Mathematics 805 Final Examination Answers


 Shannon Goodwin
 3 years ago
 Views:
Transcription
1 . 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 Lebesgue s Domined Convergence Theorem. b Se Lebesgue s Monoone Convergence Theorem. c Define [, n] f n n oherwise Noe h f n d, bu n f n. Why does his conrdic neiher he Domined nor he Monoone Convergence Theorems? Answer: Suppose h f n L nd f n f lmos everyone. Suppose furher h f n g lmos everywhere, wih g L. Then f L nd f f n. b Suppose h f n L is monoone sequence, nd suppose furher h f n is bounded. Then f n converges lmos everywhere o funcion f L, nd f f n. c This emple does no viole he Domined Convergence Theorem, becuse here is no funcion g L wih f n g. The emple does no viole he Monoone Convergence Theorem becuse he sequence f n is no monoone. 3. poins Suppose h K is compc meric spce, nd g : K R coninuous funcion, wih g > for ll K. Suppose furher h g n : K R is sequence of coninuous funcions converging uniformly o g on K. Show h here is some ineger N so h if n > N, hen g n > for ll K. Answer: Becuse K is compc, we now h g ins is minimum vlue some K. In priculr, we cn find some ɛ so h g ɛ for ll K. Becuse g n g uniformly, we cn find n ineger N so h if n N, hen g g n ɛ/ for ll K. This implies h g n ɛ/ for ll K, so g n > poins Le M, d nd M, d be meric spces, nd f : M M funcion. Define wh is men by f is uniformly coninuous. b Suppose h M, d nd M, d re meric spces, nd f : M M is uniformly coninuous. Suppose h A, B M wih d A, B. Show h d fa, fb. c Give n emple o show h i is possible for f : M M o be coninuous, wih d A, B nd d fa, fb. Answer: Given ny ɛ >, here is δ > so h if d, y < δ, hen d f, fy < ɛ. b Given ny ɛ >, we cn find δ > so h if d, y < δ, hen d f, fy < ɛ. Becuse d A, B, we cn find A nd b B so h d, b < δ. This sys h d f, fb < ɛ, nd so d fa, fb < ɛ. Becuse ɛ is rbirry, we cn conclude h d fa, fb. c Consider he funcion f : R R R defined by f, y y. Le A {n, n : n Z, n > }. Le B {n, : n Z, n > }. Then d A, B. However, fa {}, while fb {}, so d fa, fb. 5. poins Show h sin d
2 cn be defined s n improper Riemnn inegrl bu no s Lebesgue inegrl. Hin: Le, nd sin imie he proof h d cn be defined s n improper Riemnn inegrl bu no Lebesgue inegrl. Answer: Le, so h, d d, nd d d/. If, hen. Therefore, s Riemnn inegrl, we hve sin d b b b sin sin d b d Se u /, dv sin d, du 3/ d, nd v cos. We hve b sin d cos ] b b b b The firs i is cos, nd he inegrl converges, becuse Riemnn inegrl is defined. cos d. 3/ On he oher hnd, if he inegrl converged s Lebesgue inegrl, so would sme subsiuions led us o consider b n+ n sin b sin d. Bu n+ d sin d n + n 3/ d converges. Therefore, he improper n +. Becuse n + diverges, we now h he Lebesgue inegrl mus no eis. sin d. The 6. 5 poins As usul, we define e e B n n n! nd B n B n. Recll h B, B, nd B + 6. Show h B n n B n. Answer: We hve B n n! n e e B n n! n e B n + B n n e n! e + e e + e + e e e + n B n n! e / e Equing coefficiens, we see h B n + B n n B n. n B n n n! 7. poins Suppose h f C. Le m be n ineger. Prove using inducion on m h f f d + m B! f f m B m f m d
3 Answer: We firs hve he cse m. We mus show h f f d + B f f! f d f f + f d f d To chec his, we inegre by prs, seing u, du d, dv f d, nd v f. We hve f d f f + f d f d f f + ] f f f + f + f f. Now, for he inducive sep, we mus prove h m B f f m! or m m+ B! f d B m f m d f f m+ B m+ m +! f m+ d B m f m d B m+ f m f m m+ B m+ m +! m +! f m+ d Muliply hrough by m+, nd his is he sme s B m f m d m+ B m+ m +! f m f m B m+ m +! f m+ d We cn prove his by using inegrion by pr on he lefhnd side. Le u f m, du f m+ d, dv Bm d, nd v Bm+ m+!. We hve B m f m d B ] m+ m +! f m B m+ m +! f m+ d m +! B m+ m +! B m+ f m B m+ f m f m f m B m+ m +! f m+ d B m+ m +! f m+ d So i remins o show h m+ B m+ B m+. If m + is odd, his is rue becuse B m+. If m + is even, his is rivilly rue inequliy poins Here is ye noher wy o compue migh be needed o do his problem: D n : n n e i sin D n sinn + sin K n : n D n + d. Recll he following fcs, some of which
