MATH 121 A Mathematical Tools for the Physical Sciences, Study notes of Algebra

Download MATH 121 A Mathematical Tools for the Physical Sciences and more Study notes Algebra in PDF only on Docsity! University Course MATH 121 A Mathematical Tools for the Physical Sciences UC BERKELEY Spring 2004 My Class Notes Nasser M. Abbasi Spring 2004 Chapter 1 Introduction This ismyweb page for courseMATH121A,Mathematicalmethods in physical sciences, that I took in spring 2004 at UC Berkeley. Hard course. Lots of HWs. This is the syllubus 1 CHAPTER 1. INTRODUCTION Professor Fraydoun Rezakhanlou, UC Berkeley Math department, gave the course. A very good teacher. Teaches without using notes, all from memory, which I thought was really amazing. This course was lots of work. About 30-40 problems for HW each week. But only 2 random problems are graded. So it is possible to solve 39 of out 40 problems correctly and get only 50%, if the problem missed happened to be selected to grade. Exams as closed notes and closed book. No cheat sheet either. No calculators. Textbook was Mary Boas, second edition. This seems to be the standard book for this type of course are most universities, at least the ones I know about. It is a a very good book, but more detailed worked examples would have been nice. So another book such as the problem solvers type book might be useful to have. 2 CHAPTER 1. INTRODUCTION For some reason this course (and Math121B) did not have discussion period, this was unfortunate since discussion periods can be really useful. I was told this is because of budget cuts in the California university system. This below is a picture of Evans hall. It is a big tall building full of very smart people. The math department is on the 9th floor. The course was in room 75, which is on the ground floor on Evans hall. 1.0.1 Course description These are the Course outline handouts. Page 1, and page 2. it contains the problems to solve for each HW. Spring 2004 (January-May 2004) Course description: Functions of a complex variable, Fourier series, finite-dimensional linear systems. Infinite-dimensional linear systems, orthogonal expansions, special func- tions, partial differential equations arising in mathematical physics. Intended for stu- dents in the physical sciences who are not planning to take more advancedmathematics courses. Units: 4 Book: MATHEMATICAL METHODS IN PHYSICAL SCI, BOAS. 2nd edition chapters 1,2,4,7,9,14,15 Instructor: Professor Fraydoun Rezakhanlou, Math department, UC Berkeley Position: Associate Professor E-mail: rezakhan@math.berkeley.edu To reduce spam, this address is javascript en- coded. Phone: +1 (510) 642-2838 Office: 815 Evans Hall 3 2.1. Misc. notes CHAPTER 2. MY NOTES, STUDY NOTES 2. The Legendre ODE is solved using series method, assuming 𝑙 is an integer. We get one solution which is Legendre function of first kind 𝑝𝑙(𝑥). What if 𝑙 is not an integer? A: Legendre 𝑃𝑙 is only defined for integer 𝑙? YES? No, there are tables for non-integer, but these cases are not important. 3. What if we get a legendre ODE and we want to find solution for 𝑥 > 1 ? Since legendre functions are only defined for 𝑥 less than one (to have convergence). Physics example? usually 𝑥 is the cosine of an angle so it is ≤ 1. 4. What if 𝑙 is not an integer in the legendre ODE? how to get a solution? this is special cases, not important, look up handbooks. 5. problem I solved in HW#6, chapter 12, 16.3. check my solution. I claimed that the second solution is 𝑁𝑝 but since I found 𝑃NOT to be an integer, hence the second solution is one containing log and not a combination of 𝐽−𝑝. When I solved it in mathematica, I get this solution (notice complex number?), could this second solution be converted to log function? answer: OK, the solution I did will turn out to have log in it if I put p=integer and use L’hospital’s rule to evaluate. 6. When solving for equation 16.1 on page 516, we seem to only take the positive root for the variables, why? see for example page 516. 𝑏 = 2 but it is really 𝑏 = ±2 answer: OK, any of these will give a good solution, just pick one. 7. on page 528, can I just set 𝑛 = 0 always to solve for the indical equation as shown in the example? is it better to solve this using the∑ directly as shown in the example instead of setting up a table? table seems more clear, but the example method seems shorter. 8. How to solve chapter 16, 4.1 part (c) using Bayes rule? I write: Let A=event first chair is empty, let B=event second chair is empty. We need to find 𝑃(𝐴𝐵) = 𝑃(𝐴)𝑃𝐴(𝐵) = ( 110 )( 1 9 ) = 1 90 , but the answer should be 1 45 , what is it I am doing wrong? wrong. 𝑃(𝐴) = 1/5 𝑛𝑜𝑡 1/10. 9. for problem HW 12, chapter 16, 4.8, part b. It says given 2 cards drawn from deck, if you know one is an ace, what it the chance the BOTH are an ace? I know how to solve by the book. but why can I not say the following: since we KNOW that one card is an ace, then the chance that both cards are an ace is just the chance the second card being an ace (since we know the first is an ace). So this should give 3 51 6 2.2. Table summary of topics to study CHAPTER 2. MY NOTES, STUDY NOTES 10. random variable is defined as a function on the sample space. however, it is multi- valued. for example, if x= sum of 2 die throw, then more than one event can give the same random variable. is this OK? I thought a function must be single valued? answer: I am wrong. it is NOT multivalued. 11. check that my solution for chapter 16, 5.1 MATH 121B is correct, I have solution on paper. this is the last HW 2.2 Table summary of topics to study 2.2.1 Math 121A ch. title topics Exam 1 series infinite series, power series,def. of cover- gence, tests for convergence, 1 test for alternating series, power series, bi- nomial series 2 complex numbers finding circle of convergence (limit test), Euler formula 1 power and roots of complex numbers, log, inverse log 4 partial differentiation total diffenertials, chain rule, implicit dif- ferentiation 1 partial diff for max and minumum, La- grange muktipliers, change of variables Leibniz rule for differ- netiation of integrals 14 complex functions Def. of analytic fn, Cauchy-Riemann con- ditions, laplace equation, 1 contour integrals, Laurent series, Residue theorm, methods of finding residues, pole type, evaluating integrals by residue, Mapping, conformal 7 Fourier series expansion of function in sin and cosin, complex form, how to find 2 coeff, Dirichlet conditions, different inter- vals, even/odd, Parseval’s 7 2.2. Table summary of topics to study CHAPTER 2. MY NOTES, STUDY NOTES 15 Laplace/Fourier transforms Laplce transform, table, how to use Laplace to solve F ODE, Methods of finding inverse laplace, partial fraction, convolution, sum of residues, Fourier transform, sin/- consine transforms, Direc Delta Green method to solve ODE using im- pluse 9 Calculus of variations Euler equation solving, Setting up La- grange equations, KE, PE F Solving Euler with constrainsts 8 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑐𝑛 = 1 𝑇 �𝑇 𝑓(𝑥) 𝑒−𝑖𝑛𝑥𝜔 𝑑𝑥 Now, 𝜔, is the angular velocity. i.e. 𝜃 = 𝜔𝑡, so for ONE period 𝑇, 𝜃 = 2𝜋, hence 𝜔 = 2𝜋 𝑇 , so 𝑐𝑛 can be written as 𝑐𝑛 = 1 𝑇 �𝑇 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 2𝜋 𝑇 𝑑𝑥 Notice that in this chapters we use distance for period (i.e. wave length 𝜆) instead of time as period 𝑇. it does not matter, they are the same, choose one. i.e. we can say that the function repeats every 𝜆, or the function repeats every one period 𝑇. When using 𝜆 for period, say −𝑙, 𝑙 or −𝜋, 𝜋 the above equation becomes 𝑐𝑛 = 1 2𝑙 � 𝑙 −𝑙 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 2𝜋 2𝑙 𝑑𝑥 = 1 2𝑙 � 𝑙 −𝑙 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 𝜋 𝑙 𝑑𝑥 note: Above integral for 𝑐𝑛 is for negative 𝑛 as well as positive 𝑛. In non-complex expo- nential expansion, there is no negative 𝑛, only positive. note: 𝑐−𝑛 = ?̄?𝑛 note: there is a relation between the 𝑎𝑛, 𝑏𝑛, and the 𝑐𝑛 which is 𝑎𝑛 = 𝑐−𝑛 + 𝑐𝑛 and 𝑏𝑛 = 𝑖(−𝑐−𝑛 + 𝑐𝑛) IF given 𝑓(𝑥), defined over (0, 𝐿), The algorithm to find Fourier series is this: IF asked to find a(n) i.e. the COSIN series, THEN extend f(x) so that it is EVEN (this makes b(n)=0) and period now is 2L ELSE IF asked to find b(n), i.e. the SIN series, THEN extend f(x) to be ODD (this makes a(n)=0) and period now is 2L ELSE we want the standard SIN/COSIN period remains L, and use the c(n) formula (and remember to do the c(0) separatly for the DC term) END IF END IF 2.3.3.1 Parseval’s theorem for fourier series This theory gives a relation between the average of the square of 𝑓(𝑥) over a period and the fourier coefficients. Physically, it says that this: the total energy of a wave is the sum of the energies of the individual harmonics it carries Average of �𝑓(𝑥)� 2 = �12𝑎0� 2 + 1 2 ∑∞ 1 𝑎 2 𝑛 + 1 2 ∑∞ 1 𝑏 2 𝑛 over ONE period. In complex form, Average of �𝑓(𝑥)2� = ∑∞ −∞|𝑐𝑛| 2. Think of this like pythagoras theorem. For example, given 𝑓(𝑥) = 𝑥, then �𝑓(𝑥)� 2 = 1 2 ∫1 −1 𝑥2𝑑𝑥 = 1 3 , then 1 3 = ∑ ∞ −∞|𝑐𝑛| 2 In the above we used the standard formula for average of a function, which is average of 𝑓(𝑥) = 1 𝑇 ∫ 𝑇 𝑓(𝑥) 𝑑𝑥, here we should need to square 𝑓(𝑥) 11 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 2.3.4 Chapter 15. Integral transforms (Laplace and Fourier transforms) 2.3.4.1 Laplace and Fourier transforms definitions 𝐹 𝑓(𝑥) = 𝐹(𝑝) = � ∞ 0 𝑓(𝑥) 𝑒−𝑝𝑥 𝑑𝑥 𝑝 > 0 𝐹 𝑔(𝑥) = 𝑔(𝛼) = 1 2𝜋 � ∞ −∞ 𝑓(𝑥) 𝑒−𝑖𝛼𝑥 𝑑𝑥 Associate Fourier with 1 2𝜋 . (mind pic: Fourier=Fraction i.e.