I've just realised, after all this time, that a lot of the comments on my original post, when I set this blog up, would be better off as posts. This would enable Blogger to create an archive of the posts and make it easier to find relevant posts using the search facility. In the longer term (not too long I hope) I would like to replace this blog with a forum.
If you are wondering where all the original comments are, then look below at the original post with the title 'Visual Complex Analysis' which has all the previous comments.
Vasco,
ReplyDeleteMy first attempt to post appeared to fail when I clicked on preview, as my post disappeared, but I could not see any indication that it had been processed. I try again, hoping I am not redundant.
An attempt to present an answer to Exercise 24 (iv), chapter 2 of Needham by using Mathematica to draw the loops and pathways seems to provide a different picture from that in the online answers. I am wondering if the pathways in the f(z) plane were plotted from actual branch function results. The computed pathways for anti clockwise and clockwise loops around branch point 1, starting at point -1 can be seen here:
faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/CDF/multifunction.htm
Gary
Gary
DeleteHere is a document in reply to your question about Exercise 24 (iv):
https://drive.google.com/file/d/0B4HzA5QgnnvZaGl0Yno2SnFNSDA/view?usp=sharing
As far as problems posting in this blog are concerned, I have been re-thinking this and because of the difficulties in allowing other people than me to post I am thinking about a forum instead.
Vasco,
ReplyDeleteIt's a beautiful explanation and I see the source of the difficulties with the program. Unfortunately, I only see a partial solution, but I have no real need for a solution other than to satisfy my curiosity and that is not your problem. I just need to work on an algorithm that keeps track of the total angle, theta or phi, generated by an arbitrary sequence of points encircling one or the other branch point.
There seems to be a lesson here that did not emerge for me by reading Needham: That the decision to use branch cuts is optional. One can apply multifunctions without using branch cuts if it suits one's purposes. Is that correct?
It took me some time to reply because I am confused about the new posting system. I see no place to actually make a post, only places to comment or to reply. I'm glad you found my comment made on the old blog site. I was afraid I had lost contact entirely. A forum might improve ease of access if it can be managed.
After I selected a choice under "Comment as" pull down menu, the program went right to "publish" and failed to present the little security quiz. There is no indication on my screen that it actually published my message, so I am trying again, first to preview again, because I have found in the past it is sometimes the only way to obtain the quiz.
Gary
Gary
DeleteYou are certainly not forced to make branch cuts, unless you want a single-valued function over the whole complex plane.
Just to be clear I have not created a new blog. This is the old blog. All I did was add a few posts based on some of the original comments. I have now deleted these and the site is more or less the same as before except that I have left in one of the new posts (which was called Advice on using the blog) because we (You and I) have used it for some recent posts. I have changed its name (at the risk of confusing people even more) to Visual Complex Analysis continued.
I am sorry about the confusion I caused. I realise now that to allow other contributors to the blog to post would require them to give me their email address and then I would have to set the blog up to allow them to post which is too complicated. So I will leave things alone now until I am satisfied that a forum would work and then I will make it accessible from my website and have a trial period to see if it works OK.
Re. Nov 6,
ReplyDeleteVasco, thank you for the reply. This topic of multifunctions and branch cuts seems very tricky, and trying to program the behavior of the functions is even more so. My attempt seems to be functioning properly as an illustration,
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/CDF/multifunction.htm
but as an exercise in programming it is not very satisfactory, as it depends on revising the Wolfram Arg[] function and in its present state, it works with only one loop, it wouldn't work with multiple winds of the loop, nor does it go beyond the second branch. But at least the attempt has sharpened my interest in the text.
Gary
Hello can you do number 8, and number 17 chapter 8. Thanks
ReplyDeleteYes, but it will probably take a few days. I will do exercise 8 first and publish it and then exercise 17 and publish that. It's very often easier to see how to do an exercise in your head, but then writing it up to a reasonable professional standard can take a good few days. I hope you are not in too much of a hurry!
ReplyDeleteVasco
I have a couple of days. Thank you so much!
DeleteAdded exercise 8 chapter 8
ReplyDeleteAdded exercise 17 chapter 8
ReplyDeleteVasco,
ReplyDeleteI'm puzzling in understanding the proof you give for Ch.1 Ex. 28 part (ii). In your demonstration you assume that "a rotation followed by an expansion makes equal the angles that rays form with spiral". But this seems to be exactly what the author asks to demonstrate.
Thank you
Renato
Hi Renato
ReplyDeleteI think my proof is OK. Here is an explanation which I hope will be useful to you:
To start we draw the rays from the origin to intersect the spiral. At this point we do not know if the angle between the rays and the spiral is the same for all rays.
First we show that the points of intersection of the rays with the spiral, P, Q, R, S, etc. are transformed to the points Q, R, S, T, etc by the transformation F. So we can say that the angle between the ray and the spiral at P has been transformed to the angle between the ray and the spiral at Q, and similarly for the other points Q, R, S, etc.
But we know that the transformation F is a rotation followed by an expansion and we also know that rotations and expansions preserve angles between intersecting curves. It follows that the angle between P and the spiral must be equal to the angle between Q and the spiral, and so on for the other points.
I hope this explains the proof more clearly.
Please post again if you still think there is a problem.
Ciao e buone feste
Vasco
Thank you Vasco,
DeleteI should have understood it by myself by reading better your first demonstration.
Buone feste anche a te
Renato
Vasco,
ReplyDeleteYour solutions are very helpful. However, it is seriously a pain to view problems individually. Is there anyway I could get a pdf of all your solutions at once?
Thanks,
--Alex
I could add a pdf for each chapter with all the exercises for that chapter in it. Initially it would not have a contents list, but each solution would start on a new page. Would that be the sort of thing you are looking for?
DeleteI could add a contents list at a later date.
If you wish to put a number of pdf files into one single pdf file then download pdfbinder from here:
Deletehttps://code.google.com/p/pdfbinder/
It is very easy to use and it works.
Vasco
Vasco,
ReplyDeleteI was accessing the most recent comments by using the “Load More” link 3 times in succession. Now I find that that link no longer responds. Of course, I will still have access to your solutions, but I fear that you may think I have lost interest in the blog. I haven’t, and I am not likely to, so I will keep trying various ways to communicate. My email address is available at various places on my web page.
Gary
Yes, it stopped working for me for a couple of days a few days ago but it seems to be working again now. Give it another try.
DeleteThere is a useful site here:
http://www.isitdownrightnow.com/blogger.com.html
which reports on the status of Blogger and where you can report problems.
In the new year I hope to open the forum I mentioned a few weeks ago which I hope will eventually replace this blog.
Sorry you've had problems - don't worry I do know your email address.
Added exercise 7 chapter 9
ReplyDeleteAdded exercise 10 chapter 9
ReplyDeleteAdded exercise 2 chapter 9
ReplyDeleteUpdated exercise 10 chapter 1 : 'equilateral' changed to 'isosceles' and some extra explanation to try and make the proof easier to follow.
ReplyDeleteVasco
Updated exercise 9 chapter 1: improved diagram and simplified proof.
ReplyDeleteVasco
Added exercise 22 chapter 2
ReplyDeleteAdded exercise 23 chapter 2
ReplyDeleteAdded exercise 41 chapter 2
ReplyDeleteWould you be able to post exercise 11 chapter 2 please?
ReplyDeleteHi Christine
DeleteI'll have a look at it today.
Vasco
Added a partial solution to exercise 11 chapter 2. Solution shows how to calculate radius of convergence for all 3 parts and investigates convergence ON the unit circle for part (i). Convergence ON the unit circle for parts (ii) and (iii) to follow later.
DeletePublished my complete solution to exercise 11 chapter 2
DeleteCould you please do problems 9,15,17,18 of ch3. Thank you!
ReplyDeleteHi
DeleteExercise 9 is already published. I will have a look at 15,17, 18 as soon as I can.
Vasco
I apologize I just realized that. Thank you so much!
DeleteI have done exercise 18. i just need to write it up and publish it. Maybe later today or tomorrow.
DeleteVasco
Great! Thank you!
DeletePublished exercise 18 chapter 3.
DeleteIt may take a while to do the others, 15 and 17.
DeleteJust added some vertical bars to the solution to exercise 18 chapter 3 in a couple of places.
ReplyDeleteHi Gary and Missourigirl
ReplyDeleteMissouri girl asked me to look at exercise 15 chapter 3 on page 185 of the book. I am a bit mystified by the statement of the exercise and thought you might be able to help me understand what the exercise is asking us to do.
Because zqrs in the left diagram which accompanies the exercise is a cyclic quadrilateral then theta+phi is equal to pi, and in the diagram on the right phi is equal to theta, because they are angles with vertex on the circle and subtended by the same chord.
Do you think that in the left diagram z should be external to the circle and in the right diagram internal to the circle?
Vasco
Vasco,
ReplyDeleteFirst and foremost, bear in mind that my understanding of Möbius transformations is still pretty fuzzy. I'm having difficulty making sense out of (29), though I do see the rational behind the equation for [z, q, r, s]. Question 15 sends us back to (31), which seems to assume some understanding of (29) and (30). That said, (31) speaks explicitly only of points on and inside C. So my guess would be that placing z outside the circle would be straying from the assigned task unless you are working on a proof by contradiction.
I have finished a diagram for 11 and am now trying to respond to 11. (iv). Unfortunately, I don't think my diagram is showing that A inverse and B inverse are concentric. In fact, my B inverse is a line, as B passes through the center p of the circle of inversion.
Gary
Gary
DeleteThanks for the reply. I think I can explain (29) if ever you need it.
