If the series $f(z)=\sum^{\infty}_{n=0}a_n(z-i)^n$ and $g(z)=\sum^{\infty}_{n=0}b_n(z+i)^n$ both have radius of convergence $\frac{3}{2}$, how about the series of $h(z)=f(z)+g(z)$ centered at $0$?

I was thinking like say the function $f(z)=\frac{1}{1-z}$. If the center is at $i$ then the radius of convergence is $\sqrt{2}$. If the center is $0$, then the radius of convergence is 1. So we can not find some area that $overlap$ then find the radius. Really have little ideas, any help?

  • $\begingroup$ Wouldn't you first have to find a way to write the sum of the two series as a series centered at 0? Otherwise the radius of convergence isn't defined (though the domain of the resulting function is the intersections of the domains of the two summands). $\endgroup$ – Jonathan Feb 9 '18 at 17:44
  • $\begingroup$ @JonathanBrown Yes, I expanded the terms as polynomials in $z$, but the coefficients are kind of messed up... $\endgroup$ – Nan Feb 9 '18 at 18:50

Since $D(0,1/2)\subset D(i,3/2)\cap D(i,3/2),$ $f+g$ is analytic in $D(0,1/2).$ Hence the ROC of the power series of $f+g$ at $0$ is at least $1/2.$ To show the ROC can equal $1/2,$ let

$$f(z) = \frac{1}{z+i/2},\,\, g(z) = \frac{1}{z-i/2}.$$

At the other extreme, we can let

$$f(z) = \frac{1}{z+i5/2} + \frac{1}{z-i5/2},\,\, g(z) = -f(z).$$

Then $f+g=0$ in $D(0,1/2),$ so the ROC here is $\infty.$

See if you can find examples to show any $R\in (1/2,\infty)$ can be the ROC.

  • $\begingroup$ Thanks for your answer! I think I can make $f(z)=\frac{1}{z+i5/2}+\frac{1}{z-5i/2}+\frac{1}{1-z/a}$ where $|a|>3/2$ and $g(z)=-\frac{1}{z+i5/2}-\frac{1}{z-5i/2}$ to make $R \in [3/2,\infty)$. But I have a hard time finding $R \in (1/2,3/2)$ $\endgroup$ – Nan Feb 9 '18 at 21:27
  • $\begingroup$ How about $$f(z) = \frac{1}{z+i5/2} + \frac{1}{z-i5/2},\,\, g(z) = -f(z)+\frac{1}{z-ri},\,\, 1/2<r<\infty$$ $\endgroup$ – zhw. Feb 9 '18 at 22:01

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