Gary
GaryVasco
Posts: 3,352 
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Post by Gary on May 23, 2017 4:25:49 GMT
Vasco, Here is my version of 20: nh.ch7.ex20.pdf (44.03 KB) I'm already beginning to look forward to a visual approach to Cauchy's Theorem. Gary |
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Post by Admin on May 27, 2017 7:54:26 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 28, 2017 1:10:15 GMT
Vasco,
What led you to revisit the sections on analytic continuation? It's apt and interesting, but I have trouble seeing the proof of uniqueness on p 250. What does it mean to say that $g$ uniquely determines the analytic continuation of $g$ into $Q$ via $h$? Does it mean that $h$ is the only function that can be continued into $s$ and agree with $g$ on all points of $s$, even though $h$ is not the only function that can be continued into other subdomains of $P$ that don't overlap with $s$? I'm also having a bit of trouble seeing the logic of the overlapping chain by which $g$ is continued to $S$.
Gary
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Post by Admin on May 28, 2017 6:43:52 GMT
Vasco, What led you to revisit the sections on analytic continuation? Gary Gary As I worked on exercise 20 the wording (if $f=g$ on the boundary then $f=g$ everywhere inside) reminded me of paragraph 1 on page 250 where Needham talks about the eyelash in a San Francisco street. I looked back and realised that $A$ and $B$ were the same as $f$ and $g$. I didn't re-read all the other stuff about overlapping domains. I'll look at your other points as soon as I can. Vasco |
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Post by Admin on Jun 22, 2017 8:29:17 GMT
Gary
Looking at your solution again, I notice that you say in paragraph 3: "...the interior of the modular surface consists entirely of the point $|f-g|=0$." I don't think this is entirely accurate. I see the modular surface as consisting of a flat region the same shape as $\Gamma$ sitting on top of the complex plane. Maybe that's what you mean by the above phrase, but I understood it as meaning that the surface was a single point. I agree that $f-g$ is a single point in the complex plane (the origin), but the real value $|f-g|=0$ is achieved at every point of the modular surface.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 22, 2017 13:24:39 GMT
Gary Looking at your solution again, I notice that you say in paragraph 3: "...the interior of the modular surface consists entirely of the point $|f-g|=0$." I don't think this is entirely accurate. I see the modular surface as consisting of a flat region the same shape as $\Gamma$ sitting on top of the complex plane. Maybe that's what you mean by the above phrase, but I understood it as meaning that the surface was a single point. I agree that $f-g$ is a single point in the complex plane (the origin), but the real value $|f-g|=0$ is achieved at every point of the modular surface. Vasco Vasco, Thank you. Your point is valid. As I recall my thinking when I worked on that problem, I was a bit confused about it. Gary |
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