Some Inequalities variations on a common theme Lecture I, UL 2007


 Shanon Potter
 5 years ago
 Views:
Transcription
1 Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel numbers Prove h ( + b + c ) 2 + ( 2 + b 2 + c 2 ) 2 + ( 3 + b 3 + c 3 ) b 2 + c b c b c 2 3 Problem 2 Assume i, b i, c i, i =, 2, 3 re posiive numbers Prove h 3 b c b 2 c b 3 c 3 3 ( )(b + b 2 + b 3 )(c + c 2 + c 3 ), b = 2 b 2 = 3 b 3 Problem 3 Assume i, b i, c i, i =, 2, 3 re posiive numbers Prove h + b + + c 2 + b c b 3 + c 3, b +b 2 +b 3 + c +c 2 +c 3 b = 2 b 2 = 3 b 3 2 A bsic inequliy Theorem Suppose, b > 0 Then 2 b + b, > 0,
2 In oher words, 2 b = min{ + b = b : > 0} = min{s + b : s, > 0, s = } Proof + b 2 b = 2 2 b + b We deduce some consequences Theorem 2 If, b, c, d > 0 Then b + cd ( + c)(b + d), Proof For ll > 0, Hence c = b d 2[ b + cd] + b + c + d = 2[ b + cd] min{( + c) + We cn build on his in wo wys b d b + d = c = + c (b + d) = ( b) 2 (b + d) = ( + c) +, : > 0} = 2 ( + c)(b + d) Theorem 3 Suppose, b, c, d, e, f, g, h > 0 Then 4 bcd + 4 efgh 4 ( + e)(b + f)(c + g)(d + h), e = b f = c g = d h Proof Pu x = b, y = cd, u = ef, v = gh Then 4 bcd + 4 efgh = xy + uv We cn now deduce he following semen (x + u)(y + v) = ( b + ef)( cd + gh) ( + e)(b + f) c + g)(d + h) = 4 ( + e)(b + f)(c + g)(d + h) 2
3 Corollry If, b, c, e, f, g > 0, hen 3 bc + 3 efg 3 ( + e)(b + f)(c + g), e = b f = c g Proof Le d = 3 bc, h = 3 efg nd pply he heorem d + h = 4 bcd + 4 efgh 4 ( + e)(b + f)(c + g)(d + h), whence 3 bc + 3 efg = d + h 3 ( + e)(b + f)(c + g) A soluion of Problem 2 follows Exercise Prove he ls semen Exercise 2 Suppose i, b i > 0, i =, 2,, n Prove h n 2 n + n b b 2 b n n ( + b )( 2 + b 2 ) ( n + b n ), b = 2 b 2 = = n b n 3 The CuchySchwrz inequliy Theorem 2 is specil cse of disguised version of one form of perhps he mos useful inequliy in ll of Mhemics, nmely he CuchySchwrz inequliy Theorem 4 (CuchySchwrz) If x i, y i > 0, i =, 2,, n, hen n n n xi y i ( x i )( y i ), i= i= i= x y = x 2 y 2 = = x n y n Proof Invoke Theorem 2 nd use inducion We cn phrse his differenly Corollry 2 If i, i =, 2,, n re rel numbers, hen n n n 2 i = mx{ i x i : x 2 i = } i= i= i= 3
4 Proof For, if n i= x2 i =, hen n i x i nd here is equliy when i= = = n i x i i= n (i ) 2 (x i ) 2 i= n n ( 2 i )( x 2 i ) i= i= n 2 i, i= x i = i n, i =, 2,, n i= 2 i Theorem 5 If, b, c, d re rel numbers, hen ( + b)2 + (c + d) c 2 + b 2 + d 2 Proof Suppose x 2 + y 2 = Then Hence x + cy 2 + c 2, bx + dy b 2 + d 2 ( + b)x + (c + d)y x + cy + bx + dy 2 + c 2 + b 2 + d 2 I follows h ( + b)2 + (c + d) 2 = mx{ (+b)x+(c+d)y : x 2 +y 2 = } 2 + c 2 + b 2 + d 2 This is specil cse of Minkowski s inequliy Theorem 6 (Minkowski) Assume i, b i, i =, 2, 3 re rel numbers Then n ( i + b i ) 2 n 2 i + n b 2 i i= i= i= Proof Suppose n i= x2 i = Then n 2 i + n n n b 2 i i x i + b i x i i= i= 4 i= i= n ( i + b i )x i i=
