Convex Sets and Functions


 Stuart Boone
 1 years ago
 Views:
Transcription
1 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 segment connecting x nd y belongs to C Exmple B2 The convex subsets of R, +, ) re the intervls of R Regulr polygons re convex subsets of R 2 Definition B3 Let U be subset of rel liner spce L, +, ) A convex combintion of U is n element of L of the form 1 x k x k, where x 1,,x k U, i for 1 i k, nd k = 1 If the conditions i re dropped, we hve n ffine combintion of U In other words, x is n ffine combintion of U if there exist 1,, k R such tht x = 1 x k x k, for x 1,,x k U, nd k i = 1 Definition B4 Let U be subset of rel liner spce L, +, ) A subset {x 1,,x n } is ffinely dependent if = 1 x n x n such tht t lest one of the numbers 1,, n is nonzero nd i = If no such ffine combintion exists, then x 1,,x n re ffinely independent Theorem B5 The set U = {x 1,,x n } is ffinely independent if nd only if the set V = {x 1 x n,x n 1 x n } is linerly independent Proof Suppose tht U is ffinely independent but V is linerly dependent; tht is, = b 1 x 1 x n ) + + b n 1 x n 1 x n ) such tht not ll numbers b i re This implies b 1 x b n 1 x n 1 n 1 b i )x n =, which contrdicts the ffine independence of U
2 574 B Convex Sets nd Functions Conversely, suppose tht V is linerly independent but U is not ffinely independent In this cse, = 1 x n x n such tht t lest one of the numbers 1,, n is nonzero nd i = This implies n = 1 i, so = 1 x 1 x n ) + + n 1 x n 1 x n ) Observe tht t lest one of the numbers 1,, n 1 must be distinct from becuse otherwise we would hve 1 = = n 1 = n = This contrdicts the liner independence of V, so U is ffinely independent Exmple B6 Let x 1 nd x 2 be two elements of the liner spce R 2, +, ) The line tht psses through x 1 nd x 2 consists of ll x such tht x x 1 nd x x 2 re colliner; tht is, x x 1 ) + bx x 2 ) = for some, b R such tht + b Thus, we hve x = 1 x x 2, where = + b + b + b = 1, so x is n ffine combintion of x 1 nd x 2 It is esy to see tht the segment of line contined between x 1 nd x 2 is given by convex combintion of x 1 nd x 2 ; tht is, by n ffine combintion 1 x x 2 such tht 1, 2 Theorem B7 If C is convex subset of rel liner spce L, +, ), then C contins ll convex liner combintions of C Proof The proof is by induction on k 2 nd is left to the reder Theorem B8 The intersection of ny collection of convex sets of liner spce L, +, ) is convex set Proof Let C = {C i i I} be collection of convex sets nd let C = C Suppose tht x 1,,x k C, i for 1 i k, nd k = 1 Since x 1,,x k C i, it follows tht 1 x k x k C i for every i I Thus, 1 x k x k C, which proves the convexity of C Corollry B9 The fmily of convex sets of liner spce L, +, ) is closure system on PL) Proof This sttement follows immeditely from Theorem B8 by observing tht the set L is convex Corollry B9 llows us to define the convex hull of subset U of L s the closure K conv U) of U reltive to the closure system of the convex subsets of L If U R n consists of n+1 points such tht no point is n ffine combintion of the other n points, then K conv U) is n ndimensionl simplex in L Exmple B1 A twodimensionl simplex is defined strting from three points x 1,x 2,x 3 in R 2 such tht none of these points is n ffine combintion of
3 B Convex Sets nd Functions 575 the other two no point is colliner with the others two) Thus, the twodimensionl symplex generted by x 1,x 2,x 3 is the full tringle determined by x 1,x 2,x 3 In generl, n ndimensionl simplex is the convex hull of set of n + 1 points x 1,,x n+1 in R n such tht no point is n ffine combintion of the remining n points Let S be the ndimensionl simplex generted by the points x 1,,x n+1 in R n nd let x S If x S, then x is convex combintion of x 1,,x n,x n+1 In other words, there exist 1,, n, n+1 such tht 1,, n, n+1, 1), +1 i = 1, nd x = 1 x n x n + n+1 x n+1 The numbers 1,, n, n+1 