The axial dispersion model for tubular reactors at steady state can be described by the following equations: dc dz R n cn = 0 (1) (2) 1 d 2 c.

Size: px
Start display at page:

Download "The axial dispersion model for tubular reactors at steady state can be described by the following equations: dc dz R n cn = 0 (1) (2) 1 d 2 c."

Transcription

1 5.4 Applicatio of Perturbatio Methods to the Dispersio Model for Tubular Reactors The axial dispersio model for tubular reactors at steady state ca be described by the followig equatios: d c Pe dz z = 0 dc dz R c = 0 () = c dc () Pe dz z = dc dz = 0 (3) where c = C A C Ao = is the dimesioless reactat cocetratio z = Z L = is the dimesioless distace alog the reactor measured from the etrace Pe = UL D = is the axial dispersio Peclet umber L = total reactor legth U = Q A = is the liear mea velocity i the reactor (superficial velocity i packed beds) Q = volumetric flow rate through the reactor A = cross-sectioal area of the reactor D = dispersio coefficiet i the reactor (defied per total cross-sectioal area i packed beds). Note: If D is defied per cross-sectioal area uoccupied by solids i packed beds the Dε will appear i the Pe istead of D. ε is bed porosity. R = k τ C Ao - dimesioless reactio rate group (Damkohler umber). Note: I packed beds this implies that the rate was defied per uit reactor volume.

2 τ = L U - reactor space time (space time based o superficial velocity). Equatios (), () ad (3) describe the behavior of tubular reactors ad catalytic packed bed reactors uder isothermal coditios, at steady state for a -th order irreversible, sigle reactio. Applicatio of these equatios to packed bed reactors assumes that exteral ad iteral mass trasfer limitatios (i.e mass trasfer from fluid to pellets ad iside the pellets) are oexistet or have bee properly accouted for i the overall rate expressio. Solutios to eqs (-3) for a -th order reactio (( 0, ) ca be obtaied oly by umerical meas. However, the problem is a difficult oliear two-poit boudary value problem. [See: P.H. McGiis, Chem. Egr. Progr. Symp. Ser. No 55 Vol 6, p (968), Lee, E.S., AIChE J., 4(3), 490 (968), Lee, E.S., Quasiliearizatio (969)]. Sice i practical applicatios Pe umbers are quite large ( Pe 5), ad ofte Pe = O (0 ), it is of iterest to develop approximate solutios to the dispersio model for large Pe umbers. Remember, large Pe σ Pe meas small variace ad, hece, small variatio from plug flow, ad is exactly the coditio uder which the dispersio model is applicable. Such approximate solutios ca allow us to estimate well the departure from plug flow performace ad will save a lot of effort which is ecessary for umerical evaluatios of the model. We are iterested i large Pe, the Equatios (-3) ca be writte as: = ε ad ε is very small ε << Pe ( ). ε d c dz dc dz R c = 0 z = 0; = c ε dc dz () () z = ; dc dz = 0 (3) A. Outer Solutio Let us assume that a solutio of the followig form exists:

3 c = Fz ()= ε F ()= z F 0 ()+ z ε F ()+ z ε F ()+ z... (4) =0 If we ca fid such a solutio the, due to the fact that ε <<, we ca hope that the first few terms will be adequate to describe our solutio. [Strictly speakig this would be true oly if the series give by Eq (4) coverges fast. Remarkably, a good approximatio to the solutio is obtaied eve for diverget series! (For this ad further details, see Nayfeh, A.H., Perturbatio Methods, Wiley, 973) We do eed i Eq () a expressio for c. However, as we are goig to keep oly the first few terms of the series (4) raisig it to a -th power is ot that difficult. c = [ F o + ε F + ε F + ε 3 F 3 +..] = F 0 + ε F 0 F + ε ( )F 0 F + F o F +ε 3 6 ( ) ( )F 0 3 F 3 + ( )F 0 F F +F 0 F 3 + O ε 4 ( ) (5) Differetiate eq (4) oce ad twice ad substitute both derivatives ad eq (5) ito eq (): ε d F 0 dz + ε d F dz + ε d F dz + ε 3 d F 3 dz +... df 0 dz + ε df dz + df ε dz R F 0 + ε F 0 F + ε ( )F 0 F +F 0 F +... = 0 (6) Substitute also eq (4) ad its derivative ito eq (): At z = 0 = F 0 + ε F + ε F +... ε df 0 dz + ε df dz + ε df dz +... (7) Group ow the terms with equal powers of ε i eq (6) ad eq (7) together, ad require that they?? to zero: ε 0 : df 0 dz + R F 0 = 0 (8) 3