Vasco
Vasco,
ReplyDeleteThanks. I think I have it now. Both the right hand side and the left hand side send three points to the real line at 0, 1, and infinity. The equation as a whole represents a Möbius transformation z —> w = M(z). As Needham points out, we could solve for w. Presumably, we could also solve for z in terms of W to get M^(-1)(w) = z = [(s(t-v)(u-w)-r(t-u)(v-w))-rs(u-v)(t-w)] / [q(u-v)(t-w)-r(t-v)(u-w)+s(t-u)(v-w)]. (where t, u, v replace the superscripted q, r, s in Needham, p. 154; the square brackets here are ordinary brackets)
This seems to say that for any Möbius transformation that sends three points q, r, s to the real line at 0, 1, infinity, and send point z to point w, there is another Möbius transformation that sends three points t, u, v to the real line at 0, 1, infinity, and send point w to z. One needs to know all six of the points in order to solve for w or z.
Does this look correct and adequate?
Gary
RE May 2, 2015 at 10:30 PM
ReplyDeleteGary
That all looks OK to me. What makes it all work, it seems to me, is that a Möbius transformation that sends 3 given points to 3 other given points is unique.
Vasco
Vasco,
ReplyDeleteThank you for the comment of May 2, 2015. The uniqueness condition also seems right to me.
I have posted a solution for Chapter 3, Example 10. The dimensions and locations of all the elements in the figure are calculated from arbitrary assignments of origins and radii for A and B.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/020.mobius.transformations.ch3.ex10.html
I would put angle brackets on the url, but the editor rejects them.
Gary
Re May 8, 2015 at 5:20 AM
ReplyDeleteGary
I have a solution for exercise 10 chapter 3 also. I will publish it when I return from a short holiday from 10th May-17th May. Once I have published my solution I will have a look at yours.
Vasco
Vasco, Re May 8, 2015, 5:20
ReplyDeleteGood. I'm sure I will learn something new.
I noticed this morning that the equationsthat looked nice on my screen are displayed incorrectly on another computer. They are written in a Wolfram html script. I will try to find another way to display them, but they are also available via the sourcecode link in .pdf format.
Gary
Re: Gary, May 9, 2015, 5:20
ReplyDeleteVasco,
The problem with display on my web page is that recent versions of the Chrome browser do not support MathML. It should display on Firefox or Safari and perhaps Explorer.
Gary
Vasco,
ReplyDeleteI encountered some difficulties in Ch. 3, exercises 11 and beyond, so I decided to study III (1) on p. 136-137 more closely. That, too, entailed some difficulties. I leave it to others to judge whether my method of reproducing Figure [14] is mathematically sound.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/021.touching.circles.ch3.html
Gary
e, Gary, May 13, 2015 at 5:05
ReplyDeleteVasco,
This is to add that the linked page now includes an explanation for the second result in problem 1 concerning touching circles.
It’s a very instructive problem. I don’t see how one would be able to construct the orthogonal circles and the plot of the inverses given only Figure [14] other than as I have done it using C_0.
Gary
Added exercise 10 chapter 3.
ReplyDeleteGary: I will now have a look at your solution and at your method for constructing Figure [14] in the book.
Vasco
Re May 27, 2015, 5:19 PM
ReplyDeleteVasco, Thank you. Your answer to Ex. 10 is more explicit and I find that helpful.
I have also improved my answer to Ex. 11, but I have not yet been able to agree with Needham’s conclusion. Perhaps rearranging p, q, g, and h will make a difference, but I think it should not.The index for my exercises and notes is
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/
The most recent Figures are last.
More recently, I have a difficulty on p. 170 line -3, where Needham wrote “and we easily find [exercise] that the normalized matrix is [[0 i], [i 0]].” I can see that the determinant is 1, which it should be for a normalized Möbius transformation, and I am guessing that the i values in the matrix probably derive from square roots of the values in the matrix (m = -1 ) [[0 1], [1 0]] ], where the coefficients of the coordinates of 1/z are written in matrix form as [[0 1], [1, 0]]. But I can’t find a good path to that result. It would make sense if the matrix M~ is [[0 L], [L 0]], where L = lambda, because lambda in equation (33) corresponds to Sqrt(m) in equation (43). We seem to obtain a different value for lambda, L = +/- 1, by solving det([[0 1], [1 0]] - [[L 0], [0, L]] = 0. If we are allowed to assume that m = 1/L^2, then the solutions are L = -/+ i. If we use only +i, we have the matrix that Needham says is correct: [[0 i], [i 0]]. Can you provide any insight into how one goes about finding the normalized matrix?
Gary
Re Gary, May 28, 2015, 6:19 AM
ReplyDeleteVasco,
I would like to delete the final paragraph of my previous comment, but I don’t find a link to a delete function.
Regardless, I see now that [[0 i] [i 0]] is just the normalized matrix of 1/z obtained by multiplying the coefficients through by -1/Sqrt(ad - bc), from p. 150, that is, by -1/Sqrt(-1) = i, or by multiplying 1/z written as [[0 1] [1 0]] by i.
Gary
Re: Gary, May 28, 2015 at 6:19 am
ReplyDeleteGary
I spent most of yesterday looking at exercise 11 chapter 3 and drawing the diagrams in the introduction to the exercise and for parts (i) to (iii). In my opinion there is an error in the statement of the exercise:
After drawing the diagram asked for in part (iii) the 'circle' B tilde is a straight line and A tilde is a normal circle and so the request in part (iv) does not make sense.
After some thought, some study of the diagram in part (iii) and then crucially, reading the final sentence of the exercise, I realised that the original circles A and B have to be subjected to TWO inversions for the final transformation to be a Moebius transformation. It seemed logical that this second inversion should be of A tilde and B tilde in the circle centred at g to produce A' and B' - two 'proper' circles.
I tried constructing this and sure enough the circles A' and B' are concentric and centred at h'.
So in my opinion for part (iv) to continue logically from part (iii),it should read:
"....,deduce that under inversion in any circle centred at g, A tilde and B tilde are mapped to two concentric circles A' and B' centred at h'".
Now all I have to do is prove it generally by addressing part (iv).
Let me know what you think. Does this make sense to you?
Vasco
Re Vasco, May 29, 2015 at 11:14 AM
ReplyDeleteVasco,
It makes sense and it works out in the program. See the cyan and magenta circles in Figure 4.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/019.inversions.ch3.11.html
Gary
Re. Gary, May 29, 2015 at 8:41 PM
ReplyDeleteVasco,
Rereading (iv), I see that the concentric circles in my Fig 4 are not centered at h’ as they should be.
Gary
Re Vasco, May 29, 2015 at 11:14 AM
ReplyDelete"...., deduce that under inversion in any circle centred at g, A tilde and B tilde are mapped to two concentric circles A' and B' centred at h'"
I would like to suggest the following revision:
“…, deduce that under inversion in a circle of radius r_A_tilda centred at g, A tilde and B tilde are mapped to two concentric circles in A’ and B’ centered at h’.”
alternatively,
“…, deduce that under inversion in any circle centred at g, A tilde and B tilde are mapped to two concentric circles.”
Gary
Re: Gary, May 29, 2015, 10:25 PM
ReplyDeleteThe radius in the first suggested revision should be r_A, not r_A_tilda.
Gary
Re Gary, May 29, 2015 at 8:41, 9:51, 10:25, 10:30
ReplyDeleteThanks Gary for looking at my idea and for your subsequent comments. For the moment I will have to continue with my original idea, as after some experimentation I find that I disagree with your suggested changes.
Since h lies on the circle centred at g, the position of h' varies as we change the radius of this circle centred at g, whereas you are inverting in a circle whose radius depends on the radius of the original circle A, which is fixed, if I understand you correctly.
Vasco
Re: Vasco Ma6 30, 2015, 9:58 AM
ReplyDeleteVasco,
I find that you are correct. I am only starting to see my way through this problem. I looked through my program and found that I had assigned r_A to be the radius of the circle centered at G, so the inversions h’, A’, and B’ are all based on that.The r_A is only necessary to my calculations of B’ and A’ because of my choice of r_A for the circle centered at g. This suggests that I could improve my problem solving if I were to improve my programming style, in this case with an explicit variable r_g.
Gary
Re Gary, May 8, 2015 at 5:20 am
ReplyDeleteGary
I have looked at your solution to Chapter 3 exercise 10. Here are my comments:
To do the exercise it's not necessary to solve the quadratic, as it only asks to show that there exists a pair of points that are symmetric with respect to the two circles. I can see that it might be instructive to do so.
At the end you say that the centres of the two circles are F(xi+-), namely 0 and infinity, but this is not so in my opinion. F maps circles through the points of symmetry to circles through 0 and infinity, i.e. radial lines through the origin. See my solution for more details.
Vasco
Re Vasco, May 30, 2015, 6:21
ReplyDeleteVasco,
It’s not necessary to solve the quadratic. Agreed, and I should have been more explicit regarding the two solutions for xi. I tend to get more interested in seeing the plot than in the algebra and the proof. A plot with the expected results does not necessarily demonstrate that the algebra is correct.
Re. the centers of F(A) and F(B), it’s a fine point and my writing was a little sloppy, but I said “0 or Infinity”. Since Infinity doesn’t make a very practical circle center on the complex plane, it has to be 0. But I think perhaps I am missing the point. We know from the paragraph under [29] that F(z) sends the fixed points to 0 and Infinity. The circles that pass through 0 and infinity under F are the C_1 circles and not the C_2 circles that encircle the fixed points xi+-. This is shown in [29 a and b] by the positions of z -> w and z tilda -> w tilda. The C_2 circles correspond to the A and B circles in Ex. 10. Since F(A) and F(B) have centers at 0, they must be concentric. We know they have centers at 0 because, as you say in the answer, “Since F is conformal A’ and B’ must both be orthogonal to the lines through the origin”, which also pass through Infinity. Additionally, at the top of p. 164, the images of the C_2 circles are referred to as “origin centered” and “orthogonal to lines through 0”, which may be why I assumed them to be so.
