Lagrange Inversion Without Tears (Analysis) (based on Henrici) Doron ZEILBERGER

Def 1: A formal Laurent series (f.L.s) is

∑_{N}^{∞} a_{n}x^{n} ,

where a_{n} and x are symbols and N is an integer. Def 2: [x^{n}]f(x) is the coeff. of x^{n} in f(x). Def 3: Resf(x) is the coeff. of x^{−1} in f(x). Def 4: Given a sequence f_{i} of f.L.s. starting at N_{i}, their sum ∑ f_{i} makes sense if the N_{i} are bounded below and for every n, the set {[x^{n}]f_{i}(x)} has only finitely many non-zero terms, and then the coeff. of x^{n} in sum ∑ f_{i} is by definition, that sum. Def. 5

cx^{N} ∑_{M}^{∞} a_{n}x^{n} := ∑_{M}^{∞} ca_{n}x^{n+N} :

(convince yourself that the rhs makes sense) Def. 6 If f(x) = ∑_{N}^{∞} a_{n}x^{n} is a f.L.s., and so is g(x) then

f(x)g(x) := ∑_{N}^{∞} a_{n}x^{n}g(x)

(convince yourself that the rhs makes sense). Def. 7

(∑_{N}^{∞} a_{n}x^{n})′ := ∑_{N}^{∞} na_{n}x^{n−1}

Prop. 1 (Product Rule): (f(x)g(x))′ = f′(x)g(x) + f(x)g′(x) Proof: True for f(x) = x^{m}, g(x) = x^{n}, and hence in general, since both sides are linear in f(x), and in g(x). Prop. 2: (f(x)^{k})′ = kf(x)^{k−1}f′(x) Proof: True for k = 1, and then by induction on k, for positive k and for negative k by using the product rule applied to 1 = f(x)^{k}f(x)^{−k}. Prop. 3: (Chain Rule): If Φ and f(x) are f.L.s. then (Φ(f(x)))′ = Φ′(f(x))f′(x).

Proof: True for Φ = x^{k} thanks to Prop.2, now extend by linearity. Prop. 4: If f(x) is a f.L.s. then Res(f′(x)) = 0. Proof: (x^{n})′ = nx^{n−1} can never be a multiple of x^{−1}, hence true for monomials, and by linearity for all f(x). Prop. 5 (Integration by parts) If f(x) and g(x) are f.L.s. then Res(f′(x)g(x)) = −Res(f(x)g′(x)). Proof: By Prop. 1 and 4. Prop. 6 (change of variables): Let u(t) be a f.L.s starting at t (i.e. N = 1) and Ψ(z) be any f.L.s. then Res_{t}(u′(t)Ψ(u(t))) = Res_{z}Ψ(z) Proof: By linearity enough to prove it for monomials Ψ(z) = z^{k}. Both sides are 0 if k ≠ −1, the right by definition 3, the left by Prop. 4. When k = −1 the right is 1, by definition and the left is Res(u′(t)/u(t)) = (u_{1} + 2u_{2}t + …)/(u_{1}t + u_{2}t^{2} + ...) = 1/t + O(1). Theorem (Lagrange Inversion Theorem): If u(t) and Φ(t) are f.L.s. starting at t and t^{0} respectively, then u(t) = tΦ(u(t)) implies [t^{n}]u(t) = (1/n)[z^{n−1}]Φ(z)^{n} . Proof:

[t^{n}]u(t) = Res_{t}(u(t)t^{−n−1}) = Res_{t}(u(t)(t^{−n}/(−n))′) {(5) \atop =} (1/n)Res_{t}(u(t)′t^{−n}) {given \atop =}

(1/n)Res_{t}(u′(t)(Φ(u(t))/u(t))^{n}) {(6) \atop =} (1/n)Res_{z}(Φ(z)^{n}/z^{n}) = (1/n)[z^{n−1}]Φ(z)^{n} .

Added March 2, 2015: Generalizing, we have Theorem (Generalized Lagrange Inversion Theorem): If u(t) and Φ(t) are f.L.s. starting at t and t^{0} respectively, and G(t) is yet another formal power series, then u(t) = tΦ(u(t)) implies [t^{n}]G(u(t)) = (1/n)[z^{n−1}] G′(z)Φ(z)^{n} . Proof:

[t^{n}]G(u(t)) = Res_{t}(G(u(t))t^{−n−1}) = Res_{t}(G(u(t))(t^{−n}/(−n))′) {(5) \atop =} (1/n)Res_{t}(G′(u(t))u(t)′t^{−n}) {given \atop =}

(1/n)Res_{t}(u′(t)G′(u(t))(Φ(u(t))/u(t))^{n}) {(6) \atop =} (1/n)Res_{z}G′(z)(Φ(z)^{n}/z^{n}) = (1/n)[z^{n−1}]G′(z)Φ(z)^{n} .