Math 5440 Problem Set 3 Solutions
|
|
- Sophia Maria Alexander
- 5 years ago
- Views:
Transcription
1 Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 213 1: (Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping force proportionl to the velocity of the string. Obtin the dmped wve eqution u tt = c (x) 2 u xx ku t where k is the constnt dmping coefficient. Let u(x, t) be the verticl displcement of the string, nd θ(x, t) be the ngle between the tngent to the curve y = u(x, t) t x nd the positive x-xis. The steps from the book tht re the sme in this cse re: tn θ(x, t) = u x (x, t), ρ(x, t) 1+u x (x, t) 2 = ρ (x), nd T(x, t) cos(θ(x, t)) = τ(t), T(b, t) sin(θ(b, t)) T(, t) sin(θ(, t)) = τ(t)u x (x, t). The derivtion chnges in tht the eqution for conservtion of momentum is d b ρ(x, t)u t (x, t) dt 1+u x (x, t) 2 dx = T(b, t) sin(θ(b, t)) T(, t) sin(θ(, t)) b ku t (x, t)ρ(x, t) 1+u x (x, t) 2 dx, where ku t (x, t) is the dmping force per unit mss. This cn be rewritten s b ρ (x)u tt (x, t)dx = τ(t)u x (b, t) τ(t)u x (, t) b ku t (x, t)ρ (x)dx b b = τ(t) u xx (x, t)dx ku t (x, t)ρ (x)dx. Since the intervl [, b] is rbitrry, the integrnds must mtch so we obtin ρ (x)u tt (x, t) = τ(t)u xx (x, t) ku t (x, t)ρ (x). Assuming s before tht τ(t) = τ is constnt, dividing by ρ (x), nd reclling tht (c (x)) 2 τ /ρ (x), we obtin u tt (x, t) = (c (x)) 2 u xx (x, t) ku t (x, t). 1
2 2: (Logn, 1.5 # 5) Consider string whose motion is governed by the eqution ρ u tt = τ u xx for x l with boundry conditions u(, t) = nd u(l, t) =. Let the energy be defined by E(t) = l Show tht the energy is constnt for ll t. ( 1 2 ρ u 2 t + 1 ) 2 τ u 2 x dx. E (t) = l (ρ u t u tt + τ u x u xt ) dx. From ρ u tt = τ u xx, we see tht ρ u t u tt = τ u t u xx, so E (t) = l (τ u t u xx + τ u x u xt ) dx l = τ x (u tu x )dx = τ (u t (l, t)u x (l, t) u t (, t)u x (, t)) =. The lst equlity follows becuse u(, t) = for ll t implies tht u t (, t) = for ll t, nd similrly u t (l, t) =. 2
3 3: (Logn, 1.5 # 8) At the end (x = ) of long tube (x ), the density of ir chnges ccording to the formul 1 cos(2t) t, ρ(, t) =, t <. Find solution for the wve eqution ρ tt = c 2 ρ xx in the domin x > nd < t < in the form of right-trveling wve tht stisfies the given boundry conditions. Tke c = 1 nd plot the solution surfce. The generl solution to the 1-D wve eqution tells us tht ρ(x, t) = F(x ct)+g(x+ct) for some functions F nd G. The term G(x+ct) represents wve trvelling to the left, nd we drop it. Hence ρ(x, t) = F(x ct), nd our gol is to determine the fucntion F so tht the boundry condition t x = is stisfied. Since ρ(, t) = F( ct), we wnt 1 cos(2t), t, F( ct) =, t <. Letting z = ct, we see tht we wnt 1 cos( 2z/c), z, F(z) =, z >. Hence, or ρ(x, t) = F(x ct) = ρ(x, t) = 1 cos( 2(x ct)/c), x ct,, x ct >, 1 cos(2(x ct)/c), x ct,, x ct >. 3
4 t x 15 2 Figure.1: Plot of solution for x 2 nd t 1. 4: (Logn, 1.7 # 3) Suppose tht u = u(x, y, z) is solution of the Neumnn boundry vlue problem K u = f(x, y, z) (x, y, z), K u n = g(x, y, z) Show tht f nd g must stisfy the reltion f dv = (x, y, z). gda. Suppose tht solution u(x, y, z) to this problem exists. Integrte the PDE over the domin to get f(x, y, z)dv = K u dv = K u dv = Kn u da = g(x, y, z) da. Hence we see tht f(x, y, z)dv = g(x, y, z)da must be true for solution to this problem to exist. The physicl interprettion is tht for stedy-stte het flow with Neumnn boundry conditions, the net source of het in the domin equls the net flux of het cross the boundry of the domin. 4
5 5: (Logn, 1.7 # 5) Prove tht if the Dirichlet problem u = λu, (x, y, z), u =, (x, y, z), hs nontrivil solution u = u(x, y, z), then λ must be negtive. λ u 2 dv = = u u dv u u dv = ( (u u) u u) dv = n (u u) da u u dv = u u dv. Since u(x, y, z) is nonzero for some points in, the integrl multiplying λ is positive, nd so u u dv λ =. u2 dv The only wy tht the expression on the right-side could be zero is if the integrnd of the integrl in the numertor were lwys, tht is, u(x, y, z) = for ll (x, y, z). But then u would hve to be constnt for ll (x, y, z), nd since u is zero on nd u is continuous, this constnt would hve to be. Since we ssumed tht u is not identiclly zero, this cnnot hppen, nd so λ <. 5
6 6: Suppose tht u(x, t) is solution of the diffusion eqution u t = βu xx. Show tht ech of the following is lso solution of the diffusion eqution: ) v(x, t) = u x (x, t) b) v(x, t) = u t (x, t) c) v(x, t) = u(x s, t) for ny constnt s d) v(x, t) = g(s)u(x s, t)ds for ny function g. e) v(x, t) = u( x, t) for ny positive constnt. ) Let v(x, t) = u x (x, t). Then v t (x, t) = u xt (x, t) nd v xx (x, t) = u xxx (x, t), so v t βv xx = u xt βu xxx = (u t βu xx ) x = () x =. b) Let v(x, t) = u t (x, t). Then v t (x, t) = u tt (x, t) nd v xx (x, t) = u txx (x, t), so v t βv xx = u tt βu xxt = (u t βu xx ) t = () t =. c) Let v(x, t) = u(x s, t) for constnt s. Then v t (x, t) = u t (x s, t) nd v xx (x, t) = u xx (x s, t), so v t βv xx = u t (x s, t) βu xx (x s, t) =, since u stisfies the diffusion eqution for ll sptil points. d) Let v(x, t) = g(s)u(x s, t)ds. Then v t(x, t) = g(s)u t(x s, t)ds nd v xx (x, t) = g(s)u xx(x s, t)ds, so v t βv xx = g(s)(u t(x s, t) βu xx (x s, t))ds = ds =. e) Let v(x, t) = u( x, t) for constnt >. Then, v t (x, t) = u t ( x, t), v x (x, t) = u x ( x, t), nd v xx (x, t) = u xx ( x, t). So v t βv xx = (u t βu xx ) =. 6
Math 5440 Problem Set 3 Solutions
Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 25 1: Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping
More informationConsequently, the temperature must be the same at each point in the cross section at x. Let:
HW 2 Comments: L1-3. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the
More informationPDE Notes. Paul Carnig. January ODE s vs PDE s 1
PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More information1 2-D Second Order Equations: Separation of Variables
Chpter 12 PDEs in Rectngles 1 2-D Second Order Equtions: Seprtion of Vribles 1. A second order liner prtil differentil eqution in two vribles x nd y is A 2 u x + B 2 u 2 x y + C 2 u y + D u 2 x + E u +
More information1 1D heat and wave equations on a finite interval
1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion
More informationConservation Law. Chapter Goal. 5.2 Theory
Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very
More informationMath 124A October 04, 2011
Mth 4A October 04, 0 Viktor Grigoryn 4 Vibrtions nd het flow In this lecture we will derive the wve nd het equtions from physicl principles. These re second order constnt coefficient liner PEs, which model
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationHeat flux and total heat
Het flux nd totl het John McCun Mrch 14, 2017 1 Introduction Yesterdy (if I remember correctly) Ms. Prsd sked me question bout the condition of insulted boundry for the 1D het eqution, nd (bsed on glnce
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 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 informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationPHYSICS 116C Homework 4 Solutions
PHYSICS 116C Homework 4 Solutions 1. ( Simple hrmonic oscilltor. Clerly the eqution is of the Sturm-Liouville (SL form with λ = n 2, A(x = 1, B(x =, w(x = 1. Legendre s eqution. Clerly the eqution is of
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 informationKinematic Waves. These are waves which result from the conservation equation. t + I = 0. (2)
Introduction Kinemtic Wves These re wves which result from the conservtion eqution E t + I = 0 (1) where E represents sclr density field nd I, its outer flux. The one-dimensionl form of (1) is E t + I
More informationMath Fall 2006 Sample problems for the final exam: Solutions
Mth 42-5 Fll 26 Smple problems for the finl exm: Solutions Any problem my be ltered or replced by different one! Some possibly useful informtion Prsevl s equlity for the complex form of the Fourier series
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
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 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& 152 Section Integration by Parts
Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationName Solutions to Test 3 November 8, 2017
Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier
More informationAM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h
AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationSTURM-LIOUVILLE BOUNDARY VALUE PROBLEMS
STURM-LIOUVILLE 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 information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Theorem Suppose f is continuous
More information(4.1) D r v(t) ω(t, v(t))
1.4. Differentil inequlities. Let D r denote the right hnd derivtive of function. If ω(t, u) is sclr function of the sclrs t, u in some open connected set Ω, we sy tht function v(t), t < b, is solution
More informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationGreen function and Eigenfunctions
Green function nd Eigenfunctions Let L e regulr Sturm-Liouville opertor on n intervl (, ) together with regulr oundry conditions. We denote y, φ ( n, x ) the eigenvlues nd corresponding normlized eigenfunctions
More informationMATH , Calculus 2, Fall 2018
MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationSpace Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More informationConservation Law. Chapter Goal. 6.2 Theory
Chpter 6 Conservtion Lw 6.1 Gol Our long term gol is to unerstn how mthemticl moels re erive. Here, we will stuy how certin quntity chnges with time in given region (sptil omin). We then first erive the
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
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 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 informationMath 473: Practice Problems for the Material after Test 2, Fall 2011
Mth 473: Prctice Problems for the Mteril fter Test, Fll SOLUTION. Consider the following modified het eqution u t = u xx + u x. () Use the relevnt 3 point pproximtion formuls for u xx nd u x to derive
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function
More informationDifferential Equations 2 Homework 5 Solutions to the Assigned Exercises
Differentil Equtions Homework Solutions to the Assigned Exercises, # 3 Consider the dmped string prolem u tt + 3u t = u xx, < x , u, t = u, t =, t >, ux, = fx, u t x, = gx. In the exm you were supposed
More informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus
ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More information(9) P (x)u + Q(x)u + R(x)u =0
STURM-LIOUVILLE THEORY 7 2. Second order liner ordinry differentil equtions 2.1. Recll some sic results. A second order liner ordinry differentil eqution (ODE) hs the form (9) P (x)u + Q(x)u + R(x)u =0
More informationTaylor Polynomial Inequalities
Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?
