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Post by Admin on Mar 16, 2016 6:22:10 GMT
Gary
As a result of our recent exchanges I would like to make some changes to my solution for parts (i) to (iv), to correct one or two things and also to hopefully clarify other things.
I think I mentioned before that I was writing a personal commentary on section IX and I have just yesterday completed it on paper at least. I have some of the earlier parts of it in electronic form and so it should not take me too long to complete it and I will then publish it. It may be of use to other people, I'm not sure. It will include my solutions to the two suggested exercises at the top of page 241, which may or may not be relevant to part (vi) of exercise 17.
I have not given much thought to solving this part of the exercise. I was hoping that after publishing the commentary I would maybe be able to tackle it.
So I will finish my commentary, publish it, and then look at part (vi) and also your comments in the above post and then get back to you with some thoughts.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 16, 2016 15:22:51 GMT
Vasco,
I look forward to seeing it.
I did a derivation of $\overline{\hat{\xi}}\mathcal{K}$ (p. 241) but haven't had time to publish it while trying to understand 5:17. I will do so if I get the time before you have published yours. It was easier than I expected, just a few lines.
Gary
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Post by Admin on Mar 16, 2016 15:40:59 GMT
Gary
Yes that one's an easy one, but I found the other one $\mathcal{E}=\boldsymbol{\widehat{\xi}\times\mathcal{K}}$ to be considerably more difficult.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 16, 2016 16:29:48 GMT
Vasco,
I haven't tried that one yet. It did look difficult. But I did do (just now) the proportionalities at the bottom of p. 239. They are the first item in the .pdf of item #20 on my web page if you ever want to look at them.
Gary
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Post by Admin on Mar 18, 2016 15:12:06 GMT
I believe I have resolved all my questions regarding Needham’s questions and your answers to the exercises 17: (i)-(iv). I have accepted the idea that when Needham says “Which?” he may have allowed for us to try out different directions of rather than requiring a correct answer immediately. My answers might differ in a few respects, but I think the basic reasoning is the same as yours. Gary I will have a look at your document as soon as I can. This all seems too complicated. Do you have any thoughts on this? Gary I do have some thoughts on this and I am writing a short document on my approach which I will publish and make available on this forum. It involves using the approach based on figure 21 on page 240 and the equations at the top of page 241. Vasco |
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Post by Admin on Mar 22, 2016 7:26:23 GMT
Gary
I have spent some time reading your solution and it seems fine apart from figure 6. I wondered why the circles were not placed on a circle about the central shape on the RHS of your diagram. The document I have published about my approach to part (vi) explains my thinking about this. I will attempt to produce figure 21 for the four mappings in exercise 17.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 22, 2016 15:58:29 GMT
Vasco,
Thank you. I am glad you are getting into (vi), as I hate to proceed with the other functions without some confidence that I'm on the right track. Actually, I have tried log(z) with mixed results, in that the amplifications conform to (31), but the rotations are different.
The reason the perimeter circles on the RHS don't fall on a circle is that I applied $e^z$ to all the circles because the description on p. 239, 240 states that Q becomes $\tilde{Q}$ "under an analytic mapping f", which is applied as Q moves along $\hat{\xi}$, which in this case is any ray from Q to $\tilde{Q}$. Actually, I just applied f to all the circle centers and then plotted the circles using the amplification of the radius. The distribution of the circles is what the computer produced. I'll recheck it.
Gary
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Post by Admin on Mar 22, 2016 17:00:43 GMT
Gary
The discs on the RHS must lie on a circle because $f$ is an analytic mapping and so when $Q$ translates by $\xi$, $\widetilde{Q}$ translates by $f'\xi$ and $f'$ is evaluated at the same point for every $\xi$, i.e. the point where the shape $Q$ is situated.
