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Post by Admin on Dec 4, 2017 15:22:31 GMT
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Gary
GaryVasco
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Post by Gary on Dec 4, 2017 15:34:51 GMT
Vasco, I didn't know about the algebra you presented in the first sentence after (1), but it's just what I needed. Otherwise, it appears to me that the answers are similar. Gary |
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Post by Admin on Dec 4, 2017 16:57:28 GMT
Gary
In your answer, what is the reason for using $f(z)=1/z^n$ on page 1? As I understand it, the only restriction on $f(z)$ is the one stipulated in the exercise.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 4, 2017 20:24:34 GMT
Gary In your answer, what is the reason for using $f(z)=1/z^n$ on page 1? As I understand it, the only restriction on $f(z)$ is the one stipulated in the exercise. Vasco Vasco, No good reason. There are other problems, too, so I am reworking it. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 4, 2017 22:42:42 GMT
Gary In your answer, what is the reason for using $f(z)=1/z^n$ on page 1? As I understand it, the only restriction on $f(z)$ is the one stipulated in the exercise. Vasco Vasco, It is rewritten with the general $f(z)$. I'm not sure I agree that the inequalities demonstrating that $|csc(\pi z)| \rightarrow 0$ with sufficiently large |z| (or N) can be omitted. I read the last sentence on p. 440 as restricted to $f(z) cot(\pi z)$. Gary |
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Post by Admin on Dec 5, 2017 0:11:26 GMT
Gary
I didn't intend to give that impression, I just left it to the reader to prove because it's so easy and so similar to the proofs for $\cot (\pi z)$ and I thought it would be a distraction from the main thrust of the argument.
I can add it as an addendum at the end of the exercise if you think it is better not left to the reader.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 5, 2017 0:59:22 GMT
Gary I didn't intend to give that impression, I just left it to the reader to prove because it's so easy and so similar to the proofs for $\cot (\pi z)$ and I thought it would be a distraction from the main thrust of the argument. I can add it as an addendum at the end of the exercise if you think it is better not left to the reader. Vasco Vasco, I can see your point. You make it clear. For this reader, it was a good thing to practice. It was easier than the cotangent, but not entirely trivial. I wondered if it would have been better form to call it "done" as soon as it was shown that $|csc(\pi z)| \rightarrow 0$ and omit the integral <= M (perimeter) part. Gary |
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Post by Admin on Dec 5, 2017 16:49:00 GMT
Gary
Are you referring to your own answer here?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 6, 2017 15:20:22 GMT
Gary Are you referring to your own answer here? Vasco Vasco, Yes, my own. Gary |
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Post by Admin on Dec 6, 2017 18:35:47 GMT
Gary Are you referring to your own answer here? Vasco Vasco, Yes, my own. Gary Gary I suppose you could have, but it doesn't make the proof that much longer to include it, so I would say write it the way you feel most comfortable with. In the last paragraph of your document you say that $-1$ and $1$ are poles of $1/(z^2+1)$. I think this should be $i$ and $-i$ unless you meant to write $1/(z^2-1)$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 6, 2017 21:52:05 GMT
Gary I suppose you could have, but it doesn't make the proof that much longer to include it, so I would say write it the way you feel most comfortable with. In the last paragraph of your document you say that $-1$ and $1$ are poles of $1/(z^2+1)$. I think this should be $i$ and $-i$ unless you meant to write $1/(z^2-1)$. Vasco Vasco, Thank you. It must be $1/(z^2-1)$, as $i$ and $-i$ are not instances of n. Gary |
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