re the bricentric coordintes of x reltive to the simplex S nd re uniquely determined by x Indeed, if we hve x = 1 x n x n + n+1 x n+1 = b 1 x b n x n + b n+1 x n+1, nd i b i for some i, this implies 1 b 1 )x n b n )x n + n+1 b n+1 )x n+1 =, which contrdicts the ffine independence of x 1,,x n+1 The next sttement plys centrl role in the study of convexity We reproduce the proof given in [59] Theorem B11 Crthéodory s Theorem) If U is subset of R n, then for every x K conv U) we hve x = +1 ix i, where x i U, i for 1 i n + 1, nd +1 i = 1 Proof Consider x K conv U) We cn write x = p+1 ix i, where x i U, i for 1 i p + 1, nd p+1 i = 1 Let p be the smllest number which llows this kind of expression for x We prove the theorem by showing tht p n Suppose tht p n+1 Then, the set {x 1,,x p+1 } is ffinely dependent, so there exist b 1,, b p+1 not ll zero such tht = p+1 b ix i nd p+1 b i = Without loss of generlity, we cn ssume b p+1 > nd p+1 b p+1 i b i for ll i such tht 1 i p nd b i > Define i c i = b i ) p+1 b i b p+1 for 1 i p We hve c i = i p+1 b p+1 b i = 1 Furthermore, c i for 1 i p Indeed, if b i, then c i i ; if b i >, then c i becuse p+1 b p+1 i b i for ll i such tht 1 i p nd b i > Thus, we hve
4 576 B Convex Sets nd Functions c i x i = i ) p b i x i = b p i x i = x, which contrdicts the choice of p A finite set of points P in R 2 is convex polygon if no member p of P lies in the convex hull of P {p} Theorem B12 A finite set of points P in R 2 is convex polygon if nd only if no member p of P lies in twodimensionl simplex formed by three other members of P Proof The rgument is strightforwrd nd is left to the reder s n exercise Theorem B13 Rdon s Theorem) Let P = {x i R n 1 i n + 2} be set of n + 2 points in R n Then, there re two disjoint subsets R nd Q of P such tht K conv R) K conv Q) Proof Since n+2 points in R n re ffinely dependent, there exist 1,, n+2 not ll equl to such tht n+2 i x i = B1) nd +2 i = Without loss of generlity, we cn ssume tht the first k numbers re positive nd the lst n + 2 k re not Let = k i > nd let b j = j for 1 j k Similrly, let c l = l for k + 1 l n + 2 Equlity B1) cn now be written s k b j x j = j=1 n+2 l=k+1 c l x l Since the numbers b j nd c l re nonnegtive nd k j=1 b j = +2 l=k+1 c l = 1, it follows tht K conv {x 1,,x k }) K conv {x k+1,,x n+2 }) Theorem B14 Klein s Theorem) If P R 2 is set of five points such tht no three of them re colliner, then P contins four points tht form convex qudrilterl Proof Let P = {x i 1 i } If these five points form convex polygon, then ny four of them form convex qudrilterl If exctly one point is in the interior of convex qudrilterl formed by the remining four points, then the desired conclusion is reched Suppose tht none of the previous cses occur Then, two of the points, sy x p,x q, re locted inside the tringle formed by the remining points x i,x j,x k Note tht the line x p x q intersects two sides of the tringle x i x j x k,
5 B Convex Sets nd Functions 577 x i x q x p x k x j Fig B1 A fivepoint configurtion in R 2 sy x i x j nd x i x k see Figure B1) Then x p x q x k x j is convex qudrilterl A function f : R R is convex if its grph on n intervl is locted below the chord determined by the endpoints of the intervl More formlly, we hve the following definition Definition B15 A function f : R R is convex if ftx + 1 t)y) tfx) + 1 t)fy) for every x, y Domf) nd t [, 1] The function g : R R is concve if g is convex Theorem B16 If f : R R is convex function nd < b c, then fb) f) b fc) f) c Proof Since < b c, we cn write b = t + 1 t)c, where t = c b c, 1] The convexity of f yields the inequlity fb) c b c f) + b c fc), which is esily seen to be equivlent with the desired inequlity A similr result follows Theorem B17 If f : R R is convex function nd b < c, then fc) f) c fc) fb) c b
