Questions & Answers
Each e-mailed question will be posted to this page, followed by the professor’s answer. When e-mailing questions, the students should specify whether or not they would like to remain anonymous.
Wednesday, January 20, 2016
Q) I have a question related to the example you have in class talking about the insignificance of rational numbers. It was the example of the area of the cube of all the rational and irrational numbers between 0 and 1. We had a small valued epsilon, that with each rational number divded the weight that the rational number contributed by to. So we had (if E is epsilon): E + E/2 + E/4 + …=2E. I was confused as to why the weight of each additional rational number had half the weight of the previous.
On another note this finding has nice timing in relation to our recent lecture:
http://www.npr.org/sections/thetwo-way/2016/01/20/463725028/ready-for-prime-time-a-number-with-20-million-plus-digits
A) We’re trying to see how much space is occupied on the real axis by the rational numbers. A point by itself has a width of zero, so we try something that has a nonzero width for each number: the first one gets a width of E, the second one gets a width of E/2, and so on. No matter how many times you divide E by 2, you still have a number, namely, E/2^n, which is greater than zero.
We don’t want to give too much width to each rational, because, for example, if we gave them all the same width of E, then the total width would go to infinity. By assigning them the widths of E, E/2, E/4, etc., we make sure that they each get a non-zero width, albeit not too large a width to cause the overall width of the rational numbers to become infinite. In this way, we make sure that each rational number has a non-zero width, and yet the total width of the rational numbers is less than 2E – I say “less than,” because there is a lot of overlap between the little lines assigned to each rational number.
We thus see that the overall weight (or width) of all the rational numbers is still very small. (Remember that we can make E itself as small as we like, so long as it’s non-zero, it qualifies for the width that one can assign to the first rational number.)
Thank you for sharing the link to the largest prime number, which has recently been found.
Monday, January 18, 2016
Q1) This question has to do with the brain teaser (form 4 identical equilateral triangles using 6 matches). I believe that I’ve found a couple of solutions, but it’s been through a process of shooting in the dark, arranging some pieces of paper until I got something that works. That’s sort of dissatisfying, since you can’t be sure that you’ve found all the solutions. Do you know if there’s an analytical way to solve this kind of problem?
A1) I don’t know of any “general” solution to the problem of constructing four equilateral triangles out of six identical matches; you’ll just have to try different things and see what works and what doesn’t work. The best solution I know for this problem is a pyramid with a triangular base.
Q2) In physics, there was an assumption of continuity, i.e. that variables could assume any of an infinite number of values, that was found to be fundamentally untrue with the advent of quantum mechanics. Shouldn’t math, which is a reflection of the world, have some sort of fundamental discreteness to it just like physics does? Why then would there be an infinite number of numbers on the number line?
A2) Continuum mathematics can handle discrete problems that arise in quantum mechanics. For instance, the solution to the Schrodinger equation for the hydrogen atom has discrete energy levels, even though the wave-functions involved are continuous. If this does not answer your question, then there are other ways to interpret your question. For instance, group theory is a mathematical formalism that can handle sets of discrete objects.
Q3) In the lecture you mention that the sum 1 + x + x^2 + … + x^N has a closed-form expression, namely [1-x^(N+1)] / (1-x), which is valid for any x (rational, irrational, real, complex, etc.). However, it seems like it isn’t valid for x = 1, which produces division by 0. I think it’s possible to get that case by setting y = 1 – x, expanding (y – 1)^(N + 1) in the numerator (which produces 1 – (N+1)y + … from the multinomial theorem) and then finding the limit as y -> 0, resulting in N + 1.
A3) Your observation is correct. For x = 1, one basically has to assume that x is slightly different from 1, say, 1+ epsilon, where epsilon is a small number. One then applies the L’Hopital rule (which is pretty much what you’ve done in your e-mail), and find the value of the series in the limit when epsilon goes to zero. Of course, for x = 1, none of this is necessary, because the series is 1+1+1+ …+1 (N+1 times), and the sum is simply N+1. Thank you for pointing it out.
Sunday, May 10, 2015
Q) In reviewing for the test, I am going back over the Cauchy Integral Theorem and Formula. I’m noticing a resemblance to something we covered in Opti 501 and wanted to see if I’m on-track in my thinking. For analytic function with point(s) of non-analytic behavior (z-naughts) the Cauchy Integral Theorem seems to resemble Gauss’s Law for a point charge inside a shell/cylinder. The magnitude of the charge of the point(s) is given by surface integral of the E-field, much in the same way that the result of the Cauchy Integral is 2*pi*i*f(z-naught). In both, I’m seeing that the entire integral of a contour/line is given by a point in space enclosed by the perimeter. First, I’m wondering if I’m mis-interpreting that by chance? Second, do you know if either Cauchy or Gauss inspired the other, or is there another fount of thinking their ideas originated from?
A) The resemblances between Cauchy’s and Gauss’s theorems are superficial. They belong to different branches of mathematics (complex analysis and vector algebra, respectively), and the principles behind them are very different. Cauchy and Gauss were contemporaries. They were both first-rate mathematicians, and very famous during their lifetimes. I am sure they knew about each other’s work. But again, these two theorems require very different kinds of mathematics. Gauss’s theorem is rooted in physics (Coulomb’s law) — although it goes well beyond the properties of electrical charges and those of three-dimensional space. Cauchy’s theorem is a purely mathematical result with no basis in physics. It is rooted in properties of complex numbers and the concept of analyticalally of complex-valued functions defined in the complex plane.
(Follow-up comment by the student) Certainly, cosmetic at best. I think it does speak to a larger manner of thought or problem solving technique, and just so happens to be possible in diverse systems (physical and numerical) in nature. Perhaps more philosophically, it demonstrates the importance of pulling perspective backwards from a source or an in congruence, to see the effect on the whole and account for it separately.
Saturday, May 09, 2015
Q) This is question 86:
f “(x) / f(x) = -a (negative constant) –> f(x) = sin(n*pi*x/Lx),
g”(y) / g(y) = -b (negative constant) –> g(y) = sin(m*pi*y/Ly).
Why it has to equal to negative constant. If it equal to positive constant the result is same?
A) If the separation constant is positive, then the solutions will *not* be sin(…) and cos(…) but, rather, hyperbolic sine and cosine (i.e., sinh and cosh). Then the boundary conditions cannot be matched.
Tuesday, May 05, 2015
Q1) I checked the following website, and they told me that Jordan’s lemma assumes that s is not equal to zero: http://en.wikipedia.org/wiki/Jordan%27s_lemma. Thus, we should be careful when we do Fourier Transform for the function which does not converge to zero faster than 1/x.
A1) The Fourier transform F(s) of the function f(x) = 1/sqrt(x) provides a good example for the application of Jordan’s lemma. Note that, in the above function, the square-root operator (sqrt) acts on x not on |x|. Jordan’s lemma in this case works for s < 0, but not for s > 0, nor for s = 0. Complex-plane integration gives F(s) = 0 for s < 0. Pay attention to the branch-cut, which may be chosen to be on the negative x-axis (or anywhere in the lower-half of the complex plane). You may want to use Problem 43 of the Problem Set to find that F(s) = (1 – i)/sqrt(s) for s> 0.
Q2) This example is really interesting. I followed your explanation and studied it. I would like to have one comment. I think Jordan’s lemma works for s > 0 as well. Jordan’s lemma says only that a contour integral along the lower half circle is zero when the radius R goes to infinity. This is valid for s > 0 as well. The problem is how to calculate Fourier transform. When s > 0, we have to choose a detour path. Let’s assume that the branch-cut is on the negative x-axis. So we can choose an integral path as following. The first path is -R+q*i ->R+q*i along x-axis, the second path is R+q*i -> -R-q*i along the lower half circle, the third path is -R-q*i -> -q*i along x-axis, the fourth path is -q*i -> q*i along the small circle around the origin, and the fifth path is q*i -> -R + q*i along x-axis, where R is the radius of the lower half circle, q is small and real number, i is imaginary unit. When R goes to infinity, the first integral gives us F(s) and the second integral goes to zero because of Jordan’s lemma. When q goes to zero, the forth integral becomes zero. Finally, the sum of the third and fifth integral gives us the form of F(s). I checked that this is (1-i)/sqrt(s) for s>0. So, I think we can use Jordan’s lemma for the Fourier transform of 1/sqrt(x) for s>0 as well. However, the easier way is following the way in Problem 43. This detour method takes long time. I hope that this question is not in the final exam, otherwise I cannot finish it in time.
A2) Thank you for the interesting comment. It seems to me that you’re evaluating the integrals on the third and fifth legs of the contour directly, without using any complex-plane techniques. It is also the case that your fifth integral cancels out one half of your first integral. So, the use of Jordan’s lemma for s > 0 is only telling us that the conjugate of F(s) for s < 0 is equal to zero.
Unless I’ve misunderstood your technique, I believe one could evaluate F(s) for s > 0 in essentially the same way as you suggest, but without venturing into the complex plane (i.e., by direct evaluation of the original Fourier integral along the x-axis).
Q3) Thank you for the comment. Yes, we do not need a complex-plane technique for the Fourier transform for s > 0. This was a really interesting topic. Thank you for your help and discussing with me.
A3) Your question made me think about some issues related to Jordan’s lemma and complex-plane integration in general. I think you’ll find the discussion in the attached file interesting (see the attached pdf file“Calculating A Fourier Transform in The Complex Plane” at the bottom of this page).
Q4) Thank you for a new example. This is interesting and really instructive for me. I got deep understanding on complex-plane technique for a function which Jordan’s lemma is applicable to but branch-cut is needed for. The function f(x) = 1/|x|^v has Fourier transform when 0 <= v <= 1, otherwise, f(x) does not have Fourier transform due to the singularity at x = 0 ( v > 1) or divergence at x = +-infinity (v < 0). So, when we do Fourier transform for v < 0 or v > 1, is it the only way to use numerical calculation?
A4) You may use the differentiation theorem to find a lot of other Fourier Transform pairs once you know that of f(x) = 1/ |x|^nu. This is because the Fourier Transform (F.T.) of the derivative is equal to the Fourier transform of the original function multiplied by i*2*pi*s. Also, note that the first, second, third, etc., derivatives of the delta-function have interesting Fourier transforms. You may also want to consider the Step-function and its integrals (taken from minus infinity to x); to find the F.T. of these functions you can divide the F.T. of the Step function by i*2*pi*s. There are many interesting F.T. pairs that can be obtained in this way.
