I I M O I S K J H G. b gb g. Chapter 8. Problem Solutions. Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 8

Size: px
Start display at page:

Download "I I M O I S K J H G. b gb g. Chapter 8. Problem Solutions. Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 8"

Transcription

1 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos Cher 8 rolem oluos 8. he fwr s e ex f The e ex f e e f ex () () f f f f l G e f f ex f 59.9 m 60 m 0 9. m m 8. e ex we c wre hs s e ex h l e reverse s, s eve, 090., we hve ( ) ( ) l m 8. Comuer lo 8.4 The cross-secol re s x 5 x 0 We hve 4 ex ex h.5x / We c wre e Q \ We w x x 5 x 7.07x x 4.47x whch yels 4.4 ow x.5 6x 9 5x ( 4.4) 5 x 5 x We f 7.x 4. x 6

2 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8.5 () e e e G We hve µ µ.4 0. G ( ) G () Us Ese s relo, we c wre eµ eµ eµ eµ eµ e µ We hve σ eµ σ eµ l The σ σ σ σ slco juco, e x 5. x 94. x 5 The 5 x 050. ex 94. ex We w 9.54x 7 We o The 095. e e e x µ m ( ) 0. x 6 µ m ( )

3 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos whch yels G 008. () -se: E E l l G 5 5 x l 5. x ( ) G E E 09. e -se: E E l G l 5. x E E e () We c f ( 50)( 0 059).4. / s ( 40)( 0 059) 9.. / s ow e ( ) G x. x.4 5 x 4.48x / The 4.48x We f Q x 5 G ex x ex µ (c) The hole curre s rool o e x 5. x 76. x 6 The x x 8.9 ex oe, e e x. x 6 6.9x 9 () 0., x ex µ () 0., x ex x 9.9

4 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8. slco oe e x x 8. x 5 () ex x 8. ex 4.6x 7 () x 5 8. () We f µ ( 480)( ).4 / s e (.4) 0. x 6. µ m l 5. x 5.5x 5 The 9 5 e 6. x (.4).5x 4. x 4.0x / 4, he 4.0x 4 () We hve µ / s e ( )( ) l x 6 The 7.4 µ m ( ) 5. x 4 4.5x 5 5 x 9 4 e 6. x ( 5) 4.5x x 674. x / 4, he (c) 674. x 5 G l ( 0059). l 067. The f 009. We f e ex 5.5x ex 5 x x x () The ol curre s e ex x 6 74x 7.x 9 The hole curre s. ex

5 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos f ex ex e x x The elecro curre s ve y 7.x 4.0x 9 4 ex ex x x x x 9. ex 4. x 9 x x f 8. () The excess hole cocero s ve y δ e x ex ex We f The δ 5. x.5x x 6.8 µ m. ().5x4 06. ex ex 0059 δ 4 x 8. x ex.8x 4 () We hve e δ f x 4 e x 8. x ex.8x.8x x x 4, x.8x 4 () 9 4 x x ex 4.8x / (c) We hve e e ex We c eerme h 4.5x. 7 µ m The 9 x x ex 4. 7x / We c l f 7. / The, x µ m, ( ) f f µ m µ m f. / µ m 9 8. () rom rolem 8.9 (Ge oe) ow jeco mes () ( ) ow.4x 576. x 6 We hve () 0 G ex () 0 5 ( ) G l l 576. x 05. () rolem 8. ( oe) 5

6 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos ( ) (. ) The 5. x.5x 6 () 5 ( ) G l 4 0 l x The excess elecro cocero s ve y δ e x ex ex The ol umer of excess elecros s x z δ 0 We my oe h x ex ex x z G x 0 0 The e ex We c f 5 / s 59. µ m l 5. x.8x 5 8 x The x.8x e 066. ex 4 ex e The we f he ol umer of excess elecros he -reo o e: () 0., 78. x 4 () 04., 846. x 5 (c) 05., 4.0 x 7 mlrly, he ol umer of excess holes he -reo s fou o e: e ex We f h.4 / s. µ m l x x 6 The e x.50 ex o () 0.,.68 x () 04., 7. x 5 (c) 05., 605. x e E e ex ex ex The e E ex e E ex e E ex e e E E ex We hve x E ex... 6 x ex E The 6

7 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos E ( ) l 8.7 whch yels Comuer lo E e 8.6 () We hve e whch c e wre he fm C C C () Tk he ro T 00 CT ex ex T T ex ex E E Q E E T G ex E T T 00, , 8. 6 T 400, , () Germum, E 066. e 400 ex ( 066. )( ) 00 8 () lco, E. e 400 ex (. )( ) x 8.8 e coo: f r e ex e ex e l l T ( ) e 00 whch yels T 69 eco coo: e e Q E G ex C.. x x x 5 E G ex 5 x E 4.66x e e whch ecomes ex E. e, E. l4.66x l4.66x T e ( ) The T 59 Ths seco coo yels smller emerure, he mxmum emerure s T 59 7

8 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8.9 () We c wre f he -reo δf δ 0 x The eerl luo s δ ex x Bex x The oury coo x x ves δf x ex ex x B x ex he oury coo x x W ves δ x W f 0 f ex f ex f ex x W B x W rom hs equo, we hve B x W The, from he frs oury coo, we o ex G f ex ex exx ex W x W ex ex f exw ex W ex ex f exw ex W Bex x W Bex x Bex x W We he o B whch c e wre he fm l B The luo c ow e wre s x W T δ ex W sh x W xf x W xf ex ex flly, x W x sh δ ex W sh () e δ f x The 8.0 e x x ex W cosh W sh x W x e coh ex G U W x ex he emerure re 00 T 0, elec he che C o E e ex ex E e ex Tk he ro of curres, u m cos, we hve x 8

9 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos ex ex E e E e E e E e We hve T 00, e , e T e , e T 0 e , e o, f T, whch yels T 0, whch yels Comuer lo 8. e x l e C We hve 6 s x The 6 x C ( ) C 86. x 8 The Y jω C Y j 86. x 8 8. oe ω Q Q, C ow 86. x C 9 9. x ( ) ow jωc Z Y jωc ω C We hve ω πf, We f: f kz: Z 5. 9 j f 0 kz: Z 5. 9 j0. 84 f z: Z. 6 j7.4 f z: Z.8 j () Two ccces wll e equl me fwrs vole. fwr-s vole, he juco ccce s e C j The ffu ccce s / f f C where e G ex 9

10 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos e G ex We f ( 0)( 0 059) / s ( )( ) / s 7 5 5x ( ) l x ow, we o 9 4 6x ( 7) 8 85x 4 C j... C j f 6 / x f x 5x Q 5 x x 5 7 We l o l G 78. x 6 ex x 5. x 5 x 77. x 5 ex We c ow wre G ex 5 6 G G ex / C ( ). 78x x ex G C 658. x 0 ex We w o se C C j o 6 / x 076. By rl err, we f 046. hs vole, C C 8. j x ex 8.5 oe, >>, he C ow The G.5x 6 / ( ) x 6. x 7 s m, C 6.5x C.5x 9 G e () C oe-se oe, >>, he C e f

11 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos m () e G ex We f 58. µ m 7.5x The 058. x. f x ( 5).5x G ex x x 570. x ex We f (c) e r r x 50 Ω r 8.7 () -reo ρ σ e µ x ( 480) 6 Ω -reo ρ σ e µ f x 9 ( 50) Ω The ol seres ressce s Ω () 0. ( 7.) 8. m 8.8 ρ () ρ () ( ) ( ) ( ) ( ) x x 50 Ω We c wre G l () () m. l ( ) ( ) G () m. l x x ( ) ( ) G () 0 () m ( ) G l 047. () m x ( ) G l 0477.