6 578 B Convex Sets nd Functions Proof The rgument is similr to the proof of Theorem B16 Corollry B18 Let f : R R be convex function nd let p, q, p, q be four numbers such tht p p < q q We hve the inequlity fq) fp) q p fq ) fp ) q p B2) Proof By Theorem B16 pplied to the numbers p, q, q, we hve fq) fp ) q p fq ) fp ) q p Similrly, by pplying Theorem B17 to p, p, q, we obtin fq) fp) q p fq) fp ) q p The inequlity of the corollry cn be obtined by combining the lst two inequlities From Corollry B18, it follows tht if f : R R is convex nd differentible everywhere, then its derivtive is n incresing function The converse is lso true; nmely, if f is differentible everywhere nd its derivtive is n incresing function, then f is convex Indeed, let, b, c be three numbers such tht < b < c By the men vlue theorem, there is p, b) nd q b, c) such tht f p) = fb) f) b Since f p) f q), we obtin nd f q) = fc) fb) c b fb) f) b fc) fb), c b which implies fb) c b c f) + b c fc); tht is, the convexity of f Thus, if f is twice differentible everywhere nd its second derivtive is nonnegtive everywhere, then it follows tht f is convex Clerly, under the sme conditions of differentibility s bove, if the second derivtive is nonpositive everywhere, then f is concve The functions listed in the Tble B1, defined on the set R, provide exmples of convex or concve) functions Theorem B19 Jensen s Theorem) Let f be function tht is convex on n intervl I If t 1,,t n [, 1] re n numbers such tht t i = 1, then
7 B Convex Sets nd Functions 579 n ) f t i x i t i fx i ) for every x 1,, x n I Proof The rgument is by induction on n, where n 2 The bsis step, n = 2, follows immeditely from Definition B15 Suppose tht the sttement holds for n, nd let u 1,, u n, u n+1 be n + 1 numbers such tht +1 u i = 1 We hve fu 1 x u n 1 x n 1 + u n x n + u n+1 x n+1 ) = f u 1 x u n 1 x n 1 + u n + u n+1 ) u ) nx n + u n+1 x n+1 u n + u n+1 By the inductive hypothesis, we cn write fu 1 x u n 1 x n 1 + u n x n + u n+1 x n+1 ) ) un x n + u n+1 x n+1 u 1 fx 1 ) + + u n 1 fx n 1 ) + u n + u n+1 )f u n + u n+1 Next, by the convexity of f, we hve ) un x n + u n+1 x n+1 u n f fx n ) + u n+1 fx n+1 ) u n + u n+1 u n + u n+1 u n + u n+1 Combining this inequlity with the previous inequlity gives the desired conclusion Of course, if f is concve function nd t 1,,t n [, 1] re n numbers such tht t i = 1, then n ) f t i x i t i fx i ) B3) Tble B1 Exmples of convex or concve functions Function Second Convexity Derivtive Property x r for rr 1)x r 2 concve for r < 1 r > convex for r 1 ln x 1 x 2 concve xln x 1 x convex e x e x convex
8 58 B Convex Sets nd Functions Exmple B2 We sw tht the function fx) = ln x is concve Therefore, if t 1,, t n [, 1] re n numbers such tht t i = 1, then n ) ln t i x i t i lnx i This inequlity cn be written s n ) n ln t i x i ln x ti i, or equivlently n t i x i n x ti i, for x 1,,x n, ) In the specil cse where t 1 = = t n = 1 n, we hve the inequlity tht reltes the rithmetic to the geometric verge on n positive numbers: x x n n n x i ) 1 n B4) Let w = w 1,,w n ) R n be such tht w i = 1 For r, the wweighted men of order r of sequence of n positive numbers x = x 1,, x n ) R n > is the number n ) 1 r µ r w x) = w i x r i Of course, µ r w x) is not defined for r = ; we will give s specil definition µ wx) = lim r µ r wx) We hve lim r lnµr w x) = lim r = lim r n = = ln ln w ix r i r w ix r i lnx i w ix r i w i lnx i n x wi i