→ 1 2𝜋) and Fourier goes from −∞ to +∞ (mind pic: Fourier=whole Floor), Fourier imaginary exponent, Laplace real exponent. Note: Laplace transform is linear operator, hence 𝐿�𝑓(𝑡) + 𝑔(𝑡)� = 𝐿𝑓(𝑡) + 𝐿𝑔(𝑡) and 𝐿�𝑐 𝑓(𝑡)� = 𝑐 𝐿𝑓(𝑡) 2.3.4.2 Inverse Fourier and Laplace transform formulas (We do not really use the inverse Laplace formula directly (called Bromwich integral), we find inverse Laplace using other methods, see below) 𝑓(𝑥) = 1 2𝜋𝑖 � 𝑐+𝑖 ∞ 𝑐−𝑖 ∞ 𝐹(𝑧) 𝑒𝑧𝑡 𝑑𝑧 𝑡 > 0 Inverse laplace 𝑓(𝑥) = 1 2𝜋 � ∞ −∞ 𝑔(𝛼) 𝑒𝑖𝛼𝑥 𝑑𝛼 Inverse fourier The Fourier transform has 2 other siblings to it (which Laplace does not), these are the sin and cos transform and inverse transform. I’ll add these later but I do not think we will get these in the exam. Note: To get the inverse Laplace transform the main methods are 1. using partial fractions to break the expression to smaller ones we can lookup in tables 2. Use Convolution. i.e. given 𝑌 = 𝐿�𝑓1� 𝐿�𝑓2� → 𝑦 = ∫ 𝑡 0 𝑔(𝑡 − 𝜏) 𝑓(𝜏) 𝑑𝜏 = 𝑔 ⊗ 𝑓 use this as an alternative to partial fraction decomposition if easier. mind pic: 𝑡 one time, 𝜏 2 times. 3. Use the above integral (Bromwich) directly (hardly done) 4. To find 𝑓(𝑡) from the Laplace transform, instead of using the above formula, we can write 𝑓(𝑡) = sum of residues of 𝐹(𝑧)𝑒𝑧𝑡 at all poles. For example given 𝐹(𝑧), we multiply it by 𝑒𝑧𝑡, and then find all the poles of the resulting function (i.e. the zeros of the denominator), then add these. Note: To find Fourier transform , 𝑔(𝛼), must carry the integration (i.e. apply the integral directly, no tables like with Laplace). Note: we use Laplace transform as a technique to solve ODE. Why do we need Fourier transform? To represent an arbitrary function (must be periodic or extend to be period if not) as a sequence of sin/cosine functions. And why do we do this? To make it easier to analyze it and find what frequency components it has. For continuous function, use fourier transform (integral). note: Function must satisfy Dirichlet conditions to use in fourier transform or Fourier series. 12 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES note: Fourier series expansion of a functionwill accurately fit the function asmore terms are added. But in places where there is a jump, it will go to the average value of the function at the jump. question:When dowe use fourier series, andwhen to use fourier transform?Why dowe need F. transform if we can use F. Series?We use F. transform for continuous frequencies. What does this really mean? 2.3.4.3 Using Laplace transform to solve ODE Remember 𝐿(𝑦) = 𝑌 𝐿�𝑦′� = 𝑝𝑌 − 𝑦0 𝐿�𝑦′′� = 𝑝2𝑌 − 𝑝𝑦0 − 𝑦′0 note: 𝑝 has same power as order of derivative. do not mix up where the 𝑝 goes in the 𝑦′′ equation. remember the 𝑦′0 has no 𝑝with it. mind pic: think of the 𝑦0 as the senior guy since coming from before so it is the one who gets the 𝑝. note: if 𝑦0 = 𝑦′0 = 0 (whichmost HWproblemwas of this sort), then the above simplifies to 𝐿�𝑦′� = 𝑝𝑌 𝐿�𝑦′′� = 𝑝2𝑌 So given an ODE such as 𝑦′′ + 4𝑦′ + 𝑦 = 𝑓(𝑡) → �𝑝2 + 4𝑝 + 4�𝑌 = 𝐿(𝑓(𝑡)) i.e. just replace 𝑦′′ by 𝑝2, etc... This saves lots of time in exams. Now we get an equation with 𝑌 in terms of 𝑝, now solve to find 𝑦(𝑡) from 𝑌 using tables. Notice that solution of ODE this way gives a particular solution, since we used the boundary conditions already. For an ODE such as 𝐴𝑦′′ + 𝐵𝑦′ + 𝐶𝑦 = ℎ(𝑡) its Laplace transform can be written immediately as 𝐴𝑝2𝑌 + 𝐵𝑝𝑌 + 𝐶𝑌 = 𝐿 ℎ(𝑡) 𝑌 = 𝐿 ℎ(𝑡) 𝐴𝑝2 + 𝐵𝑝 + 𝐶 whenever the B.C. are 𝑦′0 = 0 and 𝑦0 = 0 2.3.4.4 Partial fraction decomposition When denominator is linear time quadratic or quadratic time quadratic PFD is probably needed. This is how to do PFD for common cases 13 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 2.3.4.7.1 Solving an ODE using green method Here we are given an ODE, with a forcing function (i.e. nonhomogeneous ODE). And given 2 solutions to it, and asked to find the particular solution. Example, 𝑦′′ − 𝑦 = 𝑓(𝑡) and solutions are 𝑦1, 𝑦2 then the particular solution is 𝑦𝑝 = 𝑦2∫ 𝑦1 𝑓(𝑡) 𝑊 𝑑𝑡 − 𝑦1∫ 𝑦2 𝑓(𝑡) 𝑊 𝑑𝑡where𝑊 = � � 𝑦′1 𝑦′2 𝑦1 𝑦2 � � 2.3.5 Chapter 2. Complex Numbers note: When given a problem such as evaluate (−2 − 2𝑖) 1 5 , always start by finding the length of the complex number, then extract it out before converting to the 𝑟𝑒𝑖𝑛𝜃 form. For example, (−2 − 2𝑖) 1 5 = 2√2 � −1 √2 − 𝑖 √2 � , the reason is that now the stuff inside the brackets has length ONE. So we now get 2√2 � −1 √2 − 𝑖 √2 � = 2√2 𝑒− 3 4𝜋𝑖 and only now apply the last raising of power to get �2√2 𝑒− 3 4𝜋𝑖� 1 5 = 2 3 10 𝑒 3 4𝜋𝑖+2𝑛𝜋 5 for 𝑛 = 0, 1, 2, 3,⋯ make sure not to forget the 2𝑛𝜋, I seem to forget that. 2.3.6 Chapter 9. Calculus of variations 2.3.6.1 Euler equation How to construct Euler equation 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � − 𝜕𝐹 𝜕𝑦 = 0. If integrand does not depend on 𝑥 then change to 𝑦. Example ∫ 𝑥1 𝑥2 𝑦′2 𝑦 𝑑𝑥 → ∫𝑦1 𝑦2 1 𝑥′2 𝑦 �𝑥 ′ 𝑑𝑦� → ∫𝑦1 𝑦2 1 𝑥′ 𝑦 𝑑𝑦 this is done by making the substitution 𝑦′ = 1 𝑥′ and 𝑑𝑥 = 𝑥′ 𝑑𝑦. Now Euler equation changes from 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � − 𝜕𝐹 𝜕𝑦 = 0 to 𝑑 𝑑𝑦 � 𝜕𝐹 𝜕𝑥′ � − 𝜕𝐹 𝜕𝑥 = 0. Normally, 𝜕𝐹 𝜕𝑦 will be zero. Hence we end up with 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � = 0 and this means 𝜕𝐹 𝜕𝑦′ = 𝑐, and so we only need to do ONE integral (i.e. solve a first order ODE). If I find myself with a 2 order ODE (for this course!) , I have done something wrong since all problems we had are of this sort. 2.3.6.2 Lagrange equations are just Euler equations, but one for each dimension. 𝐹 is now called 𝐿. where 𝐿 = 𝑇 − 𝑉 where 𝑇 = 𝐾.𝐸. and 𝑉 = 𝑃.𝐸., 𝑇 = 1 2𝑚𝑣 2, 𝑉 = 𝑚𝑔ℎ So given a problem, need to construct 𝐿 ourselves. Then solve the Euler-Lagrange equa- tions 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑥 = 0 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑦 = 0 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑧 = 0 The tricky part is finding 𝑣2 for different coordinates. This is easy if you know 𝑑𝑠2, so just remember those 16 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 (polar) 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑑𝑧2 (cylindrical) 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃𝑑𝜙2 (spherical) So to find 𝑣2 just divide by 𝑑𝑡2 and it follows right away the following 𝑣2 = ?̇?2 + 𝑟2?̇?2 (polar) 𝑣2 = ?̇?2 + 𝑟2?̇?2 + ?̇?2 (cylindrical) 𝑣2 = ?̇?2 + 𝑟2?̇?2 + 𝑟2 sin2 𝜃 ?̇?2 (spherical) To help remember these: Note 𝑑𝑠2 all start with 𝑑𝑟2 + 𝑟2𝑑𝜃2 for each coordinates system. So just need to remember the third terms. (think of polar as subset to the other two). Also see that each variable is squared. So the only hard think is to remember the last term for the spherical. Remember that in a system with particles, need to find the KE and PE for each particle, and then sum these to find the whole system KE and PE, and this will give one 𝐿 for the whole system before we start using the Lagrange equations. 2.3.6.3 Solving Euler-Lagrange with constraints The last thing to know is this chapter is how to solve constraint problems. This is just like solving for Euler, expect now we have an additional integral to deal with. So in these problems we are given 2 integrals instead of one. One of these will be equal to some number say 𝑙. Soweneed tominimize 𝐼 = ∫ 𝑥1 𝑥2 𝐹(𝑥, 𝑦′, 𝑦) 𝑑𝑥 subject to constraint that 𝑔 = ∫ 𝑥1 𝑥2 𝐺(𝑥, 𝑦′, 𝑦) 𝑑𝑥 = 𝑙 Follow the same method as Euler, but now we write 𝑑 𝑑𝑥� 𝜕 𝜕𝑦′ (𝐹 + 𝜆𝐺)� − 𝜕 𝜕𝑦 (𝐹 + 𝜆𝐺) = 0 So replace 𝐹 by 𝐹 + 𝜆𝐺 This will give as equation with 3 unknowns, 2 for integration constants, and one with 𝜆, we solve for these given the Boundary conditions, and 𝑙 but we do not have to do this, just need to derive the equations themselves. Some integrals useful to know in solving the final integrals for the Euler problems are these � 𝑐 �𝑦2 − 𝑐2 𝑑𝑦 = 𝑐 cosh−1� 𝑦 𝑐 � + 𝑘 � 𝑐 �1 − 𝑐2 𝑦2 𝑑𝑦 = 𝑐 sin−1�𝑐 𝑦� + 𝑘 � 𝑐 𝑦�𝑦2 − 𝑐2 𝑑𝑦 = 1 𝑐 cos−1� 𝑐 𝑦 � + 𝑘 17 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 2.4 MATH 121B Notes 2.4.1 Chapter 12. Series solution of ODE and special functions Bessel ODE 𝑥2 𝑦′′ + 𝑥 𝑦′ + �𝑥2 − 𝑝2�𝑦 = 0 defined for INTEGER and NON integer 𝑃 first solution 𝑦1 = 𝐽𝑝(𝑥) = ∑ −1𝑛 Γ(𝑛+1)Γ�𝑛+𝑝+1� �𝑥 2 � 2𝑛+𝑝 (for 𝑝 an integer or not) second solution 𝑦2 = 𝑁𝑝(𝑥) = 𝑌𝑝(𝑥) = cos�𝜋𝑝�𝐽𝑝(𝑥)−𝐽−𝑝 sin𝜋𝑝 (note: 𝑝 here is NOT an integer) second solution 𝑦𝑥 contains a log function. note: 𝑝 here IS an integer. Orthogonality ∫1 0 𝑥 𝐽𝑝(𝑎𝑥) 𝐽𝑝(𝑏𝑥) 𝑑𝑥 = ⎧⎪⎪⎪⎨ ⎪⎪⎪⎩ 0 𝑖𝑓 𝑎 ≠ 𝑏 1 2 𝐽 2 𝑝+1(𝑎) = 1 2 𝐽 2 𝑝−1(𝑎) = 1 2 𝐽 ′2 𝑝 (𝑎) 𝑖𝑓 𝑎 = 𝑏 𝑎, 𝑏 are zeros of 𝐽𝑝 recursive formula 𝑑 𝑑𝑥 �𝑥𝑝𝐽𝑝� = 𝑥𝑝𝐽𝑝−1, 𝑑 𝑑𝑥� 1 𝑥𝑝 𝐽𝑝� = − 1 𝑥𝑝 𝐽𝑝+1, 𝐽𝑝−1 + 𝐽𝑝+1 = 2𝑝 𝑥 𝐽𝑝, 𝐽𝑝−1 − 𝐽𝑝+1 = 2𝐽′𝑝 𝐽′𝑝 = − 𝑝 𝑥 𝐽𝑝 + 𝐽𝑝−1 = 𝑝 𝑥 𝐽𝑝 − 𝐽𝑝+1 NOTICE: No Rodrigues formula for Bessel func, since not polyn. notes: We used a generalized power series method to find the solutions. IF 𝑝 is NOT an integer, then 𝐽𝑝 and 𝐽−𝑝 (or𝑁𝑝) are two independent solutions IF 𝑝 is an integer, then 𝐽𝑝 and 𝐽−𝑝 are NOT two independent solutions, use log for 𝑦2 𝐽𝑝 is called Bessel function of first kind, and 𝑌𝑝 is called second kind. 