Gary
GaryVasco
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Post by Gary on Jun 9, 2016 22:58:35 GMT
Vasco,
Would one be correct to assume that the intersecting arcs C and D possessing Schwarz functions may be arcs of any smooth curves possessing Schwarz functions, or must they be arcs of circles? The intent seems to be to allow any analytic segment of a curve.
I have completed an answer to the problem and I note that it differs in one important respect. The difference hinges on how one interprets "reflection (inversion) in C maps D into itself" (from the problem, p. 266). If we can read it as "maps every point in D to itself", then I think my answer is OK and yours is not wrong. If we must read it as "maps every point in D to some point $\mathfrak{R}_D(\tilde{d})$, where $\tilde{d}$ may or may not equal d, then I think my answer is wrong and yours is still correct. I am arguing for the interpretation that favors my answer on the basis of the fact that the Schwarz function sends the points of K to their conjugates (p. 255, line 5), so $\mathfrak{R}_K$ would send the conjugates back to the source. What do you think?
(The exercise has been removed, because Mathematica failed to preserve the tilde's (a non-analytic function, I guess I will repost.).
Gary
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Post by Admin on Jun 10, 2016 6:17:12 GMT
Vasco, Would one be correct to assume that the intersecting arcs C and D possessing Schwarz functions may be arcs of any smooth curves possessing Schwarz functions, or must they be arcs of circles? The intent seems to be to allow any analytic segment of a curve. I have completed an answer to the problem and I note that it differs in one important respect. The difference hinges on how one interprets "reflection (inversion) in C maps D into itself" (from the problem, p. 266). If we can read it as "maps every point in D to itself", then I think my answer is OK and yours is not wrong. If we must read it as "maps every point in D to some point $\mathfrak{R}_D(\tilde{d})$, where $\tilde{d}$ may or may not equal d, then I think my answer is wrong and yours is still correct. I am arguing for the interpretation that favors my answer on the basis of the fact that the Schwarz function sends the points of K to their conjugates (p. 255, line 5), so $\mathfrak{R}_K$ would send the conjugates back to the source. What do you think? View AttachmentGary Gary I agree that the word circular should not be there in my part (i). The solution does not assume they are circular. I will remove the word circular and republish. It would be a strange situation if Schwarzian reflection in $C$ were the identity mapping. Applied twice it is the identity. So if $\widetilde{d}=\mathfrak{R}_C(d)$ then $\mathfrak{R}_C(\widetilde{d})=d$. I think you mean $\mathfrak{R}_C(\tilde{d})$ here. Is that right? I think my solution is OK as it stands, without the "circular" of course. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 10, 2016 14:13:28 GMT
Vasco, Would one be correct to assume that the intersecting arcs C and D possessing Schwarz functions may be arcs of any smooth curves possessing Schwarz functions, or must they be arcs of circles? The intent seems to be to allow any analytic segment of a curve. I have completed an answer to the problem and I note that it differs in one important respect. The difference hinges on how one interprets "reflection (inversion) in C maps D into itself" (from the problem, p. 266). If we can read it as "maps every point in D to itself", then I think my answer is OK and yours is not wrong. If we must read it as "maps every point in D to some point $\mathfrak{R}_D(\tilde{d})$, where $\tilde{d}$ may or may not equal d, then I think my answer is wrong and yours is still correct. I am arguing for the interpretation that favors my answer on the basis of the fact that the Schwarz function sends the points of K to their conjugates (p. 255, line 5), so $\mathfrak{R}_K$ would send the conjugates back to the source. What do you think? Gary Gary I agree that the word circular should not be there in my part (i). The solution does not assume they are circular. I will remove the word circular and republish. It would be a strange situation if Schwarzian reflection in $C$ were the identity mapping. Applied twice it is the identity. So if $\widetilde{d}=\mathfrak{R}_C(d)$ then $\mathfrak{R}_C(\widetilde{d})=d$. I think you mean $\mathfrak{R}_C(\tilde{d})$ here. Is that right? I think my solution is OK as it stands, without the "circular" of course. Vasco Vasco, I agree completely