9 B Convex Sets nd Functions 581 Thus, if we define µ w x) = n xwi i, the weighted men of order r becomes function continuous everywhere with respect to r For w 1 = = w n = 1 n, we hve µ 1 nx 1 x n w x) = x 2 x n + + x 1 x n 1 the hrmonic verge of x), µ wx) = x 1 x n ) 1 n the geometric verge of x), µ 1 wx) = x x n n the rithmetic verge of x) Theorem B21 If p < r, we hve µ p wx) µ r wx) Proof There re three cses depending on the position of reltive to p nd r In the first cse, suppose tht r > p > The function fx) = x r p is convex, so by Jensen s inequlity pplied to x p 1,, xp n, we hve which implies ) r p w i x p i n w i x r i, ) 1 p w i x p i w i x r i ) 1 r, which is the inequlity of the theorem If r > > p, the function fx) = x r p is gin convex becuse f x) = r r p p )x 1 r p 2 Thus, the sme rgument works s in the previous cse Finlly, suppose tht > r > p Since < r p < 1, the function fx) = x r p is concve Thus, by Jensen s inequlity, ) r p w i x p i n w i x r i Since 1 r <, we obtin gin ) 1 p w i x p i w i x r i ) 1 r
10
11 C Useful Integrls nd Formuls C1 Euler s Integrls The integrls B, b) = Γ) = x 1 1 x) b 1 dx, x 1 e x dx, re known s Euler s integrl of the first type nd Euler s integrl of the second type, respectively We ssume here tht nd b re positive numbers to ensure tht the integrls re convergent Replcing x by 1 x yields the equlity B, b) = which shows tht B is symmetric Integrting B, b) by prts, we obtin B, b) = = 1 x 1 1 x) b 1 dx 1 x) b 1 d x 1 = x 1 x) 1 b) = b 1 = b 1 1 x) 1 x) b 1 dx = Bb, ), + b 1 x 1 x) b 2 dx x 1 1 x) b 2 dx b 1 b 1 B, b 1) B, b), x 1 1 x) b 1 dx
Best 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 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 information8 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
More informationMath 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
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
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 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 informationOn the Generalized Weighted QuasiArithmetic Integral Mean 1
Int. Journl of Mth. Anlysis, Vol. 7, 2013, no. 41, 20392048 HIKARI Ltd, www.mhikri.com http://dx.doi.org/10.12988/ijm.2013.3499 On the Generlized Weighted QusiArithmetic Integrl Men 1 Hui Sun School
More informationLecture 14: Quadrature
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 relvlues nd smooth The pproximtion of n integrl by numericl
More informationUniversitaireWiskundeCompetitie. 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
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationWe 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
More informationOrthogonal Polynomials and LeastSquares Approximations to Functions
Chpter Orthogonl Polynomils nd LestSqures Approximtions to Functions **4/5/3 ET. Discrete LestSqures Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny
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 informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
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 informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationSections 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)
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
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 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 informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
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 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 information4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationFor a continuous function f : [a; b]! R we wish to define the Riemann integral
Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This
More informationVariational Techniques for SturmLiouville Eigenvalue Problems
Vritionl Techniques for SturmLiouville Eigenvlue Problems Vlerie Cormni Deprtment of Mthemtics nd Sttistics University of Nebrsk, Lincoln Lincoln, NE 68588 Emil: vcormni@mth.unl.edu Rolf Ryhm Deprtment
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 informationDiscrete Leastsquares Approximations
Discrete Lestsqures Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More informationNumerical 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
More informationChapter 4. Lebesgue Integration
4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.
