Gary
GaryVasco
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Post by Gary on Mar 18, 2017 20:49:41 GMT
Vasco,
I think you have finally gotten through.
Yes, the quote shows that Needham intended it as a function of $z$, so I see you were right about this. I don't know why that didn't register earlier. Juggling too many balls, I think, and just taking the meaning from "an explicit prescription for gradually deforming $L$ into $\widehat{L}$". And it also makes sense of the the whole paragraph containing (4). That said, I think I would like it better if it were set up as $\widehat{w} = \mathcal{L}_s \circ \mathcal{L}(z)$. I will have to try and rethink my answer to the question.
Gary
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Post by Admin on Mar 20, 2017 5:28:05 GMT
Vasco, To continue the previous, I have reread your solution to (i). It looks good to me, especially the explanation of how the preimages relate to the horizontal lines. The only question I have is how in paragraph 1 you can jump right to calculating the preimages as $z = e^{i\theta}$, which puts us back at the beginning of the progression z -> w -> $\widehat{w}$. I just noticed the lightly drawn unit circle in [6] for the first time. Gary Gary Presumably you are OK with this now. Am I right? Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 20, 2017 8:56:41 GMT
Vasco, To continue the previous, I have reread your solution to (i). It looks good to me, especially the explanation of how the preimages relate to the horizontal lines. The only question I have is how in paragraph 1 you can jump right to calculating the preimages as $z = e^{i\theta}$, which puts us back at the beginning of the progression z -> w -> $\widehat{w}$. I just noticed the lightly drawn unit circle in [6] for the first time. Gary Gary Presumably you are OK with this now. Am I right? Vasco Vasco, Yes. Gary |
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Post by Admin on Mar 20, 2017 17:24:09 GMT
Gary
I have started reading your revised document. On page 2 in the paragraph which starts "But the standardized loop...", you say that the mapping
$\widehat{w}=\widehat{\mathcal{L}}(e^{i\theta})=e^{i\Phi(\theta)}$ is not 1-1, and is an exponential function.
In my opinion both of these statements are incorrect, because $\theta\longrightarrow\Phi\longrightarrow\widehat{w}$ is a 1-1 mapping and the mapping is not an exponential function.
A value of $\theta$ produces one value for $\Phi$ which in turn produces one value for $\widehat{w}$ and the functions which determine the mapping are $\Phi(\theta)$ and $R(\theta)$, and all we know about them is that they are continuous.
The reason is that $\widehat{w}$ is associated with one value of $\Phi$. I agree that $|\widehat{w}|$ can take on the same value for different values of $\Phi$.
A way of convincing yourself of this is to remember what Needham writes at the very beginning of subsection 2 on page 342: Let $C$ be a rubber band. If we deform a rubber band then a point on the rubber band cannot go to more than once place at the same time.
Also substituting for $\theta$ gives $\widehat{w}=\widehat{\mathcal{L}}(e^{ia})=e^{i\Phi(a)}$ not $\widehat{w}=\widehat{\mathcal{L}}(e^{iA})=e^{i\Phi(A)}$. In the subsequent development you have also used capital A when it should be lower case a.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 21, 2017 6:02:33 GMT
Vasco,
I trust your experience in these exercises, but I have questions.
I don't understand why $\widehat{w} = e^{i\Phi(\theta)}$ is not an exponential function. It seems to fit the standard definitions $f(x) = b^x$ in which the constant $b$ is frequently $e$ and the argument, according to Wikipedia, can be any real or complex number.
Then should we imagine multiple $z$'s at the same location in $C$?
True, but as we see in [6b], many points on the rubber band can go to the same place, i.e. $\widehat{w}$ on $\widehat{L}$. That indicates a many-to-one relation.
But I think my attempt to define $\Phi(\theta)$ was infelicitous (fancy way of saying "wrong") and overly complicated. If we turn the focus to the generation of L, we can think of $\theta$ rotating counterclockwise and clockwise on $C$ and affecting $\Phi$ on $L$, while $R(\theta)$ moves the loop in and out wrt 0 on $L$, but in this exercise, Needham makes no mention of variation in $R$, and $\Phi$ is defined in terms of constants ($A, n, 2\pi$). If $0 < A \leq 2\pi$, n is the number of winds. So leaving aside my attempt to define $\Phi$, we can substitute these values of $\Phi$ to get an infinite series of $w$'s all aligned on a ray at angle $A$. They all pile up on the same point. In fact, we don't even have to consider the effect of $R(\theta)$ or anything apart from what happens on unit circles.