5 Hence n n ( i + b i ) 2 = mx{ ( i + b i )x i : i= i= n x 2 i = } n 2 i + n b 2 i i= i= i= 4 Discussion of Theorem 2 The produc b represens he re of recngle A wih dimensions, b, nd cd is he re of recngle B wih dimensions c, d And + c, b + d re he dimensions of recngle C he sum of A, B Thus re(a) + re(b) re(c), A, B, C re scled versions of ech oher Exercise 3 Give geomeric inerpreion of Problem 2 We cn cs he previous semen in differen wy Wih ech poin (vecor) x = (x, x 2 ) is he firs qudrn of R 2, le Then, if y = (y, y 2 ), g 2 (x) = x x 2 g 2 (x) + g 2 (y) g 2 (x + y) Thus g 2 is superddiive on R 2 Noice, oo, h, for every sclr λ > 0, λx = (λx, λx 2 ) nd g 2 (λx) = λg 2 (x), x R 2, ie, g 2 is homogeneous of degree Hence, g 2 is concve on R 2 Anlyiclly, his mens h x, y R 2, nd 0 λ, hen λg 2 (x) + ( λ)g 2 (y) g 2 (λx + ( λ)y) To ge n ide of wh i mens, geomericlly, ry nd picure he surfce in 3spce whose equion is z = xy, x, y 0 Exercise 4 Apropos Problem 2, drw similr conclusion bou he funcion g 3 (x) = 3 x x 2 x 3, x = (x, x 2, x 3 ) R 3, x i 0, i =, 2, 3 In fc, if x = (x, x 2,, x n ) is vecor in nspce nd is coordines re posiive, nd g n (x) = n x x 2 x n, hen g n is superddiive nd is homogeneous of degree on Hence, g n is concve on his region {x R n : x i 0, i =, 2,, n} 5
6 5 Discussion of Theorem 5 I s fmilir h 2 + b 2 is he eucliden disnce from he origin o he poin (, b) in R 2 Thus Theorem 5 is simply n nlyicl semen equivlen o he ringle inequliy! To pu his noher wy, le l 2 (x) = x 2 + x 2 2, x = (x, x 2 ) R 2, hen l 2 (x + y) l 2 (x) + l 2 (y), x, y R 2 In oher words, l 2 is subddiive Bu i s lso homogeneous of degree one Hence, i is convex on R 2, ie, x, y R 2, nd 0 λ, hen l 2 (λx + ( λ)y) λl 2 (x) + ( λ)l 2 (y) To ge n ide of wh his mens, geomericlly, ry nd picure he region of poins (x, y, z) in 3spce consrined by he condiions 0 z x 2 + y 2, x, y R [Hin: your nex mel exmine he sugr bowl, sy] A soluion of Problem follows once we cn show h l 3 (x) = x 2 + x x 2 3, x = (x, x 2, x 3 ) R 3, hs he sme propery Exercise 5 Esblish h l 3 is subddiive nd convex on R 3, nd solve Problem 6 Soluion of Problem 3 By now i will hve occurred o you h wh is wned o solve Problem 3 is n nlogue of Corollry for he hrmonic men of hree posiive numbers (This hs properies similr o he geomeric men, bu doesn ge s much enion, nd doesn feure in oo mny IMO problems) Nmely, we wn o show h he funcion h 3 defined by 3 h 3 (x) = x + x 2 +, x = (x, x 2, x 3 ) R 3, x, x 2, x 3 > 0 x 3 is superddiive, nd so concve, on R 3 Theorem 7 If, b, c > 0, hen + + = min{x + by + cz : 0 < x, y, z, x + y + z = } b c 6
