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Post by Admin on Feb 16, 2017 6:56:27 GMT
I was a bit confused at first when reading paragraph one of subsection 3 on page 343 where Needham writes:
"As $z$ travels once round $C$ at unit speed, $\widehat{w}$ travels once round $\widehat{J_{\nu}}$ with speed $|\nu|$, completing $|\nu|$ circuits of the unit circle"
Surely $\widehat{w}$ goes $|\nu|$ times round $\widehat{J_{\nu}}$, I thought, but no, it goes once round the loop $\widehat{J_{\nu}}$ and $|\nu|$ times round the unit circle.
So, I agree with what is written in the book.
Vasco
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Post by mondo on Apr 2, 2022 0:20:23 GMT
I am also confused by this fragment. First of all I don't get what Jνˆ is and why wˆ would travel it only once? Can you elabore a bit please.
PS: How to add math expressions here?
Thank you
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Post by Admin on Apr 2, 2022 9:57:05 GMT
I am also confused by this fragment. First of all I don't get what Jνˆ is and why wˆ would travel it only once? Can you elabore a bit please. PS: How to add math expressions here? Thank you Hi mondo To use math expressions here use dollar signs. So for example $w^2$ is written "dollarw^2dollar". Just use Latex. For more detail look in "more information about the forum" on the initial board of the forum. Here is a link to it vcaneedham.freeforums.net/thread/31/use-dollar-include-maths-posts$\widehat{\mathcal{J}_v}(z)$ is the mapping from the unit circle $C$ to the loop $\widehat{J_v}$. For example, the loop $L$ on the righthand side of figure 5 is produced by the mapping $\mathcal{L}$ applied to $C$. It's easy to see from the figure that the winding number of $L$ is $\nu=2$. Thinking of $L$ as an elastic band and the unit circle as the boundary of a solid cylinder, we put the cylinder through the inside of the loop and allow the elastic band to contract onto the cylinder and we can see that in this case it wraps round the cylinder twice. The archetypal mapping of degree $\nu$ wraps round the cylinder $|\nu|$ times, but goes round the elastic band once. Vasco |
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Post by mondo on Apr 2, 2022 19:11:48 GMT
Thank you, I think it makes sense now. There is one more thing, an exercise that I have no idea how to do - a second to last paragraph of page 346 says "The conclusion can be refined... It follows by a simple calculation that the first $(n-1)$ derivatives of P vanish at a..." How to show that?
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Post by Admin on Apr 2, 2022 20:01:24 GMT
Thank you, I think it makes sense now. There is one more thing, an exercise that I have no idea how to do - a second to last paragraph of page 346 says "The conclusion can be refined... It follows by a simple calculation that the first $(n-1)$ derivatives of P vanish at a..." How to show that? Mondo Here is hint: if $P=A^n\Omega(z)$ where $A=(z-a)$ then differentiating the product we have $P'=nA^{n-1}\Omega(z)+A^n\Omega'(z)=0$, when $z=a$ and so on for further derivatives. Try doing this by differentiating say m times and then show that the derivative is zero for $m=1...(n-1)$. Vasco |
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Post by mondo on Apr 2, 2022 22:14:00 GMT
Thank you for the hint. Before I will calculate next derivatives I would like to ask why do you say first derivative is 0 when $z = a$? In the book they say the opposite - If $P(z)$ has multiplicity $n$ then $P$ may be factorized as $A^n\Omega(z)$ where $\Omega(a) \ne 0$.
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Post by Admin on Apr 2, 2022 23:57:41 GMT
Mondo
Think. When $z=a$ then $A=0$ and $A^n=0$ and $A^{n-1}=0$ and so $P'=0$.
Vasco
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Post by mondo on Apr 3, 2022 0:33:32 GMT
Right, I forgot for a moment what A is, sorry abou that. However that also means $A^n\Omega(z) = 0$ when $z = a$. More so any higher level derivative i.e $A^{n+1}$ will also become $0$ when $z = a$. So how can we say the order is $(n-1)$ if it is always zero as we approach $a$?
