$f(z)=sum_n=1^infty a_n(z-z_0)^n$ such that $sum_n=0^infty f^n(a)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$sum|a_n|^2<infty$ and $sum a_n<infty$ implies $sumlog(1+a_n)$ convergesConvergence of $prod_n=1^infty(1+a_n)$Is $sum_n=2^inftylogleft(1+frac(-1)^nsqrtnright)$ convergent?Prove that $sum_n=1^infty fraca_nn^z$ converges absolutely and uniformlyIf $sum_n=1^infty |a_n|^2<infty$, Then : $sum_n=1^infty a_n$ Converges $Leftrightarrow prod_n=1^infty(1+a_n)$ConvergesIf $(a_n)$ is a complex sequence such that $sum_n=1^inftyfraca_nk^n = 0,forall kin mathbbN$ then $a_n = 0$ for all $ n$Prove that a power series $sum_n=0^inftya_nz^n$ which converges for any $z in mathbbN$, converges for any $z in mathbbC$.If the complex series $sum_n=0^inftya_n$ converges, show that there exists a positive number $A$ such that $|a_n| leq A$ for all $n$.Show that $sum_n=0^infty(sum_j=0^n a_jb_n-j)$ converges to $(sum_n=0^inftyb_n)(sum_n=0^inftya_n)$.Prove that $sum_n=0^inftya_nz^n$ converges absolutely and uniformly in $D$.
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$f(z)=sum_n=1^infty a_n(z-z_0)^n$ such that $sum_n=0^infty f^n(a)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$sum|a_n|^2<infty$ and $sum a_n<infty$ implies $sumlog(1+a_n)$ convergesConvergence of $prod_n=1^infty(1+a_n)$Is $sum_n=2^inftylogleft(1+frac(-1)^nsqrtnright)$ convergent?Prove that $sum_n=1^infty fraca_nn^z$ converges absolutely and uniformlyIf $sum_n=1^infty |a_n|^2<infty$, Then : $sum_n=1^infty a_n$ Converges $Leftrightarrow prod_n=1^infty(1+a_n)$ConvergesIf $(a_n)$ is a complex sequence such that $sum_n=1^inftyfraca_nk^n = 0,forall kin mathbbN$ then $a_n = 0$ for all $ n$Prove that a power series $sum_n=0^inftya_nz^n$ which converges for any $z in mathbbN$, converges for any $z in mathbbC$.If the complex series $sum_n=0^inftya_n$ converges, show that there exists a positive number $A$ such that $|a_n| leq A$ for all $n$.Show that $sum_n=0^infty(sum_j=0^n a_jb_n-j)$ converges to $(sum_n=0^inftyb_n)(sum_n=0^inftya_n)$.Prove that $sum_n=0^inftya_nz^n$ converges absolutely and uniformly in $D$.
$begingroup$
Let $f(z)=sum_n=1^infty a_n(z-a)^n$ such that $sum_n=0^infty f^n(a)$ converges then is it necessary that $f(z) $ is entire.
I have tried this by giving counter example but all in vain , Any hints leading to answer are deeply appreciated
complex-analysis analytic-functions
asked Mar 26 at 8:51
Good HeartGood Heart
20418
$endgroup$
add a comment |
$begingroup$
Let $f(z)=sum_n=1^infty a_n(z-a)^n$ such that $sum_n=0^infty f^n(a)$ converges then is it necessary that $f(z) $ is entire.
I have tried this by giving counter example but all in vain , Any hints leading to answer are deeply appreciated
complex-analysis analytic-functions
asked Mar 26 at 8:51
Good HeartGood Heart
20418
$endgroup$
1
$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53
add a comment |
$begingroup$
Let $f(z)=sum_n=1^infty a_n(z-a)^n$ such that $sum_n=0^infty f^n(a)$ converges then is it necessary that $f(z) $ is entire.
I have tried this by giving counter example but all in vain , Any hints leading to answer are deeply appreciated
complex-analysis analytic-functions
asked Mar 26 at 8:51
Good HeartGood Heart
20418
$endgroup$
Let $f(z)=sum_n=1^infty a_n(z-a)^n$ such that $sum_n=0^infty f^n(a)$ converges then is it necessary that $f(z) $ is entire.
I have tried this by giving counter example but all in vain , Any hints leading to answer are deeply appreciated
complex-analysis analytic-functions
complex-analysis analytic-functions
asked Mar 26 at 8:51
Good HeartGood Heart
20418
asked Mar 26 at 8:51
Good HeartGood Heart
20418
asked Mar 26 at 8:51
Good HeartGood Heart
20418
asked Mar 26 at 8:51
Good HeartGood Heart
20418
asked Mar 26 at 8:51
Good HeartGood Heart
20418
20418
1
$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53
add a comment |
1
$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53
1
1
$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53
$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $f(z)=sumlimits_k=1^infty a_n(z-a)^n$ in some neighborhood of $a$ then $a_n =frac f^(n) (a) n!$. Since $f^(n) (a)to 0$ it follows (by comparison with exponential series) that the series converges for all $z$ and $f$ extends to an entire function.
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
If $f(z)=sumlimits_k=1^infty a_n(z-a)^n$ in some neighborhood of $a$ then $a_n =frac f^(n) (a) n!$. Since $f^(n) (a)to 0$ it follows (by comparison with exponential series) that the series converges for all $z$ and $f$ extends to an entire function.
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
add a comment |
$begingroup$
If $f(z)=sumlimits_k=1^infty a_n(z-a)^n$ in some neighborhood of $a$ then $a_n =frac f^(n) (a) n!$. Since $f^(n) (a)to 0$ it follows (by comparison with exponential series) that the series converges for all $z$ and $f$ extends to an entire function.
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
add a comment |
$begingroup$
If $f(z)=sumlimits_k=1^infty a_n(z-a)^n$ in some neighborhood of $a$ then $a_n =frac f^(n) (a) n!$. Since $f^(n) (a)to 0$ it follows (by comparison with exponential series) that the series converges for all $z$ and $f$ extends to an entire function.
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
If $f(z)=sumlimits_k=1^infty a_n(z-a)^n$ in some neighborhood of $a$ then $a_n =frac f^(n) (a) n!$. Since $f^(n) (a)to 0$ it follows (by comparison with exponential series) that the series converges for all $z$ and $f$ extends to an entire function.
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 8:55
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
add a comment |
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
How did you get $ a_n =frac f^(n) (a) n!$
$endgroup$
– Good Heart
Mar 26 at 9:03
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
$begingroup$
and how did you expand it for all $z$
$endgroup$
– Good Heart
Mar 26 at 9:04
1
1
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
That comes from a basic theorem about power series. Any book on Complex Analysis contains a proof.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:05
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
$begingroup$
The series $sum frac f^(n)(a) n!(z-a)^n$ converges for all $z$ because $f^(n)(a)$ is a bounded sequence. But this is same as the original series so the original series converges for all $z$.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 9:12
add a comment |
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$begingroup$
My guess is that you meant $f^(n)$, instead of $f^n$.
$endgroup$
– José Carlos Santos
Mar 26 at 8:53