7 Proof This is consequence of Theorem 4 For, if x, y, z > 0 nd x+ y+ z =, hen Hence, = ( x + by b + cz c ) 2 (x + by + cz)( + b + c ) + + x + by + cz, b c whenever 0 < x, y, z, x + y + z =, nd here is equliy here when x = 2 ( + +, y = b c )2 b 2 ( + +, z = b c )2 c 2 ( + + b c )2 Corollry 3 If, b, c, d, e, f > 0, hen h 3 (, b, c) + h 3 (d, e, f) h 3 ( + d, b + e, c + f) There is equliy iff d = b e = c f Proof For, if x, y, z > 0 nd x + y + z =, hen h 3 (, b, c)+h 3 (d, e, f) 3(x+by+cz)+3(dx+ey+fz) = 3[(+d)x+(b+e)y+(c+f)z], nd so h 3 (, b, c) + h 3 (d, e, f) doesn exceed 3 min{(+d)x+(b+e)y+(c+f)z : 0 < x, y, z, x+ y+ z = } = h 3 (+d, b+e, c+f) Moreover, (?) here is equliy here iff nd x = ( + + ) = d( + + ) = ( + d)( + + ), b c d e f +d b+e c+f y = b( + + ) = e( + + ) = (b + e)( + + ), b c d e f +d b+e c+f z = c( + + ) = f( + + ) = (c + f)( + + ) b c d e f +d b+e c+f In oher words, (?) iff Exercise 6 Now solve Problem 3 d = b e = c f 7
8 7 Exercises Suppose, b > 0 Prove h = b + g 2 (, b) 2 g 2 (, + b 2 ), 2 Se nd prove n nlogue of Theorem 7 nd is corollry for h 2, where 3 Suppose, b > 0 Prove h 2 h 2 (x) = x +, x = (x, x 2 ) R 2, x i, x 2 > 0 x 2 = b + h 2 (, b) 2 h 2 (, + b 2 ), 4 Suppose, b, c > 0 Prove h g 2 (, b) + g 2 (b, c) + g 2 (c, ) + b + c, = b = c 5 Suppose, b, c > 0 Prove h h 2 (, b) + h 2 (b, c) + h 2 (c, ) + b + c, = b = c 6 Suppose, b, c > 0 Prove h 3 3 bc = min{x + by + cz : 0 < x, y, z, xyz = } Deduce Corollry 7 More generlly, if x i 0, i =, 2,, n, x = (x, x 2,, x n ), nd g n (x) = n x x 2 x n Prove h n ng n (x) = min{ x i y i : y y 2 y n =, y i > 0, i =, 2, n} i= Now redo Exercise 2 in Secion 2 8
9 8 Suppose, b, c > 0 Prove h = b = c + g 2 (, b) + g 3 (, b, c) 3g 3 (, + b 2, + b + c ), 3 9 Suppose, b, c > 0 Prove h + h 2 (, b) + h 3 (, b, c) 3h 3 (, + b 2, + b + c ), 3 = b = c 0 Suppose, b, c, d > 0 Prove h b + b + + cd c + d + Suppose, b, c > 0 Prove h < ( + c)(b + d) + b + c + d + g 2 (, b) + g 2 (b, c) + g 2 (c, ) 3g 3 ( + b 2, b + c 2, c + 2 ) + b + c, = b = c 2 Suppose, b, c > 0 Prove h h 2 (, b) + h 2 (b, c) + h 2 (c, ) 3h 3 ( + b 2, b + c 2, c + 2 ) + b + c, = b = c 9
5.1The InitialValue Problems For Ordinary Differential Equations
5.The IniilVlue Problems For Ordinry Differenil Equions Consider solving iniilvlue problems for ordinry differenil equions: (*) y f, y, b, y. If we know he generl soluion y of he ordinry differenil
More informationContraction Mapping Principle Approach to Differential Equations
epl Journl of Science echnology 0 (009) 4953 Conrcion pping Principle pproch o Differenil Equions Bishnu P. Dhungn Deprmen of hemics, hendr Rn Cmpus ribhuvn Universiy, Khmu epl bsrc Using n eension of
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples
More informationMTH 146 Class 11 Notes
8. Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples
More informationREAL ANALYSIS I HOMEWORK 3. Chapter 1
REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs
More informationThe solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.
