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The coefficient identity $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k$

Statement

For every $m \geq 0$ , in the formal series $Q (z) = z / T (z) = z / \tanh z$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , $[z^{m}] (\frac{z}{\tanh z})^{m + 1} = {\begin{cases} 1, & m even, \\ 0, & m odd. \end{cases}$ Equivalently, for every $k \geq 0$ , $[z^{2 k}] (\frac{z}{\tanh z})^{2 k + 1} = 1, [z^{2 n + 1}] (\frac{z}{\tanh z})^{2 n + 2} = 0 (n \geq 0),$ and the formal residue ${res}_{z} (\tanh z)^{- (2 k + 1)} d z$ equals $1$ .

Facts & Assumptions

Given: The series $T$ and $A$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , the integer $m \geq 0$ , and the Bernoulli-free coefficient computation below.

[F1]

$T = x S$ with $S$ an even power series of constant term $1$ , so $S$ is a unit and $Q = x / T = 1 / S$ ; in particular $T$ has order $1$ and linear coefficient $1$ (The formal hyperbolic tangent series and the even series $x / \tanh x$ ).

[F2]

For a field $K$ , $K ((x))$ consists of the Laurent series with support bounded below, with finite convolution in each degree, derivative $D (x^{n}) = n x^{n - 1}$ , and residue ${res}_{x} (f) = [x^{- 1}] f$ (Formal Laurent series $K ((x))$ , their order, derivative, and residue).

[F3]

The inverse-series lemma gives $T \circ A = z$ , $A \circ T = x$ , and $A^{'} (z) = 1 / (1 - z^{2}) = \sum_{j \geq 0} z^{2 j}$ (The formal hyperbolic tangent and artanh series are inverse, with the artanh derivative).

[F4]

Over a field $K$ containing $Q$ , for $g \in x K ⟦ x ⟧$ with nonzero linear coefficient and $F \in K ((x))$ for which $F \circ g$ is formed by Laurent substitution, ${res}_{x} ((F \circ g) D g) = {res}_{x} (F)$ (Formal residues satisfy integration by parts, logarithmic differentiation, and change of variables).

[F5]

A power series is a unit exactly when its constant coefficient is a unit, and inverses are unique; $(1 - z^{2}) \sum_{j \geq 0} z^{2 j} = 1$ (A formal power series is a unit exactly when its constant coefficient is a unit).

[F6]

Summable families may be regrouped and reindexed, and coefficient extraction is the functional $[z^{n}]$ of Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Proof

technique · direct; reduce the coefficient to a residue and change variables by the inverse series

1.1givenF1F2F5F6algebra

Since $T = u S$ with $S$ a unit of $Q ⟦ u ⟧$ by [F1], the series $F (u) : = T (u)^{- (m + 1)} = u^{- (m + 1)} S (u)^{- (m + 1)}$ is a well-defined element of $Q ((u))$ of order $- (m + 1)$ . As $(u / T (u))^{m + 1} = u^{m + 1} T (u)^{- (m + 1)}$ in Laurent series, $F (z) = z^{- (m + 1)} (z / T (z))^{m + 1}$ and therefore ${res}_{z} F (z) = [z^{- 1}] z^{- (m + 1)} (z / T (z))^{m + 1} = [z^{m}] (z / T (z))^{m + 1}$ .

2.1step 1.1F1F2F3F5

The composition $F \circ A$ is formed by Laurent substitution: $A = z U$ with $U$ a unit, so $A^{- (m + 1)} = z^{- (m + 1)} U^{- (m + 1)}$ has finitely many terms in each degree and $S^{- (m + 1)} \circ A$ is a unit power series, and every coefficient of $F \circ A$ is a finite sum. By [F3], $T \circ A = z$ , hence $F \circ A = (T \circ A)^{- (m + 1)} = z^{- (m + 1)}$ ; and $D A = A^{'} = 1 / (1 - z^{2}) = \sum_{j \geq 0} z^{2 j}$ , the last identity by [F3] and [F5].

3.1step 1.1step 2.1F3F4F5F6

The change-of-variables identity [F4] applies over $K = Q$ with $g = A$ , whose linear coefficient is $1 \neq 0$ : ${res}_{z} ((F \circ A) A^{'}) = {res}_{z} F$ . By step 2.1 the left side is ${res}_{z} (z^{- (m + 1)} (1 - z^{2})^{- 1}) = [z^{- 1}] z^{- (m + 1)} \sum_{j \geq 0} z^{2 j} = [z^{m}] \sum_{j \geq 0} z^{2 j}$ , which is $1$ when $m$ is even and $0$ when $m$ is odd, since $z^{2 j}$ has degree $2 j$ . Combining with step 1.1 gives $[z^{m}] (z / T (z))^{m + 1} = 1$ for even $m$ and $0$ for odd $m$ .

4.1step 1.1step 3.1given∎

Taking $m = 2 k$ gives $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k \geq 0$ ; taking $m = 2 n + 1$ gives $[z^{2 n + 1}] (z / \tanh z)^{2 n + 2} = 0$ ; and step 1.1 with $m = 2 k$ identifies ${res}_{z} T (z)^{- (2 k + 1)} = [z^{2 k}] (z / T (z))^{2 k + 1} = 1$ , the residue form asserted. The value $k = 0$ reads $[z^{0}] Q = 1$ , the constant term of the unit series $Q$ . All computations are coefficientwise identities between formal Laurent series over $Q$ ; no analytic contour and no choice principle is involved.

Depends on

Used by

The L-genus of complex projective space of even complex dimension is one Lemma

Dependency tree · two levels

28 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

John Milnor and James Stasheff, Characteristic Classes (re-typeset scan; original pagination) (standard reference, not scraped)
Tom Weston, An Introduction to Cobordism Theory (lecture notes, Stanford) (standard reference, not scraped)
Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)
Jacob Lurie, The Hirzebruch Signature Formula (Lecture 25, Harvard Math 287x notes) (standard reference, not scraped)

The coefficient identity [z2k](z/tanh⁡z)2k+1=1[z2k](z/tanhz)2k+1=1 for every kk