𝑝 is called the ORDER. IF 𝑝 = 𝑛 + 1 2 , a special case, we get spherical bessel functions 𝑗𝑛(𝑥) and 𝑦𝑛(𝑥) 𝑗𝑛(𝑥) = � 𝜋 2𝑥 𝐽𝑛+ 1 2 (𝑥) = 𝑥𝑛�−1 𝑥 𝑑 𝑑𝑥 � 𝑛 � sin 𝑥 𝑥 � 18 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑃(𝐴𝐵) = 𝑃(𝐴) (𝐵) If A,B independent Probability that A OR B will happen is: 𝑃(𝐴 + 𝐵) = 𝑃(𝐴) + (𝐵) − 𝑃(𝐴𝐵) 𝑃(𝐴 + 𝐵) = 𝑃(𝐴) + (𝐵) IF A,B are mutually exclusive This means that 𝑃(𝐴𝐵) = 0 if they are mutually exclusive (obvious) 𝑃(𝐴 + 𝐵 + 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − {𝑃(𝐴𝐵) + 𝑃(𝐴𝐶) + 𝑃(𝐵𝐶)} + 𝑃(𝐴𝐵𝐶) note: 𝑃𝑛𝑟 = number of permutations (arrangements) or 𝑛 things taken 𝑟 at a time. 𝑃𝑛𝑟 = 𝑛! (𝑛−𝑟)! Here order is important. i.e. ABC is DIFFERENT from CAB, hence this number will be larger than the one below. ⎛ ⎜⎜⎜⎜⎜⎝ 𝑛 𝑟 ⎞ ⎟⎟⎟⎟⎟⎠ = 𝐶 𝑛 𝑟 = 𝑛! (𝑛 − 1)! 𝑟! Number of combinations OR selections of 𝑛 things 𝑟 at a time. here order is NOT impor- tant. so ABC is counted the same as CAB, hence this number will be smaller. note: In how many ways can 10 people be seated on a bench with 4 seats? A) ⎛ ⎜⎜⎜⎜⎜⎝ 10 4 ⎞ ⎟⎟⎟⎟⎟⎠4! = 10! 6!4!4! = 10! 6! = 10 × 9 × 8 × 7 To understand this: ⎛ ⎜⎜⎜⎜⎜⎝ 10 4 ⎞ ⎟⎟⎟⎟⎟⎠ is the number of ways 4 people can be selected out of 10. ONCE those 4 people have been selected, then there are 4! different ways they can be arranged on the bench. Hence the answer is we multiply these together. note: Find number of ways of putting 𝑟 particles in 𝑛 boxes according to the 3 kinds of statistics. Answer 1. For Maxwell-Blotzman (MB) it is 𝑛𝑟 2. For Fermi-Dirac (FM), it is 𝑛𝐶𝑟 3. For Bose-Einstein (BE) it is 𝑛+1𝐶𝑟 note: If asked this: there is box A which has 5 red balls and 6 black balls, and box B which has 5 red balls and 8 white balls, what is the prob. of picking a red ball? Answer: P(pick box A) P(pick red ball from it) + P(pick box B) P(pick red ball from it) note: If we get a problem such as 2 boxes A,B, and more than more try picking balls, it is easier to draw a tree diagram and pull the chances out the tree than having to calculate them directly in the exam. Tree can be drawn in 2 minutes and will have all the info I need. note: write down the cancer chance problem. note: random variable 𝑥 is a function defined on the sample space (for the example, the sum of 2 die throw). The probability density is the probability of each random variable. average or mean of a random variable 𝜇 = ∑𝑥𝑖 𝑃𝑖 where 𝑃𝑖 is the probability of the random variable. The Variance Var measures the spread of the random variable around the average, also called dispersion defined as 21 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑉𝑎𝑟(𝑥) = ��𝑥𝑖 − 𝜇� 2 𝑃𝑖 Standard deviation is another measure of the dispersion, defined as 𝜎(𝑥) = √𝑉𝑎𝑟(𝑥) Distribution function is just a histogram of the probability density. it tells one what the probability of a random variable being less than a certain 𝑥 value. see page 711. 2.4.3 Chapter 13. PDE 𝑃𝐷𝐸 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑛𝑜𝑡𝑒𝑠 Laplace 𝑢�𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 0 describes steady state (no time) of region with no source for example, gravitional potential with nomatter, electrostatic potential with no charge, or steady state Temp. dis- tribution Poisson 𝑢�𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 𝑓�𝑥, 𝑦, 𝑧� Same as Laplace, i.e. sescribes steady state, how- evere here the source of the field is present. 𝑓�𝑥, 𝑦, 𝑧� is called the source density. i.e. it is a function that de- scribes the density distribution of the source of the potential. Diffusion or 𝑢�𝑡, 𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 1 𝛼2 𝜕𝑢 𝜕𝑡 Here 𝑢 is usually the temperature 𝑇 function. Now time heat equation is involved. So this equation is alive. Wave equation 𝑢�𝑡, 𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 1 𝑣2 𝜕2𝑢 𝜕𝑡2 Here 𝑢 is the position of a point on the wave at time 𝑡. Notice the wave equation has second derivative w.r.t. time while the diffusion is first derivative w.r.t. time Helmholtz 𝐹�𝑥, 𝑦, 𝑧� ∇ 2𝐹 + 𝑘2𝐹 = 0 The diffusion and wave equation generate this. This is the equation SPACE only solution of the wave and heat equa- tions. i.e. 𝑢 = 𝐹�𝑥, 𝑦, 𝑧�𝑇(𝑡) is the solution for both heat and wave eq. Each of these equations has a set of candidate solutions, which we start with and try to fit the boundary and initial condition into to eliminate some solution of this set that do not fit until we are left with the one candidate solution. We then use this candidate 22 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES solution to find the general solution, which is a linear combination of it. We use fourier series expansion in this part of the solution. In table below I show for each equation what the set of candidate solutions are. Use these to start the solution with unless the question asks to start at an earlier stage, which is the separation of variables. So the algorithm for solving these PDE is Select THE PDE to use ----> Obtain set of candidate solution ----> Eliminate those that do not fit -----> obtain the general solution by linear combination (use orthogonality principle here) PDE candidate solutions notes ∇ 2𝑢 = 0 𝑢�𝑥, 𝑦� = ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎩ 𝑒𝑘𝑦 cos 𝑘𝑥 𝑒𝑘𝑦 sin 𝑘𝑥 𝑒−𝑘𝑦 cos 𝑘𝑥 𝑒−𝑘𝑦 sin 𝑘𝑥 for 2 dimensions ∇ 2𝑢 = 𝑓�𝑥, 𝑦, 𝑧� 𝑢�𝑥, 𝑦, 𝑧� = − 1 4𝜋 ∫∫∫ 𝑓�𝑥′,𝑦′,𝑧′� � (𝑥−𝑥′)2+�𝑦−𝑦′� 2 +(𝑧−𝑧′)2+ 𝑑𝑥′ 𝑑𝑦′ 𝑑𝑧′ 𝑓�𝑥′, 𝑦′, 𝑧′� is a function that describes mass density distribution evaluated at point 𝑥′, 𝑦′, 𝑧′. The point 𝑥, 𝑦, 𝑧 is where we are calculating the potential 𝑢 itself ∇ 2𝑢 = 1 𝛼2 𝜕𝑢 𝜕𝑡 𝑢(𝑡, 𝑥) = ⎧⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩ 𝑒−𝑘2𝛼2𝑡 cos 𝑘𝑥 𝑒−𝑘2𝛼2𝑡 sin 𝑘𝑥 for one space dimension 23 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES 2.4.3.3 Wave equation in polar coordinates 2.5 General equations sin 𝑛𝑥 = 𝑒𝑖𝑛𝑥 − 𝑒−𝑖𝑛𝑥 2𝑖 cos 𝑛𝑥 = 𝑒𝑖𝑛𝑥 + 𝑒−𝑖𝑛𝑥 2 � sin 𝑥 cos 𝑥 𝑑𝑥 = ln(sin 𝑥) csc 𝑥 = 1 sin 𝑥 average value of 𝑓(𝑥) over [𝑏, 𝑎] = ∫𝑏 𝑎 𝑓(𝑥)𝑑𝑥 𝑏 − 𝑎 cos2 𝑘𝑥 = 1 + cos(2𝑘) 2 sin2 𝑘𝑥 = 1 − cos(2𝑘) 2 sin𝐴 sin𝐵 = 1 2 [ cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)] cos𝐴 cos𝐵 = 1 2 [ cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵)] sin𝐴 cos𝐵 = 1 2 [ sin(𝐴 − 𝐵) + sin(𝐴 + 𝐵)] I need a geometric way to visualize these equations, but for now for the exam remember them as follows: they all start with 𝐴 − 𝐵 , and when the functions being multiplied are different on the LHS, we get sin on the RHS, else we get cos (think of cos as nicer, since even function :). � tanh(𝑥) = ln(cosh 𝑥) � tan 𝑥 = − ln(cos 𝑥) ∫𝑏 𝑎 cos2 𝑘𝑥 𝑑𝑥 = 𝑏−𝑎 2 . if 𝑘(𝑏 − 𝑎) is an integer multiple of 𝜋. (the same for sin2 𝑘𝑥), for example ∫ 1 −1 cos2 𝜋𝑥 𝑑𝑥 = 1 , ∫ 1 −1 cos2 5𝜋𝑥 𝑑𝑥 = 1, ∫ 1 −5 cos2 7𝜋𝑥 𝑑𝑥 = 3 , ∫ 1 −1 sin2 𝜋𝑥 𝑑𝑥 = 1, etc... this can be very useful so remember it! ∫𝑏 𝑎 cos 𝑘𝑥 𝑑𝑥 = 0 if over a complete period. same for sin 𝑥, for example ∫ 𝜋 −𝜋 cos 𝑘𝑥 𝑑𝑥 = 0 sinh 𝑥 = −𝑖 sin(𝑖𝑥) cosh 𝑥 = cos(𝑖𝑥) tanh 𝑥 = −𝑖 tan(𝑖𝑥) 𝑒ln 𝑧 = 𝑧 𝑧𝑏 = 𝑒𝑏 ln 𝑧 26 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES 1 1 − 𝑥 = 1 + 𝑥 + 𝑥2 + 𝑥3 +⋯ 1 1 + 𝑥 = 1 − 𝑥 + 𝑥2 − 𝑥3 +⋯ arctan 𝑥 = 𝑥 − 𝑥3 3 + 𝑥5 5 −⋯ sin 𝑥 = 𝑥 − 𝑥3 3! + 𝑥5 5! −⋯ cos 𝑥 = 1 − 𝑥2 2! + 𝑥4 4! −⋯ sinh 𝑥 = 𝑥 + 𝑥3 3! + 𝑥5 5! +⋯ cosh 𝑥 = 1 + 𝑥2 2! + 𝑥4 4! +⋯ 𝑒𝑥 = 1 + 𝑥 + 𝑥2 2! + 𝑥3 3! −⋯ ln(1 + 𝑥) = 𝑥 − 𝑥2 2 + 𝑥3 3 −⋯ − 1 < 𝑥 ≤ 1 (1 + 𝑥)𝑝 = 1 + 𝑝𝑥 + 𝑝�𝑝 − 1� 2! 