4 z = 0, F 0 = (8a) ε : df dz + R F 0 F = d F 0 (9) dz z = 0, F = df 0 dz (9a) ε : df dz + R F 0 F = d F R dz ( )F 0 F (0) ε 3 : etc. z = 0, F = df dz (0a) Notice at this poit that the differetial equatios (D.E.s) (eq 8-, etc.) are first order D.E.s, while the origial equatio () was secod order. Eq (8) is first order for F 0. Oce eq (8) is solved ad F 0 determied, the right had side of eq (9) is kow ad the left had side is a first order D.E. for F, etc. Thus, we ca successively determie all the F i s. However, because all of these are first order equatios, we ca oly satisfy oe of the origial boudary coditios. If we tried to satisfy the boudary coditio at the reactor exit, give by eq (3), we would have that df i dz = 0 at z = for all i, implyig that all F i s are costat due to the form of D.E.s ( eq 8- ). Thus, we would ot be able to get ay iformatio. This idicates that we have to satisfy the coditio at the reactor etrace give by eq () as idicated by eq (7). This results i a set of coditios give by eqs(8a - 0a). The solutio of eq (8) with I.C. (iitial coditio) (8a) is readily obtaied: [ ] F 0 ()= z + R ( )z () Verify that this ideed is a cocetratio profile i a plug flow reactor for a -th order reactio! Differetiate eq () twice, substitute ito eq (9) ad solve eq (9) with I.C. eq (9a): F z [ ( )z] ()= R + R l + R ( )z [ ] () Differetiate eq () twice ad substitute together with eq () ad eq () o the right had side of eq (0). Solve eq (0) with I.C. eq (0a): 4

5 F ()= z R where u = + R ( - ) z 3 u + u ( ) l u l u (3) (4) Now we have foud F 0, F, F ad we could write our solutio as F 0 + ε F + ε F. However, we realize that such a solutio oly satisfies the B.C. eq () at the reactor ilet ad does ot satisfy coditio eq (3) at the reactor outlet. Ultimately we are iterested ot i the cocetratio profile alog the reactor but i its value (cocetratio value) at the reactor exit. Sice the B.C. at the exit is ot satisfied we have a lot of reasos to doubt the validity of our solutio at the exit ad must fid ways of improvig it. I perturbatio theory the solutio give by eq (4) is called a "outer" solutio B. Ier Solutio c 0 ()= z F 0 ()+ z ε F ( z) + ε F ( z) + ε 3 F 3 ( z)+...c O outer solutio (5) We have see before that the reaso why the "outer" solutio caot satisfy the B.C. at the exit (z= ) give by eq (3) is that all the terms F i s of the "outer" solutio were obtaied from first order D.E.s. Now we must expad ("stretch") our coordiate system i such a way that we ca take a "closer" look at what happes i the very viciity (i the "boudary layer") ext to z =. Let η = z ε α α > 0 (6) Sice ε << eq (6) gives reasoably large values of η for z very close to. Thus, we are stretchig the coordiate system ear z =. By chai rule we have: dc dz = dc dη dη dz = ε α dc dη (7) d c dz = d c ε α dη (8) Substitutig eq (7) ad eq (8) ito eq () we get: 5

6 ε α d c dc + ε α dη dη R c = 0 (9) We require ow that the derivatives should stad by equal powers of ε i.e α = α ad plus that we will get a d order D.E. This gives α =. Hece, ad becomes: d c ε dη d c dη η = z ε + dc ε dη R c = 0 (0) + dc dη ε R c = 0 () We assume ow that the solutio to eq () is the "ier" solutio c i ad that it ca be represeted by a power series i ε : c i = = 0 ε c i = c 0 i + ε c i + ε c i + ε 3 c 3 i +... () Remember agai that we are developig the ier solutio i the very viciity of z =, so the "outer" solutio i that regio is so close to that it ca be represeted by a Taylor series expasio aroud z = i.e η = 0 c o ()= z F 0 ()+ F 0 ()z + ε F + ε F ( )+ F 0 ( )+ F ()+ F ()z ()+ F ()z ( )+ F ( ) ( z ) +... ( z ) +... ( z ) O ε3 ( ) (3) where F i = df i dz Now from eq (0) z = εη ; ( z ) = ε η (4) Substitute eq (4) ito eq (3) ad group together the terms with the same powers of ε : 6