Thanks for looking at it. Do you still see a problem?
Gary
Re Unknown, May 30, 2015 at 12:14 am
ReplyDeleteGary
I have made a few more comments on your solution to exercise 3-10, in the following document:
https://drive.google.com/file/d/0B4HzA5QgnnvZZnhnYVVNOXdZazQ/view?usp=sharing
I hope you find it a bit clearer than my last post.
Vasco
Gary
ReplyDeleteI think I have managed to do exercise 11 of chapter 3. I will write it up properly and publish it as soon as I can.
Vasco
Re Vasco, May 30, 2015 at 12:14 am
ReplyDeleteVasco,
Thank you very much for the follow up. I put something like your comment at the bottom of my web page on the problem last night. Then I went to sleep on it and I realized this morning that you are right: for the purposes of solving the problem, it really doesn’t matter that F(xi) has a zero value that could be the center of the circle. The proof is as you say, the orthogonality of the circles and the lines that derives from the fact that the Möbius transformation preserves circles and orthogonality with lines through the origin under F(C_1), which must be drawn or imagined in the exercise. The new comment seems to me to be just right.
The only basis I can see for maintaining that the zero values of F(xi+-) for both solutions provide the centers is (1) it is given that 0 is one of the values of F(xi+-), (2) on p. 164 it is implied that the C_2 circles are origin-centered, and (3) on p. 164 it is reasoned that the C_2 circles are orthogonal to the C_1 circles. Then, because we found in part (i) that the fixed points are symmetric to both circles, we can see that the situation in Ex. 10 is analogous to that in [29]. The zero value of F(xi+-) and the origin are equivalent. Given the analogy, it seems justifiable to get the center directly from the two zero values of F(xi+-), but it does not make a very instructive answer.
Re May 31, 2015 at 4:00 PM
Ex. 11 is a tough one, but our discussion of Ex. 10 has helped. I won’t look at yours until I have it.
Gary
Vasco,
ReplyDeleteI have posted a solution to Ex. 11 at
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/019.inversions.ch3.11.html
I do not see why it is necessary to appeal directly to the anticonformal nature of inversion, but perhaps you will be able to elucidate it. I have appealed instead to the results of Ex. 10 i and ii.
Plotting the orthogonal circles will be tricky, so a new figure will follow later.
Gary
Re Gary June 1, 2015 at 12:43 am
ReplyDeleteI will not look at your solution until after I have posted mine.
However I think the point of exercise 11 is to show that the result of 10 can be obtained by a "more intuitive proof", as Needham puts it in the statement of exercise 11. So doing it by using the result of exercise 10 doesn't seem right to me. Drawing the sequence of diagrams suggested in Exercise 11 and then appealing to the anti-conformal nature of inversion leads to the same conclusion as in exercise 10.
There is no need to construct orthogonal circles - just draw the suggested diagrams and then follow our amended version of part (iv).
The way I see it is that appealing to the anticonformal nature of inversion means using the fact that if two curves intersect, then the angle between them will be the same after inversion.
Vasco
Re June 1, 2015, at 9:21 AM
ReplyDeleteVasco,
I agree with your assessment that there is little need to use 10 explicitly, yet I think that having done 10 (i) gave me a clue that helped with 11. I also agree there is no need to construct the orthogonal circles. I reread pp. 130-133 a couple of hours ago (It’s 11:07 am here right now) and realized that Needham probably only meant to imply that inversion preserves magnitude and not direction of angles, but since it preserves magnitude it preserves orthogonality. The proof was not affected, but thanks for reinforcing that clarification. Previously, I was worried that he might have some argument based on the actual reversal of direction, and I couldn’t see how that could be involved, as all the inversions line up on L. Were I doing this problem over, I would simplify plotting by putting A and B on the real line.
Gary
Added exercise 11 chapter 3
ReplyDeleteGary
ReplyDeleteYes, I have to keep reminding myself that you are in Nevada and 8 hours behind me here in the UK. I remember once in London at Hyde Park corner, which is traditionally called Speakers' Corner because people stand up and speak freely about anything they want, there was a guy making a humorous speech taking the mickey out of Americans and he said:"...those Americans...they couldn't even organise the Boston tea party...they're 8 hours behind us and they're never going to catch up..."
Yes the diagrams in my published solution to exercise 11 were all drawn on the real line, and then at the end I just did a rotation on each diagram to make it more general.
Vasco
Vasco,
ReplyDeleteStill trying to catch up.
I'm actually in Nevada City, CA, but I was formerly in Nevada.
Re. Ch. 3, Ex. 11
It appears to me that our proofs are quite close, though I think yours is closer to the spirit of the question. I was more concerned with how to get from A to A’ without also worrying about how intuitive my proof was. And your diagrams are better. I could fiddle with a couple of parameters and rearrange my own. On the other hand, it is interesting to see that one gets the same final result even though there are (poorly motivated) reversals in my assignments of p-q and g-h.
Typo, I think: line 4 under Part(iv): “A’ in figure 4” should read “A’ in figure 5”, and you might want to add a label to K’ in figure 5.
The 1998 version contains the same error. Is it possible we are the first to work on this problem, or lat least the first to get as far as (iv)?
I am still a little puzzled by Needham’s result statement in Ex. 10: “Any two non-intersecting, non-concentric circles can be mapped to concentric circles by means of a suitable Möbius transformation.”
We have shown that the two inversions are a suitable Möbius transformation in this case. It seems we could say something more that would help a student, perhaps that a suitable Möbius transformation depends on finding a circle of inversion orthogonal to both A tilde and B tilde, which we do here by setting the center of the circle of inversion to the intersection with L of the inverted circle that is actually a line (B tilde in your Figure 3) and by setting the radius |q tilde - g| of the circle of inversion K to make K orthogonal to the other inverted circle (A tilde). I presume this can only work if the original circles are first inverted in the prescribed manner. Both the first and second inversions are based on circles of inversion centered at intersections of circles with L. The center of K is chosen from a point in B tilde, so it is an inverted point from B, but the opposite L-intersecting point from the original circle of inversion centered at p. Probably only some of this would have helped me when I first encountered the problem, but it seems to have a nice symmetry.
Gary
Re Gary June 3, 2015 at 12:49 AM
ReplyDeleteGary
Thanks a lot for reading my solution to exercise 11 chapter 3 and for the detailed feedback. I will post a reply to the various points you raise in the next few days and also publish an amended version of my solution.
Vasco
Re Gary June 3, 2015 at 12:49 AM
ReplyDeleteGary
Here is a link to a document in reply to your comments on exercise 11 chapter 3.
https://drive.google.com/file/d/0B4HzA5QgnnvZOUljak95SnJES3M/view?usp=sharing
Vasco
Re Vasco, June 4, 2015 at 3:42 PM
Delete“In your blog post your description of how to construct the concentric circles
mentions K as the circle of inversion, but I don’t think it is.”
Vasco,
Yes, you are quite right. I think my head was swimming with points and circles when I wrote that. Your remarks are very nice and clear. I think it is right on my web page.
Your remarks on the relation of exercises 10 and 11 are interesting, but it may take me a few days to absorb them as I will be busy for a time.
Thanks,
Gary
Re: Vasco, June 4, 2015 at 3:42 PM and my reply at 11:23
ReplyDeleteVasco,
Rereading your explanation of Ex. 11 and the remarks on the relations between 10 and 11 and find myself in agreement with all of it and would certainly recommend your new comments for inclusion. I have just one comment regarding “The composition of the two transformations is still a Möbius and means that we can put the centre of the concentric circles anywhere we wish” (p. 1). I agree with this, but I don’t think I am quite at the point of being able to put them wherever I wish, and I say this after having gone through Ex. 12 as well. In 11, for K to be orthogonal to A (or B) it must have a certain radius, which means that h’ will go where it pleases as determined by r_K and the location of the center of K, which was in turn determined by the location of intersections of L with A inverse or B inverse, which was determined by the initial locations of A and B as well as their radii. So I suppose one can solve an equation for the M transformation for h’, but it will involve a few new variables. So I wondered if you have an approach worked out.
An answer to Ex. 12 is posted at
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/022.touching.circles.ch3.ex12.html
Gary
I have posted a revision of Ch. 3, Ex. 11 with new diagrams and a more orthodox sequence of lettering. This answer was doubtless influenced by a previous reading of Vasco's answer.
ReplyDeletehttps://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/019.inversions.ch3.11.html
Added exercise 12 chapter 3.
ReplyDeleteGary
I have just looked quickly at your solution to this exercise and it looks different to mine. I think a good way of describing what touching means for circles is to say that they have a common tangent at the point where they touch.
I am going away on holiday tomorrow for a week. I will look at it again when I return.
Re Gary, June 11, 2015 at 6:40 AM
Thanks for the positive comments. When I say we can put the centre of the concentric circles anywhere we wish I was only really referring to exercise 10, where we can do a translation after doing the F transformation and centre the circles where we wish. Exercise 11 is more complicated from this point of view as you say. I just wanted to point out that although in exercise 10 by using F we always end up with the circles centred at the origin, a slight modification of F to include the translation means that the general result (with the concentric circles centred anywhere), quoted at the start of exercise 10, is possible. I hope that last sentence makes sense.
Vasco
Regarding Chapter 3, Exercise 12
ReplyDeleteVasco,
I follow your very direct explanation. While mine seems to wander, I think the crux of the explanation in both rests on the conformality of Möbius transformations with the preservation of angles. I didn’t change mine, but, before seeing yours, I added two more figures in which the enclosed circles in the concentric arrangement are rotated, leading to chained circles of different sizes and positions. This is doing it backwards, but the inverse Möbius transformation provides an inverse. It is at least a demonstration. I will be putting up new source code soon.