12 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8.9 r 48 Ω We hve e ( )( ) 059. m l G ex l x ( ) 059. G l x x x 6 l 4 ( ) 8 () 6.45µ x ( 00). 4x 4 4. x () 6 x. 4x µ m m The.45x x 8 / everse-se eero curre esy e W e 8.0 () r G G ex 000. ex r whch yels r. x Ω () 000., 000. ex r r 56. x Ω 8. el reverse-suro curre esy e e We f 6 8. x 4. x 6 4 We hve G l W The ( 0059). l x / f 4 x e (. ) 885. ( 6. 5) 9 6. x Q 6 6 Q 6 6 W 4. x x 8. x 4. x 8 e 9. x 9 / e Geero curre omes Gs reversese jucos. /

13 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8. () We c wre e x x x 85. x We l hve e W e 5, we f W 4. x 4 o e Whe 7 5 x 8. x 7 e x 4. x, e x 8. x whch yels 9.88x We hve E ex C The 9 9 T 9.88x.8 x 4x. 00. ex ( ) T 00 By rl err, we f T 505 hs emerure 7 8. x 9.88x e Q 8. x / e f () ex e ex G T 00 5x. 85. x 4.6x / 7 5. x 8. x e.7x 7 / e The we c wre 7 x e.7 ex 656. x 4.6x h ( ) l 6. 56x () We c wre e We f ( )( ) Q / s ( 00)( 0 059) 58.. / s The x 8. x 575. x 9 / 5 75x x We l hve e W e ow

14 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos l W G l ( 0059). l x / f 4 x e (. ) 885. ( 8. 5) 9 6. x 7 7 Q 7 7 / W 047. x x x x e x e The ol reverse-s curre 5. 75x 65. x e 65. x wr Bs: el ffu curre 0. x 0. ex 575. ex x x ex 9. x ecomo curre 0. : W The rec 4 7 (. ) 885. x ( ) x x G W 069. x 4 e W G ex x 8. x 069. x x rec 05. W The ex (. ) 4 7 (. ) 885. x ( ) x x G W 050. x x 8. x 05. x 8 rec 6. x 9 rec Tol fwr-s curre: 0. ; 67. x 7.96x 7.96x x. 6x 9 ex x 9 () everse-s; ro of eero o el ffu curre: e 65. x 575. x (. ) Q / / 4

15 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos o 7. x 9 wr s: o of recomo o el ffu curre: x rec x o 9. x x rec 9. x o.4x Comuer lo 8.5 Comuer lo 8.6 Comuer lo 8.7 We hve h ( ) ( ) e We c wre E E ex E E ex E E f E E e E E e E E f e E E f ex e E E f ex ex e η E E η We l hve h The efe The he recomo re c e wre s η η η e e e η η η e e e e η e e e η η η To f he mxmum recomo re, se η 0 η e e e e x η e η η η 0 ( ) e e e e e e whch smlfes o 0 η η η η e e e e η η η e e e The eom s o zero, we hve η η η e e e 0 η η η η η η η η e e η η The he mxmum recomo re ecomes η e mx η η η e e e η e η η e e η e mx η e whch c e wre s e ex mx e ex f >> e, he we c elec he (-) erm he umer he () erm he eom we flly hve 5

16 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos 8.8 We hve W e z0 mx e ex egx Q.E.. hs cse, G 4x 9 s, h s cos hrouh he sce chre reo. The e W e We f G l x 5 x ( ) l x W / f 4 x e 7 (. ) 885. ( ) 9 6. x x 5 x Q x 5 x W.5x 4 The 6. x 4x.5x 5. x / e e 9 6. x 5. x e x Q / 64. x / ow G ex l 0 G 0 5x 64. x ex 5 9 whch yels ex. x l 5. x B 0 Ε e cr x 4x G B x B ( ) 9 whch yels 7. x 6 B 8.4 he rekow vole, we ee x 5 f hs o, we f µ 40 / s. The ( 40)( 0 059) 4.. / s he juco, The e 9 6x 5x x 7. x / 6

17 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos G ex 065. x 7. x ex lly 99. x Gs,, rom ure B 8.4 e Ε mx x We c wre Ε x mx e x( 7. ) 885. x x 5x x 58. x 5 We f 6 6 x x ( ) l x ow x e G / f (. ). x 58. x 9 6. x x f x x 5 x 9.68x 9. x f whch yels Q 8.44 slco juco wh 5 x 5 0 B elec comre o B x / B 4.. x x 5x e ( ) ( ) ( ) 509 µ x m. m Q / 8.45 We f 8 8 ( ) l x ow e Ε mx x x x 6 4 ( 7. ) 8. 85x whch yels x 647. x 6 ow x e G / f (. ). x 647. x 9 6. x f 8 Q The whch yels ssume slco: juco x / f e 7

18 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos ssume << () x 75 µ m The 7 ( x 75x ) x whch yels 4.5x () x 50 µ m, we f 74. x4 rom ure 8.5, he rekow vole s roxmely 00. o, ech cse, rekow s reche frs mury re 8 x 4 x rom he fure 5 B 8.48 () f 0. The we hve erf erf We f () f erf s , he whch yels We w The erf erf 0. where erf 0. erf ( ) 047. We o We hve erf ex π By rl err, ( ) G 8.50 C 8 j 0 C 4. j We hve 7 s, m m o 7 l l. x 7 s l C v The me cos s 8

19 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos C v 4. x. x 7 s ow Tur-off me (.. ) 7 s r.x 7 s x ( ) l x We f W / f e 4 7 (. ) 885. x ( ) 9 6. x whch yels 8.5 kech 5 x 5 x x 7 W 69. x Q / 9

20 emcouc hyscs evces: Bsc rcles, r eo Cher 8 oluos ul rolem oluos (e lef lk)

I K J K J. Chapter 1. Problem Solutions. Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 1