7 c oi c oi c oi c o ()= z c oi η ()= () + ε F () F 0 ()η F 0 [ ] + ε F () F ()η + F 0 ()η (5) This is ofte called a "ier-outer expasio", c oi, i.e the expasio of the outer solutio i the ier (boudary) layer. Now the ier solutio give by the series represeted by eq () ca be represeted by the sum of two series: oe from the ier-outer expasio ad the other from the true ier solutio udetectable by the outer oe. c i η c i ()= ε =O = ε c oi η = O [ ()+ G () η ] = F 0 ()+ G 0 ()+ η ε F () F 0 + ε F [ ()η +G () η ] () F ()η + F 0 ()η +G () η +O( ε3) (6) Differetiate eq (6) oce ad twice, raise it to the -th power ad substitute everythig ito eq () ad require that the sum of the terms with equal powers of ε be zero: d G 0 dη + dg 0 dη = 0 (7) d G dη + dg dη = F 0 ()+ R F 0 [ ] (8) ()+G 0 d G dη + dg dη = F 0 ()+ F () F 0 ()η + R F 0 [ ] F ()+G 0 [ () F 0 ()η + G ] (9) This is a set of d order D.E.s for the G i s. Whe the G i s are determied successively the right had side of the above equatios is always kow. For d order equatios we eed B.C.s. Oe B.C. is obtaied from eq (3) i.e. which leads to dc i = 0 at η = 0 (30) dη 7

8 dg 0 dη = 0 at η = 0 (7a) dg dη = F 0 () at η = 0 (8a) dg dη = F () at η = 0 (9a) Remember that away from the boudary at z = we do ot sese or detect the G i s as F i s icely satisfy the B.C. at z = 0 ad are probably a good represetatio of the actual solutio for >z 0 but ot at z =. This implies that as we move away from z =, i.e as η icreases, all G i s go to zero. This is the d boudary coditio: η G i 0 for i = 0,,,3,... (7b,8b,9b) The solutio of eq (7) with eq (7a) is G 0 ( η)= 0 (3) The solutio of eq (8) with eq (8a) ad eq (8b) is: [ ] G ( η)= R + R ( ) e η (3) The solutio of eq (9) with eq (9a) ad eq (9b) is: [ ( )] G ()= R η + R 3+ η +l + R [ ( )] If we substitute eq (0) ito eqs (3-33) we will get G i s i terms of z. e η (33) The cocetratio profile close to the reactor exit is give by equatio (6) (where η ca be substituted i terms of z). We are especially iterested i the outflow cocetratio at the exit, i.e at z =, η= 0. Sice ε = /Pe we get: [ ( )] c exit = x A = F 0 ( z =)+ Pe F z = ( )+G η = 0 + [ F Pe ( z =)+G ( η = 0) ]+O Pe 3 (34) 8

9 Let ρ = [ + ( )R ] c exit = x A = ρ + Pe R ρ l ρ + (35) 3 Pe + ρ lρ + ρ + + O Pe 3 (36) After some additioal algebra: 9

10 c exit = ρ + Pe ρ R lρ ρ + R Pe lρ + ( ) ρ + R +O Pe 3 (37) where R = kt C 0 ad ρ = + ( )R Notice that the leadig term is a solutio for the plug flow reactor, thus as Pe, plug flow occurs. Straightforward algebra (although admittedly tedious), would produce higher order terms 3, etc. However, there is o eed for that. Experiece from fluid mechaics, heat Pe trasfer, solid mechaics ad reactio egieerig shows that perturbatio series are ofte ot uiformly coverget ad that d order perturbatio solutios, as the oe give by Eq (37), or eve first order perturbatio solutios (droppig the term with ) provide satisfactory Pe aswers which are ofte better at lower Pe tha higher order solutios. The perturbatio solutio for the dispersio model has bee worked out i detail by Burghardt ad Zasleski, Chem. Eg. Sci., 3, (968), but their secod order term cotais a error due to improper matchig of perturbatio solutios. See their paper for compariso of approximate perturbatio solutios with umerical solutios. Let us ow examie the coditios uder which the deviatio of the tubular reactor legth from the legth required by plug flow is less tha 5%. I other words we are lookig for the coditios uder which the tubular reactor, the legth of which is withi 5% of the PFR, would give the same coversio as the PFR. L L p L = L p 0.05 (38) L For a fixed give flow rate ( ) p L p L = τ R p τ = = R ( ) p ρ ρ 0

