TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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An order-raising recursive specification has a unique solution

Statement

Let $R$ be a commutative ring, and let

$F : R ⟦ x ⟧ \to R ⟦ x ⟧$

satisfy

${ord}_{x} (F (f) - F (g)) \geq {ord}_{x} (f - g) + 1$

for all formal series $f, g$ . Then there is a unique formal series $y$ with $F (y) = y$ .

Equivalently, once a commutative coefficient ring is fixed, every order-raising recursive specification has a unique generating-function solution.

Facts & Assumptions

Given: A commutative ring $R$ and an operator $F : R ⟦ x ⟧ \to R ⟦ x ⟧$ satisfying the displayed order-raising inequality.

[L1]

Formal order is non-Archimedean under sums: in particular, ${ord}_{x} (f + g) \geq \min ({ord}_{x} f, {ord}_{x} g)$ (Formal order is non-Archimedean under sums and additive under products over a domain).

[L2]

Every $x$ -adically Cauchy sequence in $R ⟦ x ⟧$ has a unique $x$ -adic limit ( $R ⟦ x ⟧$ is complete in the $x$ -adic topology and $R [x]$ is dense by truncation).

Proof

technique · direct

1.1

Define a sequence by $f_{0} : = 0$ and $f_{j + 1} : = F (f_{j})$ . Then ${ord}_{x} (f_{j + 1} - f_{j}) \geq j$ for every $j$ : the case $j = 0$ is automatic, and if it holds at $j$ then ${ord}_{x} (f_{j + 2} - f_{j + 1}) = {ord}_{x} (F (f_{j + 1}) - F (f_{j})) \geq {ord}_{x} (f_{j + 1} - f_{j}) + 1 \geq j + 1$ .

given

2.1

For $p > q$ , write $f_{p} - f_{q} = (f_{p} - f_{p - 1}) + \dots + (f_{q + 1} - f_{q})$ . Step 1.1 and [L1] give ${ord}_{x} (f_{p} - f_{q}) \geq q$ , so $(f_{j})$ is $x$ -adically Cauchy.

step 1.1L1

3.1

By [L2], the sequence $(f_{j})$ has a unique $x$ -adic limit; call it $y$ .

step 2.1L2choose

4.1

The order-raising hypothesis applied to $f_{j}$ and $y$ gives ${ord}_{x} (F (f_{j}) - F (y)) \geq {ord}_{x} (f_{j} - y) + 1$ , so $F (f_{j}) \to F (y)$ in the $x$ -adic topology. But $F (f_{j}) = f_{j + 1}$ , and $f_{j + 1} \to y$ as well, hence $F (y) = y$ .

step 3.1given

5.1

If $z$ is another fixed point and $z \neq y$ , put $p : = {ord}_{x} (y - z)$ . Then $p = {ord}_{x} (F (y) - F (z)) \geq p + 1$ , impossible. Hence $z = y$ .

step 4.1givenassume-contradischarge-contradiction

6.1

Step 4.1 gives existence of a fixed point and step 5.1 gives uniqueness, so the recursive specification has exactly one solution.

step 4.1step 5.1∎

Depends on

Used by

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)
Stephen Melczer, An Invitation to Enumeration, Chapter 5: Combinatorial Constructions (standard reference, not scraped)