Sunday, May 3, 2015
Q1) In your discussion of Bessel functions and various modes of waves propagating, I started thinking about the use of short range lasers using multi-mode fiber in optical communications systems. Based on the graphical representations, where the rho-mn crossings spread to the right as m->infinity, I’m wondering if Bessel equations explain the physical mechanism that is at play in a multi-mode fiber system, where the peaks and crossings essentially are non-overlapping (deconstructing/interfering)? If so, is that physically occurring due to the laser or due to the fiber?
A1) Fibers are in general more complicated than vibrating membranes, because the boundary at the core-cladding interface does not force the fields (E and B) to stay at zero. You should read the exact solution of Maxwell’s equations for step-index fibers to see how the modes are structured in these fibers. I attach a copy of Chapter 9 of my electrodynamics book to this e-mail, although you can find similar discussions in any textbook in optical fibers.
Q2) I wanted to ask for your guidance on how best to accelerate my pacing on the 503A exam? I found on the mid-term that while the concepts were clear, that my pacing was far too slow, which resulted in me only completing about 65% of the exam (which did not translate well on my grade). I know you have recommended single sheet of paper with key equations for 501, which I tried for the mid-term, but I just don’t feel like my math skills are that fast these days. I plod along through the problems and eventually get there, but it’s anything but graceful. I’m actively going over old exams and am very appreciative for all that you do to help us students prepare, I hope my request is not seen in a negative light. I’m sure most of the issue stems from not being a full-time student and being at a distance.
A2) I don’t know what to suggest other than practice, practice, practice. Going over the HW problems once again, and also looking at the previous years’ exams and solutions is a good preparation for the upcoming exam. Also, make sure you know where various things are in your notebook. During the exam, when you need to find something that we went over in the class, it saves a lot of time if you know exactly where to look for it in your notebook.
Tuesday, March 17, 2015
Q) I was just wondering why singularities are called poles in the complex-plane integration we have been studying these past couple of weeks. What is the distinction between a pole and a singularity?
A) When a polynomial of degree n, such as P_n(Z) = An*Z^n + A_(n-1)*Z^(n-1) + … A1*Z + A0, is set to zero, it will have n solutions, Z1, Z2, Z3,…,Zn. Now suppose you have a function in which the denominator is a polynomial of degree n, say, f(Z) = g(Z)/P_n(Z). At each of these solutions (i.e., when Z = Z1, or Z = Z2, …, or Z = Zn), the denominator of f(Z) will be zero, and the function is said to have a pole-type singularity at that point. In other words, the technical term used to refer to a root of an nth-order polynomial in the denominator of a given function is “pole.” Thus Z1, Z2, …Zn are the poles of the function f(Z) = g(Z)/P_n(Z). Each pole will be a first-order pole if they are all distinct. When two poles overlap, we have a second-order pole. When three poles overlap, we have a third-order pole, and so on.
Of course, a function may have a singularity of a different kind. For instance, ln(Z) has a singularity at Z =0, but it is not called a pole — just a singularity. Also, all the points on a branch-cut of the function ln(Z) are points of singularity – again not poles, just singularities, because the function does not have a derivative at those points. Similarly, f(Z) = 1/sqrt(Z – Z0) has a singularity at Z = Z0 but this singularity is not a pole.
Monday, March 2, 2015
Q) Should the branch-cut be introduced to differentiate solutions to 1/z^n when n is non-integer?
A) The integral is zero only when n is an integer (greater than 1). If n happens to be non-integer, then a branch-cut will exist. Think, for example, of 1/z^(1/2). The square root will *not* be uniquely defined over the entire complex plane and, therefore, the introduction of a branch-cut is necessary. The same argument goes for any non-integer power of Z.
Monday, March 2, 2015
Q) In today’s class, you showed the loop integral of 1/z^n is zero when z in not equal to 1 by writing down the whole summation like (z2^1-n – z1^1-n) + (z3^1-n – z2^1-n) +… I think that zn^1-n is not equal to z1^1-n when n is not integer, because we need to define branch-cut when n is not integer. Is my understanding correct?
A) Your understanding is correct. The integral is zero only when n is an integer, otherwise there will have to be a branch-cut.
Thursday, January 30, 2014
Q) Please see the attached file “Question_January 30_2014.pdf.”
A) Excellent question; you get a point for asking it. Your understanding is correct. In your Eq.(1), the distance between elements that are being cancelled out of the two series is increasing again (as you cancel out first 2, then 4, then 6, etc.).
The general rule is this: Let the upper limit of the summation be N (rather than infinity). Then if you separate into two series, both series must go up to the same N (not to infinity). Whatever operation (addition, subtraction, multiplication, etc.) is legitimate in the two (or more) series that have N as their common upper limit, that operation would also be allowed for the infinite series. In other words, you should always think of the infinite series as the LIMIT of a finite series with N as the maximum value of n, then perform your operations. Only after your operation has been completed should you allow N to increase to infinity.
Friday, March 2, 2013
Q1) I just got finished going through the HW#5 solutions and notice that Jordan’s Lemma and residues were used to solve Problem 31. I believe that we didn’t see Jordan’s Lemma or residues in lecture until Wednesday after we turned in the homework. Is that correct or did I miss something in an earlier lecture?
A1) You are right about Jordan’s lemma, although we had talked about the residue theorem without mentioning it by name. Some of these HW problems are meant for you to go through and do as much as you can, then see the difficulties encountered, and wait to see how the difficulties are resolved once you receive the solutions. Most people just assume that the integral on the large semi-circle goes to zero, without knowing the justification. Jordan’s lemma is the justification. Simply by looking at the values of the function on the semi-circle, you can tell that the integral is getting smaller as R gets larger, but you can’t say so with mathematical certainty until you have seen Jordan’s lemma.
Q2) At the beginning of Lecture 13, when performing the inverse Fourier transform integral of exp(+i2*pi*s*x), how do I know to use the upper semi-circle contour in the +i complex plane for x>0, vs. the lower semi-circle contour in the -i complex plane for x<0? Is it because the sign of the (+i2*pi*s*x) is driven by x? And then if (+i2*pi*s*x) is negative use the -i complex plane contour, and if positive use the +i complex plane contour? What about s?
A2) You are correct about the sign of x and the + or minus signs in the exponent. Before starting the integral, you should check the sign and see if you should use the upper or the lower semi-circle. The parameter s is the variable of the integral; it is the thing that is replaced by Z.
Q3) I was able to show the zeros of f(z) are +/-exp(+/-itheta) and I have a question interpreting them. They aren’t point singularities, but are contour singularities, correct? Basically the function is undefined on the unit circle in the complex plane because theta could be anywhere between 0 and 2pi? If that is the case, then the contour I want to integrate should be a large circle of radius R with a diversion to exclude the small circle of radius > 1. The large circle will go to zero, so the small circle is the one I’m really interested in and it should integrate to pi/(2*sin(theta)). Or am I on the wrong track here?
A3) Theta is a constant. There are only four poles: Z1, Z2, Z3, and Z4. Think of theta as some fixed angle, say, 30 degrees.
Sunday, February 26
Q) I am having trouble getting started with homework problem 29. I have studied your solution to problem 32 and have tried to follow a similar process. Is this the technique I should be using? If so, can you point me towards the integral I should begin with. I have made a few attempts with different integrals but none seem to yield anything significant.
A) You may want to wait until after Today’s lecture to tackle that problem. I wanted the students to try it first and see how difficult it is, before I discussed it in the class. You may send in your HW on Wednesday.
Friday, February 17
Q1) On problem 25, part (a), just before the solution, how does one know y=0? Is it because there is no real component in exp(2*i*n*pi)? Therefore exp(-2*y) must equal 1, so exp(0)=exp(-2*y), y=0.
A1) On the right-hand-side of the equation, 1.0 is replaced by exp(i*2*pi*n). The magnitude of this number is unity, and its phase is 2*pi*n. One the left-hand-side of the equation, the magnitude is exp(-2*y), which must therefore be set equal to 1.0. Thus y = 0.
Q2) On problem 21a, you take the du/dx of u(x,y) in a single step. This is a lengthy derivative to show all the steps of. Is it acceptable on the exam to show the derivative of a function in a single step as you do here (when it is an intermediate step to the solution)?
A2) There are different ways of doing this derivative. I use a rule that says the derivative of a ratio such as f(x) / g(x) is given by [f ‘(x)*g(x) – f(x)*g'(x)] / [g(x)]^2. That way, there is only one step to differentiation, which is what I showed in the solution. If you are following a different rule, it is best to show the intermediate steps, because you will need them to carry out the differentiation anyway. If you don’t show the steps and the answer is wrong, you will not get any points, but if you do show the steps, you’ll get points for those parts that you did right.
Q3) I have a question on Problem 5 part (e). Integrate from 0 to inf [(x^2)*(exp(-pi*x^2))] . I am letting f(x) = x^2, and g(x)=exp(-pi*x^2). Next, f'(x)=2x, integral(g(x))dx= -[exp(-pi*x^2)]/(2*pi*x). When I do integration by parts, my first term does go to zero as in your solution, but I am having trouble with the second term which comes from: -integral(0 to inf)[f'(x)*integral(g(x)dx)]dx = -integral(0 to inf)[2x*(-[exp(-pi*x^2)]/(2*pi*x)]dx = +integral(0 to inf)[(exp(-pi*x^2)/pi]dx. As you can see, the 2 cancels, and my result comes out to 1/pi * 1/2. Can you point me at what I’m doing wrong?
A3) Your integral of g(x) is wrong. You have created a function, -[exp(-pi*x^2)]/(2*pi*x), which has x in the denominator. If you differentiate this function properly, you will not get your g(x) back. The correct way to do this problem is to set f(x) = x and g(x) = x*exp(-pi*x^2). That way, g(x) will have an integral.
Q) In today’s lecture when you were showing how to get the individual terms of the integral of f(z) to go to zero, you mentioned that f(zo) is a constant and can be pulled out to the front of the first term to get f(zo){integral(dz)}. Doesn’t it follow then that f'(zo) is the derivative of a constant which = 0? Then wouldn’t the second term of the integral of f(z) which is f'(z0){integral(zdz)} immediately go to zero as well?
