Associated graded ring and completion Let $ R $ be a ring and $I$ an ideal. This gives a decreasing filtration of $R$ by $ I^n$. Consider the associated graded ring $ \operatorname{gr}_I(R) =\bigoplus I^n/I^{n+1}$. As it is graded, there is a natural decreasing filtration. Hence we can take the corresponding completion $ \widehat{\operatorname{gr}_I(R)}$.
1) Is there any convenient description of $ \widehat{\operatorname{gr}_I(R)} $ ?
2) More generally, is there any relation between the  $I$-adic completion $ \hat{R} $ and the associated graded  ring $ \operatorname{gr}_I(R)$ ?
 A: I don't know the answer to this question in general, but there is a related geometric story which makes things a bit clearer.
Let $X$ be a variety and $x \in X $ a non singular point. Assume that $ {\rm dim} \mathscr{O}_x = n $. Let $f_1, \dots, f_n$ be a system of parameters which generated the maximal ideal $ \mathfrak{m}_x \subseteq \mathscr{O}_x$. Then there is a non singular open neighborhood $ {\rm Spec} A$ of $x$ on which each $ f_1, \dots, f_n$ is defined and we have $ \{ x \} = V(f_1, \dots, f_n) $ scheme theoretically. Let $ I = (f_1, \dots, f_n) \subseteq A $. Then we have a closed embedding $ {\rm Bl}_x {\rm Spec} A \subseteq \mathbb{P}^{n-1}_A $ defined by the map $ A[X_1, \dots, X_n] \to A \oplus I \oplus I^2 \oplus I^3\oplus \dots $ which sends $ X_i $ to $ f_i $ in degree $ 1 $. Taking the fiber of this map over $x $  gives us a closed embedding $ {\rm Proj} (A / I \oplus I / I^2 \oplus I^2 / I^3 \oplus \dots) \to \mathbb{P}^{n-1}_k $ under which $ X_i $ pulls back to $f_i $ in degree $1$ Since the exceptional divisor has dimension $ n- 1 $ this must be an isomorphism, so $ A / I \oplus I / I^2 \oplus \dots $ is a polynomial ring generated by $ f_1, \dots, f_n $. Since localizing does not affect the associated graded ring we see that $ {\rm gr}_{\mathfrak{m}_x} \mathscr{O}_x $ is a polynomial ring.
Now we want to relate the $ {\rm gr}_{\mathfrak{m}_x} \mathscr{O}_x $ to the completion of $ \mathscr{O}_{x} $ at $ \mathfrak{m}_x$. Notice that we have the following commutative diagram with exact rows:
$$
\require{AMScd}
\begin{CD}
0 @>>> (X_i)^p / (X_i)^{p+1} @>>> k[X_i] / (X_i)^{p+1} @>>> k[X_i] / (X_i)^p @>>> 0\\
@VVV @VVV @VVV  @VVV @VVV \\
0 @>>>\mathfrak{m}^p / \mathfrak{m}^{p+1} @>>> \mathscr{O} / \mathfrak{m}^{p+1} @>>> \mathscr{O} / \mathfrak{m}^p @>>> 0
\end{CD}
$$
Since the associated graded ring is a polynomial ring, the left vertical map is an isomorphism. Therefore by induction and the 5-lemma it follows that the completion of $ \mathscr{O} $ is a formal power series ring. To summarize, the difference between the associated graded ring and a polynomial ring influences the difference between $\hat{\mathscr{O}} $ and a formal power series ring.
