1
$\begingroup$

I am trying assignment in complex analysis of an institute in which I don't study because the instructor who taught me was terrible didn't bothered to give any assignment and also was poor in teaching concepts . So , I try problems of a different institute .

Question : If $f :\mathbb{C} \to \mathbb{C}$ has a power series expansion around each point then does it have a single power series expansion valid on all of $\mathbb{C}$ ?

If $f :\mathbb{R} \to \mathbb{R}$ has a power series expansion around each point then does it have a single power series expansion valid on all of $\mathbb{R}$ ?

I have no clue on which concept should be used . Kindly tell what concepts / results should be used give a hint or two (if necessary) and rest I would like to work by myself .

Thanks!!

$\endgroup$

1 Answer 1

2
$\begingroup$

The answer to the first question is affirmative. In fact a standard Complex Analysis theorem says that if $f\colon D(\subset\Bbb C)\longrightarrow\Bbb C$ is analytic and if $D_r(a)\subset D$, then the radius of convergence of the power series of $f$ centered at $a$ is at least $r$. Since, in your case, $D=\Bbb C$, then for each $a\in\Bbb C$, the radius of convergence of the power series of $f$ centered at $a$ is $\infty$. That is, it converges everywhere. And its sum is $f(z)$ (this follows from the identity theorem).

However, this is false in $\Bbb R$: if $f(x)=\dfrac1{1+x^2}$, then the radius of convergence of the Taylor series of $f$ centered at $a$ is $\sqrt{a^2+1}<\infty$.

$\endgroup$
5
  • $\begingroup$ @ Jose Carlos Santos What is $D_r(A)$ in second line of your answer? $\endgroup$
    – Avenger
    Sep 16, 2020 at 11:55
  • 1
    $\begingroup$ @Ben It was a typo. I meant $D_r(a)$, which is the open disk with center $a$ and radius $r$. $\endgroup$ Sep 16, 2020 at 21:07
  • $\begingroup$ You have answered some of my questions and I am really thankful to you for that.Can you also please answer this question if you have some spare time?math.stackexchange.com/questions/3810924/… $\endgroup$
    – Avenger
    Sep 20, 2020 at 4:38
  • $\begingroup$ by using which result did you found out radius of convergence in case of reals? $\endgroup$
    – Avenger
    Sep 20, 2020 at 4:50
  • 1
    $\begingroup$ @Ben The radius of convergence of the Taylor series centered at $a$ is the distance from $a$ to the nearest pole or essential singularity. In this case, it's the distance from $a$ to $\pm i$, which is $\sqrt{a^2+1}$. $\endgroup$ Sep 20, 2020 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.