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Post by Admin on Mar 4, 2017 7:58:15 GMT
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Gary
GaryVasco
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Post by Gary on Mar 4, 2017 23:06:43 GMT
Vasco,
I have read your post on this section and it was helpful, but I still had problems digesting some of the remainder, so here is my attempt at expanding and clarifying. The last part of subsection 3, from the top of p. 350 seemed particular hard to follow. I wondered why "all the points on the real axis are critical points as a result of crushing in the vertical direction" (line 3).
This seems to be answered on the previous page. Formula (9) "is vacuous if det[J(a)] = 0. Geometrically this means that the transformation is locally crushing at $a$; just as for analytic mappings, such a place is called a critical point." The example at the top of p. 350 is just such a nonanalytic mapping where det[J(a)] = 0. The information on critical points and degrees of crushing from p. 204 is not much help to us here, because our example $f(x+iy) = x - iy^3$ is nonanalytic.
The italicized sentence on p. 350 was hard to process.
The critical points of an analytic mapping can be distinguished purely on the basis of topological multiplicity; those of a nonanalytic mapping cannot.
Given the following sentence, "For analytic functions, we have seen that $\nu(a) = +1$ if and only if $a$ is not critical", we may also read it as, "If for an analytic function $\nu(a)$ equals something other than +1, then $a$ is critical.", so if the result of the topological equation $\nu(a) = \nu[h(\Gamma_a), p]$ (p. 348) applied to an analytic function is +1, then $a$ must be critical. I am assuming that the general ("merely continuous") topological equation works for both analytic and nonanalytic functions.
But in this subsection (3 What's topologically special about analytic functions), we are, for the most part, analyzing nonanalytic functions. These are like analytic functions in having restricted possibilities for $\nu(a)$ of non-critical points, but they may be $\pm1$. But critical points may also have $\nu(a) = \pm1$, so critical points of nonanalytic functions are not restricted to $\nu(a)$ equals something other than $\pm1$ in the way that critical points of analytic functions are restricted to $\nu(a)$ equals something other than $+1$. In fact, it seems possible from the above that critical points of nonanalytic functions may take any integer value or $0$ for $\nu(a)$. Critical and non-critical points of nonanalytic functions can not be distinguished if $\nu(a) = \pm1$. If $\nu(a) \ne \pm1$, then we may be sure that $a$ is critical.
Now I realize a little table would make it clear:
$\nu(a)$ for critical and non-critical points in analytic and nonanalytic functions
Analytic functions
$\hspace{3em}$non-critical points: $+1$
$\hspace{3em}$critical points: positive integer other than 1, but not zero
Nonanalytic functions
$\hspace{3em}$non-critical points: $\pm1$
$\hspace{3em}$critical points: $0$, $\pm1$, other integer, positive or negative
Gary
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Post by Admin on Mar 5, 2017 16:36:41 GMT
Gary
However the subsection entitled "The Geometry of Linear Transformations" that Needham refers to on page 349 at the beginning of paragraph three is useful.
I think your post sums things up pretty well and the table is a very useful summary for reference.
I find it useful to think of the different types of functions as sets, and then we can define
1.) The set of analytic functions
2.) The set of functions which are differentiable in the real sense
3.) The set of continuous functions
4.) The set of all functions
If we draw four non-intersecting, very roughly circular shapes, each one smaller than the previous one and contained within it then we can label the innermost shape Analytic Functions, the next largest Functions Differentiable in The Real Sense, the next largest Continuous Functions and the outermost All Functions, then this illustrates that the properties of differentiable functions are shared by by analytic functions, and that the properties of continuous functions are shared by differentiable functions and by analytic functions etc. Also the non-analytic functions are the functions left when members of the "innermost" set are excluded. This makes it easier for me to see the correctness of for example the last sentence of subsection 2 at the bottom of page 348.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 5, 2017 21:52:25 GMT
Vasco,
Could you elaborate a bit on this sentence? The linear transformations appear to apply to continuous functions, but I don't quite see the application to the problem of crushing. I see the dough getting rolled out flat and folded over, but not crushed (pastry analogy from p. 207).
It's a handy ordered list. I'm a little uncertain how to read "in the real sense" (2). Does it just mean "we can actually differentiate them as we would an analytic function", or does it have something to do with real numbers? I don't find it in the index, so perhaps it it not a formal analytic phrase.
Gary
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Post by Admin on Mar 6, 2017 12:22:40 GMT
On page 208 subsection 2 discusses the geometry of linear transformations, which is relevant to the current discussion in subsection 3 on page 349: "What's topologically special about analytic functions".
Analytic functions are the only functions we can differentiate with respect to $z$. When Needham says in paragraph two on page 349 "...let us restrict ourselves to nonanalytic mappings that are at least differentiable in the real sense" he means that in general the mappings can be written as a complex function $f(x+iy)=u(x,y)+iv(x,y)$ where $x$ and $y$ are real and $u$ and $v$ are differentiable real functions with respect to $x$ and $y$. An example is the mapping $f(x+iy)=x-iy^3$ introduced on page 350 at the top. Both $u=x$ and $v=-y^3$ are differentiable in the real sense, with respect to $x$ and $y$. However they are not differentiable with respect to $z$, as in complex differentiation.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 6, 2017 17:07:17 GMT
Vasco,
Thank you. Can we say that functions which are differentiable in the real sense but are not differentiable with respect to z have that status because they are not readily expressible as powers of z?
