Gary
GaryVasco
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Post by Gary on Feb 21, 2016 5:13:15 GMT
Vasco,
I have spent a few days on exercise 17 and I'm not sure whether I have made progress or not. It doesn't quite make sense to me. It is easy enough to calculate the complex curvature, but it's not so easy to look at an analytic function and decide which direction of S yields an image f(S) having vanishing curvature at f(p). I did my best and I am hoping that you will see my error, as it affects the answers to (i)-(iv). I was not able to look at the function and come up with a formula that matches the formula for $\mathcal{K} = i \bar{f''}/\bar{f'} |f'|$. My answer to (v) also partially contradicted (31).
Gary
Postscripts:
After discussions with Vasco, I gave this an extensive rewrite. The figures remain the same.
With the next day came a new perspective on the problem and another rewrite. This one seems right in that it is sufficiently general to apply to all three functions $e^z, log(z)$, and $z^m$.
Revision based on Vasco's comments of Feb 24 moved to new message, will be posted below.
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Post by Admin on Feb 23, 2016 6:52:48 GMT
Gary
In your document (which by the way seems to be attached here in a pre-compiled form and needs the Wolfram software), in your answer to part (iv) you say:
but this is not correct. The angle $\theta$ is the angle the complex number $z$ makes with the real axis, and this varies with $p$ as you will see immediately if you draw a diagram. The angle the directed line segment makes with the horizontal is constant, but not $\theta=arg\{z\}$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 23, 2016 16:08:35 GMT
Gary In your document (which by the way seems to be attached here in a pre-compiled form and needs the Wolfram software), in your answer to part (iv) you say: but this is not correct. The angle $\theta$ is the angle the complex number $z$ makes with the real axis, and this varies with $p$ as you will see immediately if you draw a diagram. The angle the directed line segment makes with the horizontal is constant, but not $\theta=arg\{z\}$. Vasco Vasco, I think you are correct that I have misconstrued the problem, in that p should be arbitrary, where I have taken it to be any point on the directed line. I'll work on it some more. Sorry about the notebook file. Gary |
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Post by Admin on Feb 23, 2016 16:13:02 GMT
Gary
You are right $p$ is any point in the complex plane, but then we draw an $S$ which passes through $p$ and go from there. $S$ makes any angle we like with the horizontal, say $\phi$, while $z=p=re^{i\theta}$ say.
Vasco
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Post by Admin on Feb 23, 2016 16:25:51 GMT
Gary
We can choose any point $p$ in the complex plane and draw a line segment $S$ through it. Then keeping $p$ fixed we can rotate $S$ about $p$. Alternatively we can keep $S$ fixed and move $p$ along $S$.
If you look at the left of figure 17 on page 235, there are four points, which all have different arguments (values of $\theta$), while the line segment $S$ makes an angle $\phi$ with the horizontal.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 23, 2016 17:27:52 GMT
Gary We can choose any point $p$ in the complex plane and draw a line segment $S$ through it. Then keeping $p$ fixed we can rotate $S$ about $p$. Alternatively we can keep $S$ fixed and move $p$ along $S$. If you look at the left of figure 17 on page 235, there are four points, which all have different arguments (values of $\theta$), while the line segment $S$ makes an angle $\phi$ with the horizontal. Vasco Vasco, Yes, agreed with this and the previous. I think the new version takes all this into account. Gary |
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Post by Admin on Feb 24, 2016 8:55:26 GMT
Gary
On page 2 of your document on lines 5-6 you write
In my opinion this is not correct. $S$ can point in any direction - it does not affect $\mathcal{K}$ in any way.
At the bottom of page 4 you write
I think it would be better to say "...parallel to $z$ would produce vanishing curvature for $f(S)$ at $p$". In this exercise we are always looking at the curvature at $f(p)$ on $f(S)$ not on $S$. The direction of $\mathcal{K}$ is then at right angles to $z$.
On page 5 I think you should be finding that $|\mathcal{K}|=1$.
I am also puzzled on line 3 by your reference to the real line here and further down near the middle of the page. I don't think the real line has any significance here does it?
According to me using (28) we find that $\mathcal{K}=(-iz/|z|)$ showing that $\mathcal{K}$ is perpendicular to $z$ with magnitude $1$.
