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Post by Admin on Jan 16, 2016 7:31:31 GMT
When I read part (ii) of this exercise for the first time I was a bit confused by the wording. I would like therefore to submit a rewording which I think makes it a lot clearer what the author intended:
(ii) The complex curvature $\mathcal{K}$ must therefore point in one of the two directions which are orthogonal to the direction of $S$ chosen in part (i). Which? By considering the image of $S$ when it ($S$ that is) points in the direction you decided on in part (i), deduce the value of $|\mathcal{K}|$, and thereby conclude that $\mathcal{K}(p)=ie^{-x}$.
Vasco
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Post by Admin on Jan 18, 2016 19:59:10 GMT
After a lot of work trying to do exercise 17 I have now changed my ideas again. This time I feel confident that I have found the essence of the problem. Here is my rewriting of part (ii):
The complex curvature $\mathcal{K}$ must therefore point in one of the two directions which are orthogonal to the direction of $S$ obtained in part (i). Which? By considering the curvature of the image of $S$ when $S$ points in the direction you have just decided on for $\mathcal{K}$, deduce the value of $|\mathcal{K}|$, and thereby conclude that $\mathcal{K}=ie^{-x}$.
In my opinion the three words in green above are missing from the exercise statement in the book.
This formulation has enabled me to find a solution to part (ii) when $f(z)=e^z$. I will now hopefully be able to carry on and complete the whole exercise.
Vasco
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Gary
GaryVasco
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Post by Gary on Feb 21, 2016 15:11:33 GMT
Vasco,
I had forgottten to look in the errata, so I had not seen your suggestions for a rewrite of 17. I agree with your rewrite of Jan 18, and I think you will find that my attempt is based upon the same approach. However, I find that I am still unable to use that information make a confident deduction of the value of |$\mathcal{K}$| and proceed from that to $\mathcal{K}(p) = ie^{-x}$. So I look forward to your answers.
Gary
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Post by Admin on Feb 21, 2016 19:39:30 GMT
Gary See if the attached document helps. documentVasco
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Gary
GaryVasco
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Post by Gary on Feb 21, 2016 20:28:10 GMT
Vasco,
Once again, you have presented the key that I missed in the reading. I am trying to read more carefully, but I’m evidently not quite as sharp as I should be. I seem to follow biases that distract me from the correct interpretations. The key is
I had read the passage to mean point S in the direction orthogonal to $\mathcal{K}$, whatever the direction of $\mathcal{K}$ might be, so I was unable to deduce the curvature at p. Now it is clear! I will redo my answer to take this into account. For the magnitude, I had written $\kappa(e) = |\mathcal{K}|cos(Pi/2)$, but it should be $\kappa(e) = |\mathcal{K}| cos(0) = |\mathcal{K}|$.
In “The method of part (ii)”, you finish the first paragraph with “we now know the magnitude of K”, but we know it only in a general way. Do you agree?
I also concluded that S had to be on the real axis (or rather, as you say, be parallel to it).
After some further consideration, it seems to me that we have not quite arrived at an answer to the problem. The document does not deduce the value $\mathcal{K} = i/e^x$ from geometrical considerations without calculation. We know the direction, but how do we deduce that $|\mathcal{K}| = 1/e^x$? This seems especially difficult since we agree that S is parallel to the real line and $e^x$ is also parallel to the real line, which makes it appear as though $\mathcal{K}$ = 0 and the radius of curvature = $\infty$. The value $|\mathcal{K}| = 1/e^x$? makes it appear that there is a circle of curvature of radius $e^x$. This is a point of confusion for me.
This document is very nice, especially the introduction. It would be good to include some of it in your answer.
Gary
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Post by Admin on Feb 21, 2016 22:24:10 GMT
Gary
The exercise only asks us to do part (i) without calculation. Also I think you are confusing $\tilde{\kappa}$, the curvature, which is a real number and $\mathcal{K}$ the complex curvature. These are two different things. If you read the document again you will see that $\tilde{\kappa}$ is calculated as $e^{-x}$ not $\mathcal{K}$, $\tilde{\kappa}$ is only equal to $|\mathcal{K}|$ when $S$ and $\mathcal{K}$ are in the same direction. Did you notice that the document has two pages? The explanation of how to calculate the curvature is explained on page 2.
Vasco
PS In the document $\kappa$ should be $\tilde{\kappa}$. I will correct this in my document.
