Homework 4 , (1) 1+( NA +N D , (2)
|
|
- Angelina Harriet Hopkins
- 5 years ago
- Views:
Transcription
1 Homework 4. Problem. Find the resistivity ρ (in ohm-cm) for piece of Si doped with both cceptors (N A = 9 cm 3 ) nd donors (N D = 6 cm 3 ). Since the electron nd hole mobilities depend on the concentrtion of the dopnts, use the following empriricl expressions to evlute them: µ n = ( NA +N D 8 6 ).9, () µ p = where the mobilities re expressed in cm 2 /V s nd N A nd N D in cm 3. Solution. Subtituting for the vlues of N A nd N D in the given equtions, nd +( NA +N D.3 6 ).76, (2) µ n = cm 2 /Vs, µ p =5.84 cm 2 /Vs. Since σ = en D µ n + en A µ p nd ρ =/σ, wehve ρ =.22 2 Ωcm. 2. Problem. Consider smple of p-type Si doped with N A = 8 cm 3 nd N D =. Over length of µm the electron concentrtion drops from 6 cm 3 to 3 cm 3. Using Eq. () bove, clculte the current density due to diffusion lone. ECE69 Spring 2
2 Solution. The electron current due to diffusion lone is: J diff = ed n dn dx. Since from Einstein reltion ed n = k b Tµ n, ssuming dn/dx 6 cm 3 / 4 cm nd using µ n from the eqution given in the first problem, we hve µ n = cm 2 /Vs nd J diff = 4.65 A/cm Problem. A p-n Si junction is formed between n-type region with N D = 2 8 cm 3 nd p-type region with N A = 5 6 cm 3. Find: () the width of the depletion region on the n side, x n (in µm), (b) the width of the depletion region on the p side, x p (in µm), (c) the built-in potentil V bi (in ev), (d) the frction of V bi which drops over the n-side of the junction, (e) the frction of V bi which drops over the p-side of the junction. Finlly, using the results just obtined, plot (to scle s ccurtely s you cn) the bnd digrm of the junction t equilibrium. Solution. () From ( ) ( 2ɛkB T NA N D W = e 2 ln + ) /2, N A N D we get W =.54 µm nd (b) Similrly, n 2 i x n = WN A N A + N D =.375 µm. x p = WN D N A + N D =.5 µm. ECE69 Spring 2 2
3 (c) The built-in potentil is given by: V bi = k BT e ln ( ) NA N D n 2 i =.892 V. (d) The potentil drop on the n side of the junction is: V bi,n = V bix n W =.2 V. (d) On the other end, the potentil drop on the p side of the junction is: V bi,p = V bix p W =.87 V. 4. Problem. ASi p-n junction hs sturted reverse current I s = 4 At 3 K. Determine the forwrd bis required to get current of () 5 µand (b) 2.5 ma. Solution. () From the expression for the diode current ( ) ev I = I s exp k B T nd from the fct tht I s = 4 Awe hve V =.5784 V when I =5µA. (b) Similrly, for I =2.5mA we hve V =.6797 V. 5. Problem. The sturted reverse current of GAs p-n junction is 5 2 A. Clculte the current under forwrd bis of ().8 V nd (b).2 V. Solution. () As in the previous problem, we hve I =.29 µafor V =.8Vnd(b) I =.66AforV =.2 V. ECE69 Spring 2 3
4 6. Problem. Derive Eq. (49) of the Lecture Notes, Prt 2. Solution. We require: J(x = l n ) = J = J { + dx α p exp Now let s multiply both sides of this eqution by: l J exp n, } exp (3) so tht Eq. (3) becomes: exp = + dx α p exp, (4) or = exp dx α p exp. (5) Now let s dd nd substrct α n in the integrnd of the lst term: = exp + dx (α n α p ) exp dx α n exp. (6) ECE69 Spring 2 4
5 The second term on the right-hnd side cn be integrted immeditely, since in generl: b ) ( dx f(x) exp dx f(x ) = exp = exp ( ( ) dx f(x b ) b ) dx f(x ). So, Eq. (6) becomes: = exp dx α n exp + exp = dx α n exp (7) so tht l n x = dx α n exp, (8) which is Eq. (49) of the Lecture Notes, Prt 2. Note tht the result bove is completely generl. We hve not mde use of the ssumption α n = α p,norof ny rbitrry ssumption of the type x exp dx(α n α p ) = exp (α n α p )(x + l p ), which is only vlid if α n nd α p do not depend on x, nd, finlly, it is vlid lso fr from brekdown, so it does not require ny condition of the sort ln l dxα =. p 7. Problem. (). Plot the Zener tunneling current J Zener given by Eq. (425) of the Lecture Notes s function of pplied bis V using m =.32 m el = kg nd E G =.ev (the gp of Si). Assume, ECE69 Spring 2 5
6 for the field F the pproximte vlue F F mx, where the mximum field in the junction, F mx, is given by Eq. (364). Assume N A = 7 cm 3 nd N D = 9 cm 3 to estimte the width l p nd l n of the depletion regions. (b). Repet the clcultion, but now using E G =.64 ev (the gp of Ge) nd m =.22 m el. (c). Repet the clcultions in () nd (b), but now ssuming higher doping, N A = 9 cm 3 nd N D = 2 cm 3. Wht hppens? Why? (d). Considering tht the current VLSI technolgy requires incresingly higher doping, cn you foresee ny problems with the possible use of Ge (insted of Si) when the width of the depletion regions shrinks? Solution. Higher doping enhnces tunneling. Ge exhibits much lrger Zener tunneling current thn Si becuse of its smller bnd-gp. This drwbck will become more severe s the doping concentrtions increse. The figure below shows the result: ECE69 Spring 2 6