[~ o o : o o ill] i 1. Mrices, Vecors, nd GussJordn Eliminion 1 x y = =  z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries
More informationEXERCISE  01 CHECK YOUR GRASP
UNIT # 09 PARABOLA, ELLIPSE & HYPERBOLA PARABOLA EXERCISE  0 CHECK YOUR GRASP. Hin : Disnce beween direcri nd focus is 5. Given (, be one end of focl chord hen oher end be, lengh of focl chord 6. Focus
More informationMathematics 805 Final Examination Answers
. 5 poins Se he Weiersrss Mes. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se
More informationHermiteHadamardFejér type inequalities for convex functions via fractional integrals
Sud. Univ. BeşBolyi Mh. 6(5, No. 3, 355 366 HermieHdmrdFejér ype inequliies for convex funcions vi frcionl inegrls İmd İşcn Asrc. In his pper, firsly we hve eslished Hermie HdmrdFejér inequliy for
More informationENGR 1990 Engineering Mathematics The Integral of a Function as a Function
ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under
More informatione t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls  hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
More informationMotion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Moion Accelerion Pr : Consn Accelerion Accelerion Accelerion Accelerion is he re of chnge of velociy. = v  vo = Δv Δ ccelerion = = v  vo chnge of velociy elpsed ime Accelerion is vecor, lhough in onedimensionl
More information0 for t < 0 1 for t > 0
8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside
More informationGENERALIZATION OF SOME INEQUALITIES VIA RIEMANNLIOUVILLE FRACTIONAL CALCULUS
 TAMKANG JOURNAL OF MATHEMATICS Volume 5, Number, 75, June doi:5556/jkjm555 Avilble online hp://journlsmhkueduw/    GENERALIZATION OF SOME INEQUALITIES VIA RIEMANNLIOUVILLE FRACTIONAL CALCULUS MARCELA
More information4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informationA Simple Method to Solve Quartic Equations. Key words: Polynomials, Quartics, Equations of the Fourth Degree INTRODUCTION
Ausrlin Journl of Bsic nd Applied Sciences, 6(6): 6, 0 ISSN 99878 A Simple Mehod o Solve Quric Equions Amir Fhi, Poo Mobdersn, Rhim Fhi Deprmen of Elecricl Engineering, Urmi brnch, Islmic Ad Universi,
More informationSeptember 20 Homework Solutions
College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum
More informationAverage & instantaneous velocity and acceleration Motion with constant acceleration
Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission
More informationPARABOLA. moves such that PM. = e (constant > 0) (eccentricity) then locus of P is called a conic. or conic section.
wwwskshieducioncom PARABOLA Le S be given fixed poin (focus) nd le l be given fixed line (Direcrix) Le SP nd PM be he disnce of vrible poin P o he focus nd direcrix respecively nd P SP moves such h PM
More information( ) ( ) ( ) ( ) ( ) ( y )
8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll
More informationSolutions from Chapter 9.1 and 9.2
Soluions from Chaper 9 and 92 Secion 9 Problem # This basically boils down o an exercise in he chain rule from calculus We are looking for soluions of he form: u( x) = f( k x c) where k x R 3 and k is
More informationGreen s Functions and Comparison Theorems for Differential Equations on Measure Chains
Green s Funcions nd Comprison Theorems for Differenil Equions on Mesure Chins Lynn Erbe nd Alln Peerson Deprmen of Mhemics nd Sisics, Universiy of NebrskLincoln Lincoln,NE 685880323 lerbe@@mh.unl.edu
More informationA LIMITPOINT CRITERION FOR A SECONDORDER LINEAR DIFFERENTIAL OPERATOR IAN KNOWLES
A LIMITPOINT CRITERION FOR A SECONDORDER LINEAR DIFFERENTIAL OPERATOR j IAN KNOWLES 1. Inroducion Consider he forml differenil operor T defined by el, (1) where he funcion q{) is relvlued nd loclly
More informationINTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).
INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely
More informationEXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECONDORDER ITERATIVE BOUNDARYVALUE PROBLEM
Elecronic Journl of Differenil Equions, Vol. 208 (208), No. 50, pp. 6. ISSN: 072669. URL: hp://ejde.mh.xse.edu or hp://ejde.mh.un.edu EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECONDORDER ITERATIVE
More information3. Renewal Limit Theorems
Virul Lborories > 14. Renewl Processes > 1 2 3 3. Renewl Limi Theorems In he inroducion o renewl processes, we noed h he rrivl ime process nd he couning process re inverses, in sens The rrivl ime process
More informationConvergence of Singular Integral Operators in Weighted Lebesgue Spaces
EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 2, 2017, 335347 ISSN 13075543 www.ejpm.com Published by New York Business Globl Convergence of Singulr Inegrl Operors in Weighed Lebesgue
