If $f$ is entire and $left|fleft(frac1ln(n+2)right)right|<frac1n$ for every positive integer $n$ then $f=0$
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Let $f(z)$ be an entire function satisfying
$$left|fleft(frac1ln(n+2)right)right|<frac1n$$ for every $ninmathbbN).$
Show that $f(z)=0.$
I need some help for this question, I am looking for some hints which helps to solve this problem (not for possible solution).
I know pretty much popular theorems about entire functions, such as Liuville theorem, Casorati-Weierstrass and some nameless theorems. However, for this question, I am not sure what approach is the key.
If I am not mistaken $fleft(frac1ln(z+2)right)$ has essential singularity at $z=-1$, so if I am right then maybe using Casorati-Weierstrass theorem is useful.
I can also say as $ntoinfty$, I get $|f(0)|leq 0$, so $f(0)=0$. Hence, If I can show $f$ is constant it must be identically $0$.
complex-analysis holomorphic-functions analytic-functions entire-functions
edited Aug 11 at 8:03
Did
242k23208443
asked Aug 11 at 4:58
Parisina
799921
add a comment |Â
up vote
3
down vote
favorite
Let $f(z)$ be an entire function satisfying
$$left|fleft(frac1ln(n+2)right)right|<frac1n$$ for every $ninmathbbN).$
Show that $f(z)=0.$
I need some help for this question, I am looking for some hints which helps to solve this problem (not for possible solution).
I know pretty much popular theorems about entire functions, such as Liuville theorem, Casorati-Weierstrass and some nameless theorems. However, for this question, I am not sure what approach is the key.
If I am not mistaken $fleft(frac1ln(z+2)right)$ has essential singularity at $z=-1$, so if I am right then maybe using Casorati-Weierstrass theorem is useful.
I can also say as $ntoinfty$, I get $|f(0)|leq 0$, so $f(0)=0$. Hence, If I can show $f$ is constant it must be identically $0$.
complex-analysis holomorphic-functions analytic-functions entire-functions
edited Aug 11 at 8:03
Did
242k23208443
asked Aug 11 at 4:58
Parisina
799921
1
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
1
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f(z)$ be an entire function satisfying
$$left|fleft(frac1ln(n+2)right)right|<frac1n$$ for every $ninmathbbN).$
Show that $f(z)=0.$
I need some help for this question, I am looking for some hints which helps to solve this problem (not for possible solution).
I know pretty much popular theorems about entire functions, such as Liuville theorem, Casorati-Weierstrass and some nameless theorems. However, for this question, I am not sure what approach is the key.
If I am not mistaken $fleft(frac1ln(z+2)right)$ has essential singularity at $z=-1$, so if I am right then maybe using Casorati-Weierstrass theorem is useful.
I can also say as $ntoinfty$, I get $|f(0)|leq 0$, so $f(0)=0$. Hence, If I can show $f$ is constant it must be identically $0$.
complex-analysis holomorphic-functions analytic-functions entire-functions
edited Aug 11 at 8:03
Did
242k23208443
asked Aug 11 at 4:58
Parisina
799921
Let $f(z)$ be an entire function satisfying
$$left|fleft(frac1ln(n+2)right)right|<frac1n$$ for every $ninmathbbN).$
Show that $f(z)=0.$
I need some help for this question, I am looking for some hints which helps to solve this problem (not for possible solution).
I know pretty much popular theorems about entire functions, such as Liuville theorem, Casorati-Weierstrass and some nameless theorems. However, for this question, I am not sure what approach is the key.
If I am not mistaken $fleft(frac1ln(z+2)right)$ has essential singularity at $z=-1$, so if I am right then maybe using Casorati-Weierstrass theorem is useful.