𝑥2 + ⋯+ |𝑥| < 1 Leibinz rule for differentiation of integrals 𝑑 𝑑𝑥 � 𝑣(𝑥) 𝑢(𝑥) 𝑓(𝑥, 𝑡) 𝑑𝑡 = 𝑓(𝑥, 𝑣(𝑥)) 𝑑 𝑑𝑥 𝑣(𝑥) − 𝑓(𝑥, 𝑢(𝑥)) 𝑑 𝑑𝑥 𝑢(𝑥) +� 𝑣 𝑢 𝜕 𝜕𝑥 𝑓(𝑥, 𝑡) 𝑑𝑡 example: 𝑑 𝑑𝑥 � 2𝑥 𝑥 𝑒𝑥𝑡 𝑡 𝑑𝑡 = 𝑒𝑥(2𝑥) 2𝑥 𝑑 𝑑𝑥 (2𝑥) − 𝑒𝑥(𝑥) 𝑥 𝑑 𝑑𝑥 (𝑥) +� 2𝑥 𝑥 𝜕 𝜕𝑥� 𝑒𝑥𝑡 𝑡 � 𝑑𝑡 = 𝑒2𝑥2 𝑥 − 𝑒𝑥2 𝑥 +� 2𝑥 𝑥 𝑡𝑒𝑥𝑡 𝑡 𝑑𝑡 = 𝑒2𝑥2 𝑥 − 𝑒𝑥2 𝑥 + � 𝑒𝑥𝑡 𝑥 � 2𝑥 𝑥 To help remember the above 2 formulas, notice that when +𝑥 we get a − shown (i.e. terms flip flop), but when we have −𝑥 the series is all positive terms. These are very important to remember for problems when finding Laurent expansion of a function. Expansion of cos and sin around a point different than 0 expand cos(𝑧) around 𝑎, we get ⎛ ⎜⎜⎜⎝cos(𝑎) − cos(𝑎)(𝑧 − 𝑎)2 2! + cos(𝑎)(𝑧 − 𝑎)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + ⎛ ⎜⎜⎜⎝− sin(𝑎)(𝑧 − 𝑎) + sin(𝑎)(𝑧 − 𝑎)3 3! ⋯ ⎞ ⎟⎟⎟⎠ For example to expand cos(𝑥) about 𝜋we get 27 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES ⎛ ⎜⎜⎜⎝cos(𝜋) − cos(𝜋)(𝑧 − 𝜋)2 2! + cos(𝜋)(𝑧 − 𝜋)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + =0 ���������������������������������������������������⎛ ⎜⎜⎜⎝− sin(𝜋)(𝑧 − 𝜋) + sin(𝑎)(𝑧 − 𝜋)3 3! ⋯ ⎞ ⎟⎟⎟⎠ = −1 + 1 2 (𝜋 − 𝑧)2 − 1 24 (𝜋 − 𝑧)4 + ⋯ so above is easy to remember. The cos(𝑧) part is the same as around zero, but it has cos(𝑎)multiplied to it, and the sin part is the same as the sin(𝑧) about zero but has sin(𝑎) multiplied to it, and the signs are reversed. For expansion of sin(𝑧) use ⎛ ⎜⎜⎜⎝sin(𝑎) − sin(𝑎)(𝑧 − 𝑎)2 2! + sin(𝑎)(𝑧 − 𝑎)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + ⎛ ⎜⎜⎜⎝cos(𝑎)(𝑧 − 𝑎) − cos(𝑎)(𝑧 − 𝑎)3 3! ⋯ ⎞ ⎟⎟⎟⎠ This is the same as the expansion of cos(𝑧) but the roles are reversed and notice the cos part start now with positive not negative term. SO all what I need to remember is that expansion of cos(𝑧) starts with cos(𝑎) termswhile expansion of sin(𝑧) start with the sin(𝑎) term. This is faster than having to do Taylor series expansion to find these series in the exam. Γ� 1 2� = √𝜋 Γ(𝑃 + 1) = 𝑃Γ(𝑃) 28 3.2. HW 1 CHAPTER 3. HWS 3.2 HW 1 31 3.2. HW1 ea CHAPTER 3. HWS o = ie = fede re dk 4a rel Phe Cmrvaiget caul Doe em ao Oe BP A; te 2 Mem Mno82ern = oog+ = Foe = $2 ,3 . Sti . jos * Goo =~ Goo = = (nt® net Gr! - Wy a, Go, Ge, fay! * 4t é? at . " (ere y = 2&z a Gre) Oe ea at K 7 Gye Ceneewarz) Farner Sy 003d 2 | ae | ap I * | +o t+ q5 : 32 3.2. HW1 CHAPTER 3. HWS n se henefafr ao 2 ] %). pn } Sesice < septa hone sever 2 | Se | n=O | a fr crmge, aco pe 2 Lt 2fe)73 > Re] soy 2 = ae at. 21) +2 cd 2 wee — 2 = Ahi ole 21) +2 ~ Ts he3 > 2.8 ai Bute 9 étee- 33  3.2. HW1 CHAPTER 3. HWS. @ = r ) fr atk, S60 eet ¢(i-%) 4 (I-x% = 6-24 994944 74Y¥ oe . 55 2.56 x lo ~# whic 5 smalls them & = 1S ol 36 3.2. HW1 CHAPTER 3. HWS chapter | )4.2 3 ae at nav MF <. amvenpetl sina Soe pase pect , fhe saree S a nos). a < 3A re ~ fre. eur as smaller blo So T-5: fe th Sere the tc Tenens wobie& 19 6-5, fr eh or™ a a tee arate) _ a-arar” [ro ior ler er en he = ae 4 “ ict Zi (Sn vu t. " N agi | ar | < ei, 2 Nene need ier = ~ { =J let E&= la for rome Sere, Gri, 1 Ay ‘ yf t ou E < fat b * x - (NV < w* of tim et 2” > ia” s 4 fog 2 7 + “n> ~ ny @3-& a3 37 © 3.2. HW1 CHAPTER 3. HWS ® cheptes ft we prehmiaey, tet te cle elete if diver ect mow ‘ reqeted 5.1 seit eey pile ® oft 8 ee 7 Z * je i7 es Ze IF te Pre fiminng Zeoalh, f+ a,» if Gn ee - Hood : wal. & au flor divmgent , elt wire teelony ol. f if egrare gent 3 oe atl oy f Tan abd Series =a C1) : i Hel we aye a peme es ee bdivarcest | L- eet - tn (n) fu lege on b—- “ if Jn(n) grows piwch pet oe i l/ dhan on. see OS™ Ph | heate fim Aye [n(aj hae co ints [mera tashn§ 6 needed b fee if fori e ¢ end 38 University Course MATH 121 A Mathematical Tools for the Physical Sciences UC BERKELEY Spring 2004 My Class Notes Nasser M. Abbasi Spring 2004 Chapter 1 Introduction This ismyweb page for courseMATH121A,Mathematicalmethods in physical sciences, that I took in spring 2004 at UC Berkeley. Hard course. Lots of HWs. This is the syllubus 1 CHAPTER 1. INTRODUCTION Professor Fraydoun Rezakhanlou, UC Berkeley Math department, gave the course. A very good teacher. Teaches without using notes, all from memory, which I thought was really amazing. This course was lots of work. About 30-40 problems for HW each week. But only 2 random problems are graded. So it is possible to solve 39 of out 40 problems correctly and get only 50%, if the problem missed happened to be selected to grade. Exams as closed notes and closed book. No cheat sheet either. No calculators. Textbook was Mary Boas, second edition. This seems to be the standard book for this type of course are most universities, at least the ones I know about. It is a a very good book, but more detailed worked examples would have been nice. So another book such as the problem solvers type book might be useful to have. 2 CHAPTER 1. INTRODUCTION For some reason this course (and Math121B) did not have discussion period, this was unfortunate since discussion periods can be really useful. I was told this is because of budget cuts in the California university system. This below is a picture of Evans hall. It is a big tall building full of very smart people. The math department is on the 9th floor. The course was in room 75, which is on the ground floor on Evans hall. 1.0.1 Course description These are the Course outline handouts. Page 1, and page 2. it contains the problems to solve for each HW. Spring 2004 (January-May 2004) Course description: Functions of a complex variable, Fourier series, finite-dimensional linear systems. Infinite-dimensional linear systems, orthogonal expansions, special func- tions, partial differential equations arising in mathematical physics. Intended for stu- dents in the physical sciences who are not planning to take more advancedmathematics courses. Units: 4 Book: MATHEMATICAL METHODS IN PHYSICAL SCI, BOAS. 2nd edition chapters 1,2,4,7,9,14,15 Instructor: Professor Fraydoun Rezakhanlou, Math department, UC Berkeley Position: Associate Professor E-mail: rezakhan@math.berkeley.edu To reduce spam, this address is javascript en- coded. Phone: +1 (510) 642-2838 Office: 815 Evans Hall 3 2.1. Misc. notes CHAPTER 2. MY NOTES, STUDY NOTES 2. The Legendre ODE is solved using series method, assuming 𝑙 is an integer. We get one solution which is Legendre function of first kind 𝑝𝑙(𝑥). What if 𝑙 is not an integer? A: Legendre 𝑃𝑙 is only defined for integer 𝑙? YES? No, there are tables for non-integer, but these cases are not important. 3. What if we get a legendre ODE and we want to find solution for 𝑥 > 1 ? Since legendre functions are only defined for 𝑥 less than one (to have convergence). Physics example? usually 𝑥 is the cosine of an angle so it is ≤ 1. 4. What if 𝑙 is not an integer in the legendre ODE? how to get a solution? this is special cases, not important, look up handbooks. 5. problem I solved in HW#6, chapter 12, 16.3. check my solution. I claimed that the second solution is 𝑁𝑝 but since I found 𝑃NOT to be an integer, hence the second solution is one containing log and not a combination of 𝐽−𝑝. When I solved it in mathematica, I get this solution (notice complex number?), could this second solution be converted to log function? answer: OK, the solution I did will turn out to have log in it if I put p=integer and use L’hospital’s rule to evaluate. 6. When solving for equation 16.1 on page 516, we seem to only take the positive root for the variables, why? see for example page 516. 𝑏 = 2 but it is really 𝑏 = ±2 answer: OK, any of these will give a good solution, just pick one. 7. on page 528, can I just set 𝑛 = 0 always to solve for the indical equation as shown in the example? is it better to solve this using the∑ directly as shown in the example instead of setting up a table? table seems more clear, but the example method seems shorter. 8. How to solve chapter 16, 4.1 part (c) using Bayes rule? I write: Let A=event first chair is empty, let B=event second chair is empty. We need to find 𝑃(𝐴𝐵) = 𝑃(𝐴)𝑃𝐴(𝐵) = ( 110 )( 1 9 ) = 1 90 , but the answer should be 1 45 , what is it I am doing wrong? wrong. 𝑃(𝐴) = 1/5 𝑛𝑜𝑡 1/10. 9. for problem HW 12, chapter 16, 4.8, part b. It says given 2 cards drawn from deck, if you know one is an ace, what it the chance the BOTH are an ace? I know how to solve by the book. but why can I not say the following: since we KNOW that one card is an ace, then the chance that both cards are an ace is just the chance the second card being an ace (since we know the first is an ace). So this should give 3 51 6 2.2. Table summary of topics to study CHAPTER 2. MY NOTES, STUDY NOTES 10. random variable is defined as a function on the sample space. however, it is multi- valued. for example, if x= sum of 2 die throw, then more than one event can give the same random variable. is this OK? I thought a function must be single valued? answer: I am wrong. it is NOT multivalued. 11. check that my solution for chapter 16, 5.1 MATH 121B is correct, I have solution on paper. this is the last HW 2.2 Table summary of topics to study 2.2.1 Math 121A ch. title topics Exam 1 series infinite series, power series,def. of cover- gence, tests for convergence, 1 test for alternating series, power series, bi- nomial series 2 complex numbers finding circle of convergence (limit test), Euler formula 1 power and roots of complex numbers, log, inverse log 4 partial differentiation total diffenertials, chain rule, implicit dif- ferentiation 1 partial diff for max and minumum, La- grange muktipliers, change of variables Leibniz rule for differ- netiation of integrals 14 complex functions Def. of analytic fn, Cauchy-Riemann con- ditions, laplace equation, 1 contour integrals, Laurent series, Residue theorm, methods of finding residues, pole type, evaluating integrals by residue, Mapping, conformal 7 Fourier series expansion of function in sin and cosin, complex form, how to find 2 coeff, Dirichlet conditions, different inter- vals, even/odd, Parseval’s 7 2.2. Table summary of topics to study CHAPTER 2. MY NOTES, STUDY NOTES 15 Laplace/Fourier transforms Laplce transform, table, how to use Laplace to solve F ODE, Methods of finding inverse laplace, partial fraction, convolution, sum of residues, Fourier transform, sin/- consine transforms, Direc Delta Green method to solve ODE using im- pluse 9 Calculus of variations Euler equation solving, Setting up La- grange equations, KE, PE F Solving Euler with constrainsts 8 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑐𝑛 = 1 𝑇 �𝑇 𝑓(𝑥) 𝑒−𝑖𝑛𝑥𝜔 𝑑𝑥 Now, 𝜔, is the angular velocity. i.e. 