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Post by Admin on Apr 2, 2016 14:42:48 GMT
Gary
Did you mean $k(\xi)$ as defined under subsection 3 on pages 238 and 239 rather than $\kappa(\xi)$?
If you do then I agree that these are all expressions for the same thing and are all defined in subsection 3 on pages 238-241.
The expression $\mathcal{K} \cdot \hat{\xi}$ is useful as it shows that if we know $\mathcal{K}$, which is intrinsic to the function $f$, then we can use this to calculate the curvature under the mapping $f$ of the image of a straight line, as a function of its direction $\widehat{\xi}$, as in figure 19 on page 239.
The expression $\mathcal{R}$ reminds us that the curvature is the rate of rotation of an infinitesimal shape with respect to the distance it moves and together with $\mathcal{E}$ and $\mathcal{K}$ enables us to make calculations such as those required to draw diagrams like (21).
Writing $k(\xi)$ into equation 30 on page 239 allows us to see that the curvature $\widetilde{\kappa}$ can be calculated by combining in a simple way the curvature $\kappa$ on the curve $K$ and the curvature introduced by the mapping $f$.
Vasco
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Gary
GaryVasco
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Post by Gary on Apr 3, 2016 7:29:41 GMT
Vasco, I’ve been going over your documents and trying to get a good match between the math and the plot. Sometimes I thought my math was right, but the plot didn’t look right. At other times, I’m wasn’t sure about my math but the plot looked good. Regarding the plot for $e^z$, I have a combination now where the math seems right and the plot corresponds well to yours. The first plots were disappointing because I chose c = 0, where c is the center of $\mathcal{Q}$. I changed to c = i and that helped to give the right hand side some rotation, but the angles of the bars were still wrong. I was getting too much rotation. I was treating the problem as though the twist was caused by the rotation implied by the angles of the $\xi_n$, as though those angles were due to the twist of f(c), so I was adding those angles into the rotation. When I finally realized that the only sources of twist for the $\tilde{\zeta_n}$ are f’(c) and $\alpha$, the plot suddenly looked better. I rewrote the answer according to my own perception of it with indices for the various $\tilde{Q}_n$. I’m still a little puzzled by the set up of the problem, especially the arrangement of all the $\xi$ on the horizontal, but I concede that it does result in a heuristic plot. The other plots should go more quickly now. Source code available on request. In the attached plot, I added the vectors for $Exp(\tilde{c’_1})$ and $Exp(\tilde{c’_1} + \zeta_1)$. They are not congruent with $\tilde{c} + \tilde{\xi_1}$ and $\tilde{c} + \tilde{\xi_1} + \tilde{\zeta}$. I have not figured out why. I'll get back to the problem of proliferating k's tomorrow. Gary nh.ch5.ex17.vi.eToz.pdf (143.19 KB) |
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Post by Admin on Apr 3, 2016 17:52:53 GMT
Gary
Some initial thoughts - more to follow:
In case it helps you to check if your results are the same as mine, here is the data I used:
I positioned $Q$ at $0.5+i$ and used $\xi=.5$ and the radius of $Q$ as $0.5$.
I'm not sure what you mean here, are you speaking about the fact that all the levers are horizontal on the LHS?
Vasco
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Gary
GaryVasco
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Post by Gary on Apr 3, 2016 18:06:24 GMT
Gary Some initial thoughts - more to follow: In case it helps you to check if your results are the same as mine, here is the data I used: I positioned $Q$ at $0.5+i$ and used $\xi=.5$ and the radius of $Q$ as $0.5$. I'm not sure what you mean here, are you speaking about the fact that all the levers are horizontal on the LHS? Vasco Sorry, that should have read "all of the $\zeta$ on the horizontal." I have c = i, $\xi = 1$, r = .1, $\zeta = 2r$. Are you sure you didn't mean r of $Q$ = $0.05$? Mine resembles yours with those values. I added a bit of discussion to the document and made one or two small revisions. Gary |
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Post by Admin on Apr 3, 2016 18:40:15 GMT
Gary
Sorry, but I meant that the radius of $Q$ in my calculation is $0.25$.
Vasco
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Gary
GaryVasco
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Post by Gary on Apr 3, 2016 19:41:09 GMT
Gary Sorry, but I meant that the radius of $Q$ in my calculation is $0.25$. Vasco I think your constants make a better illustration, but I want to move on and see what I can do with Log(z). |
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Post by Admin on Apr 4, 2016 6:55:55 GMT
Gary
In your document you write Let $\xi_n=e^{n\pi/3}$, $n=1..12$.
I think this should be $\xi_n=e^{in\pi/6}, n=1..12$, since $\pi/3=60^{\circ}$.
