Easy and LTTR.

Let \(n\in \mathbb {N}\). We need to check that the coefficient \(\left[x^{n}\right]\left(\prod _{i\in M}\mathbf{a}_{i}\right)\) does not depend on \(M\). In other words, we need to check that if \(M_{1}\) and \(M_{2}\) are two finite subsets of \(I\) that each determine the \(x^{n}\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i\in I}\), then

So let us prove this. Since \(M_{1}\) determines the \(x^{n}\)-coefficient, every finite subset \(J\) of \(I\) with \(M_{1}\subseteq J\) satisfies \(\left[x^{n}\right]\left(\prod _{i\in J}\mathbf{a}_{i}\right) =\left[x^{n}\right]\left(\prod _{i\in M_{1}}\mathbf{a}_{i}\right)\). Applying this to \(J=M_{1}\cup M_{2}\), we obtain

The same argument (with the roles of \(M_{1}\) and \(M_{2}\) swapped) yields

Since \(M_{1}\cup M_{2}=M_{2}\cup M_{1}\), the left hand sides are equal, whence \(\left[x^{n}\right]\left(\prod _{i\in M_{1}}\mathbf{a}_{i}\right) =\left[x^{n}\right]\left(\prod _{i\in M_{2}}\mathbf{a}_{i}\right)\).

Argue that \(I\) itself is a subset of \(I\) that determines all coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i\in I}\). See Section A.1 for a detailed proof.

This follows from the Cauchy product formula: the \(x^m\)-coefficient of a product \(ac\) depends only on the coefficients of \(a\) and \(c\) up to degree \(m\).

First show that \(c^{-1}\overset {x^n}{\equiv } d^{-1}\) using \(x^n\)-equivalence of inverses, then apply Lemma 1.346 to \(a\cdot c^{-1}\) and \(b\cdot d^{-1}\).

The FPS \(af\) is a multiple of \(f\) (since \(af=fa\)). Hence, Lemma 1.132 yields that \(\left[x^{m}\right]\left(af\right)=0\) for each \(m\in \left\{ 0,1,\ldots ,n\right\} \). Now, for each \(m\in \left\{ 0,1,\ldots ,n\right\} \), we have

This is just Lemma 1.348, applied \(\left|J\right|\) many times. See Section A.1 for a detailed proof.

This is an easy consequence of Lemma 1.349. See Section A.1 for a detailed proof.

LTTR. (See Section A.1 for a detailed proof.)

The empty set determines every coefficient (the product over \(\emptyset \) is \(1\)).

The empty set determines all coefficients.

The single-element set determines all coefficients.

Any finite set containing \(j\) determines all coefficients (the product is \(0\)).

\(I\) itself determines all coefficients.

This is a special case of Proposition 1.345.

The empty set is an \(x^n\)-approximator for every \(n\), and \(\prod _{i\in \emptyset }\mathbf{a}_i = 1\).

By induction on \(m\), using the identity \(\left(1-a\right)\left(1+a\right)=1-a^{2}\).

Apply Theorem 1.350: the family \(\left(x^{2^{i}}\right)_{i\in \mathbb {N}}\) is summable (for each \(n\), at most one \(i\) satisfies \(2^i = n\) since \(i\mapsto 2^i\) is injective).

For each \(n\), the set \(\{ 0,1,\ldots ,n\} \) determines the \(x^n\)-coefficient: any factor \(\left(1+x^{2^i}\right)\) with \(i {\gt} n\) has \(2^i {\gt} n\), so it does not affect coefficients \(\le n\). The finite product for \(i\in \{ 0,\ldots ,n\} \) equals \(\left(1-x^{2^{n+1}}\right)/(1-x)\) by Lemma 1.359, and its \(x^n\)-coefficient equals that of \(1/(1-x)\) since \(2^{n+1} {\gt} n\).

This is an easy consequence of the fact that a union of finitely many finite sets is finite. A detailed proof can be found in Section A.1.