𝜃 = 𝜔𝑡, so for ONE period 𝑇, 𝜃 = 2𝜋, hence 𝜔 = 2𝜋 𝑇 , so 𝑐𝑛 can be written as 𝑐𝑛 = 1 𝑇 �𝑇 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 2𝜋 𝑇 𝑑𝑥 Notice that in this chapters we use distance for period (i.e. wave length 𝜆) instead of time as period 𝑇. it does not matter, they are the same, choose one. i.e. we can say that the function repeats every 𝜆, or the function repeats every one period 𝑇. When using 𝜆 for period, say −𝑙, 𝑙 or −𝜋, 𝜋 the above equation becomes 𝑐𝑛 = 1 2𝑙 � 𝑙 −𝑙 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 2𝜋 2𝑙 𝑑𝑥 = 1 2𝑙 � 𝑙 −𝑙 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 𝜋 𝑙 𝑑𝑥 note: Above integral for 𝑐𝑛 is for negative 𝑛 as well as positive 𝑛. In non-complex expo- nential expansion, there is no negative 𝑛, only positive. note: 𝑐−𝑛 = ?̄?𝑛 note: there is a relation between the 𝑎𝑛, 𝑏𝑛, and the 𝑐𝑛 which is 𝑎𝑛 = 𝑐−𝑛 + 𝑐𝑛 and 𝑏𝑛 = 𝑖(−𝑐−𝑛 + 𝑐𝑛) IF given 𝑓(𝑥), defined over (0, 𝐿), The algorithm to find Fourier series is this: IF asked to find a(n) i.e. the COSIN series, THEN extend f(x) so that it is EVEN (this makes b(n)=0) and period now is 2L ELSE IF asked to find b(n), i.e. the SIN series, THEN extend f(x) to be ODD (this makes a(n)=0) and period now is 2L ELSE we want the standard SIN/COSIN period remains L, and use the c(n) formula (and remember to do the c(0) separatly for the DC term) END IF END IF 2.3.3.1 Parseval’s theorem for fourier series This theory gives a relation between the average of the square of 𝑓(𝑥) over a period and the fourier coefficients. Physically, it says that this: the total energy of a wave is the sum of the energies of the individual harmonics it carries Average of �𝑓(𝑥)� 2 = �12𝑎0� 2 + 1 2 ∑∞ 1 𝑎 2 𝑛 + 1 2 ∑∞ 1 𝑏 2 𝑛 over ONE period. In complex form, Average of �𝑓(𝑥)2� = ∑∞ −∞|𝑐𝑛| 2. Think of this like pythagoras theorem. For example, given 𝑓(𝑥) = 𝑥, then �𝑓(𝑥)� 2 = 1 2 ∫1 −1 𝑥2𝑑𝑥 = 1 3 , then 1 3 = ∑ ∞ −∞|𝑐𝑛| 2 In the above we used the standard formula for average of a function, which is average of 𝑓(𝑥) = 1 𝑇 ∫ 𝑇 𝑓(𝑥) 𝑑𝑥, here we should need to square 𝑓(𝑥) 11 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 2.3.4 Chapter 15. Integral transforms (Laplace and Fourier transforms) 2.3.4.1 Laplace and Fourier transforms definitions 𝐹 𝑓(𝑥) = 𝐹(𝑝) = � ∞ 0 𝑓(𝑥) 𝑒−𝑝𝑥 𝑑𝑥 𝑝 > 0 𝐹 𝑔(𝑥) = 𝑔(𝛼) = 1 2𝜋 � ∞ −∞ 𝑓(𝑥) 𝑒−𝑖𝛼𝑥 𝑑𝑥 Associate Fourier with 1 2𝜋 . (mind pic: Fourier=Fraction i.e.→ 1 2𝜋) and Fourier goes from −∞ to +∞ (mind pic: Fourier=whole Floor), Fourier imaginary exponent, Laplace real exponent. Note: Laplace transform is linear operator, hence 𝐿�𝑓(𝑡) + 𝑔(𝑡)� = 𝐿𝑓(𝑡) + 𝐿𝑔(𝑡) and 𝐿�𝑐 𝑓(𝑡)� = 𝑐 𝐿𝑓(𝑡) 2.3.4.2 Inverse Fourier and Laplace transform formulas (We do not really use the inverse Laplace formula directly (called Bromwich integral), we find inverse Laplace using other methods, see below) 𝑓(𝑥) = 1 2𝜋𝑖 � 𝑐+𝑖 ∞ 𝑐−𝑖 ∞ 𝐹(𝑧) 𝑒𝑧𝑡 𝑑𝑧 𝑡 > 0 Inverse laplace 𝑓(𝑥) = 1 2𝜋 � ∞ −∞ 𝑔(𝛼) 𝑒𝑖𝛼𝑥 𝑑𝛼 Inverse fourier The Fourier transform has 2 other siblings to it (which Laplace does not), these are the sin and cos transform and inverse transform. I’ll add these later but I do not think we will get these in the exam. Note: To get the inverse Laplace transform the main methods are 1. using partial fractions to break the expression to smaller ones we can lookup in tables 2. Use Convolution. i.e. given 𝑌 = 𝐿�𝑓1� 𝐿�𝑓2� → 𝑦 = ∫ 𝑡 0 𝑔(𝑡 − 𝜏) 𝑓(𝜏) 𝑑𝜏 = 𝑔 ⊗ 𝑓 use this as an alternative to partial fraction decomposition if easier. mind pic: 𝑡 one time, 𝜏 2 times. 3. Use the above integral (Bromwich) directly (hardly done) 4. To find 𝑓(𝑡) from the Laplace transform, instead of using the above formula, we can write 𝑓(𝑡) = sum of residues of 𝐹(𝑧)𝑒𝑧𝑡 at all poles. For example given 𝐹(𝑧), we multiply it by 𝑒𝑧𝑡, and then find all the poles of the resulting function (i.e. the zeros of the denominator), then add these. Note: To find Fourier transform , 𝑔(𝛼), must carry the integration (i.e. apply the integral directly, no tables like with Laplace). Note: we use Laplace transform as a technique to solve ODE. Why do we need Fourier transform? To represent an arbitrary function (must be periodic or extend to be period if not) as a sequence of sin/cosine functions. And why do we do this? To make it easier to analyze it and find what frequency components it has. For continuous function, use fourier transform (integral). note: Function must satisfy Dirichlet conditions to use in fourier transform or Fourier series. 12 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES note: Fourier series expansion of a functionwill accurately fit the function asmore terms are added. But in places where there is a jump, it will go to the average value of the function at the jump. question:When dowe use fourier series, andwhen to use fourier transform?Why dowe need F. transform if we can use F. Series?We use F. transform for continuous frequencies. What does this really mean? 2.3.4.3 Using Laplace transform to solve ODE Remember 𝐿(𝑦) = 𝑌 𝐿�𝑦′� = 𝑝𝑌 − 𝑦0 𝐿�𝑦′′� = 𝑝2𝑌 − 𝑝𝑦0 − 𝑦′0 note: 𝑝 has same power as order of derivative. do not mix up where the 𝑝 goes in the 𝑦′′ equation. remember the 𝑦′0 has no 𝑝with it. mind pic: think of the 𝑦0 as the senior guy since coming from before so it is the one who gets the 𝑝. note: if 𝑦0 = 𝑦′0 = 0 (whichmost HWproblemwas of this sort), then the above simplifies to 𝐿�𝑦′� = 𝑝𝑌 𝐿�𝑦′′� = 𝑝2𝑌 So given an ODE such as 𝑦′′ + 4𝑦′ + 𝑦 = 𝑓(𝑡) → �𝑝2 + 4𝑝 + 4�𝑌 = 𝐿(𝑓(𝑡)) i.e. just replace 𝑦′′ by 𝑝2, etc... This saves lots of time in exams. Now we get an equation with 𝑌 in terms of 𝑝, now solve to find 𝑦(𝑡) from 𝑌 using tables. Notice that solution of ODE this way gives a particular solution, since we used the boundary conditions already. For an ODE such as 𝐴𝑦′′ + 𝐵𝑦′ + 𝐶𝑦 = ℎ(𝑡) its Laplace transform can be written immediately as 𝐴𝑝2𝑌 + 𝐵𝑝𝑌 + 𝐶𝑌 = 𝐿 ℎ(𝑡) 𝑌 = 𝐿 ℎ(𝑡) 𝐴𝑝2 + 𝐵𝑝 + 𝐶 whenever the B.C. are 𝑦′0 = 0 and 𝑦0 = 0 2.3.4.4 Partial fraction decomposition When denominator is linear time quadratic or quadratic time quadratic PFD is probably needed. This is how to do PFD for common cases 13 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 2.3.4.7.1 Solving an ODE using green method Here we are given an ODE, with a forcing function (i.e. nonhomogeneous ODE). And given 2 solutions to it, and asked to find the particular solution. Example, 𝑦′′ − 𝑦 = 𝑓(𝑡) and solutions are 𝑦1, 𝑦2 then the particular solution is 𝑦𝑝 = 𝑦2∫ 𝑦1 𝑓(𝑡) 𝑊 𝑑𝑡 − 𝑦1∫ 𝑦2 𝑓(𝑡) 𝑊 𝑑𝑡where𝑊 = � � 𝑦′1 𝑦′2 𝑦1 𝑦2 � � 2.3.5 Chapter 2. Complex Numbers note: When given a problem such as evaluate (−2 − 2𝑖) 1 5 , always start by finding the length of the complex number, then extract it out before converting to the 𝑟𝑒𝑖𝑛𝜃 form. For example, (−2 − 2𝑖) 1 5 = 2√2 � −1 √2 − 𝑖 √2 � , the reason is that now the stuff inside the brackets has length ONE. So we now get 2√2 � −1 √2 − 𝑖 √2 � = 2√2 𝑒− 3 4𝜋𝑖 and only now apply the last raising of power to get �2√2 𝑒− 3 4𝜋𝑖� 1 5 = 2 3 10 𝑒 3 4𝜋𝑖+2𝑛𝜋 5 for 𝑛 = 0, 1, 2, 3,⋯ make sure not to forget the 2𝑛𝜋, I seem to forget that. 2.3.6 Chapter 9. Calculus of variations 2.3.6.1 Euler equation How to construct Euler equation 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � − 𝜕𝐹 𝜕𝑦 = 0. If integrand does not depend on 𝑥 then change to 𝑦. Example ∫ 𝑥1 𝑥2 𝑦′2 𝑦 𝑑𝑥 → ∫𝑦1 𝑦2 1 𝑥′2 𝑦 �𝑥 ′ 𝑑𝑦� → ∫𝑦1 𝑦2 1 𝑥′ 𝑦 𝑑𝑦 this is done by making the substitution 𝑦′ = 1 𝑥′ and 𝑑𝑥 = 𝑥′ 𝑑𝑦. Now Euler equation changes from 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � − 𝜕𝐹 𝜕𝑦 = 0 to 𝑑 𝑑𝑦 � 𝜕𝐹 𝜕𝑥′ � − 𝜕𝐹 𝜕𝑥 = 0. Normally, 𝜕𝐹 𝜕𝑦 will be zero. Hence we end up with 𝑑 𝑑𝑥 � 𝜕𝐹 𝜕𝑦′ � = 0 and this means 𝜕𝐹 𝜕𝑦′ = 𝑐, and so we only need to do ONE integral (i.e. solve a first order ODE). If I find myself with a 2 order ODE (for this course!) , I have done something wrong since all problems we had are of this sort. 2.3.6.2 Lagrange equations are just Euler equations, but one for each dimension. 𝐹 is now called 𝐿. where 𝐿 = 𝑇 − 𝑉 where 𝑇 = 𝐾.𝐸. and 𝑉 = 𝑃.𝐸., 𝑇 = 1 2𝑚𝑣 2, 𝑉 = 𝑚𝑔ℎ So given a problem, need to construct 𝐿 ourselves. Then solve the Euler-Lagrange equa- tions 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑥 = 0 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑦 = 0 𝑑 𝑑𝑡� 𝜕𝐿 𝜕?̇? � − 𝜕𝐿 𝜕𝑧 = 0 The tricky part is finding 𝑣2 for different coordinates. This is easy if you know 𝑑𝑠2, so just remember those 16 2.3. Math 121 A notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 (polar) 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑑𝑧2 (cylindrical) 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃𝑑𝜙2 (spherical) So to find 𝑣2 just divide by 𝑑𝑡2 and it follows right away the following 𝑣2 = ?̇?2 + 𝑟2?̇?2 (polar) 𝑣2 = ?̇?2 + 𝑟2?̇?2 + ?̇?2 (cylindrical) 𝑣2 = ?̇?2 + 𝑟2?̇?2 + 𝑟2 sin2 𝜃 ?