Gary
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Post by Admin on Mar 7, 2017 17:14:13 GMT
Vasco, Thank you. Can we say that functions which are differentiable in the real sense but are not differentiable with respect to z have that status because they are not readily expressible as powers of z? Gary Gary If you look at this Wikipedia page you will see that you can express a real or complex-valued function f as a Taylor series about a real or complex point $a$ if f is infinitely differentiable in a neighbourhood of $a$ (in the real sense or in the complex sense). I'm not sure if this answers your question. For f(z) to be differentiable at $z$ it is necessary that the four partial derivatives $\partial_xu$ etc exist and satisfy the Cauchy-Riemann equations. This is one way to see that this function $f=x-iy^3$ is not differentiable with respect to $z$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 8, 2017 17:08:39 GMT
Vasco,
I'm not sure either, but I see that your answer is sufficient to the determination of the differentiability of f.
Gary
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Post by mondo on Apr 3, 2022 23:55:25 GMT
Two questions from me:
1. In the second paragraph of page 349 Needham says "...consider $h(z) = \overline{z}$. The unique preimage of $p$ is $a = \overline{p}$.." isn't that a mistake? I think $p$ is a preimage of $\overline{p}$?
2. Vasco in your addendum file you say "We can see that $f(x+iy) = x -iy^3$ is not analytic" - can you see it just by looking at the function or head to do some calculations?
Thank you.
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Post by Admin on Apr 4, 2022 14:33:06 GMT
Two questions from me: 1. In the second paragraph of page 349 Needham says "...consider $h(z) = \overline{z}$. The unique preimage of $p$ is $a = \overline{p}$.." isn't that a mistake? I think $p$ is a preimage of $\overline{p}$? 2. Vasco in your addendum file you say "We can see that $f(x+iy) = x -iy^3$ is not analytic" - can you see it just by looking at the function or head to do some calculations? Thank you. 1. No it's correct. If $h(z)=p$ then $\overline{z}=p$ and so $z=\overline{p}$. But what you say is also true because the conjugate operation is involutory: if $\overline{z}=w$ then $\overline{w}=z$ 2. Because $f(x+iy) = x -iy^3$ is such a simple function you can see it immediately by using the Cauchy-Riemann equations, result (6) on page 210 of the book and post #6 above in this thread. From your questions it seems to me extremely likely that you have not studied the previous chapters of the book and done the exercises. This can be a big disadvantage as each chapter of the book relies very heavily on previous chapters. Vasco |
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Post by mondo on Apr 6, 2022 23:11:07 GMT
Two questions from me: 1. In the second paragraph of page 349 Needham says "...consider $h(z) = \overline{z}$. The unique preimage of $p$ is $a = \overline{p}$.." isn't that a mistake? I think $p$ is a preimage of $\overline{p}$? 2. Vasco in your addendum file you say "We can see that $f(x+iy) = x -iy^3$ is not analytic" - can you see it just by looking at the function or head to do some calculations? Thank you. 1. No it's correct. If $h(z)=p$ then $\overline{z}=p$ and so $z=\overline{p}$. But what you say is also true because the conjugate operation is involutory: if $\overline{z}=w$ then $\overline{w}=z$ 2. Because $f(x+iy) = x -iy^3$ is such a simple function you can see it immediately by using the Cauchy-Riemann equations, result (6) on page 210 of the book and post #6 above in this thread. From your questions it seems to me extremely likely that you have not studied the previous chapters of the book and done the exercises. This can be a big disadvantage as each chapter of the book relies very heavily on previous chapters. Vasco You are partially right - it's not that I haven't studied the previous chapters but rather sometimes I had to make breaks of a few months due to other responsibilities. So once I went back to reading I forgot plenty of things and yes it is as I haven't read that. So now I am motivated to read constantly, every day to eliminate this problem. As for 1. - The index dosn't list "preimage" so I think the standard definition is assumed here where a preimage is just a domain of some mapping. For example in $h(z) = \overline{z}$ then we say $z$ is a preimage of $\overline{z}$ right? So now going back to the quote from the book, they define $h(z) = \overline{z}$ and then say $\overline{p}$ is a preimage but it is confusing as when you plug $p$ into $h(z)$ you will get $\overline{p}$ so it is evident that $p$ is a preimage here. |
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Post by Admin on Apr 6, 2022 23:22:48 GMT
1. No it's correct. If $h(z)=p$ then $\overline{z}=p$ and so $z=\overline{p}$. But what you say is also true because the conjugate operation is involutory: if $\overline{z}=w$ then $\overline{w}=z$ 2. Because $f(x+iy) = x -iy^3$ is such a simple function you can see it immediately by using the Cauchy-Riemann equations, result (6) on page 210 of the book and post #6 above in this thread. From your questions it seems to me extremely likely that you have not studied the previous chapters of the book and done the exercises. This can be a big disadvantage as each chapter of the book relies very heavily on previous chapters. Vasco You are partially right - it's not that I haven't studied the previous chapters but rather sometimes I had to make breaks of a few months due to other responsibilities. So once I went back to reading I forgot plenty of things and yes it is as I haven't read that. So now I am motivated to read constantly, every day to eliminate this problem. As for 1. - The index dosn't list "preimage" so I think the standard definition is assumed here where a preimage is just a domain of some mapping. For example in $h(z) = \overline{z}$ then we say $z$ is a preimage of $\overline{z}$ right? So now going back to the quote from the book, they define $h(z) = \overline{z}$ and then say $\overline{p}$ is a preimage but it is confusing as when you plug $p$ into $h(z)$ you will get $\overline{p}$ so it is evident that $p$ is a preimage here. Mondo But if you plug $\overline{p}$ into $h(z)$ you get $p$ and so it is also true that $\overline{p}$ is a preimage of $p$. So both are equally true. For the function $h$ we can say that $p$ is a preimage of $\overline{p}$ and vice versa. Vasco |
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