Gary
GaryVasco
Posts: 3,352 
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Post by Gary on May 17, 2017 4:53:48 GMT
Vasco,
How would you interpret "root of order n"?
Gary
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Post by Admin on May 17, 2017 5:30:13 GMT
Vasco, How would you interpret "root of order n"? Gary Gary If $f$ has a root $p$ of order $n$ then $f=(z-p)^ng(z)$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 17, 2017 17:02:16 GMT
Vasco, How would you interpret "root of order n"? Gary Gary If $f$ has a root $p$ of order $n$ then $f=(z-p)^ng(z)$. Vasco Vasco, Thank you. That is what I concluded. Here is a similar question: Regarding Ex. 19 (iv), the problem statement appears to contradict the discussion of critical points on p. 205, $\mathbb{S}$ 2 Breakdown of Conformality, where we find "We quantify the degree of this strange behavior [of $z^m$] by saying that z = 0 is a critical point of order (m-1)." Problem (iv) introduces a "critical point of order m" and "the case m = 1". Part (i) is a little ambiguous as to whether f is analytic, but I think Needham intends it to be analytic. I'm having difficulty determining just what is signified by "the case m = 1". Assuming for simplicity that the function is a simple power function, would the question refer to $f = z$ or to $f = z^2$. If I had to guess, I would pick $f = z^2$? What do you think? Gary |
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Post by Admin on May 18, 2017 8:26:54 GMT
Gary If $f$ has a root $p$ of order $n$ then $f=(z-p)^ng(z)$. Vasco Vasco, Thank you. That is what I concluded. Here is a similar question: Regarding Ex. 19 (iv), the problem statement appears to contradict the discussion of critical points on p. 205, $\mathbb{S}$ 2 Breakdown of Conformality, where we find "We quantify the degree of this strange behavior [of $z^m$] by saying that z = 0 is a critical point of order (m-1)." Problem (iv) introduces a "critical point of order m" and "the case m = 1". Part (i) is a little ambiguous as to whether f is analytic, but I think Needham intends it to be analytic. I'm having difficulty determining just what is signified by "the case m = 1". Assuming for simplicity that the function is a simple power function, would the question refer to $f = z$ or to $f = z^2$. If I had to guess, I would pick $f = z^2$? What do you think? Gary Gary As I understand it, a root of order m is a critical point of order m-1 since the first m-1 derivatives at the root are zero. So I'd go with $f^2$. I don't have access to the book at the moment as you know. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 18, 2017 21:14:34 GMT
Vasco, Thank you. That is what I concluded. Here is a similar question: Regarding Ex. 19 (iv), the problem statement appears to contradict the discussion of critical points on p. 205, $\mathbb{S}$ 2 Breakdown of Conformality, where we find "We quantify the degree of this strange behavior [of $z^m$] by saying that z = 0 is a critical point of order (m-1)." Problem (iv) introduces a "critical point of order m" and "the case m = 1". Part (i) is a little ambiguous as to whether f is analytic, but I think Needham intends it to be analytic. I'm having difficulty determining just what is signified by "the case m = 1". Assuming for simplicity that the function is a simple power function, would the question refer to $f = z$ or to $f = z^2$. If I had to guess, I would pick $f = z^2$? What do you think? Gary Gary As I understand it, a root of order m is a critical point of order m-1 since the first m-1 derivatives at the root are zero. So I'd go with $f^2$. I don't have access to the book at the moment as you know. Vasco Vasco, Thank you. I take it you meant to write $f = z^2$. I think there is more to discuss when you have the book in hand. Gary |
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Post by Admin on May 18, 2017 21:32:49 GMT
Gary
Yes I meant $f=z^2$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 20, 2017 16:42:40 GMT
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