More informationJournal of Inequalities in Pure and Applied Mathematics
Journl of Inequlities in Pure nd Applied Mthemtics GENERALIZATIONS OF THE TRAPEZOID INEQUALITIES BASED ON A NEW MEAN VALUE THEOREM FOR THE REMAINDER IN TAYLOR S FORMULA volume 7, issue 3, rticle 90, 006.
More informationEulerMaclaurin Summation Formula 1
Jnury 9, EulerMclurin Summtion Formul Suppose tht f nd its derivtive re continuous functions on the closed intervl [, b]. Let ψ(x) {x}, where {x} x [x] is the frctionl prt of x. Lemm : If < b nd, b Z,
More informationSOLUTIONS 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.
More informationNOTES AND PROBLEMS: INTEGRATION THEORY
NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents
More informationLecture 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
More informationCzechoslovak 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 HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationLecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature
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
More informationSTURMLIOUVILLE BOUNDARY VALUE PROBLEMS
STURMLIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationProblem Set 3
14.102 Problem Set 3 Due Tuesdy, October 18, in clss 1. Lecture Notes Exercise 208: Find R b log(t)dt,where0
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 informationMath 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
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 informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More information1.3 Regular Expressions
56 1.3 Regulr xpressions These hve n importnt role in describing ptterns in serching for strings in mny pplictions (e.g. wk, grep, Perl,...) All regulr expressions of lphbet re 1.Ønd re regulr expressions,
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationInnerproduct spaces
Innerproduct spces Definition: Let V be rel or complex liner spce over F (here R or C). An inner product is n opertion between two elements of V which results in sclr. It is denoted by u, v nd stisfies:
More informationCS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018
CS 301 Lecture 04 Regulr Expressions Stephen Checkowy Jnury 29, 2018 1 / 35 Review from lst time NFA N = (Q, Σ, δ, q 0, F ) where δ Q Σ P (Q) mps stte nd n lphet symol (or ) to set of sttes We run n NFA
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More informationParallel Projection Theorem (Midpoint Connector Theorem):
rllel rojection Theorem (Midpoint onnector Theorem): The segment joining the midpoints of two sides of tringle is prllel to the third side nd hs length onehlf the third side. onversely, If line isects
More informationNUMERICAL 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
More informationMATH 423 Linear Algebra II Lecture 28: Inner product spaces.
MATH 423 Liner Algebr II Lecture 28: Inner product spces. Norm The notion of norm generlizes the notion of length of vector in R 3. Definition. Let V be vector spce over F, where F = R or C. A function
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 informationWeek 7 Riemann Stieltjes Integration: Lectures 1921
Week 7 Riemnn Stieltjes Integrtion: Lectures 1921 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More information63. 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 =
More informationNew Integral Inequalities for ntime Differentiable Functions with Applications for pdfs
Applied Mthemticl Sciences, Vol. 2, 2008, no. 8, 353362 New Integrl Inequlities for ntime Differentible Functions with Applictions for pdfs Aristides I. Kechriniotis Technologicl Eductionl Institute
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationIII. Lecture on Numerical Integration. File faclib/dattab/lecturenotes/numericalinter03.tex /by EC, 3/14/2008 at 15:11, version 9
III Lecture on Numericl Integrtion File fclib/dttb/lecturenotes/numericalinter03.tex /by EC, 3/14/008 t 15:11, version 9 1 Sttement of the Numericl Integrtion Problem In this lecture we consider the
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions
ENGI 44 Engineering Mthemtics Five Tutoril Exmples o Prtil Frctions 1. Express x in prtil rctions: x 4 x 4 x 4 b x x x x Both denomintors re liner nonrepeted ctors. The coverup rule my be used: 4 4 4
More informationx = 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
More informationMATH 409 Advanced Calculus I Lecture 22: Improper Riemann integrals.