Gary
Re Gary June 16,2015 at 1:10AM
ReplyDeleteGary
I agree with you that the crux of both our explanations is the fact that a Möbius transformation maps circles to circles and is conformal. Your explanation using the inverse seems fine to me and is just another way of proving the result, and which also provides some extra insights.
I will leave mine as it is since Needham stipulates the use of the result of exercise 10.
I will also update my version of exercise 11 to include the extra explanations we discussed previously.
Vasco
Added a slightly amended version of exercise 11 chapter 3, which provides a bit more explanation.
ReplyDeleteAdded exercise 14 chapter 3.
ReplyDeleteVasco,
ReplyDeleteThank you for your comments of June 22, 3:35 PM. I am glad to see the additions, including 14 and I will read them with interest.
Right now, I am completely baffled by 13(ii). I can’t picture an arrangement in which all six of the S_i could simultaneously touch A, B, and C, so I think perhaps that S_i considered as a unit has every one of its constituents touching A, B *OR* C. Then, perhaps they could be arranged so that the centers of all the spheres lie in the same plane.
Another arrangement that occurs to me is that A and B are internal to C. The chain is then wrapped around the intersection of A and B so that all S_i touch A, B, and/or C.
Do you have an insight into this problem?
Gary
Vasco,
ReplyDeleteIt occurred to me to look up Soddy’s Hexlet in Wikipedia, so I see the intended arrangement.
Gary
Gary
ReplyDeleteI had the same problem with 13(ii). It's not easy to visualise. I found the Wikipedia page, took one look at it and decided to leave it for another day! I will read it again properly now that you have found it useful, and see if I can follow it. At first sight it looks as though the proof will be a similar approach to the proof for 12: reduce the problem to the much simpler situation in 13(i).
Vasco
Re Ch. 3, Ex 13 (ii).
ReplyDeleteVasco
I have posted an answer, which is, as you suggested, similar to the proof for 12. In fact, I think it can be simplified to 12 as a special case by giving A and B the same diameters and placing q on a plane passing through the centers of all the circles, including C, but not A and B. The bit that still escapes me is how much latitude we have in placing q in the configuration described in part (i) in order to produce the general configuration of part (ii).
Plot to follow. This won’t be an easy one. The Wikipedia image is a nice target.
Gary
Re Gary, June 30, 2015, 12:53 AM, missing URL
ReplyDeletehttps://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/023.Soddys.hexlet.ch3.ex13ii.html
Re Gary june 30,2015 at 12:53AM and 12:56 AM
ReplyDeleteGary
I will have a go at the exercise myself now. I will look at your solution afterwards..
Vasco
Vasco,
ReplyDeleteI have posted an answer to Ch. 3, Ex. 14. It was harder than it first appeared to be, so I consulted your answer. There are small variations of notation and approach. I was puzzled by the notation Ia, Ib, etc. as I wasn’t sure whether it referred to absolute values of line segments. I concluded that it does, but I wrote it a little differently. No improvement, I’m sure.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/024.perspective.drawing.ch3.ex14.html
Gary
Gary,
DeleteI can see why you found my notation confusing. I will modify my solution to make it clearer as soon as I get a moment.
I still haven't had a go at Ch. 3, Ex 13, as I have been working on Ch. 3, Ex. 15 which has given me a few headaches. I think this is another exercise which may not be quite correct as stated in the book. After reading it a few times and referring to (31) on page 155, I felt that z in the diagrams accompanying the exercise should not be shown as lying on the circle, but at a more general point inside or outside the circle.
I think I now have a reasonable solution to the exercise but I am not 100% confident. I would appreciate your comments on the exercise itself - do you think it makes sense as written in the book?
I am going away on holiday for a week on 16th July and will not be posting during that period, but I will check to see if you have any comments on the exercise.
Vasco
Re Vasco, July 14, 2015 3:13 PM
ReplyDeleteVasco,
I'll have a look. Thanks for the heads-up.
Gary
Re Vasco, July 14, 2015 3:13 PM
ReplyDeleteVasco,
I have made some progress on 15, but my plane geometry skills are insufficient to make the final deduction. See what you think:
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/024.nh.ch3.ex14-15.html
Note that the link for exercise 14 has changed to this link.
Gary
Re Vasco, July 14, 2015 3:13 PM
ReplyDeleteVasco,
I think 15 is correct as written. I think I have it now.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/024.nh.ch3.ex14-15.html
Gary
Gary,
DeleteThanks. I'll have a detailed look after I've published my version.
Vasco
Vasco,
ReplyDeleteI have posted an answer to Chapter 3, Ex. 17.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/025.nh.ch3.ex17.mobius.t.unit.circle.html
I’m not sure if my handling of the ratios of angles is legal. There is no hurry for answers as I realize you are vacationing.
Gary
Re Gary July 17, 2015 at 10:32AM
ReplyDeleteGary
I have now published my solution to exercise 3-17 and have now looked at your solution. Mine is very different to yours and uses the hints given by Needham. I think that your argument about the third side of a triangle being proportional to the angle between the other two sides is not correct. I thought about it for a while and then used a similar approach using the cosine rule which leads to the same result. There is no need to construct the two extra circles for this proof, just the lines between the given points, as in your diagram.
Here is a link to my proof using the cosine rule:
https://drive.google.com/file/d/0B4HzA5QgnnvZT29nOFFxS1VrRXc/view?usp=sharing
I would appreciate your comments on both proofs.
Vasco
PS I recently went to the Royal Greenwich Observatory and the National Maritime museum. I had not been there for a good few years and as well as seeing the meridian line again I enjoyed seeing the four clocks (H1-H4) invented by John Harrison for use on ships to enable them to calculate longitude accurately at sea. Amazing devices. You can read all about them on Wikipedia.
Re Vasco, July 28, 2015 at 10:43 AM
ReplyDeleteVasco,
Your proof looks very nice on first inspection and rightly take advantage of the hints. Regarding the second proof, I began with the cosine rule, but got bogged down in the algebra, so I tried to simplify. Since my proof leads to the same outcome, I can't help thinking that the ratios must be correct, but I don't command a theorem that would justify the approach. I am unfortunately bogged down in some responsibilities for a week, so I will take a close look at your proofs and think some more about my own and get back to you, hopefully with more clarity.
Meanwhile, I have discovered that Google Chrome does not recognize the MathML notation and does a poor job of displaying my web pages. I hope this hasn’t caused you or other readers too much inconvenience. I was using the Foxfire browser, which does recognize MathML. One can always inspect the explanation in the introduction to the source code that I usually provide. I am going to abandon MathML and fix as many of the problems as I can find time for.
The Harrison clocks appear to be a fascinating topic. I enjoyed seeing the HMS Victory at Portsmouth a few years ago. The National Maritime Museum must be engrossing.
Gary
Re Gary May 9, 2015 at 4:59PM and Unknown July 29,2015 at 5:52AM
DeleteThanks Gary. I have been using Firefox to look at your solutions since your comments in May this year about the Chrome browser's lack of support.
When I was looking at your solution to exercise 17 chapter 3, I first thought of Apollonius' theorem from my school days, and looked it up in Wikipedia, which mentions a generalisation of it called Stewart's theorem, which is also in Wikipedia and which applies directly to this situation. Looking at the proof for Stewart's theorem, I saw that it could be proved using the cosine rule and so I decided to use the cosine rule directly, since I think it is more well known than Stewart's theorem.
Vasco
Re Vasco, July 28, 2015 at 10:43 AM and July 29 at 7:12 AM
ReplyDeleteVasco,
I have gone over your proofs carefully and I agree with you on all counts. The argument from the two lines orthogonal to K’ is the most responsive to the question. I would only add that a’ should fall below the real line because the orientation of b, c, d is positive on the circle and a lies outside the circle containing b, c, d.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/025.nh.ch3.ex17.mobius.t.unit.circle.html
As you determined, my first attempt is unfounded. I like the proof using the cosine rule.
Gary
Gary
DeleteThanks for the feedback. I will amend the figure as you suggest.
Vasco
Vasco,
ReplyDeleteI have posted an answer to Ch 3, Ex. 19. The only problematic part was (iii), but it was very problematic. I would like to know your thoughts on it.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/027.nh.ch3.ex19.html
Gary
Gary
ReplyDeleteI will try the exercise myself and then have a look at your solution.
Vasco
Vasco,
ReplyDeleteI have posted Chapter 3, Exercise 20, parts (i-iii). I found your part (ii) very helpful, but I wonder in (ii) if we are testing whether the identity matrix itself is in the set or whether the result of the identity transformation is in the set. I ask because the identity matrix {{1 0}, {0, 1}} works with any 2x2 matrix. I used Mathematica to apply the criteria for group membership.
I have not altered my answer to part 1, which is similar to yours.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/
Gary
Re Gary August 13, 2015 at 6:38 AM
ReplyDeleteGary
All Möbius transformations, where (ad-bc) is not equal to 0, form a group (see 3 The Group Property on page 150-151) and the particular form of the transformation in exercise 20 forms a subgroup of this group of all Möbius transformations. So my answer to your question is that in my opinion we are testing whether the identity matrix itself is part of the group. Any subgroup of the 'big' group will contain this identity matrix. I hope that makes sense.
I am preparing to publish my answer to Chapter 3 , Exercise 16, so I have not had a detailed look at exercise 19 yet. It's next on the agenda,
When I was in London I went to Foyle's bookshop and found a new printing of the book. There were no corrections, but the diagrams were better quality I thought. I remember you said that some of the diagrams in your copy were hard to read because parts were indistinct. I particularly looked at Chapter 3 exercise 11, but it had not been changed. Needless to say I didn't buy it.
