Examples of different type of entire functions
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I try to answer following question. I would like to find a general approach for part b, since I think any entire function with $n $ roots is either polynomial of degree $n$ or is a product of polynomial with another entire function with no roots.
Give an example of an entire function that is not a polynomial hand has
a) no root;
b) exactly $n$ roots;
c) infinitely many roots.
For part a) I think $e^z$ works.
Part b) I think $f(z)=e^zP(z)$, where $P(z)$ is a polynomial of degree $n$. However,I would like to know if there is any entire function not constructed by any polynomial.
For part c) I think $g(z)=sin(z)$, which $z=npi$, where $n$ is an integer.
complex-analysis entire-functions
asked May 20 at 22:39
Parisina
797921
add a comment |Â
up vote
1
down vote
favorite
I try to answer following question. I would like to find a general approach for part b, since I think any entire function with $n $ roots is either polynomial of degree $n$ or is a product of polynomial with another entire function with no roots.
Give an example of an entire function that is not a polynomial hand has
a) no root;
b) exactly $n$ roots;
c) infinitely many roots.
For part a) I think $e^z$ works.
Part b) I think $f(z)=e^zP(z)$, where $P(z)$ is a polynomial of degree $n$. However,I would like to know if there is any entire function not constructed by any polynomial.
For part c) I think $g(z)=sin(z)$, which $z=npi$, where $n$ is an integer.
complex-analysis entire-functions
asked May 20 at 22:39
Parisina
797921
2
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
@MarkViola Thanks!
– Parisina
May 20 at 22:50
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I try to answer following question. I would like to find a general approach for part b, since I think any entire function with $n $ roots is either polynomial of degree $n$ or is a product of polynomial with another entire function with no roots.
Give an example of an entire function that is not a polynomial hand has
a) no root;
b) exactly $n$ roots;
c) infinitely many roots.
For part a) I think $e^z$ works.
Part b) I think $f(z)=e^zP(z)$, where $P(z)$ is a polynomial of degree $n$. However,I would like to know if there is any entire function not constructed by any polynomial.
For part c) I think $g(z)=sin(z)$, which $z=npi$, where $n$ is an integer.
complex-analysis entire-functions
asked May 20 at 22:39
Parisina
797921
I try to answer following question. I would like to find a general approach for part b, since I think any entire function with $n $ roots is either polynomial of degree $n$ or is a product of polynomial with another entire function with no roots.
Give an example of an entire function that is not a polynomial hand has
a) no root;
b) exactly $n$ roots;
c) infinitely many roots.
For part a) I think $e^z$ works.
Part b) I think $f(z)=e^zP(z)$, where $P(z)$ is a polynomial of degree $n$. However,I would like to know if there is any entire function not constructed by any polynomial.
For part c) I think $g(z)=sin(z)$, which $z=npi$, where $n$ is an integer.
complex-analysis entire-functions
asked May 20 at 22:39
Parisina
797921
edited Aug 10 at 18:56
asked May 20 at 22:39
Parisina
797921
asked May 20 at 22:39
Parisina
797921
asked May 20 at 22:39
Parisina
797921
797921
2
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
@MarkViola Thanks!
– Parisina
May 20 at 22:50
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36
add a comment |Â
2
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
@MarkViola Thanks!
– Parisina
May 20 at 22:50
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36
2
2
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
@MarkViola Thanks!
– Parisina
May 20 at 22:50
@MarkViola Thanks!
– Parisina
May 20 at 22:50
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your answers are fine. As for your separate question: let $f$ be entire with only $n$ roots $z_i, i=1ldots n$. If $alpha_i$ is the order of the root $z_i$, then $$g(z)=fracf(z)prod_i=1^n (z-z_i)^alpha_i$$
is an entire function with no roots. Therefore $f$ is the product of a polynomial and an entire function with no roots.
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your answers are fine. As for your separate question: let $f$ be entire with only $n$ roots $z_i, i=1ldots n$. If $alpha_i$ is the order of the root $z_i$, then $$g(z)=fracf(z)prod_i=1^n (z-z_i)^alpha_i$$
is an entire function with no roots. Therefore $f$ is the product of a polynomial and an entire function with no roots.
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
add a comment |Â
up vote
1
down vote
accepted
Your answers are fine. As for your separate question: let $f$ be entire with only $n$ roots $z_i, i=1ldots n$. If $alpha_i$ is the order of the root $z_i$, then $$g(z)=fracf(z)prod_i=1^n (z-z_i)^alpha_i$$
is an entire function with no roots. Therefore $f$ is the product of a polynomial and an entire function with no roots.
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your answers are fine. As for your separate question: let $f$ be entire with only $n$ roots $z_i, i=1ldots n$. If $alpha_i$ is the order of the root $z_i$, then $$g(z)=fracf(z)prod_i=1^n (z-z_i)^alpha_i$$
is an entire function with no roots. Therefore $f$ is the product of a polynomial and an entire function with no roots.
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
Your answers are fine. As for your separate question: let $f$ be entire with only $n$ roots $z_i, i=1ldots n$. If $alpha_i$ is the order of the root $z_i$, then $$g(z)=fracf(z)prod_i=1^n (z-z_i)^alpha_i$$
is an entire function with no roots. Therefore $f$ is the product of a polynomial and an entire function with no roots.
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
answered May 20 at 23:01
Arnaud Mortier
19.3k22159
19.3k22159
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2789338%2fexamples-of-different-type-of-entire-functions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Take any non-polynomial, entire function $f(z)$ that has no zeroes. Then, $g(z)=prod_k=1^n(f(z)-f(z_k))$ is entire, has $n$ zeroes, and is not a polynomial.
– Mark Viola
May 20 at 22:45
@MarkViola Thanks!
– Parisina
May 20 at 22:50
You're welcome. My pleasure
– Mark Viola
May 21 at 0:36