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Post by Admin on Mar 21, 2016 13:11:44 GMT
I have produced a short document which outlines a possible approach to solving part (vi) of exercise 17. Here is a link to it. I would appreciate any feedback. Note that this is not a solution for the four mappings, just a possible way of achieving a solution. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 23, 2016 16:08:07 GMT
I have produced a short document which outlines a possible approach to solving part (vi) of exercise 17. Here is a link to it. I would appreciate any feedback. Note that this is not a solution for the four mappings, just a possible way of achieving a solution. Vasco Vasco, Here are my thoughts on the document. “Stage 2 Apply the mapping f to this shape to produce its image $\tilde{Q}$. This produces a similar shape by amplitwisting the shape $Q$. In general the size of the shape will be changed and it will be rotated about its centre.” This sounds right. Needham definitely implies that $\tilde{Q}$ is the result of $f(Q)$. “Stage 3 We now translate $Q$ by $\xi$. …” Yes, but we could have done that immediately. One would only have to postpone it if one were applying f to the whole of $\mathbb{C}$, but we are only applying it to $\tilde{Q}$. Oh, I see later on that you are going about this one ray at a time. That seems like a good idea. “$\xi$ will be amplitwisted by f to an arrow emanating from $\tilde{Q}$, and this is the translation referred to in the introductory paragraph above”. Sounds right. In Needham’s diagram, $\xi$ looks larger than the little tangents we have been drawing for infinitesimal numbers, and that was misleading for me. Also, it’s important that $\xi$ be drawn un-amplitwisted on the LHS and amplitwisted on the RHS. I think it is important to state that we are taking $f’(q)\xi$ and plotting the arrow from $\tilde{q}$, where $\tilde{q}$ is the center of $\tilde{Q}$. Then, when we take $f’(q)\epsilon$, where $\epsilon$ is the bar or lever on $Q$, we plot $\tilde{\epsilon}$ from $\tilde{q} + \tilde{\xi}$. Programming this is tricky. The most straightforward way to do it would be to plot the small circles as points (or lines drawn on recalculated points), but that gets cumbersome managing all the lists of points. Since the function is analytic and conformal, all the points of a circle (the ends of rays) have the same twist, so the result is a circle and one can just calculate the amplification on a radius and plot the circle at the correct center point. The bar will reveal the rotation. I need more time to digest the last paragraph. Gary |
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Post by Admin on Mar 23, 2016 19:39:43 GMT
I have produced a short document which outlines a possible approach to solving part (vi) of exercise 17. Here is a link to it. I would appreciate any feedback. Note that this is not a solution for the four mappings, just a possible way of achieving a solution. Vasco Vasco, Here are my thoughts on the document. “Stage 2 Apply the mapping f to this shape to produce its image $\tilde{Q}$. This produces a similar shape by amplitwisting the shape $Q$. In general the size of the shape will be changed and it will be rotated about its centre.” This sounds right. Needham definitely implies that $\tilde{Q}$ is the result of $f(Q)$. “Stage 3 We now translate $Q$ by $\xi$. …” Yes, but we could have done that immediately. One would only have to postpone it if one were applying f to the whole of $\mathbb{C}$, but we are only applying it to $\tilde{Q}$. Oh, I see later on that you are going about this one ray at a time. That seems like a good idea. “$\xi$ will be amplitwisted by f to an arrow emanating from $\tilde{Q}$, and this is the translation referred to in the introductory paragraph above”. Sounds right. In Needham’s diagram, $\xi$ looks larger than the little tangents we have been drawing for infinitesimal numbers, and that was misleading for me. Also, it’s important that $\xi$ be drawn un-amplitwisted on the LHS and amplitwisted on the RHS. I think it is important to state that we are taking $f’(q)\xi$ and plotting the arrow from $\tilde{q}$, where $\tilde{q}$ is the center of $\tilde{Q}$. Then, when we take $f’(q)\epsilon$, where $\epsilon$ is the bar or lever on $Q$, we plot $\tilde{\epsilon}$ from $\tilde{q} + \tilde{\xi}$. Programming this is tricky. The most straightforward way to do it would be to plot the small circles as points (or lines drawn on recalculated points), but that gets cumbersome managing all the lists of points. Since the function is analytic and conformal, all the points of a circle (the ends of rays) have the same twist, so the result is a circle and one can just calculate the amplification on a radius and plot the circle at the correct center point. The bar will reveal the rotation. I need more time to digest the last paragraph. Gary Gary Thanks for looking at my document and for your comments. I don't think we can calculate the position of the bar/lever in the way you suggest because $f’(q)\epsilon$ is $\tilde{\epsilon}$ when $\epsilon$ emanates from $q$ not when it emanates from $q+\epsilon$. I think the way to do this is using the ideas in the last paragraph of my document. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 23, 2016 22:54:35 GMT
