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Post by telemeter on Mar 5, 2019 9:20:17 GMT
Vasco in the errata considers the first sentence of 33 ii) to be an error, saying w=cos z should read w = cos-1 z.
I do not agree as I completed the exercise as stated in the text...but maybe my last step was invalid. Thoughts appreciated.
Let w = cos z = 0.5*(eiz + e-iz)
Multiply through by eiz to get a quadratic :
e2iz - 2*eiz*w +1 = 0
eiz = w +/- (w2 - 1)^(.5)
Then take Log of both sides
z = cos-1w = -i*Log[w +/- sqrt(w2-1)]
Here's the key bit... w is just a complex number so we can use another symbol for it, choose z, so equally
cos-1 z = -i*Log[z +/- sqrt(z2-1)] as per the book, (apart from justifying the removal of the +/-)
In other words, the z in the initial equation was just a dummy and has precious little to do with the z in the derived formula which is true for any complex number.
Alternatively, if we re-write the question as z= cos w (as per Vasco), then the next paragraph would have to ask for a quadratic in eiw rather than eiz, which would be admittedly a bit more elegant.
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Gary
GaryVasco
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Post by Gary on Mar 5, 2019 16:19:23 GMT
Vasco in the errata considers the first sentence of 33 ii) to be an error, saying w=cos z should read w = cos -1 z. I do not agree as I completed the exercise as stated in the text...but maybe my last step was invalid. Thoughts appreciated. Let w = cos z = 0.5*(e iz + e -iz) Multiply through by e iz to get a quadratic : e 2iz - 2*e iz*w +1 = 0 e iz = w +/- (w 2 - 1)^(.5) Then take Log of both sides z = cos -1w = -i*Log[w +/- sqrt(w 2-1)] Here's the key bit... w is just a complex number so we can use another symbol for it, choose z, so equally cos -1 z = -i*Log[z +/- sqrt(z 2-1)] as per the book, (apart from justifying the removal of the +/-) In other words, the z in the initial equation was just a dummy and has precious little to do with the z in the derived formula which is true for any complex number. Alternatively, if we re-write the question as z= cos w (as per Vasco), then the next paragraph would have to ask for a quadratic in e iw rather than e iz, which would be admittedly a bit more elegant. One can swap the roles of z and w at the beginning of the derivation or at the end, so it looks as though it was not a mistake. But it might be conceptually easier to see the z's in their final positions earlier rather than later. But in later chapters, Needham often defines w and z planes where w = f(z). I wonder if one would still write $cos^{-1}z = -i Log[z \pm \sqrt{z^2 - 1}]$ given $w = cos z$.
Gary
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Post by Admin on Mar 5, 2019 19:49:48 GMT
Vasco in the errata considers the first sentence of 33 ii) to be an error, saying w=cos z should read w = cos -1 z. I do not agree as I completed the exercise as stated in the text...but maybe my last step was invalid. Thoughts appreciated. Let w = cos z = 0.5*(e iz + e -iz) Multiply through by e iz to get a quadratic : e 2iz - 2*e iz*w +1 = 0 e iz = w +/- (w 2 - 1)^(.5) Then take Log of both sides z = cos -1w = -i*Log[w +/- sqrt(w 2-1)] Here's the key bit... w is just a complex number so we can use another symbol for it, choose z, so equally cos -1 z = -i*Log[z +/- sqrt(z 2-1)] as per the book, (apart from justifying the removal of the +/-) In other words, the z in the initial equation was just a dummy and has precious little to do with the z in the derived formula which is true for any complex number. Alternatively, if we re-write the question as z= cos w (as per Vasco), then the next paragraph would have to ask for a quadratic in e iw rather than e iz, which would be admittedly a bit more elegant. Hi When looking at exercises with several parts as in this case, and having done many of Needham's exercises like this, one finds that there is almost always a theme or connection between the parts. Sometimes one finds (as in this case) that even though the parts are meant to be connected some inconsistencies in notation occur and can cause confusion. By attempting one part of the exercise in isolation we can miss these inconsistences. Looking at part (i) of this exercise (33 on page 119), we see that we start by looking at the multifunction $\cos^{-1}(z)$. In my opinion, it's almost certain that, given part (i), Needham intended to write the same equation in part (ii). The starting point of part (ii), $w=\cos z$ is inconsistent with the other parts and so it seems like a reasonable thing to do to consider this an error since the change I suggest makes the exercise as a whole totally consistent. However if we look at part (i) with its reference to figure 26 on page 88 we find that we also need to swap all occurrences of $w$ and $z$ in figure 26. These changes seem to me to be the best way to make the exercise and its associated diagram 26 consistent. I do not claim that it is the only rewrite possible of this exercise, but its the one I prefer, and the exercise certainly cannot be acceptable the way it is printed in the current versions of the book. Vasco PS Since Needham uses $\text{Log}$ with a capital "L" for the principal branch of the logarithm I think that here we should use $\log$, the multifunction logarithm.
