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Post by Admin on Jul 6, 2016 9:57:36 GMT
Gary I have finally managed to produce a commentary which I think is in a fit state to be published. It is written primarily for my benefit and so might not be as neat and tidy as the exercises hopefully are, but it may nevertheless be useful to other people. Anything in it is open for discussion. Here is a link to itVasco |
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Gary
GaryVasco
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Post by Gary on Jul 11, 2016 0:21:03 GMT
Vasco,
I have a few questions and comments on your commentary.
3 Uniqueness
You mention "some specified image points". There must be a minimum number. Is it two or three?
4 Preservation of identities
Interesting and powerful. But Needham has not given us a proof has he? If he had, I think he would have referred to it. He refers to it as a demonstration.
5 Analytic Continuation via Reflection
This is useful. I think I finally understand what this is about now. We have an analytic function from P to Q, but not from P to $\overline{Q}$ or from $\overline{P}$ to $\overline{Q}$.
$f*$ gives us the analytic function from $\overline{P}$ to $\overline{Q}$.
I read the remainder quite carefully and it all looks right. It seems like quite a good set of principles to become conversant with.
Gary
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Post by Admin on Aug 10, 2016 15:33:57 GMT
Vasco, I have a few questions and comments on your commentary. 3 Uniqueness You mention "some specified image points". There must be a minimum number. Is it two or three? 4 Preservation of identities Interesting and powerful. But Needham has not given us a proof has he? If he had, I think he would have referred to it. He refers to it as a demonstration. 5 Analytic Continuation via Reflection This is useful. I think I finally understand what this is about now. We have an analytic function from P to Q, but not from P to $\overline{Q}$ or from $\overline{P}$ to $\overline{Q}$. $f*$ gives us the analytic function from $\overline{P}$ to $\overline{Q}$. I read the remainder quite carefully and it all looks right. It seems like quite a good set of principles to become conversant with. Gary Gary I just remembered that I hadn't replied to your post. 3 Uniqueness As I understand it, the important thing is that the points lie on a curve (so at least two). Specifying the image of one point would not be enough. 4 Preservation of identities I can't see where Needham refers to this as a demonstration. To me it seems that he proves it for power series and for the identity $e^x\cdot e^y=e^{x+y}$. At the end of the subsection he writes "It should be clear that this reasoning extends to any identity, even one involving more than two variables". 5 Analytic Continuation via Reflection I agree with what you have written above. Vasco |
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Gary
GaryVasco
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Post by Gary on Aug 10, 2016 17:24:12 GMT
Vasco,
Thanks for the comments.
I can't find it either, and I looked several pages back and forward. I don't know what I was thinking.
Gary
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Post by telemeter on Mar 3, 2019 17:36:09 GMT
Excellent commentary on the most compact and hence most opaque section of the book so far. Thank you.
I agree on the errors you noted re the F's. As dagger and ++ refer to the extended domain while F is defined as covering the entire space it is very hard to see what F dagger and F ++ could refer to. Your correction I concur with.
You do have a typo I think on the second correction one at the bottom of page 8. You say the book on p257 should read F(z)=F(z tilde) bar. Given your notation, should this be F(z) =F(z tilde) circumflex reflecting (no pun intended!) the reflection in L circumflex?
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Post by Admin on Mar 8, 2019 7:12:56 GMT
Excellent commentary on the most compact and hence most opaque section of the book so far. Thank you. I agree on the errors you noted re the F's. As dagger and ++ refer to the extended domain while F is defined as covering the entire space it is very hard to see what F dagger and F ++ could refer to. Your correction I concur with. You do have a typo I think on the second correction one at the bottom of page 8. You say the book on p257 should read F(z)=F(z tilde) bar. Given your notation, should this be F(z) =F(z tilde) circumflex reflecting (no pun intended!) the reflection in L circumflex? Hi Telemeter Yes I agree that there is typo at the bottom of page 8: it should read "...$F(z)=\widetilde{F(\widetilde{z})}$. This is Schwarz's Symmetry Principle." I also noticed another typo higher up on the same page: "Reflect the result across the curve $\widetilde{L}$" should be "Reflect the result across the curve $\widehat{L}$" and also on page 2 second paragraph it should read "... such that $f(z)=ze^{i\phi}$." I have made these changes and reposted. Thanks for pointing this out. Vasco |
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Post by Admin on Mar 14, 2019 15:39:12 GMT
I have made a few further changes to my document at the top of page 5 - in figure 1 I have removed the arrows and made a change in paragraph 2 to avoid confusion - $f(1/\bar{a})$ can be anywhere in the complex plane.