PS: I noticed the latex formating doesn't work on chrome for Android..I see raw $ equations.
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Post by Admin on Apr 3, 2022 7:19:32 GMT
Right, I forgot for a moment what A is, sorry abou that. However that also means $A^n\Omega(z) = 0$ when $z = a$. More so any higher level derivative i.e $A^{n+1}$ will also become $0$ when $z = a$. So how can we say the order is $(n-1)$ if it is always zero as we approach $a$? PS: I noticed the latex formating doesn't work on chrome for Android..I see raw $ equations. Mondo The formatting does work with Chrome on an Android phone. All the formatting is done by the forum software and has nothing to do with Chrome or Android. Sometimes you need to refresh the forum however, when using Previous and Next. Here $A^2$, $A^3$, $A^n$ etc do not mean second derivative, 3rd derivative etc, they mean powers of $A$. The notation $A^4$ for example means $(x-a)^4=(z-a)(z-a)(z-a)(z-a)$ The notation for derivatives is $A^{(2)}$, $A^{(3)}$, $A^{(n)}$ etc. Vasco |
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Post by mondo on Apr 3, 2022 20:48:31 GMT
Here $A^2$, $A^3$, $A^n$ etc do not mean second derivative, 3rd derivative etc, they mean powers of $A$. The notation $A^4$ for example means $(x-a)^4=(z-a)(z-a)(z-a)(z-a)$ The notation for derivatives is $A^{(2)}$, $A^{(3)}$, $A^{(n)}$ etc. Vasco Yes, my notation was confusing, sorry. I think I have figured it out - The $(n-1)$ derivatives at $a$ are 0 because each of them has a sum of elements multiplied by $A^x$ where $x>0$ and this is always $0$. However as soon as we calculate the Nth derivatives we get $A^{n-n}$ and this is just 1 so Nth derivative is the first non-vanishing one. However I don't get why they define the the order of a critical point as $(n-1)$ then. Shouldn't the order be just $n$ as this is the first one that is non zero? As for the formatting it is strange, it should be, as you said processed by forum and just output to the browser... Don't you have some condition there that checks browser version compatibility and that is why I see this -> i2.paste.pics/9e6173c1bc424c8b9066ce3a201a2e5a.png ? |
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Post by Admin on Apr 4, 2022 15:06:43 GMT
Here $A^2$, $A^3$, $A^n$ etc do not mean second derivative, 3rd derivative etc, they mean powers of $A$. The notation $A^4$ for example means $(x-a)^4=(z-a)(z-a)(z-a)(z-a)$ The notation for derivatives is $A^{(2)}$, $A^{(3)}$, $A^{(n)}$ etc. Vasco Yes, my notation was confusing, sorry. I think I have figured it out - The $(n-1)$ derivatives at $a$ are 0 because each of them has a sum of elements multiplied by $A^x$ where $x>0$ and this is always $0$. However as soon as we calculate the Nth derivatives we get $A^{n-n}$ and this is just 1 so Nth derivative is the first non-vanishing one. However I don't get why they define the the order of a critical point as $(n-1)$ then. Shouldn't the order be just $n$ as this is the first one that is non zero? As for the formatting it is strange, it should be, as you said processed by forum and just output to the browser... Don't you have some condition there that checks browser version compatibility and that is why I see this -> i2.paste.pics/9e6173c1bc424c8b9066ce3a201a2e5a.png ? Mondo Again the definition of order with a critical point is explained in previous chapters: On page 205 subsection 2 first paragraph which references footnote 4. If you had searched in the index to the book you would have found a reference to this page. If you do not intend to read the previous chapters of the book in their entirety I would suggest you consult the index in the book as a first step towards solving any problem you encounter. Often the explanation is in the earlier chapters of the book. Of course if you then still have a problem then you are of course welcome to post it here on the forum. Vasco |
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