Vasco, Here are my thoughts on the document. “Stage 2 Apply the mapping f to this shape to produce its image $\tilde{Q}$. This produces a similar shape by amplitwisting the shape $Q$. In general the size of the shape will be changed and it will be rotated about its centre.” This sounds right. Needham definitely implies that $\tilde{Q}$ is the result of $f(Q)$. “Stage 3 We now translate $Q$ by $\xi$. …” Yes, but we could have done that immediately. One would only have to postpone it if one were applying f to the whole of $\mathbb{C}$, but we are only applying it to $\tilde{Q}$. Oh, I see later on that you are going about this one ray at a time. That seems like a good idea. “$\xi$ will be amplitwisted by f to an arrow emanating from $\tilde{Q}$, and this is the translation referred to in the introductory paragraph above”. Sounds right. In Needham’s diagram, $\xi$ looks larger than the little tangents we have been drawing for infinitesimal numbers, and that was misleading for me. Also, it’s important that $\xi$ be drawn un-amplitwisted on the LHS and amplitwisted on the RHS. I think it is important to state that we are taking $f’(q)\xi$ and plotting the arrow from $\tilde{q}$, where $\tilde{q}$ is the center of $\tilde{Q}$. Then, when we take $f’(q)\epsilon$, where $\epsilon$ is the bar or lever on $Q$, we plot $\tilde{\epsilon}$ from $\tilde{q} + \tilde{\xi}$. Programming this is tricky. The most straightforward way to do it would be to plot the small circles as points (or lines drawn on recalculated points), but that gets cumbersome managing all the lists of points. Since the function is analytic and conformal, all the points of a circle (the ends of rays) have the same twist, so the result is a circle and one can just calculate the amplification on a radius and plot the circle at the correct center point. The bar will reveal the rotation. I need more time to digest the last paragraph. Gary Gary Thanks for looking at my document and for your comments. I don't think we can calculate the position of the bar/lever in the way you suggest because $f’(q)\epsilon$ is $\tilde{\epsilon}$ when $\epsilon$ emanates from $q$ not when it emanates from $q+\epsilon$. I think the way to do this is using the ideas in the last paragraph of my document. Vasco Vasco, I didn't write $q+\epsilon$, but I'll have another look at the last paragraph. Gary |
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Post by Admin on Mar 24, 2016 19:57:50 GMT
Vasco, I didn't write $q+\epsilon$ but I'll have another look at the last paragraph. Gary Gary Sorry, my mistake. Vasco |
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Post by Admin on Mar 24, 2016 20:37:21 GMT
Gary
I think I have managed to produce the equivalent of 21 for $f=e^z$ and $f=\log z$ and am currently working on $f=z^m$ and the Mobius transformation. I will publish them as soon as I can.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 24, 2016 20:47:06 GMT
Vasco,
I am using $\xi$ for the spokes and $\epsilon$ for the bars/levers. No matter, because I'm still floundering around with this problem. I now have a sensible looking plot for log(z) in (vi) done with amplitwist on both the RHS spokes and bars, but I can't say I thoroughly understand it. But at least it amplifies and rotates in the right places. I am least sure about what point to send as an argument to f'(p) for amplitwist when calculating spokes and bars/levers. I'm thinking about the last paragraph of your document but I haven't quite got the sense of it yet. That is, why $|f'||\xi|\mathcal{R}$ and $1 + |f'||\xi|\mathcal{E}$? The first parts of both --- $|f'||\xi|$ --- look like amplitwist on $|\xi|$ and what you say in your document suggests that you would use $f'(q)$ for calculating both spokes and bars. If this is right, the question becomes, why is this multiplied by the rate of rotation in the first case and by the linear size in the second.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 24, 2016 21:48:36 GMT
Thoughts and questions re. Exercise 17 (vi)
On Mar 21, Vasco posted a suggested approach. I don't fully understand it yet, so I am taking a different approach trying to use amplitwist and getting results that satisfy (31) on p. 240, but I'm not satisfied that I understand my own results.
