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On the bottom of page 10 of this paper (/!\ It's in french !) they talk about the 'differential $\mathbb{C}$-algebra of convergent series $\{C(x);x^2d/dx\}$' and about the 'differential $\mathbb{C}$-algebra of the germ of functions holomorphic at the origin $\{\mathcal{O}_0;x^2d/dx\}$'

What are those ?

Hippalectryon
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    From the footnote: "Une $\mathbb{C}$-algèbre différentielle est une $\mathbb{C}$-algèbre munie d'un endomorphisme $\mathbb{C}$-linéaire $\delta$ qui est une dérivation: $\delta(ab)=a\delta(b)+\delta(a)b$". So presumably your question is what $\delta$ is for these examples? – mdp Oct 07 '14 at 16:12
  • @MattPressland Indeed. I've never seen 'differental' algebras before. – Hippalectryon Oct 07 '14 at 16:13
  • I think al it means is that they're including an operation $\delta$ which acts as a derivation i.e. satisfies the product rule. If memory serves that's quite natural once one starts asymptotics for differential equations. – Semiclassical Oct 07 '14 at 16:18
  • @Semiclassical But what would ${C(x);x^2d/dx}$ exactly stand for ? Where does that notation come from ? – Hippalectryon Oct 07 '14 at 16:19
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    Ah, the curly braces confused me at first, but now I think that $x^2 d/dx$ is actually supposed to be $\delta$; so your extra structure is the map "differentiate with respect to $x$ and multiply by $x^2$". You should be able to check that this is a derivation in each case - it will follow simply from $d/dx$ being a derivation (whence the name). – mdp Oct 07 '14 at 16:32

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A differential algebra $A$ over a field $k$ is an algebra over $k$ equipped with a differential $d$; that is, a linear map $d: A \to A$ such that $d(fg) = fdg + gdf$. The name is motivated by the algebra of smooth functions $\Bbb R \to \Bbb R$ whose differential is, well, taking the derivative.\

Let's talk about your two algebras. First, we have the algebra of convegent power series, whose differential is $d=x^2\frac{d}{dx}$; explicitly, this means $$d: \sum_{n=0}^\infty a_nx^n \mapsto \sum_{n=2}^\infty (n-1)a_{n-1}x^n.$$ (We get this by first taking the derivative, then multiplying by $x^2$.) It's tedious but not hard to see that this satisfies the product rule $d(fg)=fdg+gdf$.

Note that this is all over $\Bbb C$, and is taking place in the complex plane, rather than on the real line.

Talking about germs of functions is a bit more advanced, but once you understand what germs of holomorphic functions are, seeing that the latter algebra has the structure of a differential algebra is just a special case of seeing that the former does.