A) f(zo) is the value of the function at a fixed point, z = zo. As such, it is a constant. This doesn’t mean that f(z) is a constant. If you go from zo to a nearby point, say, z, then the value of the function will change. That is why the derivative of f(z) evaluated at zo, which we have denoted by f ‘(zo), is not zero.
Q1) I have a question on the last step of problem #4. How do you get from this expression: Integral (-inf. to +inf.) exp[-pi*(x-xo)^2)] dx to this expression: Integral (-inf. to +inf.) exp(-pi*x^2) dx? I follow the steps leading up to this point, but I don’t understand what happened to the exp(-pi*xo^2) term, or the exp(2*pi*x*xo) term.
A1) I used a change of variable from x-xo to x. (Perhaps if I called the new variable x’ instead of x, you would not have had this difficulty). Another way to see this is to note that the area under the curve exp[-pi*(x-xo)^2] must be the same as the area under the curve exp(-pi*x^2), because shifting along the x-axis (by xo in this case) cannot possibly change the area under the curve.
A2) The exams are open books, open notes, open everything. You may bring your computer and calculator as well.
Q) Thanks for revealing that complex numbers aren’t really “imaginary”. I’ve never seen the topic presented in such a straightforward manner before. If real numbers are on a line and complex numbers are on a plane, what kind of numbers fall into a three dimensional space?
A) Great question; you get a point for asking it. As far as numbers are concerned, complex plane is the limit. All polynomial equations have their solutions contained in the complex plane (even polynomials whose coefficients are complex-valued!). Therefore, people did not need to extend the complex plane into three dimensions in order to find more numbers. However, higher-dimensional spaces are still being explored in linear algebra and vector algebra. Hamilton also studied “quaternions,” which, if I remember correctly, are a form of higher-dimensional numbers. But at this point in time, the most general numbers that we have are the complex numbers. They contain ALL the solutions of ALL polynomial equations of any order.
A) You cannot pull out the (-1) factor. This is a double sum, first over n, then over m. You should reverse the order of summation (i.e., do the double sum first over m then over n). You will then have to evaluate a geometric series.
Q1) Problem # 13 on zeta function: The zeta function was defined to have a constant power ‘s’ on varying integer n (1—> infinity or 2—-> infinity). Problem 13 seems to change from that definition as the power is varying as same as integer n, that is : sum[zeta(m)-1] = [1/(m^m) -1]
Is my understanding not correct? (I’m coming with sum[zeta(m)-1] = [1/(m^m) -1] = -1 instead of 1 )
A1) Problem 13 is a double sum. Think of the xy-plane. On the x-axis the discrete variable n varies from 1 to infinity. On the y-axis, the discrete variable m varies from 2 to infinity. The trick to solve this problem is to change the order of summation, so that instead of first summing over n then over m, you change the order and first sum over m then over n. Along the way, you should use the formula for geometric series to simplify the inner sum and get to the final result.
Q2) Problem # 12: The example you did in class regarding “wrong way” and “rigth way ” to solve the series, These problem 12 do not have jump between ‘n’ (n and 2*n), and my partial fractions seem to be right but I can’t get the end result. Is there something I should consider in the calculation?
Q) I notice that in your homework solutions, you commonly use d(tanx)/dx = 1+(tan^2)(x). I understand that this is the same as (sec^2)(x), but why do you default to that notation? Does it tend to make other portions of the problems more easily identifiable?
A) It is a personal preference; I don’t use sec and cosec, because I always forget which is which. The trig functions I always use are sin, cos, and tan.
Q1) During Monday’s lecture at 0:37:41 on the .wmv file, it was unclear to me if you stated that the points on the line and the points in the square are countably infinite, or if you said “un”countably infinite (I think there was a glitch in the recording that sounded a lot like the prefix “un”!). From the rest of your lecture, I’m pretty sure they are countably infinite (1 to 1 correspondence between points on the line and point in the square). Just making sure I understood correctly.
A1) I said “Uncountably infinite.” The line-segment is uncountably infinite, as I explained in the Cantor’s proof. Since it is in one-to-one correspondence with the square, that makes the square uncountably infinite as well. Recall that for a set to be countably infinite, it has to be in one-to-one correspondence with the set of “Integers.” You seem to be mixing the set of integers with the set of all numbers on the line-segment between zero and one.
Q2) Opti 503A, Problems 2 and 5: It has been a while since I looked at differentiation notation. Are these problems written in Leibniz or Euler notation? I think seeing the notation in type font vs hand writing is throwing me off.
A2) I don’t know the difference between Leibnitz and Euler notations; I’m using the “standard” notation for differentiation. I am just trying to see if the students remember the chain rule of differentiation. If you don’t, don’t worry about it; just wait until I post the solutions on Monday.
Q1) I’m curious why today’s lecture was shorter than usual? Also, is the final next week going to cover the material from the entire course or just the material since the last exam?
A1) The first 15 minutes of today’s lecture were used by the students filling evaluation forms for the course. This is a routine that is followed every semester in every class at the Optical Sciences. Usually the camera runs silently during these 15 minutes, but today the operator stopped the camera and started it again when I returned to the classroom.
The final exam will cover everything that has been taught in Opti 503 from the first to the last lecture. Some emphasis, of course, will be given to the material covered during the final month, as the previous exams covered the first 3 months of the semester. I hope the students will take this last opportunity to review the course material from the beginning to the end, to see the connections and relations among various subjects that they may have missed when they learned the material at first.
Q2) Do you have any solved problems related to the past 3 lectures for us? I am studying the notes but can you suggest any additional problems to do in preparation for the finals?
A2) The University of Arizona has a policy that no homework should be assigned during the dead week; that is the last week of the semester before the final exam. This is why I am not assigning any homework for the last two lectures (that is, today’s lecture plus the lecture that I will deliver on Wednesday 5/4/11). The 86 problems that you have been given in the form of a problem set, plus the 10 problems given in the first and second midterm exams, cover the entire subject matter of the course except for the last two lectures. I expect the students to be able to answer simple questions about these last two lectures during the final exam, but I won’t ask any detailed questions that go deeper than the lectures themselves.
And, by the way, the final exam is going to be comprehensive, covering everything that has been taught from the beginning of the semester to the end. This does NOT mean that there will be one question on every topic, but it means that the students should prepare for the test treating each and every topic as the subject of a potential exam question.
A1) You are right about the sum and the integral. In the case of finite dimensions, the full solution is a sum over individual modes, represented by discrete values of n_x and n_y. In the limit when the dimensions of the vibrating membrane go to infinity, the sum is replaced by an integral, and the Fourier series becomes a Fourier transform.
Q2) When solving a linear homogeneous 2D PDE in Cartesian format using the separation of variables techniques, we usually conclude that the separation coefficient is negative. For example, the wave equation:
PDE: d^2(z(x,y,t))/dx^2 + d^2(z(x,y,t))/dy^2 = d^2(z(x,y,t))/dt^2 + alpha*d(z(x,y,t))/dt
Assume z(x,y,t) = f1(x)*f2(y)*g(t), then the separation coefficient is negative i.e.
d^2(z(x,y,t))/dx^2 + d^2(z(x,y,t))/dy^2 = -C^2 & d^2(z(x,y,t))/dt^2 + alpha*d(z(x,y,t))/dt = -C^2
because the functions f1(x) and f2(y) need to be oscillating to satisfy the boundary conditions.
For the cylindrical case, assume z(r,phi,t) = f1(r)*f2(phi)*g(t). then
PDE: d^2(z(r,phi,t))/dr^2 + (1/r)*d(z(r,phi,t))/dr + (1/r^2)*d^2(z(r,phi,t))/dphi^2 =
d^2(z(r,phi,t))/dt^2 + alpha*d(z(r,phi,t))/dt
What is the justification for the separation coefficient to be negative? Can we use the argument that on the boundary, in the limit as r -> R (constant) the PDE reduces to a PDE with constant coefficients as in the Cartesian case? Therefore, as in the Cartesian case, the DE with f1(r) needs to be oscillating to satify the boundary conditions?
A2) The justification in the case of cylindrical coordinates is similar to that in the case of Cartesian coordinates. The vibration amplitude must be zero at the boundaries; if you choose the separation constant to be negative, you’ll end up with Bessel functions of the first and second kinds, J_m(k*r) and Y_m(k*r), which oscillate as a function of r and have an infinite number of zero-crossings, each of which may be made to coincide with the boundary at r = R. If, however, you choose a positive value for the separation constant, you will end up with the modified Bessel functions, I_m and K_m, which behave similarly to the exponential functions (the latter appearing in the case of Cartesian coordinates). In essence, the modified Bessel functions are obtained by replacing the argument k*r of ordinary Bessel functions by i*k*r; they do not have zero crossings and, therefore, are unacceptable as solutions for the radial component of the wave function.
Q3) For question Q2 above, I’m assuming that since the end pts are fixed, we also have the boundary condition g'(0) = g'(L) = 0. Using these boundary conditions to solve g”(t) + alpha*g'(t) + (n*v0*pi/L)^2*g(t) = 0, we get g(t) = exp(-alpha/2*t)*[A*exp(h(n)*t) + B*exp(-h(n)*t)] such that h(n) = sqrt(alpha/2)^2 – (v0*n*pi/L)^2). Plugging the boundary conditions outlined above, I get either A = B = 0 or if either A or B are non zero, then n = 0. In either case, z(x,t) = 0, which seems incorrect. I was wondering if you knew what I was doing wrong?
Q1) In reviewing my notes, it appears as if a factor of pi is missing from the radical in the impulse response. When the width of the Gaussian was finite, the factor of pi multiplied the sum of wo^2 and 4Dt. However, when you let wo go to 0 for the delta function, the factor of pi disappeared. Please let me know if my notes are inaccurate. (I am referring to the coefficient of the Gaussian. I understand why the factor of pi disappears in the exponent.)
A1) There is a factor of pi in the coefficient; if I didn’t write it on the board, it is my mistake. In my notes, I do have a factor of pi under the square root in the denominator of the coefficient of the Gaussian.