More information1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation
1 1.1. Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem. 1.1.1. Overview
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More information4181H Problem Set 11 Selected Solutions. Chapter 19. n(log x) n 1 1 x x dx,
48H Problem Set Selected Solutions Chpter 9 # () Tke f(x) = x n, g (x) = e x, nd use integrtion by prts; this gives reduction formul: x n e x dx = x n e x n x n e x dx. (b) Tke f(x) = (log x) n, g (x)
More informationdf dt f () b f () a dt
Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
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 informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationu t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx
Chpter 9: Green s functions for time-independent problems Introductory emples One-dimensionl het eqution Consider the one-dimensionl het eqution with boundry conditions nd initil condition We lredy know
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
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 informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationRAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties
M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5. - 5.3) Remrks on the course. Slide Review: Sec. 5.-5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
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 informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2.
Mth 43 Section 6. Section 6.: 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
More informationProblem set 1: Solutions Math 207B, Winter 2016
Problem set 1: Solutions Mth 27B, Winter 216 1. Define f : R 2 R by f(,) = nd f(x,y) = xy3 x 2 +y 6 if (x,y) (,). ()Show tht thedirectionl derivtives of f t (,)exist inevery direction. Wht is its Gâteux
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More informationEnergy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon
Phys463.nb 49 7 Energy Bnds Ref: textbook, Chpter 7 Q: Why re there insultors nd conductors? Q: Wht will hppen when n electron moves in crystl? In the previous chpter, we discussed free electron gses,
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 informationMath 115 ( ) Yum-Tong Siu 1. Lagrange Multipliers and Variational Problems with Constraints. F (x,y,y )dx
Mth 5 2006-2007) Yum-Tong Siu Lgrnge Multipliers nd Vritionl Problems with Constrints Integrl Constrints. Consider the vritionl problem of finding the extremls for the functionl J[y] = F x,y,y )dx with
More informationNote 16. Stokes theorem Differential Geometry, 2005
Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion
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 informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationChapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1
Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationIndefinite Integral. Chapter Integration - reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing 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 informationIntegrals - Motivation
Integrls - Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution
More informationLine Integrals. Chapter Definition
hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xy-plne. It
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More information( ) 2. ( ) is the Fourier transform of! ( x). ( ) ( ) ( ) = Ae i kx"#t ( ) = 1 2" ( )"( x,t) PC 3101 Quantum Mechanics Section 1
1. 1D Schrödinger Eqution G chpters 3-4. 1.1 the Free Prticle V 0 "( x,t) i = 2 t 2m x,t = Ae i kxt "( x,t) x 2 where = k 2 2m. Normliztion must hppen: 2 x,t = 1 Here, however: " A 2 dx " " As this integrl
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
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 informationPhysics 215 Quantum Mechanics 1 Assignment 2
Physics 15 Quntum Mechnics 1 Assignment Logn A. Morrison Jnury, 16 Problem 1 Clculte p nd p on the Gussin wve pcket α whose wve function is x α = 1 ikx x 1/4 d 1 Solution Recll tht where ψx = x ψ. Additionlly,
More informationThe area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationEvaluating Definite Integrals. There are a few properties that you should remember in order to assist you in evaluating definite integrals.
Evluting Definite Integrls There re few properties tht you should rememer in order to ssist you in evluting definite integrls. f x dx= ; where k is ny rel constnt k f x dx= k f x dx ± = ± f x g x dx f
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 information2. THE HEAT EQUATION (Joseph FOURIER ( ) in 1807; Théorie analytique de la chaleur, 1822).
mpc2w4.tex Week 4. 2.11.2011 2. THE HEAT EQUATION (Joseph FOURIER (1768-1830) in 1807; Théorie nlytique de l chleur, 1822). One dimension. Consider uniform br (of some mteril, sy metl, tht conducts het),
More informationF (x) dx = F (x)+c = u + C = du,
35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil
More informationMath Advanced Calculus II
Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More information