Gary
GaryVasco
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Post by Gary on Mar 17, 2017 22:50:16 GMT
Vasco, I plowed ahead, but I didn't want to post this until we resolved some of the problems in Exercise 2. I think we are OK there. I found (i) in exercise 3 to be puzzling, so I am interested to see what you will do with it. nh.ch7.ex3.pdf (612.28 KB) nh.ch7.notes.notation.pdf (123.79 KB) Gary |
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Post by Admin on Mar 17, 2017 23:03:54 GMT
Vasco, I plowed ahead, but I didn't want to post this until we resolved some of the problems in Exercise 2. I think we are OK there. I found (i) in exercise 3 to be puzzling, so I am interested to see what you will do with it. View AttachmentGary Gary I have a solution to exercise 3, I just haven't published it yet, because of the graphics involved. Our solutions to exercise 2 are very different don't you think? Vasco |
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Post by Admin on Mar 20, 2017 6:58:18 GMT
Gary
I have looked at your solution to exercise 3 part (i)
i seems to me that you have not been able to verify the Argument Principle for this part. Looking at my solution, I think the reason for this is that you have not found all the $p$-points lying inside the disc $|z|\leq(4/3)$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 22, 2017 1:38:17 GMT
Gary I have looked at your solution to exercise 3 part (i) i seems to me that you have not been able to verify the Argument Principle for this part. Looking at my solution, I think the reason for this is that you have not found all the $p$-points lying inside the disc $|z|\leq(4/3)$. Vasco Vasco, That sounds right. I'll look for some more. Thanks. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 22, 2017 8:01:59 GMT
Gary I have looked at your solution to exercise 3 part (i) i seems to me that you have not been able to verify the Argument Principle for this part. Looking at my solution, I think the reason for this is that you have not found all the $p$-points lying inside the disc $|z|\leq(4/3)$. Vasco Vasco, That sounds right. I'll look for some more. Thanks. Gary Four p-points found and exercise reposted. Gary |
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Post by Admin on Mar 23, 2017 19:39:51 GMT
Gary
In part (ii) of your document near the end you say:
"A double wind around the origin can be seen in Figure 2b."
But we have to consider winding about the point $p=1$ not the origin in this case.
In part (iii) you say that all derivatives of $f$ are $0$ at the origin. I didn't find this to be the case in my solution.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 23, 2017 21:27:11 GMT
Gary In part (ii) of your document near the end you say: "A double wind around the origin can be seen in Figure 2b." But we have to consider winding about the point $p=1$ not the origin in this case. In part (iii) you say that all derivatives of $f$ are $0$ at the origin. I didn't find this to be the case in my solution. Vasco Vasco, Regarding your first point, I was getting it confused with a polynomial, as in [8], where [10] would have been a better reference. Thank you for spotting that. I'm puzzled by your second point, it appears to me that all the derivatives of $f$ to infinity have a coefficent $b a^n$. If a = 0, then the derivative is zero for all. I have now expanded the Taylor series a bit to illustrate this. If you have a different solution, I think I will have to await your posting. I also find that I am unable to determine the number of winds by tracing or counting intersections due to the complexity of the curve and the vast difference in resolution between the smallest and largest elements. Gary |
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Post by Admin on Mar 23, 2017 23:13:28 GMT
Gary
Can you explain how you see a double wind about 1 in figure 2(b)?
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 5:02:19 GMT
Gary Can you explain how you see a double wind about 1 in figure 2(b)? Vasco Vasco, I blame it on an extra glass of wine. I think it is fixed now. Gary |
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Post by Admin on Mar 24, 2017 5:36:06 GMT
Gary Can you explain how you see a double wind about 1 in figure 2(b)? Vasco Vasco, I blame it on an extra glass of wine. I think it is fixed now. Gary Gary I'm envious but it's too early in the morning here. What do you get when you calculate the topological multiplicity from the Taylor series, rather than looking at the diagram? Vasco |
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Post by Admin on Mar 24, 2017 12:00:40 GMT
Gary Here is my solution to exercise 3. It contains links to an Open Office spreadsheet with three sheets, one for each of the three parts of the exercise, and provides the graphical verification asked for by the exercise. It allows the scale of the graphic to be changed to facilitate checking near the origin. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 16:51:14 GMT
Gary Here is my solution to exercise 3. It contains links to an Open Office spreadsheet with three sheets, one for each of the three parts of the exercise, and provides the graphical verification asked for by the exercise. It allows the scale of the graphic to be changed to facilitate checking near the origin. Vasco Vasco, It looks good. Regarding (ii), I had also used the Taylor's series at one point to arrive at $nu[f(a), p] = 2$, and I saw the apparent contradiction, but I did not arrive at your solution of the two congruent winds. Regarding (iii), I didn't realize I could apply Taylor's series to finding the degrees for the function of a critical point. So that clears up two of my difficulties. The links to the spreadsheet did not appear in the .pdf file. Gary |
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Post by Admin on Mar 24, 2017 17:05:31 GMT
Gary
Are you absolutely certain? When I download it the links are not visible but if you hover over the word spreadsheet you can see the cursor change to a hand and the spreadsheet can be downloaded.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 24, 2017 20:36:44 GMT
Gary Are you absolutely certain? When I download it the links are not visible but if you hover over the word spreadsheet you can see the cursor change to a hand and the spreadsheet can be downloaded. Vasco Vasco, The links are there, but not highlighted, so I didn't see them. I am in the process of reading your answers and going over the whole exercise again. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 20:46:33 GMT
Vasco,
The links all go to the plot of (iii). Was that your intention?
In Ch 7, Ex 3 (ii)
I wondered why you made the long argument with hyperbolic trig functions to establish that z = 0, when solutions to cos(z) = 1 are well known to be $n 2Pi, n = 0, \pm1, \pm2, ...$. Same question for (iii).
You used the Taylor's series form of cos(z). How does this reveal the multiplicity? ($\frac{z^2}{!2}$ = 0 at 0 and $d(\frac{z^2}{!2}) = 0$ at 0)
Gary
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