|
|
Post by Admin on Apr 17, 2017 10:52:18 GMT
|
|
Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on Apr 22, 2017 5:28:49 GMT
Vasco, My answer: nh.ch7.ex12.pdf (108.79 KB) I agree that loop would be better than curve. Oops, I didn't see part (iii). I'll add it tomorrow. (Now included) Gary |
|
|
|
Post by Admin on Apr 22, 2017 8:00:34 GMT
Gary
Reading your solution makes me think that your $H$ looks more 'typical' than mine. I will probably change mine so that it looks more typical.
In your Figure 1 shouldn't the caption be: $H(z)=1+g(z)/f(z)$, and in your last sentence shouldn't that be "...$f+g$ has the same number of roots inside $\Gamma$ as $f$."
In the second-to-last sentence before the figure, shouldn't the first "are" be "and"?
I wouldn't call the $\nu[..,..]$ arguments since they are not angles but winding numbers. Multiplying them by $2\pi$ would make them arguments I suppose.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Apr 22, 2017 16:09:05 GMT
Gary Reading your solution makes me think that your $H$ looks more 'typical' than mine. I will probably change mine so that it looks more typical. In your Figure 1 shouldn't the caption be: $H(z)=1+g(z)/f(z)$, and in your last sentence shouldn't that be "...$f+g$ has the same number of roots inside $\Gamma$ as $f$." In the second-to-last sentence before the figure, shouldn't the first "are" be "and"? I wouldn't call the $\nu[..,..]$ arguments since they are not angles but winding numbers. Multiplying them by $2\pi$ would make them arguments I suppose. Vasco Vasco, I like your diagram as a general loop, but I suppose in math problems one more typically sees multiple loops. Thank you for the corrections and the hint. It was not as self-evident as I thought at first. I have taken the hint and rewritten. Gary |
|
|
|
Post by Admin on Apr 23, 2017 15:48:10 GMT
Gary
In the latest version of your document, I notice that you did not correct the caption to figure 1.
Also, I do not understand your answer to part (iii). Why do you say that $f$ and $g$ are analytic?
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Apr 23, 2017 18:17:18 GMT
Gary In the latest version of your document, I notice that you did not correct the caption to figure 1. Also, I do not understand your answer to part (iii). Why do you say that $f$ and $g$ are analytic? Vasco Vasco, I wonder if you have an early version of my answer to ex 12. My last two versions at least have used the capital H in figure 1, or is there another problem I am not seeing? (Oh, I see now it should be $1+\frac{g(z)}{f(z)}$.) Thank you. In part III, it was not my intent to say that $f$ and $g$ are analytic. It was to say that in the event that $f+g \neq 0$, either $f$ and $g$ are analytic OR they are cancelling functions. I will try to improve the sentence. Gary |
|
|
|
Post by Admin on Apr 23, 2017 18:40:07 GMT
Gary In the latest version of your document, I notice that you did not correct the caption to figure 1. Also, I do not understand your answer to part (iii). Why do you say that $f$ and $g$ are analytic? Vasco Vasco, I wonder if you have an early version of my answer to ex 12. My last two versions at least have used the capital H in figure 1, or is there another problem I am not seeing? (Oh, I see now it should be $1+\frac{g(z)}{f(z)}$.) Thank you. In part III, it was not my intent to say that $f$ and $g$ are analytic. It was to say that in the event that $f+g \neq 0$, either $f$ and $g$ are analytic OR they are cancelling functions. I will try to improve the sentence. Gary Gary I still don't understand your argument: "...that in the event that $f+g \neq 0$, either $f$ and $g$ are analytic OR they are cancelling functions". Can you explain the basis of this? Vasco PS Just noticed that the caption on figure 1 is still incorrect. |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Apr 24, 2017 5:45:51 GMT
Vasco, I wonder if you have an early version of my answer to ex 12. My last two versions at least have used the capital H in figure 1, or is there another problem I am not seeing? (Oh, I see now it should be $1+\frac{g(z)}{f(z)}$.) Thank you. In part III, it was not my intent to say that $f$ and $g$ are analytic. It was to say that in the event that $f+g \neq 0$, either $f$ and $g$ are analytic OR they are cancelling functions. I will try to improve the sentence. Gary Gary I still don't understand your argument: "...that in the event that $f+g \neq 0$, either $f$ and $g$ are analytic OR they are cancelling functions". Can you explain the basis of this? Vasco PS Just noticed that the caption on figure 1 is still incorrect. Vasco, It is explained on p. 352, paragraph 2: In general, note that $\nu[h(\Gamma), p] = 0$ merely implies that either there are no preimages inside $\Gamma$ or the preimages have cancelling multiplicities, as above. However, if $f$ is analytic and $\nu[f(\Gamma), p] = 0$ then the conclusion is quite definite: there are no preimages inside $\Gamma$.