This then carries over into the formula for $\widetilde{\xi}_n$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 4, 2016 14:44:58 GMT
Gary In your document you write Let $\xi_n=e^{n\pi/3}$, $n=1..12$. I think this should be $\xi_n=e^{in\pi/6}, n=1..12$, since $\pi/3=60^{\circ}$. This then carries over into the formula for $\widetilde{\xi}_n$. Vasco Vasco, Quite right. Thanks. You don't get 12 circles with nPi/3. The document has been corrected. Gary |
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Gary
GaryVasco
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Post by Gary on Apr 4, 2016 15:30:17 GMT
Vasco,
I plotted the Log(z) version of [21] and got some funny angles. Half look right, but the other half look questionable, so I will have to figure out just what I want to see. I recall your warning me about the angles with the Log function.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 4, 2016 16:19:23 GMT
Gary Did you mean $k(\xi)$ as defined under subsection 3 on pages 238 and 239 rather than $\kappa(\xi)$? If you do then I agree that these are all expressions for the same thing and are all defined in subsection 3 on pages 238-241. The expression $\mathcal{K} \cdot \hat{\xi}$ is useful as it shows that if we know $\mathcal{K}$, which is intrinsic to the function $f$, then we can use this to calculate the curvature under the mapping $f$ of the image of a straight line, as a function of its direction $\widehat{\xi}$, as in figure 19 on page 239. The expression $\mathcal{R}$ reminds us that the curvature is the rate of rotation of an infinitesimal shape with respect to the distance it moves and together with $\mathcal{E}$ and $\mathcal{K}$ enables us to make calculations such as those required to draw diagrams like (21). Writing $k(\xi)$ into equation 30 on page 239 allows us to see that the curvature $\widetilde{\kappa}$ can be calculated by combining in a simple way the curvature $\kappa$ on the curve $K$ and the curvature introduced by the mapping $f$. Vasco Vasco, Regarding sentence 1, the k in $k(\hat{\xi})$ in [19] and the $\kappa$ (kappa) in (30) are almost indistinguishable to me. I distinguish them by the brackets for the argument of the function $k(\hat{\xi})$. Sentence 3 makes a point that I should have had foremost in mind when beginning problem 17(vi). Curvature of f at p equals intrinsic (complex) curvature of f at p dotted with direction of f at p. Sentence 4 raises the possibility of defining $\sigma$ as $|f'||\xi|k(\hat{\xi})$, but I think that $k(\hat{\xi})$ is only the first term of $\tilde{\kappa}$ (p. 238) and intermediate step in the derivation of $K\cdot\hat{\xi}$, so there is no point to using it that way. Gary |
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Post by Admin on Apr 4, 2016 16:38:46 GMT
Gary
Just to reply to your first point:
$k(\xi)$ and $\kappa$ are completely different quantities. Equation (30) could be written as $\widetilde{\kappa}=k(\xi)+\kappa/|f'|$.
Vasco
PS
Can you give a reference to the corrected document please? The original reference points to the uncorrected version.
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Gary
GaryVasco
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Post by Gary on Apr 4, 2016 17:56:17 GMT
Gary Just to reply to your first point: $k(\xi)$ and $\kappa$ are completely different quantities. Equation (30) could be written as $\widetilde{\kappa}=k(\xi)+\kappa/|f'|$. Vasco Vasco, Yes, thank you. I realize that. That is why I use the parenthesis on the $k(\xi)$ function to tell them apart. It's the k characters themselves that I have trouble distinguishing when reading. Anyway, I have no problem using them correctly. It's just the letters. Gary |
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Post by Admin on Apr 4, 2016 18:03:12 GMT
Gary Just to reply to your first point: $k(\xi)$ and $\kappa$ are completely different quantities. Equation (30) could be written as $\widetilde{\kappa}=k(\xi)+\kappa/|f'|$. Vasco Vasco, Yes, thank you. I realize that. That is why I use the parenthesis on the $k(\xi)$ function to tell them apart. It's the k characters themselves that I have trouble distinguishing when reading. Anyway, I have no problem using them correctly. It's just the letters. Gary Gary Sorry, I misunderstood you. I thought you meant "see" in the mathematical sense. Now I'll have a look at your other comments from the same post. Vasco |
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Gary
GaryVasco
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Post by Gary on Apr 4, 2016 19:19:25 GMT
Gary Just to reply to your first point: $k(\xi)$ and $\kappa$ are completely different quantities. Equation (30) could be written as $\widetilde{\kappa}=k(\xi)+\kappa/|f'|$. Vasco PS Can you give a reference to the corrected document please? The original reference points to the uncorrected version. Vasco, Not sure how that happened. nh.ch5.ex17.vi.eToz.pdf (143.64 KB) Gary |
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Post by Admin on Apr 6, 2016 15:54:27 GMT
I have amended my solution to part (vi) of exercise 17 chapter 5. The main changes are for the cases $f=z^m$ and the Mobius case, where there are some changes and additions to the maths and to the diagrams. You can download the new version here. This document may still need amending. I hope to soon include it in my version of the complete solution to exercise 17, all parts, (i) - (vi). Vasco |
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