors of (n + )!, with n!/3 d k < d k+ (n + )!/3 Let d k d k = d k+d k+ = (n + )! Then d k+ nd d k re two successive divisors of (n + )! with 3 d k+ < d k 3(n + ) As for n 5 we hve 3(n + ) < n!/3, this suffices to conclude the proof Lemm Let n [9, 56] be n integer Then n cn be represented s sum of different squres d + + d k, where d < < d k 0 Proof The proof cn be found by the enumertion of 8 representtions There is slightly shorter proof which will be left to the reder Lemm Let n N, n Then every integer [9,σ (n!) n! 9] cn be represented s = d k, where the d k re different divisors of n! Here σ m () = d d m Proof Let L kn = [9, t] be the longest intervl in [9, ) whose integers cn ll be represented s sum of different squres of some of the first k divisors of n! Let l kn = L kn, the length of the intervl In the proof the nottion will be bbrevite to l k = L k if it is cler which n is ment In the second lemm we sw tht l 0 = 8 Notice tht l = 49 (= 8 + ) Any 56 is represented by the divisors lesser thn or equl to 0, while the integers re represented using We will show in generl tht l k+ = l k + d k+ by induction, s long s d k+ < n!/ The proof will be given in two steps In the first step we prove tht d k+ < l k+ given d k < l k nd l k+ = l k + d k+ In the second step we will prove tht l k+ = l k + d k+ given d k < l k Using these two steps nd the bsic ssumption (k = ) we cn prove for rbitrry k tht l k+ = l k + d k+ First step Given d k < l k nd l k+ = l k + d k+ we find tht d k+ < l k+ Proof The first lemm tells us tht d k+ < d k Therefore we hve d k+ < d k+ + d k < d k+ + l k < l k+ Second step Given d k < l k we find tht l k+ = l k + d k+ Proof The proof is comprble to the proof bove For L k, it is cler tht L k+ s well, while for the numbers L k+ \L k notice tht l k + 8 < l k + d k+ + 8 If we use the number d k+ to represent the sum, we find for the rest y = d k+ tht l k d k+ + 8 < y l k + 8 Using Lemm gin we hve l k d k+ + 8 > l k d k + 8 > 8 Therefore y L k We cn rewrite the results l k+ = l k + d k+ s for k m, where d m = n!/3 l k = k di 8, i=
5 Problemen/UWC NAW 5/7 nr juni In order to complete the proof of Lemm 3 we need to prove tht l (m+)n = l mn + d n!/, where d m+ = n!/ Notice tht 4 < Therefore for rbitrry L (m+)n, we find either L mn or 4 n! L mn Now we find l k = k di 8, i= for k m +, where d m+ = n!/ This concludes the proof of this lemm Theorem For every q Q such tht 0 < q < π 6 k, finite subset K N eists, such tht Proof Let q Q with 0 < q < π 6 We cn find n n N fulfilling ech of the three following properties by choosing n sufficiently lrge Moreover ech of these properties is monotonic, mening tht if it is true for some n 0, it will be true for ll n > n 0 n, If q = /b, where nd b hve no common divisors, then b divides n!, If q = /(n!), then n is chosen such tht 8 < < σ (n!) (n!) 8 To prove the eistence of 3) notice tht = q σ lim (n!) (n!) 8 n n! = π 6 Now Lemm 3 my be pplied, showing tht cn be represented s sum of squres of different divisors of n! This gives us the sought for representtion of q Remrk A solution of the Str Problem turns out to hve been published in Ron Grhm s On Finite Sums of Unit Frctions, Proc London Mth Soc(4), 964, pp The bsic ides behind the two solutions re similr Grhm strts with multiplictive set S, which in the Str Problem is the set of squres Grhm then defines P(S), the set of sums of elements of S Using the nottion S for the set of inverses of the elements of S, Grhm shows tht if P(S) contins ll positive integers, up to finite number, S is finite, nd s n+ /s n is bounded, then p q P(S ) whenever q s for some s S Moreover, for every ɛ > 0 there is n s P(S ) such tht s p q < ɛ
padic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
More informationW. 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)
More informationA BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int
A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure
More informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationAQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system
Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationMath 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:
More informationUNIFORM 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,
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationLecture 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
More informationMath 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
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationThe 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
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationMath 4310 Solutions to homework 1 Due 9/1/16
Mth 4310 Solutions to homework 1 Due 9/1/16 1. Use the Eucliden lgorithm to find the following gretest common divisors. () gcd(252, 180) = 36 (b) gcd(513, 187) = 1 (c) gcd(7684, 4148) = 68 252 = 180 1
More informationMATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35
MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 9. Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules.