I K J K J. Chapter 1. Problem Solutions. Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 1 Semcouc hyscs evces: sc rcles, r eo her. () cc: 8 cer oms /8 om 6 ce oms ½ oms ol o 4 oms er u cell () cc: 8 cer oms /8 om eclose om om ol o oms er u cell mo: 8 cer oms /8 om 6 ce oms ½ oms 4 eclose oms

More information

P a g e 3 6 of R e p o r t P B 4 / 0 9

P a g e 3 6 of R e p o r t P B 4 / 0 9 P a g e 3 6 of R e p o r t P B 4 / 0 9 p r o t e c t h um a n h e a l t h a n d p r o p e r t y fr om t h e d a n g e rs i n h e r e n t i n m i n i n g o p e r a t i o n s s u c h a s a q u a r r y. J

More information

b gb g L N b gb gb g Chapter 13 Problem Solutions Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter

b gb g L N b gb gb g Chapter 13 Problem Solutions Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter Semicouct hysics Devices: Bsic riciples, r eitio Chpter Solutios ul rolem Solutios Chpter rolem Solutios Sketch Sketch p-chel JE Silico 9 4 e 0 0 5 0 0 4 7 ( ) 8850 579 ow 8 500 0059 l 0 5 0 0884 so 579

More information

Repeated Root and Common Root

Repeated Root and Common Root Repeted Root d Commo Root 1 (Method 1) Let α, β, γ e the roots of p(x) x + x + 0 (1) The α + β + γ 0, αβ + βγ + γα, αβγ - () (α - β) (α + β) - αβ (α + β) [ (βγ + γα)] + [(α + β) + γ (α + β)] +γ (α + β)

More information

X-Ray Notes, Part III

X-Ray Notes, Part III oll 6 X-y oe 3: Pe X-Ry oe, P III oe Deeo Coe oupu o x-y ye h look lke h: We efe ue of que lhly ffee efo h ue y ovk: Co: C ΔS S Sl o oe Ro: SR S Co o oe Ro: CR ΔS C SR Pevouly, we ee he SR fo ye hv pxel

More information

176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s

176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s A g la di ou s F. L. 462 E l ec tr on ic D ev el op me nt A i ng er A.W.S. 371 C. A. M. A l ex an de r 236 A d mi ni st ra ti on R. H. (M rs ) A n dr ew s P. V. 326 O p ti ca l Tr an sm is si on A p ps

More information

Parameter Estimation and Hypothesis Testing of Two Negative Binomial Distribution Population with Missing Data

Parameter Estimation and Hypothesis Testing of Two Negative Binomial Distribution Population with Missing Data Avlble ole wwwsceceeccom Physcs Poce 0 475 480 0 Ieol Cofeece o Mecl Physcs Bomecl ee Pmee smo Hyohess es of wo Neve Boml Dsbuo Poulo wh Mss D Zhwe Zho Collee of MhemcsJl Noml UvesyS Ch zhozhwe@6com Absc

More information

University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences.

University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences. Uversty of Clfor t Berkeley College of Egeerg et. of Electrcl Egeerg Comuter Sceces EE 5 Mterm I Srg 6 Prof. Mg C. u Feb. 3, 6 Gueles Close book otes. Oe-ge formto sheet llowe. There re some useful formuls

More information

Decompression diagram sampler_src (source files and makefiles) bin (binary files) --- sh (sample shells) --- input (sample input files)

Decompression diagram sampler_src (source files and makefiles) bin (binary files) --- sh (sample shells) --- input (sample input files) . Iroduco Probblsc oe-moh forecs gudce s mde b 50 esemble members mproved b Model Oupu scs (MO). scl equo s mde b usg hdcs d d observo d. We selec some prmeers for modfg forecs o use mulple regresso formul.

More information

Calculation of Effective Resonance Integrals

Calculation of Effective Resonance Integrals Clculo of ffecve Resoce egrls S.B. Borzkov FLNP JNR Du Russ Clculo of e effecve oce egrl wc cludes e rel eerg deedece of euro flux des d correco o e euro cure e smle s eeded for ccure flux deermo d euro

More information

T h e C S E T I P r o j e c t

T h e C S E T I P r o j e c t T h e P r o j e c t T H E P R O J E C T T A B L E O F C O N T E N T S A r t i c l e P a g e C o m p r e h e n s i v e A s s es s m e n t o f t h e U F O / E T I P h e n o m e n o n M a y 1 9 9 1 1 E T

More information

March Algebra 2 Question 1. March Algebra 2 Question 1

March Algebra 2 Question 1. March Algebra 2 Question 1 March Algebra 2 Question 1 If the statement is always true for the domain, assign that part a 3. If it is sometimes true, assign it a 2. If it is never true, assign it a 1. Your answer for this question

More information

Differential Equation of Eigenvalues for Sturm Liouville Boundary Value Problem with Neumann Boundary Conditions

Differential Equation of Eigenvalues for Sturm Liouville Boundary Value Problem with Neumann Boundary Conditions Ierol Reserc Jorl o Aled d Bsc Sceces 3 Avlle ole www.rjs.co ISSN 5-838X / Vol 4 : 997-33 Scece Exlorer Plcos Derel Eqo o Eevles or Sr Lovlle Bodry Vle Prole w Ne Bodry Codos Al Kll Gold Dere o Mecs Azr

More information

-HYBRID LAPLACE TRANSFORM AND APPLICATIONS TO MULTIDIMENSIONAL HYBRID SYSTEMS. PART II: DETERMINING THE ORIGINAL

-HYBRID LAPLACE TRANSFORM AND APPLICATIONS TO MULTIDIMENSIONAL HYBRID SYSTEMS. PART II: DETERMINING THE ORIGINAL UPB Sc B See A Vo 72 I 3 2 ISSN 223-727 MUTIPE -HYBRID APACE TRANSORM AND APPICATIONS TO MUTIDIMENSIONA HYBRID SYSTEMS PART II: DETERMININ THE ORIINA Ve PREPEIŢĂ Te VASIACHE 2 Ace co copeeă oă - pce he

More information

(1) Cov(, ) E[( E( ))( E( ))]

(1) Cov(, ) E[( E( ))( E( ))] Impac of Auocorrelao o OLS Esmaes ECON 3033/Evas Cosder a smple bvarae me-seres model of he form: y 0 x The four key assumpos abou ε hs model are ) E(ε ) = E[ε x ]=0 ) Var(ε ) =Var(ε x ) = ) Cov(ε, ε )

More information

Laplace Transform. Definition of Laplace Transform: f(t) that satisfies The Laplace transform of f(t) is defined as.

