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







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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















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







share|cite|improve this question







New contributor




Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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













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







share|cite|improve this question







New contributor




Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






share|cite|improve this question







New contributor




Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







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Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked Nov 16 at 8:08









Dillain Smith

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New contributor





Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Dillain Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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










  • 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










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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.






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    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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 16 at 8:28









        Kemono Chen

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