More informationf t f a f x dx By Lin McMullin f x dx= f b f a. 2
Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes
More informationON NEW INEQUALITIES OF SIMPSON S TYPE FOR FUNCTIONS WHOSE SECOND DERIVATIVES ABSOLUTE VALUES ARE CONVEX
Journl of Applied Mhemics, Sisics nd Informics JAMSI), 9 ), No. ON NEW INEQUALITIES OF SIMPSON S TYPE FOR FUNCTIONS WHOSE SECOND DERIVATIVES ABSOLUTE VALUES ARE CONVEX MEHMET ZEKI SARIKAYA, ERHAN. SET
More informationP441 Analytical Mechanics  I. Coupled Oscillators. c Alex R. Dzierba
Lecure 3 Mondy  Deceber 5, 005 Wrien or ls upded: Deceber 3, 005 P44 Anlyicl Mechnics  I oupled Oscillors c Alex R. Dzierb oupled oscillors  rix echnique In Figure we show n exple of wo coupled oscillors,
More information3 Motion with constant acceleration: Linear and projectile motion
3 Moion wih consn ccelerion: Liner nd projecile moion cons, In he precedin Lecure we he considered moion wih consn ccelerion lon he is: Noe h,, cn be posiie nd neie h leds o rie of behiors. Clerl similr
More informationForms of Energy. Mass = Energy. Page 1. SPH4U: Introduction to Work. Work & Energy. Particle Physics:
SPH4U: Inroducion o ork ork & Energy ork & Energy Discussion Definiion Do Produc ork of consn force ork/kineic energy heore ork of uliple consn forces Coens One of he os iporn conceps in physics Alernive
More informationPredator  Prey Model Trajectories and the nonlinear conservation law
Predaor  Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories
More informationNew Inequalities in Fractional Integrals
ISSN 17493889 (prin), 17493897 (online) Inernionl Journl of Nonliner Science Vol.9(21) No.4,pp.493497 New Inequliies in Frcionl Inegrls Zoubir Dhmni Zoubir DAHMANI Lborory of Pure nd Applied Mhemics,
More informationChapter Direct Method of Interpolation
Chper 5. Direc Mehod of Inerpolion Afer reding his chper, you should be ble o:. pply he direc mehod of inerpolion,. sole problems using he direc mehod of inerpolion, nd. use he direc mehod inerpolns o
More informationwhite strictly far ) fnf regular [ with f fcs)8( hs ) as function Preliminary question jointly speaking does not exist! Brownian : APA Lecture 1.
Am : APA Lecure 13 Brownin moion Preliminry quesion : Wh is he equivlen in coninuous ime of sequence of? iid Ncqe rndom vribles ( n nzn noise ( 4 e Re whie ( ie se every fm ( xh o + nd covrince E ( xrxs
More informationMagnetostatics Bar Magnet. Magnetostatics Oersted s Experiment
Mgneosics Br Mgne As fr bck s 4500 yers go, he Chinese discovered h cerin ypes of iron ore could rc ech oher nd cerin mels. Iron filings "mp" of br mgne s field Crefully suspended slivers of his mel were
More informationLecture notes. Fundamental inequalities: techniques and applications
Lecture notes Fundmentl inequlities: techniques nd pplictions Mnh Hong Duong Mthemtics Institute, University of Wrwick Emil: m.h.duong@wrwick.c.uk Februry 8, 207 2 Abstrct Inequlities re ubiquitous in
More informationCALDERON S REPRODUCING FORMULA FOR DUNKL CONVOLUTION
Avilble online hp://scik.org Eng. Mh. Le. 15, 15:4 ISSN: 499337 CALDERON S REPRODUCING FORMULA FOR DUNKL CONVOLUTION PANDEY, C. P. 1, RAKESH MOHAN AND BHAIRAW NATH TRIPATHI 3 1 Deprmen o Mhemics, Ajy
More informationChapter 2. First Order Scalar Equations
Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.
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 informationAvailable Online :
fo/u fopjr Hh# u] ugh vjehs de] foifr ns[ NsMs rqjr e/;e eu dj ';ea iq#" flg ldyi dj] lgrs foifr vusd] ^cu^ u NsMs /;s; ds] j?qcj j[s VsdAA jfpr% euo /ez iz.sr ln~xq# Jh j.nsmnlh egj STUDY PACKAGE Subjec
More informationMATH1050 CauchySchwarz Inequality and Triangle Inequality
MATH050 CuchySchwrz Inequlity nd Tringle Inequlity 0 Refer to the Hndout Qudrtic polynomils Definition (Asolute extrem for relvlued functions of one rel vrile) Let I e n intervl, nd h : D R e relvlued
More informationThink of the Relationship Between Time and Space Again
Repor nd Opinion, 1(3),009 hp://wwwsciencepubne sciencepub@gmilcom Think of he Relionship Beween Time nd Spce Agin Yng Fcheng Compny of Ruid Cenre in Xinjing 15 Hongxing Sree, Klmyi, Xingjing 834000,
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 information3D Transformations. Computer Graphics COMP 770 (236) Spring Instructor: Brandon Lloyd 1/26/07 1
D Trnsformions Compuer Grphics COMP 770 (6) Spring 007 Insrucor: Brndon Lloyd /6/07 Geomery Geomeric eniies, such s poins in spce, exis wihou numers. Coordines re nming scheme. The sme poin cn e descried
More informationON NEW INEQUALITIES OF SIMPSON S TYPE FOR FUNCTIONS WHOSE SECOND DERIVATIVES ABSOLUTE VALUES ARE CONVEX.