7 2 Ge J Zener (A/cm 2 ) 2 Si high doping Ge low doping Si V pplied (V) Note tht the current is nonzero only under reverse bis such tht V < (E G V bi ) or under forwrd bis V >E G + V bi. For bis of smller mgnitude, V, tunneling cnnot occurr becuse there re no finl sttes vilble. The plot show only the current in reverse bis, since this mtters in devices. (Under forwrd bis the diode current is very lrge nywy nd tunneling is not concern.) NOTE: The most importnt dependence of the Zener tunneling current on the pplied reverse bis V is vi the depletion width l p ppering in the expression for the field F mx = en A l p /ɛ s. Mny of you hve forgotten this crucil dependence. 8. Problem. The current trends in VLSI technology lso demnd reduced pplied bises (i.e., smller V ). Without doing ny clcultion, do you think tht the strictest brekdown limittions will be due to impct ioniztion or to Zener tunneling? ECE69 Spring 2 7
8 Solution. Zener tunneling will dominte, since the energy threshold for impct-ioniztion is E G, while for Zener tunneling is E G ev bi,wherev bi is the built-in potentil of the drin/body junction in MOSFET. 9. Problem. Using Eqns. (45) nd (45), plot the totl chrge t the GAs-side of the GAs-Al x G x As heterojunction s function of the interfce potentil ψ i = eψ i /(k B T ). Indicte clerly the ccumultion, depletion, inversion, nd strong inversion regions. Refer to the discussion on pge 35 of the Notes for help. Solution. The equtions you were told to use ccount only for the chrge of the mjority crriers (electrons) nd donors. They do not ccount for minority crriers (holes), so tht it is impossible to obtin the chrge in inversion (wek or strong), unless the hole chrge-density is dded. The chrge density of the minority crriers is included in the equtions for the chrge t the interfce of n MOS cpcitor. So, one could use similr eqution, remembering tht in tht cse we considered p-type substrte, while here we re deling with n n-type substrte, so the plot should be flipped (s in the mirror imge ψ i ψ i ).. Problem. Derive Eq. (457) from Eq. (458). Solution. The potentil energy is The mximum of the potentil energy is given by e2 V im (z) = 6πɛ s z ef zz. (9) = dv im(z) dz The vlue of z t which this occurs is thus: = e 2 6πɛ s z 2 ef z. () z = ( ) e /2. () 6πɛ s F z ECE69 Spring 2 8
9 Inserting this into Eq. (9) we get: e2 V im (z ) = ef z z = e 6πɛ s z ( efz 4πɛ s ) /2, (2) so tht the brier-lowering (in V) will be Φ B =ef z /(4πɛ s ) /2.. Problem. Using the derivtion of Eq. (48) s guide, derive Eq. (482). Solution. The expression given in clss, Eq. (475) of the Notes, Prt 2, for the tunneling current from the metl to the semiconductor is: j MS = em M 2π 2 h 3 deef M (E) f S (E) exp { zt 2 } dz κ(z). (3) Approximting the potentil brrier with φ(z) φ B F zz, we hve for the tunneling distnce z t = (eφ B E)/(eF z) nd: zt dz κ z (z) = (2m S )/2 h zt dz (eφ B ef z E) /2. (4) Using the new integrtion vrible y = eφ B ef z E we get: so tht zt j MS = em M 2π 2 h 3 which is Eq. (482). dz κ z (z) = (2m S )/2 e hf z eφ B E dy y /2 = 2(2m S )/2 3e hf z (eφ B E)3/2, (5) { } deef M (E) f S (E) exp 4(2m S )/2 (eφ B 3e hf E)3/2 z, (6) ECE69 Spring 2 9
Lecture 13 - Linking E, ϕ, and ρ
Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
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 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 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 informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More 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 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 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 informationLecture 10: PN Junction & MOS Capacitors
Lecture 10: P Junction & MOS Cpcitors Prof. iknej eprtment of EECS Lecture Outline Review: P Junctions Therml Equilibrium P Junctions with Reverse Bis (3.3-3.6 MOS Cpcitors (3.7-3.9: Accumultion, epletion,
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
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 informationMath 113 Exam 2 Practice
Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.-3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This
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 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 informationNumerical integration
2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter
More 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 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 informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
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 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 informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More 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 information221B Lecture Notes WKB Method
Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using
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 informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationJim Lambers MAT 169 Fall Semester Lecture 4 Notes
Jim Lmbers MAT 169 Fll Semester 2009-10 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of
More 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 information1.3 The Lemma of DuBois-Reymond
28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBois-Reymond We needed extr regulrity to integrte by prts nd obtin the Euler- Lgrnge eqution. The following result shows tht, t lest sometimes, the extr
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationExam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B
PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
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 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 informationProblem Set 3 Solutions
Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner
More informationFinal Exam - Review MATH Spring 2017
Finl Exm - Review MATH 5 - Spring 7 Chpter, 3, nd Sections 5.-5.5, 5.7 Finl Exm: Tuesdy 5/9, :3-7:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
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 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 informationFundamentals of Analytical Chemistry
Homework Fundmentls of nlyticl hemistry hpter 9 0, 1, 5, 7, 9 cids, Bses, nd hpter 9(b) Definitions cid Releses H ions in wter (rrhenius) Proton donor (Bronsted( Lowry) Electron-pir cceptor (Lewis) hrcteristic
More informationProblem Set 3 Solutions
Chemistry 36 Dr Jen M Stndrd Problem Set 3 Solutions 1 Verify for the prticle in one-dimensionl box by explicit integrtion tht the wvefunction ψ ( x) π x is normlized To verify tht ψ ( x) is normlized,
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
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 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 informationDIRECT CURRENT CIRCUITS
DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationState space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies
Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response
More 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 informationTHE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationFig. 1. Open-Loop and Closed-Loop Systems with Plant Variations
ME 3600 Control ystems Chrcteristics of Open-Loop nd Closed-Loop ystems Importnt Control ystem Chrcteristics o ensitivity of system response to prmetric vritions cn be reduced o rnsient nd stedy-stte responses
More informationChapter 3 Polynomials
Dr M DRAIEF As described in the introduction of Chpter 1, pplictions of solving liner equtions rise in number of different settings In prticulr, we will in this chpter focus on the problem of modelling
More informationMAT 168: Calculus II with Analytic Geometry. James V. Lambers
MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd
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 informationLecture 19: Continuous Least Squares Approximation
Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationJackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero
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 information1. Weak acids. For a weak acid HA, there is less than 100% dissociation to ions. The B-L equilibrium is:
th 9 Homework: Reding, M&F, ch. 15, pp. 584-598, 602-605 (clcultions of ph, etc., for wek cids, wek bses, polyprotic cids, nd slts; fctors ffecting cid strength). Problems: Nkon, ch. 18, #1-10, 16-18,
More informationTheoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4