Vasco
Re: Vasco, August 13, 2015 at 6:38 AM
ReplyDeleteVasco,
Thank you. Yes, I see your point now. One not only has to show that there is an identity matrix, but also that the matrix belongs to the group. That seems valid. But isn’t it the case that
{{1 0}, {0, 1}}.{{a b}, {c d}} = {{a b}, {c d}} for all a, b, c, and d, so {{1 0}, {0, 1}}
must be the identity matrix for all Möbius transformations in general when written as matrices, or at least those for which the determinant is non-zero? Also, it seems that the language of p. 151, (i) is ambiguous, or at least I am not fully tuned in: “(i) the identity mapping E(z) = z belongs to the set;”. Does “mapping” refer to E = {{1 0}, {0 1}} or to the result of applying E?
To support your interpretation, once E is established as a member of the set, criterion (ii) would also apply to a composition of E and any matrix M. Both bases are covered.
As for the errors or deficiencies, at least it is nice to know that my printing is compatible with later printings! But it would also be nice to have a complete list of errors/problems. Perhaps we should compile a list. Something for a rainy day.
Gary
Small change to chapter 3 exercise 17 to make figure 1 and figure 2 consistent. Thanks to Gary's post on August 4, 2015 at 10:46 AM.
ReplyDeleteRe Gary August 13, 2015 at 9:02PM
ReplyDeleteGary
If we think of the non-singular Möbius transformations as a set, then this set forms a group under composition (as stated on page 151 of the book).
So I would say: yes, the unit 2x2 matrix is the identity transformation for the group of ALL non-singular Möbius transformations. If you are saying that you find this problematical then I don't think it is. The set of Möbius transformations defined in exercise 20 is a subset of the set of 'ALL' Möbius transformations, and one of the elements of the 'ALL' set is the identity matrix, which is also a member of the exercise 20 subset. This is also true of the subset defined in exercise 21. Every subset of the 'ALL' subset, which is also a group under composition, must contain the identity element. Generally speaking, any given Möbius transformation can belong to any number of subsets, which may or may not be groups.
I'm not sure if I've addressed your concerns or not.
As far as the language used on page 151 is concerned then I would tend to think of E as being a transformation and E(z)=z as being a mapping, but I think the two words are used almost interchangeably by most authors. Personally I tend to think of E as describing what is to be done by the transformation, and E(z) as representing the values, (often called w), corresponding to a set of points {z} after they have been transformed by E, if you follow me.
Vasco
Re Vasco August 14, 2015 at 5:28 PM
ReplyDeleteVasco,
I have been rereading the relevant sections and I reached an understanding fully in accord with both your paragraphs, but thank you for putting it succinctly. Your answer to the previous helped me to see that the identity matrix is also a member of the subset described in 21.
Gary
Re: Gary August 13, 2015 at 9:02 PM
ReplyDeleteGary
I agree with you that it would be nice to have a list of errors in the book. What do you think would be the best way to do it? There are not a lot of them. In the chapters I have looked at: 1, 2, 3, 4, 5, 8 and 9, there are at most 20 I would guess. We could both create a list of the errors etc that we find (and have already found) as we work through the book, and which we could later merge and publish on the site with the solutions to the exercises - maybe that would work? What do you think?
Rather than it being just a list of errors, in appropriate cases we could include a discussion, such as the one we had about exercise 11 chapter 3.
Vasco
Re Vasco, August 18, 2015 at 8:57
ReplyDeleteVasco,
I will gather what I have at my first opportunity and either include it in a posting here or on my site. I don't have a great many, but I have made marks in the book when I find them. I favor including the discussion where appropriate.
Gary
Vasco,
ReplyDeleteI have posted an answer to Chapter 3, Exercise 22. I think the answer is not entirely wrong, but I was unable to make use of the hint. I am also wondering if the language in sentence three is precise.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/
Gary
Errata for Needham, Tristan. Visual Complex Analysis. clarendon, Oxford. Paperback, 1998.
ReplyDeletep. 119, Ex. 33 (ii) I believe that you suggested a bit of rewriting of the problem replacing z with w at appropriate points.
p. 119, Ex. 34 replace the plus sign before 19z^6/5760 with a minus sign.
This is all I can find for now. Anything else must be in our postings and discussions on your blog.
Gary
Added exercise 16 chapter 3.
ReplyDeleteRe Gary August 22, 2015 at 12:10 AM
ReplyDeleteGary
I have done exercise 22 myself, using the hint. To me the language in sentence three (What is the image of this line under complex inversion, w--->(1/w)?) seems precise. It is saying: apply complex inversion to the result of the w^2 mapping and see what happens to the line. I tried to look at your solution but you had already removed it, so I'll publish my solution at some point and wait for you to re-publish.
Vasco
Re Gary August 22, 2015 at 12:44 AM
ReplyDeleteGary
Thanks for the errata. Your printing (1998 with corrections) is earlier than mine (2000 with corrections) and so presumably contains more errors. We'll obviously have to make this distinction between the two printings in our list.
Vasco
Re: Gary, August 22, 12:10 AM
ReplyDeleteI have reposted an answer that makes use of the hint. I agree the language is precise.
Gary
Nice diagrams in 3:16!
ReplyDeleteGary
Re: Ch. 3, p. 178, possible typo
ReplyDeleteVasco,
on p. 178, para. -1, line 2 I find “orthogonal to C will map D to itself, the two regions into which D is divided by J”
I think the D’s should be replaced with C’s.
“orthogonal to C will map C to itself, the two regions into which C is divided by J”
There is no circle D in [6] or [38] and it makes more sense in the context of [38]. Do you agree?
Gary
Question
ReplyDeleteVasco,
I have reached a state of confusion. Needham asserts that z -> 1/z is elliptic on p. 169 and in Ex. 23. An elliptic transformation is a pure rotation, which can be written Exp(i alpha) z, which appears not to change the absolute value of z. Circles rotate into themselves. But we can write z = r Exp[i theta] => 1/z = (1/r) Exp[-i theta]. This does change the absolute value, so that |z| != |1/z| (!= is "not equal"). Furthermore, Figure[1](a) indicates that 1/z is not simply a rotation about the origin. How can these two perspectives be reconciled? I have tried to superimpose the problem onto [29] without much success.
Gary
Gary
Re Gary August 26 2015 at 11:53 PM & August 27 2015 at 1:29 AM
ReplyDeleteGary
I'm away from home at the moment without access to the book. I may be able to reply to some of your comments before I get back, otherwise it will be this Sunday when I get back home.
Vasco
Re Gary August 26, 2015 at 11:53PM
ReplyDeleteGary
I agree with you.
Vasco
Re Gary August 27, 2015 at 1:29 AM
ReplyDeleteGary
I can see why you are confused by Needham's explanation's of the different types of Möbius transformations. I think I understand it now, after reading and re-reading sections VII, VIII, and IX several times. I will write a document by way of explanation and publish it as soon as I can. All this Möbius stuff is new to me and I am struggling to understand it too!
Vasco
Re: Vasco, Aug 31, 2015 at 3:32 PM
ReplyDeleteI look forward to the document, and I will also be doing a few re-readings. The coffee cups are piling up on my desk.
Gary
Re: Gary, September 1, 2015 at 3:18 AM
ReplyDeleteGary
You can download my document here:
https://drive.google.com/file/d/0B4HzA5QgnnvZNTVzSXRKVzBNcm8/view?usp=sharing
It doesn't go into a lot of mathematical detail, but is just an attempt to clarify some of the ideas in the text. I hope it will enable us to move forward and maybe find other things which need clarification!
I hope you find it useful.
Vasco
Re: Vasco, September 1, 2015 at 3:18 AM
ReplyDeleteVasco,
I look forward to reading the document. Thanks for posting.
I have posted answers to Chapter 3, Exercise 24 at the usual location:
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/
Gary
Re: Vasco, September 1, 2015 at 3:18 AM, document
ReplyDeleteVasco,
Your discussion helps a great deal. I had also noticed the sentence ”We call M(z) an elliptic Mobius transformation if M~ is elliptic, …”. This morning I made a calculation to determine whether F(1/z), from p. 163, could be called elliptic (getting the fixed points from p. 169). It leads to (z - 1)/(z + 1), which does not appear to be a pure rotation or even a similarity transformation. So my confusion lingers, mainly on the topic of 1/z.
Gary
Re Gary, September 1, 2015 at 3:18 AM
ReplyDeleteGary
I have added another page to my document to try and explain how Needham calls 1/z an elliptical transformation. I am still confused about it myself but I think I understand the mechanical procedure. Have a look and let me know what you think.
Vasco
Re Vasco, Sept 2, 2015 at 2:56 PM
ReplyDeleteVasco, I will take a close look at first opportunity. Gary
Re Vasco September 2, 2015 at 3:18 AM document
ReplyDeleteVasco,
Your interpretation makes good sense to me. The light has finally dawned. When I read p. 164 the first time (and a few times thereafter), I thought, "Why is Needham taking so much trouble to define M~?" I never put it together that we would need M~ to characterize M. Perhaps if Needham had introduced it with "Even though M may fall into one of the categories of elliptic, hyperbolic, etc., we can not always determine this by an inspection of M itself if the fixed points of M are not 0 and infinity. In this case, we will need to send the fixed points to 0 and infinity, as in the RHS of [29] and find a formula for the corresponding (?symmetrical) transformation M~ on the RHS." Then on to paragraph 2 on p. 164.
See if you think this is correct:
The RHS of [29] can be considered as a copy of the complex plane C. M~ on the complex plane (label it C~) induces a transformation on the Riemann sphere that enables one to visually determine whether M~ on C~ is elliptic, hyperbolic, etc. Then one would also know the character of M on C.
