is it possible to simplify the square root of an infinite summation?

Question

So I was solving a physics problem and stumbled upon a mathematical problem that I couldn't solve.
I know that for $X$ and $Y$, where both are positive and $X>Y$:
$$\sqrt{X\pm Y}=\sqrt{\frac{X+\sqrt{{X^2}+{Y^2}}}{2}}\pm\sqrt{\frac{X-\sqrt{{X^2}+{Y^2}}}{2}}$$
Can this be extended for an infinite summation? Is there a closed expression for the square root of a taylor series: $$\sqrt{\sum_{n=0}^\infty{A_nx^n}}$$ If not - are there some special cases where this is possible?
In any case - Is there a way to expand some of the coefficients can be derived?

Answer 1

Start with the special case $A_0=1$, where we can expect the square root to have constant term $1$ as well. Expand the $\left(\sum_n B_nx^n\right)^2$ (with $B_0=1$) and equate like powers, so $$\begin{align} 1&=1\\ 2B_1&=A_1\\ 2B_2+B_1^2&=A_2\\ 2B_3+2B_1B_2&=A_3\\ 2B_4+2B_1B_3+B_2^2&=A_4\\ &\vdots \end{align} $$ You always find $$B_n=\frac12A_n-\frac12\sum_{k=0}^{n-1}B_kB_{n-1-k}$$ as a way to recursively express the "next" $B$-term in terms of $A_n$ and lower terms.

More generally, if $A_0>0$, first divide by $A_0$ to arrive at the above special case, and afterwards multiply the result by $\sqrt{A-0}$.

Even more generally, if $A_0=\ldots=A_{n-1}=0$ and $A_n>0$, fist divide by $x^n$ and afterwards multiply by $x^{n/2}$.