Figure [18] shows the curve K and derivatives of a function f’ with the purpose of illustrating the derivation of $\tilde{\kappa}$. In Ex. 17 (vi) we have the function (actually several), but there is no curve K. We just have infinitesimal shape $\mathcal{Q}$ at some unspecified point. Under the analytical mapping f, this is translated to a similar shape $\tilde{\mathcal{Q}}$. So it seems a case in which we can’t know $\kappa$, but $\tilde{\kappa}$, $\mathcal{K}$, and amplitwist can be calculated for f.
What puzzles me is, what are we doing when we move $\mathcal{Q}$ along some infinitesimal spoke $\xi$? For log(z), if this were centered at the origin, it would move a line in a horizontal direction. I moved it off the origin. Two times the radius of $\mathcal{Q}$ seems to be enough to obtain a circular image for $\mathcal{Q}$.
In previous exercises, we have applied f to a curve (or line) to observe the change in curvature, but in this problem, there is no preexisting curve and all the lines are infinitesimal. That suggests that we should plot all the spokes and bars/levers on the RHS by amplitwisting the ones on the LHS. If that is correct, the main question becomes how to choose p for f’(p) and |f’(p)|. Does one use different values of p when applying f’ to the spokes and |f’| to the bars? My first thought was to use f'(q) throughout on the theory that the amplification and twist should be the same for the spokes ($\xi$) and the radii of the peripheral circles, but that didn't produce coherent results.
Gary
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Post by Admin on Mar 25, 2016 7:05:06 GMT
Vasco, I am using $\xi$ for the spokes and $\epsilon$ for the bars/levers. No matter, because I'm still floundering around with this problem. I now have a sensible looking plot for log(z) in (vi) done with amplitwist on both the RHS spokes and bars, but I can't say I thoroughly understand it. But at least it amplifies and rotates in the right places. I am least sure about what point to send as an argument to f'(p) for amplitwist when calculating spokes and bars/levers. I'm thinking about the last paragraph of your document but I haven't quite got the sense of it yet. That is, why $|f'||\xi|\mathcal{R}$ and $1 + |f'||\xi|\mathcal{E}$? The first parts of both --- $|f'||\xi|$ --- look like amplitwist on $|\xi|$ and what you say in your document suggests that you would use $f'(q)$ for calculating both spokes and bars. If this is right, the question becomes, why is this multiplied by the rate of rotation in the first case and by the linear size in the second. Gary Gary The definitions of $\mathcal{R}$ and $\mathcal{E}$ say that they are rates of rotation and expansion with respect to the distance travelled by $\widetilde{Q}$. So to obtain the actual rotation and expansion we need to multiply by the distance travelled by $\widetilde{Q}$, namely $|f'||\xi|$, which as you say is the amplification of $|\xi|$ under $f$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 25, 2016 15:06:09 GMT
Vasco,
Ah yes. But I am wondering now if this is equivalent to a calculation done with just amplitwist on infinitesimal spokes and bars. It seems we should be able to make some equation between the two methods. Or is there something about this example that makes it wrong to use the usual amplitwist approach?
Gary
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Post by Admin on Mar 25, 2016 15:35:09 GMT
Gary
I have almost finished a document which explains in detail how to make the calculations necessary to produce 21. The maths is analogous to the maths required to prove (23). I hope to be able to publish it in the next 2-3 hours. If not, then early tomorrow - 6-9am UK time.
The approach is an amplitwist approach.