̇?2 (spherical) To help remember these: Note 𝑑𝑠2 all start with 𝑑𝑟2 + 𝑟2𝑑𝜃2 for each coordinates system. So just need to remember the third terms. (think of polar as subset to the other two). Also see that each variable is squared. So the only hard think is to remember the last term for the spherical. Remember that in a system with particles, need to find the KE and PE for each particle, and then sum these to find the whole system KE and PE, and this will give one 𝐿 for the whole system before we start using the Lagrange equations. 2.3.6.3 Solving Euler-Lagrange with constraints The last thing to know is this chapter is how to solve constraint problems. This is just like solving for Euler, expect now we have an additional integral to deal with. So in these problems we are given 2 integrals instead of one. One of these will be equal to some number say 𝑙. Soweneed tominimize 𝐼 = ∫ 𝑥1 𝑥2 𝐹(𝑥, 𝑦′, 𝑦) 𝑑𝑥 subject to constraint that 𝑔 = ∫ 𝑥1 𝑥2 𝐺(𝑥, 𝑦′, 𝑦) 𝑑𝑥 = 𝑙 Follow the same method as Euler, but now we write 𝑑 𝑑𝑥� 𝜕 𝜕𝑦′ (𝐹 + 𝜆𝐺)� − 𝜕 𝜕𝑦 (𝐹 + 𝜆𝐺) = 0 So replace 𝐹 by 𝐹 + 𝜆𝐺 This will give as equation with 3 unknowns, 2 for integration constants, and one with 𝜆, we solve for these given the Boundary conditions, and 𝑙 but we do not have to do this, just need to derive the equations themselves. Some integrals useful to know in solving the final integrals for the Euler problems are these � 𝑐 �𝑦2 − 𝑐2 𝑑𝑦 = 𝑐 cosh−1� 𝑦 𝑐 � + 𝑘 � 𝑐 �1 − 𝑐2 𝑦2 𝑑𝑦 = 𝑐 sin−1�𝑐 𝑦� + 𝑘 � 𝑐 𝑦�𝑦2 − 𝑐2 𝑑𝑦 = 1 𝑐 cos−1� 𝑐 𝑦 � + 𝑘 17 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 2.4 MATH 121B Notes 2.4.1 Chapter 12. Series solution of ODE and special functions Bessel ODE 𝑥2 𝑦′′ + 𝑥 𝑦′ + �𝑥2 − 𝑝2�𝑦 = 0 defined for INTEGER and NON integer 𝑃 first solution 𝑦1 = 𝐽𝑝(𝑥) = ∑ −1𝑛 Γ(𝑛+1)Γ�𝑛+𝑝+1� �𝑥 2 � 2𝑛+𝑝 (for 𝑝 an integer or not) second solution 𝑦2 = 𝑁𝑝(𝑥) = 𝑌𝑝(𝑥) = cos�𝜋𝑝�𝐽𝑝(𝑥)−𝐽−𝑝 sin𝜋𝑝 (note: 𝑝 here is NOT an integer) second solution 𝑦𝑥 contains a log function. note: 𝑝 here IS an integer. Orthogonality ∫1 0 𝑥 𝐽𝑝(𝑎𝑥) 𝐽𝑝(𝑏𝑥) 𝑑𝑥 = ⎧⎪⎪⎪⎨ ⎪⎪⎪⎩ 0 𝑖𝑓 𝑎 ≠ 𝑏 1 2 𝐽 2 𝑝+1(𝑎) = 1 2 𝐽 2 𝑝−1(𝑎) = 1 2 𝐽 ′2 𝑝 (𝑎) 𝑖𝑓 𝑎 = 𝑏 𝑎, 𝑏 are zeros of 𝐽𝑝 recursive formula 𝑑 𝑑𝑥 �𝑥𝑝𝐽𝑝� = 𝑥𝑝𝐽𝑝−1, 𝑑 𝑑𝑥� 1 𝑥𝑝 𝐽𝑝� = − 1 𝑥𝑝 𝐽𝑝+1, 𝐽𝑝−1 + 𝐽𝑝+1 = 2𝑝 𝑥 𝐽𝑝, 𝐽𝑝−1 − 𝐽𝑝+1 = 2𝐽′𝑝 𝐽′𝑝 = − 𝑝 𝑥 𝐽𝑝 + 𝐽𝑝−1 = 𝑝 𝑥 𝐽𝑝 − 𝐽𝑝+1 NOTICE: No Rodrigues formula for Bessel func, since not polyn. notes: We used a generalized power series method to find the solutions. IF 𝑝 is NOT an integer, then 𝐽𝑝 and 𝐽−𝑝 (or𝑁𝑝) are two independent solutions IF 𝑝 is an integer, then 𝐽𝑝 and 𝐽−𝑝 are NOT two independent solutions, use log for 𝑦2 𝐽𝑝 is called Bessel function of first kind, and 𝑌𝑝 is called second kind. 𝑝 is called the ORDER. IF 𝑝 = 𝑛 + 1 2 , a special case, we get spherical bessel functions 𝑗𝑛(𝑥) and 𝑦𝑛(𝑥) 𝑗𝑛(𝑥) = � 𝜋 2𝑥 𝐽𝑛+ 1 2 (𝑥) = 𝑥𝑛�−1 𝑥 𝑑 𝑑𝑥 � 𝑛 � sin 𝑥 𝑥 � 18 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑃(𝐴𝐵) = 𝑃(𝐴) (𝐵) If A,B independent Probability that A OR B will happen is: 𝑃(𝐴 + 𝐵) = 𝑃(𝐴) + (𝐵) − 𝑃(𝐴𝐵) 𝑃(𝐴 + 𝐵) = 𝑃(𝐴) + (𝐵) IF A,B are mutually exclusive This means that 𝑃(𝐴𝐵) = 0 if they are mutually exclusive (obvious) 𝑃(𝐴 + 𝐵 + 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − {𝑃(𝐴𝐵) + 𝑃(𝐴𝐶) + 𝑃(𝐵𝐶)} + 𝑃(𝐴𝐵𝐶) note: 𝑃𝑛𝑟 = number of permutations (arrangements) or 𝑛 things taken 𝑟 at a time. 𝑃𝑛𝑟 = 𝑛! (𝑛−𝑟)! Here order is important. i.e. ABC is DIFFERENT from CAB, hence this number will be larger than the one below. ⎛ ⎜⎜⎜⎜⎜⎝ 𝑛 𝑟 ⎞ ⎟⎟⎟⎟⎟⎠ = 𝐶 𝑛 𝑟 = 𝑛! (𝑛 − 1)! 𝑟! Number of combinations OR selections of 𝑛 things 𝑟 at a time. here order is NOT impor- tant. so ABC is counted the same as CAB, hence this number will be smaller. note: In how many ways can 10 people be seated on a bench with 4 seats? A) ⎛ ⎜⎜⎜⎜⎜⎝ 10 4 ⎞ ⎟⎟⎟⎟⎟⎠4! = 10! 6!4!4! = 10! 6! = 10 × 9 × 8 × 7 To understand this: ⎛ ⎜⎜⎜⎜⎜⎝ 10 4 ⎞ ⎟⎟⎟⎟⎟⎠ is the number of ways 4 people can be selected out of 10. ONCE those 4 people have been selected, then there are 4! different ways they can be arranged on the bench. Hence the answer is we multiply these together. note: Find number of ways of putting 𝑟 particles in 𝑛 boxes according to the 3 kinds of statistics. Answer 1. For Maxwell-Blotzman (MB) it is 𝑛𝑟 2. For Fermi-Dirac (FM), it is 𝑛𝐶𝑟 3. For Bose-Einstein (BE) it is 𝑛+1𝐶𝑟 note: If asked this: there is box A which has 5 red balls and 6 black balls, and box B which has 5 red balls and 8 white balls, what is the prob. of picking a red ball? Answer: P(pick box A) P(pick red ball from it) + P(pick box B) P(pick red ball from it) note: If we get a problem such as 2 boxes A,B, and more than more try picking balls, it is easier to draw a tree diagram and pull the chances out the tree than having to calculate them directly in the exam. Tree can be drawn in 2 minutes and will have all the info I need. note: write down the cancer chance problem. note: random variable 𝑥 is a function defined on the sample space (for the example, the sum of 2 die throw). The probability density is the probability of each random variable. average or mean of a random variable 𝜇 = ∑𝑥𝑖 𝑃𝑖 where 𝑃𝑖 is the probability of the random variable. The Variance Var measures the spread of the random variable around the average, also called dispersion defined as 21 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES 𝑉𝑎𝑟(𝑥) = ��𝑥𝑖 − 𝜇� 2 𝑃𝑖 Standard deviation is another measure of the dispersion, defined as 𝜎(𝑥) = √𝑉𝑎𝑟(𝑥) Distribution function is just a histogram of the probability density. it tells one what the probability of a random variable being less than a certain 𝑥 value. see page 711. 2.4.3 Chapter 13. PDE 𝑃𝐷𝐸 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑛𝑜𝑡𝑒𝑠 Laplace 𝑢�𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 0 describes steady state (no time) of region with no source for example, gravitional potential with nomatter, electrostatic potential with no charge, or steady state Temp. dis- tribution Poisson 𝑢�𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 𝑓�𝑥, 𝑦, 𝑧� Same as Laplace, i.e. sescribes steady state, how- evere here the source of the field is present. 𝑓�𝑥, 𝑦, 𝑧� is called the source density. i.e. it is a function that de- scribes the density distribution of the source of the potential. Diffusion or 𝑢�𝑡, 𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 1 𝛼2 𝜕𝑢 𝜕𝑡 Here 𝑢 is usually the temperature 𝑇 function. Now time heat equation is involved. So this equation is alive. Wave equation 𝑢�𝑡, 𝑥, 𝑦, 𝑧� ∇ 2𝑢 = 1 𝑣2 𝜕2𝑢 𝜕𝑡2 Here 𝑢 is the position of a point on the wave at time 𝑡. Notice the wave equation has second derivative w.r.t. time while the diffusion is first derivative w.r.t. time Helmholtz 𝐹�𝑥, 𝑦, 𝑧� ∇ 2𝐹 + 𝑘2𝐹 = 0 The diffusion and wave equation generate this. This is the equation SPACE only solution of the wave and heat equa- tions. i.e. 𝑢 = 𝐹�𝑥, 𝑦, 𝑧�𝑇(𝑡) is the solution for both heat and wave eq. Each of these equations has a set of candidate solutions, which we start with and try to fit the boundary and initial condition into to eliminate some solution of this set that do not fit until we are left with the one candidate solution. We then use this candidate 22 2.4. MATH 121B Notes CHAPTER 2. MY NOTES, STUDY NOTES solution to find the general solution, which is a linear combination of it. We use fourier series expansion in this part of the solution. In table below I show for each equation what the set of candidate solutions are. Use these to start the solution with unless the question asks to start at an earlier stage, which is the separation of variables. So the algorithm for solving these PDE is Select THE PDE to use ----> Obtain set of candidate solution ----> Eliminate those that do not fit -----> obtain the general solution by linear combination (use orthogonality principle here) PDE candidate solutions notes ∇ 2𝑢 = 0 𝑢�𝑥, 𝑦� = ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎩ 𝑒𝑘𝑦 cos 𝑘𝑥 𝑒𝑘𝑦 sin 𝑘𝑥 𝑒−𝑘𝑦 cos 𝑘𝑥 𝑒−𝑘𝑦 sin 𝑘𝑥 for 2 dimensions ∇ 2𝑢 = 𝑓�𝑥, 𝑦, 𝑧� 𝑢�𝑥, 𝑦, 𝑧� = − 1 4𝜋 ∫∫∫ 𝑓�𝑥′,𝑦′,𝑧′� � (𝑥−𝑥′)2+�𝑦−𝑦′� 2 +(𝑧−𝑧′)2+ 𝑑𝑥′ 𝑑𝑦′ 𝑑𝑧′ 𝑓�𝑥′, 𝑦′, 𝑧′� is a function that describes mass density distribution evaluated at point 𝑥′, 𝑦′, 𝑧′. The point 𝑥, 𝑦, 𝑧 is where we are calculating the potential 𝑢 itself ∇ 2𝑢 = 1 𝛼2 𝜕𝑢 𝜕𝑡 𝑢(𝑡, 𝑥) = ⎧⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩ 𝑒−𝑘2𝛼2𝑡 cos 𝑘𝑥 𝑒−𝑘2𝛼2𝑡 sin 𝑘𝑥 for one space dimension 23 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES 2.4.3.3 Wave equation in polar coordinates 2.5 General equations sin 𝑛𝑥 = 𝑒𝑖𝑛𝑥 − 𝑒−𝑖𝑛𝑥 2𝑖 cos 𝑛𝑥 = 𝑒𝑖𝑛𝑥 + 𝑒−𝑖𝑛𝑥 2 � sin 𝑥 cos 𝑥 𝑑𝑥 = ln(sin 𝑥) csc 𝑥 = 1 sin 𝑥 average value of 𝑓(𝑥) over [𝑏, 𝑎] = ∫𝑏 𝑎 𝑓(𝑥)𝑑𝑥 𝑏 − 𝑎 cos2 𝑘𝑥 = 1 + cos(2𝑘) 2 sin2 𝑘𝑥 = 1 − cos(2𝑘) 2 sin𝐴 sin𝐵 = 1 2 [ cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)] cos𝐴 cos𝐵 = 1 2 [ cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵)] sin𝐴 cos𝐵 = 1 2 [ sin(𝐴 − 𝐵) + sin(𝐴 + 𝐵)] I need a geometric way to visualize these equations, but for now for the exam remember them as follows: they all start with 𝐴 − 𝐵 , and when the functions being multiplied are different on the LHS, we get sin on the RHS, else we get cos (think of cos as nicer, since even function :). � tanh(𝑥) = ln(cosh 𝑥) � tan 𝑥 = − ln(cos 𝑥) ∫𝑏 𝑎 cos2 𝑘𝑥 𝑑𝑥 = 𝑏−𝑎 2 . if 𝑘(𝑏 − 𝑎) is an integer multiple of 𝜋. (the same for sin2 𝑘𝑥), for example ∫ 1 −1 cos2 𝜋𝑥 𝑑𝑥 = 1 , ∫ 1 −1 cos2 5𝜋𝑥 𝑑𝑥 = 1, ∫ 1 −5 cos2 7𝜋𝑥 𝑑𝑥 = 3 , ∫ 1 −1 sin2 𝜋𝑥 𝑑𝑥 = 1, etc... this can be very useful so remember it! ∫𝑏 𝑎 cos 𝑘𝑥 𝑑𝑥 = 0 if over a complete period. same for sin 𝑥, for example ∫ 𝜋 −𝜋 cos 𝑘𝑥 𝑑𝑥 = 0 sinh 𝑥 = −𝑖 sin(𝑖𝑥) cosh 𝑥 = cos(𝑖𝑥) tanh 𝑥 = −𝑖 tan(𝑖𝑥) 𝑒ln 𝑧 = 𝑧 𝑧𝑏 = 𝑒𝑏 ln 𝑧 26 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES 1 1 − 𝑥 = 1 + 𝑥 + 𝑥2 + 𝑥3 +⋯ 1 1 + 𝑥 = 1 − 𝑥 + 𝑥2 − 𝑥3 +⋯ arctan 𝑥 = 𝑥 − 𝑥3 3 + 𝑥5 5 −⋯ sin 𝑥 = 𝑥 − 𝑥3 3! + 𝑥5 5! −⋯ cos 𝑥 = 1 − 𝑥2 2! + 𝑥4 4! −⋯ sinh 𝑥 = 𝑥 + 𝑥3 3! + 𝑥5 5! +⋯ cosh 𝑥 = 1 + 𝑥2 2! + 𝑥4 4! +⋯ 𝑒𝑥 = 1 + 𝑥 + 𝑥2 2! + 𝑥3 3! −⋯ ln(1 + 𝑥) = 𝑥 − 𝑥2 2 + 𝑥3 3 −⋯ − 1 < 𝑥 ≤ 1 (1 + 𝑥)𝑝 = 1 + 𝑝𝑥 + 𝑝�𝑝 − 1� 2! 