Gary
GaryVasco
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Post by Gary on Mar 25, 2017 17:56:13 GMT
Gary I think that with complex-valued equations that can have multiple solutions and with multifunctions, you need to make sure you don't miss any solutions. Even though it is clear that a complex equation like $\cos z=1$ has some real solutions, you need to make sure you pick up any complex solutions, as in part (iii) where a real number to the power $(1/4)$ has two imaginary roots, as well as 2 real roots. Vasco Vasco, I see now. Since I started with the root notation rather than the exponential, I have now corrected and added the complex roots with the root notation. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 25, 2017 19:05:51 GMT
Vasco,
On p. 5 of your answer to (iii) did you mean to say that f has period $\pi/2$?
Gary
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Post by Admin on Mar 25, 2017 19:28:21 GMT
Gary
Yes, do you disagree?
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 25, 2017 19:37:34 GMT
Vasco,
I followed your lead here, but arrived at a different answer. I have just reposted. See what you think.
But forget the Pi/4. That was just a moment of confusion between Pi and 2Pi, which leads to confusion between Pi/2 and Pi/4. It's like a graphical typo.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 26, 2017 22:03:47 GMT
Vasco, I got a big surprise today regarding (iii). With figures (c)-(i) of my answer printed at full page size, I traced the curve all the way from a $w_0$ back to itself. The initial point was approximately -.3. The trace looped around p and crossed the real line 10 times on the left of $w_0$, and then it returned to home for a total of 11 windings. I won't claim total confidence in this number, but along the way I numbered all the endpoints and checked them against the axis scale and I used a different colored number to mark the crossings of the x-axis. As you must know, it's a tedious exercise, so replications are welcome. In case anyone would like to try this, I have put the images in a .pdf file which is about 590 KB in size. It can be obtained from Dropbox. Unfortunately, Dropbox has changed their access policy, so you have to sign up with them in order to download the files, but it is free. I could also post the full file here on request. Images for (iii)I have also reposted the revisions at the original location in the first message in this thread. Gary |
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Post by Admin on Mar 27, 2017 7:23:52 GMT
Gary Thanks for that - very useful. I came to the same conclusion about the winding number in part (iii) by a different route which is detailed in my revised solution. Between us we seem to have homed in on the correct solution. Vasco PS Managed to download without signing up to Dropbox. |
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Gary
GaryVasco
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Post by Gary on Mar 27, 2017 17:30:53 GMT
Gary Thanks for that - very useful. I came to the same conclusion about the winding number in part (iii) by a different route which is detailed in my revised solution. Between us we seem to have homed in on the correct solution. Vasco PS Managed to download without signing up to Dropbox. Vasco, Moving the winding points out and gradually reintroducing them is a neat way to see their effect. I think we are homing in, but I'm not quite there yet. I also arrived at 44 winds this morning, but I couldn't believe my eyes as it contradicted our calculations of 24 winds. I just now posted a revision and then found your message. Here is part of what I added to my answer: Working with the computer generated plot itself, I find that 11 winds can be produced as z traverses an angle of only $\frac{\pi}{2}$. This suggests that the periodicity of $sin(z^4)$ is actually $\frac{\pi}{2}$ as you proposed, which is twice the periodicity of the cos(z) (See ii). But this would mean that the observed number of 11 windings is repeated four times as z goes round D to make a total of 44, which contradicts the 24 p-points. The periodicity of $\frac{\pi}{2}$ should indicate that the number of visible windings of $sin(z^4)$ is 24/4 = 6, which is what you observed or calculated. I have rechecked the visible windings and I still arrive at 11. I don't know how to resolve this. I will take a careful look at your answer and graphs later today. Gary |
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Post by Admin on Mar 27, 2017 17:35:15 GMT
Gary Thanks for that - very useful. I came to the same conclusion about the winding number in part (iii) by a different route which is detailed in my revised solution. Between us we seem to have homed in on the correct solution. Vasco PS Managed to download without signing up to Dropbox. Vasco, Moving the winding points out and gradually reintroducing them is a neat way to see their effect. I think we are homing in, but I'm not quite there yet. I also arrived at 44 winds this morning, but I couldn't believe my eyes as it contradicted our calculations of 24 winds. I just now posted a revision and then found your message. Here is part of what I added to my answer: Working with the computer generated plot itself, I find that 11 winds can be produced as z traverses an angle of only $\pi/2$. This suggests that the periodicity of $sin(z^4)$ is actually $\pi/2$ as you proposed, which is twice the periodicity of the cos(z) (See ii). But this would mean that the observed number of 11 windings is repeated four times as z goes round D to make a total of 44, which contradicts the 24 p-points. The periodicity of $\pi/2$ should indicate that the number of visible windings of $sin(z^4)$ is 24/4 = 6, which is what you observed or calculated. I have rechecked the visible windings and I still arrive at 11. I don't know how to resolve this. I will take a careful look at your answer and graphs later today. Gary Gary If you look at my solution you will see that I now have 44 $p$-points. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 27, 2017 18:15:37 GMT
