Gary
GaryVasco
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Post by Gary on Mar 7, 2017 2:22:48 GMT
Vasco, I'm floundering wildly in exercise 2, but I'm posting it just to get the discussion started. nh.ch7.ex3.pdf (587.9 KB) (now much revised) Gary |
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Post by Admin on Mar 7, 2017 10:41:15 GMT
Vasco, I'm floundering wildly in exercise 2, but I'm posting it just to get the discussion started. View AttachmentGary Gary I haven't attempted exercise 2 yet so I won't look at yours for the moment. I'll concentrate on producing my attempt at a solution as soon as possible. Vasco |
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Post by Admin on Mar 13, 2017 16:11:57 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 13, 2017 21:42:45 GMT
I've printed it out; It looks interesting. |
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Post by Admin on Mar 15, 2017 7:00:31 GMT
Gary
In your answer to part (i), you say near the bottom of page 1:
"...the preimages of $w=e^{iA}$ are $z=e^{ia}$, so the preimages can be written as $z=e^{i\Phi(A)}$..."
I don't follow this reasoning.
Vasco
PS I sent you a message by mistake about this.
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Gary
GaryVasco
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Post by Gary on Mar 15, 2017 16:06:28 GMT
Gary In your answer to part (i), you say near the bottom of page 1: "...the preimages of $w=e^{iA}$ are $z=e^{ia}$, so the preimages can be written as $z=e^{i\Phi(A)}$..." I don't follow this reasoning. Vasco PS I sent you a message by mistake about this. Vasco, The reasoning was not great, but it was simple. In the problem statement, the preimages are referred to as $z=e^{ia}$. I am just rewriting this to preserve the relation of $w$ to $z$ as image to preimage and asserting that the preimages are a group of loops that have been drawn and tightened onto the unit circle. I should add that I don't understand why Needham wrote $z=e^{ia}$ with "$a$". Gary |
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Post by Admin on Mar 15, 2017 17:59:33 GMT
Gary In your answer to part (i), you say near the bottom of page 1: "...the preimages of $w=e^{iA}$ are $z=e^{ia}$, so the preimages can be written as $z=e^{i\Phi(A)}$..." I don't follow this reasoning. Vasco PS I sent you a message by mistake about this. Vasco, The reasoning was not great, but it was simple. In the problem statement, the preimages are referred to as $z=e^{ia}$. I am just rewriting this to preserve the relation of $w$ to $z$ as image to preimage and asserting that the preimages are a group of loops that have been drawn and tightened onto the unit circle. I should add that I don't understand why Needham wrote $z=e^{ia}$ with "$a$". Gary Gary The way I see it is that $z=e^{i\theta}$ on the unit circle and the loop itself is $\widehat{w}=e^{i\Phi(\theta)}$. $\Phi$ is a function of $\theta$ and therefore varies as $z$ moves round the unit circle. If $\theta=a$ is a specific value of $\theta$ then the corresponding value of $\Phi$ is $A=\Phi(a)$, so that when $z=e^{ia}$ the value of $\widehat{w}=e^{iA}$. The preimages all lie on the same unit circle it seems to me, because as $z$ moves round the unit circle once, $\widehat{w}$ moves once round the loop and $\nu$ times round the image unit circle. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 18, 2017 5:05:17 GMT
Vasco, I plowed ahead, but I didn't want to post this until we resolved some of the problems in Exercise 2. I think we are OK there. I found (i) in exercise 3 to be puzzling, so I am interested to see what you will do with it. Gary Gary I have a solution to exercise 3, I just haven't published it yet, because of the graphics involved. Our solutions to exercise 2 are very different don't you think?Vasco Vasco, Our answers are different, but I think they are based on the same understanding of the situation. I thought that the notation in the question varied in its meaning from that in the text, and it would be difficult to sort out actual mistakes. Your correction of the notation on the three w's would help. I have not yet incorporated that development into my answer. I think my answer treated them as though the w's actually are on $\widehat{L}$. If you find an error in my answer, I would be happy to learn about it, or if you want to dig into it some more, that's fine, too. My understanding of Ex. 2 (i) now is that $\widehat{w} = e^{iA}$ is an image on $\widehat{L}$. Before "taking up the slack", there are also an infinite number of positive and negative $\Phi$ preimages sharing the space on $\widehat{L}$. These can be represented as $\widehat{w}_\Phi = e^{\Phi(A)} = e^{A \pm n 2\pi i}$. These are converted into their common image $\widehat{w} = e^{iA}$ by means of $\mathcal{L}_{t=1} = e^{i\Phi_t(A)}$. The horizontal lines represent the infinite real values that A could take, but I think all the preimages fall on $\widehat{L}$ along with the image. To get back to a preimage $w$ in $L$ would require a function $R(A)$ (p. 342), which we don't have, so for each $\widehat{w}_\Phi$, there must be an infinite number of $w$ on a ray from the origin. In that very abstract sense the horizontal lines represent the complete set of preimages. Having written this, I have a feeling that I have missed the point and there is some more simple approach. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 18, 2017 5:44:54 GMT