WiSe 1 8.1.1 Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Mtthis Sb m Lehrstuhl für Theoretische Physik I Deprtment für Physik Friedrich-Alexnder-Universität Erlngen-Nürnberg Theoretische
More informationInfinite Geometric Series
Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More information13: Diffusion in 2 Energy Groups
3: Diffusion in Energy Groups B. Rouben McMster University Course EP 4D3/6D3 Nucler Rector Anlysis (Rector Physics) 5 Sept.-Dec. 5 September Contents We study the diffusion eqution in two energy groups
More informationThe Velocity Factor of an Insulated Two-Wire Transmission Line
The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
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 informationMath 120 Answers for Homework 13
Mth 12 Answers for Homework 13 1. In this problem we will use the fct tht if m f(x M on n intervl [, b] (nd if f is integrble on [, b] then (* m(b f dx M(b. ( The function f(x = 1 + x 3 is n incresing
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
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 information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationCHM Physical Chemistry I Chapter 1 - Supplementary Material
CHM 3410 - Physicl Chemistry I Chpter 1 - Supplementry Mteril For review of some bsic concepts in mth, see Atkins "Mthemticl Bckground 1 (pp 59-6), nd "Mthemticl Bckground " (pp 109-111). 1. Derivtion
More informationMath 31S. Rumbos Fall Solutions to Assignment #16
Mth 31S. Rumbos Fll 2016 1 Solutions to Assignment #16 1. Logistic Growth 1. Suppose tht the growth of certin niml popultion is governed by the differentil eqution 1000 dn N dt = 100 N, (1) where N(t)
More information13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes
The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce
More informationADVANCEMENT OF THE CLOSELY COUPLED PROBES POTENTIAL DROP TECHNIQUE FOR NDE OF SURFACE CRACKS
ADVANCEMENT OF THE CLOSELY COUPLED PROBES POTENTIAL DROP TECHNIQUE FOR NDE OF SURFACE CRACKS F. Tkeo 1 nd M. Sk 1 Hchinohe Ntionl College of Technology, Hchinohe, Jpn; Tohoku University, Sendi, Jpn Abstrct:
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationIntro to Nuclear and Particle Physics (5110)
Intro to Nucler nd Prticle Physics (5110) Feb, 009 The Nucler Mss Spectrum The Liquid Drop Model //009 1 E(MeV) n n(n-1)/ E/[ n(n-1)/] (MeV/pir) 1 C 16 O 0 Ne 4 Mg 7.7 14.44 19.17 8.48 4 5 6 6 10 15.4.41
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More 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 informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationMath 259 Winter Solutions to Homework #9
Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658-659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
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 informationNumerical Analysis: Trapezoidal and Simpson s Rule
nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =
More informationDiscrete Least-squares Approximations
Discrete Lest-squres Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More informationScientific notation is a way of expressing really big numbers or really small numbers.
Scientific Nottion (Stndrd form) Scientific nottion is wy of expressing relly big numbers or relly smll numbers. It is most often used in scientific clcultions where the nlysis must be very precise. Scientific
More informationMath Solutions to homework 1
Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationMath 61CM - Solutions to homework 9
Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More information8 Laplace s Method and Local Limit Theorems
8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved
More informationMath 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 informationMassachusetts Institute of Technology Quantum Mechanics I (8.04) Spring 2005 Solutions to Problem Set 6
Msschusetts Institute of Technology Quntum Mechnics I (8.) Spring 5 Solutions to Problem Set 6 By Kit Mtn. Prctice with delt functions ( points) The Dirc delt function my be defined s such tht () (b) 3
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 informationx = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is
Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According
More informationLecture 20: Numerical Integration III
cs4: introduction to numericl nlysis /8/0 Lecture 0: Numericl Integrtion III Instructor: Professor Amos Ron Scribes: Mrk Cowlishw, Yunpeng Li, Nthnel Fillmore For the lst few lectures we hve discussed
More information