If each \(k\in \{ 0,1,\ldots ,n\} \) satisfies \([x^k]f=[x^k]g\), then in particular each \(k\in \{ 0,1,\ldots ,m\} \subseteq \{ 0,1,\ldots ,n\} \) satisfies \([x^k]f=[x^k]g\).

This is immediate from the definition, since \(\{ 0,1,\ldots ,0\} =\{ 0\} \).

Immediate from the definition: \([x^m]f=[x^m]f\) for all \(m\).

If \([x^m]f=[x^m]g\) for all \(m\leq n\), then \([x^m]g=[x^m]f\) by symmetry of equality.

If \([x^m]f=[x^m]g\) and \([x^m]g=[x^m]h\) for all \(m\leq n\), then \([x^m]f=[x^m]h\) by transitivity of equality.

All of these properties are analogous to familiar properties of integer congruences, except for Theorem 1.328 (d), which is moot for integers (since there are not many integers that are invertible in \(\mathbb {Z}\)).

(a) Reflexivity, symmetry and transitivity follow directly from the corresponding properties of equality of coefficients.

(b) For addition: If \(a\overset {x^{n}}{\equiv }b\) and \(c\overset {x^{n}}{\equiv }d\), then for each \(m\in \{ 0,1,\ldots ,n\} \) we have \([x^m](a+c)=[x^m]a+[x^m]c=[x^m]b+[x^m]d=[x^m](b+d)\). Subtraction is analogous.

For multiplication: We have \([x^m](ac)=\sum _{i=0}^{m}[x^i]a\cdot [x^{m-i}]c\). Since \(m\leq n\), each \(i\leq m\) satisfies \(i\leq n\) and \(m-i\leq n\), so \([x^i]a=[x^i]b\) and \([x^{m-i}]c=[x^{m-i}]d\). Thus \([x^m](ac)=[x^m](bd)\).

(c) For each \(m\leq n\), we have \([x^m](\lambda a)=\lambda \cdot [x^m]a=\lambda \cdot [x^m]b=[x^m](\lambda b)\).

(d) Let \(a,b\in K[[x]]\) be two invertible FPSs satisfying \(a\overset {x^{n}}{\equiv }b\). We want to show that \(a^{-1}\overset {x^{n}}{\equiv }b^{-1}\).

The FPS \(a\) is invertible; thus, its constant term \([x^{0}]a\) is invertible in \(K\) (by Proposition 1.111).

We prove that each \(m\in \{ 0,1,\ldots ,n\} \) satisfies \([x^{m}](a^{-1})=[x^{m}](b^{-1})\) by strong induction on \(m\). Fix some \(k\in \{ 0,1,\ldots ,n\} \), and assume (as induction hypothesis) that \([x^{m}](a^{-1})=[x^{m}](b^{-1})\) for each \(m\in \{ 0,1,\ldots ,k-1\} \).

We know that

(here, we have split off the addend for \(i=0\) from the sum). Thus,

We can solve this equation for \([x^{k}](a^{-1})\) (since \([x^{0}]a\) is invertible), and thus obtain

The same argument (applied to \(b\) instead of \(a\)) yields

The right hand sides of the latter two equalities are equal (since each \(i\in \{ 1,2,\ldots ,k\} \) satisfies \([x^{i}]a=[x^{i}]b\) as a consequence of \(a\overset {x^{n}}{\equiv }b\), and satisfies \([x^{k-i}](a^{-1})=[x^{k-i}](b^{-1})\) as a consequence of the induction hypothesis, and since we have \([x^{0}]a=[x^{0}]b\) as a consequence of \(a\overset {x^{n}}{\equiv }b\)). Hence, the left hand sides must also be equal. In other words, \([x^{k}](a^{-1})=[x^{k}](b^{-1})\), which is precisely what we wanted to prove. Thus, the induction step is complete, so that \(a^{-1}\overset {x^{n}}{\equiv }b^{-1}\) is proved.

(e) Combine parts (b) and (d): we have \(a\overset {x^{n}}{\equiv }b\) and \(c^{-1}\overset {x^{n}}{\equiv }d^{-1}\) (by (d)), so \(ac^{-1}\overset {x^{n}}{\equiv }bd^{-1}\) (by (b)).

(f) The sum case follows by induction on \(|S|\) using part (b) for the induction step (adding one element at a time). The product case is analogous, using the multiplication part of (b).

For each \(m\leq n\), we have \([x^m](-a)=-[x^m]a=-[x^m]b=[x^m](-b)\).

