|
|
Post by Admin on Dec 25, 2016 13:17:24 GMT
|
|
Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on Jan 11, 2017 1:58:01 GMT
Vasco, Here is my take. There might be something on interest in (i). My part (ii) followed yours pretty closely. The web page on Beltrami looks very interesting. nh.ch6.ex16.pdf (233.07 KB) Gary |
|
|
|
Post by Admin on Jan 11, 2017 17:14:34 GMT
Gary
I don't understand your Figure 3 on page 1, which presumably should be labelled figure 1 since I think you refer to it on line 2 of paragraph 3 page 1. Why have you drawn a hemisphere? Are you using the hemisphere model here? Also, I think that when Needham talks about tractrix generators he is always talking about the pseudosphere.
I don't see part (i) of this question as a tautology. Tractrix generators are represented as vertical lines on the map, but in itself that doesn't mean other geodesics on the pseudosphere do not go all the way up to the top of the pseudosphere.
In part (ii) the reason we can write $-\ln(\sin\alpha)$ as $|\ln(\sin\alpha)|$ is because since $\alpha$ is acute $\sin\alpha$ is always $\leq 1$ and so $\ln(\sin\alpha)$ is always negative and therefore $-\ln(\sin\alpha)$ is always positive. The value of $\sigma$ can never be negative.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Jan 11, 2017 22:13:54 GMT
Gary I don't understand your Figure 3 on page 1, which presumably should be labelled figure 1 since I think you refer to it on line 2 of paragraph 3 page 1. Why have you drawn a hemisphere? Are you using the hemisphere model here? Also, I think that when Needham talks about tractrix generators he is always talking about the pseudosphere. I don't see part (i) of this question as a tautology. Tractrix generators are represented as vertical lines on the map, but in itself that doesn't mean other geodesics on the pseudosphere do not go all the way up to the top of the pseudosphere. In part (ii) the reason we can write $-\ln(\sin\alpha)$ as $|\ln(\sin\alpha)|$ is because since $\alpha$ is acute $\sin\alpha$ is always $\leq 1$ and so $\ln(\sin\alpha)$ is always negative and therefore $-\ln(\sin\alpha)$ is always positive. The value of $\sigma$ can never be negative. Vasco Vasco, The figure was mislabeled. It should be Figure 1 rather than 3. The reason that I used the hemisphere model (see p. 235, $\mathbb{P}$ 1 and Figure [41]) is that the failure to reach the top ($\infty$) seemed too obvious with the Poincare upper half plane. The question specified that the extensions should be verified "mathematically." The previous problems involved projections and the hemisphere model requires projections, so I thought that might be applicable to this problem. I think it is applicable, though quite possibly not necessary. I haven't studied your answer yet, but will do so soon. Having since done part (ii), I can now imagine what form the mathematical verification might take on the UHP alone. Regarding your point: "I don't see part (i) of this question as a tautology. Tractrix generators are represented as vertical lines on the map, but in itself that doesn't mean other geodesics on the pseudosphere do not go all the way up to the top of the pseudosphere.": It appeared to me that the pseudosphere extends infinitely in the Z axis, which becomes the y axis in the map (p. 297). Geodesics other than the tractrix generators are (in the map) semicircles that have some finite point at their maximum y value. If that maximum point were to be $\infty$, they would become (?two) tractrix generators. Point re. $-\ln(\sin\alpha)$ is noted. Thank you. Gary |
|
|
|
Post by Admin on Jan 12, 2017 7:57:37 GMT
Gary
As a result of re-reading my solution to exercise 16, while reading your solution, I decided that my explanation of part (i) was not watertight and so I have rephrased paragraph 3 of part (i) and republished.
Vasco
|
|
|
|
Post by Admin on Jan 12, 2017 8:24:16 GMT
Gary
I agree that the PD model and the hemisphere model and others can all be applied to this problem. It's just that I don't think they are suited to finding a solution to the exercise. Thinking about the PD model I found that the h-lines which correspond to half-lines in the PUHP all emanate from $-i$ and remind me of the ridges on a cockle shell all emanating from one point. However I don't see a way to the solution using this or its projection onto the hemisphere.
I think that if the 'maximum point' on a semicircular h-line goes to infinity then it becomes a half-line and corresponds to a tractrix generator on the pseudo sphere as before [see the last paragraph on page 302: "Before we..." and especially the last three lines of the paragraph at the top of page 303.]
Vasco.
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Jan 15, 2017 8:38:45 GMT
Vasco,
I would agree with respect to the PD, but applying the hemisphere to the problem seemed reasonably straightforward.
Gary
|
|
|
|
Post by Admin on Jan 15, 2017 15:46:46 GMT
Vasco, I would agree with respect to the PD, but applying the hemisphere to the problem seemed reasonably straightforward. Gary Gary I have read part (i) in your document again and I now see what you mean. For some reason I missed it on first reading. Vasco |
|