Q2) I understand the general concept of the Frobenius method but there are some details in the HW solutions that I need clarification on please.
a) On page 2 of HW #74, you write that:
s = 0: f(x) = sum[A(k) * x^k, k=0:Inf] = A0*x^0 + A1*x^1 + A2*x^2 + A3*x^3 + …
s = 1: f(x) = sum[A(k-1) * x^k, k=1:Inf] = A0*x^1 + A1*x^2 + A2*x^3 + A3*x^4 + … = A0*x^1 + A2*x^3 + A4*x^5 + … (since the odd A’s are zero)
s = -1: f(x) = sum[A(k+1) * x^k, k=0:Inf] = A1*x^0 + A2*x^1 + A3*x^2 + A4*x^3 + … = A1*x^0 + A3*x^2 + A5*x^4 + … (since the even A’s are zero)
Here, you mention that “Clearly the index k is shifted by one unit” for both the s = 1 and s = -1 cases. This is not obvious to me. Can you please explain what you mean by this? I see that for s = 0, we have A(k); for s = 1 we have A(k-1); and finally for s = -1 we have A(k+1) … but I don’t truly understand what you mean … I am confused
b) On page 2 of HW #74, you mention that the solutions obtained for s = 1 and s = -1 are inherent in the solutions for s = 0. As I have summarized above, we have:
s = 0: f(x) = A0*x^0 + A1*x^1 + A2*x^2 + A3*x^3 + …
s = 1: f(x) = A0*x^1 + A2*x^3 + A4*x^5 + …
s = -1: f(x) = A1*x^0 + A3*x^2 + A5*x^4 + …
Inspecting the equations, it is not obvious to me that the solutions obtained for s = 1 and s = -1 are inherent in the solutions for s = 0. Can you please explain what you mean by this? Is this because any linear combination of a solution is still a solution (principle of superposition)?
c) On page 3 of HW #74 for the A(2m) coefficient, you state “Note that if n is an even integer the above series terminates at k = 2m = n, whereas for odd-integer values of n, the series continues indefinitely”.
This is not obvious to me. Can you please explain what you mean by this?
d) On page 3 of HW #74 for the A(2m+1) coefficient, you state “Note that if n is an odd integer the above series terminates at k = 2m+1 = n, whereas for even-integer values of n, the series continues indefinitely”.
This is not obvious to me. Can you please explain what you mean by this?
A2) These are long questions and are best discussed on the phone. Before getting on the phone, however, please try the following method to convince yourself that the statements I made in the solutions are correct. Put numbers into the expressions, by starting with k=0 or k=1, then deriving formulas for A0, A1, A2, A3, A4, etc. Then put the expressions for A_k that you’ll find in this way into the Frobenius expression, and see for yourself that the answers you get for s = 0 are the same as those you get for s = 1 and s = -1.
Tuesday, April 19
Q1) whenever you do a Fourier Transform of the function f(x), if f(x) is “unbounded” at x = -Inf and/ or x = Inf, you HAVE to apply a decaying function, correct? I know you mentioned this but I would like to confirm so I can etch it into my brain. For instance, if f(x) = x or x^2 then when I do the F.T., I would just multiply f(x) by exp(-a * |x|), and apply the normal F.T. equation, correct? Then at the very end, I let “a” approach 0 and “simplify”, correct?
A1) Yes, when the function is such that its Fourier integral is divergent, you need to modify the function by multiplying, for instance, with exp(-alpha*x), finding the F.T., then letting alpha go to zero. Alternative methods exist as well. Sometimes you find the F.T. of a function, such as Step(x), using the aforementioned scheme. Then you use one of the F.T. theorems (e.g., differentiation theorem) to find the F.T. of other badly-behaving functions, such as f(x) = x Step(x).
Q2) I know that Frobenius is a genius, but how would a normal guy like me know how to assume a power series when encountering Bessel’s equation? Not only that, he knew to have the x^s term too … The reason I ask this is practical: as I encounter new ODE’s in school and work; I’ll try to rearrange them and make them fit into the already known sets of equations out there. How hard should I try before just resorting to numerical methods? I guess everything boils down to experience and hard work.
A2) The Frobenius method is a standard method of solving ODEs these days. True, he was a genius for discovering the method, but the rest of us can now use the method as a recipe for solving ODEs. It is like everything else in mathematics: some genius discovers a method, then we learn it and use it, and, after a while, it becomes part of our standard tool set so much so that we no longer recognize the difficulties that people must have had before the discovery of the tool. Do you think the decimal number system, addition and multiplication, integration, differentiation, etc., are any different than the Frobenius technique of solving ODEs?
Q3) Like we’ve done over the last couple of weeks, we see that the solution to ODEs with constant coefficients is a linear combination of complex exponentials. You may have already mentioned this in class, but COULD you use the Frobenius method on ODEs with constant coefficients? Would you get a power series that is identically the complex exponentials?
A3) The Frobenius method can be used to solve differential equations with constant coefficients. I think I already mentioned this in class yesterday. You will then get the power-series representing real or complex solutions to these equations. Try, for instance, solving the equation f “(x) + f(x) = 0 using the Frobenius method. The two solutions that you’ll get represent the Taylor series expansions of cos(x) and sin(x). The general solution is then A0*cos(x) + A1*sin(x). You can then choose A0 and A1 to obtain exp(+ix) and exp(-ix) as another possible pair of solutions of the differential equation.
Q4) You’ve mentioned that this is the first time that OPTI 503A is being offered. Are you planning to offer an OPTI 503B course? This would be extremely useful for a person like me who’s been away from school for over 10 years. I’d love to take a OPTI 503B class before diving into the core optics curriculum.
A4) There are no plans for teaching a follow-up course to Opti 503. I hope I have been able to show the students how to go about reading a book such as Arfken’s, and solving problems on their own. There is a lot more to learn in “Mathematical Methods for Physicists,” of course, but hopefully the path through the book should be clear by now.
Q5) Problem 74/75: I have plugged in the functions into the equation under investigation and yielded (k + s)(k+s-1) SUM A(k) x^(k + s – 2) for the last term. I defined a new variable k’ as k’= k – 2. Question 1, when k=0, k’= -2, therefore do I start the summation for this term at k=-2 ?
A5) Separate the terms corresponding to k = 0 and k = 1, then write the rest of the series with k starting at 2. This way, when you switch to the new variable k’, the count will start at k’ = 0.
Q1) I was revieweing the solution to the 1-D transverse wave problem. The final solution is z(x,t)=A*sin(n*pi*x/L)*g(t). Does that imply that the initial condition (or intial function) have to be a sine series i.e. odd function for a solution to exist?
A1) Yes, the initial condition for the shape of the string must be an odd function. However, the actual string only exists between x = 0 and x = L. So, defining its shape outside this region, say, in the (-L, 0) interval, does not have any physical significance. It is just a mathematical trick to determine the Fourier series that represents the initial shape of the string in the (0, L) interval.
Q2) What are the conditions required to use the separation of variables technique? Does this technique always find every solution? Are there cases when this technique is not appropriate?
A2) The separation of variables works as long as the equations are linear. Think of a simple nonlinear PDE, say, [df(x,t)/dx]^2 + f(x,t) = df(x,t)/dt, and see for yourself why the separation of variables is useless in this case.
Q3) I believe someone asked a similar question in class, however, the response was a little unclear. When solving the 2-dimensional wave equation in Cartesian coordinates, the solution has the form z0*sin(nx…)*sin(ny…)*g(t). Are z0, nx and ny determined from the Fourier transform of the initial condition?
A3) The initial conditions are usually given in terms of the initial position of the membrane, z(x, y, t=0) and its initial velocity, z’(x, y, t=0). The general solution is written as Sum{zo(t)*sin(n_x*pi*x/L_x)* sin(n_y*pi*y/L_y)} — here I have merged zo and g(t) into a single expression, and called it zo(t).
Q1) I just went over your solutions to the exam and then went through my notes again. Regarding problem #1, I can’t find anything from the lectures where you talked about the differentiation theorem. I used a lot of my time during the exam trying to figure out what that was. After reviewing your solution, of course it makes sense, but we never covered that in class. If we did, can you please point out in which lecture or homework problem that was?
A1) Take a look at Problems 53(b) and 70; in particular, the first line of my solution to Problem 70. Also, when I discussed in the class how to Fourier transform a differential equation (such as that governing the mass-and-spring system), I showed how to obtain the F.T. of first and second derivatives of a function. The Fourier transform integral and its inverse being essentially identical (aside from a minus sign in the exponent of the integrand), this differentiation procedure works both ways, i.e., in going from the F.T. of f(x) to that of f ‘(x), or in going from the Inverse F.T. of F(s) to that of F'(s).
Monday, April 11
Q1) I just finished the second exam and once again I feel completely frustrated. I feel I know the material, but I just can’t finish in 75 minutes. I realize you design the tests to be that way and considering that at least one person got a perfect score on the first exam, maybe it’s just me. But when would I ever need to do this type of math so quickly? Given more time, I could “easily” do it all. Is this a true test of whether or not we have mastered the material? If I had another half hour to an hour, it would all be done. I just can’t think of a situation in the work force where I would have to rush through this type of math. Sorry, I had to vent.
A1) I understand your concern, and agree that in the real world there is usually enough time to solve problems (unless you are an Apollo 13 astronaut!). However, the University rules require us to have exams, and to give grades in a way that reflects the student’s performance in the course. If I were to give twice as much time to the students, then almost everyone would end up getting a perfect grade, unless I made the test twice as hard; the result would then be exactly the same.
Q1) I was curious if it would be possible for you to email out copies of your lecture notes for future lectures? The reason that I ask is that it seems to take way longer than 75 minutes to watch each lecture because I’m constantly having to pause to either catch up writing my notes or to wait for the camera to catch up to you. In previous grad work I’ve had professors who gave out the notes and it made it so much easier to pay attention during the lectures because I wasn’t constantly writing. Actually I wish I had thought to ask this early in the semester. Let me know if this is something you are able to do.
A1) I am afraid my notes are not going to be of much help; they are sketchy and disorganized, as I am putting this course together for the first time. In future years, perhaps, I’ll have my notes organized and typed for distribution to the class. For now, however, I am afraid to say that the notes that you take from the whiteboard are the best and most comprehensive notes that are possible under the circumstances.