4 cosn + K n n + cos sinn + / n + sin/ D n d cos sin / K n d Suppose h g is Riemnninegrble on [, b] wih g M. Show h b cos b M g d nd conclude h b Show h b cos g d. cos 4 sin d c Le f 4 sin, < Show h i is possible o define f so h f is coninuous nd bounded on [, ]. d Show h cos f d. e Show h nd hen f Le y nd conclude h g Finlly, inegre by prs o show h cos d sin / d 4. / sin sin d d. sin sin d d. Answer: This problem is en from An elemenry mehod for evluing in infinie inegrl, by M.R. Spiegel, The Americn Mhemicl Monhly, 58:8, Oc. 95, pp This is rivil: b cos g d b cos b g d M Mb d.
5 The i follows immediely. b We hve K d K d cos d cos cos sin d / cos 4 sin d / c Recll h sin y y y This mens h sin/ nd sin / Therefore, f 4 sin 4 sin 4 sin / + 4 / Therefore, he funcion is cully nlyic, if we define f o be. d Becuse f sisfies he hypoheses of pr, his follows immediely from. e We hve cos 4 sin d cos cos d 4 sin d cos d cos d d sin d sin d 4 f Le y, so y, nd hen d dy. Noe h if, hen y. We hve 4 sin sin sin dy dy sin y y dy sin / dy sin y y dy g Inegrion by prs is no quie rivil. We se u sin, dv d, du sin cos d sin d, nd v. We hve sin ] d sin sin + d sin y y dy
6 sin Now, we cn compue h nd hen dy d, so d dy: sin, nd s well. For he res, we subsiue y, sin d sin dy sin y y dy
INTEGRALS. 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 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 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 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 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 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 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 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 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 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 informationMAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017
MAT 66 Clculus for Engineers II Noes on Chper 6 Professor: John Quigg Semeser: spring 7 Secion 6.: Inegrion by prs The Produc Rule is d d f()g() = f()g () + f ()g() Tking indefinie inegrls gives [f()g
More information5.1The 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 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 informationIntegral Transform. Definitions. Function Space. Linear Mapping. Integral Transform
Inegrl Trnsform Definiions Funcion Spce funcion spce A funcion spce is liner spce of funcions defined on he sme domins & rnges. Liner Mpping liner mpping Le VF, WF e liner spces over he field F. A mpping
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 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 informationSOLUTIONS TO ASSIGNMENT 2  MATH 355. with c > 3. m(n c ) < δ. f(t) t. g(x)dx =
SOLUTIONS TO ASSIGNMENT 2  MATH 355 Problem. ecall ha, B n {ω [, ] : S n (ω) > nɛ n }, and S n (ω) N {ω [, ] : lim }, n n m(b n ) 3 n 2 ɛ 4. We wan o show ha m(n c ). Le δ >. We can pick ɛ 4 n c n wih
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 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 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 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 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 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 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 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 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 informationSome Inequalities variations on a common theme Lecture I, UL 2007
Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel
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 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 informationf(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2
Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe
More informationChallenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k
Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,
More information2k 1. . And when n is odd number, ) The conclusion is when n is even number, an. ( 1) ( 2 1) ( k 0,1,2 L )
Scholrs Journl of Engineering d Technology SJET) Sch. J. Eng. Tech., ; A):86 Scholrs Acdemic d Scienific Publisher An Inernionl Publisher for Acdemic d Scienific Resources) www.sspublisher.com ISSN X