All of these $w$'s can be drawn at the same angle $\theta$, so this angle must be $A$ and we could place the image on either $L$ or $\widehat{L}$. I would stop at L (a unit circle in this exercise). There is no need to apply either $\mathcal{L}_s$ or $\widehat{\mathcal{L}}_t$. We are not trying to distort $L$. We only want to see how to explain the horizontal lines. Each of the horizontal lines represents the angle A plus or minus n winds.
Judging from the introduction and from (ii), Needham appears to expect the image to appear on $\widehat{L}$, so perhaps he is assuming the $L$ is not limited to the unit circle, or that there is no $L$ in this case, and the progression is $C \rightarrow \widehat{L}$.
Does this sound right?
As for relating the angles to $a$ in $z = e^{ia}$ and $w = e^{\Phi(a)}$, I still don't understand why he used $a$ instead of $\theta$ unless the point was just that $\theta$ is given as a constant. How would you define $\Phi(a)$?
I am no longer sure that the $w$ in (i) should be $\widehat{w}$.
Gary
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Post by Admin on Mar 21, 2017 17:05:22 GMT
Gary I have produced a document which I hope answers some of your questions. Here's the linkVasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 21, 2017 20:03:04 GMT
Gary I have produced a document which I hope answers some of your questions. Here's the linkVasco Vasco, Your document resolved all my questions and reservations. My view of this section has finally converged on your various points. Another rewrite in progress. Gary |
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Post by Admin on Mar 22, 2017 9:41:35 GMT
Gary
I have looked at your revised document. I notice that you still have expressions with things like $\theta=A$ and $\Phi(A)$. I thought we now agreed that $A$ represents a value of $\Phi$ not $\theta$?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 22, 2017 15:25:28 GMT
Gary I have looked at your revised document. I notice that you still have expressions with things like $\theta=A$ and $\Phi(A)$. I thought we now agreed that $A$ represents a value of $\Phi$ not $\theta$? Vasco Vasco, I did abandon my old definition of $\phi$, but retained this more specific equality at $\theta = A$. I'm trying to understand this point. I was thinking that $A$ is both a value of $\theta$ and one of the values of $\Phi$. Figure [7] shows that where $\theta = 2\pi$, $\Phi = 2\pi \nu$. Where $\nu = 1$, then $\Phi(2\pi) = 2\pi$. In Figure 1, $\Phi = A$ at $A$. But in this problem $\nu$ may be +1 or -1, so I'm not sure whether this reasoning is correct. If $\Phi(A) \neq A$, then how would all the preimages land on the same point of L with the given values of $\Phi$? If $\Phi(A) \neq A$, then Figure 1 must also be wrong. But now I see that writing $\Phi(A)$ in $\Phi_t(\theta)$ with t = 1 reduces it to the trivial $A + A - A$ = $A$ and for other values of $\theta$, the "slack" function $e^{i\theta_t(\phi)}$ applied to this problem does not seem to send all the $z$'s to the same $\widehat{w}$. So I am still missing something. I will reread your answer. Gary |
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Post by Admin on Mar 22, 2017 20:36:50 GMT
Gary I have looked at your revised document. I notice that you still have expressions with things like $\theta=A$ and $\Phi(A)$. I thought we now agreed that $A$ represents a value of $\Phi$ not $\theta$? Vasco Vasco, I did abandon my old definition of $\phi$, but retained this more specific equality at $\theta = A$. I'm trying to understand this point. I was thinking that $A$ is both a value of $\theta$ and one of the values of $\Phi$. Figure [7] shows that where $\theta = 2\pi$, $\Phi = 2\pi \nu$. Where $\nu = 1$, then $\Phi(2\pi) = 2\pi$. In Figure 1, $\Phi = A$ at $A$. But in this problem $\nu$ may be +1 or -1, so I'm not sure whether this reasoning is