Prove that $f(z)$ is analytic in $z_{0}$, then $f(z)in mathcal{C}^{infty}$ for some $mathcal{N}(z_{0})$.
$begingroup$
Given $f(z)$ a function of complex variable. Prove that if $f(z)$ is analytic in $z_{0}$, then $f(z)in mathcal{C}^{infty}$ for some neighborhood $mathcal{N}(z_{0})$.
Ok I've an idea, but I don't understand what excactly implies "$f(z)in mathcal{C}^{infty}$" or what I need to prove.
I know $mathcal{C}^{infty}$ means that $f$ is infinitely differentiable.
Someone can help me?
complex-analysis
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
$endgroup$
add a comment |
$begingroup$
Given $f(z)$ a function of complex variable. Prove that if $f(z)$ is analytic in $z_{0}$, then $f(z)in mathcal{C}^{infty}$ for some neighborhood $mathcal{N}(z_{0})$.
Ok I've an idea, but I don't understand what excactly implies "$f(z)in mathcal{C}^{infty}$" or what I need to prove.
I know $mathcal{C}^{infty}$ means that $f$ is infinitely differentiable.
Someone can help me?
complex-analysis
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
$endgroup$
1
$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
3
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51
add a comment |
$begingroup$
Given $f(z)$ a function of complex variable. Prove that if $f(z)$ is analytic in $z_{0}$, then $f(z)in mathcal{C}^{infty}$ for some neighborhood $mathcal{N}(z_{0})$.
Ok I've an idea, but I don't understand what excactly implies "$f(z)in mathcal{C}^{infty}$" or what I need to prove.
I know $mathcal{C}^{infty}$ means that $f$ is infinitely differentiable.
Someone can help me?
complex-analysis
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
$endgroup$
Given $f(z)$ a function of complex variable. Prove that if $f(z)$ is analytic in $z_{0}$, then $f(z)in mathcal{C}^{infty}$ for some neighborhood $mathcal{N}(z_{0})$.
Ok I've an idea, but I don't understand what excactly implies "$f(z)in mathcal{C}^{infty}$" or what I need to prove.
I know $mathcal{C}^{infty}$ means that $f$ is infinitely differentiable.
Someone can help me?
complex-analysis
complex-analysis
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
edited Jan 13 at 7:04
Eric Wofsey
189k14216347
189k14216347
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
asked Jan 13 at 6:54
tajiri_numero_1tajiri_numero_1
115
115
1
$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
3
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51
add a comment |
1
$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
3
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51
1
1
$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
3
3
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51
add a comment |
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$begingroup$
The proof-writing tag for when you have a proof but have a question about how best to write it out. It is not for questions just about finding a proof.
$endgroup$
– Eric Wofsey
Jan 13 at 7:06
3
$begingroup$
"Ok I've an idea" Good, where is it?
$endgroup$
– Did
Jan 13 at 12:56
$begingroup$
Oh yes mmh my idea is: if $f$ can be wrote like a taylor series, i.e $f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^n$, then $f(z_{0})in mathcal(N)(z_{0})$. But I need prove that if $f$ is analytic then can be expressed like a taylor series and $f^{(n)}$ are smooth for every n.
$endgroup$
– tajiri_numero_1
Jan 13 at 15:51