11 L p L = ρ ρ p ρ p ρ = ρ 0.05 (39) ρ ρ p 0.05 (40) ρ ρ I the plug flow reactor we have x Ap = ρ p (4) For the axial dispersio model we will use the first order perturbatio solutio: x A = ρ + Pe R lρ ρ = x Ap (4) Therefore: ρ p ρ =+ Pe R lρ ρ ρ p = + R lρ ρ Pe ρ (43) The usig eq. 40) we get ρ p = + R lρ ρ Pe ρ 0.05 (44) ρ + R lρ Pe ρ 0.05 (45) ρ For > + R lρ Pe ρ ρ ρ (46a)

12 For 0 < < + R lρ Pe ρ ρ ρ (46b) For > (From eq. (46a)) R lρ 0.05 Pe ρ ρ (47a) For 0 < < (From eq. (46b)) R lρ 0.05 Pe ρ ρ (47b) Multiply (47a) ad (47b) o both sides by (-). I (47a) - > 0 ad the iequality is preserved, i (47b) - < 0 ad whe multiplyig with the egative umber the iequality sig is reversed. Thus, for all,. Pe R lρ ρ 0.05 (48) ρ which gives Pe 0 lρ (49) Thus, whe Pe umber satisfies iequality (49) the differece i the legth of a tubular reactor which ca be described by the axial dispersio model ad plug flow reactor, both givig the same coversio, will be withi 5%. Eq. (49) may be writte as: I the last approximatio (usig x A ρ ) we get: Pe 0 lρ (50) Pe 0 l (5) x A

13 Formula (5) clearly shows that the magitude of the Pe umber which will guaratee small discrepacies betwee tubular reactors ad plug flow depeds ot oly o the reactio order but also o coversio! For packed beds Bo = Ud p D (5) Equatios (49) ad (5) become: L d p 0 Bo( ) lρ (53) L 0 d p Bo l (54) x A How log the reactor should be with respect to the size of packig i order to elimiate dispersio effects is depedet o Bo (Bodestei) umber, reactio order,, ad dimesioless reactio umber R (Damkohler umber) or coversio. Similarly oe may develop a criterio that will show how large Pe should be i order to guaratee that the exit cocetratio from the tubular reactor will be withi 5% of the oe predicted by plug flow desig. c exit c p c p 0.05 (55) ρ + Pe R ρ lρ ρ ρ R lρ 0.05 Pe ρ 0.05 (56) Pe 0 R lρ ( )ρ (57) The reactor legth to particle diametger ratio i packed beds that will guaratee maximum 5% deviatio i exit cocetratio based o plug flow desig is: 3

14 L d p B o 0 R lρ ( (58) )ρ For referece ad idustrial usage of the above see:. D.E. Mears, Chem. Eg. Sci., 6, 36 (97). D.E. Mears, Id. Eg. Chem. Fudametals, 5, 0 (976) 4

Finally, we show how to determine the moments of an impulse response based on the example of the dispersion model.

Finally, we show how to determine the moments of an impulse response based on the example of the dispersion model. 5.3 Determiatio of Momets Fially, we show how to determie the momets of a impulse respose based o the example of the dispersio model. For the dispersio model we have that E θ (θ ) curve is give by eq (4).