correct. If $\Phi(A) \neq A$, then how would all the preimages land on the same point of L with the given values of $\Phi$? If $\Phi(A) \neq A$, then Figure 1 must also be wrong. But now I see that writing $\Phi(A)$ in $\Phi_t(\theta)$ with t = 1 reduces it to the trivial $A + A - A$ = $A$ and for other values of $\theta$, the "slack" function $e^{i\theta_t(\phi)}$ applied to this problem does not seem to send all the $z$'s to the same $\widehat{w}$. So I am still missing something. I will reread your answer. Gary Gary When $\nu=1$ then as you say when $\theta=2\pi$ then $\Phi=2\pi$, but that doesn't mean that $\Phi=\theta$ for all $\theta$. Even when mapping the unit circle to another unit circle it doesn't mean that when $\theta=a$, say, that $\Phi$ is also equal to $a$, because the circle could be stretched and folded back on itself by the mapping $\Phi(\theta)$ (as in 6b). The unit circles look identical but a point on one circle at a given angle is not mapped to a point on the second circle at the same angle. The only restriction is that when you have gone round one circle you must have also gone round the other, which is what "when $\theta=2\pi$ then $\Phi=2\pi$" means. Of course in the archetypal mapping of degree 1 it is true that $\Phi(\theta)=\theta$ for all values of $\theta$. So if $\Phi=A$ when $\theta=a$ then $\Phi(a)=A=a$. So you could write, as you have, $A=\Phi(a)=\phi(A)$, but only if we make it clear that this is when $\nu=1$ and $t=1$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 23, 2017 21:42:48 GMT
Vasco, I did abandon my old definition of $\phi$, but retained this more specific equality at $\theta = A$. I'm trying to understand this point. I was thinking that $A$ is both a value of $\theta$ and one of the values of $\Phi$. Figure [7] shows that where $\theta = 2\pi$, $\Phi = 2\pi \nu$. Where $\nu = 1$, then $\Phi(2\pi) = 2\pi$. In Figure 1, $\Phi = A$ at $A$. But in this problem $\nu$ may be +1 or -1, so I'm not sure whether this reasoning is correct. If $\Phi(A) \neq A$, then how would all the preimages land on the same point of L with the given values of $\Phi$? If $\Phi(A) \neq A$, then Figure 1 must also be wrong. But now I see that writing $\Phi(A)$ in $\Phi_t(\theta)$ with t = 1 reduces it to the trivial $A + A - A$ = $A$ and for other values of $\theta$, the "slack" function $e^{i\theta_t(\phi)}$ applied to this problem does not seem to send all the $z$'s to the same $\widehat{w}$. So I am still missing something. I will reread your answer. Gary Gary When $\nu=1$ then as you say when $\theta=2\pi$ then $\Phi=2\pi$, but that doesn't mean that $\Phi=\theta$ for all $\theta$. Even when mapping the unit circle to another unit circle it doesn't mean that when $\theta=a$, say, that $\Phi$ is also equal to $a$, because the circle could be stretched and folded back on itself by the mapping $\Phi(\theta)$ (as in 6b). The unit circles look identical but a point on one circle at a given angle is not mapped to a point on the second circle at the same angle. The only restriction is that when you have gone round one circle you must have also gone round the other, which is what "when $\theta=2\pi$ then $\Phi=2\pi$" means. Of course in the archetypal mapping of degree 1 it is true that $\Phi(\theta)=\theta$ for all values of $\theta$. So if $\Phi=A$ when $\theta=a$ then $\Phi(a)=A=a$. So you could write, as you have, $A=\Phi(a)=\phi(A)$, but only if we make it clear that this is when $\nu=1$ and $t=1$. Vasco Vasco, I agree with all this. I think I will just have to try to make the thoughts more clear. I was assuming that Needham intended for L to be on the unit circle, so there would be no looping back and forth. That is $R(\theta)$ = 1 for all $\theta$. All the $\Phi$'s would determine continuous winds of varying degrees in one direction. Gary |
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