This follows easily from Definition 1.362 and Definition 1.343 (b). See Section A.1 for a detailed proof.

Fix \(n\in \mathbb {N}\). Lemma 1.363 (applied to \(J\) instead of \(I\)) shows that there exists an \(x^{n}\)-approximator \(U\) for \(\left(\mathbf{a}_{i}\right)_{i\in J}\). Lemma 1.363 shows that there exists an \(x^{n}\)-approximator \(V\) for \(\left(\mathbf{a}_{i}\right)_{i\in I\setminus J}\). Note that \(U\cup V\) is finite. Now, \(U\cup V\) is an \(x^{n}\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i\in I}\). Hence, the \(x^{n}\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i\in I}\) is finitely determined. This proves part (a). Part (b) follows using Proposition 1.364 (b).

See Section A.1 for the details.

Fix \(n\in \mathbb {N}\). Lemma 1.363 shows that there exists an \(x^{n}\)-approximator \(U\) for \(\left(\mathbf{a}_{i}\right)_{i\in I}\). Lemma 1.363 (applied to \(\mathbf{b}_{i}\) instead of \(\mathbf{a}_{i}\)) shows that there exists an \(x^{n}\)-approximator \(V\) for \(\left(\mathbf{b}_{i}\right)_{i\in I}\). Note that \(U\cup V\) is finite. Now, \(U\cup V\) is an \(x^{n}\)-approximator for \(\left(\mathbf{a}_{i}\mathbf{b}_{i}\right)_{i\in I}\). From here, proceed as in the proof of Proposition 1.365.

See Section A.1 for the details.

This is similar to the proof of Proposition 1.366, but using Lemma 1.347 (\(x^n\)-equivalence of quotients) instead of Lemma 1.346. See Section A.1 for the details.

Let \(T\supseteq U\cap J\) be a finite subset of \(J\). Then \(T\cup (U\setminus J)\) and \((U\cap J)\cup (U\setminus J) = U\) are both finite supersets of \(U\), so their products are \(x^n\)-equivalent to \(\prod _{i\in U}\mathbf{a}_i\) by the approximator property. Writing each product as a product over the \(J\)-part times a product over \(U\setminus J\), and canceling the invertible common factor \(\prod _{i\in U\setminus J}\mathbf{a}_i\), we conclude \(\prod _{i\in T}\mathbf{a}_i \overset {x^n}{\equiv } \prod _{i\in U\cap J}\mathbf{a}_i\).

We must show that \((\mathbf{a}_i)_{i\in J}\) is multipliable whenever \(J \subseteq I\). The idea is to show that if \(U\) is an \(x^{n}\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i\in I}\), then \(U\cap J\) determines the \(x^{n}\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i\in J}\) (and, in fact, is an \(x^{n}\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i\in J}\)). This uses Lemma 1.368 and relies on the invertibility of \(\prod _{i\in U\setminus J}\mathbf{a}_{i}\); this is why the FPS \(\mathbf{a}_{i}\) are required to be invertible in the proposition.

See Section A.1 for the details.

This is a trivial consequence of the definitions of multipliability and infinite products.

By induction on \(|V|\). The base case \(V = \emptyset \) is trivial. In the inductive step, use \(x^n\)-equivalence of products: \(\prod _{v\in V \cup \{ w\} } c_v = c_w \cdot \prod _{v\in V} c_v\) and similarly for \(d\), then apply the fact that \(x^n\)-equivalence is preserved under multiplication.

This can be derived from the analogous property of finite products, since all coefficients in a multipliable product are finitely determined (conveniently using \(x^{n}\)-approximators, which determine several coefficients at the same time). See Section A.1 for the details.

The first equality follows by applying Proposition 1.372 to \(S=I\times J\) and \(W=I\) and \(f(i,j)=i\) (and appropriately reindexing the products). The second equality is analogous. See Section A.1 for the details.

See Section A.1.