1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.