Q2) Do you have electronic copies of additional HW problems that you’ve done in the past or just general math notes? I’d like to have it for my future reference (studies, work, etc).
A2) Since this is the first time I am teaching this course, it is hard enough to come up with new problems for each homework assignment. In future years, I’ll make up more problems, but, for now, you’ve got all the problems that I have thought of (and for which I have written the solutions) on the course website.
Q3) Problem 73 part b: How did you know to just keep the sqrt(c) root for f(x)? When I re-solved this problem, I was very excited to get both roots: f(x) = A*exp(sqrt(c)*x) + B*exp(-sqrt(c)*x). I see your explanation and it makes sense; but if left on my own, I wouldn’t have known to throw away the other root. In other words, is there a step by step “procedure” that I should use when solving ODE’s and PDE’s?
A3) Both + and – sqrt(c) are acceptable solutions. It is just that one of them corresponds to the wave propagating to the right, the other one propagating to the left. The sign of sqrt(c) that tells you which wave is propagating in which direction depends on the sign that one chooses for the excitation frequency, namely, whether it is exp(+i*w*t) or exp(-i*w*t). Once you have solved a few wave-propagation problems, it becomes easy to see which sign of the square root to pick for a given direction of propagation. The procedure for solving ordinary as well as partial differential equations is pretty straightforward. But you need some experience to decide which coefficients to pick here and there. Like everything else in life, experience counts for something here too.
Q4) Problem 73 part b: Given that f(x) = A*exp(sqrt(c)*x). How did you get A = 1?
A4) I put the coefficient “A” in front of the “final” solution for zeta(x, t). Instead of “A,” I’ve called it “xo” in this problem.
Q5) Problem 73 part d: In your solutions, you wrote: z0(f) = F{z0(t)} (1).
Then, you wrote: z0(t) = integral{ z0(f) * exp(-i*2*pi*f*t) df } (2).
I found z0(t) by taking the inverse Fourier Transform of z0(f) (see Eq. 1 above):
z0(t) = F^-1{z0(f)} = integral{ z0(f) * exp(+i*2*pi*f*t) df } (3)
Clearly my Eq. 3 does not match your Eq. 2 (positive versus negative exponents). So where did I go wrong?
A5) You didn’t do anything wrong; I just “defined” my Fourier transform for this problem with the + and – signs reversed. There is nothing “sacred” about the plus and minus signs, in general. What is important is to use “opposite” signs for the Fourier transform and its inverse. To avoid confusion, I generally recommend to use the minus sign in the exponent of the forward F.T., and the plus sign in the exponent of the inverse F.T. But, in this particular problem, since we had started with the time-dependent factor being exp(-i*2*pi*fo*t), it made sense to switch the signs. (In the literature, you’ll see people defining the F.T. sometime with the plus sign and other times with the minus sign; but, of course, they always use the opposite sign for the inverse transform. Problem 73 shows you that there is nothing wrong with using either definition depending on the circumstances of the problem under consideration.)
Q6) Problem 73 part d: In your solutions, you wrote: z(x,t) = integral{ z0(f) * exp(-beta…) * exp(i*2*pi*f…) df }. How did you get this? It’s a crucial step and I got lost.
A6) In part (d) I am just using the results of part (c). Look for the similarity between the individual Fourier components (i.e., what you have under the integral sign), and the previous parts of Problem 73. It’s all there!
Q) I have several questions regarding the previous homework:
a) HW 67: can you please explain in more detail the part about “a low-pass filter with transfer function 1/sinc(as) will thus recover f(x) … I need help to understand this please
b) HW 68 part a: I don’t see how you arrived at the conclusion that a0 = F(s = sigma_x/lambda, z = 0+). I see that s = sigma_x/lambda but don’t follow the steps you provided to arrive at a0
c) HW 68 part c: in my class notes I have the factor sqrt(2/(n * |g”(x0)|)) but your solution has sqrt(2*pi/(n * |g”(x0)|)). There’s an extra PI in your solutions. Which one is correct?
d) HW 68 part c: why do you have the “i” in the factor 1/sqrt(i*lambda*z0)? I get 1/sqrt(lambda*z0). This is because we take |g”(x0)|, correct?
e) HW 68 part c: what happened to the exp(-i*pi/4) factor in the final solution? It seemed to have disappeared.
f) HW 68 part c: how did you get (cos(theta))^3/2? I get (cos(theta))^3/4
A) Here are the answers to your questions.
a) Electronic filters have an “impulse response” and a “transfer function.” If you excite the filter with an impulse (the input), what you’ll get at the output is the “impulse response.” The Fourier transform of the impulse-response is called the “transfer function.” When any other signal is used as an input, the corresponding output signal will be the convolution between the input signal and the impulse-response. (This is something that people usually see in their undergrad courses in linear system theory; we have not proven any of this in the class, but I assumed the students are familiar with the concept.) In any event, the convolution theorem then shows that the F.T. of the output is always given by the F.T. of the input multiplied by the F.T. of the impulse-response, the latter being called the transfer function. Thus if the F.T. of the sampled function in Problem 67 contains the desired spectrum multiplied by sinc(as), the filter that will restore the original signal must have a transfer function equal to 1/sinc(as).
b) Here I simply set z equal to zero in the expression for a plane-wave, a(x, z), the reason being that the initial distribution considered here is located in the plane at z=0 (same thing as z=0+, because of the continuity of the function a(x, z)).
c) In the lecture notes I also have the coefficient sqrt[2*pi/eta*|g”(xo)]. The sqrt(pi) came about because the final integral, Int{exp(+ix^2)dx} or Int{exp(-ix^2)dx}, was found in Problem 32 to be equal to sqrt(pi)*exp(+i*pi/4) or sqrt(pi)*exp(-i*pi/4). This is the reason why you must have a sqrt(pi) in the final answer when you do the stationary-phase approximation of one-dimensional integrals.
d) I absorbed exp(-i*pi/4) into the square-root in the denominator. Recall that exp(-i*pi/4) = 1/sqrt(i).
e) See the answer to part (d) above.
f) zo^2/(zo^2 + x^2) is equal to cos^2(theta), not cos(theta). Therefore, when raised to the power of 3/4 it yields cos^3/2(theta).
Q1) I’m having some difficulty with problem #72. Initially, I believed it was supposed to be solved using the Fourier transform method. Part a) suggests otherwise (unless I’m incorrect in this assumption). I’m assuming all parts should not be solved using the Fourier transform method to compare with the solution obtained in class. For part b), I’m trying to solve using the variation of parameters method, however, when I try to evaluate the integral, they diverge (See attached). At first, I thought I incorrectly used the quadratic equation when evaluating the characteristic roots for the homogeneous equation. Comparison with last weeks notes states they are supposed to have a negative real part (i.e. a, b < 0 in the attached document). Would it be possible to get advice on how to tackle the problem?
A1) Your homogeneous solution is correct. For part (b), i.e., the particular solution, the method of variation of parameters is NOT the right approach. One usually makes a guess as to what the particular solution should look like. In the case, the excitation function is in the form of Fo*exp(i*2*p*fo*t), so the best guess for the particular solution is A*exp(i*2*p*fo*t), where A is the constant to be determined upon solving the equation for this choice of the solution. Note that I did not include the step function in the assumed particular solution. So, the solution that you’ll find is going to be valid for t > 0 only.
Then, for t < 0, we know that the entire solution must be zero. We are now left with a solution for t <0 (i.e., zero all the way from -infinity to 0-), and another solution for t>0 (i.e., the sum of homogeneous and particular solutions). At this point the matching of the boundary conditions must be invoked to determine the two unknown coefficients, namely, the amplitudes of the two (exponentially decaying) homogeneous solutions. This will produce the complete solution to the problem, which is valid for t from -infinity to + infinity.
Q2) In the 1-D wave equation derivation, when we did the separation of variables when we got to the step:
f”(x) = c/v^2 * f(x), why did we have to assume c is a negative constant? When c is negative, we get the correct solution. But why is c > 0 a bad solution? It would yield:
f(x) = A * exp (+lambda * x) + B * exp (-lambda * x), where lambda = sqrt(c/v^2)
Is it because we have an exponentially growing component: A * exp (+lambda * x) ? This is not physically possible?
Q3) Why would you prefer Fourier analysis of a system (e.g. spring-mass) over the Laplace analysis method? Laplace analysis seems more intuitive when it comes to determining damping (i.e. over, under, critical) than in Fourier analysis. Additionally, by using partial fraction expansion and Tables one can usually avoid “messy” inverse integrations. Nevertheless, using the Fourier analysis methods gives us practice in another analysis method and in complex plane integration.
Q1) The camera resolution is sometimes not that crisp. Could you please post the solution for z(t) that you worked out for the underdamped solution? I could not quite make out the details.
A1) My lecture notes are attached (click on “Attachments” at the bottom of this page). These are not as neatly written as my HW solutions, but I think you can extract from them the information that you need.
Q2) HW #62: Can you please explain why we have the following “property” in your solution:
a*sinc(as)*[sum{1/p * Delta(s – n/p)] = a/p * [sum{sinc(an/p) * Delta(s – n/p)].
Q3) Lecture notes on Monday 3/28 regarding spring/mass system: For the underdamped case, we said that for t < 0, we must use the lower semicircle to do the contour integration. The reason you cited was that this is because the “complex exponential has the right properties for Jordan’s Lemma.” I never really truly understood Jordan’s Lemma, even after having reviewed the notes and reading Arfken’s short description. Can you please explain this? That is, how do we know which semicircle (lower vs. upper) to use when t > 0 and t < 0, and how Jordan’s Lemma is invoked?
Q) This is more of an academic question than a question focused towards what we are learning in the course. In discussing diffraction theory, are there methods for estimating or measuring the electromagnetic function in the near field (i.e at z=0+) without calculating it using electromagnetic theory?
A) The usual problem in measuring the electromagnetic field is that the measuring device (e.g., photographic plate, near-field probe, CCD camera, etc.) will modify the field that it is trying to measure. Immediately behind the aperture at z = 0+, the problems are compounded by the fact that the material medium that defines the aperture is usually excited by the incident beam in such a way as to produce evanescent fields at and around the aperture. Placing the probe close to this medium will turn the evanescent waves into propagating waves, which are then detected by the probe itself. All in all, it is best to not attempt to measure the field at z = 0+, but, instead, calculate it using a Maxwell solver, or estimate it using the standard (although approximate) methods of the classical theory of diffraction.
