A263646 - OEIS

A263646

Coefficients for an expansion of the Schwarzian derivative of a power series.

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 5, 1, 1, 1, 6, 1, 1, 1, 7, 1, 1, 1, 1, 8, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1

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OFFSET

1,3

COMMENTS

Coefficients for an expansion of the Schwarzian derivative of a power series f(x), with f'(0) = 1, expressed in terms of an expansion of the natural logarithm of the derivative of the function G(x) = log(D f(x)) = Sum_{n >= 1} -F(n) x^n/n.

LINKS

Table of n, a(n) for n=1..98.

FORMULA

Schwarzian{f(x)} = S{f(x)} = (D^3 f(x)) / (D f(x)) - (3/2) [(D^2 f(x)) / D f(x)]^2 = D [(D^2 f(x)) / D f(x)] - (1/2) [(D^2 f(x)) / D f(x)]^2 = D^2 log[D f(x)] - (1/2) [D log[D f(x)]]^2.

Then, with G(x) = log[D f(x)], S{f(x)} = D^2 G(x) - (1/2) [D G(x)]^2.

With f'(0) = 1, G(x) = log[D f(x)] = sum[n >= 1, -F(n) * x^n/n], and F(n) as Fn,

S{f(x)} = -[(F2 + F1^2/2) + (2 F3 + F1 F2) x + (3 F4 + F1 F3 + F2^2/2) x^2 + (4 F5 + F1 F4 + F2 F3) x^3 + (5 F6 + F1 F5 + F2 F4 + F3^2/2) x^4 + (6 F7 + F1 F6 + F2 F5 + F3 F4) x^5 + (7 F8 + F1 F7 + F2 F6 + F3 F5 + F4^2/2) x^6 + ...] .

This entry's a(m) are the numerators of the coefficients of the binary partitions in the brackets. For the singular partition of the integer n, the coefficient is (n-1); for the symmetric partition, 1/2; and for the rest, 1.

More symmetrically, x^2 S{f(x)}= - sum{n>=2, x^n [(n-1)F(n) + (1/2) sum(k=1 to n-1, F(n-k) F(k))]}.

With f(x)= c(0) + x + c(2) x^2 + ... , F(n) are given by the Faber polynomials of A263916: F(n) = Faber(n,2c(2),3c(3),..,(n+1)c(n+1)).

EXAMPLE

Partitions by powers of x^n:

n=0: -(F2 + F1^2/2)

n=1: -(2 F3 + F1 F2)

n=2: -(3 F4 + F1 F3 + F2^2/2) = -[3 F4 + (F1 F3 + F2 F2 + F3 F1) / 2]

n=3: -(4 F5 + F1 F4 + F2 F3) = -[4 F5 + (F1 F4 + F2 F3 + F3 F2 + F4 F1) / 2]

n=4: -(5 F6 + F1 F5 + F2 F4 + F3^2/2)

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Example series:

f(x)= (1/2) / (1-x)^2 = 1/2 + x + (3/2) x^2 + 2x^3 + (5/2)x^4 + ... .

log(f'(x)) = log(1 + 3x + 6x^2 + 10x^3 + ...) = 3x + 3 x^2/2 + 3 x^3/3 + ... .

Then F(n) = -3 for n>=1, and the Schwarzian derivative series is

S{f(x)} = - [(-3 + 3^2/2) + (-2*3 + 3^2) x + (-3*3 + 3^2 + 3^2/2) x^2 + ...] = -3/2 - 3x - (9/2)x^2 - 6x^3 - (15/2)x^4 - ... .

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The Schwarzian vanishes if and only if acting on a Moebius, or linear fractional, transformation. This corresponds to F(n) = (-1)^(n+1) 2 * d^n, where d is an arbitrary constant.

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Example polynomial:

f(x) = (x-x1)(x-x2)/-(x1+x2)

log(f'(x)) = log[1 - 2x/(x1+x2)] = Sum_{n>= 1} -(2/(x1+x2))^n x^n/n.

Then F(n) = (2/(x1+x2))^n, and the Schwarzian derivative series is

S{f(x)} = (Sum_{n>= 0} -6(n+1) 2^n (x/(x1+x2))^n) / (x1+x2)^2 = -6 / (x1+x2-2*x)^2 (cf. A001787 and A085750).

PROG

(Python) print(sum(([n]+[1]*((n+1)//2) for n in range(1, 18)), [])) # Andrey Zabolotskiy, Mar 07 2024

CROSSREFS

Cf. A263916, A001787, A085750, A051340.

Sequence in context: A130296 A126705 A367849 * A113924 A335124 A367579

Adjacent sequences: A263643 A263644 A263645 * A263647 A263648 A263649