ON NEW INEQUALITIES OF SIMPSON S TYPE FOR FUNCTIONS WHOSE SECOND DERIVATIVES ABSOLUTE VALUES ARE CONVEX. MEHMET ZEKI SARIKAYA?, ERHAN. SET, AND M. EMIN OZDEMIR Asrc. In his noe, we oin new some ineuliies
More informationCBSE 2014 ANNUAL EXAMINATION ALL INDIA
CBSE ANNUAL EXAMINATION ALL INDIA SET Wih Complee Eplnions M Mrks : SECTION A Q If R = {(, y) : + y = 8} is relion on N, wrie he rnge of R Sol Since + y = 8 h implies, y = (8 ) R = {(, ), (, ), (6, )}
More informationMATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)
MATH 14 AND 15 FINAL EXAM REVIEW PACKET (Revised spring 8) The following quesions cn be used s review for Mh 14/ 15 These quesions re no cul smples of quesions h will pper on he finl em, bu hey will provide
More informationPhysics 2A HW #3 Solutions
Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 33 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen
More informationTwo Coupled Oscillators / Normal Modes
Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own
More informationMA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions
MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firsorder linear ODE is: An inegraing facor is given by
More information22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 9: The High Beta Tokamak
.65, MHD Theory of Fusion Sysems Prof. Freidberg Lecure 9: The High e Tokmk Summry of he Properies of n Ohmic Tokmk. Advnges:. good euilibrium (smll shif) b. good sbiliy ( ) c. good confinemen ( τ nr )
More information1.0 Electrical Systems
. Elecricl Sysems The ypes of dynmicl sysems we will e sudying cn e modeled in erms of lgeric equions, differenil equions, or inegrl equions. We will egin y looking fmilir mhemicl models of idel resisors,
More informationFM Applications of Integration 1.Centroid of Area
FM Applicions of Inegrion.Cenroid of Are The cenroid of ody is is geomeric cenre. For n ojec mde of uniform meril, he cenroid coincides wih he poin which he ody cn e suppored in perfecly lnced se ie, is
More informationY 0.4Y 0.45Y Y to a proper ARMA specification.
HG Jan 04 ECON 50 Exercises II  0 Feb 04 (wih answers Exercise. Read secion 8 in lecure noes 3 (LN3 on he common facor problem in ARMAprocesses. Consider he following process Y 0.4Y 0.45Y 0.5 ( where
More informationA Kalman filtering simulation
A Klmn filering simulion The performnce of Klmn filering hs been esed on he bsis of wo differen dynmicl models, ssuming eiher moion wih consn elociy or wih consn ccelerion. The former is epeced o beer
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 informationHow to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.
How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels:96377679,735 Emil:hf@scsne.org Commens: 3 ges SubjClss: Funcionl nlsis, comle
More information1. Find a basis for the row space of each of the following matrices. Your basis should consist of rows of the original matrix.
