How does negative power of laurent series vanish?
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Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $a_n$ is zero for negative $n$. I apply Cauchy-Goursat thereom since $f(z)(z)^n$ for positive n is analytic on the annulus except for one point. Is this valid?
complex-analysis
asked Nov 26 at 6:46
user43529463
16118
add a comment |
up vote
0
down vote
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Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $a_n$ is zero for negative $n$. I apply Cauchy-Goursat thereom since $f(z)(z)^n$ for positive n is analytic on the annulus except for one point. Is this valid?
complex-analysis
asked Nov 26 at 6:46
user43529463
16118
Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $a_n$ is zero for negative $n$. I apply Cauchy-Goursat thereom since $f(z)(z)^n$ for positive n is analytic on the annulus except for one point. Is this valid?
complex-analysis
asked Nov 26 at 6:46
user43529463
16118
Suppose the function is an analytic function on ${omega={z:0<|z|le 1}$}.Moreover $f$ satisfies $ |f(z)| le frac{1}{|z|^{1/2}}|sin(frac{1}{|z|})|$ for all $z$ in $omega$. $a_n$ is the coefficient of Laurent series. I want to show that $a_n$ is zero for negative $n$. I apply Cauchy-Goursat thereom since $f(z)(z)^n$ for positive n is analytic on the annulus except for one point. Is this valid?
complex-analysis
complex-analysis
asked Nov 26 at 6:46
user43529463
16118
asked Nov 26 at 6:46
user43529463
16118
edited Nov 26 at 7:41
asked Nov 26 at 6:46
user43529463
16118
asked Nov 26 at 6:46
user43529463
16118
asked Nov 26 at 6:46
user43529463
16118
16118
Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04
add a comment |
Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04
Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04
add a comment |
1 Answer
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up vote
2
down vote
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Let $r in (0,1)$ and let $n in mathbb Z$ such that $n le -1$. We have
$a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| le r^{-n-1/2} sin(1/r)$.
Since $r in (0,1)$ was arbitrary and $sin(1/r)$ is bounded, we get with $r to 0$ that $a_n=0$.
answered Nov 26 at 6:58
Fred
43.6k1644
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $r in (0,1)$ and let $n in mathbb Z$ such that $n le -1$. We have
$a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| le r^{-n-1/2} sin(1/r)$.
Since $r in (0,1)$ was arbitrary and $sin(1/r)$ is bounded, we get with $r to 0$ that $a_n=0$.
answered Nov 26 at 6:58
Fred
43.6k1644
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
add a comment |
up vote
2
down vote
accepted
Let $r in (0,1)$ and let $n in mathbb Z$ such that $n le -1$. We have
$a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| le r^{-n-1/2} sin(1/r)$.
Since $r in (0,1)$ was arbitrary and $sin(1/r)$ is bounded, we get with $r to 0$ that $a_n=0$.
answered Nov 26 at 6:58
Fred
43.6k1644
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $r in (0,1)$ and let $n in mathbb Z$ such that $n le -1$. We have
$a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| le r^{-n-1/2} sin(1/r)$.
Since $r in (0,1)$ was arbitrary and $sin(1/r)$ is bounded, we get with $r to 0$ that $a_n=0$.
answered Nov 26 at 6:58
Fred
43.6k1644
Let $r in (0,1)$ and let $n in mathbb Z$ such that $n le -1$. We have
$a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| le r^{-n-1/2} sin(1/r)$.
Since $r in (0,1)$ was arbitrary and $sin(1/r)$ is bounded, we get with $r to 0$ that $a_n=0$.
answered Nov 26 at 6:58
Fred
43.6k1644
edited Nov 26 at 12:43
answered Nov 26 at 6:58
Fred
43.6k1644
answered Nov 26 at 6:58
Fred
43.6k1644
answered Nov 26 at 6:58
Fred
43.6k1644
43.6k1644
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
add a comment |
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Is it because the series expansion has to be valid for all r and is unique in the annulus , so you can take limit r tends to zero and get the answer?
– user43529463
Nov 26 at 7:40
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
Yes, we have for each(!) $r in (0,1)$ that $a_n= frac{1}{2 pi i} int_{|z|=r} frac{f(w)}{w^{n+1}}.$
– Fred
Nov 26 at 7:43
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
what is the meaning of(!)?
– user43529463
Nov 26 at 7:55
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
It is an emphasis on the word "each".
– Fred
Nov 26 at 7:57
add a comment |
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Define $a_n$. Are you talking about Laurent series expansion?
– Kavi Rama Murthy
Nov 26 at 7:23
yes. it is laurent series expansion
– user43529463
Nov 26 at 8:04