# Divide D-finite power series by x, still D-finite?

A power series $f \in K[[x_1,...,x_n]]$ is called D-finite if all partial derivations of f lie in a finite-dimensional vector space over $K(x_1,...,x_n)$. For one variable this is equivalent to: f satisfies a linear differential equation with polynomial coefficients.

Let f and g be power series in one variable $x$ and let f be D-finite. If $g \cdot x^k = f$ for some k does it then follows(and if so why?) that g is D-finite? The same question with several variables: $f,g \in K[[x_1,...,x_n]]$ and $g \cdot x_n^k = f$, does it follow that g is D-finite?

Thanks

-
I assume that $K(x)$ should be $K(x_1,\ldots,x_n)$ in the $n$-variable case. –  Robert Israel Apr 12 '11 at 19:46
For the one-variable case: if $\sum_{j=0}^m c_j(x) f^{(j)}(x) = 0$ (with $c_m \ne 0$) is a linear differential equation with polynomial coefficients satisfied by $f$, then just substitute $f = x^k g$ and you get a linear differential equation with polynomial coefficients satisfied by $g$. Note that this is nontrivial because the coefficient of $g^{(m)}$ is c_m(x) x^k$. In the several-variable case, suppose all partial derivatives$D^\beta(f)$of$f$are linear combinations over$K(x_1,\ldots,x_n)$of$D^{\alpha}(f)$with$|\alpha|<=m$. I claim this is also true for$g$. The proof is by induction on$|\beta|$, using the fact that$D^\beta(x^k g) - x^k D^\beta(g)$is a linear combination of$D^\gamma(g)$with$|\gamma| < |\beta|\$.