I was reading examples to find the radius of convergence for power series. The power series is defined as $\displaystyle\sum\limits_{n=0}^\infty c_n(z-z_0)^n$. And to find the radius of convergence $R$ we use $\displaystyle\limsup\limits_{n\rightarrow\infty} |c_n|^\frac{1}{n}=\frac{1}{R}$.

For $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}z^n$, how did they get that $\displaystyle\limsup\limits_{n\rightarrow\infty} |c_n|^\frac{1}{n} = \limsup\limits_{n\rightarrow\infty}(2^{-n^2})^\frac{1}{n}$? I can see they took $c_n = 2^{-n^2}$, but how does $c_n$ actually relate to $(2^{-n^2})$? And how does $(z-z_0)^n$ relate to $z^n$?

As in, im confused as how $\displaystyle\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ is related to $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}z^n$. What is $z^n$ and how is it linked to $(z-z_0)^n$? What is $c_n$ and how is it related to $(2^{-n^2})$

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    $\begingroup$ It's likely a typo; they meant $z^n$. Once you find $R$ then the series will converge whenever $|z-z_0|<R$. $\endgroup$ – David Mitra May 24 '12 at 2:00
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    $\begingroup$ Who or what is "they"? $\endgroup$ – Jonas Meyer May 24 '12 at 2:01
  • $\begingroup$ @DavidMitra Sorry, it was my typo. I fixed the question. It makes slightly more sense, but how exactly is $z^n$ and $(z-z_0)^n$ linked? @ JonasMeyer Sorry, should have said it was from my notes, the examples. $\endgroup$ – Derrick May 24 '12 at 2:06
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    $\begingroup$ @Derrick: In this case, $z_0=0$. In general, $z_0$ is some fixed complex number, and it is the base point for the series. $\endgroup$ – Jonas Meyer May 24 '12 at 2:11
  • $\begingroup$ Thanks @JonasMeyer, maybe it was not explained in my notes, but what exactly is $c_n$ and how is that linked to $2^{-n^2}$? $\endgroup$ – Derrick May 24 '12 at 2:14

$z_0 =0$. In general, $z_0$ is a fixed complex number that is the base point for the expansion. $(z-0)^n=z^n$, so it sort of hides in this case.

$c_n=2^{-n^2}$, because that is the coefficient of $z^n=(z-0)^n$. Generally, $c_n$ is just whatever number is multiplied by $(z-z_0)^n$ in the power series. So $c_n$ relates to $2^{-n^2}$ by being equal to it.

  • $\begingroup$ so when finding the radius of convergence, it does not 'really matter' what the $z_0$ would be? For example, the radius of convergence for $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}z^n$ would be the same, as say $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}(z-4i)^n$? $\endgroup$ – Derrick May 24 '12 at 12:53
  • $\begingroup$ @Derrick: Yes, it is determined by the coefficients. Notice that if $\displaystyle f(z)=\sum\limits_{n=0}^\infty2^{-n^2}z^n$ and $\displaystyle g(z)=\sum\limits_{n=0}^\infty2^{-n^2}(z-4i)^n$, then $g(z)=f(z-4i)$, so if the series for $f$ converges when $|z|<R$, then the series for $g$ converges when $|z-4i|<R$. However, if you are considering one fixed function but with different base points for the expansions, the coefficients with be different. E.g. if $f(z)=\frac{1}{1-z}$, then at $z_0=0$ you have $f(z)=\sum\limits_{n=0}^\infty z^n$, with radius of convergence $1$. But if $z_0=3$... $\endgroup$ – Jonas Meyer May 24 '12 at 17:10
  • $\begingroup$ ...then $f(z)=\sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{2^{n+1}}(z-3)^n$, with radius of convergence $2$. $\endgroup$ – Jonas Meyer May 24 '12 at 17:17

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