𝑥2 + ⋯+ |𝑥| < 1 Leibinz rule for differentiation of integrals 𝑑 𝑑𝑥 � 𝑣(𝑥) 𝑢(𝑥) 𝑓(𝑥, 𝑡) 𝑑𝑡 = 𝑓(𝑥, 𝑣(𝑥)) 𝑑 𝑑𝑥 𝑣(𝑥) − 𝑓(𝑥, 𝑢(𝑥)) 𝑑 𝑑𝑥 𝑢(𝑥) +� 𝑣 𝑢 𝜕 𝜕𝑥 𝑓(𝑥, 𝑡) 𝑑𝑡 example: 𝑑 𝑑𝑥 � 2𝑥 𝑥 𝑒𝑥𝑡 𝑡 𝑑𝑡 = 𝑒𝑥(2𝑥) 2𝑥 𝑑 𝑑𝑥 (2𝑥) − 𝑒𝑥(𝑥) 𝑥 𝑑 𝑑𝑥 (𝑥) +� 2𝑥 𝑥 𝜕 𝜕𝑥� 𝑒𝑥𝑡 𝑡 � 𝑑𝑡 = 𝑒2𝑥2 𝑥 − 𝑒𝑥2 𝑥 +� 2𝑥 𝑥 𝑡𝑒𝑥𝑡 𝑡 𝑑𝑡 = 𝑒2𝑥2 𝑥 − 𝑒𝑥2 𝑥 + � 𝑒𝑥𝑡 𝑥 � 2𝑥 𝑥 To help remember the above 2 formulas, notice that when +𝑥 we get a − shown (i.e. terms flip flop), but when we have −𝑥 the series is all positive terms. These are very important to remember for problems when finding Laurent expansion of a function. Expansion of cos and sin around a point different than 0 expand cos(𝑧) around 𝑎, we get ⎛ ⎜⎜⎜⎝cos(𝑎) − cos(𝑎)(𝑧 − 𝑎)2 2! + cos(𝑎)(𝑧 − 𝑎)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + ⎛ ⎜⎜⎜⎝− sin(𝑎)(𝑧 − 𝑎) + sin(𝑎)(𝑧 − 𝑎)3 3! ⋯ ⎞ ⎟⎟⎟⎠ For example to expand cos(𝑥) about 𝜋we get 27 2.5. General equations CHAPTER 2. MY NOTES, STUDY NOTES ⎛ ⎜⎜⎜⎝cos(𝜋) − cos(𝜋)(𝑧 − 𝜋)2 2! + cos(𝜋)(𝑧 − 𝜋)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + =0 ���������������������������������������������������⎛ ⎜⎜⎜⎝− sin(𝜋)(𝑧 − 𝜋) + sin(𝑎)(𝑧 − 𝜋)3 3! ⋯ ⎞ ⎟⎟⎟⎠ = −1 + 1 2 (𝜋 − 𝑧)2 − 1 24 (𝜋 − 𝑧)4 + ⋯ so above is easy to remember. The cos(𝑧) part is the same as around zero, but it has cos(𝑎)multiplied to it, and the sin part is the same as the sin(𝑧) about zero but has sin(𝑎) multiplied to it, and the signs are reversed. For expansion of sin(𝑧) use ⎛ ⎜⎜⎜⎝sin(𝑎) − sin(𝑎)(𝑧 − 𝑎)2 2! + sin(𝑎)(𝑧 − 𝑎)4 4! −⋯ ⎞ ⎟⎟⎟⎠ + ⎛ ⎜⎜⎜⎝cos(𝑎)(𝑧 − 𝑎) − cos(𝑎)(𝑧 − 𝑎)3 3! ⋯ ⎞ ⎟⎟⎟⎠ This is the same as the expansion of cos(𝑧) but the roles are reversed and notice the cos part start now with positive not negative term. SO all what I need to remember is that expansion of cos(𝑧) starts with cos(𝑎) termswhile expansion of sin(𝑧) start with the sin(𝑎) term. This is faster than having to do Taylor series expansion to find these series in the exam. Γ� 1 2� = √𝜋 Γ(𝑃 + 1) = 𝑃Γ(𝑃) 28 3.2. HW 1 CHAPTER 3. HWS 3.2 HW 1 31 3.2. HW1 ea CHAPTER 3. HWS o = ie = fede re dk 4a rel Phe Cmrvaiget caul Doe em ao Oe BP A; te 2 Mem Mno82ern = oog+ = Foe = $2 ,3 . Sti . jos * Goo =~ Goo = = (nt® net Gr! - Wy a, Go, Ge, fay! * 4t é? at . " (ere y = 2&z a Gre) Oe ea at K 7 Gye Ceneewarz) Farner Sy 003d 2 | ae | ap I * | +o t+ q5 : 32 3.2. HW1 CHAPTER 3. HWS n se henefafr ao 2 ] %). pn } Sesice < septa hone sever 2 | Se | n=O | a fr crmge, aco pe 2 Lt 2fe)73 > Re] soy 2 = ae at. 21) +2 cd 2 wee — 2 = Ahi ole 21) +2 ~ Ts he3 > 2.8 ai Bute 9 étee- 33  3.2. HW1 CHAPTER 3. HWS. @ = r ) fr atk, S60 eet ¢(i-%) 4 (I-x% = 6-24 994944 74Y¥ oe . 55 2.56 x lo ~# whic 5 smalls them & = 1S ol 36 3.2. HW1 CHAPTER 3. HWS chapter | )4.2 3 ae at nav MF <. amvenpetl sina Soe pase pect , fhe saree S a nos). a < 3A re ~ fre. eur as smaller blo So T-5: fe th Sere the tc Tenens wobie& 19 6-5, fr eh or™ a a tee arate) _ a-arar” [ro ior ler er en he = ae 4 “ ict Zi (Sn vu t. " N agi | ar | < ei, 2 Nene need ier = ~ { =J let E&= la for rome Sere, Gri, 1 Ay ‘ yf t ou E < fat b * x - (NV < w* of tim et 2” > ia” s 4 fog 2 7 + “n> ~ ny @3-& a3 37 © 3.2. HW1 CHAPTER 3. HWS ® cheptes ft we prehmiaey, tet te cle elete if diver ect mow ‘ reqeted 5.1 seit eey pile ® oft 8 ee 7 Z * je i7 es Ze IF te Pre fiminng Zeoalh, f+ a,» if Gn ee - Hood : wal. & au flor divmgent , elt wire teelony ol. f if egrare gent 3 oe atl oy f Tan abd Series =a C1) : i Hel we aye a peme es ee bdivarcest | L- eet - tn (n) fu lege on b—- “ if Jn(n) grows piwch pet oe i l/ dhan on. see OS™ Ph | heate fim Aye [n(aj hae co ints [mera tashn§ 6 needed b fee if fori e ¢ end 38 3.2. HW1 CHAPTER 3. HWS ©) cafes |] leo s pepe «tN eer prison E.5%)| Jat fe corr ne. 2 i,4d,t-.te ; = — - yp’ AR BR are , ¥ - - : (od 2 fe Te hoe Ete Ee vt pl Bi de Ja gn Bere a. > a : tn 4 / Lh C22} Shu fed Syrue xt di fen pet Sas probes Seb Syne cack Pls Saavphig Phat ¢ ‘ i . Cortes par S15 Pain i 275 tame in el t nr Cor ey Se I ! Lg oh gol el 1 = 2) ae sa _—_— + + \s" So fF Oe Ind wt “peer las nee oem 7 i - ! a u I i ogame, compre * 27 tT 3S te Set ee fact. fawn on st V4 Larger Than each es ; den of lero: cen By a pes: 8. Bee cpa hence =z h er n ia tna) Irin) 18 smaller Tren fre : ’ all fiers Kae ns. 41 3.2. HW1 CHAPTER 3. HWS Be Chapt ' : — fest use inlegont tt & Fee if Seren duverns Geergee z | ! ! i . _ a n fnty ° ela aed + Te Dar Seiten aft fay on eile awed Guy in ¢ Can caee ivteped? Tact . L f a in(Inem)| = In (In(w)) = ln (yee = wo Tal “ st divertes { = 7 foley ihe Tatesrad diverpee , Ba Z 42 CHAPTER 3. HWS @) 3.2. HW1 chapter | Ht jeu ) Ss —— oo. ' (n HS) ‘ey Te bd ay oe eee Jae: «= samy cates) fern Coverings | token oe eo) I-—s = heute. | eh é e.1s fees! ten enfle 5 rieg Past. lees | vse integral fat fo preve the felimoing Sen ted prse ai of Ped - Breer tent M\s "Ms | nf ) C divegant if fel eo AO ox -Ftl Le J = dem = Poe = = / = s -P+l ite ace OX fihen PIU, then rm ss 5 grave herve excl la-par ee oc eee — here Tso. hena [ereges| ; . [ tinen Pel. then (PP 93 Mepis, Aa Po og = clenemene far 209 neo. So idnemas sine hon ce eiveipa vet cease / whee PE Pos ec ea I dieepeo: eee I-p “a have. | Seq vents diver ces aon 43 3.2. HW1 CHAPTER 3. HWS | © crap ten Seren Cetera at fe. 18) ee re) “7 & fed ‘f dlivenpe oo a = 2 =— ? zl " 7” Aig > ‘aS * ey rd ) a> ?* . 2 ar in * — 7 neste 2 = arg) jan = pa => 2 See deere ater hens os ty! lwettenny th Cen vegan ———— . 2 x be Lin py Ann 3 , Moe feat me (one = = o yo A aye ae =| ——*; law pane Thee 7 <= % Fn Cae SE jranre | een een gnu ly _— ke . ny y inesy! . ot Be - (ae t) 6.2) = 5 (a!) &,: ( a, are ae loyal fan)! (2le+ " tery is @ SC inert) fon)! aS ant Ine}! 2" é 1); oy (ee vee he sia ee a BC ‘ 1 S (rn!) ( ary! mye¢ alia Mine dipbeyy : = |vl¥ engines) (nw n! . * f ' 4 Az © Cavey Gl seis’ feast ty (aed! 7 nee Cerntey feria dafi ave (#6 i fenre,! > olney BA es vian (2 { i S(aei) fen)! m iGo ) € t41) (aoe) noe ga ea S fn etna => (tay! (eee) Ciner) TEA FE 46 CHAPTER 3. HWS Adie: by ae 2 @ : df hee [eer [< +27 vie ra tie feo wad " = oe ce yee nel : Joe joo 3 Ga os 1 Ants a A Cari) tee a att lee a at) lee n x | odes | Jere. (nays Jee dpe A a | Pame ; xf nore prey : eof 1 - joe pe y | Sa in A se (Joe) (1) = n> ( qc ‘ Yaw dé | divege \ 47 3.2. HW1 CHAPTER 3. HWS cheptes f By , iiss fo oe Lomcten 6 feo te qin @ : 3} wie ipod fen yore a « — = vw L ast g -4a a . Mest 0 Gaus Kirt wed te tend Whe empeee 4 r eat fhe leg jp wehar o"_ ny wo WAR oo aul loeb ot Ss leg 2" 2 née 2 " w + 3 far fey ¥ ie 2 2 in ete ov Sipe lef phen oh See loon Frees e 4 2” in te orem Cer at eu dh = 4 Jee deronatt: <2 : \ 2" f a:> : % with red 4 | nt fevecent lemvente » Bete Gearews } 2 } Heme vite tesp (al 2 ts arene Le OO 7 es ra ee we « ae few : = p op eae 1 a ee a roe 1. naw 2% i ue fg “ fo . _ . . Ohh ee YO jrese 6 ume Hae. 2 hi jhe Hhiia « Cinite New, 4. ( oo , = 2" n”~ { ees | nF enawees 48 CHAPTER 3. HWS 3.2. HW1 cheytn ! F pn G3 [7. 2] tet te follows aberrant Sermon eee iy - = (2) ‘ Fist fee if aes Gneegotl tee if fe, nel ae fle paced BH dg wien leat fitte an aftepare ate Semaeg fgagi ii ths: Aen Cot int 1s temteicentd | . 