Mixing
Analytic continuation of $sum _{n=0}^{infty} frac{z^{2^n}} {2^n}$ beyond the unit disc
0
$begingroup$
The function :
$sum _{n=0}^{infty} frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $overline{D_1(0)}$.
I need to prove that f can't be exteneded to any domain $Omega$ such that $overline{D_1(0)}subseteqOmega$.
Any ideas?
complex-analysis holomorphic-functions analytic-continuation
asked Jan 22 at 14:17
user3708158user3708158
1307
$endgroup$
add a comment |
0
$begingroup$
The function :
$sum _{n=0}^{infty} frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $overline{D_1(0)}$.
I need to prove that f can't be exteneded to any domain $Omega$ such that $overline{D_1(0)}subseteqOmega$.
Any ideas?
complex-analysis holomorphic-functions analytic-continuation
asked Jan 22 at 14:17
user3708158user3708158
1307
$endgroup$
add a comment |
0
0
0
$begingroup$
The function :
$sum _{n=0}^{infty} frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $overline{D_1(0)}$.
I need to prove that f can't be exteneded to any domain $Omega$ such that $overline{D_1(0)}subseteqOmega$.
Any ideas?
complex-analysis holomorphic-functions analytic-continuation
asked Jan 22 at 14:17
user3708158user3708158
1307
$endgroup$
The function :
$sum _{n=0}^{infty} frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $overline{D_1(0)}$.
I need to prove that f can't be exteneded to any domain $Omega$ such that $overline{D_1(0)}subseteqOmega$.
Any ideas?
complex-analysis holomorphic-functions analytic-continuation
complex-analysis holomorphic-functions analytic-continuation
asked Jan 22 at 14:17
user3708158user3708158
1307
asked Jan 22 at 14:17
user3708158user3708158
1307
asked Jan 22 at 14:17
user3708158user3708158
1307
asked Jan 22 at 14:17
user3708158user3708158
1307
asked Jan 22 at 14:17
user3708158user3708158
1307
1307
add a comment |
add a comment |
1 Answer
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If it has an analytical continuation on some domain $Omega supset overline{D}(0,1)$, then $Omega$ contains some $overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.
answered Jan 22 at 14:41
MindlackMindlack
4,920211
$endgroup$
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
If it has an analytical continuation on some domain $Omega supset overline{D}(0,1)$, then $Omega$ contains some $overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.
answered Jan 22 at 14:41
MindlackMindlack
4,920211
$endgroup$
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
|
show 2 more comments
2
$begingroup$
If it has an analytical continuation on some domain $Omega supset overline{D}(0,1)$, then $Omega$ contains some $overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.
answered Jan 22 at 14:41
MindlackMindlack
4,920211
$endgroup$
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
|
show 2 more comments
2
2
2
$begingroup$
If it has an analytical continuation on some domain $Omega supset overline{D}(0,1)$, then $Omega$ contains some $overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.
answered Jan 22 at 14:41
MindlackMindlack
4,920211
$endgroup$
If it has an analytical continuation on some domain $Omega supset overline{D}(0,1)$, then $Omega$ contains some $overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.
answered Jan 22 at 14:41
MindlackMindlack
4,920211
answered Jan 22 at 14:41
MindlackMindlack
4,920211
answered Jan 22 at 14:41
MindlackMindlack
4,920211
answered Jan 22 at 14:41
MindlackMindlack
4,920211
4,920211
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
|
show 2 more comments
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges.
$endgroup$
– user3708158
Jan 22 at 14:45
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain.
$endgroup$
– Mindlack
Jan 22 at 14:46
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
I am not sure, for example you have the function $f(z)=frac {1}{1-z} = sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain.
$endgroup$
– user3708158
Jan 22 at 14:48
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
That is why I wrote: converges normally on any closed disc that is a subset of the domain. $overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z longmapsto (1-z)^{-1}$.
$endgroup$
– Mindlack
Jan 22 at 14:50
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
$begingroup$
Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge?
$endgroup$
– user3708158
Jan 22 at 14:52
|
show 2 more comments
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