My attempt:
For an indeterminate $Y$ we know there exists $a_k \in \mathbb{R}$ with $$\sqrt{1+Y} = \sum_{k=0}^{\infty} a_k Y^k.$$ If we would take an element $f(X) = \sum_{k=m}^{\infty} b_k X^k \in \mathbb{R}((X)$ for $b_k \in \mathbb{R}$ and $m \in \mathbb{Z}$ we could substitute $f(X)$ for $Y$. Then we would get $$ \sqrt{1+f(X)} = \sum_{k=0}^{\infty} a_k \left( f(X) \right)^k. $$ But the last sum is not an element in $\mathbb{R}((X))$ in general. If we would take $f(X) = X^{-1}$ for example then there exists no finite $n \in \mathbb{Z}$ and $c_k \in \mathbb{R}$ with $\sum_{k=0}^{\infty} a_k \left( X^{-1} \right)^k = \sum_{k=n}^{\infty} c_k X^k$. For $m \geq 0 $ otherwise we would get an element in $\mathbb{R}((X))$. But this would imply (for $f(X) = -1 + X$) that X is a square in $\mathbb{R}((X))$ which is false. Thus I think I cannot substitute simply $f(X)$ for $Y$. So my questions are:
- Why is this substitution not allowed?
- How can you find all the squares in $\mathbb{R}((X))$?
Nothing mysterious. $$\Bbb{R}((X))^\times = X^\Bbb{Z}\ \times\ \Bbb{R}^\times\times\ 1\!+\!X \Bbb{R}[[X]]$$ $$\Bbb{R}((X))^{\times 2} = X^{2\Bbb{Z}}\ \times\ \Bbb{R}_{>0}\times\ 1\!+\!X \Bbb{R}[[X]]$$ In other words, an element of $\Bbb{R}((X))^\times$ is a square if and only if it leading term is of the form $a_kX^k$ where $k$ is an even integer (possibly negative) and $a_k>0$.