Q) We are able to derive (using the definition of the Fourier Transform) that: F{Tri(x)} = sinc^2(s). What formula allows us to easily say that (see HW Problem #60): F{sinc^2(x)} = Tri(s). I know that proper application of the Fourier Transform formula always gets us the answer; but is there some “easy” trick / shortcut? In general it seems to me that:
F{Tri(x)} = sinc^2(s) <=> F{sinc^2(x)} = Tri(s)
F{Rect(x)} = sinc(s) <=> F{sinc(x)} = Rect(s)
F{1} = Delta(s) <=> F{Delta(x)} = 1
In other words, it APPEARS like we have the following “property”: F{h(x)} = g(s) <=> F{g(x)} = h(s). Is there such a property? Or am I completely wrong? Is it a coincidence that the functions above are even?
A) The difference between forward and reverse Fourier transforms is just a minus sign in the exponent of exp(i*2*pi*x*s). So, if you exchange the roles of x and s, and make a change of variable, either from +x to -x, or from +s to -s, you will see that the roles of f(x) and F(s) get reversed, except that either f(x) becomes f(-x) or F(s) becomes F(-s). When a function such as f(x) is even, we’ll have f(x) = f(-x); therefore, nothing changes. For odd function, f(-x) = -f(x), thus the forward and reverse Fourier transforms are not identical; they differ by a minus sign. Try this latter case on an odd function such as f(s) = Sign(x), or f(x) = sin(x), and see the result for yourself.
Friday, March 25
Q) I am learning a lot from your HW solutions. It is clearing up a lot of confusion I had. Thanks very much for that. Do you have any additional references on the COMB function? I would like to read more about it and would like to collect more example problems. Like I said before, an internet search doesn’t yield much and the books that I have also don’t yield much either.
A) There is not much else one can lean about the comb function. You have already seen its properties in the class and its applications in the homework problems. It is the combination of comb with other functions that usually leads to interesting results. But what you already know about the comb function, the convolution theorem, Fourier series, and the sampling theorem should be plenty for any application.
Wednesday, March 23
Q) I am in the process of scheduling time off from work. The course syllabus states we have two midterms and 1 final. I seem to recall the final in mid May and I thought we had a second midterm in mid April. However, I was not able to find the exact dates. In particular, my assumption about a second midterm may be false since, it appears we have a taped lecture on every Monday and Wednesday from now until the last day. Would it be possible to clarify?
A) Please look at the course website under “Exams and Solutions.” The information you are looking for has been there since the beginning of the semester.
Q1) I thought I understood your response last night … but now that I review my notes, there are still three “conflicting” examples in the notes for me:
a) When computing the FT of RECT(x), we do NOT apply exp(-alfa*x)
b) When computing the FT of STEP(x), we do apply exp(-alfa*x)
c) When computing the FT of exp(i*2*pi*so*x), we do apply exp(-alfa*|x|)
In general I don’t understand when to apply the various forms of exp(-alfa*x). Also, I still don’t understand the difference between a) and b) above. The RECT(x) function is applied directly to the FT definition whereas we have to apply the additional exp(-alfa*x) to STEP(x) when computing its FT.
A1) The rect function has upper and lower bounds that are finite. So, when you integrate the Fourier transform integral, the bounds of the integral will be finite. There is nothing ambiguous about these bounds. However, with the step function and the other function that you mentioned, there is an infinity in the integral. The integrated function is indefinite; that is why you multiply with exp(-alpha*x), so that the integral becomes well-defined. Once you have the value of the integral at hand, you can let alpha go to zero.
Q2) am struggling with the homework problems involving the comb function. I’ve reviewed the lecture but I am not having much luck applying it to the HW. Bracewell has a short section on it, Arfken seems to not have anything, Stephenson/Radmore has nothing either. I look forward to your solutions. But to fully understand it, can you recommend some additional references? I would like to understand this before the next exam. Is it just me who is struggling?
A2) With the comb function, I wanted you to work a bit on your own to figure things out. I said enough in the class last week to enable you to think about this problem, but if you don’t feel like you have enough information, do your best and I’ll discuss it a little more tomorrow. Also, the solutions will be posted later tomorrow, so that should clarify other questions that you might have about the comb function. In any case, please feell free to call by all means.
Sunday, March 20
Q1) In attempting problem 61, can I evaluate the integral of the Lorentzian(x)Comb(x) using complex techniques? If so, is the comb function nonzero only when x is a real integer or only for integers that are imaginary? Might I also be on the wrong track? I derived the left hand side through the Fourier transform and the integral of f(x) at s=0. It is not clear to me how to arrive at the exponential expression that you gave in the solutions so I thought that performing the integral directly would shed some light.
A1) Problem 61 does not require complex operations. The Fourier transform of the Lorentzian is something that I derived in the class a while ago [ exp(-a*|s|) ]. The rest relies on the convolution theorem and the fact that the Fourier transform of comb(x) is comb(s).
Q2) I have not been submitting solutions to the brain teasers. Are we required to submit solutions? I have been under the assumption that they are not, so I spend some time thinking about them, but relatively not a lot compared to the homework problems.
A2) There is no need to submit solutions for the brain-teasers; they are just for fun.
Q1) I followed your derivation of the Fourier Transform of the step function but I don’t quite understand why we had to take this approach: F{step(x)} = F{step(x) * exp(-alfa * x} and let alfa go to zero. Why can’t I just “plug-in” step(x) into the Fourier Transform formula? This yields -i / (2 * pi *s) which is apparently the incorrect solution.
A1) If the step function is inserted into the integral directly, the upper limit of the integral will become indefinite, because the complex exponential keeps oscillating as x goes to infinity. That is the reason for introducing the parameter alpha into the equations.
Q2) Can you please rederive the Fourier Transform of the truncated comb(x) function in class? I tried to rederive what you did but keep getting a different answer in an intermediate step which then gets me stuck. The step that I can’t rederive is where you show that: [sum(exp(-i*2*pi*s), n=-N to N] = 2*i*sin[2*pi*(N+1)*s] / (exp(i*2*pi*s) – 1). I know that this is correct … but I can’t rederive the two steps preceding this step. Everything after this step makes sense.
A2) I made a couple of algebraic mistakes in the class last week when deriving the F.T. of the comb function. I have now posted a detailed solution to this problem on the web. Please see Problem 63 of the Problem_Set and its posted solution.
From: Luca Caucci caucci@email.arizona.edu
I have a couple of questions about the HW due on Monday.
A2) Problem 61 part (b) requires Taylor series expansion up to terms in a^2, but it does not require too much algebra. My solution of this part of the problem is done in less than half a page. You may want to break up the functions involved into simpler ones, then do the Taylor series expansions for these simpler functions; just make sure that you keep track of all the terms that will end up contributing to the coefficients of “a” and “a^2,” etc.
Q) Regarding the convolution integral of problem 54, are the bounds for the convolution integral finite due to the Rect function or infinite because the Lorentzian function overlaps the Rect function from infinitely far away? I chose finite bounds and didn’t get the correct answer and I think that infinite bounds would yield pi. Maybe I’m doing something more fundamentally wrong.
A) In problem 54 the bounds on the integral are finite due to the Rect function. After integration, you’ll have to use the trig identity tan (x-y) = [tan(x) – tan(y)] / [1 + tan(x)*tan(y)] to simplify the final result.
Monday, March 14
Q1) I have a question about problem 53. Part (a) asks us to find the F.T. of the derivative of Tri(x). I arrived at -i2πs(sinc(s))^2. This makes conceptual sense because the sign of the F.T. follows the sign of the derivative of Tri, i.e., negative values of x correspond to a positive derivative of Tri and positive values of the FT for negative s values. The converse holds too. However, when applying the theorem of section (b), the F.T. of Tri is found to be -(sinc(s))^2, rather than (sinc(s))^2. Might I have a sign wrong in the first part? If so, this would make the FT of part (a) not connect with what one would expect from performing a F.T. on essentially two displaced, oppositely signed Rect functions. Can you clarify this?
A1) Your answer to part (a) is not correct; it should be +i*2*pi*s*sinc(s)^2. I don’t understand your argument about the sign of the F.T. following the sign of the derivative of Tri; there is no such relationship as far as I know. Fourier transforming two displaced, oppositely-signed Rect functions will yield +i*2*pi*s*sinc(s)^2 as well; please try it.
Q2) Is there a physical interpretation for negative frequencies, or are negative frequencies conceptually similar to imaginary numbers, in that they are required simply to solve certain problems?
Tuesday, March 8
Q1) I’m reading Chapter 15 (Integral Transforms) of Arfken and I’m curious why they write the definition of the Fourier Transform as: 1/sqrt(2*Pi) * integral (-inf to inf) of f(x)*e^(i*s*x) dx.
A1) There are different ways to define the Fourier integral. For each way, however, the inverse Fourier transform must be defined such that the original function is recovered after a F.T. and then an inverse F.T.
The method I’ve used in the class is probably the easiest one to remember in terms of the symmetry of the forward and inverse operations. It is also the method used by Bracewell. You can pick any method that you feel comfortable with and stick with it.
Q2) If you did not state the exam results in class, could you please post the class average and standard deviation, if such a thing is even statistically valid given such a small sample size.
A2) The average was 14.4 (out of a total of 25), and the standard deviation was 3.0. The total number of students taking the course is 19; the highest grade was 25 and the lowest was 1.0.
Thursday, March 3, 2011
Q1) I realize that it’s been only a day since the exam, but I was wondering if you can provide some indication about when you will post our grades. Also, will the distance learning students receive our graded exams?
Q2) I just solved HW #37. I find it peculiar/fascinating that the x^2 term in the integrand does not matter to the final solution. This is the same observation that we made with HW #33. Other powers of x would matter … Is this a coincidence? Or, can you please provide a mathematical explanation?
A3) The integral you started with is correct. Just “complete the square” in the exponent of the integrand, and you’ll see that what you end up with is the same integral as in Prob. 32 (after a simple change of variable).