Mh 7 Exm  Prcice Prolem Solions. Find sis for he row spce of ech of he following mrices. Yor sis shold consis of rows of he originl mrix. 4 () 7 7 8 () Since we wn sis for he row spce consising of rows
More informationON THE OSTROWSKIGRÜSS TYPE INEQUALITY FOR TWICE DIFFERENTIABLE FUNCTIONS
Hceepe Journl of Mhemics nd Sisics Volume 45) 0), 65 655 ON THE OSTROWSKIGRÜSS TYPE INEQUALITY FOR TWICE DIFFERENTIABLE FUNCTIONS M Emin Özdemir, Ahme Ock Akdemir nd Erhn Se Received 6:06:0 : Acceped
More information1. Introduction. 1 b b
Journl of Mhemicl Inequliies Volume, Number 3 (007), 45 436 SOME IMPROVEMENTS OF GRÜSS TYPE INEQUALITY N. ELEZOVIĆ, LJ. MARANGUNIĆ AND J. PEČARIĆ (communiced b A. Čižmešij) Absrc. In his pper some inequliies
More informationMath 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:
Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial
More informationHow to prove the Riemann Hypothesis
Scholrs Journl of Phsics, Mhemics nd Sisics Sch. J. Phs. Mh. S. 5; (B:56 Scholrs Acdemic nd Scienific Publishers (SAS Publishers (An Inernionl Publisher for Acdemic nd Scienific Resources *Corresonding
More informationHamilton J acobi Equation: Explicit Formulas In this lecture we try to apply the method of characteristics to the HamiltonJacobi equation: u t
M ah 5 2 7 Fall 2 0 0 9 L ecure 1 0 O c. 7, 2 0 0 9 Hamilon J acobi Equaion: Explici Formulas In his lecure we ry o apply he mehod of characerisics o he HamilonJacobi equaion: u + H D u, x = 0 in R n
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 informationPHYSICS 1210 Exam 1 University of Wyoming 14 February points
PHYSICS 1210 Em 1 Uniersiy of Wyoming 14 Februry 2013 150 poins This es is opennoe nd closedbook. Clculors re permied bu compuers re no. No collborion, consulion, or communicion wih oher people (oher
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 information23.5. HalfRange Series. Introduction. Prerequisites. Learning Outcomes
HalfRange Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion
More information1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.
. Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.
More informationAn integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.
Improper Inegrls To his poin we hve only considered inegrls f(x) wih he is of inegrion nd b finie nd he inegrnd f(x) bounded (nd in fc coninuous excep possibly for finiely mny jump disconinuiies) An inegrl
More informationANSWERS TO EVEN NUMBERED EXERCISES IN CHAPTER 2
ANSWERS TO EVEN NUMBERED EXERCISES IN CHAPTER Seion Eerise : Coninuiy of he uiliy funion Le λ ( ) be he monooni uiliy funion defined in he proof of eisene of uiliy funion If his funion is oninuous y hen
More informationOn Hadamard and FejérHadamard inequalities for Caputo kfractional derivatives
In J Nonliner Anl Appl 9 8 No, 698 ISSN: 868 elecronic hp://dxdoiorg/75/ijn8745 On Hdmrd nd FejérHdmrd inequliies for Cpuo frcionl derivives Ghulm Frid, Anum Jved Deprmen of Mhemics, COMSATS Universiy
More informationHamilton J acobi Equation: Weak S olution We continue the study of the HamiltonJacobi equation:
M ah 5 7 Fall 9 L ecure O c. 4, 9 ) Hamilon J acobi Equaion: Weak S oluion We coninue he sudy of he HamilonJacobi equaion: We have shown ha u + H D u) = R n, ) ; u = g R n { = }. ). In general we canno
More informationStochastic models and their distributions
Sochasic models and heir disribuions Couning cusomers Suppose ha n cusomers arrive a a grocery a imes, say T 1,, T n, each of which akes any real number in he inerval (, ) equally likely The values T 1,,
More informationAnnouncements: Warmup Exercise:
Fri Apr 13 7.1 Sysems of differenial equaions  o model mulicomponen sysems via comparmenal analysis hp//en.wikipedia.org/wiki/mulicomparmen_model Announcemens Warmup Exercise Here's a relaively simple
More informationAn Introduction to Malliavin calculus and its applications
An Inroducion o Malliavin calculus and is applicaions Lecure 5: Smoohness of he densiy and Hörmander s heorem David Nualar Deparmen of Mahemaics Kansas Universiy Universiy of Wyoming Summer School 214
More informationWe just finished the ErdősStone Theorem, and ex(n, F ) (1 1/(χ(F ) 1)) ( n
Lecure 3  KövariSósTurán Theorem Jacques Versraëe jacques@ucsd.edu We jus finished he ErdősSone Theorem, and ex(n, F ) ( /(χ(f ) )) ( n 2). So we have asympoics when χ(f ) 3 bu no when χ(f ) = 2 i.e.