MATH 409 Advned Clulus I Leture 22: Improper Riemnn integrls. Improper Riemnn integrl If funtion f : [,b] R is integrble on [,b], then the funtion F(x) = x f(t)dt is well defined nd ontinuous on [,b].
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs
ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs A.I. KECHRINIOTIS AND N.D. ASSIMAKIS Deprtment of Eletronis Tehnologil Edutionl Institute of Lmi, Greee EMil: {kehrin,
More informationA new algorithm for generating Pythagorean triples 1
A new lgorithm for generting Pythgoren triples 1 RH Dye 2 nd RWD Nicklls 3 The Mthemticl Gzette (1998; 82 (Mrch, No. 493, pp. 86 91 http://www.nicklls.org/dick/ppers/mths/pythgtriples1998.pdf 1 Introduction
More informationMath 4200: Homework Problems
Mth 4200: Homework Problems Gregor Kovčič 1. Prove the following properties of the binomil coefficients ( n ) ( n ) (i) 1 + + + + 1 2 ( n ) (ii) 1 ( n ) ( n ) + 2 + 3 + + n 2 3 ( ) n ( n + = 2 n 1 n) n,
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
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 informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More information3.4 Numerical integration
3.4. Numericl integrtion 63 3.4 Numericl integrtion In mny economic pplictions it is necessry to compute the definite integrl of relvlued function f with respect to "weight" function w over n intervl [,
More informationProblem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:
(x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
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 informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More information38 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 xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationMATH 573 FINAL EXAM. May 30, 2007
MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.
More informationMATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.
MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints
More informationAMATH 731: Applied Functional Analysis Fall Some basics of integral equations
AMATH 731: Applied Functionl Anlysis Fll 2009 1 Introduction Some bsics of integrl equtions An integrl eqution is n eqution in which the unknown function u(t) ppers under n integrl sign, e.g., K(t, s)u(s)
More informationPractice final exam solutions
University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If
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 informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationThinPlate Splines. Contents
ThinPlte Splines Dvid Eberly, Geometric Tools, Redmond WA 98052 https://www.geometrictools.com/ This work is licensed under the Cretive Commons Attribution 4.0 Interntionl License. To view copy of this
More informationConducting Ellipsoid and Circular Disk
1 Problem Conducting Ellipsoid nd Circulr Disk Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 (September 1, 00) Show tht the surfce chrge density σ on conducting ellipsoid,
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationNecessary and Sufficient Conditions for Differentiating Under the Integral Sign
Necessry nd Sufficient Conditions for Differentiting Under the Integrl Sign Erik Tlvil 1. INTRODUCTION. When we hve n integrl tht depends on prmeter, sy F(x f (x, y dy, it is often importnt to know when
More informationA GENERAL INTEGRAL RICARDO ESTRADA AND JASSON VINDAS
A GENERAL INTEGRAL RICARDO ESTRADA AND JASSON VINDAS Abstrct. We define n integrl, the distributionl integrl of functions of one rel vrible, tht is more generl thn the Lebesgue nd the DenjoyPerronHenstock
More informationKeywords : Generalized Ostrowski s inequality, generalized midpoint inequality, Taylor s formula.
Generliztions of the Ostrowski s inequlity K. S. Anstsiou Aristides I. Kechriniotis B. A. Kotsos Technologicl Eductionl Institute T.E.I.) of Lmi 3rd Km. O.N.R. LmiAthens Lmi 3500 Greece Abstrct Using
More informationset is not closed under matrix [ multiplication, ] and does not form a group.
Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed
More informationChapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1
Chpter 5. Let g ( e. on [, ]. The derivtive of g is g ( e ( Write the slope intercept form of the eqution of the tngent line to the grph of g t. (b Determine the coordinte of ech criticl vlue of g. Show
More informationChapter 4. Additional Variational Concepts
Chpter 4 Additionl Vritionl Concepts 137 In the previous chpter we considered clculus o vrition problems which hd ixed boundry conditions. Tht is, in one dimension the end point conditions were speciied.
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More information