Laplace Transform. Definition of Laplace Transform: f(t) that satisfies The Laplace transform of f(t) is defined as. Lplce Trfor The Lplce Trfor oe of he hecl ool for olvg ordry ler dfferel equo. - The hoogeeou equo d he prculr Iegrl re olved oe opero. - The Lplce rfor cover he ODE o lgerc eq. σ j ple do. I he pole o

More information

OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9

OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9 OH BOY! O h Boy!, was or igin a lly cr eat ed in F r en ch an d was a m a jor s u cc ess on t h e Fr en ch st a ge f or young au di enc es. It h a s b een s een by ap pr ox i ma t ely 175,000 sp ect at

More information

this is the indefinite integral Since integration is the reverse of differentiation we can check the previous by [ ]

this is the indefinite integral Since integration is the reverse of differentiation we can check the previous by [ ] Atervtves The Itegrl Atervtves Ojectve: Use efte tegrl otto for tervtves. Use sc tegrto rules to f tervtves. Aother mportt questo clculus s gve ervtve f the fucto tht t cme from. Ths s the process kow

More information

THIS PAGE DECLASSIFIED IAW EO 12958

THIS PAGE DECLASSIFIED IAW EO 12958 THIS PAGE DECLASSIFIED IAW EO 2958 THIS PAGE DECLASSIFIED IAW EO 2958 THIS PAGE DECLASSIFIED IAW E0 2958 S T T T I R F R S T Exhb e 3 9 ( 66 h Bm dn ) c f o 6 8 b o d o L) B C = 6 h oup C L) TO d 8 f f

More information

Technical Appendix for Inventory Management for an Assembly System with Product or Component Returns, DeCroix and Zipkin, Management Science 2005.

Technical Appendix for Inventory Management for an Assembly System with Product or Component Returns, DeCroix and Zipkin, Management Science 2005. Techc Appedx fo Iveoy geme fo Assemy Sysem wh Poduc o Compoe eus ecox d Zp geme Scece 2005 Lemm µ µ s c Poof If J d µ > µ he ˆ 0 µ µ µ µ µ µ µ µ Sm gumes essh he esu f µ ˆ > µ > µ > µ o K ˆ If J he so

More information

Interval Estimation. Consider a random variable X with a mean of X. Let X be distributed as X X

Interval Estimation. Consider a random variable X with a mean of X. Let X be distributed as X X ECON 37: Ecoomercs Hypohess Tesg Iervl Esmo Wh we hve doe so fr s o udersd how we c ob esmors of ecoomcs reloshp we wsh o sudy. The queso s how comforble re we wh our esmors? We frs exme how o produce

More information

Chapter 5 Transient Analysis

Chapter 5 Transient Analysis hpr 5 rs Alyss Jsug Jg ompl rspos rs rspos y-s rspos m os rs orr co orr Dffrl Equo. rs Alyss h ffrc of lyss of crcus wh rgy sorg lms (ucors or cpcors) & m-ryg sgls wh rss crcus s h h quos rsulg from r

More information

Hidden Markov Model. a ij. Observation : O1,O2,... States in time : q1, q2,... All states : s1, s2,..., sn

Hidden Markov Model. a ij. Observation : O1,O2,... States in time : q1, q2,... All states : s1, s2,..., sn Hdden Mrkov Model S S servon : 2... Ses n me : 2... All ses : s s2... s 2 3 2 3 2 Hdden Mrkov Model Con d Dscree Mrkov Model 2 z k s s s s s s Degree Mrkov Model Hdden Mrkov Model Con d : rnson roly from

More information

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o u l d a l w a y s b e t a k e n, i n c l u d f o l

More information

Continuous Time Markov Chains

Continuous Time Markov Chains Couous me Markov chas have seay sae probably soluos f a oly f hey are ergoc, us lke scree me Markov chas. Fg he seay sae probably vecor for a couous me Markov cha s o more ffcul ha s he scree me case,

More information

Homework 2 solutions

Homework 2 solutions Sectio 2.1: Ex 1,3,6,11; AP 1 Sectio 2.2: Ex 3,4,9,12,14 Homework 2 solutios 1. Determie i ech uctio hs uique ixed poit o the speciied itervl. gx = 1 x 2 /4 o [0,1]. g x = -x/2, so g is cotiuous d decresig

More information

Optimality of Strategies for Collapsing Expanded Random Variables In a Simple Random Sample Ed Stanek

Optimality of Strategies for Collapsing Expanded Random Variables In a Simple Random Sample Ed Stanek Optmlt of Strteges for Collpsg Expe Rom Vrles Smple Rom Smple E Stek troucto We revew the propertes of prectors of ler comtos of rom vrles se o rom vrles su-spce of the orgl rom vrles prtculr, we ttempt

More information

P a g e 5 1 of R e p o r t P B 4 / 0 9

P a g e 5 1 of R e p o r t P B 4 / 0 9 P a g e 5 1 of R e p o r t P B 4 / 0 9 J A R T a l s o c o n c l u d e d t h a t a l t h o u g h t h e i n t e n t o f N e l s o n s r e h a b i l i t a t i o n p l a n i s t o e n h a n c e c o n n e

More information

Instruction Sheet COOL SERIES DUCT COOL LISTED H NK O. PR D C FE - Re ove r fro e c sed rea. I Page 1 Rev A

Instruction Sheet COOL SERIES DUCT COOL LISTED H NK O. PR D C FE - Re ove r fro e c sed rea. I Page 1 Rev A Instruction Sheet COOL SERIES DUCT COOL C UL R US LISTED H NK O you or urc s g t e D C t oroug y e ore s g / as e OL P ea e rea g product PR D C FE RES - Re ove r fro e c sed rea t m a o se e x o duct

More information

Numerical Methods using the Successive Approximations for the Solution of a Fredholm Integral Equation

Numerical Methods using the Successive Approximations for the Solution of a Fredholm Integral Equation ece Advce Appled d eorecl ec uercl eod u e Succeve Approo or e Soluo o Fredol Ierl Equo AIA OBIŢOIU epre o ec d opuer Scece Uvery o Peroş Uvery Sree 6 Peroş OAIA rdorou@yoo.co Arc: pper pree wo eod or

More information

Nonlocal Boundary Value Problem for Nonlinear Impulsive q k Symmetric Integrodifference Equation

Nonlocal Boundary Value Problem for Nonlinear Impulsive q k Symmetric Integrodifference Equation OSR ol o Mec OSR-M e-ssn: 78-578 -SSN: 9-765X Vole e Ve M - A 7 PP 95- wwwojolog Nolocl Bo Vle Poble o Nole lve - Sec egoeece Eo Log Ceg Ceg Ho * Yeg He ee o Mec Yb Uve Yj PR C Abc: A oe ole lve egoeece

More information

ONE APPROACH FOR THE OPTIMIZATION OF ESTIMATES CALCULATING ALGORITHMS A.A. Dokukin

ONE APPROACH FOR THE OPTIMIZATION OF ESTIMATES CALCULATING ALGORITHMS A.A. Dokukin Iero Jor "Iforo Theore & co" Vo 463 ONE PPROH FOR THE OPTIIZTION OF ETITE UTING GORITH Do rc: I h rce he ew roch for ozo of eo ccg gorh ggeed I c e ed for fdg he correc gorh of coexy he coex of gerc roch

