Mixing
Continuity of $sum_{ngeq 0} a_n x^n$ on $(-1,1)$
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If the series $sum_{ngeq 0} a_n$ converges, does it follow that the power series $sum_{ngeq 0} a_n x^n$ converge to a continuous function on $(-1,1)$?
Will showing uniform convergence of $sum_{ngeq 0} a_n x^n$ help?
real-analysis continuity power-series
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add a comment |
$begingroup$
If the series $sum_{ngeq 0} a_n$ converges, does it follow that the power series $sum_{ngeq 0} a_n x^n$ converge to a continuous function on $(-1,1)$?
Will showing uniform convergence of $sum_{ngeq 0} a_n x^n$ help?
real-analysis continuity power-series
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$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
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– Mindlack
Jan 26 at 23:28
1
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@Mindlack Abel says that if it defined on 1, it is left continuous there
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– Holo
Jan 26 at 23:30
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@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31
add a comment |
$begingroup$
If the series $sum_{ngeq 0} a_n$ converges, does it follow that the power series $sum_{ngeq 0} a_n x^n$ converge to a continuous function on $(-1,1)$?
Will showing uniform convergence of $sum_{ngeq 0} a_n x^n$ help?
real-analysis continuity power-series
$endgroup$
If the series $sum_{ngeq 0} a_n$ converges, does it follow that the power series $sum_{ngeq 0} a_n x^n$ converge to a continuous function on $(-1,1)$?
Will showing uniform convergence of $sum_{ngeq 0} a_n x^n$ help?
real-analysis continuity power-series
real-analysis continuity power-series
asked Jan 26 at 23:25
user397197
$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
$endgroup$
– Mindlack
Jan 26 at 23:28
1
$begingroup$
@Mindlack Abel says that if it defined on 1, it is left continuous there
$endgroup$
– Holo
Jan 26 at 23:30
$begingroup$
@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31
add a comment |
$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
$endgroup$
– Mindlack
Jan 26 at 23:28
1
$begingroup$
@Mindlack Abel says that if it defined on 1, it is left continuous there
$endgroup$
– Holo
Jan 26 at 23:30
$begingroup$
@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31
$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
$endgroup$
– Mindlack
Jan 26 at 23:28
$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
$endgroup$
– Mindlack
Jan 26 at 23:28
1
1
$begingroup$
@Mindlack Abel says that if it defined on 1, it is left continuous there
$endgroup$
– Holo
Jan 26 at 23:30
$begingroup$
@Mindlack Abel says that if it defined on 1, it is left continuous there
$endgroup$
– Holo
Jan 26 at 23:30
$begingroup$
@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31
$begingroup$
@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31
add a comment |
1 Answer
1
active
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For $|x| leq 1-epsilon$ the series $sum |a_n| |x^{n}|$ is dominated by the convergent series $C(1-epsilon)^{n}$ where $C=sup ]{|a_n|:ngeq 1}$. By M-test, the given series converges uniformly for $|x| leq 1-epsilon$ and hence its sum is continuous there. Since $epsilon $ is arbitrary we are done. Note that $C <infty$ because $a_n to 0$.
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
$endgroup$
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But we do not have that $a_n$ is absolutely convergent
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– user397197
Jan 26 at 23:33
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He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
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– Clayton
Jan 26 at 23:34
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@Matematleta The question has nothing to do with convergence at $x=-1$.
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– Kavi Rama Murthy
Jan 26 at 23:52
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Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
$begingroup$
For $|x| leq 1-epsilon$ the series $sum |a_n| |x^{n}|$ is dominated by the convergent series $C(1-epsilon)^{n}$ where $C=sup ]{|a_n|:ngeq 1}$. By M-test, the given series converges uniformly for $|x| leq 1-epsilon$ and hence its sum is continuous there. Since $epsilon $ is arbitrary we are done. Note that $C <infty$ because $a_n to 0$.
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
$endgroup$
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
add a comment |
$begingroup$
For $|x| leq 1-epsilon$ the series $sum |a_n| |x^{n}|$ is dominated by the convergent series $C(1-epsilon)^{n}$ where $C=sup ]{|a_n|:ngeq 1}$. By M-test, the given series converges uniformly for $|x| leq 1-epsilon$ and hence its sum is continuous there. Since $epsilon $ is arbitrary we are done. Note that $C <infty$ because $a_n to 0$.
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
$endgroup$
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
add a comment |
$begingroup$
For $|x| leq 1-epsilon$ the series $sum |a_n| |x^{n}|$ is dominated by the convergent series $C(1-epsilon)^{n}$ where $C=sup ]{|a_n|:ngeq 1}$. By M-test, the given series converges uniformly for $|x| leq 1-epsilon$ and hence its sum is continuous there. Since $epsilon $ is arbitrary we are done. Note that $C <infty$ because $a_n to 0$.
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
$endgroup$
For $|x| leq 1-epsilon$ the series $sum |a_n| |x^{n}|$ is dominated by the convergent series $C(1-epsilon)^{n}$ where $C=sup ]{|a_n|:ngeq 1}$. By M-test, the given series converges uniformly for $|x| leq 1-epsilon$ and hence its sum is continuous there. Since $epsilon $ is arbitrary we are done. Note that $C <infty$ because $a_n to 0$.
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
edited Jan 26 at 23:35
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
answered Jan 26 at 23:32
Kavi Rama MurthyKavi Rama Murthy
69.2k53169
69.2k53169
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
add a comment |
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
But we do not have that $a_n$ is absolutely convergent
$endgroup$
– user397197
Jan 26 at 23:33
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
He doesn't use that $sum a_n$ converges absolutely anywhere (instead, he proves that $sum a_nx^n$ converges absolutely and uniformly for $|x|leq1-varepsilon$).
$endgroup$
– Clayton
Jan 26 at 23:34
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
@Matematleta The question has nothing to do with convergence at $x=-1$.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:52
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
$begingroup$
Yes, I misunderstood the question. I deleted my comment. I thought the OP was asking about convergence at the boundary. The more interesting fact is that if $sum a_n$ converges then $f(x)to sum a_n$ as $xto 1.$
$endgroup$
– Matematleta
Jan 26 at 23:54
add a comment |
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$begingroup$
Yes and yes. I am quite sure you can easily prove uniform convergence on $[-r,r]$, $1>r >0$. And from Abel’s theorem (iirc) the series is defined and left continuous at $x=1$.
$endgroup$
– Mindlack
Jan 26 at 23:28
1
$begingroup$
@Mindlack Abel says that if it defined on 1, it is left continuous there
$endgroup$
– Holo
Jan 26 at 23:30
$begingroup$
@Mindlack I don't know of Abel's theorem. Can it be proved in some other way?
$endgroup$
– user397197
Jan 26 at 23:31