The proof proceeds by strong induction on the coefficient index, analogous to the proof of Theorem 1.328 (d), using the equality of unit values \(u=v\) and the \(x^n\)-equivalence of \(f\) and \(g\).

The constant terms of \(c\) and \(d\) agree (since \(c\overset {x^{n}}{\equiv }d\)), so Lemma 1.330 gives \(c^{-1}\overset {x^{n}}{\equiv }d^{-1}\). Then \(a \cdot c^{-1}\overset {x^{n}}{\equiv }b \cdot d^{-1}\) by the multiplication rule of Theorem 1.328 (b).

By induction on \(k\). The base case \(k=0\) gives \(a^0=1=b^0\), which is trivially \(x^n\)-equivalent to itself. For the induction step, \(a^{k+1}=a^k\cdot a \overset {x^{n}}{\equiv }b^k\cdot b=b^{k+1}\) by the induction hypothesis and (53).

The truncation \(\operatorname {trunc}_{n+1}(f)\) is the polynomial \(\sum _{k=0}^{n}([x^k]f)\cdot x^k\). Two such truncations are equal if and only if all their coefficients at degrees \(0,1,\ldots ,n\) agree, which is exactly the condition \(f\overset {x^{n}}{\equiv }g\).

Take \(p=\sum _{k=0}^{n}([x^k]f)\cdot x^k\). Then for each \(m\leq n\), \([x^m]p=[x^m]f\).

A simple consequence of Lemma 1.130.

\((\Longrightarrow )\): Assume \(f\overset {x^{n}}{\equiv }g\). Then for each \(m{\lt}n+1\) (i.e., \(m\leq n\)), we have \([x^m](f-g)=[x^m]f-[x^m]g=0\). Thus the first \(n+1\) coefficients of \(f-g\) are \(0\), so \(f-g\) is a multiple of \(x^{n+1}\) by Lemma 1.130.

\((\Longleftarrow )\): Assume \(x^{n+1}\mid (f-g)\). Then for each \(m{\lt}n+1\), we have \([x^m](f-g)=0\) (by the power series analogue of Lemma 1.130), so \([x^m]f=[x^m]g\). Since this holds for each \(m\leq n\), we obtain \(f\overset {x^{n}}{\equiv }g\).

This is simply a restatement of Proposition 1.335, since “\(f-g\) is a multiple of \(x^{n+1}\)” means exactly that there exists such a \(q\).

Since \([x^0]f=0\), the order of \(f\) is at least \(1\), and thus the order of \(f^k\) is at least \(k{\gt}m\). Hence \([x^m](f^k)=0\).

Write \(a\) and \(b\) as \(a=\sum _{i\in \mathbb {N}}a_{i}x^{i}\) and \(b=\sum _{i\in \mathbb {N}}b_{i}x^{i}\) (with \(a_{i},b_{i}\in K\)). Then, \(a\overset {x^{n}}{\equiv }b\) means that \(a_{i}=b_{i}\) for all \(i\leq n\). Combine this with \(c^{i}\overset {x^{n}}{\equiv }d^{i}\) (which holds for all \(i\in \mathbb {N}\) as a consequence of \(c\overset {x^{n}}{\equiv }d\) and of (57)) to obtain the relation \(a_{i}c^{i}\overset {x^{n}}{\equiv }b_{i}d^{i}\) for all \(i\leq n\). But this relation also holds for all \(i{\gt}n\), since all such \(i\) satisfy \([x^{m}](c^{i})=[x^{m}](d^{i})=0\) for all \(m\in \{ 0,1,\ldots ,n\} \) (a consequence of Lemma 1.337 using \([x^{0}]c=0\) and \([x^{0}]d=0\)). Thus, the relation \(a_{i}c^{i}\overset {x^{n}}{\equiv }b_{i}d^{i}\) holds for all \(i\in \mathbb {N}\). Summing over all \(i\), we find \(a\circ c\overset {x^{n}}{\equiv }b\circ d\).

\((\Longrightarrow )\): If \(f=g\), then \(f\overset {x^{n}}{\equiv }g\) for every \(n\) by reflexivity.

\((\Longleftarrow )\): If \(f\overset {x^{n}}{\equiv }g\) for all \(n\), then in particular for each \(m\in \mathbb {N}\), taking \(n=m\) gives \([x^m]f=[x^m]g\) (since \(m\leq m\)). Thus \(f=g\) by extensionality (Lemma 1.86).