More information

H STO RY OF TH E SA NT

H STO RY OF TH E SA NT O RY OF E N G L R R VER ritten for the entennial of th e Foundin g of t lair oun t y on ay 8 82 Y EEL N E JEN K RP O N! R ENJ F ] jun E 3 1 92! Ph in t ed b y h e t l a i r R ep u b l i c a n O 4 1922

More information

5 Operations on Multiple Random Variables

5 Operations on Multiple Random Variables EE360 Random Signal analysis Chapter 5: Operations on Multiple Random Variables 5 Operations on Multiple Random Variables Expected value of a function of r.v. s Two r.v. s: ḡ = E[g(X, Y )] = g(x, y)f X,Y

More information

Asymptotic Dominance Problems. is not constant but for n 0, f ( n) 11. 0, so that for n N f

Asymptotic Dominance Problems. is not constant but for n 0, f ( n) 11. 0, so that for n N f Asymptotc Domce Prolems Dsply ucto : N R tht s Ο( ) ut s ot costt 0 = 0 The ucto ( ) = > 0 s ot costt ut or 0, ( ) Dee the relto " " o uctos rom N to R y g d oly = Ο( g) Prove tht s relexve d trstve (Recll:

More information

A L A BA M A L A W R E V IE W

A L A BA M A L A W R E V IE W A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N

More information

Example: Two Stochastic Process u~u[0,1]

Example: Two Stochastic Process u~u[0,1] Co o Slo o Coco S Sh EE I Gholo h@h. ll Sochc Slo Dc Slo l h PLL c Mo o coco w h o c o Ic o Co B P o Go E A o o Po o Th h h o q o ol o oc o lco q ccc lco l Bc El: Uo Dbo Ucol Sl Ab bo col l G col G col

More information

Chapter 2. Review of Hydrodynamics and Vector Analysis

Chapter 2. Review of Hydrodynamics and Vector Analysis her. Ree o Hdrodmcs d Vecor Alss. Tlor seres L L L L ' ' L L " " " M L L! " ' L " ' I s o he c e romed he Tlor seres. O he oher hd ' " L . osero o mss -dreco: L L IN ] OUT [mss l [mss l] mss ccmled h me

More information

P-Convexity Property in Musielak-Orlicz Function Space of Bohner Type

P-Convexity Property in Musielak-Orlicz Function Space of Bohner Type J N Sce & Mh Res Vol 3 No (7) -7 Alble ole h://orlwlsogocd/deh/sr P-Coey Proery Msel-Orlcz Fco Sce o Boher ye Yl Rodsr Mhecs Edco Deree Fcly o Ss d echology Uerss sl Neger Wlsogo Cerl Jdoes Absrcs Corresodg

More information

Solution: f( 1) = 3 1)

Solution: f( 1) = 3 1) Gateway Questions How to Evaluate Functions at a Value Using the Rules Identify the independent variable in the rule of function. Replace the independent variable with big parenthesis. Plug in the input

More information

UNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006)

UNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006) UNIVERSITY OF BRISTOL Exmitio for the Degrees of B.Sc. d M.Sci. (Level C/4) ANALYSIS B, SOLUTIONS MATH 6 (Pper Code MATH-6) My/Jue 25, hours 3 miutes This pper cotis two sectios, A d B. Plese use seprte

More information

Ionization Energies in Si, Ge, GaAs

Ionization Energies in Si, Ge, GaAs Izt erges S, Ge, GAs xtrsc Semcuctrs A extrsc semcuctrs s efe s semcuctr whch ctrlle muts f secfc t r murty tms hve bee e s tht the thermlequlbrum electr hle ccetrt re fferet frm the trsc crrer ccetrt.

More information

Math 61CM - Solutions to homework 9

Math 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 information

AC 2-3 AC 1-1 AC 1-2 CO2 AC 1-3 T CO2 CO2 F ES S I O N RY WO M No.

AC 2-3 AC 1-1 AC 1-2 CO2 AC 1-3 T CO2 CO2 F ES S I O N RY WO M No. SHEE OES. OVE PCE HOSS SSOCE PPUCES. VE EW CORO WR. S SE EEVO S EXS. 2. EW SSORS CCOS. S SE EEVO S HOSS. C 2-3 C - C -2 C 2- C -3 C 4- C 2-2 P SUB pproved Filename: :\\2669 RP Performing rts Center HVC\6-C\s\2669-3.dwg

More information

On Several Inequalities Deduced Using a Power Series Approach

On Several Inequalities Deduced Using a Power Series Approach It J Cotemp Mth Sceces, Vol 8, 203, o 8, 855-864 HIKARI Ltd, wwwm-hrcom http://dxdoorg/02988/jcms2033896 O Severl Iequltes Deduced Usg Power Seres Approch Lored Curdru Deprtmet of Mthemtcs Poltehc Uversty

More information

". :'=: "t',.4 :; :::-':7'- --,r. "c:"" --; : I :. \ 1 :;,'I ~,:-._._'.:.:1... ~~ \..,i ... ~.. ~--~ ( L ;...3L-. ' f.':... I. -.1;':'.

. :'=: t',.4 :; :::-':7'- --,r. c: --; : I :. \ 1 :;,'I ~,:-._._'.:.:1... ~~ \..,i ... ~.. ~--~ ( L ;...3L-. ' f.':... I. -.1;':'. = 47 \ \ L 3L f \ / \ L \ \ j \ \ 6! \ j \ / w j / \ \ 4 / N L5 Dm94 O6zq 9 qmn j!!! j 3DLLE N f 3LLE Of ADL!N RALROAD ORAL OR AL AOAON N 5 5 D D 9 94 4 E ROL 2LL RLLAY RL AY 3 ER OLLL 832 876 8 76 L A

More information

Linford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4)

Linford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4) Liford 1 Kyle Liford Mth 211 Hoors Project Theorems to Alyze: Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl (A4) Theorem 2.8 The Squeeze Theorem (A5) Theorem 2.9 The Limit of Si(x)/x = 1 (p. 85) Theorem

More information

Week 8 Lecture 3: Problems 49, 50 Fourier analysis Courseware pp (don t look at French very confusing look in the Courseware instead)

Week 8 Lecture 3: Problems 49, 50 Fourier analysis Courseware pp (don t look at French very confusing look in the Courseware instead) Week 8 Lecure 3: Problems 49, 5 Fourier lysis Coursewre pp 6-7 (do look Frech very cofusig look i he Coursewre ised) Fourier lysis ivolves ddig wves d heir hrmoics, so i would hve urlly followed fer he

More information

1. Consider an economy of identical individuals with preferences given by the utility function

1. Consider an economy of identical individuals with preferences given by the utility function CO 755 Problem Se e Cbrer. Cosder ecoomy o decl dduls wh reereces e by he uly uco U l l Pre- rces o ll hree oods re ormled o oe. Idduls suly ood lbor < d cosume oods d. The oerme c mose d lorem es o oods