More information

ChE 471 Lecture 10 Fall 2005 SAFE OPERATION OF TUBULAR (PFR) ADIABATIC REACTORS

ChE 471 Lecture 10 Fall 2005 SAFE OPERATION OF TUBULAR (PFR) ADIABATIC REACTORS SAFE OPERATION OF TUBULAR (PFR) ADIABATIC REACTORS I a exothermic reactio the temperature will cotiue to rise as oe moves alog a plug flow reactor util all of the limitig reactat is exhausted. Schematically

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

Algebra of Least Squares

Algebra of Least Squares October 19, 2018 Algebra of Least Squares Geometry of Least Squares Recall that out data is like a table [Y X] where Y collects observatios o the depedet variable Y ad X collects observatios o the k-dimesioal

More information

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t = Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,

More information

SNAP Centre Workshop. Basic Algebraic Manipulation

SNAP Centre Workshop. Basic Algebraic Manipulation SNAP Cetre Workshop Basic Algebraic Maipulatio 8 Simplifyig Algebraic Expressios Whe a expressio is writte i the most compact maer possible, it is cosidered to be simplified. Not Simplified: x(x + 4x)

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you

More information

Numerical Methods in Fourier Series Applications

Numerical Methods in Fourier Series Applications Numerical Methods i Fourier Series Applicatios Recall that the basic relatios i usig the Trigoometric Fourier Series represetatio were give by f ( x) a o ( a x cos b x si ) () where the Fourier coefficiets

More information

Mark Lundstrom Spring SOLUTIONS: ECE 305 Homework: Week 5. Mark Lundstrom Purdue University

Mark Lundstrom Spring SOLUTIONS: ECE 305 Homework: Week 5. Mark Lundstrom Purdue University Mark udstrom Sprig 2015 SOUTIONS: ECE 305 Homework: Week 5 Mark udstrom Purdue Uiversity The followig problems cocer the Miority Carrier Diffusio Equatio (MCDE) for electros: Δ t = D Δ + G For all the

More information

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT TR/46 OCTOBER 974 THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION by A. TALBOT .. Itroductio. A problem i approximatio theory o which I have recetly worked [] required for its solutio a proof that the

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

Series III. Chapter Alternating Series

Series III. Chapter Alternating Series Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with

More information

Chapter Vectors

Chapter Vectors Chapter 4. Vectors fter readig this chapter you should be able to:. defie a vector. add ad subtract vectors. fid liear combiatios of vectors ad their relatioship to a set of equatios 4. explai what it

More information

SOLUTIONS: ECE 606 Homework Week 7 Mark Lundstrom Purdue University (revised 3/27/13) e E i E T

SOLUTIONS: ECE 606 Homework Week 7 Mark Lundstrom Purdue University (revised 3/27/13) e E i E T SOUIONS: ECE 606 Homework Week 7 Mark udstrom Purdue Uiversity (revised 3/27/13) 1) Cosider a - type semicoductor for which the oly states i the badgap are door levels (i.e. ( E = E D ). Begi with the

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

TEACHER CERTIFICATION STUDY GUIDE

TEACHER CERTIFICATION STUDY GUIDE COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra

More information

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1. SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad

More information

Zeros of Polynomials

Zeros of Polynomials Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree

More information

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.

More information

Kinetics of Complex Reactions

Kinetics of Complex Reactions Kietics of Complex Reactios by Flick Colema Departmet of Chemistry Wellesley College Wellesley MA 28 wcolema@wellesley.edu Copyright Flick Colema 996. All rights reserved. You are welcome to use this documet

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

Comparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series

Comparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series Applied Mathematical Scieces, Vol. 7, 03, o. 6, 3-337 HIKARI Ltd, www.m-hikari.com http://d.doi.org/0.988/ams.03.3430 Compariso Study of Series Approimatio ad Covergece betwee Chebyshev ad Legedre Series

More information

Lecture 4 Conformal Mapping and Green s Theorem. 1. Let s try to solve the following problem by separation of variables

Lecture 4 Conformal Mapping and Green s Theorem. 1. Let s try to solve the following problem by separation of variables Lecture 4 Coformal Mappig ad Gree s Theorem Today s topics. Solvig electrostatic problems cotiued. Why separatio of variables does t always work 3. Coformal mappig 4. Gree s theorem The failure of separatio

More information

Math 257: Finite difference methods

Math 257: Finite difference methods Math 257: Fiite differece methods 1 Fiite Differeces Remember the defiitio of a derivative f f(x + ) f(x) (x) = lim 0 Also recall Taylor s formula: (1) f(x + ) = f(x) + f (x) + 2 f (x) + 3 f (3) (x) +...

