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Rooted plane trees satisfy $T (x) = x / (1 - T (x))$

Statement

Let $T (x)$ be the generating function of rooted plane trees, specified by

$T = Z \times SEQ (T) .$

Then $T (x)$ is the unique formal power series with zero constant coefficient satisfying

$T (x) = \frac{x}{1 - T (x)} .$

Facts & Assumptions

Given: The recursive specification $T = Z \times SEQ (T)$ .

[L1]

If $A$ has no size-zero objects then $SEQ (A)$ has generating function $1 / (1 - A (x))$ (If $A$ has no size-zero objects then $SEQ (A)$ has generating function $1 / (1 - A (x))$ ).

[L2]

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

[L3]

Formal order is additive under multiplication by $x$ , and a unit has order $0$ (Formal order is non-Archimedean under sums and additive under products over a domain, A formal power series is a unit exactly when its constant coefficient is a unit).

[L4]

The atomic class $Z$ has generating function $x$ , and every object of $Z$ has size $1$ (The neutral class $E$ and the atomic class $Z$ ).

Proof

technique · direct

1.1

On the set $x Z ⟦ x ⟧$ of series with zero constant coefficient, define $F (Y) = x / (1 - Y)$ . This is well defined because $1 - Y$ has constant coefficient $1$ . For $U, V$ in this set, $F (U) - F (V) = x (U - V) / ((1 - U) (1 - V))$ ; both denominators are units of order $0$ , so [L3] gives ${ord}_{x} (F (U) - F (V)) \geq {ord}_{x} (U - V) + 1$ . Moreover $F (Y)$ again has zero constant coefficient.

L3algebra

1.2

Every object of $T$ has a root from the atomic class $Z$ , so every tree has size at least $1$ . Thus $T$ has no size-zero objects.

givenL4

2.1

Define $T_{0} : = 0$ and $T_{j + 1} : = F (T_{j})$ . Step 1.1 gives ${ord}_{x} (T_{j + 1} - T_{j}) \geq j$ by induction on $j$ , and [L3] then shows that $(T_{j})$ is $x$ -adically Cauchy. By [L2] it has an $x$ -adic limit $T$ , whose constant coefficient is $0$ .

step 1.1L2L3choose

3.1

Step 1.1 also gives $F (T_{j}) \to F (T)$ . Since $F (T_{j}) = T_{j + 1}$ and the shifted sequence has the same limit $T$ , uniqueness of limits from [L2] yields $F (T) = T$ .

step 1.1step 2.1L2

4.1

If $U, V \in x Z ⟦ x ⟧$ are distinct fixed points and $p : = {ord}_{x} (U - V)$ , then step 1.1 gives $p = {ord}_{x} (F (U) - F (V)) \geq p + 1$ , a contradiction. Thus $T$ is the unique zero-constant fixed point.

step 1.1step 3.1assume-contradischarge-contradiction

5.1

Applying [L1] to $SEQ (T)$ and using [L4] for the root factor shows that $SEQ (T)$ has generating function $1 / (1 - T (x))$ , while $Z$ contributes $x$ . Therefore the defining equation of $T$ reads $T (x) = x / (1 - T (x))$ , and step 4.1 gives the asserted uniqueness in the zero-constant class.

step 1.2step 4.1L1L4∎

Depends on

Used by

Dependency tree · two levels

13 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)

Rooted plane trees satisfy T(x)=x/(1−T(x))T(x)=x/(1−T(x))