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Gary
GaryVasco
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Post by Gary on Mar 5, 2019 21:43:10 GMT
Vasco in the errata considers the first sentence of 33 ii) to be an error, saying w=cos z should read w = cos -1 z. I do not agree as I completed the exercise as stated in the text...but maybe my last step was invalid. Thoughts appreciated. Let w = cos z = 0.5*(e iz + e -iz) Multiply through by e iz to get a quadratic : e 2iz - 2*e iz*w +1 = 0 e iz = w +/- (w 2 - 1)^(.5) Then take Log of both sides z = cos -1w = -i*Log[w +/- sqrt(w 2-1)] Here's the key bit... w is just a complex number so we can use another symbol for it, choose z, so equally cos -1 z = -i*Log[z +/- sqrt(z 2-1)] as per the book, (apart from justifying the removal of the +/-) In other words, the z in the initial equation was just a dummy and has precious little to do with the z in the derived formula which is true for any complex number. Alternatively, if we re-write the question as z= cos w (as per Vasco), then the next paragraph would have to ask for a quadratic in e iw rather than e iz, which would be admittedly a bit more elegant. Hi When looking at exercises with several parts as in this case, and having done many of Needham's exercises like this, one finds that there is almost always a theme or connection between the parts. Sometimes one finds (as in this case) that even though the parts are meant to be connected some inconsistencies in notation occur and can cause confusion. Looking at part (i) of this exercise (33 on page 119), we see that we start by looking at the multifunction $\cos^{-1}(z)$. Then in part (ii) we end up looking at the same function as in part (i) and also in part (iii). However the starting point of part (ii), $w=\cos z$ is inconsistent with the other parts and so it seems like a reasonable thing to do to consider this an error since the change I suggest makes the exercise as a whole totally consistent. However if we look at part (i) with its reference to figure 26 on page 88 we find that we also need to swap all occurrences of $w$ and $z$ in figure 26. These changes seem to me to be the best way to make the exercise and its associated diagram 26 consistent. I do not claim that it is the only rewrite possible of this exercise, but its the one I prefer, and the exercise certainly cannot be acceptable the way it is printed in the current versions of the book. Vasco
If we try to make the problem consistent with Figure [26], which seems a good idea, then we would keep $w = cos\,z$, but parts (i) and (ii) would be rewritten as (i) Use [26] to discuss the branch points of the multifunction $cos^{-1}(w)$. (ii) Rewrite the equation $w = cos\ z$ as a quadratic in $e^{iz}$. By solving this equation, deduce that $cos^{-1}(w) = -i\ log[w + \sqrt{w^2 - 1}]$. Since it is now $w$ that is traversing the loop, $z$’s would also be replaced with $w$’s in (iii).
Gary
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Post by Admin on Mar 5, 2019 23:12:50 GMT
Yes I agree that this is also a way of re-writing the exercise and it could be argued that this way is better than my initial proposal since it leaves the diagram unchanged. So I think each person can do their own re-write. The main thing is to be aware that as it appears in the book it is inconsistent and requires a re-write.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 6, 2019 2:05:09 GMT
Yes I agree that this is also a way of re-writing the exercise and it could be argued that this way is better than my initial proposal since it leaves the diagram unchanged. So I think each person can do their own re-write. The main thing is to be aware that as it appears in the book it is inconsistent and requires a re-write. Vasco
I think Vasco anticipated my suggested rewrite. I did some poking around in my files and found the following note in my solution from Jan 1, 2015, which I never posted. I will do so, but it contains some other material that I need to revisit. At that time, I didn't see the solution with the quadratic, but reached the correct answer by using Euler's equation.
“Vasco says: Rewrite the equation $w = cos\ z$ as a quadratic in $e^{iz}$. By solving this equation, deduce that $cos^{-1}\ w = -i\ log[w + \sqrt{w^2-1}]$. This would mean that we should also rewrite part (iii) as: Show that as w travels along a loop that goes once round either 1 or -1 (but not both), the value of $[w + \sqrt{w^2-1}]$ changes to $\frac{1}{[w + \sqrt{w^2-1}]}$.”
Gary
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Post by telemeter on Mar 7, 2019 11:32:21 GMT
Interesting discussion. I must admit that by the time I had got to the end of my post I had changed my mind...the jarring I felt when I did the question should have been a big enough clue that this was not as Needham intended. Now on Chapter 6, I am impressed how every part of this book weaves together as a whole and agree with Vasco that the elegance evident everywhere was absent from the phrasing of this part.
Yes my Log should have been a log...carelessness.
