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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?


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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
up vote 1 down vote accepted

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 $|\gamma| < |\beta|$.

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