4 = i wnt fu ok ot og” we hod Woe e a fincre leg , Thon by eos ales 2? sine on gerd fasts esate al lad gros plegny | #6 ¢ wg lacy? oo S ee 7 Le eee o 2 Coe Ajvevfes, Sa fermen ~ net ae — ot gay stil levvercet] haweret ° An alten Semry Greer fen of } any } Z| Seiad jar | tim ay == n= Ant 2 | 2" a) = 2" | ati eel 2 Jas] | nt /4 | “ wt : (2-1) ¢ wr) ( Ly | </ ey '| Forme! > | PEN |. LB | aT wm pes abe 3 & Tew te | aa | ° >| ¥ Made: 51 3.2. HW1 CHAPTER 3. HWS ao Cheptet ] 17-3} bie affeinete Senee Et Su net > a yt Finch cle Phe cabselm te Comacgance tea lao at = tl te oa cquvee cant Servs br in Fegened feat (fee 6th) - oe — } oD a at} hence since abs. Gorngemt, EEL iy [lonvercerl] ee el ? n “2 jie t fee at S weit mee tevin a \ = A Sime SL pimp L "n bwce 9, hom te dls wy nef abealnt Cvs cont =) eet loo le «t [ @nrey |—~ ; = __Intieny | Mnf) beep bee ay A )4-) a porer | | | sey) ) anti| a \ a, | Se are tole at BO 4, NSe Sos fiw LG. hem [eonnnjee Gnditinelly | nm In) "g “ een neerreamensis incr 52 3.2. HW1 CHAPTER 3. HWS Che pte | S. 6 } fost altemate Sera ~ al \ 2 = (-0) 4 : wait nil na . . [hee alk Aes a we? tosebte Con vangel , Teme ioe we I de wie | a. | —— | ie { nt! | (hott fntip(nts! _ nt gna & | (nyo) ater | ‘we o) | - & [ant I ig [ann | Se baa) \ a~ | hewn ex [divin | av stm Eo Fi] tat te Grenson dene in aa = ‘ | wet Are) (n+) gin ee 4-d ° a hi = r = =) pret tty fond ee On > go ANE G anes 1+ S>8) 9 Se put Teh peers dee ra fre ast a . fis ia _ eter. vdmvators) np bn ee " Jowp 7 ee esi)” AB pen Gaal? foots y Z “Dols :-yT_ 53 2) : 3.2. HW1 CHAPTER 3. HWS chapler | st fen)! e ! = ae ye” feat fim gig hen nae we Pa -_) : cd we “ pa ; tina 5 a> ~ fe : naw SP siieeer 2 fogiee them ” hence tee ad ware fen Pag Hs Che bytes = £ Bren Ff fen) = Boerne 2 an aq st? n) 7 Vi, 7? BU fat) % ' Pease i faery ct Msbets e st etm: no; EE OR coe: CE secs a eta | eek yt 2 (ie amen” fla erate, oy = oo ue “3 1 = levescarse _sotrmeow bane ne Ciers ny i* - i See i eal a7 we: ‘ ' 3 ( ny we er ? fins we othe = = ‘ + Gera)x(nea}ae oe tn a fama, faaes Cnr) (tnti - oie EGE os =, L ‘ = z (ars fintty (ret) 3. ant ra eters A te G4 +h iss d ?! any [rs > % apel" ad Sat send 56 3.2. HW1 CHAPTER 3. HWS Chegte« 4 fegt foe Camrtngemes or diverges Co &«) la 2 oF = ! —_— 2 net oat cect beg vehapud : fer: ephic hie Cente Ahan pest ere ewes | eimweepioon] fe: aftmi he + an Ee cet erie ‘| | ap _. A 5 a 4 oes eet ot Prd wet | Ges tes yg t a” 2 4] fi (= wis Ie 443 ay “42 ty = 4 3 2 \ & ‘ =e he Gt = 2 443 ert 7 Oy “f 4 3 z | ce i 4 ve? B+ > oath gry Uy i t ! Mi. wo | a Se | = CURT easy) | ° 57 3.2. HW1 CHAPTER 3. HWS at 3 . a) (an#3) Hh poate R20 mr pnts (becenn denn ster ta lenges Paw Lhe se pres foot fe oa pometatet So mace{ vent testes ° 58 3.2. HW1 CHAPTER 3. HWS . Qe chap I fof Onell 4 tay | fats i a aA — (n-2) — Be 4! as (x-2). rn “ fae ja ye -| aaa i. Ge). 3" fT eae | re an o = i C2-r) | a fin ( x { i =F | naa fa * Zz 61 3.2. HW1 cha | peel where jammo pave rfrt a [do. 14 | —— = F fo 3 = [4 late CHAPTER 3. HWS 62 3.2. HW1 CHAPTER 3. HWS Cheetos ‘ a spfinalpnse & [1a - 20} feet foe Comyn oy A cy én = 2 —_— a Bee if = (-0 = ey Mtn “o hn iy Lag & . (-\) => => pr kh ge eH fe See : ha! Bari) Sie num | Caer} 2 i i ~ é ! 2 » s f2 oS wi | | 24 | nigeas KA MT, nei fd ee at Gy wif fe - a ho value Fo f ff te tevemgen foe on men TO, bat ge x4! in 63 3.2. HW1 CHAPTER 3. HWS } Ea Che gts eeu Tat) Feed pew 700 fr om 4 & ¥ ~@e rine, Ke EF ee ye ef gt dence ( La VCe~ x) =| 7 ane See eH = fot Gun +4, x4 x 4 Gy 2TH oe Bhan (OY a4, 24 HN iG wi oR BP al = | on é 2 Go +a Me ew +2 3" + Oo+ Om 4 a = a, s. Mae Meee + Ble 3 = 4 / de a )+ ond ey + Aix at @a-¥)e> (3 4) Ht et by tpapersd dep, han te de = { aj se 4, -t2 oa _ fa= a z 4-% = ay =o bea Se) 2? 3 ! 5 Ge ce A ee de a +4420 == Fy a) oS nty zy a —_. 1 | + a 5 ip ‘s "be & ie. =e + — ee See (2) ay 66 3.2. HW1 CHAPTER 3. HWS ch pls f inicinc Woe = ot yg” wo. weave, & agp eo ae Beas + 2 4 é # fBdie Bos Oe ae BP a 67 3.2. HW1 CHAPTER 3. HWS che phe ‘ B Pelé] pad pom fee f« ~ sunf In Citx) J 2 Inflae) = aie ey les, y . ~|42<! 2 3° + mally es ae EE. oe awe . Ly ata lef Indl) = 9» ? © Sieg - 24 4. > o een: ° alingr yo lpn oh 2 Tenth jase 7, f+ |= ae x? - Jo L 2: Ee ut z 3 ae Boy 7 ~ 3 dm (ie wate th as Maracte , og ‘ pm fb feta Crp tess = fetievy= Eales ft 4 ! L * + lowe ender fy ta (en) +1 da (1** | L By s jotiong at ah 1 femme ie fel i= See y+d CegeEley (e-gre) fr mY) 2 ies ja oy. Pe x Be) ; ete a ae g iel ete TES I sehr) = (wag) (Ing, (Ee dO) s (eae (j- gee dee Gt a ss 68 3.2. HW 1 CHAPTER 3. HWS 71 3.2. HW1 CHAPTER 3. HWS cA \y 13-42 | i: aad foo= wA pon! 4 Vx Y — V Mors v + ees = wns tes = AS Cl t=) =5 t —— avd « 2¢ Ye a exp . < + — : l + z= ete 4 P * +z)" = (#8) = beret LPNs, ODES) xz - 3! ‘ 4g) a(ity is Inger 1G) 2, at 72 3.2. HW1 CHAPTER 3. HWS Abie anethe wy to solve (734A) 4s Che gts. az 2S” — [3.40 } expend fide fe abewt pee? eee , " a ot pest fred paces Few ¢ fix) = a £(a)- f(x) z -t x f"fa) = & Eo -% "a, 2 - la e fe peer Sonn j- x? at tts 2 meciesitt i a) ey ss finiz flat (ma) £itay 7 gy beng” Fe gy Or a ae 2 oye eck + fm-13) (-e =) 41 fea) 2 = fe + (eb Ge + FH (ed) (5 AEP pers) ie) 2o4 &9) _ as Ge 2! ; Te alana ce ERE Yrerwe 3 - ° ZaQacer ~——_- —----- _- eases - li“ , I 7 - = cowl Gets} ut -is) 4 e- BS “ x § aor ye (4) / Barer ee, 73 3.3. HW 2 CHAPTER 3. HWS ch! @ [4.6] Shs phat bh Ui-x) ex «th aq ere fens Blan a2 a-ocre fe [x] <a} expord Ly (1-x) “We Pewnr Stee flan Ci-x) = wx - 2. x? _x? Fiat fem Se ) erm / </ @y* | x go co2TT me Jeol ow BM 2 < 03-0056. @ED . 76 3.3. HW 2 CHAPTER 3. HWS aa @ i on M4) Find sum of 3 Sere #2 toil —(trtete- a 2.t lL hate Serine = vrata Pr a nine) oT tet = ti Jit +4-l - fied-~ rfp cee 8 | hence |S mw 2ay = ! Mew te frre th Renindac, feurithe <4 it a1 #1 4 ! ot £ z *3 = +2 -2 + F a es Fe tpl Ramm 1413 otal mee aug Bhat far ellarrahur Semes , ty Femindar atts teres 75 S| Gaen™ teem. / Aane , Be 14,5,] fap. js LL pom 1, An then E : ot pec02 4B a Hy wr ASioe , an = Hitae) = Zeo(tni) = 6: oe arise eohile a ghs @eeFd7s eshed >a,7 Aenea net Pediat Coppeact, a merch smeddec - 77 3.3. HW 2 CHAPTER 3. HWS ch) ®@ Use prowat Bening comma The poelin wt tha Sram 4 a nite) at mead jx Jaf pee Senne Lo Clr?) Ln ( [+x) = x= By ae a = -2y-- 4 Molnee. by x? 2 3 & + wm he (itn) = i ar, a = STR ting} so f= ae oxi Gu? i = + =- ae © Feg¢ 4 * ~~ Sem" 5 gx? _ i325! 2 = 3 = to = 3 tox” | go ® Iyx0 x 4 * ae f =O = 360x™ =z s + SYA" siege 2, 4 ce 4 2 flared) = —(@2) 362 4 law> Bey try? ete 3 [* —6- 6950 Chtier Mg senna nebhet » ate etm juts Mere te ade complicate Alpers tin wey Finn . 78 3.3. HW 2 CHAPTER 3. HWS wim L a xe (Ire a Tipe )-l me a * st ( "ap Tk d~™ di ticle by x in aL . we 2 + BF 7 / THe wee 1 @ 81 3.3. HW 2 CHAPTER 3. HWS © ce t+t,--. tit 7 G «+ tee TO ene | bjeagee Growl Bhe pretlen wits the pong 2 that mh enrash Pepe 5 bee considered . Tia guvotion it ! Po > it (odes te ke drew Fo Pot few, ot try fe we get ? ? af a wo ty ‘o> 3a lege Jeo ane we cee that it fs at dre . ole Feme dag “pews f" he emges is fay id fl Bm 82 3.3. HW 2 CHAPTER 3. HWS ch | = ® 16:19 | Find’ Maclean Sere F arctan x= J He oo fput Page bnctsatomee Sere, os Pagter das expoarialned about Ae @ expend a in pee Tagte Sane a) ject we fide => fle) = / 1 Wr ws ¢ , we we (x)= 2a => fl) =° Cte” 4° c $Me) : Bu® 2% => fle) a =4 ie (ite? (ita) ( Pye 180 ge we HM) = OO C4et "easy? iM(a)2 3894 2g¢ 4? a ve 24¢ ant Cray? + me => £ (j=24 Se ‘4) = - |} tee - a ~,> fli = fli)+ FC) lade fa (a~a) ofa) ea) ft cua)! > . ¥! 2/ = l +0 -24" 4 a al +o 4 ate 83 3.3. HW 2 CHAPTER 3. HWS CAL @ Ve} put 2 2 Cz (sin E) — . _ ave: fz As): ur. 2(4) RR MO) Rink » 2(h) LIE i Taso = (ngl-Z iE -/ieb = Uy =2 ach > (res (*E) alee Tee sie £) “2 = 56} Pit Ge) = GS 3 bey. 2) ghey ' a4, fis Me pom pate [47 wa TF selemnt trent vette mem 7 86 3.3. HW 2 Ch 2 (3) Put r 6 > wo Kea) = Te@): 5 gu - 4755 = > er ae Jal = | a) + Inka) - Ss wos 7 5 awa |. 54 " CHAPTER 3. HWS @ 5 (Geo 2 +7 se) (ey = (1s, 478) par ta 4s (qe) = (SE) aur 5 (ae ees a TZ) se* 5. 22 | gun Ba 2- be’ | (2%) (Sr lzay - t_ ($98) (ase) 1 2S Wei L . _) Geigy Find in rectamgeler ore Cat tb) Fowa-4 Ya | a pLeey—y* — ett BM (sy x Gaye nedbie D bo plex congenet. =F -1hy Fry ort yu (le on. - * s|—= Ft apie As elHy 6g ro aA ! Cary qxey ) = (tee) ) (eegt-a os) L a3 = — 7. 4 (py ped uty Ty OD" 87 3.3. HW 2 CHAPTER 3. HWS che vt ® 5.32 fod wank «f (2-34) lz}~ re [yey = fae ee fh lal Jat = Y @-aiy Ceres 2) 4a = Pt llelel” 5 ert (St 2 (isyen) =f[169 o Bhs means (+4) "| = [ Gers] ™ ‘toe Situs balla seen from crel®) t. Ge (4a lecg th. [5-39 Sele Tie pooh ptasse Bt an mbes > 9. atre4Z = Tee away { rte m4 ali amk- = 49 aint ef J x ka long oo may o> en ae files tne (mreg) = Ca-tg)” + Ta . sha ' wee + lxyd = mote J any a "a way ey te - my” = a” > ee XTC, arg red “ea = 4 ad mr 88 3.3. HW 2 CHAPTER 3. HWS ey nt pot Jex| +[24=4 go = Gey rey genetee: the aris Cert y ont @O ue Vc tg + Jeena" ke ee peed ‘te tele The thew wae oe a - _ -) e4* Jagsg = F7 Ve0* Sqvare both sideo ty vers th. J 2 aufty = (6- awe) (ent +9 = 649- )€ Vasey ™ + Gea ty *) ~ vs Leue a i+ z pehe Bx +it7 = 64 - e 4) +g +% 2KT 4 4u = 64 — Cf Genreg ey Gay yt = 69-4 => 4yayeyy = 16-% Pqvare Coen 16 ((x-09 ) = (ie-x) le( tase +! ty ) = 276-32 TH - u joy [om — Bam tle gy” = WE - 3TH TH ; isx*41037 = 240 Bot 20 als «| pe fle on egmltm fan ipe Z+%& 7 = 91 3.3. HW 2 CHAPTER 3. HWS & 4 tha) cu oe Se a= = b- ie The fret cee a=tVarb (ie) amd (- ol, 92 3.3. HW 2 CHAPTER 3. HWS ) wie ch 2 Te) wtp oes SUS cons fest - tat for cnengene =f Cuplen semen sete fe (t+) marl ey niy” (. oe oF }| | eee oy ersiy” -! t-e) 2 . | werd |, |=" led word |- iss (rae) cre) 24 e |-# | ef. le] EE : \ =e 2 \ z fee 4 we “ Rom . | \-i| = pret = fz ?/ pre ee = nse IT sme >}, been Gann | ret rmgent ] EQ x cy ~afF : lel hel A =r . (es (ints) (lec) (a a Un" eet)" [= j-ivd | 1- . Get Cie) y | pee rie CY) ve we Cet) Cv) Cirive) | 1 | le (0, : : (vB + 2 ith) | . | (28) (28) = dy G8) eine df (is we)edityeed) = eV? sida = gk AE ef tm bee I 93