Complex Analysis. Showing absolute convergence of principal part of Laurent series for a holomorphic...
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I am given that $f: D(P, r) setminus { P } to mathbb{C}$ is holomorphic. I want to show that the principal part of the Laurent series expansion for $f$ converges absolutely on $mathbb{C} setminus {P}$. Confused on how to proceed.
I know that the convergence is absolute on $D(P, r) setminus { P }$. I don't see how to extend the principal part to $mathbb{C}setminus{P}$.
sequences-and-series complex-analysis
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I am given that $f: D(P, r) setminus { P } to mathbb{C}$ is holomorphic. I want to show that the principal part of the Laurent series expansion for $f$ converges absolutely on $mathbb{C} setminus {P}$. Confused on how to proceed.
I know that the convergence is absolute on $D(P, r) setminus { P }$. I don't see how to extend the principal part to $mathbb{C}setminus{P}$.
sequences-and-series complex-analysis
$endgroup$
add a comment |
$begingroup$
I am given that $f: D(P, r) setminus { P } to mathbb{C}$ is holomorphic. I want to show that the principal part of the Laurent series expansion for $f$ converges absolutely on $mathbb{C} setminus {P}$. Confused on how to proceed.
I know that the convergence is absolute on $D(P, r) setminus { P }$. I don't see how to extend the principal part to $mathbb{C}setminus{P}$.
sequences-and-series complex-analysis
$endgroup$
I am given that $f: D(P, r) setminus { P } to mathbb{C}$ is holomorphic. I want to show that the principal part of the Laurent series expansion for $f$ converges absolutely on $mathbb{C} setminus {P}$. Confused on how to proceed.
I know that the convergence is absolute on $D(P, r) setminus { P }$. I don't see how to extend the principal part to $mathbb{C}setminus{P}$.
sequences-and-series complex-analysis
sequences-and-series complex-analysis
asked Dec 7 '18 at 0:29
user623740
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1 Answer
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The principal part gets better as you get farther from the pole, since it has only negative powers of $(z - P)$. Apply a comparison test, using the fact that the series is convergent at some point $z ne P$.
In particular, you can use the fact that
$$left|frac{a_n}{(z - P)^n}right| le frac{|a_n|}{r^n}$$
for $z notin mathbb{D}(P, r)$.
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
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Thank you. I will try this out.
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– user623740
Dec 7 '18 at 0:36
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So basically take the limit of the right-hand side as $r to infty$?
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– user623740
Dec 7 '18 at 1:33
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No, because $r$ is just some fixed number.
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– T. Bongers
Dec 7 '18 at 1:34
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Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The principal part gets better as you get farther from the pole, since it has only negative powers of $(z - P)$. Apply a comparison test, using the fact that the series is convergent at some point $z ne P$.
In particular, you can use the fact that
$$left|frac{a_n}{(z - P)^n}right| le frac{|a_n|}{r^n}$$
for $z notin mathbb{D}(P, r)$.
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
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$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
add a comment |
$begingroup$
The principal part gets better as you get farther from the pole, since it has only negative powers of $(z - P)$. Apply a comparison test, using the fact that the series is convergent at some point $z ne P$.
In particular, you can use the fact that
$$left|frac{a_n}{(z - P)^n}right| le frac{|a_n|}{r^n}$$
for $z notin mathbb{D}(P, r)$.
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
add a comment |
$begingroup$
The principal part gets better as you get farther from the pole, since it has only negative powers of $(z - P)$. Apply a comparison test, using the fact that the series is convergent at some point $z ne P$.
In particular, you can use the fact that
$$left|frac{a_n}{(z - P)^n}right| le frac{|a_n|}{r^n}$$
for $z notin mathbb{D}(P, r)$.
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
$endgroup$
The principal part gets better as you get farther from the pole, since it has only negative powers of $(z - P)$. Apply a comparison test, using the fact that the series is convergent at some point $z ne P$.
In particular, you can use the fact that
$$left|frac{a_n}{(z - P)^n}right| le frac{|a_n|}{r^n}$$
for $z notin mathbb{D}(P, r)$.
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:34
T. BongersT. Bongers
23.1k54662
23.1k54662
$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
add a comment |
$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
Thank you. I will try this out.
$endgroup$
– user623740
Dec 7 '18 at 0:36
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
So basically take the limit of the right-hand side as $r to infty$?
$endgroup$
– user623740
Dec 7 '18 at 1:33
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
$begingroup$
No, because $r$ is just some fixed number.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:34
add a comment |
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