More information(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer
Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.
More informationHandout: Natural deduction for first order logic
MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More information4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve
Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions
More informationPARTIAL FRACTION DECOMPOSITION
PARTIAL FRACTION DECOMPOSITION LARRY SUSANKA 1. Fcts bout Polynomils nd Nottion We must ssemble some tools nd nottion to prove the existence of the stndrd prtil frction decomposition, used s n integrtion
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More information1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...
Contents 1 Sets 1 1.1 Functions nd Reltions....................... 3 1.2 Mthemticl Induction....................... 5 1.3 Equivlence of Sets nd Countbility................ 6 1.4 The Rel Numbers..........................
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationAdvanced 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
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σfinite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationIntroduction to Group Theory
Introduction to Group Theory Let G be n rbitrry set of elements, typiclly denoted s, b, c,, tht is, let G = {, b, c, }. A binry opertion in G is rule tht ssocites with ech ordered pir (,b) of elements
More informationRead section 3.3, 3.4 Announcements:
Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationIntegral points on the rational curve
Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin  Mrinette 750 W. Byshore Street Mrinette, WI 5443453 Also: Konstntine Zeltor P.O. Box
More informationOn Error Sum Functions Formed by Convergents of Real Numbers
3 47 6 3 Journl of Integer Sequences, Vol. 4 (), Article.8.6 On Error Sum Functions Formed by Convergents of Rel Numbers Crsten Elsner nd Mrtin Stein Fchhochschule für die Wirtschft Hnnover Freundllee
More informationf(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
More informationMORE 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
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationMTH 505: Number Theory Spring 2017
MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c
More informationReview 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.
More informationON THE EXCEPTIONAL SET IN THE PROBLEM OF DIOPHANTUS AND DAVENPORT
ON THE EXCEPTIONAL SET IN THE PROBLEM OF DIOPHANTUS AND DAVENPORT Andrej Dujell Deprtment of Mthemtics, University of Zgreb, 10000 Zgreb, CROATIA The Greek mthemticin Diophntus of Alexndri noted tht the
More informationA basic logarithmic inequality, and the logarithmic mean
Notes on Number Theory nd Discrete Mthemtics ISSN 30 532 Vol. 2, 205, No., 3 35 A bsic logrithmic inequlity, nd the logrithmic men József Sándor Deprtment of Mthemtics, BbeşBolyi University Str. Koglnicenu
More informationa n+2 a n+1 M n a 2 a 1. (2)
Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside
More informationMath 130 Midterm Review
Mth 130 Midterm Review April 6, 2013 1 Topic Outline: The following outline contins ll of the mjor topics tht you will need to know for the exm. Any topic tht we ve discussed in clss so fr my pper on the
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationIn Section 5.3 we considered initial value problems for the linear second order equation. y.a/ C ˇy 0.a/ D k 1 (13.1.4)
678 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions 13.1 TWOPOINT BOUNDARY VALUE PROBLEMS In Section 5.3 we considered initil vlue problems for the liner second order eqution
More informationa n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.
MAS221(21617) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by
More informationJim Lambers MAT 169 Fall Semester Lecture 4 Notes
Jim Lmbers MAT 169 Fll Semester 200910 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of
More informationAnonymous Math 361: Homework 5. x i = 1 (1 u i )
Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define
More informationSection 3.2: Negative Exponents
Section 3.2: Negtive Exponents Objective: Simplify expressions with negtive exponents using the properties of exponents. There re few specil exponent properties tht del with exponents tht re not positive.