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 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 informationMATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence
MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,
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 informationFURTHER GENERALIZATIONS. QI Feng. The value of the integral of f(x) over [a; b] can be estimated in a variety ofways. b a. 2(M m)
Univ. Beogrd. Pul. Elekroehn. Fk. Ser. M. 8 (997), 79{83 FUTHE GENEALIZATIONS OF INEQUALITIES FO AN INTEGAL QI Feng Using he Tylor's formul we prove wo inegrl inequliies, h generlize K. S. K. Iyengr's
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 informationBernoulli numbers. Francesco Chiatti, Matteo Pintonello. December 5, 2016
UNIVERSITÁ DEGLI STUDI DI PADOVA, DIPARTIMENTO DI MATEMATICA TULLIO LEVICIVITA Bernoulli numbers Francesco Chiai, Maeo Pinonello December 5, 206 During las lessons we have proved he Las Ferma Theorem
More informationMATH 351 Solutions: TEST 3B 23 April 2018 (revised)
MATH Soluions: TEST B April 8 (revised) Par I [ ps each] Each of he following asserions is false. Give an eplici counereample o illusrae his.. If H: (, ) R is coninuous, hen H is unbounded. Le H() =
More informationES.1803 Topic 22 Notes Jeremy Orloff
ES.83 Topic Noes Jeremy Orloff Fourier series inroducion: coninued. Goals. Be able o compue he Fourier coefficiens of even or odd periodic funcion using he simplified formulas.. Be able o wrie and graph
More informationSolutions to Assignment 1
MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we
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 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 informationHint: There's a table of particular solutions at the end of today's notes.
Mah 8 Fri Apr 4, Finish Wednesday's noes firs Then 94 Forced oscillaion problems via Fourier Series Today we will revisi he forced oscillaion problems of las Friday, where we prediced wheher or no resonance
More informationMATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion
More informationChapter 2. Motion along a straight line. 9/9/2015 Physics 218
Chper Moion long srigh line 9/9/05 Physics 8 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy
More informationProperties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)
Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =
More informationPositive and negative solutions of a boundary value problem for a
Invenion Journl of Reerch Technology in Engineering & Mngemen (IJRTEM) ISSN: 24553689 www.ijrem.com Volume 2 Iue 9 ǁ Sepemer 28 ǁ PP 7383 Poiive nd negive oluion of oundry vlue prolem for frcionl, difference
More information23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes
Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals
More informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be onetoone, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
More informationWeek #13  Integration by Parts & Numerical Integration Section 7.2
Week #3  Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by HughesHalle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission
More informationSMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.
SMT Calculus Tes Soluions February 5,. Le f() = and le g() =. Compue f ()g (). Answer: 5 Soluion: We noe ha f () = and g () = 6. Then f ()g () =. Plugging in = we ge f ()g () = 6 = 3 5 = 5.. There is a
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 informationFractional Calculus. Connor Wiegand. 6 th June 2017
Frcionl Clculus Connor Wiegnd 6 h June 217 Absrc This pper ims o give he reder comforble inroducion o Frcionl Clculus. Frcionl Derivives nd Inegrls re defined in muliple wys nd hen conneced o ech oher
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 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 informationEchocardiography Project and Finite Fourier Series
Echocardiography Projec and Finie Fourier Series 1 U M An echocardiagram is a plo of how a porion of he hear moves as he funcion of ime over he one or more hearbea cycles If he hearbea repeas iself every
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 informationEE 315 Notes. Gürdal Arslan CLASS 1. (Sections ) What is a signal?