More information

ENGI Series Page 6-01

ENGI Series Page 6-01 ENGI 3425 6 Series Page 6-01 6. Series Cotets: 6.01 Sequeces; geeral term, limits, covergece 6.02 Series; summatio otatio, covergece, divergece test 6.03 Stadard Series; telescopig series, geometric series,

More information

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai

More information

x a x a Lecture 2 Series (See Chapter 1 in Boas)

x a x a Lecture 2 Series (See Chapter 1 in Boas) Lecture Series (See Chapter i Boas) A basic ad very powerful (if pedestria, recall we are lazy AD smart) way to solve ay differetial (or itegral) equatio is via a series expasio of the correspodig solutio

More information

6.003 Homework #3 Solutions

6.003 Homework #3 Solutions 6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the

More information

Fall 2013 MTH431/531 Real analysis Section Notes

Fall 2013 MTH431/531 Real analysis Section Notes Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters

More information

Ma 530 Introduction to Power Series

Ma 530 Introduction to Power Series Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power

More information

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?

More information

Notes on iteration and Newton s method. Iteration

Notes on iteration and Newton s method. Iteration Notes o iteratio ad Newto s method Iteratio Iteratio meas doig somethig over ad over. I our cotet, a iteratio is a sequece of umbers, vectors, fuctios, etc. geerated by a iteratio rule of the type 1 f

More information

1 Approximating Integrals using Taylor Polynomials

1 Approximating Integrals using Taylor Polynomials Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................

More information

Math 142, Final Exam. 5/2/11.

Math 142, Final Exam. 5/2/11. Math 4, Fial Exam 5// No otes, calculator, or text There are poits total Partial credit may be give Write your full ame i the upper right corer of page Number the pages i the upper right corer Do problem

More information

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x

More information

Section 5.5. Infinite Series: The Ratio Test

Section 5.5. Infinite Series: The Ratio Test Differece Equatios to Differetial Equatios Sectio 5.5 Ifiite Series: The Ratio Test I the last sectio we saw that we could demostrate the covergece of a series a, where a 0 for all, by showig that a approaches

More information

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b) Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig

More information

1 Adiabatic and diabatic representations

1 Adiabatic and diabatic representations 1 Adiabatic ad diabatic represetatios 1.1 Bor-Oppeheimer approximatio The time-idepedet Schrödiger equatio for both electroic ad uclear degrees of freedom is Ĥ Ψ(r, R) = E Ψ(r, R), (1) where the full molecular

More information

Chapter 14: Chemical Equilibrium

Chapter 14: Chemical Equilibrium hapter 14: hemical Equilibrium 46 hapter 14: hemical Equilibrium Sectio 14.1: Itroductio to hemical Equilibrium hemical equilibrium is the state where the cocetratios of all reactats ad products remai

More information

Castiel, Supernatural, Season 6, Episode 18

Castiel, Supernatural, Season 6, Episode 18 13 Differetial Equatios the aswer to your questio ca best be epressed as a series of partial differetial equatios... Castiel, Superatural, Seaso 6, Episode 18 A differetial equatio is a mathematical equatio

More information

Recursive Algorithms. Recurrences. Recursive Algorithms Analysis

Recursive Algorithms. Recurrences. Recursive Algorithms Analysis Recursive Algorithms Recurreces Computer Sciece & Egieerig 35: Discrete Mathematics Christopher M Bourke cbourke@cseuledu A recursive algorithm is oe i which objects are defied i terms of other objects

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

U8L1: Sec Equations of Lines in R 2

U8L1: Sec Equations of Lines in R 2 MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie

More information

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 116C. Problem Set 4. Benjamin Stahl. November 6, 2014

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 116C. Problem Set 4. Benjamin Stahl. November 6, 2014 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 6C Problem Set 4 Bejami Stahl November 6, 4 BOAS, P. 63, PROBLEM.-5 The Laguerre differetial equatio, x y + ( xy + py =, will be solved