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More informationInfinite Geometric Series
Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to
More informationHomework 11. Andrew Ma November 30, sin x (1+x) (1+x)
Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 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
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationJournal of Inequalities in Pure and Applied Mathematics
Journl o Inequlities in Pure nd Applied Mthemtics http://jipm.vu.edu.u/ Volume 6, Issue 4, Article 6, 2005 MROMORPHIC UNCTION THAT SHARS ON SMALL UNCTION WITH ITS DRIVATIV QINCAI ZHAN SCHOOL O INORMATION
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More informationAdvanced 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
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationThe Algebra (aljabr) of Matrices
Section : Mtri lgebr nd Clculus Wshkewicz College of Engineering he lgebr (ljbr) of Mtrices lgebr s brnch of mthemtics is much broder thn elementry lgebr ll of us studied in our high school dys. In sense
More informationA 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
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationHomework Solution  Set 5 Due: Friday 10/03/08
CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution  et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte nonfinl.
More informationThe usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1
Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be nonnegtive. By introducing
More informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationNatural examples of rings are the ring of integers, a ring of polynomials in one variable, the ring
More generlly, we define ring to be nonempty set R hving two binry opertions (we ll think of these s ddition nd multipliction) which is n Abelin group under + (we ll denote the dditive identity by 0),
More informationCoalgebra, Lecture 15: Equations for Deterministic Automata
Colger, Lecture 15: Equtions for Deterministic Automt Julin Slmnc (nd Jurrin Rot) Decemer 19, 2016 In this lecture, we will study the concept of equtions for deterministic utomt. The notes re self contined
More informationf(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
More informationState space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies
Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More information7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series
7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts
More informationThe 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
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationCHAPTER 4 MULTIPLE INTEGRALS
CHAPTE 4 MULTIPLE INTEGAL The objects of this chpter re fivefold. They re: (1 Discuss when sclrvlued functions f cn be integrted over closed rectngulr boxes in n ; simply put, f is integrble over iff
More informationList all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.
Mth Anlysis CP WS 4.X Section 4.4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether  is root of 0. Show
More informationGoals: 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
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationImproper 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
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationMAT 215: Analysis in a single variable Course notes, Fall Michael Damron
MAT 215: Anlysis in single vrible Course notes, Fll 2012 Michel Dmron Compiled from lectures nd exercises designed with Mrk McConnell following Principles of Mthemticl Anlysis, Rudin Princeton University
More informationRiemann 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
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More informationConsolidation Worksheet
Cmbridge Essentils Mthemtics Core 8 NConsolidtion Worksheet N Consolidtion Worksheet Work these out. 8 b 7 + 0 c 6 + 7 5 Use the number line to help. 2 Remember + 2 2 +2 2 2 + 2 Adding negtive number is
More informationMATH FIELD DAY Contestants Insructions Team Essay. 1. Your team has forty minutes to answer this set of questions.
MATH FIELD DAY 2012 Contestnts Insructions Tem Essy 1. Your tem hs forty minutes to nswer this set of questions. 2. All nswers must be justified with complete explntions. Your nswers should be cler, grmmticlly
More informationA HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction
Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127956z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationFor convenience, we rewrite m2 s m2 = m m m ; where m is repeted m times. Since xyz = m m m nd jxyj»m, we hve tht the string y is substring of the fir
CSCI 2400 Models of Computtion, Section 3 Solutions to Homework 4 Problem 1. ll the solutions below refer to the Pumping Lemm of Theorem 4.8, pge 119. () L = f n b l k : k n + lg Let's ssume for contrdiction
More informationWHEN IS A FUNCTION NOT FLAT? 1. Introduction. {e 1 0, x = 0. f(x) =
WHEN IS A FUNCTION NOT FLAT? YIFEI PAN AND MEI WANG Abstrct. In this pper we prove unique continution property for vector vlued functions of one vrible stisfying certin differentil inequlity. Key words:
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationModule 6: LINEAR TRANSFORMATIONS
Module 6: LINEAR TRANSFORMATIONS. Trnsformtions nd mtrices Trnsformtions re generliztions of functions. A vector x in some set S n is mpped into m nother vector y T( x). A trnsformtion is liner if, for
More informationIntroduction to Mathematical Reasoning, Saylor 111
Frction versus rtionl number. Wht s the difference? It s not n esy question. In fct, the difference is somewht like the difference between set of words on one hnd nd sentence on the other. A symbol is
More information