Vasco, Moving the winding points out and gradually reintroducing them is a neat way to see their effect. I think we are homing in, but I'm not quite there yet. I also arrived at 44 winds this morning, but I couldn't believe my eyes as it contradicted our calculations of 24 winds. I just now posted a revision and then found your message. Here is part of what I added to my answer: Working with the computer generated plot itself, I find that 11 winds can be produced as z traverses an angle of only $\pi/2$. This suggests that the periodicity of $sin(z^4)$ is actually $\pi/2$ as you proposed, which is twice the periodicity of the cos(z) (See ii). But this would mean that the observed number of 11 windings is repeated four times as z goes round D to make a total of 44, which contradicts the 24 p-points. The periodicity of $\pi/2$ should indicate that the number of visible windings of $sin(z^4)$ is 24/4 = 6, which is what you observed or calculated. I have rechecked the visible windings and I still arrive at 11. I don't know how to resolve this. I will take a careful look at your answer and graphs later today. Gary Gary If you look at my solution you will see that I now have 44 $p$-points. Vasco Vasco, Yes, I saw that. It was a nice surprise. Correction! I see you said 44 $p$-points. I read that as windings. I still need to investigate how you arrived at 44. Above, I was just indicating what was in my recent posting. Gary |
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Post by Admin on Mar 28, 2017 3:34:29 GMT
Gary If you look at my solution you will see that I now have 44 $p$-points. Vasco Vasco, Yes, I saw that. It was a nice surprise. Correction! I see you said 44 $p$-points. I read that as windings. I still need to investigate how you arrived at 44. Above, I was just indicating what was in my recent posting. Gary Gary You are right it should be 44 windings about 0. From 41 $p$-points, one (zero) having multiplicity 4. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 28, 2017 4:39:04 GMT
Vasco, Yes, I saw that. It was a nice surprise. Correction! I see you said 44 $p$-points. I read that as windings. I still need to investigate how you arrived at 44. Above, I was just indicating what was in my recent posting. Gary Gary You are right it should be 44 windings about 0. From 41 $p$-points, one (zero) having multiplicity 4. Vasco Vasco, I was right about the 44 windings, but I still hadn't realized that there had to be 44 p-points so I should look harder for them. I have now revised and will repost. Oh, I see from your message that there are only 41 p-points, but the 0 p-point has multiplicity four. Why not just say there are four p-points at 0? Gary |
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Post by Admin on Mar 28, 2017 5:55:38 GMT
Gary You are right it should be 44 windings about 0. From 41 $p$-points, one (zero) having multiplicity 4. Vasco Vasco, I was right about the 44 windings, but I still hadn't realized that there had to be 44 p-points so I should look harder for them. I have now revised and will repost. Oh, I see from your message that there are only 41 p-points, but the 0 p-point has multiplicity four. Why not just say there are four p-points at 0? Gary Gary Yes that's what it amounts to really, but I just wanted it to be clear what I was saying. Vasco |
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Post by Admin on Mar 28, 2017 8:10:37 GMT
Gary
In the latest version of your document: March 27, 2017, 9:30 pm PST in part (iii) on page 8 at the top, you say:
"...topological multiplicity of $\nu[\sin z^4,p]=44$" and "...the algebraic multiplicity $N$."
But 44 is not the topological multiplicity and the formula is the formula for the winding number of $\sin z^4$ round $p=0$ as $z$ goes round $|z|=2$.
As far as I understand Needham's notation in this chapter:
$N$ is the number of $p$-points counted with their multiplicities (=44)
$\nu[f(\Gamma),p]$ is the winding number of $f(\Gamma)$ round $p$ (=44)
$\nu(a)$ is the multiplicity of a $p$-point $a$.
It is true that the Topological Argument Principle tells us that the numerical values of $N$ and $\nu[f(\Gamma),p]$ are equal, but the notations refer to different concepts.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 28, 2017 15:39:42 GMT
Vasco,
Thanks. I'll take another look at the notation. I've never been very clear on the correct notation for p-points, algebraic multiplicity, topological multiplicity, and winding numbers. I have the habit of equating the last two. Having gone through this complex problem (iii), I think it will become more clear.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 28, 2017 22:45:00 GMT
Vasco,
I have added a note on notation to my posting at the top of this thread.
Gary
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