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Uniform Convergence of Laurent Series using Weierstrass M-test
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$G=z in mathbbC: 1<$. Suppose $sum_n=-infty^infty alpha_nz^n$ converges absolutely pointwise on some open neighborhood of $barG=z in mathbbC: 1 leq $. We need to show that $sum_n=-infty^infty alpha_nz^n$ converges unifomly on $barG=z in mathbbC: 1 leq $.
Absolute convergence of series gave that $sum|alpha_n|$ converges. Now I wish to use the Weierstrass M-test. The uniform convergence of $sum_n=1^infty fracalpha_-nz^n$ is done. I am not able to show the uniform convergence of positive power series.
complex-analysis
asked Aug 6 at 23:22
Arindam
193113
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up vote
0
down vote
favorite
$G=z in mathbbC: 1<$. Suppose $sum_n=-infty^infty alpha_nz^n$ converges absolutely pointwise on some open neighborhood of $barG=z in mathbbC: 1 leq $. We need to show that $sum_n=-infty^infty alpha_nz^n$ converges unifomly on $barG=z in mathbbC: 1 leq $.
Absolute convergence of series gave that $sum|alpha_n|$ converges. Now I wish to use the Weierstrass M-test. The uniform convergence of $sum_n=1^infty fracalpha_-nz^n$ is done. I am not able to show the uniform convergence of positive power series.
complex-analysis
asked Aug 6 at 23:22
Arindam
193113
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$G=z in mathbbC: 1<$. Suppose $sum_n=-infty^infty alpha_nz^n$ converges absolutely pointwise on some open neighborhood of $barG=z in mathbbC: 1 leq $. We need to show that $sum_n=-infty^infty alpha_nz^n$ converges unifomly on $barG=z in mathbbC: 1 leq $.
Absolute convergence of series gave that $sum|alpha_n|$ converges. Now I wish to use the Weierstrass M-test. The uniform convergence of $sum_n=1^infty fracalpha_-nz^n$ is done. I am not able to show the uniform convergence of positive power series.
complex-analysis
asked Aug 6 at 23:22
Arindam
193113
$G=z in mathbbC: 1<$. Suppose $sum_n=-infty^infty alpha_nz^n$ converges absolutely pointwise on some open neighborhood of $barG=z in mathbbC: 1 leq $. We need to show that $sum_n=-infty^infty alpha_nz^n$ converges unifomly on $barG=z in mathbbC: 1 leq $.
Absolute convergence of series gave that $sum|alpha_n|$ converges. Now I wish to use the Weierstrass M-test. The uniform convergence of $sum_n=1^infty fracalpha_-nz^n$ is done. I am not able to show the uniform convergence of positive power series.
complex-analysis
asked Aug 6 at 23:22
Arindam
193113
asked Aug 6 at 23:22
Arindam
193113
asked Aug 6 at 23:22
Arindam
193113
asked Aug 6 at 23:22
Arindam
193113
193113
add a comment |Â
add a comment |Â
1 Answer
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Any open neighborhood of $overline G$ contains a point with $|z| >2$ so $sum |alpha_n| (2+epsilon)^n <infty$ for some $epsilon >0$. Now you can apply M-test.
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Any open neighborhood of $overline G$ contains a point with $|z| >2$ so $sum |alpha_n| (2+epsilon)^n <infty$ for some $epsilon >0$. Now you can apply M-test.
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
add a comment |Â
up vote
2
down vote
Any open neighborhood of $overline G$ contains a point with $|z| >2$ so $sum |alpha_n| (2+epsilon)^n <infty$ for some $epsilon >0$. Now you can apply M-test.
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Any open neighborhood of $overline G$ contains a point with $|z| >2$ so $sum |alpha_n| (2+epsilon)^n <infty$ for some $epsilon >0$. Now you can apply M-test.
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
Any open neighborhood of $overline G$ contains a point with $|z| >2$ so $sum |alpha_n| (2+epsilon)^n <infty$ for some $epsilon >0$. Now you can apply M-test.
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
answered Aug 6 at 23:33
Kavi Rama Murthy
21.2k2830
21.2k2830
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
add a comment |Â
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
It is some open neighborhood in $barG$, why does it contain a point with $|z|>2$?
– Arindam
Aug 6 at 23:37
1
1
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Let $U$ be a neigborhood of $overline G$. $z=2$ is in $overline Gsubset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w in U$. In particular $2+frac r 2 in U$.
– Kavi Rama Murthy
Aug 6 at 23:40
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
Oh, I was considering an open set inside $barG$. Thank you.
– Arindam
Aug 6 at 23:44
add a comment |Â
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