Furthermore, given M~ on C~ with fixed points at 0 and infinity, one should be able to generate interesting M_E+/- by treating the fixed points as variables in F^(-1)(M~(z)). I haven't tried this. Perhaps it will show up in a problem.
Gary
Re Gary August 7, 2015 at 10:06 PM
ReplyDeleteGary
I have now posted my solution to exercise 19 chapter 3, and as a result I have looked at your solution as requested in your post. It seems to me to be mainly OK. You got sidetracked by the equation for the circle I_M which can be re-written as |z+d/c|=1/|c| and so you can see immediately that it is centred at -(d/c) with radius 1/|c|.
You might find my solution interesting, but I see from your solution that you arrived at this conclusion in your own way.
Vasco
Re Gary September 3, 2015 at 5:24 PM
ReplyDeleteGary
"The RHS of [29] can be considered as a copy of the complex plane C. M~ on the complex plane (label it C~) induces a transformation on the Riemann sphere that enables one to visually determine whether M~ on C~ is elliptic, hyperbolic, etc. Then one would also know the character of M on C."
This seems right to me.
"Furthermore, given M~ on C~ with fixed points at 0 and infinity, one should be able to generate interesting M_E+/- by treating the fixed points as variables in F^(-1)(M~(z)). I haven't tried this. Perhaps it will show up in a problem."
Not quite sure what you mean by this. Could you explain in a bit more detail?
Vasco
Re: Gary August 26, 2015 at 11:53 PM, RETRACTION AND OTHER TOPICS
ReplyDeleteVasco,
I wrote
“on p. 178, para. -1, line 2 I find “orthogonal to C will map D to itself, the two regions into which D is divided by J”
I think the D’s should be replaced with C’s.
“orthogonal to C will map C to itself, the two regions into which C is divided by J”
There is no circle D in [6] or [38] and it makes more sense in the context of [38]. Do you agree?”
I have to retract my comment, something that many working scientists seem to be doing these days. I should have retraced my steps a couple of pages before commenting. On page 176, Needham wrote “Let D denote the unit disc (including C)”. I thought that he was trying to contrast C with J, but the contrast actually was C and D.
Re: Vasco, September 5, 2014 at 4:01 PM
Thank you for looking at 3:19. Your solutions are generally more direct than mine, so I certainly will study yours.
Re: Vasco, September 5, 2014 at 8:05 PM
In the comment which you quoted
"Furthermore, given M~ on C~ with fixed points at 0 and infinity, one should be able to generate interesting M_E+/- by treating the fixed points as variables in F^(-1)(M~(z)). I haven't tried this. Perhaps it will show up in a problem."
I probably should have written F^(-1)(M~(z~)), using z~ instead of z for more clarity. I will try to work up an interesting example. But the idea is just that if one is given only the right hand side of [29], one can send 0 and infinity to different fixed points to get different families of orthogonal circles on the LHS.
Gary
Just corrected a couple of typos in exercise 19 chapter 3.
ReplyDeleteAdded exercise 22 chapter 3.
ReplyDeleteRe Gary September 6, 2015 at 12:05 AM
ReplyDeleteGary
You wrote:
"I probably should have written F^(-1)(M~(z~)), using z~ instead of z for more clarity. I will try to work up an interesting example. But the idea is just that if one is given only the right hand side of [29], one can send 0 and infinity to different fixed points to get different families of orthogonal circles on the LHS."
OK, I think I get it now. It would be interesting to see an example.
Vasco
Re: Vasco, September 5, 2015 at 4:01 PM
ReplyDeleteVasco,
Your solution to 19:3 is much nicer. I was flailing around, because I need more work with absolute value calculations.
I also liked the before and after arrangement of diagrams in section (ii).
Gary
Gary
DeleteGlad you found my solution to 19:3 helpful. Have you tried Exercise 3 chapter 2, which is referred to on page 178 in chapter 3 in the last sentence of subsection 2.? It is a good exercise in using complex conjugates and absolute values and is concerned with the group of Möbius transformations which are automorphisms of the unit disc.
Vasco
Added my complete solution (all three parts) to exercise 20 chapter 3. Previously the version on my website only contained parts (i) and (ii).
ReplyDeleteRe: Vasco September 10, 2015 at 10:29 AM
ReplyDeleteVasco, I will work through 2:3 again.
Gary
Re: Unknown September 10, 2015 at 6:50 PM
ReplyDeleteVasco,
2:3 was instructive. I don't know why I didn't keep a record of it. But it was the trick of pulling a factor out of the absolute brackets that I had evidently forgotten in regards to 3:19. I must have encountered it at some time.
In going over your solution to 2:3, Part (iii) I found it hard to see at first why the third line from the bottom should be true: 1 - (|z-a|^2) / (|Conj(a)z - 1|^2) > 0. At that point, I wrote
(1-|a|^2)(1-|z|^2) >= 0
and therefore
|Conj(a)z- 1|^2 - |z-a|^2 >= 0
|Conj(a)z- 1|^2 >= |z-a|^2
and
|z-a|^2 / |Conj(a)z- 1|^2 <= 1 (I think you need this to write the statement that I had difficulty understanding.)
Hence
|M(z)|^2 = |(z-a)/(Conj(a)z- 1)|^2 = |(z-a)|/|(Conj(a)z- 1)|^2 <= 1
|M(z)| <= 1
I made a correction to 3:24 and posted an answer to 3:25(i).
Gary
Re Gary September 11, 2015 at 8:02AM
ReplyDeleteGary
You wrote:
"In going over your solution to 2:3, Part (iii) I found it hard to see at first why the third line from the bottom should be true: 1 - (|z-a|^2) / (|Conj(a)z - 1|^2) > 0."
What I did there was to notice that if I divided both sides by (Conj(a)z- 1)|^2, a positive real number, then this was like saying
if c^2-d^2>=0 and c>0 then I can divide both sides by c^2 and obtain:
1-(d^2/c^2)>=0
I think our methods are equivalent.
Vasco
Re Vasco, September 11, 2015 at 9:38 AM
ReplyDeleteVasco,
Quite right. I'm embarrassed that I didn't see it.
Gary
Re Vasco, September 8, 2015 at 7:05 AM
ReplyDelete" It would be interesting to see an example."
Vasco,
A description of the experiment with F^-1 on Figure [29] is available at the following link:
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
Gary
Gary
DeleteWill take a look as soon as I can.
Vasco
Re Vasco September 14, 2015 at 8:57 AM
ReplyDeleteVasco,
I made a few changes.
Gary
Re Gary September 17, 2015 at 11:24PM
ReplyDeleteGary,
Just back from another week away from home! I will now be able to have a good look at your experiment with F^-1. On a first quick reading I noticed that you refer to figure [29] on page 165 of the book. Is it really on page 165 of your edition? It's on page 163 of mine.
Will post again when I've had a good read.
Vasco
Re Vasco September 17, 2015 at 11:24 PM
ReplyDeleteVasco,
I have also just returned from a trip.
The correct page for [29] in my copy is also 163.
Gary
Re Gary September 12th 2015 at 10:04PM
ReplyDeleteGary
I have looked at the link:
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
and what you have written seems fine to me. Since F(z)=(z-\xi_+)/(z-\xi_-) then |F(z)|=|(z-\xi_+)|/|(z-\xi_-)|. As z moves along the imaginary axis, |(z-\xi_+)| = |(z-\xi_-)| and so |F(z)|=1, and so w lies on the unit circle. If z lies to the left of the imaginary axis, |(z-\xi_+)| < |(z-\xi_-)| and so |F(z)|<1 and so w lies inside the unit circle and vice versa if z lies to the right of the imaginary axis.
I THINK that the reason your red arrows go in opposite directions is because you have chosen a circle OUTSIDE the unit circle. If you were to choose a circle inside the unit circle your arrows would go in the same direction, I think.
When you say you tried using (27) on page 152 (without success), to calculate the fixed points, which Möbius transformation were you trying to find the fixed points of?
Vasco
Re Vasco September 22, 2015
ReplyDeleteVasco,
Re: “I THINK that the reason your red arrows go in opposite directions is because you have chosen a circle OUTSIDE the unit circle. If you were to choose a circle inside the unit circle your arrows would go in the same direction, I think.”
Yes, I agree. That’s what I was trying to say somewhat vaguely I admit.
Thank you for the little tutorial on seeing the side of the axis in the absolute brackets.
I posted a new attempt at calculating the fixed points, this one also a dud.
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
By the way, the unit circle in Fig 1a is the second from the center. It didn’t show the dashing well.
Gary
Re Gary September 23, 2015 at 1:05AM
ReplyDeleteGary
Re your discussion document on fixed points at the following link:
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
I have produced a short document which addresses some of the issues raised by your discussion. Here is a link to it:
https://drive.google.com/file/d/0B4HzA5QgnnvZMXhNUkdmVUZFQmM/view?usp=sharing
Vasco
Re Vasco September 23, 2015 at 7:58 PM
ReplyDeleteVasco,
What you say makes sense. In fact, my first attempt was using the matrix that you suggest and I think I normalized it, so I will have see if I can reconstruct that attempt. I found your latest message just as I was going to post the following. It will be interesting now to redo the calculations as you suggest and see what results.
Gary
Re Vasco, September 22, 2015 at 4:52 PM and Gary September 23, 2015 at 1:05 AM
Vasco,
In trying to determine the fixed points for M(z) in the experiment reversing [29], I found that Needham says on p. 152 that the fixed points of a rotation e^{i alpha} z on the complex plane are 0 and infinity. This would apply to the W plane in my Figure 1a. To take this to the Z plane, apply F^-1 to w = 0 and w = infinity.
