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I need to find the radius of convergence about the origin for this function

$$ G(z) = \left(\frac{1 - \sqrt{1-4abz^2}}{2}\right) $$ where, $$ \\ a+b = 1, \\ 0 < b < \frac 12 $$

I'm finding it hard to convert $G(z)$ to a power series and if I use the Taylor's expansion of $(1+x)^\frac12$, I get $R=\infty$ which I don't believe is correct.

Can someone please give me a hint on the right way to approach this question.

My steps so far (I'm not sure if they're correct),

\begin{align} G(z) & = \left(\frac{1 - \sqrt{1-4abz^2}}{2}\right) \\ & = \frac{2abz^2}{1 + \sqrt{1-4abz^2}} \\ & = \frac{2abz^2}{1 - \left(-\sqrt{1-4abz^2}\right)} \\ & = 2abz^2\sum_{j=0}^\infty (-1)^j\left(\sqrt{1-4abz^2}\right)^j & \text{if}\; \sqrt{1-4abz^2} & < 1 \\ \end{align}

Now, \begin{align} \text{if} \; & \sqrt{1-4abz^2} & < 1 \\ \implies & 1 - 4abz^2 & < 1 \\ \implies & 4abz^2 & > 0 \\ \implies & z \in \mathbb R \\ \implies & R = \infty \\ \end{align}

Srinivas Eswar
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  • The Taylor expansion of $(1+x)^{1/2}$ only has finite radius (if expanded about $0$ then, the radius is $1$). Why don't you show your calculation leading to $R=\infty$? – Erick Wong Sep 01 '15 at 04:32
  • Just curious, but is $z$ complex or real? – Mark Viola Sep 01 '15 at 04:42
  • @ErickWong I've added the steps to my answer. – Srinivas Eswar Sep 01 '15 at 04:51
  • @Dr.MV $G(z)$ is supposed to be a generating function for some random variable. There is no constraint on $z$ I suppose but I always think of $z \in [0,1]$. – Srinivas Eswar Sep 01 '15 at 04:54
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    Hadamard's theorem is useful here: the radius of convergence of the Taylor series is the distance to the nearest singularity of the function in the complex plane. For this problem, that singularity is where the square root is 0. – Paul Sinclair Sep 01 '15 at 04:54
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    To add to the comment from @PaulSinclair, there are two branch points, equidistant from the origin. – Mark Viola Sep 01 '15 at 05:15

1 Answers1

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The series you present is not a Taylor series for G(z) about the origin. A Taylor series would be of the form $G(x)= a_0+a_1 x+a_2 x^2+...$. It would be easier to compute the radius of convergence of $1-2 G(x)=(1-4abz^2)^{-1/2}$ as it's the same radius as for $G(x)$. A special Taylor series discovered by Newton, is the generalized binomial theorem : If $|x|<1$ then for any real number $r$ we have $$(1+x)^r= \sum_{n=0}^ {\infty} C(n,r) x^n $$ where $C(0,r)=1$ and $C(n+1,r)=C(n,r)(r-n)/(n+1)$..... I don't know whether you have enough background to finish this. Newton's series converges for $|x|<1$.For $|x|>1$ and $r$ not a natural number and not $0$ the individual terms of the series do not even converge to $0$.The Hadamard formula for the radius $r$ of convergence of $a_0+a_1 x +a_2 x^2 +..$ is given by $$1/r = \lim_{n \to \infty} [ \sup_{m>n} |a_m|^{1/m}].$$ The radius of convergence for $G$ is $1/ 2 \sqrt{a b}$ ...( Newton's series with $x=2 a b z^2$).

DanielWainfleet
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