$\xi$ is not a ray from $Q$ to $\widetilde{Q}$, but a ray from $Q$ to any one of the copies of $Q$ on the LHS. $f'\xi$ is a ray from $\widetilde{Q}$ to any of its copies on the RHS. If you think of $\xi$ as being the spokes of a wheel from the hub ($Q$) to the copies of $Q$ on the rim of the wheel, then these spokes are amplitwisted to the spokes on the wheel with hub $\widetilde{Q}$. The wheel size is amplified and the wheel is turned about its hub.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 22, 2016 18:10:42 GMT
Gary The discs on the RHS must lie on a circle because $f$ is an analytic mapping and so when $Q$ translates by $\xi$, $\widetilde{Q}$ translates by $f'\xi$ and $f'$ is evaluated at the same point for every $\xi$, i.e. the point where the shape $Q$ is situated. $\xi$ is not a ray from $Q$ to $\widetilde{Q}$, but a ray from $Q$ to any one of the copies of $Q$ on the LHS. $f'\xi$ is a ray from $\widetilde{Q}$ to any of its copies on the RHS. If you think of $\xi$ as being the spokes of a wheel from the hub ($Q$) to the copies of $Q$ on the rim of the wheel, then these spokes are amplitwisted to the spokes on the wheel with hub $\widetilde{Q}$. The wheel size is amplified and the wheel is turned about its hub. Vasco Vasco, I didn't mean to say from $Q$ to $\widetilde{Q}$; I meant to say from $Q$ to peripheral circles on the LHS, so I agree with the rest of the second sentence. Regarding the first part, I think I was having trouble thinking of $\xi$ as an infinitesimal, when clearly it is, as shown in Figure [18] and explained elsewhere. I am still having trouble reconciling the fact that if we call a spoke an infinitesimal, we multiply by f', but if we don't call it an infinitesimal, we would just apply f. Gary |
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Post by Admin on Mar 22, 2016 18:56:37 GMT
Gary The discs on the RHS must lie on a circle because $f$ is an analytic mapping and so when $Q$ translates by $\xi$, $\widetilde{Q}$ translates by $f'\xi$ and $f'$ is evaluated at the same point for every $\xi$, i.e. the point where the shape $Q$ is situated. $\xi$ is not a ray from $Q$ to $\widetilde{Q}$, but a ray from $Q$ to any one of the copies of $Q$ on the LHS. $f'\xi$ is a ray from $\widetilde{Q}$ to any of its copies on the RHS. If you think of $\xi$ as being the spokes of a wheel from the hub ($Q$) to the copies of $Q$ on the rim of the wheel, then these spokes are amplitwisted to the spokes on the wheel with hub $\widetilde{Q}$. The wheel size is amplified and the wheel is turned about its hub. Vasco Vasco, I didn't mean to say from $Q$ to $\widetilde{Q}$; I meant to say from $Q$ to peripheral circles on the LHS, so I agree with the rest of the second sentence. Regarding the first part, I think I was having trouble thinking of $\xi$ as an infinitesimal, when clearly it is, as shown in Figure [18] and explained elsewhere. I am still having trouble reconciling the fact that if we call a spoke an infinitesimal, we multiply by f', but if we don't call it an infinitesimal, we would just apply f. Gary Gary We apply $f'$ to infinitesimal arrows or vectors and $f$ to points. If we think of $\xi$ as a point at a distance $|\xi|$ from the origin, then $f(\xi)$ is the corresponding point in the mapped plane. If we think of $\xi$ (not infinitesimal) as a complex number (arrow) emanating from a point $p$ anywhere in the plane and ending somewhere else say $q$ in the plane, then we can calculate $f(p)$ and $f(q)$ and we can then draw an arrow from $f(p)$ to $f(q)$. The mapping $f$ has transformed $\xi$ or $q-p$ into $f(q)-f(p)$. If $\xi$ is infinitesimal then we can say $f(q)-f(p)=f'(p)(q-p)=f'(p)\xi$. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 22, 2016 20:29:49 GMT
