Gary
GaryVasco
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Post by Gary on Nov 25, 2017 17:13:43 GMT
Vasco,
I can't get past 8 (ii). If I read the problem correctly, it appears that $\sum_{n=2}^\infty(-\frac{p}{n}) = \frac{1}{i \, n \, sin(\frac{\pi}{n})}$, where $n = 2, 4, 6, 8, \cdot \cdot \cdot$. One could rewrite the RHS as $\frac{1}{i n}$ times a cosecant series but I don't get anywhere with it. The sentence strikes me as ambiguous between $F_n(z)$ in the UHP and $z$ in the UHP. How does the UHP constraint come into play? $p$ is a pole at every $n$. Are the $p$'s associated with some of these even $n$'s ineligible as poles?
Gary
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Post by Admin on Nov 26, 2017 14:13:25 GMT
Gary
Here are a couple of hints that might help you see a way through:
1. Notice that $F_n(z)$ has a finite number $n$, of poles, so your summation should not be an infinite series.
2. Look at the result just over halfway down on page 24 about roots of a polynomial with real coefficients and then think about the roots of unity and then rewrite the RHS of the result in part (i).
If these don't help, then come back with more questions.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 27, 2017 19:23:44 GMT
Gary Here are a couple of hints that might help you see a way through: 1. Notice that $F_n(z)$ has a finite number $n$, of poles, so your summation should not be an infinite series. 2. Look at the result just over halfway down on page 24 about roots of a polynomial with real coefficients and then think about the roots of unity and then rewrite the RHS of the result in part (i). If these don't help, then come back with more questions. Vasco Vasco, The hints were a great help. Here is my answer: nh.ch9.ex8.pdf (382.39 KB). Gary |
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Post by Admin on Nov 28, 2017 17:10:24 GMT
Gary
I haven't looked at your answer in detail yet, but from a quick look it would seem that you have not split the integral round $L+J$ into two integrals and shown that one of them is zero, which I think is the way it has to be done. If you read subsection 3 on pages 436-437 you will see what I mean.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 28, 2017 17:18:29 GMT
Gary I haven't looked at your answer in detail yet, but from a quick look it would seem that you have not split the integral round $L+J$ into two integrals and shown that one of them is zero, which I think is the way it has to be done. If you read subsection 3 on pages 436-437 you will see what I mean. Vasco OK, I'll take another look. |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 28, 2017 23:02:53 GMT
Gary I haven't looked at your answer in detail yet, but from a quick look it would seem that you have not split the integral round $L+J$ into two integrals and shown that one of them is zero, which I think is the way it has to be done. If you read subsection 3 on pages 436-437 you will see what I mean. Vasco Vasco, I have revised my answer and reposted. Gary |
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Post by Admin on Nov 29, 2017 16:30:14 GMT
Gary
I have now published my answer to exercise 8. I will post again soon with some comments on your latest answer.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 29, 2017 17:01:15 GMT
Gary I have now published my answer to exercise 8. I will post again soon with some comments on your latest answer. Vasco Vasco, I have read your answer and I can see what you did, which all looks good to me. I am going to revise mine, so I will be interested in your comments. Shouldn't the approach in this problem be termed "roots of negative unity"? Gary |
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Post by Admin on Nov 30, 2017 8:57:08 GMT
Gary
That's a great title! Negative unity has a lot of interesting and amusing connotations.
I noticed that if you have the roots of either 1 or -1 for a given value of $n$ then the other roots can be obtained by multiplying by $e^{i\pi/n}$.
If you draw the $n$ roots of either 1 or -1 on the unit circle then the above multiplication amounts to rotating the circle through an angle $\pm(\pi/n)$ to get the other roots. So each time you rotate through this angle you flick between one set of roots and the other.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 30, 2017 18:05:39 GMT
Vasco,
It's a handy point of reference and the following a nice insight regarding the rotation, but did you mean to write $e^{i2\pi/n}$ multiplied successively to the previous root? it's the ratio for the geometric series for both the roots and the residues. Multiplying by $e^{i\pi/n}$ would give 2n roots on the unit circle.
Gary
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Post by Admin on Nov 30, 2017 18:41:38 GMT
Gary
No, I'm talking about having a set of $n$ roots say for $z^n+1$, and multiplying or rotating each one by $e^{i\pi/n}$ to obtain the set of $n$ roots for $z^n-1$.
Multiplying by $e^{i2\pi/n}$ changes a given root to the next root in the same set.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 1, 2017 1:26:29 GMT
Gary No, I'm talking about having a set of $n$ roots say for $z^n+1$, and multiplying or rotating each one by $e^{i\pi/n}$ to obtain the set of $n$ roots for $z^n-1$. Multiplying by $e^{i2\pi/n}$ changes a given root to the next root in the same set. Vasco Got it. Interesting. root_test.bmp (916.04 KB) |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 1, 2017 6:10:07 GMT
question withdrawn
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Post by Admin on Dec 1, 2017 6:15:08 GMT
Gary
Thanks for that. It illustrates it perfectly!
Vasco
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