More information

I n t e r n a t i o n a l E l e c t r o n i c J o u r n a l o f E l e m e n t a r y E.7 d u, c ai ts is ou n e, 1 V3 1o-2 l6, I n t h i s a r t

I n t e r n a t i o n a l E l e c t r o n i c J o u r n a l o f E l e m e n t a r y E.7 d u, c ai ts is ou n e, 1 V3 1o-2 l6, I n t h i s a r t I n t e r n a t i o n a l E l e c t r o n i c J o ue rlne am l e not fa r y E d u c a t i o n, 2 0 1 4, 1 37-2 ( 16 ). H o w R e a d i n g V o l u m e A f f e c t s b o t h R e a d i n g F l u e n c y

More information

Presentation Problems 5

Presentation Problems 5 Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

More information

Solutions Manual for Polymer Science and Technology Third Edition

Solutions Manual for Polymer Science and Technology Third Edition Solutos ul for Polymer Scece d Techology Thrd Edto Joel R. Fred Uer Sddle Rver, NJ Bosto Idols S Frcsco New York Toroto otrel Lodo uch Prs drd Cetow Sydey Tokyo Sgore exco Cty Ths text s ssocted wth Fred/Polymer

More information

1. Linear second-order circuits

1. Linear second-order circuits ear eco-orer crcut Sere R crcut Dyamc crcut cotag two capactor or two uctor or oe uctor a oe capactor are calle the eco orer crcut At frt we coer a pecal cla of the eco-orer crcut, amely a ere coecto of

More information

We will begin by supplying the proof to (a).

We will begin by supplying the proof to (a). (The solutios of problem re mostly from Jeffrey Mudrock s HWs) Problem 1. There re three sttemet from Exmple 5.4 i the textbook for which we will supply proofs. The sttemets re the followig: () The spce

More information

a n+2 a n+1 M n a 2 a 1. (2)

a 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 information

Introduction to Laplace Transforms October 25, 2017

Introduction to Laplace Transforms October 25, 2017 Iroduco o Lplc Trform Ocobr 5, 7 Iroduco o Lplc Trform Lrr ro Mchcl Egrg 5 Smr Egrg l Ocobr 5, 7 Oul Rvw l cl Wh Lplc rform fo of Lplc rform Gg rform b gro Fdg rform d vr rform from bl d horm pplco o dffrl

More information

f(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that

f(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that Uiversity of Illiois t Ur-Chmpig Fll 6 Mth 444 Group E3 Itegrtio : correctio of the exercises.. ( Assume tht f : [, ] R is cotiuous fuctio such tht f(x for ll x (,, d f(tdt =. Show tht f(x = for ll x [,

More information

Continuous Random Variables

Continuous Random Variables 1 / 24 Continuous Random Variables Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay February 27, 2013 2 / 24 Continuous Random Variables

More information

Lecture 23 - Frequency Resp onse of Amplifiers (I) Common-Source Amplifier. May 6, 2003

Lecture 23 - Frequency Resp onse of Amplifiers (I) Common-Source Amplifier. May 6, 2003 6.0 Microelectronic Devices and Circuits Spring 003 Lecture 3 Lecture 3 Frequency Resp onse of Amplifiers (I) CommonSource Amplifier May 6, 003 Contents:. Intro duction. Intrinsic frequency resp onse of

More information

The Poisson Process Properties of the Poisson Process

The Poisson Process Properties of the Poisson Process Posso Processes Summary The Posso Process Properes of he Posso Process Ierarrval mes Memoryless propery ad he resdual lfeme paradox Superposo of Posso processes Radom seleco of Posso Pos Bulk Arrvals ad

More information

The Basic Properties of the Integral

The Basic Properties of the Integral The Bsic Properties of the Itegrl Whe we compute the derivtive of complicted fuctio, like x + six, we usully use differetitio rules, like d [f(x)+g(x)] d f(x)+ d g(x), to reduce the computtio dx dx dx

More information

SOLUTIONS TO CONCEPTS CHAPTER 6

SOLUTIONS TO CONCEPTS CHAPTER 6 SOLUIONS O CONCEPS CHAPE 6 1. Let ss of the block ro the freebody digr, 0...(1) velocity Agin 0 (fro (1)) g 4 g 4/g 4/10 0.4 he co-efficient of kinetic friction between the block nd the plne is 0.4. Due

More information

Analysis of the Preference Shift of. Customer Brand Selection. and Its Matrix Structure. -Expansion to the second order lag

Analysis of the Preference Shift of. Customer Brand Selection. and Its Matrix Structure. -Expansion to the second order lag Jourl of Compuo & Modellg vol. o. 6-9 ISS: 79-76 (pr) 79-88 (ole) Scepre Ld l of he Preferece Shf of Cuomer Brd Seleco d I Mr Srucure -Epo o he ecod order lg Kuhro Teu rc I ofe oerved h coumer elec he

More information

CH 39 USING THE GCF TO REDUCE FRACTIONS

CH 39 USING THE GCF TO REDUCE FRACTIONS 359 CH 39 USING THE GCF TO EDUCE FACTIONS educig Algeric Frctios M ost of us lered to reduce rithmetic frctio dividig the top d the ottom of the frctio the sme (o-zero) umer. For exmple, 30 30 5 75 75

More information

8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before

8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before 8.1 Arc Legth Wht is the legth of curve? How c we pproximte it? We could do it followig the ptter we ve used efore Use sequece of icresigly short segmets to pproximte the curve: As the segmets get smller

More information

ICS141: Discrete Mathematics for Computer Science I

ICS141: Discrete Mathematics for Computer Science I Uversty o Hw ICS: Dscrete Mthemtcs or Computer Scece I Dept. Iormto & Computer Sc., Uversty o Hw J Stelovsy bsed o sldes by Dr. Be d Dr. Stll Orgls by Dr. M. P. Fr d Dr. J.L. Gross Provded by McGrw-Hll

More information

_ J.. C C A 551NED. - n R ' ' t i :. t ; . b c c : : I I .., I AS IEC. r '2 5? 9

_ J.. C C A 551NED. - n R ' ' t i :. t ; . b c c : : I I .., I AS IEC. r '2 5? 9 C C A 55NED n R 5 0 9 b c c \ { s AS EC 2 5? 9 Con 0 \ 0265 o + s ^! 4 y!! {! w Y n < R > s s = ~ C c [ + * c n j R c C / e A / = + j ) d /! Y 6 ] s v * ^ / ) v } > { ± n S = S w c s y c C { ~! > R = n

More information

Integral Equations and their Relationship to Differential Equations with Initial Conditions

