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Post by Admin on Nov 28, 2016 14:30:22 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Jan 26, 2017 1:30:51 GMT
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Post by Admin on Jan 26, 2017 6:56:45 GMT
Gary
Just on your first point in your document:
...But if taken literally, that would leave $A$ and $B$ unrotated...". This doesn't make sense to me.
Look at figure 27 on page 308. Imagine it being animated as the arrows suggest. Figure 1 in your document and the one on page 336 are the same as this diagram, it's just that the diagram provided for the exercise and yours have less detail. Notice the two h-lines labelled $L_1$ and $L_2$ in figure 27. As illustrated, under an h-rotation $L_1$ becomes $L_2$, and as you can see, the points $A$ and $B$ in your diagram move eastwards along the horizon as a result of the movement of h-line $AB$ analogous to the movement of $L_1$. Eventually, as the h-rotation continues, $L_1$ becomes the vertical line through the centre of rotation and so, in your diagram, all the points $A, a, b, B$ on the original h-line now lie on the vertical h-line in the original order going up the line, with B at infinity. With further rotation the vertical h-line $L_1$ returns to its original position with the points in reverse order.
So the hint is fine as it is in the book I think, and having reviewed my solution based on this hint, I feel confident that it is correct.
Regarding your note on page 2: "We are apparently mapping $b$ to $i$,...". I don't think this is the case, since we are not using the same points as on page 154.
As for part (ii), you have sown the seeds of doubt in my mind, so I am currently having another look at it.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 26, 2017 17:31:34 GMT
Gary Just on your first point in your document: ...But if taken literally, that would leave $A$ and $B$ unrotated...". This doesn't make sense to me. Look at figure 27 on page 308. Imagine it being animated as the arrows suggest. Figure 1 in your document and the one on page 336 are the same as this diagram, it's just that the diagram provided for the exercise and yours have less detail. Notice the two h-lines labelled $L_1$ and $L_2$ in figure 27. As illustrated, under an h-rotation $L_1$ becomes $L_2$, and as you can see, the points $A$ and $B$ in your diagram move eastwards along the horizon as a result of the movement of h-line $AB$ analogous to the movement of $L_1$. Eventually, as the h-rotation continues, $L_1$ becomes the vertical line through the centre of rotation and so, in your diagram, all the points $A, a, b, B$ on the original h-line now lie on the vertical h-line in the original order going up the line, with B at infinity. With further rotation the vertical h-line $L_1$ returns to its original position with the points in reverse order. So the hint is fine as it is in the book I think, and having reviewed my solution based on this hint, I feel confident that it is correct. Regarding your note on page 2: "We are apparently mapping $b$ to $i$,...". I don't think this is the case, since we are not using the same points as on page 154. As for part (ii), you have sown the seeds of doubt in my mind, so I am currently having another look at it. Vasco Vasco, I didn't express myself very clearly. I do understand how the solution relates to the diagram on p. 308, and I do agree that your solution is correct. In fact, I saw immediately that it was correct and that I would have to try to understand why. I concluded it wasn't just b rotating around a, it was a rotation of the entire plane as in p. 308. It seemed to me that the hint had misled me. Having thought it over a bit more, I see that the fault was my own. Whenever one thing in the plane is rotated by a $M\ddot{o}bius$ transformation, everything rotates. We discussed this before. I should have realized immediately that A and B would be rotated into alignment by the same rotation that brought b into the position directly above a. Since I have made this mistake repeatedly, I have to wonder why. There are many situations in which a plot requires rotating a vector z to a certain angle without changing the remainder of the plot. The new vector often takes the form of $z r e^{i\theta} + p$. Perhaps one gets habituated to this kind of thinking and $M\ddot{o}bius$ transformations require a more global way of thinking. So my take away here is: When solving problems involving $M\ddot{o}bius$ transformations, if one thing rotates, everything rotates. But even some problems involving M transformations can be solved without requiring consideration of the effect on the whole plane. On pp. 312, there is a discussion of Figure [32], which illustrates an M transformation decomposable into four reflections. It shows the rotation of $d\tilde{z}$ to $dw$ about $w$ without showing the effect on $m$, $z$, and $dz$, which must move to new locations under the rotation about $w$, but are no longer in focus because they are no longer needed for illustration. So it seems to me that situations like this differ from situations like that in exercise 22, where the algebraic solution requires a global view of the rotation of several points. You are right. My figure 1 placed point b by finding h{a, b} and plotting (a + H{a, b}i) using the equation on p. 303. Unfortunately, I mistook the inversion tildes for conjugation signs, so the calculation made it appear that b was at or near $i$. I will replace it with the new calculation, which places $\mathcal{R}(b)$ higher. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 27, 2017 16:00:58 GMT
Vasco,
I posted a new answer to part (ii).
Gary
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Post by Admin on Jan 27, 2017 16:26:21 GMT
Gary
I won't look until I've sorted mine out.
Vasco
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Post by Admin on Jan 28, 2017 7:55:24 GMT
Gary
I have re-done my proof of part (ii) of the exercise and re-published. Thanks for pointing out that my previous attempt was incorrect/incomplete.
When Needham splits an exercise into parts he almost always makes connections between them, as in this case via the cross-ratio.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 28, 2017 16:52:16 GMT
Vasco,
The new version looks correct to me and I think it satisfies the question. I did it differently because I formed the impression that he wanted (ii) solved in somewhat the same way that (i) was solved. I have no strong opinion on the issue. Your solution has the advantage of economy. Mine, if correct, demonstrates that it can be done within the PD.
Gary
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Post by Admin on Jan 30, 2017 6:30:33 GMT
Vasco, The new version looks correct to me and I think it satisfies the question. I did it differently because I formed the impression that he wanted (ii) solved in somewhat the same way that (i) was solved. I have no strong opinion on the issue. Your solution has the advantage of economy. Mine, if correct, demonstrates that it can be done within the PD. Gary Gary Your solution seems fine to me. Vasco |
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