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Post by Admin on Sept 5, 2016 9:50:20 GMT
Gary I have now published my solution to exercise 2 here.One thing that confused me for a while when doing this exercise was the following sentence in the exercise It seems to me that it can be interpreted in 2 ways: 1. that the word 'restriction' applies to the points that are to be transformed, or 2. that it applies to the points after the transformation. I have decided that Needham intends the meaning in 1. If you don't interpret it this way it can lead to confusion, at least for me! Vasco |
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Gary
GaryVasco
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Post by Gary on Sept 5, 2016 12:50:17 GMT
Vasco,
I think that works for me, but I'll have to study it a bit later.
I took a look at it and I agree with you that the restriction to $\Sigma$ applies to the points to be transformed, but I don't think it makes much difference, because the starting point and the induced point are restricted to $\Sigma$ or to $\mathbb{C}$. On p. 142, in paragraph 3, Needham has If K is the sphere of radius $\sqrt{2}$ centred at N, then stereographic projection is the restriction to $\mathbb{C}$ or $\Sigma$ of inversion in K. Two of the three items that we are to transform, $a$ and $R_{\Pi}(a)$ are on $\Sigma$. With $\Pi$ I chose points on $\Sigma$ to meet the restriction.
Gary
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Post by Admin on Sept 7, 2016 15:51:28 GMT
Gary
Which means that these are the points on the circle of intersection of $\Pi$ with $\Sigma$. Agreed?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 7, 2016 19:08:15 GMT
Vasco,
Yes, as you can see in my figures which are posted.
Gary
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Post by Admin on Sept 8, 2016 6:22:53 GMT
Gary
I have just finished studying your solution and the very useful 3-D diagrams. They give me confidence that my own solution is OK.
On page 2 second to last paragraph I'm not sure what $\mathcal{I}_{\mathbb{C}}(\widetilde{a})$ means.
Also on page 3 paragraph (c) you say that the two reflections do not equal a conformal transformation because $\mathcal{I}_K$ is conformal. I thought that it was anticonformal, making the two reflections combination conformal.
I think there is a typo in the last paragraph of the document line 1: It should read "..., because $\mathcal{I}_K$ is a..."
Vasco
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Gary
GaryVasco
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Post by Gary on Sept 8, 2016 17:04:23 GMT
Vasco,
I think I missed the fact that you had posted this exercise. I will reply to your comments first and then take a look.
I actually wrote $\mathcal{I}_{C}(\hat{a})$. It is just inversion in the unit circle on $\mathbb{C}$. I think I should have written the inversion as, you did: $\mathcal{I}_{C}$ I don't think I put a tilde directly over any "a", so if there is one in the document you are consulting, it must be an old document. I'm looking at one I downloaded from the posting this morning. It's dated Sept 6, 4:45 pm.
See the quote after the first paragraph on p. 141 and accompanying explanation. I intended to cite this and forgot.
Noted. Will fix. Thanks. I also noticed that in Mathematica, if $\hat{a}$ is italicized (which is done automatically when the hat is added and must be undone), it looks like a grave. I have tried to fix them all.
Now I have to hunt down and fix every page where I wrote "stereoscopic" instead of "stereographic"!
Gary
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Gary
GaryVasco
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Post by Gary on Sept 8, 2016 22:39:20 GMT
Vasco,
After reading your answer, I added a note to the following effect to my own:
Afterward: After reading Vasco's answer, I see that I missed the point. Vasco wrote: Now apply the three-dimensional preservation of symmetry result (13). [p. 135] Since a and $\mathfrak{R}_{\Pi}(a)$ are symmetric with respect to $\Pi$, their stereographic images $\mathcal{I}_K(a)$ and and $\mathcal{I}_K(\mathfrak{R}_{\Pi}(a))$ will be symmetric with respect to the stereographic image of $\mathcal{I}_K(\Pi)$. I think what I wrote above was not incorrect, but I had forgotten the wording of the three-dimensional preservation of symmetry principle, which would have made it easier. My looser reasoning was more along these lines: Since all the operations that produced $\hat{a}$ and $\widehat{\mathfrak{R}}_{\Pi}(a)$ were symmetric, then these two points lying on opposite sides of the induced blue circle in $\mathbb{C}$ must be symmetric. When I tested the idea empirically, I found that they are indeed.
Gary
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Post by Admin on Sept 9, 2016 7:44:51 GMT
Gary
I'm not quite sure why you missed the fact that I had published my solution to exercise 2. In future, just to be sure, I'll send you a private message.
The reason I used the 'preservation of symmetry argument' is because in exercise 2 Needham asks us to "Explain (18) by generalizing the argument that was used to obtain the special case (17) on p. 143." - my italics.
Vasco
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Post by Admin on Sept 9, 2016 7:59:43 GMT
Gary
My mistake putting a tilde over the $a$, so it's just a matter of changing $\mathcal{I}_{\mathbb{C}}$ to $\mathcal{I}_{C}$.
I can't find the quote on page 141(?) that you refer to, but on page 131 Needham points out that inversion in a circle is anticonformal.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 9, 2016 21:48:51 GMT
Vasco,
Also on page 3 paragraph (c) you say that the two reflections do not equal a conformal transformation because K
My mistake this time. The quote is on p. 142, but the discussion begins on the last paragraph of page 141. Perhaps this is a matter of perspective. Any inversion in a sphere is also an inversion in circle, which would be anticonformal in the plane of the circle, but on this page we find that stereographic projection is conformal if the circle on the sphere is viewed from inside the sphere. Perhaps I was wrong to label it $\mathcal{I}_K$, which might imply inversion in a circle. What do you think?
Gary
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