More information

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

Geometry of LS. LECTURE 3 GEOMETRY OF LS, PROPERTIES OF σ 2, PARTITIONED REGRESSION, GOODNESS OF FIT

Geometry of LS. LECTURE 3 GEOMETRY OF LS, PROPERTIES OF σ 2, PARTITIONED REGRESSION, GOODNESS OF FIT OCTOBER 7, 2016 LECTURE 3 GEOMETRY OF LS, PROPERTIES OF σ 2, PARTITIONED REGRESSION, GOODNESS OF FIT Geometry of LS We ca thik of y ad the colums of X as members of the -dimesioal Euclidea space R Oe ca

More information

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0. THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of

More information

Number of fatalities X Sunday 4 Monday 6 Tuesday 2 Wednesday 0 Thursday 3 Friday 5 Saturday 8 Total 28. Day

Number of fatalities X Sunday 4 Monday 6 Tuesday 2 Wednesday 0 Thursday 3 Friday 5 Saturday 8 Total 28. Day LECTURE # 8 Mea Deviatio, Stadard Deviatio ad Variace & Coefficiet of variatio Mea Deviatio Stadard Deviatio ad Variace Coefficiet of variatio First, we will discuss it for the case of raw data, ad the

More information

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n. 0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate

More information

Additional Notes on Power Series

Additional Notes on Power Series Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here

More information

Principle Of Superposition

Principle Of Superposition ecture 5: PREIMINRY CONCEP O RUCUR NYI Priciple Of uperpositio Mathematically, the priciple of superpositio is stated as ( a ) G( a ) G( ) G a a or for a liear structural system, the respose at a give

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES 11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS

More information

Math 113 Exam 4 Practice

Math 113 Exam 4 Practice Math Exam 4 Practice Exam 4 will cover.-.. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for

More information

Section A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics

Section A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics X0/70 NATIONAL QUALIFICATIONS 005 MONDAY, MAY.00 PM 4.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.

More information

Special Modeling Techniques

Special Modeling Techniques Colorado School of Mies CHEN43 Secial Modelig Techiques Secial Modelig Techiques Summary of Toics Deviatio Variables No-Liear Differetial Equatios 3 Liearizatio of ODEs for Aroximate Solutios 4 Coversio

More information

Recurrence Relations

Recurrence Relations Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial(-)); } Let t be the umber of multiplicatios eeded to calculate factorial(). The

More information

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at

More information

10.6 ALTERNATING SERIES

10.6 ALTERNATING SERIES 0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose

More information

Lecture 3. Digital Signal Processing. Chapter 3. z-transforms. Mikael Swartling Nedelko Grbic Bengt Mandersson. rev. 2016

Lecture 3. Digital Signal Processing. Chapter 3. z-transforms. Mikael Swartling Nedelko Grbic Bengt Mandersson. rev. 2016 Lecture 3 Digital Sigal Processig Chapter 3 z-trasforms Mikael Swartlig Nedelko Grbic Begt Madersso rev. 06 Departmet of Electrical ad Iformatio Techology Lud Uiversity z-trasforms We defie the z-trasform

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

PRELIM PROBLEM SOLUTIONS

PRELIM PROBLEM SOLUTIONS PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems

More information

LESSON 2: SIMPLIFYING RADICALS

LESSON 2: SIMPLIFYING RADICALS High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN.. 4 6 N.RN. 4. N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 80 448 4 5 6 48 00 6 6 6

More information

Polynomial Functions and Their Graphs

Polynomial Functions and Their Graphs Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively

More information

APPENDIX F Complex Numbers

APPENDIX F Complex Numbers APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios

More information

The Interval of Convergence for a Power Series Examples

The Interval of Convergence for a Power Series Examples The Iterval of Covergece for a Power Series Examples To review the process: How to Test a Power Series for Covergece. Fid the iterval where the series coverges absolutely. We have to use the Ratio or Root

More information

DETERMINATION OF MECHANICAL PROPERTIES OF A NON- UNIFORM BEAM USING THE MEASUREMENT OF THE EXCITED LONGITUDINAL ELASTIC VIBRATIONS.

DETERMINATION OF MECHANICAL PROPERTIES OF A NON- UNIFORM BEAM USING THE MEASUREMENT OF THE EXCITED LONGITUDINAL ELASTIC VIBRATIONS. ICSV4 Cairs Australia 9- July 7 DTRMINATION OF MCHANICAL PROPRTIS OF A NON- UNIFORM BAM USING TH MASURMNT OF TH XCITD LONGITUDINAL LASTIC VIBRATIONS Pavel Aokhi ad Vladimir Gordo Departmet of the mathematics

More information

Quadratic Functions. Before we start looking at polynomials, we should know some common terminology.

