Order of poles on a function
$begingroup$
How can I determine what the order of the pole on the following function is:
$$ f(z)= frac{e^{bz}}{zsinh(az)}$$
From the Laurent series, I found that the residue would be b/a or -b/a, however, I am confused whether this is a simple pole or a pole of second order.
complex-analysis roots laurent-series
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
$endgroup$
|
show 4 more comments
$begingroup$
How can I determine what the order of the pole on the following function is:
$$ f(z)= frac{e^{bz}}{zsinh(az)}$$
From the Laurent series, I found that the residue would be b/a or -b/a, however, I am confused whether this is a simple pole or a pole of second order.
complex-analysis roots laurent-series
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
$endgroup$
$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
1
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
|
show 4 more comments
$begingroup$
How can I determine what the order of the pole on the following function is:
$$ f(z)= frac{e^{bz}}{zsinh(az)}$$
From the Laurent series, I found that the residue would be b/a or -b/a, however, I am confused whether this is a simple pole or a pole of second order.
complex-analysis roots laurent-series
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
$endgroup$
How can I determine what the order of the pole on the following function is:
$$ f(z)= frac{e^{bz}}{zsinh(az)}$$
From the Laurent series, I found that the residue would be b/a or -b/a, however, I am confused whether this is a simple pole or a pole of second order.
complex-analysis roots laurent-series
complex-analysis roots laurent-series
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
edited Dec 6 '18 at 20:49
A.Γ.
22.7k32656
22.7k32656
asked Dec 6 '18 at 20:37
LechugaLechuga
105
asked Dec 6 '18 at 20:37
LechugaLechuga
105
asked Dec 6 '18 at 20:37
LechugaLechuga
105
105
$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
1
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
|
show 4 more comments
$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
1
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
1
1
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= frac{e^{bz}}{zsinh(az)} = frac{2e^{bz}}{z}left(e^z-e^{-z}right)^{-1} = frac{e^{bz}}{z}sum_{k = -infty}^infty frac{(-1)^kz}{(pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
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add a comment |
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$begingroup$
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= frac{e^{bz}}{zsinh(az)} = frac{2e^{bz}}{z}left(e^z-e^{-z}right)^{-1} = frac{e^{bz}}{z}sum_{k = -infty}^infty frac{(-1)^kz}{(pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
add a comment |
$begingroup$
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= frac{e^{bz}}{zsinh(az)} = frac{2e^{bz}}{z}left(e^z-e^{-z}right)^{-1} = frac{e^{bz}}{z}sum_{k = -infty}^infty frac{(-1)^kz}{(pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
add a comment |
$begingroup$
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= frac{e^{bz}}{zsinh(az)} = frac{2e^{bz}}{z}left(e^z-e^{-z}right)^{-1} = frac{e^{bz}}{z}sum_{k = -infty}^infty frac{(-1)^kz}{(pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= frac{e^{bz}}{zsinh(az)} = frac{2e^{bz}}{z}left(e^z-e^{-z}right)^{-1} = frac{e^{bz}}{z}sum_{k = -infty}^infty frac{(-1)^kz}{(pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 10 '18 at 12:40
Brevan EllefsenBrevan Ellefsen
11.7k31649
11.7k31649
add a comment |
add a comment |
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$begingroup$
How can you know the residue without the order of the pole?
$endgroup$
– DaveNine
Dec 6 '18 at 20:43
$begingroup$
Is it not possible? I thought the Laurent series would give me the residue at the z^-1 term.
$endgroup$
– Lechuga
Dec 6 '18 at 20:44
1
$begingroup$
Do you mean the pole at $z=0?$ Anyway, the numerator is never $0$ so the order of the pole is just the order of the zero of the denominator.
$endgroup$
– saulspatz
Dec 6 '18 at 20:44
$begingroup$
Yes I do mean the pole at z = 0.
$endgroup$
– Lechuga
Dec 6 '18 at 20:45
$begingroup$
@saulspatz does that mean that it is of order two? Since you would have a zero relating to sinh and one to z?
$endgroup$
– Lechuga
Dec 6 '18 at 20:45