The fact that the functions cos(a*x^2) and sin(a*x^2) are both real and even functions of x is important for the interpretation of these results. Both functions have real (and even) Fourier transforms, as functions of s. That’s why you can say that the real part of the F.T. of exp(i*a*x^2) is the F.T. of cos(a*x^2), and the imaginary part of the F.T. of exp(i*a*x^2) is the F.T. of sin(a*x^2).
Q1) After taking the exam a couple hours ago, I’m curious how you plan to grade it? Is it going to be a straight percentage or based on a curve? I ask because only having 25 possible points makes the difference in a couple points drastic to the percentage.
One thing I’m really curious about is if we are going to be discussing specific applications of these methods to optical systems in this course or just the straight math behind it all? I find it really helps me to apply what I’m learning to an actual optically related example.
Sorry for the rant, I’m just concerned about the exams and the future direction of the course.
So far we have only covered the basic mathematics: series, integration, complex analysis, etc. My goal is to start using the math in the context of physical systems very soon. I’ll introduce some simple systems that show the essential behavior of optical systems (e.g., wave propagation), then develop methods to solve the equations that govern these systems. You’ll see the physical applications soon. The trouble is that, until I have established a sound basis for the mathematical methods that we’ll use to analyze these systems, I can’t get to the physical models themselves.
Follow up on Q1) That’s pretty much what I figured in both questions. I could see the direction the course was heading and just wanted to make sure that physical systems were going to be part of the topics down the line.
Q1) I am reviewing the derivation of Bernoulli numbers and have a question. We carried out the first two terms in the series of x/(exp(x)-1) then moved them to the left side of the equation to get
A1) In your last expression, the final term should not be -x; it should be -1.
Q2) In problem 29 you say that cos (theta) = 1/2*(z + 1/z). I’m having trouble proving that to myself, can you please tell me how you derived that?
A2) On the unit-circle the amplitude of z is unity. Since z = exp(i*theta) and 1/z = exp(-i*theta) on the unit-circle, the sum of these gives 2*cos(theta). Division by 2 then yields cos(theta).
Q3) I don’t recall you saying so in previous lectures, but are the brain teasers actually counted as points toward our final grades? I ask because I’ve been focusing on the homework and lectures. I’ve done a few of the brain teasers, but I believe I only actually emailed you one response to them. I just want to make sure I’m not shorting myself a chance at more points in the course.
Q1) Is 0!! = 1? Or is it undefined?
Q2) I solved the problem 15 but using a completely different method. How did you get the following (2nd line of your solution): Product[ {1-4x^2/(2n)^2/PI^2} {1-4x^2/(2n-1)^2/PI^2}] = Product[ {1-(2x/n/PI)^2}]?
A2) It’s easy. If n = 1, the two terms on the left hand side have in their denominator 2n = 2 and 2n-1 = 1; if n = 2, then these denominators are 4 and 3, if n = 3, the denominators are 6 and 5, etc. You see that all the integers thus appear in the denominator of the two terms in the left-hand side. This is exactly what we have on the right-hand side.
Q3) Problem 16: You got the following two solutions for L2 (lambda 2) in the following form:
L2_1 = -a – b
L2_2 = -a + b
Where L2_1 is the first solution of L2 and L2_2 is the second solution for L2. Then the solution for L1 (lambda 1) is in the following form:
L1 = +/- L2 * c
How did you know which combination of L1’s and L2’s would ultimately yield the correct final solution? For instance, I get four possible solutions for L1’s:
L1_1 = +L2_1 * c = (-a – b)c
L1_2 = -L2_1 * c = -(-a – b)c
L1_3 = +L2_2 * c = (-a + b)c
L1_4 = -L2_2 * c = -(-a + b)c
I used all combinations of L1’s and L2’s: (L2_1, L1_1), (L2_1, L1_2), (L2_2, L1_3), (L2_2, L1_4); then solve for the various X1’s, X2’s, Y1’s, and Y2’s; and finally insert the X1’s, X2’s, Y1’s, and Y2’s into the objective function f(X1,X2,Y1,Y2) to find if they resulted in “global” versus “local” minimums and/or maximums. This was quite long and tedious. In your solution however, you seemed to know which L1 belonged with which L2 and the solution was much simpler. That is you used (L2_1, L1_1) and (L2_2, L1_4) and discarded (L2_1, L1_2), (L2_2, L1_3).
A3) I don’t understand this question. In my solution, I did “not” pick and choose L1 and L2. They were the only solutions that I could find for the problem.
Q4) Problem 17: In your solution you say that U dot V = 0 for all values of x and y. Is this true? At the origin (x=0, y=0), taking U dot V = 0/0 which is undefined. Can you please explain?
A4) Alright, at the origin things are not defined, but everywhere else U and V are orthogonal.
Q5) Problem 19: a) In your solution you say that “if f^2 > 4ed there will be two straight lines in the xy-plane”. I don’t see this so can you please explain?
b) In your solution you say that “if f^2 = 4ed … then dg/dx = 0 along one straight line and dg/dy = 0 along another straight line”. Can you please explain this too?
A5) You’ll have to go back and see the method of proof for maxima/minima that I derived in the class for this type of problem. There is a quadratic equation for Delta_y/Delta_x; when this equation has two solutions there will be two straight lines that pass through (x0, y0). My statements in the solution to problem 19 follow directly from what I discussed in the class with regard to the maximum/minimum problem.
Q6) I am struggling to understand Jordan’s Lemma and its applications. I’ve reread my lecture notes and I still cannot see how you used Jordan’s Lemma to solve problems 31 and 36. Can you please go over some of this on Monday’s lecture?
A6) Jordan’s lemma is thoroughly discussed in Arfken, in much the same way as I discussed in the class. Please read Arfken’s section on this topic. All that Jordan’s lemma says is that, under certain general conditions, the integral over the semi-circle goes to zero whe R goes to infinity. In problems 31 and 36, I invoke the lemma to ignore the contribution of the semi-circle to the contour integral.
Q7) I am trying to do Problem 33. I am getting a “z1^2” on the numerator of the first term and “z4^2” on the numerator of the second term in the solution. How did you get rid of these z^2 terms in the numerator? The original integrand is z^2 / (1+z^4).
Q1) In reviewing problem 15, there is an inconsistency in your solution that I cannot reconcile. If I substitute n=1 into the LHS of the bottom two equations, the left and right sides do not equal each other. The LHS is 1-5x^2/π^2+(4x^2/π^2)^2 which is clearly not equal to 1-4x^2/π^2. Can you clarify how the discussion resolves this?
A1) Don’t multiply pairs of terms; it will only cause confusion. Simply look at the numbers that you have on the left hand side. For n = 1, the two terms on the left have denominators that are 2 and 1, respectively. Then for n = 2, the denominators are 4 and 3. For n = 3, the denominators are 6 and 5. Clearly all the integers are covered in this way. On the right hand side, there is only one term for each n, but the denominators here are going up sequentially, 1, 2, 3, 4, 5… This is how you can see the equality of the two sides.
Q2) I am currently redoing all the HW problems to study for the exam. I am trying to resolve all of my mistakes and also pick up all the “tricks”/techniques/shortcuts that you employ in solving these problems.
Toward that end, for HW problem #13, how did you get so quickly (in one step) the following results:
a) In part a), how did you know in one step:
sum n=2 to inf [sum m=2 to inf [1 / n^m]] = sum n=2 to inf [(1/n^2)/(1-1/n)]
I got the SAME FINAL answer as you but I took MANY MORE INTERMEDIATE steps.
b) In part b), how did you know in one step:
sum n=2 to inf [sum m=2 to inf [(-1)^m / n^m]] = sum n=2 to inf [(1/n^2)/(1+1/n)]
Again, I got the SAME FINAL answer as you but I took MANY MORE INTERMEDIATE steps.
A2) In both cases I used the formula for the geometric series. Looking at the sum, I see the first term, 1/n^2, and also the coefficient that each term is multiplied with to get the next term. This coefficient is +1/n in the first case and -1/n in the second. I then use the formula for the geometric series, which is the first term of the series divided by one minus the coefficient.
Follow up on Q2) Yes, I also used the geometric series but in an “inefficient”/roundabout way. I’ve been studying/absorbing what you suggested below and I like it a lot better than my method. I think it’s true what you said on the first day of class: Math is easy/trivial after you see the solution. In other words, math is always easier when someone else has discovered the solution!
Q3) In your solution to problem 33, it looks as if the function f(zm) = zm^2 was excluded from the residue calculation. When evaluating the residues using this additional function, I get the same result, but I explicitly included f(zm) and it was not obvious to me that you did. Can you explain this?
A3) You’re right. I forgot the factor z^2 in the numerator. The final answer, however, is going to be the same. The first residue must be multiplied by +i, the second by -i. When the algebra is done with these additional factors, the end result turns out to be pi/sqrt(2) again. Thanks for pointing this out.
Q1) In the last lecture you went through a long derivation of how to compute the [Residue at (zo = i*PI)] = -exp(i*a*PI) for exp(az)/(1+exp(z)). When I use the formula from the Stephenson and Radmore book, I get: Residue = +exp(i*a*PI). Both methods seem correct, but why is there a difference in sign? Stephenson and Radmore give the formula for the residue as:
lim z->zo [ (z-zo)f(z)] =
where I have used 1+exp(z) = (z-i*PI) because zo = i*PI is the only pole in our rectangular contour. Can you please tell me where I went wrong?
A1) Your mistake is in replacing the denominator, (1 + e^z), with (z – i*pi). You must expand 1+exp(z) around the pole at z = i*pi. When you do so, you must use the derivative of exp(z) at z=i*pi as the coefficient of (z – i*pi); this according to the standard rules for Taylor series expansion. But the derivative of exp(z) is also exp(z), and so the coefficient of (z – i*pi) must be exp(i*pi), which is -1.
Q2) On the Wed 2/23 lecture, I do not see how you got the Taylor series expansion of f(z1 + a*exp(i*b)) to become f(z1) + f ‘ (z1) (a*exp(i*b)) + (1/2!) f ” (z1)(a*exp(i*b))^2 + …, where a = epsilon and b = theta.
I recognize that the Taylor series of a function f(x) that is infinitely differentiable in the neighborhood of the point a is, f(x) = f(a) + f ‘ (a) (x-a) + (1/2!) f ”(a)(x-a)^2 + … I don’t completely follow your derivation above however.