Integral Equations and their Relationship to Differential Equations with Initial Conditions Scece Refleco SR Vol 6 wwwscecereflecocom Geerl Leers Mhemcs GLM 6 3-3 Geerl Leers Mhemcs GLM Wese: hp://wwwscecereflecocom/geerl-leers--mhemcs/ Geerl Leers Mhemcs Scece Refleco Iegrl Equos d her Reloshp

More information

Lecture 3 summary. C4 Lecture 3 - Jim Libby 1

Lecture 3 summary. C4 Lecture 3 - Jim Libby 1 Lecue su Fes of efeece Ivce ude sfoos oo of H wve fuco: d-fucos Eple: e e - µ µ - Agul oeu s oo geeo Eule gles Geec slos cosevo lws d Noehe s heoe C4 Lecue - Lbb Fes of efeece Cosde fe of efeece O whch

More information

Chapter 13 Problem Solutions Computer Simulation Computer Simulation ma/ V 80. r I (120)(0.026)

Chapter 13 Problem Solutions Computer Simulation Computer Simulation ma/ V 80. r I (120)(0.026) Chapter 3 Pblem lutions 3. Computer Simulation 3. Computer Simulation 3.3 (a) ( Ri) g 0 C m T 0.0 r 80 o MΩ C 0 r 80 o MΩ C 0 0.79 m/ Ri + ( + βn) R 7 (0)(0.0) 7 5. kω 0. BE ( on) 0. C 0.030 m R 0 (0)(0.0)

More information

Area and the Definite Integral. Area under Curve. The Partition. y f (x) We want to find the area under f (x) on [ a, b ]

Area and the Definite Integral. Area under Curve. The Partition. y f (x) We want to find the area under f (x) on [ a, b ] Are d the Defte Itegrl 1 Are uder Curve We wt to fd the re uder f (x) o [, ] y f (x) x The Prtto We eg y prttog the tervl [, ] to smller su-tervls x 0 x 1 x x - x -1 x 1 The Bsc Ide We the crete rectgles

More information

In order to ensure that an overall development in service by those. of total. rel:rtins lo the wapris are

In order to ensure that an overall development in service by those. of total. rel:rtins lo the wapris are AhAY ggkhu e evue he eve us wch my be eese s esu eucs ese mbes buges hve bee ke cvu vs. Css e vse e' he w m ceges cec ec css. Dec Dgqs_1q W qge5ee.pe_s_ v V cuss ke ecy 1 hc huse ees. bse cu wc esb shmes.

More information

Years. Marketing without a plan is like navigating a maze; the solution is unclear.

Years. Marketing without a plan is like navigating a maze; the solution is unclear. F Q 2018 E Mk l lk z; l l Mk El M C C 1995 O Y O S P R j lk q D C Dl Off P W H S P W Sl M Y Pl Cl El M Cl FIRST QUARTER 2018 E El M & D I C/O Jff P RGD S C D M Sl 57 G S Alx ON K0C 1A0 C Tl: 6134821159

More information

: i; ii: i:i: I. : q t:iig l3 [i E. i;issii:i::: r p. ! s:r;e;:e;!f. as Iet5;Fgi. i EiiF;:'+3EI. : :*gsc:li-ii. ; si;;ei:i:g; .

: i; ii: i:i: I. : q t:iig l3 [i E. i;issii:i::: r p. ! s:r;e;:e;!f. as Iet5;Fgi. i EiiF;:'+3EI. : :*gsc:li-ii. ; si;;ei:i:g; . , / - H f ) - $ H A - --" G \/ - - f \f ff P) G T - ), - ) -..R '' (.) f- w (, AA - Нотная библиотека \ S R.- R \?! -! h )! - X* - Y - J. ** " { B e B Rk * f [He 1e [Y!!-- e *J! *Jee.1!.,1g -!. -! >. j.,

More information

Science & Technologies GENERAL BIRTH-DEATH PROCESS AND SOME OF THEIR EM (EXPETATION- MAXIMATION) ALGORITHM

Science & Technologies GENERAL BIRTH-DEATH PROCESS AND SOME OF THEIR EM (EXPETATION- MAXIMATION) ALGORITHM GEERAL BIRH-EAH ROCESS A SOME OF HEIR EM EXEAIO- MAXIMAIO) ALGORIHM Il Hl, Lz Ker, Ylldr Seer Se ery o eoo,, eoo Mcedo l.hl@e.ed.; lz.er@e.ed.; ylldr_@hol.co ABSRAC Brh d deh roce coo-e Mrco ch, h odel

More information

Riemann Integral and Bounded function. Ng Tze Beng

Riemann Integral and Bounded function. Ng Tze Beng Riem Itegrl d Bouded fuctio. Ng Tze Beg I geerlistio of re uder grph of fuctio, it is ormlly ssumed tht the fuctio uder cosidertio e ouded. For ouded fuctio, the rge of the fuctio is ouded d hece y suset

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

NEW FLOODWAY (CLOMR) TE TE PIN: GREENS OF ROCK HILL, LLC DB: 12209, PG: ' S67 46'18"E APPROX. FLOODWAY NEW BASE FLOOD (CLOMR)

NEW FLOODWAY (CLOMR) TE TE PIN: GREENS OF ROCK HILL, LLC DB: 12209, PG: ' S67 46'18E APPROX. FLOODWAY NEW BASE FLOOD (CLOMR) W LOOWY (LOMR) RVRWLK PKWY ROK HLL, S PPROX. LOOWY W BS LOO (LOMR) lient nformation 4 SS- RM:4 V : PV Pipe V OU: PV Pipe JB SS- RM: V OU: PV Pipe RU R " PV Pipe @. LO SPS OL SSBL GRL ORMO: S OS: M BS LOO

More information

Physics 232 Exam I Feb. 13, 2006

Physics 232 Exam I Feb. 13, 2006 Phsics I Fe. 6 oc. ec # Ne..5 g ss is ched o hoizol spig d is eecuig siple hoic oio. The oio hs peiod o.59 secods. iiil ie i is oud o e 8.66 c o he igh o he equiliiu posiio d oig o he le wih eloci o sec.

More information

AGENDA REPORT. Payroll listing conforms to the approved budget except as noted and has been paid WILLIAM A HUSTON CITY MANAGER

AGENDA REPORT. Payroll listing conforms to the approved budget except as noted and has been paid WILLIAM A HUSTON CITY MANAGER Age e 4 AGEDA RERT Reewe ge Fce Dec EETG DATE Al 2 2 T FR A A T T AAGER AEA ARED KG FAE DRETR BET RATFAT F AR AR The cl h e he e f Gee e ec 728 eee he e f f T Reeele Agec blg h e ccce wh he e bge ce e

More information

Chapter Simpson s 1/3 Rule of Integration. ( x)

Chapter Simpson s 1/3 Rule of Integration. ( x) Cper 7. Smpso s / Rule o Iegro Aer redg s per, you sould e le o. derve e ormul or Smpso s / rule o egro,. use Smpso s / rule o solve egrls,. develop e ormul or mulple-segme Smpso s / rule o egro,. use

More information

ANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists.

ANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists. ANALYSIS HW 3 CLAY SHONKWILER () Fid ll smooth fuctios f : R R with the property f(x + y) = f(x) + f(y) for ll rel x, y. Demostrtio: Let f be such fuctio. Sice f is smooth, f exists. The The f f(x + h)

More information

Parametric Methods. Autoregressive (AR) Moving Average (MA) Autoregressive - Moving Average (ARMA) LO-2.5, P-13.3 to 13.4 (skip

Parametric Methods. Autoregressive (AR) Moving Average (MA) Autoregressive - Moving Average (ARMA) LO-2.5, P-13.3 to 13.4 (skip Pmeti Methods Autoegessive AR) Movig Avege MA) Autoegessive - Movig Avege ARMA) LO-.5, P-3.3 to 3.4 si 3.4.3 3.4.5) / Time Seies Modes Time Seies DT Rdom Sig / Motivtio fo Time Seies Modes Re the esut

More information

THIS PAGE DECLASSIFIED IAW E

THIS PAGE DECLASSIFIED IAW E THS PAGE DECLASSFED AW E0 2958 BL K THS PAGE DECLASSFED AW E0 2958 THS PAGE DECLASSFED AW E0 2958 B L K THS PAGE DECLASSFED AW E0 2958 THS PAGE DECLASSFED AW EO 2958 THS PAGE DECLASSFED AW EO 2958 THS

More information

Pre-Calculus - Chapter 3 Sections Notes

Pre-Calculus - Chapter 3 Sections Notes Pre-Clculus - Chpter 3 Sectios 3.1-3.4- Notes Properties o Epoets (Review) 1. ( )( ) = + 2. ( ) =, (c) = 3. 0 = 1 4. - = 1/( ) 5. 6. c Epoetil Fuctios (Sectio 3.1) Deiitio o Epoetil Fuctios The uctio deied

More information

Section Properties of Rational Expressions

Section Properties of Rational Expressions 88 Section. - Properties of Rational Expressions Recall that a rational number is any number that can be written as the ratio of two integers where the integer in the denominator cannot be. Rational Numbers:

More information

Software Process Models there are many process model s in th e li t e ra t u re, s om e a r e prescriptions and some are descriptions you need to mode

Software Process Models there are many process model s in th e li t e ra t u re, s om e a r e prescriptions and some are descriptions you need to mode Unit 2 : Software Process O b j ec t i ve This unit introduces software systems engineering through a discussion of software processes and their principal characteristics. In order to achieve the desireable

More information

In Calculus I you learned an approximation method using a Riemann sum. Recall that the Riemann sum is

In Calculus I you learned an approximation method using a Riemann sum. Recall that the Riemann sum is Mth Sprg 08 L Approxmtg Dete Itegrls I Itroducto We hve studed severl methods tht llow us to d the exct vlues o dete tegrls However, there re some cses whch t s ot possle to evlute dete tegrl exctly I

More information

MTH 146 Class 11 Notes

MTH 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 information

ELEC 6041 LECTURE NOTES WEEK 3 Dr. Amir G. Aghdam Concordia University

ELEC 6041 LECTURE NOTES WEEK 3 Dr. Amir G. Aghdam Concordia University ecre Noe Prepared b r G. ghda EE 64 ETUE NTE WEE r. r G. ghda ocorda Uer eceraled orol e - Whe corol heor appled o a e ha co of geographcall eparaed copoe or a e cog of a large ber of p-op ao ofe dered

More information

BINOMIAL THEOREM OBJECTIVE PROBLEMS in the expansion of ( 3 +kx ) are equal. Then k =

BINOMIAL THEOREM OBJECTIVE PROBLEMS in the expansion of ( 3 +kx ) are equal. Then k = wwwskshieduciocom BINOMIAL HEOREM OBJEIVE PROBLEMS he coefficies of, i e esio of k e equl he k /7 If e coefficie of, d ems i e i AP, e e vlue of is he coefficies i e,, 7 ems i e esio of e i AP he 7 7 em

More information

Chapter 3: Maximum-Likelihood & Bayesian Parameter Estimation (part 1)

Chapter 3: Maximum-Likelihood & Bayesian Parameter Estimation (part 1) Aoucemes Reags o E-reserves Proec roosal ue oay Parameer Esmao Bomercs CSE 9-a Lecure 6 CSE9a Fall 6 CSE9a Fall 6 Paer Classfcao Chaer 3: Mamum-Lelhoo & Bayesa Parameer Esmao ar All maerals hese sles were

More information

Stat 6863-Handout 5 Fundamentals of Interest July 2010, Maurice A. Geraghty

Stat 6863-Handout 5 Fundamentals of Interest July 2010, Maurice A. Geraghty S 6863-Hou 5 Fuels of Ieres July 00, Murce A. Gerghy The pror hous resse beef cl occurreces, ous, ol cls e-ulero s ro rbles. The fl copoe of he curl oel oles he ecooc ssupos such s re of reur o sses flo.

More information

The Reimann Integral is a formal limit definition of a definite integral

The Reimann Integral is a formal limit definition of a definite integral MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl

More information

SOLUTIONS 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 (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 information

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS Geometry Of The ircle Tngents & Secnts GEOMETRY OF THE IRLE TNGENTS & SENTS www.mthletics.com.u Tngents TNGENTS nd N Secnts SENTS Tngents nd secnts re lines tht strt outside circle. Tngent touches the

More information

MAC 1147 Final Exam Review

MAC 1147 Final Exam Review MAC 1147 Final Exam Review nstructions: The final exam will consist of 15 questions plu::; a bonus problem. Some questions will have multiple parts and others will not. Some questions will be multiple

More information

Abstract Key Words: 1 Introduction

Abstract Key Words: 1 Introduction f(x) x Ω MOP Ω R n f(x) = ( (x),..., f q (x)) f i : R n R i =,..., q max i=,...,q f i (x) Ω n P F P F + R + n f 3 f 3 f 3 η =., η =.9 ( 3 ) ( ) ( ) f (x) x + x min = min x Ω (x) x [ 5,] (x 5) + (x 5)

More information

Periodic Table of Elements. EE105 - Spring 2007 Microelectronic Devices and Circuits. The Diamond Structure. Electronic Properties of Silicon

Periodic Table of Elements. EE105 - Spring 2007 Microelectronic Devices and Circuits. The Diamond Structure. Electronic Properties of Silicon EE105 - Srg 007 Mcroelectroc Devces ad Crcuts Perodc Table of Elemets Lecture Semcoductor Bascs Electroc Proertes of Slco Slco s Grou IV (atomc umber 14) Atom electroc structure: 1s s 6 3s 3 Crystal electroc

More information

Differential Equations 2 Homework 5 Solutions to the Assigned Exercises

Differential 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 information