EE 35 Noes Gürdal Arslan CLASS (Secions..2) Wha is a signal? In his class, a signal is some funcion of ime and i represens how some physical quaniy changes over some window of ime. Examples: velociy of
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 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 informationLAPLACE TRANSFORMS. 1. Basic transforms
LAPLACE TRANSFORMS. Bic rnform In hi coure, Lplce Trnform will be inroduced nd heir properie exmined; ble of common rnform will be buil up; nd rnform will be ued o olve ome dierenil equion by rnforming
More informationrank Additionally system of equation only independent atfect Gawp (A) possible ( Alb ) easily process form rang A. Proposition with Definition
Defiion nexivnol numer ler dependen rows mrix sid row Gwp elimion mehod does no fec h numer end process i possile esily red rng fc for mrix form der zz rn rnk wih m dcussion i holds rr o Proposiion ler
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationChapter 7: Solving Trig Equations
Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions
More informationSOLUTIONS TO ECE 3084
SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no
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 informationJournal of Mathematical Analysis and Applications. Two normality criteria and the converse of the Bloch principle
J. Mh. Anl. Appl. 353 009) 43 48 Conens liss vilble ScienceDirec Journl of Mhemicl Anlysis nd Applicions www.elsevier.com/loce/jm Two normliy crieri nd he converse of he Bloch principle K.S. Chrk, J. Rieppo
More informationSignals and Systems Profs. Byron Yu and Pulkit Grover Fall Midterm 1 Solutions
890 Signals and Sysems Profs. Byron Yu and Pulki Grover Fall 07 Miderm Soluions Name: Andrew ID: Problem Score Max 0 8 4 6 5 0 6 0 7 8 9 0 6 Toal 00 Miderm Soluions. (0 poins) Deermine wheher he following
More informationAn random variable is a quantity that assumes different values with certain probabilities.
Probabiliy The probabiliy PrA) of an even A is a number in [, ] ha represens how likely A is o occur. The larger he value of PrA), he more likely he even is o occur. PrA) means he even mus occur. PrA)
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 informationdt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.
Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies
More informationt + t sin t t cos t sin t. t cos t sin t dt t 2 = exp 2 log t log(t cos t sin t) = Multiplying by this factor and then integrating, we conclude that
ODEs, Homework #4 Soluions. Check ha y ( = is a soluion of he secondorder ODE ( cos sin y + y sin y sin = 0 and hen use his fac o find all soluions of he ODE. When y =, we have y = and also y = 0, so
More informationEECE 301 Signals & Systems Prof. Mark Fowler
EECE 3 Signals & Sysems Prof. Mark Fowler Noe Se # Wha are ConinuousTime Signals??? /6 ConinuousTime Signal Coninuous Time (CT) Signal: A CT signal is defined on he coninuum of ime values. Tha is:
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 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 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 informationa. Show that these lines intersect by finding the point of intersection. b. Find an equation for the plane containing these lines.
Mah A Final Eam Problems for onsideraion. Show all work for credi. Be sure o show wha you know. Given poins A(,,, B(,,, (,, 4 and (,,, find he volume of he parallelepiped wih adjacen edges AB, A, and A.
More informationcan be viewed as a generalized product, and one for which the product of f and g. That is, does
Boyce/DiPrim 9 h e, Ch 6.6: The Convoluion Inegrl Elemenry Differenil Equion n Bounry Vlue Problem, 9 h eiion, by Willim E. Boyce n Richr C. DiPrim, 9 by John Wiley & Son, Inc. Someime i i poible o wrie
More informationPhil Wertheimer UMD Math Qualifying Exam Solutions Analysis  January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
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 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 informationMATH 128A, SUMMER 2009, FINAL EXAM SOLUTION
MATH 28A, SUMME 2009, FINAL EXAM SOLUTION BENJAMIN JOHNSON () (8 poins) [Lagrange Inerpolaion] (a) (4 poins) Le f be a funcion defined a some real numbers x 0,..., x n. Give a defining equaion for he Lagrange
More informationTHE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.
