# If f has power series expansion around each point then does it have a single power series expansion on following sets

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!!

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 $$\sqrt{a^2+1}<\infty$$.
• @ Jose Carlos Santos What is $D_r(A)$ in second line of your answer? – Avenger Sep 16 '20 at 11:55
• @Ben It was a typo. I meant $D_r(a)$, which is the open disk with center $a$ and radius $r$. – José Carlos Santos Sep 16 '20 at 21:07
• @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 $\sqrt{a^2+1}$. – José Carlos Santos Sep 20 '20 at 7:50