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Post by Admin on May 30, 2016 8:03:05 GMT
Although for the most part Needham's book is excellent, in this last section of chapter 5, section XI Analytic continuation, there is a distinct lack of examples, in the body of the text, on the practical use of Schwarz's Symmetry Principle, and none of the exercises explore its practical uses either. I think it would have been an aid to understanding (what seem to me) the rather abstract ideas of section XI, if some examples and exercises had been included.
In fairness to Needham he does explore these ideas in a more practical situation in the last chapter of the book, chapter 12 in section V Flow Around an Obstacle.
Also, again in chapter 10, Section IV The Complex Curvature Revisited, he explores some practical applications for the ideas introduced in chapter 5 section IX.
Vasco
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on May 30, 2016 15:35:01 GMT
Vasco,
I agree with the point. More practical examples would be welcome. But do you agree that the point of the Symmetry Principle is to get from region A to region B in a conformal way when there is no analytic function from A to B, but there is one from $\tilde{A}$ to $\tilde{B}$? So we cross over to the other side, take the function from $\tilde{A}$ to $\tilde{B}$, and cross back. If this is correct, it should not be too hard to devise real examples. Of course, crossing back could be more difficult if the boundary has been altered, so that a new Schwarzian function is required.
Gary
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Post by Admin on Jun 3, 2016 9:50:04 GMT
Gary
I agree with your description above apart from one thing.
In my view there is a function from $A$ to $B$ - it's the one we are trying to find - the analytic continuation.
Remember the example in subsection 1 of section XI of chapter 5 on pages 247-8?
$G(z)=1+z+z^2+z^4+...$ which works inside the unit circle. Then
$H(z)=\frac{1}{2}\bigg[1+\bigg(\frac{z+1}{2}\bigg)+\bigg(\frac{z+1}{2}\bigg)^2+...\bigg]$ inside the circle of radius 2 centred at $z=-1$
and finally $M(z)=1/(1-z)$ which works everywhere except at $z=1$.
This example shows how different functions can represent the same geometric mapping in different regions. This example does not of course use the Schwarzian function ideas etc. to find these other functions, as this situation is fairly straightforward, but for more complicated situations the Schwarzian reflection is one way to find them. This is how I look at it.
It would be useful if we could find a few examples which use one or more of the methods introduced in this section, and then publish them in the forum.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 3, 2016 23:39:39 GMT
Gary I agree with your description above apart from one thing. In my view there is a function from $A$ to $B$ - it's the one we are trying to find - the analytic continuation. Remember the example in subsection 1 of section XI of chapter 5 on pages 247-8? $G(z)=1+z+z^2+z^4+...$ which works inside the unit circle. Then $H(z)=\frac{1}{2}\bigg[1+\bigg(\frac{z+1}{2}\bigg)+\bigg(\frac{z+1}{2}\bigg)^2+...\bigg]$ inside the circle of radius 2 centred at $z=-1$ and finally $M(z)=1/(1-z)$ which works everywhere except at $z=1$. This example shows how different functions can represent the same geometric mapping in different regions. This example does not of course use the Schwarzian function ideas etc. to find these other functions, as this situation is fairly straightforward, but for more complicated situations the Schwarzian reflection is one way to find them. This is how I look at it. It would be useful if we could find a few examples which use one or more of the methods introduced in this section, and then publish them in the forum. Vasco Vasco, I take your point regarding the existence of a continuation function, but it seems to lead into epistemology. Thank you for reminding me of that context for Section 5. I'm grappling a bit with the exposition. When one refers to the "same geometric mapping" or the "complete analytic continuation of the mapping" or "the entire geometric mapping", it is understood that one or another of the continued functions will map only a region of overlap (i.e. G(z)). The overlap of two or more functions occurs in both the z and w planes. The mappings are said to be "the same" because there is an identical mapping within the domain of overlap, but then H(z) extends or "continues" the domain and the range of the mapping and 1/(1-z) continues it further. Is that correct? Analytic continuation appears to have potential applications, so I second the suggestion in your last sentence. Gary |
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Post by Admin on Jun 5, 2016 15:51:39 GMT
Gary I agree with your description above apart from one thing. In my view there is a function from $A$ to $B$ - it's the one we are trying to find - the analytic continuation. Remember the example in subsection 1 of section XI of chapter 5 on pages 247-8? $G(z)=1+z+z^2+z^4+...$ which works inside the unit circle. Then $H(z)=\frac{1}{2}\bigg[1+\bigg(\frac{z+1}{2}\bigg)+\bigg(\frac{z+1}{2}\bigg)^2+...\bigg]$ inside the circle of radius 2 centred at $z=-1$ and finally $M(z)=1/(1-z)$ which works everywhere except at $z=1$. This example shows how different functions can represent the same geometric mapping in different regions. This example does not of course use the Schwarzian function ideas etc. to find these other functions, as this situation is fairly straightforward, but for more complicated situations the Schwarzian reflection is one way to find them. This is how I look at it. It would be useful if we could find a few examples which use one or more of the methods introduced in this section, and then publish them in the forum. Vasco Vasco, I take your point regarding the existence of a continuation function, but it seems to lead into epistemology. Thank you for reminding me of that context for Section 5. I'm grappling a bit with the exposition. When one refers to the "same geometric mapping" or the "complete analytic continuation of the mapping" or "the entire geometric mapping", it is understood that one or another of the continued functions will map only a region of overlap (i.e. G(z)). The overlap of two or more functions occurs in both the z and w planes. The mappings are said to be "the same" because there is an identical mapping within the domain of overlap, but then H(z) extends or "continues" the domain and the range of the mapping and 1/(1-z) continues it further. Is that correct? Analytic continuation appears to have potential applications, so I second the suggestion in your last sentence. Gary Gary What you say seems right to me. I think of the geometric mapping existing as an entity in itself and we try to find functions to describe it. Its existence does not depend on our finding these functions. In fact we may be unable to find one as Needham says in the first sentence of the introduction to section XI. They are for our convenience. In the example in the text we have three different functions describing the same geometric mapping. Each analytic continuation is unique and so is $G$. So $G$ is the only function on the unit circle which gives us the geometric mapping we want, $H$ is the only function on the larger circle which gives us the same mapping on the unit circle as before and also extends it to the larger circle. The final continuation, the Mobius transformation, is the only function on the entire complex plane which gives us the same mapping on the two circles as before and extends it to the entire complex plane. The next two subsections, 2 Rigidity and 3 Uniqueness take these two intuitive ideas from subsection 1, that the rigidity imposed by the mapping being an amplitwist means that it must grow in a unique way, and explores them further. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 5, 2016 18:04:00 GMT
Vasco,
This is a perspective I had not fully absorbed before, though I think I have seen bits of it in your previous postings. I like it.
Gary
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