As an aside, my justification for saying that there was no need to show the negative square root was that it was redundant. The negative root, only differing from the positive by and argument of pi, would be automatically covered on another branch of the log function (which repeats every 2*pi, which of course Log does not!). Have I got the right approach?
telemeter
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Gary
GaryVasco
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Post by Gary on Mar 7, 2019 21:59:38 GMT
When questions go without replies for a time, possibly days, it generally means we are puzzling over them, not that we don't want to answer them promptly.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 17, 2019 0:12:51 GMT
Interesting discussion. I must admit that by the time I had got to the end of my post I had changed my mind...the jarring I felt when I did the question should have been a big enough clue that this was not as Needham intended. Now on Chapter 6, I am impressed how every part of this book weaves together as a whole and agree with Vasco that the elegance evident everywhere was absent from the phrasing of this part. Yes my Log should have been a log...carelessness. As an aside, my justification for saying that there was no need to show the negative square root was that it was redundant. The negative root, only differing from the positive by and argument of pi, would be automatically covered on another branch of the log function (which repeats every 2*pi, which of course Log does not!). Have I got the right approach? telemeter
After rereading p. 90, paragraph 2 on the root as a multifunction and p. 99 on Log vrs log, I agree with you that it would be redundant and that is essentially the answer to Needham's question (not some clever numerical proof as I at first thought and spent some time on). But I think the second sentence doesn't quite capture it. Every branch of the log function would operate on both values of the square root, so it would not exactly be covered by a different branch of log. For the function $z = cos^{-1} w = -i log[w + \sqrt{w^2-1}]$, there would be two values of $(w + \sqrt{w^2-1})$ and an infinite number of branches of log, because $\hspace{5em} log [w + \sqrt{w^2-1}] = ln|w + \sqrt{w^2-1}| + i \, arg (w + \sqrt{w^2-1})$. The negative root would have to be considered in every branch of log, so the total number of values is potentially $2 \cdot \infty$. That is, a branch of log for every integer times two values for the square root, as in the second equation of Section VII The Logarithm Function (p. 98). This is essentially 2 values for every time that $w$ makes a complete traversal of the ellipse in Fig [26], p. 88. Now having thought about it a bit more, I can see how one can think of the two values of the square root as creating two branches of log, but then one must still describe the infinity of branches. I'm not presenting this on a stone tablet. This is just the way I see it. Gary
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Gary
GaryVasco
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Post by Gary on Mar 17, 2019 5:32:06 GMT
I attach a plot I made while trying to understand the multifunctions in this exercise:
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Post by Admin on Mar 17, 2019 18:28:17 GMT
I attach a plot I made while trying to understand the multifunctions in this exercise: Gary In your document you say that the square root function has two branches. I think it has one branch point at the origin of order 1, and an infinite number of branches. The $\log$ function has one branch point at the origin of infinite order (usually called a logarithmic branch point) and an infinite number of branches. Vasco
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Gary
GaryVasco
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Post by Gary on Mar 17, 2019 19:36:30 GMT
I attach a plot I made while trying to understand the multifunctions in this exercise:
Gary In your document you say that the square root function has two branches. I think it has one branch point at the origin of order 1, and an infinite number of branches. The $\log$ function has one branch point at the origin of infinite order (usually called a logarithmic branch point) and an infinite number of branches. Vasco Vasco, I see the error and will correct it. I was confusing the "value" of a multifunction with the number of branches. Gary
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Gary
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Post by Gary on Mar 17, 2019 20:36:13 GMT
I have edited both the document and my reply to telemeter.
Gary
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Post by Admin on Jun 12, 2019 13:22:53 GMT
Gary In your document you say that the square root function has two branches. I think it has one branch point at the origin of order 1, and an infinite number of branches. The $\log$ function has one branch point at the origin of infinite order (usually called a logarithmic branch point) and an infinite number of branches. Vasco Vasco, I see the error and will correct it. I was confusing the "value" of a multifunction with the number of branches. Gary Gary I was browsing the forum and noticed this post. I don't know why I claimed that the square root had an infinite number of branches. Surely you were right when you said it has two? Vasco
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Gary
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Post by Gary on Jun 12, 2019 14:11:43 GMT
Vasco, I see the error and will correct it. I was confusing the "value" of a multifunction with the number of branches. Gary Gary I was browsing the forum and noticed this post. I don't know why I claimed that the square root had an infinite number of branches. Surely you were right when you said it has two? Vasco Vasco,
Thank you. I don't know why I accepted the claim of infinite branches so easily except that I was feeling uncertain about my grasp of the idea of "branch". With the log, it's easy to see, as the imaginary part is an angle. But maybe you were right about the square root. This author seems to agree with you <http://mathfaculty.fullerton.edu/mathews/c2003/ComplexFunBranchMod.html>. Start with paragraph 2.20.
After thinking about Ch 12, Ex 12 some more, I realize my grasp of chapters 11 and 12 is in shambles. Another read is in order. But must go now, as the thermometer is headed for 85 F and I have to remove some brush (known around here as "fuels") from the lot.
Gary
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