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For what $c in mathbbC$ is there a entire function such that $f(1/k) = c^k$?
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For what $c in mathbbC$ is there a entire function $f$ such that $f(1/k) = c^k$?
I would like to use the identity theorem. However, the obvious choice of $c^1/z = e^log(c)/z$ is not analytic at the limit point $0$.
If $|c|>1$, by continuity we get $f(0) to infty$, but that contradicts that $f$ is entire.
If $|c|<1$, we can get $f equiv 0$. Suppose not, and choose $n$ to be such that $f^(n)(0) neq 0$.Then by looking at the series expansion of $f$ one can see $frac1z^nf(z)$ does not go to $0$ as $z to 0$. But, $f(1/k) = k^nc^k to 0$ as $k to infty$.
What about $|c|=1$? Well, if $c=1$ then $f equiv 1$ satisfies the condition. If $c = e^ifracpqpi$ for rational $p/q$, I can choose a subsequence of $f(1/k)$ that equals $1$ and apply the identity theorem to get $f equiv 1$.
I have an idea on how to handle the other $|c|=1$, but it is not pretty, and I would like others' thoughts.
complex-analysis holomorphic-functions
asked Jul 25 at 0:04
Pentaki
745
add a comment |Â
up vote
2
down vote
favorite
For what $c in mathbbC$ is there a entire function $f$ such that $f(1/k) = c^k$?
I would like to use the identity theorem. However, the obvious choice of $c^1/z = e^log(c)/z$ is not analytic at the limit point $0$.
If $|c|>1$, by continuity we get $f(0) to infty$, but that contradicts that $f$ is entire.
If $|c|<1$, we can get $f equiv 0$. Suppose not, and choose $n$ to be such that $f^(n)(0) neq 0$.Then by looking at the series expansion of $f$ one can see $frac1z^nf(z)$ does not go to $0$ as $z to 0$. But, $f(1/k) = k^nc^k to 0$ as $k to infty$.
What about $|c|=1$? Well, if $c=1$ then $f equiv 1$ satisfies the condition. If $c = e^ifracpqpi$ for rational $p/q$, I can choose a subsequence of $f(1/k)$ that equals $1$ and apply the identity theorem to get $f equiv 1$.
I have an idea on how to handle the other $|c|=1$, but it is not pretty, and I would like others' thoughts.
complex-analysis holomorphic-functions
asked Jul 25 at 0:04
Pentaki
745
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For what $c in mathbbC$ is there a entire function $f$ such that $f(1/k) = c^k$?
I would like to use the identity theorem. However, the obvious choice of $c^1/z = e^log(c)/z$ is not analytic at the limit point $0$.
If $|c|>1$, by continuity we get $f(0) to infty$, but that contradicts that $f$ is entire.
If $|c|<1$, we can get $f equiv 0$. Suppose not, and choose $n$ to be such that $f^(n)(0) neq 0$.Then by looking at the series expansion of $f$ one can see $frac1z^nf(z)$ does not go to $0$ as $z to 0$. But, $f(1/k) = k^nc^k to 0$ as $k to infty$.
What about $|c|=1$? Well, if $c=1$ then $f equiv 1$ satisfies the condition. If $c = e^ifracpqpi$ for rational $p/q$, I can choose a subsequence of $f(1/k)$ that equals $1$ and apply the identity theorem to get $f equiv 1$.
I have an idea on how to handle the other $|c|=1$, but it is not pretty, and I would like others' thoughts.
complex-analysis holomorphic-functions
asked Jul 25 at 0:04
Pentaki
745
For what $c in mathbbC$ is there a entire function $f$ such that $f(1/k) = c^k$?
I would like to use the identity theorem. However, the obvious choice of $c^1/z = e^log(c)/z$ is not analytic at the limit point $0$.
If $|c|>1$, by continuity we get $f(0) to infty$, but that contradicts that $f$ is entire.
If $|c|<1$, we can get $f equiv 0$. Suppose not, and choose $n$ to be such that $f^(n)(0) neq 0$.Then by looking at the series expansion of $f$ one can see $frac1z^nf(z)$ does not go to $0$ as $z to 0$. But, $f(1/k) = k^nc^k to 0$ as $k to infty$.
What about $|c|=1$? Well, if $c=1$ then $f equiv 1$ satisfies the condition. If $c = e^ifracpqpi$ for rational $p/q$, I can choose a subsequence of $f(1/k)$ that equals $1$ and apply the identity theorem to get $f equiv 1$.
I have an idea on how to handle the other $|c|=1$, but it is not pretty, and I would like others' thoughts.
complex-analysis holomorphic-functions
asked Jul 25 at 0:04
Pentaki
745
asked Jul 25 at 0:04
Pentaki
745
asked Jul 25 at 0:04
Pentaki
745
asked Jul 25 at 0:04
Pentaki
745
745
add a comment |Â
add a comment |Â
3 Answers
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5
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It is well known that $c^k:k geq 1$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^2pi it$ with $t$ irrational). In this case we get a contradiction to the fact that $f(frac 1 k) to f(0)$. If $c$ is a root of unity then $f(frac 1 k)=1$ for infinitely many $n$, so $f equiv 1$.
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
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up vote
3
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If $|c|=1,$ then $lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = lim c^k.$ Then $Lne 0$ and we have
$$c=fracc^k+1c^k to fracLL=1.$$
answered Jul 25 at 21:06
zhw.