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Gary
GaryVasco
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Post by Gary on Feb 22, 2016 2:01:39 GMT
, Gary The exercise only asks us to do part (i) without calculation. Also I think you are confusing $\tilde{\kappa}$, the curvature, which is a real number and $\mathcal{K}$ the complex curvature. These are two different things. If you read the document again you will see that $\tilde{\kappa}$ is calculated as $e^{-x}$ not $\mathcal{K}$, $\tilde{\kappa}$ is only equal to $|\mathcal{K}|$ when $S$ and $\mathcal{K}$ are in the same direction. Did you notice that the document has two pages? The explanation of how to calculate the curvature is explained on page 2. Vasco ... Vasco, I don't think I've been confusing $\tilde{\kappa}$ and $\mathcal{K}$ in general, though I think there is temptation to confuse the two for the reasons that you explain. I saw the second page, but I didn't follow the reasoning out completely. It appeared to me on that reading that you were merely calculating the value, which seemed to me to not comply with the instructions. I have reread it and I see now how you used the formula for $\tilde{\kappa}$ to deduce the value of $|\mathcal{K}|$. Problem solved, I think. Gary
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Post by Admin on Feb 22, 2016 8:40:22 GMT
Gary
I just realised that I hadn't replied to your query:
If you mean not as a numeric value like $3$, then I agree. In general $|\mathcal{K}|$ will be a function of $z$, as in the case when $f(z)=e^z$ and $|\mathcal{K}|=e^{-x}$. In this case, regardless of the orientation of $S$, $|\mathcal{K}|$ is just a function of the real part of $z$. The same goes for the direction of $\mathcal{K}$. In general it will be a function of $z$ only, whereas, again in the case of $f(z)=e^z$, it is constant and equal to $i$.
I am hoping to produce a solution to the last exercise of this chapter, exercise 31, over the next few days- tempting fate, it may turn out to be another '17'! I will then return to looking at exercise 17 again and hopefully also finish my commentary on section 9.
Are you happy to use "calculation" in part (ii), or do you think we should use purely geometric methods to find $|\mathcal{K}|$?
Vasco
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Gary
GaryVasco
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Post by Gary on Feb 23, 2016 16:53:23 GMT
Gary I just realised that I hadn't replied to your query: If you mean not as a numeric value like $3$, then I agree. In general $|\mathcal{K}|$ will be a function of $z$, as in the case when $f(z)=e^z$ and $|\mathcal{K}|=e^{-x}$. In this case, regardless of the orientation of $S$, $|\mathcal{K}|$ is just a function of the real part of $z$. The same goes for the direction of $\mathcal{K}$. In general it will be a function of $z$ only, whereas, again in the case of $f(z)=e^z$, it is constant and equal to $i$. I am hoping to produce a solution to the last exercise of this chapter, exercise 31, over the next few days- tempting fate, it may turn out to be another '17'! I will then return to looking at exercise 17 again and hopefully also finish my commentary on section 9. Are you happy to use "calculation" in part (ii), or do you think we should use purely geometric methods to find $|\mathcal{K}|$? Vasco Vasco, The method of part (i) in the document produces a direction for $\mathcal{K}$, but the problem requests a direction for S after $\mathcal{K}$ has been determined by the function and the choice of p. At least for the log(z) and $z^m$, I think we need to consider $\overline{f''}/\overline{f'}$ as the direction of an S that is orthogonal to $\mathcal{K}$. The wording suggests avoiding calculation to me, but I doubt one can answer the problem without at least noticing that in the case of $e^z$, f''/f' = 1. It seems the line between deduction considering the formula and actual calculation with the formula is a little fuzzy. I imagine have a long road to travel before arriving at 31, but I look forward to seeing it when I get there. I will post immediately a revised answer to 17 in the exercise folder based on sentence 1 of this message and your comments there. Gary
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Post by Admin on Feb 24, 2016 9:54:37 GMT
Referring to exercise 17 part (ii): The wording suggests avoiding calculation to me, but I doubt one can answer the problem without at least noticing that in the case of $e^z$, f''/f' = 1. It seems the line between deduction considering the formula and actual calculation with the formula is a little fuzzy. Gary You may well be right. Maybe that's why Needham says in the square brackets of part (ii): I was wondering why he used the word "see" rather than "calculate", and if you do calculate, you can obtain an exact expression. I will give this some more thought, and have a go at it using diagrams etc. as you suggest. The more I think about it, the more it makes sense with the wording of the exercise. Vasco
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Gary
GaryVasco
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Post by Gary on Feb 24, 2016 16:41:51 GMT
I was wondering why he used the word "see" rather than "calculate", and if you do calculate, you can obtain an exact expression. I will give this some more thought, and have a go at it using diagrams etc. as you suggest. The more I think about it, the more it makes sense with the wording of the exercise. Vasco Vasco, Yes, and perhaps because one can deduce the exact value for f''/f' in the case of $e^z$, but with the other two functions, one must know the value of p. Gary
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Post by Admin on Mar 7, 2016 17:40:21 GMT
Gary
After considering our previous exchanges on this exercise I have now changed my mind. As I said at the time, your comments about the wording of parts (i) and (ii), and especially (ii), made me do a complete rethink of the exercise and I do not consider my previous rewording of part (ii) to be correct, except where it clarifies what Needham has written.