THE BERNOULLI NUMBERS The Bernoulli numbers are defined here by he exponenial generaing funcion ( e The firs one is easy o compue: (2 and (3 B 0 lim 0 e lim, 0 e ( d B lim 0 d e +e e lim 0 (e 2 lim 0 2(e
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 informationReview  Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y
Review  Quiz # 1 (1) Solving Special Tpes of Firs Order Equaions I. Separable Equaions (SE). d = f() g() Mehod of Soluion : 1 g() d = f() (The soluions ma be given implicil b he above formula. Remember,
More informationgraph of unit step function t
.5 Piecewie coninuou forcing funcion...e.g. urning he forcing on nd off. The following Lplce rnform meril i ueful in yem where we urn forcing funcion on nd off, nd when we hve righ hnd ide "forcing funcion"
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 information4 Sequences of measurable functions
4 Sequences of measurable funcions 1. Le (Ω, A, µ) be a measure space (complee, afer a possible applicaion of he compleion heorem). In his chaper we invesigae relaions beween various (nonequivalen) convergences
More informationMath 106: Review for Final Exam, Part II. (x x 0 ) 2 = !
Mah 6: Review for Final Exam, Par II. Use a seconddegree Taylor polynomial o esimae 8. We choose f(x) x and x 7 because 7 is he perfec cube closes o 8. f(x) x / f(7) f (x) x / f (7) x / 7 / 7 f (x) 9
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 informationon the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?
Calculus II MAT 46 Improper Inegrals A mahemaician asked a fence painer o complee he unique ask of paining one side of a fence whose face could be described by he funcion y f (x on he inerval (x + x
More informationIX.1.1 The Laplace Transform Definition 700. IX.1.2 Properties 701. IX.1.3 Examples 702. IX.1.4 Solution of IVP for ODEs 704
Chper IX The Inegrl Trnform Mehod IX. The plce Trnform November 4, 7 699 IX. THE APACE TRANSFORM IX.. The plce Trnform Definiion 7 IX.. Properie 7 IX..3 Emple 7 IX..4 Soluion of IVP for ODE 74 IX..5 Soluion
More informationFrom Complex Fourier Series to Fourier Transforms
Topic From Complex Fourier Series o Fourier Transforms. Inroducion In he previous lecure you saw ha complex Fourier Series and is coeciens were dened by as f ( = n= C ne in! where C n = T T = T = f (e
More informationp(x) = 3x 3 + x n 3 k=0 so the right hand side of the equality we have to show is obtained for r = b 0, s = b 1 and 2n 3 b k x k, q 2n 3 (x) =
Norwegin University of Science nd Technology Deprtment of Mthemticl Sciences Pge 1 of 5 Contct during the exm: Elen Celledoni, tlf. 73593541, cell phone 48238584 PLESE NOTE: this solution is for the students
More informationSolutions for Nonlinear Partial Differential Equations By TanCot Method
IOSR Journl of Mhemics (IOSRJM) eissn: 78578. Volume 5, Issue 3 (Jn.  Feb. 13), PP 611 Soluions for Nonliner Pril Differenil Equions By TnCo Mehod Mhmood Jwd Abdul Rsool Abu AlSheer Al Rfidin Universiy
More informationSome basic notation and terminology. Deterministic Finite Automata. COMP218: Decision, Computation and Language Note 1
COMP28: Decision, Compuion nd Lnguge Noe These noes re inended minly s supplemen o he lecures nd exooks; hey will e useful for reminders ou noion nd erminology. Some sic noion nd erminology An lphe is
More informationEECE 301 Signals & Systems Prof. Mark Fowler
EECE 3 Signals & Sysems Prof. Mark Fowler Noe Se #2 Wha are ConinuousTime Signals??? Reading Assignmen: Secion. of Kamen and Heck /22 Course Flow Diagram The arrows here show concepual flow beween ideas.
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