that your answer is correct. Was just trying to make an argument that mine might also be correct, but I no longer think it is. I'll go point by point here. Actually, I didn't notice that you used the word "circular". I just thought the problem statement was a bit vague. Do you agree, or is it the normal practice in mathematics to interpret "arc" more generally? You say "It would be a strange situation if Schwarzian reflection in $C$ were the identity mapping". Quite right. I see now why my answer is obviously wrong. I think I misread page 255, where the equation directly under Figure [30] has $\tilde{q} = \overset{-}{S_K(q)} = \overset{-}{(\overset{-}{q})} = q$, which appears to be the identity mapping for a point on the curve K being reflected, in my mistaken impression, in the real line, but in fact with respect to point q, the curve K is being reflected in itself. Actually, my difficulty can be seen in the last two lines of the original question where I wrote " the Schwarz function sends the points of K to their conjugates (p. 255, line 5), so $\mathfrak{R}_K$ would send the conjugates back to the source". To make the comparison, this is the reflection of d in D rather than d in C. You are right that $\mathfrak{R}_C(\tilde{d})$ was intended, rather than $\mathfrak{R}_D(\tilde{d})$. I agree that your solution is the only correct one. I will fix mine. Gary |
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Post by Admin on Jun 10, 2016 14:34:19 GMT
Gary
Yes "arc" can mean any segment of a curve, but in the context of the exercise I agree that it could be misleading, and I think for clarity Needham could have written it as "...generalize $C$ and $D$ to intersecting arcs (not necessarily circular) possessing Schwarz....".
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 10, 2016 22:35:08 GMT
Vasco, I have rewritten my answers to parts (ii) and (iii) and reposted here without further consultation of yours beyond your comments in this forum. I have never quite understood the statement "an analytic mapping is determined by its values on a curve", so my comprehension of your answer to (iii) is not well grounded. I offered what seemed to me to be a plausible answer. nh.ch5.ex31.pdf (72.56 KB) Gary |
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Post by Admin on Jun 11, 2016 5:07:58 GMT
Vasco, I have rewritten my answers to parts (ii) and (iii) and reposted here without further consultation of yours beyond your comments in this forum. I have never quite understood the statement "an analytic mapping is determined by its values on a curve", so my comprehension of your answer to (iii) is not well grounded. I offered what seemed to me to be a plausible answer. View AttachmentGary Gary As I understand it, this is just using the uniqueness idea as explained in the first two paragraphs of subsection 3 on page 250. I will take a look at your alternative proof. Vasco |
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Post by Admin on Jun 11, 2016 14:54:37 GMT
Gary
I have looked at your solution to part (iii), and I think it is important to note that what we have proved in part (ii) is that the mappings
$(\mathfrak{R}_D\circ\mathfrak{R}_C)$ and $(\mathfrak{R}_C\circ\mathfrak{R}_D)$ are equal at points of $D$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 11, 2016 23:03:08 GMT
Gary I have looked at your solution to part (iii), and I think it is important to note that what we have proved in part (ii) is that the mappings $(\mathfrak{R}_D\circ\mathfrak{R}_C)$ and $(\mathfrak{R}_C\circ\mathfrak{R}_D)$ are equal at points of $D$. Vasco Vasco, I started over from the beginning and realized I had not visualized the problem correctly and I had missed the crucial point that $\tilde{d}$ lies on $D$! I think I see now why the point in your quote is important: that $(\mathfrak{R}_D\circ\mathfrak{R}_C)$ and $(\mathfrak{R}_C\circ\mathfrak{R}_D)$ are equal at points of $D$. Thank you for the added hint. I will replace the document immediately. Perhaps the question of orthogonality needs further discussion. Gary |
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Post by Admin on Jun 22, 2016 7:56:41 GMT
Gary
I have added some discussion of this to my solution to exercise 31. I hope it is the sort of thing you were looking for.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 23, 2016 5:20:07 GMT
Gary I have added some discussion of this to my solution to exercise 31. I hope it is the sort of thing you were looking for. Vasco Yes. It looks good. |
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