More informationPhysic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =
Mi i fd l Phsic 3 Lecure 4 Min poins of od s lecure: Emple: ddiion of elociies Trjecories of objecs in dimensions: dimensions: g 9.8m/s downwrds ( ) g o g g Emple: A foobll pler runs he pern gien in he
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 information[5] Solving Multiple Linear Equations A system of m linear equations and n unknown variables:
[5] Solving Muliple Liner Equions A syse of liner equions nd n unknown vribles: + + + nn = b + + + = b n n : + + + nn = b n n A= b, where A =, : : : n : : : : n = : n A = = = ( ) where, n j = ( ); = :
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 information(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.
Answers o Een Numbered Problems Chper. () 7 m s, 6 m s (b) 8 5 yr 4.. m ih 6. () 5. m s (b).5 m s (c).5 m s (d) 3.33 m s (e) 8. ().3 min (b) 64 mi..3 h. ().3 s (b) 3 m 4..8 mi wes of he flgpole 6. (b)
More informationInventory Analysis and Management. MultiPeriod Stochastic Models: Optimality of (s, S) Policy for KConvex Objective Functions
MuliPeriod Sochasic Models: Opimali of (s, S) Polic for Convex Objecive Funcions Consider a seing similar o he Nsage newsvendor problem excep ha now here is a fixed reordering cos (> 0) for each (re)order.
More informationMatrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24
Mtrix lger Mtrix ddition, Sclr Multipliction nd rnsposition Mtrix lger Section.. Mtrix ddition, Sclr Multipliction nd rnsposition rectngulr rry of numers is clled mtrix ( the plurl is mtrices ) nd the
More informationChapter 2 Summary. Carnegie Learning
Chaper Summary Key Terms inducion (.1) deducion (.1) counerexample (.1) condiional saemen (.1) proposiional form (.1) proposiional variables (.1) hypohesis (.1) conclusion (.1) ruh value (.1) ruh able
More informationLesson Notes: Week 40Vectors
Lesson Notes: Week 40Vectors Vectors nd Sclrs vector is quntity tht hs size (mgnitude) nd direction. Exmples of vectors re displcement nd velocity. sclr is quntity tht hs size but no direction. Exmples
More informationAnalytical 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
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More information14. The fundamental theorem of the calculus
4. The funmenl heorem of he clculus V 20 00 80 60 40 20 0 0 0.2 0.4 0.6 0.8 v 400 200 0 0 0.2 0.5 0.8 200 400 Figure : () Venriculr volume for subjecs wih cpciies C = 24 ml, C = 20 ml, C = 2 ml n (b) he
More informationSome estimates on the HermiteHadamard inequality through quasiconvex functions
Annls of University of Criov, Mth. Comp. Sci. Ser. Volume 3, 7, Pges 8 87 ISSN: 13693 Some estimtes on the HermiteHdmrd inequlity through qusiconvex functions Dniel Alexndru Ion Abstrct. In this pper
More informationdy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page
Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,
More informationChapter 3. Vector Spaces
3.4 Liner Trnsformtions 1 Chpter 3. Vector Spces 3.4 Liner Trnsformtions Note. We hve lredy studied liner trnsformtions from R n into R m. Now we look t liner trnsformtions from one generl vector spce
More informationSolutions to Problems from Chapter 2
Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5
More informationTransformations. Ordered set of numbers: (1,2,3,4) Example: (x,y,z) coordinates of pt in space. Vectors
Trnformion Ordered e of number:,,,4 Emple:,,z coordine of p in pce. Vecor If, n i i, K, n, i uni ecor Vecor ddiion +w, +, +, + V+w w Sclr roduc,, Inner do roduc α w. w +,.,. The inner produc i SCLR!. w,.,
More informationExercises: Similarity Transformation
Exercises: Similariy Transformaion Problem. Diagonalize he following marix: A [ 2 4 Soluion. Marix A has wo eigenvalues λ 3 and λ 2 2. Since (i) A is a 2 2 marix and (ii) i has 2 disinc eigenvalues, we
More informationApplication on Inner Product Space with. Fixed Point Theorem in Probabilistic
Journl of Applied Mhemics & Bioinformics, vol.2, no.2, 2012, 110 ISSN: 17926602 prin, 17926939 online Scienpress Ld, 2012 Applicion on Inner Produc Spce wih Fixed Poin Theorem in Probbilisic Rjesh Shrivsv
More informationt is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...
Mah 228 Fri Mar 24 5.6 Marix exponenials and linear sysems: The analogy beween firs order sysems of linear differenial equaions (Chaper 5) and scalar linear differenial equaions (Chaper ) is much sronger
More informationMath 426: Probability Final Exam Practice
Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by
More information