Vasco, The reasoning was not great, but it was simple. In the problem statement, the preimages are referred to as $z=e^{ia}$. I am just rewriting this to preserve the relation of $w$ to $z$ as image to preimage and asserting that the preimages are a group of loops that have been drawn and tightened onto the unit circle. I should add that I don't understand why Needham wrote $z=e^{ia}$ with "$a$". Gary Gary The way I see it is that $z=e^{i\theta}$ on the unit circle and the loop itself is $\widehat{w}=e^{i\Phi(\theta)}$. $\Phi$ is a function of $\theta$ and therefore varies as $z$ moves round the unit circle. If $\theta=a$ is a specific value of $\theta$ then the corresponding value of $\Phi$ is $A=\Phi(a)$, so that when $z=e^{ia}$ the value of $\widehat{w}=e^{iA}$. The preimages all lie on the same unit circle it seems to me, because as $z$ moves round the unit circle once, $\widehat{w}$ moves once round the loop and $\nu$ times round the image unit circle. Vasco Vasco, This is also the way I see it, but the directionality of the $\mathcal{L}$ arrow in [5] and the phrasing of (i) makes me think that we can't just send the preimages $\widehat{w}$ back to $C$, although that is effectively what we do. I am reposting, taking your comments into account, but my notation is slightly different. Gary |
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Post by Admin on Mar 18, 2017 6:01:55 GMT
Gary The way I see it is that $z=e^{i\theta}$ on the unit circle and the loop itself is $\widehat{w}=e^{i\Phi(\theta)}$. $\Phi$ is a function of $\theta$ and therefore varies as $z$ moves round the unit circle. If $\theta=a$ is a specific value of $\theta$ then the corresponding value of $\Phi$ is $A=\Phi(a)$, so that when $z=e^{ia}$ the value of $\widehat{w}=e^{iA}$. The preimages all lie on the same unit circle it seems to me, because as $z$ moves round the unit circle once, $\widehat{w}$ moves once round the loop and $\nu$ times round the image unit circle. Vasco Vasco, This is also the way I see it, but the directionality of the $\mathcal{L}$ arrow in [5] and the phrasing of (i) makes me think that we can't just send the preimages $\widehat{w}$ back to $C$, although that is effectively what we do. I am reposting, taking your comments into account, but my notation is slightly different. Gary Gary Above you say "...the preimages $\widehat{w}$..." - Is this a typo? $\widehat{w}$ is not a preimage as I see it. Vasco |
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Post by Admin on Mar 18, 2017 15:24:20 GMT
Gary I have a solution to exercise 3, I just haven't published it yet, because of the graphics involved. Our solutions to exercise 2 are very different don't you think? Vasco Vasco, Our answers are different, but I think they are based on the same understanding of the situation. I thought that the notation in the question varied in its meaning from that in the text, and it would be difficult to sort out actual mistakes. Your correction of the notation on the three w's would help. I have not yet incorporated that development into my answer. I think my answer treated them as though the w's actually are on $\widehat{L}$. If you find an error in my answer, I would be happy to learn about it, or if you want to dig into it some more, that's fine, too. My understanding of Ex. 2 (i) now is that $\widehat{w} = e^{iA}$ is an image on $\widehat{L}$. Before "taking up the slack", there are also an infinite number of positive and negative $\Phi$ preimages sharing the space on $\widehat{L}$. These can be represented as $\widehat{w}_\Phi = e^{\Phi(A)} = e^{A \pm n 2\pi i}$. These are converted into their common image $\widehat{w} = e^{iA}$ by means of $\mathcal{L}_{t=1} = e^{i\Phi_t(A)}$. The horizontal lines represent the infinite real values that A could take, but I think all the preimages fall on $\widehat{L}$ along with the image. To get back to a preimage $w$ in $L$ would require a function $R(A)$ (p. 342), which we don't have, so for each $\widehat{w}_\Phi$, there must be an infinite number of $w$ on a ray from the origin. In that very abstract sense the horizontal lines represent the complete set of preimages. Having written this, I have a feeling that I have missed the point and there is some more simple approach. Gary Gary There are two main mappings here: 1. $\mathcal{L}$ sending values of $z$ on the unit circle to values of $w$ on $L$ 2. $\mathcal{L}_s$ sending values of $z$ on the unit circle to values of $\widehat{w}$ on $\widehat{L}$ There is also an unnamed, implicit third mapping which maps values of $w$ on $L$ to values of $\widehat{w}$ on $\widehat{L}$. Mapping 2 can be thought of as the composition of mapping 1 followed by the unnamed mapping. Draw a simple picture with $C$ the unit circle, $L$ the complex loop, and $\widehat{L}$ the standardised loop on the unit circle. Don't think of this as the same unit circle as $C$, think of it as $\widehat{L}$, which just happens to be the unit circle in the image plane. Draw an arrow ($\mathcal{L}$) from $C$ to $L$, another arrow ($\widehat{L}_s$) from $C$ to $\widehat{L}$, and another (the unnamed mapping) from $L$ to $\widehat{L}$. The preimages of the mapping $\mathcal{L}$ are values of $z$ on $C$, and the preimages of the mapping $\mathcal{L}_s$ are again the values of $z$ on $C$. When we draw the horizontal lines on [7] corresponding to $\Phi=A$ etc, the values of $\theta$ corresponding to the points of intersection of these lines with the graph of $\Phi$ give us the preimages $\theta=a$ on $C$ corresponding to the values $\Phi=A$, and hence, via $z=e^{i\theta}$ they also give us the corresponding preimages $z$. We read off these preimage values of $\theta$ by dropping a line perpendicular to the $\theta$-axis from the points of intersection until it intersects the horizontal axis of [7]. By the way the inverse of $R(\theta)=1$ and the inverse of $\Phi(\theta)=\theta$. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 18, 2017 16:21:00 GMT
Vasco, This is also the way I see it, but the directionality of the $\mathcal{L}$ arrow in [5] and the phrasing of (i) makes me think that we can't just send the preimages $\widehat{w}$ back to $C$, although that is effectively what we do. I am reposting, taking your comments into account, but my notation is slightly different. Gary Gary Above you say "...the preimages $\widehat{w}$..." - Is this a typo? $\widehat{w}$ is not a preimage as I see it. Vasco Vasco, It's not a typo. In a revision I posted yesterday and amended slightly this morning, I refer to the preimages as $\widehat{w}_\Phi$ to indicate that they are the result of taking up slack and lie on $\widehat{L}$, as does the image. I find it hard to know just what Needham means by "preimage" here because there are several stages of deformation (two one way and two the other). I think one could regard what I call $\widehat{w}_\Phi$ in $\widehat{L}$ as equivalent to reversal from $w$ in $L$ to $z$ in $C$. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 18, 2017 16:51:10 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
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Gary
GaryVasco
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Post by Gary on Mar 18, 2017 17:54:27 GMT
Vasco,
The diagram you suggest is a helpful exercise, but there are a couple of points that might be revised:
(3), named $\mathcal{L}_s(z)$, is introduced for this purpose (p. 343, top)
Incidentally, do you find $\widehat{L}$ to be ambiguous between "$\widehat{L}$ is almost a unit circle and "the loop is almost $\widehat{L}$ (which is a unit circle)"? I think the last sentence before [6] argues for the second interpretation.
Agreed. This has always been my interpretation.
Is there a rationale for drawing the arrow $\widehat{L}_s$ from C to $\widehat{L}$ rather than an arrow $\mathcal{L}_s$ from L to $\widehat{L}$? This is really the same comment as the previous regarding the unnamed mapping. I'm guess also there was a typo in writing a function arrow without the script font.
In view of the previous comments, the preimages of $\mathcal{L}_s$ would be the values of $w$ on $L$. Perhaps this is the crux of the difference in our answers.
Going right from the image $\widehat{w}$ in $\widehat{L}$ to preimages $z$ on $C$ would mean skipping over $L$. To this point, we have not mentioned "taking up the slack". In view of the way Needham has developed the notation, I would suggest an arrow $\mathcal{L}_t$ from $\widehat{L}$ to itself, but what if the result of $\mathcal{L}_s$ is not yet in $\widehat{L}$. What if it is "$\widehat{L}$ (almost)"?
Gary
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Post by Admin on Mar 18, 2017 19:19:34 GMT
Gary
This is where I disagree with you. As I said in a recent previous post, this transformation $\mathcal{L}_s(z)$ is a transformation from $C$ to $\widehat{L}$. I think this is fundamental to understanding the text.
You can see this by looking at (3) on page 343.
When $s=0$ you can see that $\mathcal{L}_{s=0}(C)=\mathcal{L}(C)$, the mapping from $C$ to $L$. When $s$ lies somewhere between $0$ and $1$ then $\mathcal{L}_s(z)$ maps $C$ to somewhere between $L$ and $\widehat{L}$,
and then when $s=1$ we have $\mathcal{L}_{s=1}(C)$ becoming the mapping from $C$ to $\widehat{L}$.
This is why we can think of $\mathcal{L}_{s=1}(C)$ as a composite mapping of $\mathcal{L}_{s=0}=\mathcal{L}$ and the unnamed mapping which maps $L$ to $\widehat{L}$.
Note that the first sentence under (3) on page 343 says: "As $s$ varies from 0 to 1, $\mathcal{L}_{s}(C)$ gradually...changes from $L$ to $\widehat{L}$."
Vasco
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