In the last paragraph of page 5 when discussing $f(z)=z^m$ you again mention the real line which puzzles me again. Maybe I am missing something here?
Also the exercise specifies that $m$ is a positive integer, and so I think that an expression like $\bigg|\frac{m-1}{mz^m}\bigg|$ is better written as $\big(\frac{m}{m-1}\big)\cdot\frac{1}{|z|^m}$ or even $\big(\frac{m}{m-1}\big)\cdot\frac{1}{r^m}$ as it makes it easier to understand - only my opinion.
I have not yet looked at part (v) and part (vi).
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 24, 2016 16:59:40 GMT
Vasco, You are quite right, but when I wrote it, I had specified that the curvature must be vanishing at f(p), so in that context I think the sentence is correct. I will rewrite it to clarify. Agreed. I missed that one in the rewrite. I was only looking at p = x as a special case, exploring a bit more than was called for by a strict interpretation of the problem. I puzzled some over the best arrangement of the absolute brackets, so I am glad to have your comments on that. This version takes these comments into account and includes corrections to the plot of the circles and bars under $e^z$, which are now seen to conform to Needham's description. I have revised the attachment and moved it to a later posting, where it is still undergoing revision. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 9, 2016 22:30:05 GMT
Vasco,
I have tried to work through 5:17 once more with the benefit of your new posting to the archive. I still find I am not crystal clear regarding the criteria that one might use to determine the direction of rotation when one is making S parallel to $\mathcal{K}$. The choice of direction is critical because S is directed and its ultimate direction affects the sign of $\mathcal{K}$. I am also a little unclear on the criteria for determining $\overline{\kappa}$ by inspection of the function or the plot rather than by calculation. I wonder if I am creating problems that don’t exist, or if this still needs a bit of work.
(Attachment revised and moved to March 11)
I have also added a figure for log(z) to part (vi). It raises a similar problem in determining $\mathcal{K}$.
Gary
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Post by Admin on Mar 10, 2016 20:36:20 GMT
Vasco, I have tried to work through 5:17 once more with the benefit of your new posting to the archive. I still find I am not crystal clear regarding the criteria that one might use to determine the direction of rotation when one is making S parallel to $\mathcal{K}$. The choice of direction is critical because S is directed and its ultimate direction affects the sign of $\mathcal{K}$. I am also a little unclear on the criteria for determining $\overline{\kappa}$ by inspection of the function or the plot rather than by calculation. I wonder if I am creating problems that don’t exist, or if this still needs a bit of work. View AttachmentI have also added a figure for log(z) to part (vi). It raises a similar problem in determining $\mathcal{K}$. Gary Gary Whichever way we choose to point both $\mathcal{K}$ and $S$, the scalar product $\boldsymbol{\mathcal{K}\cdot\widehat{\xi}}=|\mathcal{K}|$, which is positive. Since this is also equal to the curvature of $f(S)$ at $p$, it means that this curvature must also be positive. If we take the case when $f=e^z$ and choose the direction of $\mathcal{K}$ to be that of decreasing values of $y$, then we can see from figure 3(b) that the curvature at $f(p)$ is negative, and so this is a contradiction. If we choose the direction of $\mathcal{K}$ to be that of increasing values of $y$, then we can see from figure 3(b) that the curvature at $f(p)$ is positive, and so this must be the direction for $\mathcal{K}$. If you apply these ideas to the other cases, you will see that it is easy to decide on the direction of $\mathcal{K}$. If $f(S)$ is a line or a circle then it is fairly easy to deduce the curvature $\widetilde{\kappa}$, but otherwise not, as we see with $f=\log z$ and $f=z^m$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 10, 2016 23:37:42 GMT
