If $f$ is analytic on $mathbb{R}$, is it necessary that $f = sum_{n = 0}^{infty} a_{n} x^{n}$ converges for...
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I have two questions regarding analyticity. They are pretty easy, and I think I have them correct, but I just want to make sure.
First, regarding the question in the title,
I think that it is necessary. By the definition of analyticity, we must have the power series in some radius of the function. So, if it is analytic on all of $mathbb{R}$, it must be within some radius (and thus convergent) for all of $mathbb{R}$.
Second, if $f$ is analytic, is it necessary for $text{exp}(f)$ to be analytic? Pretty sure that this again is necessary since $e^{x}$ is analytic, and the composition of analytic functions is analytic.
real-analysis functions analyticity
asked Nov 16 at 8:08
Dillain Smith
396
add a comment |
up vote
1
down vote
favorite
I have two questions regarding analyticity. They are pretty easy, and I think I have them correct, but I just want to make sure.
First, regarding the question in the title,
I think that it is necessary. By the definition of analyticity, we must have the power series in some radius of the function. So, if it is analytic on all of $mathbb{R}$, it must be within some radius (and thus convergent) for all of $mathbb{R}$.
Second, if $f$ is analytic, is it necessary for $text{exp}(f)$ to be analytic? Pretty sure that this again is necessary since $e^{x}$ is analytic, and the composition of analytic functions is analytic.
real-analysis functions analyticity
asked Nov 16 at 8:08
Dillain Smith
396
This is correct.
– Zanzi
Nov 16 at 8:13
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
1
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Yes. This is correct.
– Andrej
Nov 16 at 8:16
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have two questions regarding analyticity. They are pretty easy, and I think I have them correct, but I just want to make sure.
First, regarding the question in the title,
I think that it is necessary. By the definition of analyticity, we must have the power series in some radius of the function. So, if it is analytic on all of $mathbb{R}$, it must be within some radius (and thus convergent) for all of $mathbb{R}$.
Second, if $f$ is analytic, is it necessary for $text{exp}(f)$ to be analytic? Pretty sure that this again is necessary since $e^{x}$ is analytic, and the composition of analytic functions is analytic.
real-analysis functions analyticity
asked Nov 16 at 8:08
Dillain Smith
396
I have two questions regarding analyticity. They are pretty easy, and I think I have them correct, but I just want to make sure.
First, regarding the question in the title,
I think that it is necessary. By the definition of analyticity, we must have the power series in some radius of the function. So, if it is analytic on all of $mathbb{R}$, it must be within some radius (and thus convergent) for all of $mathbb{R}$.
Second, if $f$ is analytic, is it necessary for $text{exp}(f)$ to be analytic? Pretty sure that this again is necessary since $e^{x}$ is analytic, and the composition of analytic functions is analytic.
real-analysis functions analyticity
real-analysis functions analyticity
asked Nov 16 at 8:08
Dillain Smith
396
asked Nov 16 at 8:08
Dillain Smith
396
asked Nov 16 at 8:08
Dillain Smith
396
asked Nov 16 at 8:08
Dillain Smith
396
asked Nov 16 at 8:08
Dillain Smith
396
396
This is correct.
– Zanzi
Nov 16 at 8:13
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
1
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Yes. This is correct.
– Andrej
Nov 16 at 8:16
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21
add a comment |
This is correct.
– Zanzi
Nov 16 at 8:13
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
1
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Yes. This is correct.
– Andrej
Nov 16 at 8:16
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21
This is correct.
– Zanzi
Nov 16 at 8:13
This is correct.
– Zanzi
Nov 16 at 8:13
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
1
1
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Yes. This is correct.
– Andrej
Nov 16 at 8:16
Yes. This is correct.
– Andrej
Nov 16 at 8:16
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21
add a comment |
1 Answer
1
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up vote
2
down vote
Counterexample to question 1: $f(x)=frac1{1+x^2}$. The Maclaurin series of it does not converge everywhere is because $f$ is a meromorphic function in $mathbb{C}$.
If $f(z)$ is analytic on $mathbb{C}$, the series you gave is convergent everywhere.
Your explanation of question 2 is correct.
answered Nov 16 at 8:28
Kemono Chen
1,647330
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Counterexample to question 1: $f(x)=frac1{1+x^2}$. The Maclaurin series of it does not converge everywhere is because $f$ is a meromorphic function in $mathbb{C}$.
If $f(z)$ is analytic on $mathbb{C}$, the series you gave is convergent everywhere.
Your explanation of question 2 is correct.
answered Nov 16 at 8:28
Kemono Chen
1,647330
add a comment |
up vote
2
down vote
Counterexample to question 1: $f(x)=frac1{1+x^2}$. The Maclaurin series of it does not converge everywhere is because $f$ is a meromorphic function in $mathbb{C}$.
If $f(z)$ is analytic on $mathbb{C}$, the series you gave is convergent everywhere.
Your explanation of question 2 is correct.
answered Nov 16 at 8:28
Kemono Chen
1,647330
add a comment |
up vote
2
down vote
up vote
2
down vote
Counterexample to question 1: $f(x)=frac1{1+x^2}$. The Maclaurin series of it does not converge everywhere is because $f$ is a meromorphic function in $mathbb{C}$.
If $f(z)$ is analytic on $mathbb{C}$, the series you gave is convergent everywhere.
Your explanation of question 2 is correct.
answered Nov 16 at 8:28
Kemono Chen
1,647330
Counterexample to question 1: $f(x)=frac1{1+x^2}$. The Maclaurin series of it does not converge everywhere is because $f$ is a meromorphic function in $mathbb{C}$.
If $f(z)$ is analytic on $mathbb{C}$, the series you gave is convergent everywhere.
Your explanation of question 2 is correct.
answered Nov 16 at 8:28
Kemono Chen
1,647330
answered Nov 16 at 8:28
Kemono Chen
1,647330
answered Nov 16 at 8:28
Kemono Chen
1,647330
answered Nov 16 at 8:28
Kemono Chen
1,647330
1,647330
add a comment |
add a comment |
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This is correct.
– Zanzi
Nov 16 at 8:13
Only an accumuled point is required.
– Zanzi
Nov 16 at 8:14
1
Counterexample to question 1: $f(x)=frac1{1+x^2}$.
– Kemono Chen
Nov 16 at 8:14
Yes. This is correct.
– Andrej
Nov 16 at 8:16
@KemonoChen Why not an official answer even it is short?
– Paul Frost
Nov 16 at 8:21