Square Root of Formal Power Series

Question

Suppose $R(x)$ is a formal power series. If $[R(x)]^2=x+1$, must be $R(x)$ be the formal power series of $\sqrt{x+1}$, or $-\sqrt{x+1}$?

Answer 1

Assuming you mean power series centered at the origin, i.e. $\sum a_n x^n$, there is no such power series, "formal" or otherwise.

Say $R(x)=\sum a_n x^n$. If $a_0\ne0$ then $R(x)^2$ has non-zero constant term, so it's not $x$. If $a_0=0$ the series for $R(x)^2$ starts with an $x^2$ term, also not $x$.

Second Edit: The question keeps changing. I don't have time to keep revising my answer to be valid for the current version. For the record, the following is an answer to this version: Suppose $a>0$ and $R(x)=\sum c_n(x-a)^n$ is a formal power series with $R(x)^2=x$. Must $R(x)$ be a formal power series for $\sqrt x$ or $-\sqrt x$?

First Edit: It turns out we mean a series centered at $a>0$: $R(x)=\sum c_n(x-a)^n$. If the series has positive radius of convergence and $R(x)^2=x$ then in fact $R(x)$ converges to $\sqrt x$ or to $-\sqrt x$.

Which doesn't quite answer the question, I suppose. It's not at all clear to me what the phrase "a formal power series for $\sqrt x$ " means, exactly. It seems clear that if $R(x)$ is a formal power series and $R(x)^2=x$ "formally" then $R(x)$ does in fact converge to $\pm\sqrt x$.

Yes. This is because there are only two formal power series that square to $x$, and since there are two convergent such series, every such formal series actually converges. Why there are only two such formal series:

Start with $$x= a + (x-a) = (c_0+c_1(x-a)+\dots)^2$$

Write out the square on the right and equate coefficients. You get a system of equations $$c_0^2=a,\quad 2c_0c_1 =1,\quad 2c_0c_2+c_1^2=0,\dots$$

There are two possibilities for $c_0$. But once $c_0$ is determined there is only one possibility for $c_1$, then only one possibility for $c_2$, etc. The crucial observation is that for $n>1$ the first time $c_n$ appears it appears to the first power. (In fact the first time $c_n$ appears it appears in a term of the form $2c_0c_n$, which is good because $c_0\ne0$, so this does determine $c_n$.)

Answer 2

There is no such power series: Suppose there were, say, $$R(x) := a + b x + O(x^2) .$$ Computing gives $$x = [R(x)]^2 = a^2 + 2 a b x + O(x^2) ,$$ and comparing the first two coefficients gives $$a^2 = 0, \qquad 2 a b = 1,$$ but there is no solution $(a, b)$ to this system.