I can also say as $ntoinfty$, I get $|f(0)|leq 0$, so $f(0)=0$. Hence, If I can show $f$ is constant it must be identically $0$.
complex-analysis holomorphic-functions analytic-functions entire-functions
edited Aug 11 at 8:03
Did
242k23208443
asked Aug 11 at 4:58
Parisina
799921
edited Aug 11 at 8:03
Did
242k23208443
edited Aug 11 at 8:03
Did
242k23208443
edited Aug 11 at 8:03
Did
242k23208443
242k23208443
asked Aug 11 at 4:58
Parisina
799921
asked Aug 11 at 4:58
Parisina
799921
asked Aug 11 at 4:58
Parisina
799921
799921
1
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
1
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49
add a comment |Â
1
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
1
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49
1
1
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
1
1
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
By considering the limit as $nto +infty$ we have $f(0)=0$. If we assume $f(x)notequiv 0$ then $c_m x^m$, for some $minmathbbN^+$ and $c_mneq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=left|c_mright|$ and $g(x)=fracf(x)x^m$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $left|g(x)right|geq fracM2$.
On the other hand we also have
$$ left|gleft(frac1log(n+2)right)right|leq fraclog(n+2)^mn, $$
so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)equiv 0$.
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By considering the limit as $nto +infty$ we have $f(0)=0$. If we assume $f(x)notequiv 0$ then $c_m x^m$, for some $minmathbbN^+$ and $c_mneq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=left|c_mright|$ and $g(x)=fracf(x)x^m$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $left|g(x)right|geq fracM2$.
On the other hand we also have
$$ left|gleft(frac1log(n+2)right)right|leq fraclog(n+2)^mn, $$
so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)equiv 0$.
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
add a comment |Â
up vote
2
down vote
accepted
By considering the limit as $nto +infty$ we have $f(0)=0$. If we assume $f(x)notequiv 0$ then $c_m x^m$, for some $minmathbbN^+$ and $c_mneq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=left|c_mright|$ and $g(x)=fracf(x)x^m$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $left|g(x)right|geq fracM2$.
On the other hand we also have
$$ left|gleft(frac1log(n+2)right)right|leq fraclog(n+2)^mn, $$
so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)equiv 0$.
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By considering the limit as $nto +infty$ we have $f(0)=0$. If we assume $f(x)notequiv 0$ then $c_m x^m$, for some $minmathbbN^+$ and $c_mneq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=left|c_mright|$ and $g(x)=fracf(x)x^m$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $left|g(x)right|geq fracM2$.
On the other hand we also have
$$ left|gleft(frac1log(n+2)right)right|leq fraclog(n+2)^mn, $$
so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)equiv 0$.
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
By considering the limit as $nto +infty$ we have $f(0)=0$. If we assume $f(x)notequiv 0$ then $c_m x^m$, for some $minmathbbN^+$ and $c_mneq 0$, is the first non-zero term of the Maclaurin series of $f(x)$.
Let $M=left|c_mright|$ and $g(x)=fracf(x)x^m$.
$g(x)$ is entire and in a neighbourhood $U$ of the origin we have $left|g(x)right|geq fracM2$.
On the other hand we also have
$$ left|gleft(frac1log(n+2)right)right|leq fraclog(n+2)^mn, $$
so, no matter how small $M$ or $U$ are, for some large $n$ we get a contradiction.
It follows that $f(x)equiv 0$.
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
answered Aug 11 at 6:25
Jack D'Aurizio♦
271k31266632
271k31266632
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
add a comment |Â
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
Dammit! In million years I can't solve it without any help! How am I supposed to be good in this stuff! would you tell me how?
– Parisina
Aug 11 at 6:37
1
1
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
@Parisina: you just have to get accustomed to some standard tricks, like dividing/multiplying by monomials, applying Cayley transforms, identifying entire functions with the list of the coefficients of their Maclaurin series etc. It just takes a bit of practice.
– Jack D'Aurizio♦
Aug 11 at 6:39
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
You used the fact that the zeros of non zero analytic functions are of finite order, right? I can prove this one! Thanks alot!
– Parisina
Aug 11 at 6:44
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
@Parisina: exactly that. You're welcome :)
– Jack D'Aurizio♦
Aug 11 at 6:54
add a comment |Â
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1
Hint: What is the behaviour of $|f(z)|$ when $|z|to0$ if the function $f$ is analytical at $0$ and such that $f(0)=0$?
– Did
Aug 11 at 5:44
1
(The tag (cauchy-schwarz-inequality) is unrelated.)
– Did
Aug 11 at 5:49