In this case, the function is $H(z)$ rather than $h(z)$, but if $f$ and $g$ are analytic, then so is $H(z)$. And if $f$ and $g$ are not analytic, then their preimages have cancelling multiplicities, or at least the preimages of $H(z)$ have cancelling multiplicities. I think in my example with $f = z$ and $g = -\bar{z}$, there is a single preimage of 0, which is the same for $f$ and $g$, but the multiplicities (+1 and -1) cancel. I'm not sure what to say about a root for $H(z)$. I think it would be the whole real line. If so, does that explain why $\nu[H(\Gamma), 0] = 0$ better than the cancellation of the multiplicities of the preimages of $f$ and $g$? The caption: So it is! I think the correct version is now installed. At least, I have been able to download it. It is my practice to check after uploading, so I'm baffled. Gary |
|
|
|
Post by Admin on Apr 24, 2017 9:54:15 GMT
Gary
I have looked at the version of your document for April 23rd 2017, 11:30 am PST and the caption under the figure is
$H(z)=1+f(z)/g(z)$ when it should be $H(z)=1+g(z)/f(z)$
Thanks for the explanation in your last post. I now understand where you are coming from in part (iii) of the exercise, but we are being asked to show that if $|g(z)|\leq|f(z)|$ and $f+g\neq 0$ then it follows that $\nu[H(\Gamma),0]=0$. It seems to me that you are coming at it from the other direction: if $\nu[H(\Gamma),0]=0$ then ...
Also I don't see why the multiplicities of $f$ and $g$ are relevant, as they are not $p$-points of $H$.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Apr 25, 2017 21:55:50 GMT
Gary I have looked at the version of your document for April 23rd 2017, 11:30 am PST and the caption under the figure is $H(z)=1+f(z)/g(z)$ when it should be $H(z)=1+g(z)/f(z)$ Thanks for the explanation in your last post. I now understand where you are coming from in part (iii) of the exercise, but we are being asked to show that if $|g(z)|\leq|f(z)|$ and $f+g\neq 0$ then it follows that $\nu[H(\Gamma),0]=0$. It seems to me that you are coming at it from the other direction: if $\nu[H(\Gamma),0]=0$ then ... Also I don't see why the multiplicities of $f$ and $g$ are relevant, as they are not $p$-points of $H$. Vasco Vasco, Thank you. I was uneasy with my solution to (iii), so I didn't want to look at your answer until I had more time to think about it. You put your finger on the problem. I have rewritten and reposted with a different approach and I will look at your solution now. Gary |
|
|
|
Post by Admin on Apr 25, 2017 22:17:40 GMT
Gary
If $f+g\neq 0$ then $f\neq -g=e^{i\pi}g$. So as long as $f$ and $g$ are not diametrically opposite things are OK.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Apr 25, 2017 22:56:38 GMT
Gary If $f+g\neq 0$ then $f\neq -g=e^{i\pi}g$. So as long as $f$ and $g$ are not diametrically opposite things are OK. Vasco Vasco, Agreed. I will fix it. Gary |
|
|
|
Post by Admin on May 4, 2017 8:01:13 GMT
Gary
While working on exercise 15 I looked back at exercise 12 and realised that in the proof of Rouché's Theorem it is assumed that $\Gamma$ is a simple loop and so I will leave the diagram unchanged in my solution to exercise 12.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 4, 2017 13:58:01 GMT
Gary While working on exercise 15 I looked back at exercise 12 and realised that in the proof of Rouché's Theorem it is assumed that $\Gamma$ is a simple loop and so I will leave the diagram unchanged in my solution to exercise 12. Vasco Vasco, I agree. "Simple" does not normally extend to multiple winds. Gary |
|
|
|
Post by Admin on May 11, 2017 7:25:30 GMT
Gary
I have made a slight change to my solution to this exercise.
While working on exercise 15 I noticed that in the introduction to it on page 373 Needham says that he doesn't want to rely on the TAP for the proof of Brouwer's FP theorem, because he doesn't want to limit the proof to the case where $g$ is a continuous mapping having a finite number of $p$-points in a finite region.
This caused me to look back at exercise 12 and I noticed that in the proof of Rouché's theorem in part (ii) I had used the TAP in the final step. However in part (iii) of 12 we are only asked to deduce that $\nu[(f+g)(\Gamma),0]=\nu[f(\Gamma),0]$, and not take the final step and claim this as a proof of Rouché's theorem. In my answer however I had in fact made this final step and I now think that Needham was thinking about exercise 15 and so in exercise 12 he only asks us to take part (iii) so far, which enables him to use it in the proof of the first part of exercise 15. In this way the result of exercise 15 (Brouwer's FP theorem) does not rely on the TAP, which is just what Needham wants.
So I have altered my answer to 12 part (iii) so that it points out that if we want to go a step further and prove Rouché's theorem we must accept the restrictions on $g$ that the TAP brings with it.
Do you think this is right?
Vasco
|
|