Vasco, I have tried to work through 5:17 once more with the benefit of your new posting to the archive. I still find I am not crystal clear regarding the criteria that one might use to determine the direction of rotation when one is making S parallel to $\mathcal{K}$. The choice of direction is critical because S is directed and its ultimate direction affects the sign of $\mathcal{K}$. I am also a little unclear on the criteria for determining $\overline{\kappa}$ by inspection of the function or the plot rather than by calculation. I wonder if I am creating problems that don’t exist, or if this still needs a bit of work. I have also added a figure for log(z) to part (vi). It raises a similar problem in determining $\mathcal{K}$. Gary Gary Whichever way we choose to point both $\mathcal{K}$ and $S$, the scalar product $\boldsymbol{\mathcal{K}\cdot\widehat{\xi}}=|\mathcal{K}|$, which is positive. Since this is also equal to the curvature of $f(S)$ at $p$, it means that this curvature must also be positive. If we take the case when $f=e^z$ and choose the direction of $\mathcal{K}$ to be that of decreasing values of $y$, then we can see from figure 3(b) that the curvature at $f(p)$ is negative, and so this is a contradiction. If we choose the direction of $\mathcal{K}$ to be that of increasing values of $y$, then we can see from figure 3(b) that the curvature at $f(p)$ is positive, and so this must be the direction for $\mathcal{K}$. If you apply these ideas to the other cases, you will see that it is easy to decide on the direction of $\mathcal{K}$. If $f(S)$ is a line or a circle then it is fairly easy to deduce the curvature $\widetilde{\kappa}$, but otherwise not, as we see with $f=\log z$ and $f=z^m$. Vasco Vasco, Sorry to be requoting everything, but when I tried selective quoting, I was getting only fragments of my selections. Thank you for these helpful comments. However obtuse some of my questions, I do think I am seeing the scope of the problems more clearly as I look at them from different angles and return to them. I am finding the terminology to be slightly imprecise. By “direction” and “orientation” it seems that we can mean either the angle of S or $\mathcal{K}$ or a direction of movement along the line S or the curve f(S). Direction is not a problem along the x or y axis, but how does one define it on a ray that has been rotated? Is the positive direction of movement of p on rotated S the movement outward from the origin of unrotated S? I realize that you defined it in terms of the rotation of p, but I have not quite assimilated that. I will try to keep these things straight. This puzzles me, because $\mathbf{\mathcal{K} \cdot \widehat{\xi}}$ is defined as $\mathbf{a \cdot b} = |a| |b| cos \theta$ (p. 27) and since $cos \theta$ may be negative, we should have to write $|\mathbf{\mathcal{K} \cdot \overset{~}{\xi}}| = |\mathcal{K}|$. We see that $\widetilde{\kappa} = \mathcal{K} \cdot \widehat{\xi} + \kappa/|f’|$ (30), p. 239, and we are used to thinking of $\overset{~}{\kappa}$ as equivalent to 1/r, which would render it positive, but I wonder if it might be negative at times and if $\mathcal{K} \cdot \widehat{\xi}$ might also be negative at times, so that curvature of f(z) at p might also be negative. Gary |
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Post by Admin on Mar 11, 2016 0:03:38 GMT
Gary
It is midnight here and I am about to go to bed, so I will just give a quick reply to your quote and look at the rest tomorrow. I am talking about when the complex curvature and $S$ point in the same direction and so the scalar product then has the cosine being equal to 1. Sorry for that, my wording was not quite precise enough.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 11, 2016 2:05:32 GMT
Gary It is midnight here and I am about to go to bed, so I will just give a quick reply to your quote and look at the rest tomorrow. I am talking about when the complex curvature and $S$ point in the same direction and so the scalar product then has the cosine being equal to 1. Sorry for that, my wording was not quite precise enough. Vasco Vasco, I should have realized that. I have been thinking about this, trying to get to your understanding of it. Here is my present thinking on the method of (ii): It seems easiest to begin with S because, for example, in the case f(S) has zero curvature at p and S must lie on a line parallel to the real line (which we know in the case of $e^z$ from inspection of $f(z) = e^x e^{iy}$). Now I’m going to reverse your example. We $\mathbf{guess}$ that $\mathcal{K}$ is positive, which means we draw it as $i\widehat{\xi}$ (i.e., to direction Pi/2) at p and rotate S to be parallel with it. The direction of rotation of S should not matter, because all that really matters is the location of p as the argument of f(p). Now we move p in the direction of increasing y, which is also the direction of $\mathcal{K}$. Then, if our guess about $\mathcal{K}$ is correct, f(p) should display counterclockwise rotation. If f(p) displays clockwise rotation, that would be a contradiction indicating that $\mathcal{K}$ is actually negative. Applying this to the other cases seems a little murkier. In the case where f(S) has zero curvature at p and S is a ray or line