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Post by Admin on Mar 16, 2016 15:30:47 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Mar 23, 2016 22:34:18 GMT
Vasco, The commentary is just what was needed to clarify these sections. I don't think I ever realized that in Figure [18], K and f are different things. The proof of $\mathcal{E} = \hat{\xi} \times \mathcal{K}$ from $\nu$ is nice. Lots of K's to keep track of in this chapter: K curve from Figure [18], $\kappa$ curvature, $\tilde{\kappa}$ curvature, $\mathcal{K}$ complex curvature. Gary |
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Post by Admin on Mar 24, 2016 20:22:04 GMT
Vasco, The commentary is just what was needed to clarify these sections. I don't think I ever realized that in Figure [18], K and f are different things. The proof of $\mathcal{E} = \hat{\xi} \times \mathcal{K}$ from $\nu$ is nice. Lots of K's to keep track of in this chapter: K curve from Figure [18], $\kappa$ curvature, $\tilde{\kappa}$ curvature, $\mathcal{K}$ complex curvature. Gary Gary Glad you find it useful. You missed one - $\widetilde{K}$. I was thinking exactly the same when I was writing the document. Vasco |
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Post by Admin on Apr 4, 2016 14:31:20 GMT
I have published a new version of my commentary on Section IX of chapter 5. I have not changed it much - hopefully I have made one or two points clearer in the last few paragraphs of the document. The previous link will download the updated version.
Vasco
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Post by telemeter on Mar 4, 2019 19:05:59 GMT
Vasco, thank you for the commentary...very helpful.
In this section, I think Needham is at his unclearest. I was initially confused as I read the section between the curvatures k of the curve K and the complex curvature K of the mapping f. It took a little while to realise they have nothing whatsoever to do with each other, despite having very similar names and symbols. I was also stuck thinking that changes on curve K had something to do with f'!
I was struggling through the section wondering which direction is the complex curvature of f in figure [18].
What would have been more helpful to me is a discussion in the book of the direction required of the complex curvature of mapping f to maximise the change in curvature of K under the mapping. This could have been illustrated showing explicitly that the change in curvature under the mapping is maximised if the mapping's complex curvature aligns with tangent xi, while change in amplification is maximised if complex curvature is orthogonal to tangent xi. Then it would have been clear from the beginning.
For anyone else having difficulty with this section, I do not think the concepts are that difficult; the presentation is. So take heart and persevere, its not you!
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Post by mondo on Jan 13, 2023 23:45:31 GMT
I decided to look at this extra section too. The first place I am stuck is page 237 and sentence below equation (26) "If $\xi$ and $\zeta$ both underwent precisely the same twist then the turning angle $\tilde{\epsilon}$ for the images would equal the original turning angle $\epsilon$. However the twist at $q$ will differ very slightly, say by $\delta$"
1. How do we know that twist at $q$ will differ from that at $p$?
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Post by Admin on Jan 14, 2023 8:35:06 GMT
I decided to look at this extra section too. The first place I am stuck is page 237 and sentence below equation (26) "If $\xi$ and $\zeta$ both underwent precisely the same twist then the turning angle $\tilde{\epsilon}$ for the images would equal the original turning angle $\epsilon$. However the twist at $q$ will differ very slightly, say by $\delta$" 1. How do we know that twist at $q$ will differ from that at $p$? Mondo Why wouldn't the twist at $q$ differ from the twist at $p$? Vasco |
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Post by mondo on Jan 14, 2023 18:22:20 GMT
I decided to look at this extra section too. The first place I am stuck is page 237 and sentence below equation (26) "If $\xi$ and $\zeta$ both underwent precisely the same twist then the turning angle $\tilde{\epsilon}$ for the images would equal the original turning angle $\epsilon$. However the twist at $q$ will differ very slightly, say by $\delta$" 1. How do we know that twist at $q$ will differ from that at $p$? Mondo Why wouldn't the twist at $q$ differ from the twist at $p$? Vasco Heh, that's what I don't know. From both $p$ and $q$ we draw an infinitesimal vectors, also we transformed them using the same function $f(z) = e^z$, so why their twist would differ? |
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Post by Admin on Jan 14, 2023 18:37:18 GMT
Mondo
Why do you think the twist would be the same? Makes me think you still don't understand what ampltwist is.
Vasco
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Post by mondo on Jan 14, 2023 19:01:12 GMT
Mondo Why do you think the twist would be the same? Makes me think you still don't understand what ampltwist is. Vasco Because as I wrote in the previous post we have two infinitesimal vectors and the same mapping $f(z) = e^z$. Is it not the same because the Curvature $K$ varies? Also there is a lot of confusion in this paragraph - "However the twist at $q$ will differ very slightly, say by $\delta$, from that at $p$" - how can we have a twist at $p$ and $q$ if they are "preimages" before the mapping? We can only speak about twist at points $\tilde{p}$ and $\tilde{q}$ right? |
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Post by Admin on Jan 14, 2023 19:22:21 GMT
Mondo
Why do you think the mapping is $f(z)=e^z$? It is just a general analytic function. But apart from that you need to understand that the amplitwist of an analytic function $f(z)$ varies as $z$ varies.
No, there is an amplitwist at points $p$ and $q$ because the function $f$ is being applied to infinitesimal vectors coming from $p$ and $q$ not to $p$ and $q$ themselves.
Vasco
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Post by Admin on Jan 14, 2023 19:31:16 GMT
Mondo
It's always the infinitesimal vectors from the preimages which are amplitwisted to produce the image infinitesimals.
Vasco
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Post by mondo on Jan 14, 2023 19:50:45 GMT
Mondo Why do you think the mapping is $f(z)=e^z$? It is just a general analytic function. Right, my mistake, it's just some mapping we don't care what kind of exactly, all we care about is for it to be analytic. But apart from that you need to understand that the amplitwist of an analytic function $f(z)$ varies as $z$ varies. So you say for a mapping $z$ -> $z^2$ amplitwist varies as the argument varies? That does not make sense, the amplitwist/derivative should always be the same, no matter at what point we calculate it right? No, there is an amplitwist at points $p$ and $q$ because the function $f$ is being applied to infinitesimal vectors coming from $p$ and $q$ not to $p$ and $q$ themselves. I don't get it here, its like saying there is an amplitwist at point $z$ figure 17 page 235 (previous page) while no, the amplitwist is at the image of $z$, in this case a point $e^z$ right? So in the same way I would say there is a twist and in general "amplitwist" but at the image points of $p$ and $q$ which are shown on the top of image [18] and not at $p$ and $q$ itself. |
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Post by mondo on Jan 14, 2023 20:08:57 GMT
Mondo It's always the infinitesimal vectors from the preimages which are amplitwisted to produce the image infinitesimals. Vasco Yes I agree. The image infinitesimals are at $\tilde{p}$ and $\tilde{q}$ and we measure the amplitwist by comparing preimages with images. |
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Post by Admin on Jan 14, 2023 20:22:04 GMT
Mondo
What I mean when I say the amplitwist varies is that it is different for different values of $z$, so for $z^2$ it is $2z$ which varies with $z$.
No, the amplitwist is at the preimage and is applied to the infinitesimal vector to produce the image infinitesimal vector.
in figure 6 we see that the amplitwist at the point $z$ is $2re^{i\theta}$ which when applied to $\epsilon$ gives $2re^{i\theta}\epsilon$ the image infinitesimal vector.
Vasco
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