Singularities of $f(x)=\left(\sqrt{x+1}-\sqrt{x}\right)^{\alpha}$

Question

I'm working with the following function

$$f(x)=\left(\sqrt{x+1}-\sqrt{x}\right)^{\alpha}=\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)^{\alpha}$$

with $x\in[0,\infty)$ and $\alpha\in\mathbb{R}$. More specific, with the Taylor expansion $$\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)$$

I remember that $R$ the radius of convergence of this series is related with its singulary points, that is, $R$ is at least the minimum distance between $x_0$ and its singularities.

• First at all, ¿it's this true?
• An then, What are the singularities, if any, of this function?

Thanks in advance!!

Answer 1

Singularity of $\sqrt{x}$ is $x=0$.

Singularity of $\sqrt{x+1}$ is $x=-1$.

Singularity for the $\alpha$ power is where $\sqrt{x+1}-\sqrt{x} = 0$. (Unless $\alpha$ is a nonnegative integer) Are there any complex numbers $x$ that satisfy this?

So these are the only candidates for your singularity.