parallel to a ray, as in log(z), we apparently choose a ray for S and guess at the proper orientation of $\mathcal{K}$, which in any case is orthogonal to S. Let us guess that positive $\mathcal{K}$ points in the direction of positive rotation of p. We then align S with $\mathcal{K}$ by rotation of S in either direction until they are parallel. As you showed, when we move p on S, we alter arg(p). This will affect the angle of log(p) = ln|p| + i arg(p), but the relation is complex, because arg(f(p)) = arg(log(p)) = arctan(arg(p)/ln|p|. What would drive arg(log(p)) in one direction or another? It would seem to depend on the quadrant of arg(p) and whether ln|p| > 0 or < 0. There seems to be no intrinsic orientation of S. The important thing is the movement of p. As I understand your answer, if we have guessed $\mathcal{K}$ to be positive and movement of p along rotated S increases the angle of p (positive rotation), then if f(p) shows positive curvature, then we have guessed correctly about $\mathcal{K}$. I think you might have dealt with some of this in your appendix, which I intend to study. Gary |
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Post by Admin on Mar 11, 2016 10:10:35 GMT
Gary
I'm not sure what you mean by $S$ at the beginning of the above, especially where you say "...$S$ must lie on a line parallel...". The phrase "we guess that $\mathcal{K}$ is positive" should presumably be "...we guess that $|\mathcal{K}|$ is positive." or maybe "we guess that $\mathcal{K}$ points in the positive direction of $S$" Also "...indicating that $\mathcal{K}$ is actually negative." is probably better written as "indicating that $\mathcal{K}$ points in the other direction", because $\mathcal{K}$ here is in general a complex number.
I think "Let us guess that positive $\mathcal{K}$ points in the direction of positive rotation of $p$." should be written as "Let us guess that $\mathcal{K}$ points in the direction $ie^{i\theta}$" We can't say "positive $\mathcal{K}$" as $\mathcal{K}$ is a complex number, and I think using $ie^{i\theta}$ as the direction for $\mathcal{K}$ is precise and shows that if we vary $\arg p$ then this direction also varies.
As you are looking at the appendix in my document, you will see that we can be very precise about the curve $f(S)$ in figure 5(b) of my document. There is a symmetry about the point $f(p)$ and if we drew the two horizontal lines defined by $\theta=\phi\pm\pi/2$ they would be asymptotes for the curve $f(S)$. If we move $p$ around in the complex plane then the general shape of this curve remains the same as it moves around on the complex plane.
The important thing is the movement of $f(p)$ as $p$ moves along $S$ in figure 5(a). Does this movement represent positive or negative curvature?
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 11, 2016 15:31:08 GMT
That first sentence should have read: "...in the case $e^z$, f(S) has zero curvature ...". Also, omit "Now I'm going to reverse your example", as I think I only did that in the second example just to see how the logic would work out.
Vasco,
I believe I have resolved all my questions regarding Needham’s questions and your answers to the exercises 17: (i)-(iv). I have accepted the idea that when Needham says “Which?” he may have allowed for us to try out different directions of $\mathcal{K}$ rather than requiring a correct answer immediately. My answers might differ in a few respects, but I think the basic reasoning is the same as yours.
I have started working on part (vi) applied to log(z). The question seems to call for the application of the approach developed in (i)-(iv) as well as some use of amplitwist. But I’m encountering problems in seeing how (i)-(iv) would apply to this problem. In this case, f(z) applies to the infinitesimal shape $\mathscr{Q}$ and it’s clones at the perimeter, rather than to a line S. Needham speaks of “the direction of $\mathcal{K}$” (31, ln 3). In a sense, every perimeter circle represents a line S drawn from the center of Q to its own center, so one could imaging 12 different $\mathcal{K}$s, each based on one of the perimeter circles. For log(z), $\mathcal{K}$ = $-ie^{i\theta}$, so we could calculate a $\mathcal{K}(\theta)$ for the angle of each perimeter circle. Then we could move $\mathscr{Q}$ along this $\mathcal{K}$ to a perimeter circle and apply f(S), which should produce maximum curvature (seen in the angle of the bar?) for this $\mathcal{K}(\theta)$. This all seems too complicated. Do you have any thoughts on this?
Attachment moved to March 22, 2016, after correction of plot of $e^z$ in (vi) according to Vasco's suggestion.
Gary
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