I have published my answers to parts (i)-(v) with part (vi) to follow later I hope.
Part (i) I now consider that the phrase "Without calculation" in part (i) means "Without calculation of the curvature".
Part (ii) Here is a possible rewrite of this part which now contains what are for me just clarifications:
The complex curvature $\mathcal{K}$ must therefore point in one of the two directions orthogonal to the direction of $S$ decided on in part (i). Which? By considering the image of $S$ when $S$ points in the direction of $\mathcal{K}$ that you have just decided on, deduce the value of $|\mathcal{K}|$, and thereby conclude that $\mathcal{K}(p)=ie^{-x}$.
All this is of course only my own personal interpretation of what Needham intended, as are my answers to the various parts.
Thanks for your previous comments which enabled me to arrive at this particular point, and I would be interested to know what you think of my solutions if you have enough time to wade through them.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 12, 2016 22:11:07 GMT
I would be interested to know what you think of my solutions if you have enough time to wade through them. Vasco, I have waded in a few steps, but I'm trying to avoid stepping off where I don't know the depth, so these comments pertain just to the beginning of the archived answer. I will probably follow this with more, but it would be good to come to an understanding on these questions before proceeding to log(z) and $z^m$. As I think I said, on first reading of archived 17, it all looked straightforward, but as I tried to implement the ideas in a plot, it didn't seem to work out. Then I tried to figure out why. I think I can now make my question regarding 5:17 more coherent. Needham wrote in (ii) “The complex curvature $\mathcal{K}$ must therefore point in one of two directions. Which? We know it must be $\pm i\hat{\xi}$, but we don’t know which. We are clearly being asked to choose the correct direction, because the next sentence is “By considering the image of S when it points in this direction, deduce the value of $\mathcal{K}$, …”. When we are asked to choose the direction, we do not yet have the image of S when it points in that direction. So we must choose before plotting and considering the image. You wrote in the archived answer, p. 1, “The method of part (ii)”, line “We now have to decide in which of the two directions $\pm i\hat{\xi}$, the complex curvature $\mathcal{K}$ points.” But you don’t actually decide, or at least the criteria are not explicit. You write “Let’s write this direction as $e^{i\alpha}$”, but that doesn't tell us whether it is $i\hat{\xi}$ or $-i\hat{\xi}$. Then you rotate S and consider the image of f(S) on rotated S “when it points in the direction of $\mathcal{K}$”, but the direction of $\mathcal{K}$ is still undetermined. Then you calculate $\tilde{\kappa}$ as an absolute value. In the following paragraph, you write “Knowing the direction of $\mathcal{K}$ and the magnitude of $\mathcal{K}$, we can write down the expression for $\mathcal{K}$ as $\mathcal{K} = |\mathcal{K}| \cdot e^{i\alpha}$. But the direction of $\mathcal{K}$ is still undetermined or unexplained. So there is still an open question of how to pick the right direction of $\mathcal{K}$. Is it $i\hat{\xi}$ or $-i\hat{\xi}$ ? What criteria can we use before plotting f(p)? There seems to be no criterion without doing some inspection of the derivatives or the quotient of the first and second derivatives. Now consider the second part of the answer to the problem of $e^z$ on p. 4. Here you examine f(p) prior to making the decision on the direction of $\mathcal{K}$. You conclude that since p moves positively on the y axis, and f(p) moves anticlockwise around the origin, “$\mathcal{K}$ must point in the positive direction of the imaginary axis”. This seems like a reasonable conclusion but the steps are out of order. I wonder if the question is impossible to answer without first plotting or calculating a valid direction for $\mathcal{K}$. Gary
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Post by Admin on Mar 12, 2016 23:21:36 GMT
Gary
Reading your post I think that you have misunderstood what i have written. In the preamble I am just outlining the method. When I say let the direction be $e^{i\alpha}$ I am just assuming that we have found this direction so that I can go on outlining the method. This is not meant to be an answer to the exercise.
In part (ii) of my document we do know that $\mathcal{K}$ points in one of two directions - that's all I am assuming. We know everything about $S$ except which way it points.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 12, 2016 23:46:45 GMT
Vasco,
Agreed. But don't you agree that Needham is asking us to make the choice prior to considering the image of S when it points in the chosen direction?
I won't be able to reply to another post immediately.
Gary
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