Gary
GaryVasco
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Post by Gary on May 23, 2017 22:38:13 GMT
Vasco, I did the exercise, but I'm not sure if "order m" at a double or triple pole applies to the pair/triplet or to every pole. nh.ch7.notes.p366.pdf (123.51 KB) Gary |
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Post by Admin on May 24, 2017 11:02:26 GMT
Gary
$P(z)=\frac{\displaystyle 1}{\displaystyle\sin z}$ has a pole whenever $z$ is a multiple of $\pi$ and since $(\sin z)'=\cos z\neq0$ at each pole, each pole is of order 1 i.e. simple.
So $P(z)$ has an infinite number of simple poles.
$Q(z)=\frac{\displaystyle\cos z}{\displaystyle z^2}=\frac{\displaystyle\cos z}{\displaystyle(z-0)^2}$ has a double pole at $z=0$.
$R(z)=\frac{\displaystyle 1}{\displaystyle (e^z-1)^3}$ has a triple pole whenever $e^z=1$ or $z$ is a multiple of $2\pi i$. So $R(z)$ has an infinite number of triple poles.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 24, 2017 13:26:35 GMT
Gary $P(z)=\frac{\displaystyle 1}{\displaystyle\sin z}$ has a pole whenever $z$ is a multiple of $\pi$ and since $(\sin z)'=\cos z\neq0$ at each pole, each pole is of order 1 i.e. simple. So $P(z)$ has an infinite number of simple poles. $Q(z)=\frac{\displaystyle\cos z}{\displaystyle z^2}=\frac{\displaystyle\cos z}{\displaystyle(z-0)^2}$ has a double pole at $z=0$. $R(z)=\frac{\displaystyle 1}{\displaystyle (e^z-1)^3}$ has a triple pole whenever $e^z=1$ or $z$ is a multiple of $2\pi i$. So $R(z)$ has an infinite number of triple poles. Vasco Vasco, Yes, so far so good, but what if the first nonvanishing derivative of $F = 1/Q(z)$ or $F = 1/R(z)$ has order 2 or 3? As I read it again now, the order of a (one) pole equals the order of the first nonvanishing derivative of $1/f$. $Q(z)$ has a double pole and $F^{(2)}(0) = 2$, so do each of the two poles of $Q(z)$ have order 2, or do they together have order 2? Is the order of $R(z)$ equal to 3 or is it 3 x infinity, which is infinite. Or do we only speak of the order of a pole, not of a function? I think the problem here is the ambiguity of "a pole". Can "a pole" refer to "a double pole" or "a triple pole". That would make the most sense here. Gary |
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Post by Admin on May 24, 2017 14:33:11 GMT
Gary $P(z)=\frac{\displaystyle 1}{\displaystyle\sin z}$ has a pole whenever $z$ is a multiple of $\pi$ and since $(\sin z)'=\cos z\neq0$ at each pole, each pole is of order 1 i.e. simple. So $P(z)$ has an infinite number of simple poles. $Q(z)=\frac{\displaystyle\cos z}{\displaystyle z^2}=\frac{\displaystyle\cos z}{\displaystyle(z-0)^2}$ has a double pole at $z=0$. $R(z)=\frac{\displaystyle 1}{\displaystyle (e^z-1)^3}$ has a triple pole whenever $e^z=1$ or $z$ is a multiple of $2\pi i$. So $R(z)$ has an infinite number of triple poles. Vasco Vasco, Yes, so far so good, but what if the first nonvanishing derivative of $F = 1/Q(z)$ or $F = 1/R(z)$ has order 2 or 3? As I read it again now, the order of a (one) pole equals the order of the first nonvanishing derivative of $1/f$. $Q(z)$ has a double pole and $F^{(2)}(0) = 2$, so do each of the two poles of $Q(z)$ have order 2, or do they together have order 2? Is the order of $R(z)$ equal to 3 or is it 3 x infinity, which is infinite. Or do we only speak of the order of a pole, not of a function? I think the problem here is the ambiguity of "a pole". Can "a pole" refer to "a double pole" or "a triple pole". That would make the most sense here. Gary Gary $Q$ has a double pole as you say, but double here means the order of the(one) pole. There is only one pole at $z=0$ not two. Your final statement is the way things are. Poles are like roots in this respect; a function has a root which can have an order of 1,2,3 etc. Same with poles. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 24, 2017 14:55:33 GMT
Vasco, Yes, so far so good, but what if the first nonvanishing derivative of $F = 1/Q(z)$ or $F = 1/R(z)$ has order 2 or 3? As I read it again now, the order of a (one) pole equals the order of the first nonvanishing derivative of $1/f$. $Q(z)$ has a double pole and $F^{(2)}(0) = 2$, so do each of the two poles of $Q(z)$ have order 2, or do they together have order 2? Is the order of $R(z)$ equal to 3 or is it 3 x infinity, which is infinite. Or do we only speak of the order of a pole, not of a function? I think the problem here is the ambiguity of "a pole". Can "a pole" refer to "a double pole" or "a triple pole". That would make the most sense here. Gary Gary $Q$ has a double pole as you say, but double here means the order of the(one) pole. There is only one pole at $z=0$ not two. Your final statement is the way things are. Poles are like roots in this respect; a function has a root which can have an order of 1,2,3 etc. Same with poles. Vasco Vasco, Thank you. I think that clears it up. I had started to realize that I was making it too difficult and that I should just make the analogy with roots. Gary |
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