Vasco
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Post by mondo on Aug 14, 2023 5:15:29 GMT
Very useful document however I have some troubles understanding the mapping that sends point from $C$ to $\hat{C}$ on a real line (Figure [1] page 5 in this document). You say the mapping is $a => \frac{1}{\overline{a}} => f(\frac{1}{\overline{a}})$ but how does it send points on the circle boundary to a real line?
If the point lies on the circle boundary then the mapping $1/\overline{a}$ is an identity mapping, meaning it will map point $a$ to the same point $a$. So if we want to map it a real line we would have to take a real part if it, so the mapping would be $a => Re(a)$
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Post by Admin on Aug 15, 2023 5:12:30 GMT
Mondo
I am away from home for a few days. I will get back to you as soon as I can.
Vasco
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Post by mondo on Aug 15, 2023 6:11:36 GMT
Sure, thanks for letting me know. Enjoy your vacation  |
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Post by Admin on Aug 16, 2023 9:51:04 GMT
Very useful document however I have some troubles understanding the mapping that sends point from $C$ to $\hat{C}$ on a real line (Figure [1] page 5 in this document). You say the mapping is $a => \frac{1}{\overline{a}} => f(\frac{1}{\overline{a}})$ but how does it send points on the circle boundary to a real line? If the point lies on the circle boundary then the mapping $1/\overline{a}$ is an identity mapping, meaning it will map point $a$ to the same point $a$. So if we want to map it a real line we would have to take a real part if it, so the mapping would be $a => Re(a)$ Mondo The mapping $1/\overline{a}$ where $a$ is on the circle is not in general an identity mapping. It maps any point on the circle to another point on the circle, reflecting it in the real axis. It is only an identity mapping if the point is on the circle and on the real axis. Vasco |
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Post by mondo on Aug 16, 2023 21:36:49 GMT
Very useful document however I have some troubles understanding the mapping that sends point from $C$ to $\hat{C}$ on a real line (Figure [1] page 5 in this document). You say the mapping is $a => \frac{1}{\overline{a}} => f(\frac{1}{\overline{a}})$ but how does it send points on the circle boundary to a real line? If the point lies on the circle boundary then the mapping $1/\overline{a}$ is an identity mapping, meaning it will map point $a$ to the same point $a$. So if we want to map it a real line we would have to take a real part if it, so the mapping would be $a => Re(a)$ Mondo The mapping $1/\overline{a}$ where $a$ is on the circle is not in general an identity mapping. It maps any point on the circle to another point on the circle, reflecting it in the real axis. It is only an identity mapping if the point is on the circle and on the real axis. Vasco I don't agree with this. On top of page 125 we have this mapping well described - "the inversion interchanges the interior and exterior of $C$ (the unit circle), while each point on $C$ remains fixed". So I don't agree that it is reflecting in a real line - for this we would have to have $\frac{1}{a}$. However I think I got the point you make on page 5 of your document. $f$ is a mapping that sends points on $C$ to a points of a segment of a real axis. We don't know it's formula though. And then we say that analytic continuation for points outside $C$ is a reflection in this line. It is all easy to say in words but I am not sure if we can find $f$ that will do both - reflect points outside $C$ with points in its interior and in the same time map these on the boundary of $C$ to a segment of a real line. |
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Post by Admin on Aug 17, 2023 18:48:13 GMT
Mondo
You are right of course. I don't know why I wrote that. Apologies.
Find a mapping that does one thing and a mapping that does the other and then create a composition mapping.
If I remember rightly you did not study the chapter on Mobius transformations.
Vasco
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