Radius of Convergence for two power series is equal.
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Given any two power series $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ , if there is some m ∈ N such that $a_n$= $b_n$ for all n ≥ m, then $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ have the same radius of convergence.
I have found this stated in my book, but it is given without proof and I am unsure as to how this is simply proved.
complex-analysis analysis power-series
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
$endgroup$
add a comment |
$begingroup$
Given any two power series $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ , if there is some m ∈ N such that $a_n$= $b_n$ for all n ≥ m, then $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ have the same radius of convergence.
I have found this stated in my book, but it is given without proof and I am unsure as to how this is simply proved.
complex-analysis analysis power-series
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
$endgroup$
add a comment |
$begingroup$
Given any two power series $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ , if there is some m ∈ N such that $a_n$= $b_n$ for all n ≥ m, then $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ have the same radius of convergence.
I have found this stated in my book, but it is given without proof and I am unsure as to how this is simply proved.
complex-analysis analysis power-series
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
$endgroup$
Given any two power series $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ , if there is some m ∈ N such that $a_n$= $b_n$ for all n ≥ m, then $sum_{n=0}^{infty}{a_n{(z - a)}^n}$ and $sum_{n=0}^{infty}{b_n{(z - b)}^n}$ have the same radius of convergence.
I have found this stated in my book, but it is given without proof and I am unsure as to how this is simply proved.
complex-analysis analysis power-series
complex-analysis analysis power-series
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
edited Dec 1 '18 at 16:18
José Carlos Santos
156k22125227
156k22125227
asked Dec 1 '18 at 16:12
SuiriSuiri
32
asked Dec 1 '18 at 16:12
SuiriSuiri
32
asked Dec 1 '18 at 16:12
SuiriSuiri
32
32
add a comment |
add a comment |
1 Answer
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It's because the radius of convergence is$$frac1{limsup_{ninmathbb N}sqrt[n]{lvert a_nrvert}}=frac1{limsup_{ninmathbb N}sqrt[n]{lvert b_nrvert}}$$(since $a_n=b_n$ is $n$ is large enough).
A more basic approach consists in using that fact that, if you have two series $sum_{n=0}^infty z_n$ and $sum_{n=0}^infty w_n$ and if $z_n=w_n$ if $n$ is large enough, then either both series converge or both series diverge.
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
It's because the radius of convergence is$$frac1{limsup_{ninmathbb N}sqrt[n]{lvert a_nrvert}}=frac1{limsup_{ninmathbb N}sqrt[n]{lvert b_nrvert}}$$(since $a_n=b_n$ is $n$ is large enough).
A more basic approach consists in using that fact that, if you have two series $sum_{n=0}^infty z_n$ and $sum_{n=0}^infty w_n$ and if $z_n=w_n$ if $n$ is large enough, then either both series converge or both series diverge.
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
add a comment |
$begingroup$
It's because the radius of convergence is$$frac1{limsup_{ninmathbb N}sqrt[n]{lvert a_nrvert}}=frac1{limsup_{ninmathbb N}sqrt[n]{lvert b_nrvert}}$$(since $a_n=b_n$ is $n$ is large enough).
A more basic approach consists in using that fact that, if you have two series $sum_{n=0}^infty z_n$ and $sum_{n=0}^infty w_n$ and if $z_n=w_n$ if $n$ is large enough, then either both series converge or both series diverge.
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
add a comment |
$begingroup$
It's because the radius of convergence is$$frac1{limsup_{ninmathbb N}sqrt[n]{lvert a_nrvert}}=frac1{limsup_{ninmathbb N}sqrt[n]{lvert b_nrvert}}$$(since $a_n=b_n$ is $n$ is large enough).
A more basic approach consists in using that fact that, if you have two series $sum_{n=0}^infty z_n$ and $sum_{n=0}^infty w_n$ and if $z_n=w_n$ if $n$ is large enough, then either both series converge or both series diverge.
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
It's because the radius of convergence is$$frac1{limsup_{ninmathbb N}sqrt[n]{lvert a_nrvert}}=frac1{limsup_{ninmathbb N}sqrt[n]{lvert b_nrvert}}$$(since $a_n=b_n$ is $n$ is large enough).
A more basic approach consists in using that fact that, if you have two series $sum_{n=0}^infty z_n$ and $sum_{n=0}^infty w_n$ and if $z_n=w_n$ if $n$ is large enough, then either both series converge or both series diverge.
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
edited Dec 1 '18 at 16:29
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 1 '18 at 16:16
José Carlos SantosJosé Carlos Santos
156k22125227
156k22125227
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
add a comment |
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
$begingroup$
Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :)
$endgroup$
– Suiri
Dec 1 '18 at 16:31
add a comment |
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