F^-1(0) = (-Xi_(-) 0 + Xi_(+))/(-0 + 1) = Xi_(+)
F^-1(infinity) = (Xi_(-) (-infinity) + Xi_(+))/(-infinity + 1) => Xi_(-)
It appears that there are no constraints on the fixed points of M(z) on the Z plane, so perhaps the choice of Xi_(+ -) = (- +) 1 is valid.
Gary
Re Gary September 23, 2015 at 9:07 PM
DeleteGary
I agree that in the LHS of [29] Xi_(+ -) can be any two points in the plane.
Vasco
While I was working on exercise 21 of chapter 3 on page 187 of the book, I realised that my published solution to exercise 20 was incomplete. I had forgotten to show that the resulting composition matrix and the inverse both have |p|>|q|. I have now published a corrected version to account for this. Apologies for any confusion caused.
ReplyDeleteRe Vasco September 24, 2015 at 8:31 AM
ReplyDeleteVasco,
Thank you. After one more reading of VIII, 1 The Main Idea (pp. 162-4), I, too, can see why one can put the fixed points anyplace. In Figure [29], these are the fixed points of M. I thought that the choice of M must determine the fixed points, so I would have to calculate them. But I see that what I did in reversing the process was to begin with an (elliptic) M~ (RHS) with fixed points 0 and infinity. Applying F^-1 to the RHS of [29] can send 0 and infinity to whatever fixed points one chooses for the LHS.
I could not see at first why M on the LHS of [29] would “know” its new fixed points. M on the LHS is F^-1(M~), which is a composition of two Möbius transformations. Hence, it is conformal and M must also be elliptical with fixed points at the sites where the C1 circles intersect (because each one passes through both fixed points). Those points are predetermined in F^-1 and by the construction of M~.
One could presumably do something similar with a hyperbolic M~. Applied to a ray of the RHS, it should take it to an M on a C1 circle on the LHS. And it must be possible to produce a version of [32] by the same method. Another day.
Gary
Added exercise 21 chapter 3
ReplyDeleteVasco,
ReplyDeleteI have tried a couple more experiments with the inverse of the function that takes the fixed points to 0 and infinity. I am wondering what equation on the complex plane would produce the balanced loxodromic curve shown in Needham’s Figure [32]. One might investigate derivatives of spirals. By chapter 4, this should be possible.
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
Gary
Re Gary September 28, 2015 at 1:59 AM
DeleteGary
I will study your document. When I had a quick look at it I noticed on page 2 that when calculating the fixed points you used xi+- on both sides of the equations and then assumed they were the same. I think it would be better to use another symbol on the LHS to avoid problems when simplifying. Also I don't think you can assume that [M] is normalised just because [M~] is normalised. I think you would have to normalise F{-1} as well.
Vasco
Added exercise 23 chapter 3
ReplyDeleteRe Vasco September 28, 2015 at 4:28 PM
ReplyDeleteVasco,
Thanks for the ideas.
Gary
Re Vasco September 28 at 4:28 PM
ReplyDeleteVasco,
I did the long calculation of the fixed points of figure [29] inversed once again (same link) with F (and F^-1) normalized. It made the calculations more interesting, but I think they are still contradictory. However, they are so involved that I don't have a high degree of confidence that the results are correct. Don't bother with them unless the problem also interests you.
Gary
Re Gary 30 September, 2015at 4:10 AM
ReplyDeleteGary
I've had a look and it occurred to me that if you are trying to calculate the fixed points of M maybe you should be calculating [M]=[F]^{-1 } [M~] [F] rather than [F]^{-1 } [M~].
Vasco
Re Gary 30 September, 2015at 4:10 AM
ReplyDeleteGary
I calculated the matrix [M] according to my post October 1, 2015 at 8:46 AM and if I substitute the fixed points xi{+-}=-1,+1 and \theta=\pi, then I get the matrix for the Möbius transformation (1/z) as would be expected and if I use (27) on page 152 to calculate the fixed points using the values from the calculated matrix [M] I get +1 and -1 as would be expected. So I suppose that this calculated [M] represents the general elliptic matrix.
Also in this case it is not necessary to normalise F or F^-1. If M~ is normalised then so is M. See the end of subsection 5 Computing the multiplier on page 170.
Vasco
Re Vasco October 1, 2015 at 10:16 PM
ReplyDeleteGary
Please ignore the last paragraph in the post referred to above, beginning "Also in this case...". For a proper explanation read the following short document:
https://drive.google.com/file/d/0B4HzA5QgnnvZbXNMN0hoMTQ4X0U/view?usp=sharing
I hope that after reading my document you will be able to see that in your document you are using matrices, BUT instead of using the inverse matrix of [F] you are using the matrix which corresponds to the normalised inverse transformation of the transformation F, which is not the same thing. The rest of the content of my last two posts is OK I think.
So to do your calculation you can use the matrices [F^{-1}], [M~], and [F], using the 'proper' inverse matrix of [F]. and I think you will find that it works out OK.
Vasco
Re Vasco October 2, 2015 at 3:43, document
ReplyDeleteVasco,
The document is very helpful in pointing out the difference between [M^-1] and [M]^-1. I had just taken the inverse from (25), p. 180 and not bothered to check it further, so I was not using the 1/det multiplier on the inverse of the matrix. This caused me trouble in more than one problem. I'll try it all again when I get the chance.
Gary
Re Gary October 3, 2015 at 4:24 AM
ReplyDeleteGary
I have added two more pages to my document so that it now includes the calculation for M, according to me. You will probably prefer to have a go yourself before looking at my attempt.
Vasco
Hello to all visitors
ReplyDeleteI was trying to correct a problem with looking at the member profiles (top left of the blog) and in doing so I accidentally changed the design of the blog and couldn't find how to go back. I hope the new layout is not too disconcerting.
Vasco
PS I didn't manage to correct the problem with member profiles.
Re Vasco, October 34, 2015 at 3:06 PM
ReplyDeleteVasco,
I like the new look. It seems a bit more compressed, probably because of a smaller type size, but the net effect is less scrolling.
Gary
Re Vasco, Oct 3, 2015 at 8:47 AM
ReplyDeleteVasco,
I reworked the experiment according to your suggestions. It worked out nicely, so I'm very glad you pointed out the differences in the inverses between the divisor notation and the matrix notation. Since I wanted to see the answer to the problem I posed for myself originally, I worked out (27) p. 152, which came out to Xi(+/-) as it should, showing again that the fixed points can be placed anywhere.
I noticed also that Figure 2 had disappeared along with its discussion, so that has been restored here.
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.fig29inv.pdf
Gary
Vasco,
ReplyDeleteI have posted answers to Ch 3, Ex 25 on my web site. Some small bits of them may even be correct. I am eager to see what you do with this problem.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/FIGURES/
Gary
Re Gary October 5, 2015 at 8:50 PM
ReplyDeleteGary
I am working on Ch 3 Ex 24 at the moment. I'll move on to Ex 25 as soon as I can - looks interesting.
Vasco
Added exercise 24 chapter 3
ReplyDeleteVasco,
ReplyDeleteI have started chapter 4. In exercise 3, I could not see (without resorting to Cauchy-Riemann) from my diagram why Conj(z)^2/z was not an amplitwist with amplification of unity, so I consulted your answer and found this:
“The image of the arrow at right angles to z is also at right angles to Ω(z) and is rotated in a clockwise direction through an angle of 4 [Theta] + Pi.”
What is the reason for adding Pi?
I have put my answers to Exercises 1-3 at this link:
https://dl.dropboxusercontent.com/u/35393965/nh.ch4.ex1-3.pdf
I think the approach to Ch 4, Ex. 2 is a little bit more direct, for once. See what you think.
Gary
Re Gary October 12, 2015 at 12:39 AM
ReplyDeleteGary
You need to think about what happens to the point at the end of the arrow (the one with the solid black head in my diagram and the blue one in yours):
since the angle for this point is slightly larger than \theta then it is rotated clockwise through an angle slightly larger than 4 * \theta by the function and so the head of the arrow is on the OTHER side of your dotted line.
It's very important to grasp this point. I would recommend reading the explanation for z---->.conj(z) in section V on page 200.
I will post again when I've read your solution to Ch. 4 Ex. 2.
Vasco
Re Gary October 12, 2015 at 12:39 AM
ReplyDeleteGary
Yes I like your concise and direct solution to Ch. 4 Ex. 2. The only thing I would say is that at this stage (Chapter 4) I would avoid using the CR equations (unless specifically asked to in the exercise), and use the idea of the amplitwist as a way of defining analytic functions, as described on page 197. This is a much more graphic/geometric way of looking at the complex derivative, and mechanical use of the CR equations can lead to problems - see exercise 12.
Only my opinion of course.
Vasco
Re Vasco, October 12, 2015 at 12:39 AM
ReplyDeleteVasco,
Yes, that last paragraph of section V, p. 200 is the key. I realized sometime in the very early morning that if I did the conjugation first and then two additional rotations of theta, I would get a more sensible result. I got off on the wrong foot with my reading of ex. 4:3 “By writing z in polar form, find out the geometric effect of Omega.” I wrote the whole function in polar form and did the division, apparently getting a different function. (Conj(z)^2)/z is apparently a two-step function: conjugate first and then divide.
I am still puzzled about how to deal with conj(z)^2 as it affects the infinitesimal vectors. Does one apply an expansion of 2 abs(v) or does one square abs(v), where v is one of the infinitesimal vectors? I applied 2, treating it as a pseudo derivative.
The figure is new. It’s at the same address.
I'll try to focus on the amplitwist as opposed to the CR.