Vasco, I didn't mean to say from $Q$ to $\widetilde{Q}$; I meant to say from $Q$ to peripheral circles on the LHS, so I agree with the rest of the second sentence. Regarding the first part, I think I was having trouble thinking of $\xi$ as an infinitesimal, when clearly it is, as shown in Figure [18] and explained elsewhere. I am still having trouble reconciling the fact that if we call a spoke an infinitesimal, we multiply by f', but if we don't call it an infinitesimal, we would just apply f. Gary Gary We apply $f'$ to infinitesimal arrows or vectors and $f$ to points. If we think of $\xi$ as a point at a distance $|\xi|$ from the origin, then $f(\xi)$ is the corresponding point in the mapped plane. If we think of $\xi$ (not infinitesimal) as a complex number (arrow) emanating from a point $p$ anywhere in the plane and ending somewhere else say $q$ in the plane, then we can calculate $f(p)$ and $f(q)$ and we can then draw an arrow from $f(p)$ to $f(q)$. The mapping $f$ has transformed $\xi$ or $q-p$ into $f(q)-f(p)$. If $\xi$ is infinitesimal then we can say $f(q)-f(p)=f'(p)(q-p)=f'(p)\xi$. Vasco Vasco, I thought that I understood all that. In this plot, I had placed the center of $Q$ at the origin and let $|\xi|$ = 1. If $\xi_n$ is the name of a spoke or ray to a peripheral circle, then it seemed to me that $f(\xi_n)$ should be the same as $f'(0) \xi_n$. That still seems right, so perhaps I just made a mistake in the programming. In any case, I reprogrammed it using amplitwist only on both $\xi_n$ and the radii of the peripheral circles and the result was as you predicted. Afterall, if you are viewing an infinitesimal tangent and you zoom in, then it is no longer infinitesimal, in which case, it should be OK to write $f(\xi)$. It seems a bit like Alice going down the rabbit hole. nh.ch5.ex17.pdf (248.1 KB) Gary |
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Post by Admin on Mar 22, 2016 21:03:09 GMT
Gary We apply $f'$ to infinitesimal arrows or vectors and $f$ to points. If we think of $\xi$ as a point at a distance $|\xi|$ from the origin, then $f(\xi)$ is the corresponding point in the mapped plane. If we think of $\xi$ (not infinitesimal) as a complex number (arrow) emanating from a point $p$ anywhere in the plane and ending somewhere else say $q$ in the plane, then we can calculate $f(p)$ and $f(q)$ and we can then draw an arrow from $f(p)$ to $f(q)$. The mapping $f$ has transformed $\xi$ or $q-p$ into $f(q)-f(p)$. If $\xi$ is infinitesimal then we can say $f(q)-f(p)=f'(p)(q-p)=f'(p)\xi$. Vasco Vasco, I thought that I understood all that. In this plot, I had placed the center of $Q$ at the origin and let $|\xi|$ = 1. If $\xi_n$ is the name of a spoke or ray to a peripheral circle, then it seemed to me that $f(\xi_n)$ should be the same as $f'(0) \xi_n$. That still seems right, so perhaps I just made a mistake in the programming. In any case, I reprogrammed it using amplitwist only on both $\xi_n$ and the radii of the peripheral circles and the result was as you predicted. Afterall, if you are viewing an infinitesimal tangent and you zoom in, then it is no longer infinitesimal, in which case, it should be OK to write $f(\xi)$. It seems a bit like Alice going down the rabbit hole. View AttachmentGary Gary But $f(\xi_n)$ is not equal to $f'(0) \xi_n$. If $\xi_n=1$ then $f(\xi_n)=f(1)=e$ and $f'(0)\xi_n=e^{0}\cdot 1=1$ Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 22, 2016 22:58:33 GMT
Vasco,
Yes, silly mistake.
In any case, log(z) presents a problem because (log(z))' is undefined at 0. I moved Q off the origin to a center q, and I calculated the centers of the peripheral circles on the RHS with amplitwist f'(q), but I'm still seeing rotations of the bars that don't agree with (31). The bars, I think, should be calculated with amplitwist f'(p), where p is the center of a circle on the LHS.
Gary
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Post by Admin on Mar 22, 2016 23:43:12 GMT
Gary
Have you read my document outlining how I think we should calculate figure 21? If yes, do you think it is correct?
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 23, 2016 1:57:49 GMT
Vasco,
I've been going through it and trying to implement it on the computer. It this point, I'm still not sure. I'm finding Needham's description a little vague, which makes it harder to evaluate your process. I'll get back.
Gary
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