65.6k42870
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up vote
2
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Given $c in mathbbCsetminus0$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, cdots$. Write $f(z) = z^m g(z)$ for some $m geq 0$ and entire function $g(z)$ with $g(0) neq 0$. Then
$$ c^k = f(1/k) = k^-m g(1/k) $$
and hence
$$ 1 = lim_ktoinfty fracg(frac1k+1)g(frac1k) = c. $$
So $f(1/k) = 1$ for all $k = 1, 2, cdots$ and by the identity theorem, $f equiv 1$.
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
It is well known that $c^k:k geq 1$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^2pi it$ with $t$ irrational). In this case we get a contradiction to the fact that $f(frac 1 k) to f(0)$. If $c$ is a root of unity then $f(frac 1 k)=1$ for infinitely many $n$, so $f equiv 1$.
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
add a comment |Â
up vote
5
down vote
accepted
It is well known that $c^k:k geq 1$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^2pi it$ with $t$ irrational). In this case we get a contradiction to the fact that $f(frac 1 k) to f(0)$. If $c$ is a root of unity then $f(frac 1 k)=1$ for infinitely many $n$, so $f equiv 1$.
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
It is well known that $c^k:k geq 1$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^2pi it$ with $t$ irrational). In this case we get a contradiction to the fact that $f(frac 1 k) to f(0)$. If $c$ is a root of unity then $f(frac 1 k)=1$ for infinitely many $n$, so $f equiv 1$.
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
It is well known that $c^k:k geq 1$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^2pi it$ with $t$ irrational). In this case we get a contradiction to the fact that $f(frac 1 k) to f(0)$. If $c$ is a root of unity then $f(frac 1 k)=1$ for infinitely many $n$, so $f equiv 1$.
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 0:30
Kavi Rama Murthy
20.1k2829
20.1k2829
add a comment |Â
add a comment |Â
up vote
3
down vote
If $|c|=1,$ then $lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = lim c^k.$ Then $Lne 0$ and we have
$$c=fracc^k+1c^k to fracLL=1.$$
answered Jul 25 at 21:06
zhw.
65.6k42870
add a comment |Â
up vote
3
down vote
If $|c|=1,$ then $lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = lim c^k.$ Then $Lne 0$ and we have
$$c=fracc^k+1c^k to fracLL=1.$$
answered Jul 25 at 21:06
zhw.
65.6k42870
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $|c|=1,$ then $lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = lim c^k.$ Then $Lne 0$ and we have
$$c=fracc^k+1c^k to fracLL=1.$$
answered Jul 25 at 21:06
zhw.
65.6k42870
If $|c|=1,$ then $lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = lim c^k.$ Then $Lne 0$ and we have
$$c=fracc^k+1c^k to fracLL=1.$$
answered Jul 25 at 21:06
zhw.
65.6k42870
answered Jul 25 at 21:06
zhw.
65.6k42870
answered Jul 25 at 21:06
zhw.
65.6k42870
answered Jul 25 at 21:06
zhw.
65.6k42870
65.6k42870
add a comment |Â
add a comment |Â
up vote
2
down vote
Given $c in mathbbCsetminus0$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, cdots$. Write $f(z) = z^m g(z)$ for some $m geq 0$ and entire function $g(z)$ with $g(0) neq 0$. Then
$$ c^k = f(1/k) = k^-m g(1/k) $$
and hence
$$ 1 = lim_ktoinfty fracg(frac1k+1)g(frac1k) = c. $$
So $f(1/k) = 1$ for all $k = 1, 2, cdots$ and by the identity theorem, $f equiv 1$.
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
2
down vote
Given $c in mathbbCsetminus0$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, cdots$. Write $f(z) = z^m g(z)$ for some $m geq 0$ and entire function $g(z)$ with $g(0) neq 0$. Then
$$ c^k = f(1/k) = k^-m g(1/k) $$
and hence
$$ 1 = lim_ktoinfty fracg(frac1k+1)g(frac1k) = c. $$
So $f(1/k) = 1$ for all $k = 1, 2, cdots$ and by the identity theorem, $f equiv 1$.
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Given $c in mathbbCsetminus0$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, cdots$. Write $f(z) = z^m g(z)$ for some $m geq 0$ and entire function $g(z)$ with $g(0) neq 0$. Then
$$ c^k = f(1/k) = k^-m g(1/k) $$
and hence
$$ 1 = lim_ktoinfty fracg(frac1k+1)g(frac1k) = c. $$
So $f(1/k) = 1$ for all $k = 1, 2, cdots$ and by the identity theorem, $f equiv 1$.
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
Given $c in mathbbCsetminus0$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, cdots$. Write $f(z) = z^m g(z)$ for some $m geq 0$ and entire function $g(z)$ with $g(0) neq 0$. Then
$$ c^k = f(1/k) = k^-m g(1/k) $$
and hence
$$ 1 = lim_ktoinfty fracg(frac1k+1)g(frac1k) = c. $$
So $f(1/k) = 1$ for all $k = 1, 2, cdots$ and by the identity theorem, $f equiv 1$.
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
answered Jul 25 at 2:53
Sangchul Lee
85.5k12155253
85.5k12155253
add a comment |Â
add a comment |Â
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