Gary
Re Gary October 12, 2015 at 7:24 PM
ReplyDeleteGary
I think it is better to substitute z=re^{i\theta} into the expression for Omega and find the geometric effect of Omega as suggested in the exercise and then deduce what will happen to the small arrows. When I did this I found that the expression for omega was a rotation of z.
Vasco
Re Vasco Octobe4 13, 2015 at 11:32 PM
ReplyDeleteVasco,
Yes, I can see that is what Needham requested. I tried that first and found Omega = r e^{-3 Theta}, which shows that Omega is rotated by -4 Theta and the amplitude remains the same: |z|. Then one could deduce from the Conj(z) in the numerator that the orientation of the infinitesimal arrows was reversed by a reflection on the real line. I didn’t have the confidence to make that deduction in the context of the square and the division, but I can see now that I should have.
Also, one could deduce that this was no amplitwist from the last sentence of Section V on p. 200, “Thus different arrows must be rotated by different amounts (which is not a twist) and hence there is no amplitwist.” But the latter approach seemed to me to ignore the instruction that “Your picture should show this in two ways.”
I still don’t feel I have answered this question fully, because any figure that I could draw with my current understanding would only show the lack of amplitwist in one way—the reversal of orientation. Perhaps the answer requires beginning with two different z’s and plotting Omegas for both of them. That would show both reversal of orientation and two different twists, thereby satisfying the instruction to show it in two ways.
Gary
Vasco,
ReplyDeleteWhat program did you use to make the nice diagrams in ch 4, ex 4?
Gary
Re Gary October 14, 2015 at 3:22
ReplyDeleteVasco,
I have put a revision of my answer and figure for Chapter 4, Ex. 3 at this link:
https://dl.dropboxusercontent.com/u/35393965/nh.ch4.ex1-3rv.pdf
Gary
Re Gary October 14, 2015 at 3:25 AM
ReplyDeleteGary
I use TexNicCenter to create my documents using Latex and I have incorporated several so-called packages, The one I use for the graphics is PStricks, which creates postscript files which are incorporated automatically into the final .pdf file. It is all free and very powerful - I have barely touched the surface in my use of it. I took the approach of learning it as I went along. There is a pretty comprehensive .pdf manual available.
Vasco
Re Gary October 14, 2015 at 11:32 PM
ReplyDeleteGary
I am glad that you are looking at chapter 4 because I did these exercises back in 2012 and I am interested in your approach, which is making me think about them again.
In my working of exercise 3 chapter 4 I use the fact that the two arrows in my diagram are twisted different amounts as one way of showing that Omega is not an amplitwist. The fact that the angle between the arrows has the same magnitude before and after the mapping Omega, but with the opposite sense shows that Omega is not conformal and so not an amplitwist. So I have shown that Omega is not an amplitwist in two different ways.
Vasco
Re Vasco October 13, 2015 at 6:33 AM
ReplyDeleteVasco,
I think I have finally grasped your point, if it is that the orthogonal arrow twists more than the parallel arrow. It’s a twist, but not an amplitwist. OK.
Information on NexNicCenter noted. Thank you.
Gary
Re Gary October 14, 2015 at 7:42 AM and Vasco October 14, 2015 at 6:33AM
ReplyDeleteYes, that's my point exactly and also that because Omega makes the right-angle between the orthogonal arrow and the parallel arrow change sense, it is not conformal and so not an amplitwist.
I think we have to be careful with the language here, and use twist and amplification and also amplitwist in the correct way. I plead guilty to misuse of the teminology in my post referred to above.
We can only say that a function or mapping (Omega in this case) is an AMPLIFICATION if the effect of the mapping is to amplify ALL vectors (arrows) at z by the SAME amount, and we can only say that it is a TWIST if it rotates ALL vectors at z by the same amount in the same sense. Finally we can only say that the mapping is an AMPLITWIST at z if it possesses both an amplification and a twist. We must be careful not to use the word twist when we mean rotation as I did in my post referred to above. Guilty as charged!
Vasco
Re Vasco, October 15, 2015 at 6:33 AM.
ReplyDeleteVasco,
Your summary is confirmed by the last paragraph on p. 193, and had I focused just on this I probably would have answered question 3 as you did, but I became distracted. I can see now it is consistent with Needham’s use of the terms in sections VI 1 and 2. That paragraph and yours should be set in bold typeface. A pardon is in order. With the various analytic and non-analytic infinitesimal rotations and amplifications discussed in 199-202 it is easy to get led down a garden path.
Gary
Vasco,
ReplyDeleteYour answer to 4:5 is more explicit than mine and therefore substantially better, but I thought you might find something of interest in mine.
https://dl.dropboxusercontent.com/u/35393965/nh.ch4.ex5.pdf
Gary
Re Gary October 5, 2015 at 8:50 PM
ReplyDeleteGary
I have published my solution to exercise 25 chapter 3. I found part (ii) needed a lot of thought, and I wasn't sure what Needham wanted in part (iii). However I am pleased with my solution to part (ii), and I hope that part (iii) is close to what Needham is looking for.
I see that you used the Riemann mapping theorem for part (iii). I will study your answer over the next few days.
Re Gary October 15, 2015 at 8:03PM
Interesting counter example to show that f is not conformal.
Vasco
Re Vasco, October 16, 2015 at 11:45 AM
Delete*****************
(PLEASE NOTE: This post is actually Gary's post which I edited at his request - Vasco
*****************
Part (i) opened my eyes. I hadn’t realized that one could choose the q, r, s on both sides of (29), but I see on page 154, lines 1-2, “if we can find a Möbius transformation M that maps three given points q, r, s to three other given points q~, r~, s~, then M is unique”. I place the emphasis on “given”. The solution to (i) looks valid to me. My only question concerns notation. Is the M(z) in the last sentence of your answer the same as the M(z) in the first sentence of the question, or is it a sub-category that might be given it’s own label, such as M_r(z)?
Even if my misgivings regarding (ii) turn out to be unwarranted, I always learn from your approach. Degrees of freedom baffle me, so I found your answer in part (iv) very useful.
In general, I found the problems difficult. I have delayed posting 26 because I haven’t solved (v) and I don’t understand 27 at all. Needham says that (53) is taken up again in Ch. 6, so perhaps that will clarify it, eventually.
Gary
Vasco,
ReplyDeleteRe my previous post, I should have included a link. Here it is:
https://dl.dropboxusercontent.com/u/35393965/nh.ch3.ex25.pdf
The original link also works.
Gary
Re Gary October 17, 2015 at 7:16 AM
ReplyDeleteVasco,
I see your point regarding part (ii), a little late again, so pardon me for being slow and using so much blog space. I have revised my answer based on your approach, but as I reread your answer, I see that it is tighter than I realized. I would delete paragraphs 2 and 3 of my post if I could, but the delete function apparently times out.
Gary
Re Gary October 17, 2015 at 6:41 PM
ReplyDeleteGary
I was working on a short document to explain in more detail my solution to exercise 25 chapter 3, when I saw your post. I will still publish it, but as a general document for anyone who is interested.
Don't worry about the blog space - that's what it is for. Would you like me to delete the second and third paragraphs from your post of October 17, 2015 7:16 AM? As the owner of the blog I think I can do it.
Vasco
Re Vasco, October 18, 2015 at 8:27 AM
ReplyDeleteVasco,
I think those two paragraphs just add confusion, so yes, delete them at your convenience.
Gary
Re Gary October, 2015 at 5:00 AM
ReplyDeleteGary
I could not edit your post directly, so I copied it and posted an edited version as a reply to my post. Unfortunately the post then looks like mine and so I have added a note to say it is really your post. I hope this is OK for you.
Vasco
Here is link to a two-page document which attempts to provide more explanation to accompany my solution of exercise 25 chapter 3:
ReplyDeletehttps://drive.google.com/file/d/0B4HzA5QgnnvZTXdySlBoMXdvNTQ/view?usp=sharing
Added exercise 26 chapter 3.
ReplyDeleteRe Vasco, Oct 19, 2015, 7:13 AM, 3:59 PM and new posting of 4:6
ReplyDeleteVasco
The editing is fine, thanks. The document re 25:3 makes it all quite explicit.
I have posted my answers to 4:6. Regarding problem (ii), the first paragraph deals with a difficulty I have had in determining the directions of infinitesimal vectors in the most trivial cases. It may strike you as obvious, but I find it helps.
https://dl.dropboxusercontent.com/u/35393965/nh.ch4.ex6.pdf
Gary
Vasco,
ReplyDeleteI have posted my answers to Ch 4, Exercise 8. There are a couple of points of interest. I find a paradox in the answer to (ii). My answer to (iii) is a bit different.
https://dl.dropboxusercontent.com/u/35393965/nh.ch4.ex8.pdf
Gary
Added exercise 27 chapter 3
ReplyDeleteRe Gary October 20, 2015 11:28 PM and October 22,2015 at 12:03 AM
ReplyDeleteGary
I am reading through your solutions to chapter 4 and I will post my comments as soon as possible. I'm finding it interesting to see how someone else tackles these exercises.
Vasco
Gary
ReplyDeleteI have looked at your solutions to chapter 4 exercises 6 to 7 part(i) and here is a link to a document with my comments:
https://drive.google.com/file/d/0B4HzA5QgnnvZbHNHQVJQVWVla1U/view?usp=sharing
I will add to the document as I read the other solutions to 7 parts (ii) and (iii) and 8.
Vasco
Re Vasco October 23, 2015 at 7:19 PM
ReplyDeleteGary
I have updated the document to cover all parts of exercises 7 and 8.
https://drive.google.com/file/d/0B4HzA5QgnnvZbHNHQVJQVWVla1U/view?usp=sharing
Vasco
Added exercise 18 chapter 4.
ReplyDeleteAdded exercise 10 chapter 5.
ReplyDelete