Quadratic Functions. Before we start looking at polynomials, we should know some common terminology. Quadratic Fuctios I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively i mathematical

More information

SECTION 2 Electrostatics

SECTION 2 Electrostatics SECTION Electrostatics This sectio, based o Chapter of Griffiths, covers effects of electric fields ad forces i static (timeidepedet) situatios. The topics are: Electric field Gauss s Law Electric potetial

More information

NUMERICAL METHODS FOR SOLVING EQUATIONS

NUMERICAL METHODS FOR SOLVING EQUATIONS Mathematics Revisio Guides Numerical Methods for Solvig Equatios Page 1 of 11 M.K. HOME TUITION Mathematics Revisio Guides Level: GCSE Higher Tier NUMERICAL METHODS FOR SOLVING EQUATIONS Versio:. Date:

More information

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x

More information

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n = 60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece

More information

Ma 530 Infinite Series I

Ma 530 Infinite Series I Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li

More information

Solutions to Homework 1

Solutions to Homework 1 Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.

More information

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL

More information

Streamfunction-Vorticity Formulation

Streamfunction-Vorticity Formulation Streamfuctio-Vorticity Formulatio A. Salih Departmet of Aerospace Egieerig Idia Istitute of Space Sciece ad Techology, Thiruvaathapuram March 2013 The streamfuctio-vorticity formulatio was amog the first

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

(b) What is the probability that a particle reaches the upper boundary n before the lower boundary m?

(b) What is the probability that a particle reaches the upper boundary n before the lower boundary m? MATH 529 The Boudary Problem The drukard s walk (or boudary problem) is oe of the most famous problems i the theory of radom walks. Oe versio of the problem is described as follows: Suppose a particle

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces

More information

Math 116 Second Midterm November 13, 2017

Math 116 Second Midterm November 13, 2017 Math 6 Secod Midterm November 3, 7 EXAM SOLUTIONS. Do ot ope this exam util you are told to do so.. Do ot write your ame aywhere o this exam. 3. This exam has pages icludig this cover. There are problems.

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

CEE 522 Autumn Uncertainty Concepts for Geotechnical Engineering

CEE 522 Autumn Uncertainty Concepts for Geotechnical Engineering CEE 5 Autum 005 Ucertaity Cocepts for Geotechical Egieerig Basic Termiology Set A set is a collectio of (mutually exclusive) objects or evets. The sample space is the (collectively exhaustive) collectio

More information

PHY4905: Nearly-Free Electron Model (NFE)

PHY4905: Nearly-Free Electron Model (NFE) PHY4905: Nearly-Free Electro Model (NFE) D. L. Maslov Departmet of Physics, Uiversity of Florida (Dated: Jauary 12, 2011) 1 I. REMINDER: QUANTUM MECHANICAL PERTURBATION THEORY A. No-degeerate eigestates

More information

7 Sequences of real numbers

7 Sequences of real numbers 40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are

More information

The Growth of Functions. Theoretical Supplement

The Growth of Functions. Theoretical Supplement The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that

More information

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said

More information

Topic 9: Sampling Distributions of Estimators

Topic 9: Sampling Distributions of Estimators Topic 9: Samplig Distributios of Estimators Course 003, 2016 Page 0 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be

More information

The Method of Least Squares. To understand least squares fitting of data.

The Method of Least Squares. To understand least squares fitting of data. The Method of Least Squares KEY WORDS Curve fittig, least square GOAL To uderstad least squares fittig of data To uderstad the least squares solutio of icosistet systems of liear equatios 1 Motivatio Curve

More information

Salmon: Lectures on partial differential equations. 3. First-order linear equations as the limiting case of second-order equations

Salmon: Lectures on partial differential equations. 3. First-order linear equations as the limiting case of second-order equations 3. First-order liear equatios as the limitig case of secod-order equatios We cosider the advectio-diffusio equatio (1) v = 2 o a bouded domai, with boudary coditios of prescribed. The coefficiets ( ) (2)

More information