A2) Mine is exactly the same Taylor series that you have written down for f(x) around the point x = a. For the example that I did in the class, a = z1, and (x – a) = (z – z1) = a*exp(ib). What is it exactly that you don’t see in this Taylor series expansion?
Q1) This is in regards to the integral from -inf to inf of e^ax/(1+e^x). In your derivation of the contour integral around the rectangle, why couldn’t you say that the integral of the left and right “legs” cancel because their paths are in opposite directions? In regards to the integral from -inf to inf of 1/(1+x^2), for your derivation of the contour around the semicircle with a pole inside, it seems that you said that the integral of the left and right “legs” of the small circle cancel because their paths are in opposite directions. Can you please explain the difference?
A1) In the first part of your question, the integrals on the left and right legs do not cancel out. Although their directions are opposite, the values of the function on these two legs are not the same. Just try some numerical values for the function with z = R + i*y on the right leg and z = -R + i*y on the left leg, to see that the function is “not” the same on the two legs of the rectangle.
Q2) In general, you’ve shown us very nice techniques to tackle tough integrals. However, without being either (a) very clever or (b) having a lot of experience, is there a general “rule of thumb” for selecting the appropriate contour C? I ask because I OFTEN get stuck without looking at your examples in class or the homework problems.
Q1) I have finally revisited the integration by parts problem of x^2*exp(-k*x^2). I tried defining u and v two different ways and neither produce the desired result. What steps did you follow in obtaining The RHS of the equation?
A1) I suppose you’ve seen my solution to Problem 5. I chose u = x and v’ = x*exp(-k*x^2).
Q3) Will the exam also cover “number theory” and “geometry”? Some of the homework problems covered rational/irrational numbers and problems related to similar triangles, etc. Should we focus on this also or focus exclusively on the latter topics such as series, Lagrange, curvilinear coordinates, complex numbers, contour integration? I am starting to study for the exam today and would like to know where I should emphasize.
A3) The purpose of the exam is to get the students to go over everything once again and try to see the big picture. The exam cannot possibly cover everything that we discussed during the past six weeks, but I expect the students to be prepared to answer any question on the material that has been covered.
Q) Would it be possible to explain how to evaluate the integral of the square root of z on the unit circle? My main difficulty is choosing the domain of integration. In this case, -pi to pi works well, but we have to be careful if the domain of integration is taken from 0 to 2*pi. Is there a quick and dirty way to decide that the domain of integration should be taken from -pi to pi?
A) The square root of z is a multi-valued function. You must define a branch-cut before proceeding with any operations on this function. If you choose your branch-cut to be the positive real axis, then your range of integration will be from zero to 2*pi. However, you may choose the branch-cut on the negative real-axis, in which case the integral on the unit circle will go from -pi to +pi. The result of this integration over the unit-circle will depend on the choice of the branch-cut, which is understandable, given the multi-valued nature of the square-root function.
Q) I am having trouble following your steps for proving how an integral goes to zero as R goes to infinity. You did this for problems 32 and 35. Can you please take a moment to explain this in class? This seems like a very important skill to have and I want to be prepared for the upcoming test too.
A) Yes, I will discuss this in class today and on Wednesday. There is a lot more to discuss on the subject of contour integrals before all the important points are covered.
Tuesday, February 15
Q) Regarding your answer to problem 25 posted today, I approached the problem from the perspective of a product series for each expression. From this I found real and imaginary roots to each equation. However, I am a bit puzzled because there were effectively an infinite number of real and imaginary solutions. This seemed to violate the concept of placing branch cuts into the complex plane. I can see that there are an infinite number of real-valued solutions, but are there the same number of imaginary-valued solutions? From our discussion, it seems that there should be at most two imaginary solutions, and that they are orthogonal due to the nature of the problem, excluding z=0 for sin(z) and sinh(z).
A) Please wait until tomorrow afternoon, when I’ll post the solutions. There is no need for product series expansions; the problem is much simpler than that. The sine and cosine functions are periodic, with zeros at equal intervals on the real axis. The sinh and cosh functions are also periodic, but with zeros placed at equal intervals on the imaginary axis. This problem has no branch cuts.
A) Take a function such as sinh(z). Write it according to its definition; in this case,
Monday, February 14
Q) I do not quite see the objective of Problem 23. Do you want us to verify the Cauchy-Reimann condition or take the derivative or both or perhaps something else?
Sunday, February 13
Q) I have been reviewing your solutions to problems 21, 22, and 24. I have a question about 21’s solution. Part (c) looks correct, but for (a) and (b) it looks like there is a discrepency with the sign of du/dx. In both ()a and (b), it looks like du/dx = -dv/dy, not du/dx = dv/dy. The du/dy = -dv/dx parts look correct, but not the first part. Am I missing something, why is there a sign discrepency? Like I said, part (c) looks fine.
Monday, February 7
Q) In your work on elliptical coordinates, you determined that the area of the ellipse was pi/4*a*b^3. However, the area of an ellipse is pi*a*b. How do you reconcile this? The two approaches should agree independent of the coordinate system.
A) I didn’t calculate the area of the ellipse. What I calculated was the integral of y^2 over the area of the ellipse. If I were to calculate the integral of “1” over the area, I would get pi*a*b, which is the area of the ellipse. You may want to check this using the curvilinear coordinate system.
Q1) Would you consider posting the homework assignments in advance by one week? The reason I ask is sometimes things slow down at work so I would like to get to the homework well before it’s due. Conversely, sometimes work gets extremely hectic and I find myself doing homework late into the night just to meet the Wednesday deadline.
A1) This is the first time that Opti 503 is being offered. I am making up the lectures and homework problems as we go along. As soon as I have enough problems for an assignment I will post them, but I am afraid this is not going to be too far in advance of the due date. Besides, you’ll need to see the lectures before you can do the homework anyway, so you can’t run too far ahead of the lectures either.
Q2) Would you consider giving us “bonus” homework problems for extra credit (similar to asking a “good” question)?
A2) In future years, there will probably be time for me to prepare bonus problems, but this year I’ll have to struggle just to give you enough HW problems (and their solutions) on time.
Thursday, February 3
Q1) I finished homework #3, with the exception of Problem 17, part 2, where you asked: “Show that the vectors U and V which specify the directions of the local u and v coordinates are orthogonal everywhere.” Would you please give a hint?
Q2) I love the brain teasers, Can we get more brain teasers per week?
A2) I would love to give you more brain teasers per week, but I’ll quickly run out of them, because there are very few that I know which are actually clever and also have some mathematical trick about them. If you find new ones please share with me. May be next year I’ll know enough good ones to be able to give the students more than one per week.
Q) I have a question about number theory that came about from problem 9. Since I was able to obtain a rational number between two arbitrary irrational numbers, why is it that there are more rational numbers than irrational? It seems that because one can find a rational number between two irrationals that the two should be the same.
A) You can find MORE than one irrational between any two rationals. In fact, there is an infinite number of irrationals between any two rationals, no matter how close the rationals may be to each other. There is also an infinite number of rationals between any two irrationals, no matter how close. The question is: which infinity is greater? According to Cantor, the irrationals form a larger (infinite) set than the rationals.
Monday, January 31
Q) I’m looking ahead at my work schedule and the date of the first exam. I was wondering what your policy is for distance students taking exams. Do I have to take it at the exact same time as the on campus students? Or am I allowed to take it early or later? It’s impossible to predict my work schedule, I won’t know anything until we get closer to March 2nd, but I may have to work the exam in early or later than the 2nd. It’s also possible that I won’t have any problems and can take it whenever on the 2nd. I just want to clear everything up before it gets close to exam time.
Q) I was finally able to solve Problem 13, Part 1, but now I am stuck on Part 2. I tried to leverage what I did in Part 1, but to no avail. I have honestly spent at least 14 hours on this SINGLE homework problem. I am attaching my homework. Can you please review it to see where I am going wrong?
A) Part 2 is also based on the geometrical series, just as in part 1. Instead of multiplying each term by 1/n, it multiplies each term by -1/n.
A) I actually showed the basic approach to this problem in Lecture 1. You may want to review that lecture to see how I proposed to do it. There are some loose ends that need to be tied up, but the essence of the argument was presented in that lecture.
In general, I’d like the students to think about the problems as much as they can. Then, if they can’t solve the problem, wait until I post the solutions at the end of the week.
For many of the subjects that I’ll discuss in this class, Arfken would be a good reference, but for the first few things that were covered in lectures 1 and 2, I am not using any references in particular. Sometimes, you may find useful stuff on the web by googling the relevant words; in this case, for example, “proof that between two rationals there is at least one irrational” would be a good starting point for a search.
Saturday, January 22
Q) I was able to complete all the problems in HW #1 EXCEPT for problem number 6. I tried using trigonometric identities; I tried using Taylor series approximations; and finally, I tried using complex exponentials. I was unsuccessful with all these methods. I have come to a complete roadblock. Can you please help me?
A) Start with the sum exp(i*theta) + exp(i*2*theta) + … + exp(i*n*theta), which is a geometric series. Then look at the real and imaginary parts of both sides of the equation.
Q) But what if the contradiction lies in the initial assumption about the table? Can one make the argument that the initial table is not complete? Since we start with a table that is thought to be complete and derive another number, one can claim that the initial table was not complete.
Q1. About there being more irrational numbers than rational ones: How can there be more irrational numbers than rational ones if there are an infinite number of rationals between two arbitrary rationals while there is also an infinite number of irrationals? How can one infinite set be larger than another infinite set?
A1) There are different types of infinities: countable and uncountable. The rationals are countably infinite, the irrationals are not.
Q2. About Cantor’s Proof: Just because the first digit differs from the first digit of the first number in the table and the second digit differs from the second digit of the second number, and so on and so forth…who’s to say that that new number doesn’t appear somewhere in that table? I’m having trouble understanding why this proves that irrational numbers aren’t countable.
A2) The table is assumed to contain all real numbers between 0 and 1 (both rational and irrational). The assumption of countability allows us to number them: 1, 2, 3, 4, etc. Now, the newly created number differs from ALL the numbers in this table, because it differs in at least one digit from any number selected from the table; therefore, it cannot be in the table. But this is a number between 0 and 1, and, by assumption, it must be in the table, hence the contradiction.