Vasco
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Post by Admin on Mar 25, 2016 17:53:44 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 26, 2016 3:55:23 GMT
Vasco, Regarding the document “Method for calculating the numbers required for drawing figure 21 in the book, chapter 5 page 240” This is very helpful. Section IX makes more sense to me now. I will try to implement it, but it may take a couple of days. I have some remarks. If you see something wrong, I would like to know. 3. Would it be correct to write $\tilde{\xi} = f’(c) \xi$? 4. Trivial point (1): Why not draw $\zeta$ long enough to protrude from the copy of Q at c’ as Needham does? Trivial point (2): Since q’ is a circle, why not write it as Q’ as you did at the bottom of p. 1 and on p. 2 4. para. 4 identifying c and c’ with p and q I had trouble seeing this, until I remembered that $\xi$ and $\zeta$ in [18] are infinitesimal numbers. When Needham says to move Q along $\xi$, isn’t he also saying move the point of application of f(z) along the curve $\xi$ to c + $\xi$, and more importantly, move the point of application of f’ from c to c + $\xi$. (This is the point at which I applied the equivalent of f’(c’) to my equivalent to $\zeta$ (which was $\epsilon$ that I used for the bar). Here is another little discrepancy in the comparison. In [18] $\zeta$ is drawn tangent to the curve. In [21], the bar or lever, which might be taken to correspond to $\zeta$, is drawn as a horizontal line, rather than tangent to $\xi$. I suppose that does not matter, as it will be subject to the same amplitwist at any angle. Needham just describes $\zeta$ as “another small (ultimately infinitesimal) tangent complex number” (p. 235, last paragraph). On the difference in turning angles at c and c’ (or p and q). It was easy to see that the turning angles at p and q were slightly different, because K in [18] displayed variation in curvature. Perhaps that was misleading, as the spokes in [21] did not remind me of K. As we move from c to c’ along $\xi$, we should expect from the equation for $\mathcal{K}$ that $\mathcal{K}(c)$ would differ from $\mathcal{K}(c’)$. So each of the 12 $\xi$ is like a straight version of curve K, rather like Figure [19]. p. 2, last line, “to be applied to $\mathcal{Q}$ to obtain $\tilde{Q’}$.” Here I think you are speaking of the 12 circles centered at c’ and $\tilde{c’}$, but it would help if it were more explicit. This seems enough to implement a script to plot the given functions. One final comment: We now have three symbols for curvature: $\kappa(\xi)$, $\mathcal{K} \cdot \hat{\xi}$, and $\mathcal{R}$. Why the proliferation of symbols for what seems to be all the same thing? Gary |
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Post by Admin on Mar 26, 2016 10:26:48 GMT
Gary
I have updated my document to correct the typo you detected, where I typed $q'$ instead of $Q'$, and also to make it clearer in places as a result of your comments. I have also added a diagram. The link in my first post now gives access to this new document.
3 Answer: Yes
4 I do not want to refer to figure 21 in this document as it would confuse the issue. It has nothing to do with 21 other than the fact that the ideas can be used to draw 21 - if you see what I mean. You will see in my comments below and in my document I have tried to stress this point.
Corrected the typo.
In your paragraph beginning "I had trouble..." you talk about curves. It is best to forget completely about curves when reading my document. Just think about the function $f$ on the complex plane and infinitesimal shapes.
Same comments apply to the paragraph which begins " On the difference..."
Paragraph "p. 2, last line,..." I do not want to refer to the 12 circles here, as again I think it could confuse the simple idea I am trying to illustrate (i.e. that $\widetilde{Q'}$ can be obtained by applying the amplitwist which sends $Q$ to $\widetilde{Q}$ to the shape $Q'$, and then the extra amplitwist obtained from $\sigma$ and $\alpha$, but I agree that $Q'$ and $\widetilde{Q'}$ are analogous to corresponding circles on the rims of the wheels in (21), and $Q$ and $\widetilde{Q}$ are analogous to the black hubs.
I will reply to your comments on the symbols for curvature in another post, as I would like to give it some more thought before replying.
Vasco
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Post by Admin on Mar 26, 2016 17:36:53 GMT
There were two typos in my document in the last paragraph: $\mathcal{R}$ and $\mathcal{E}$ were in the wrong expressions. I have corrected this. Please download the document again to get an updated version.
Vasco
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