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Numerator coefficients of the bivariate Maclaurin series ("inverse Kepler equation") developed as Lagrange inversion E=KeplerInv(e,M) of Kepler's equation M = Kepler(e,E) = E - e*sin(E).
(history; published version)
#18 by N. J. A. Sloane at Tue Feb 26 10:53:51 EST 2019
STATUS

proposed

approved

#17 by Herbert Eberle at Sun Feb 24 07:01:51 EST 2019
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editing

proposed

#16 by Herbert Eberle at Sun Feb 24 07:00:39 EST 2019
EXAMPLE

Matrix (isosceles regular triangle) lexicographically ascending in the rows:

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proposed

editing

Discussion
Sun Feb 24
07:01
Herbert Eberle: thanx! regular!
#15 by Herbert Eberle at Sun Feb 24 04:07:24 EST 2019
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editing

proposed

Discussion
Sun Feb 24
04:36
Michel Marcus: isosceles triangle : rather regular ? see keyword tabl at https://oeis.org/eishelp2.html#RK
#14 by Herbert Eberle at Sun Feb 24 04:06:52 EST 2019
COMMENTS

The derivative dKepler/dE = 1 - e*cos(E) goes to zero at E = i*arccosh(1/e) in the complex plane. Thus dKeplerInv/dM goes to infinity at M = i*(arccosh(1/e) - sqrt(1-e^2), ), so that the radius of convergence of KeplerInv(e,M) is arccosh(1/e) - sqrt(1-e^2). KeplerInv(e,M) converges linearly within the circle of convergence |M| < arccosh(1/e) - sqrt(1-e^2).

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proposed

editing

Discussion
Sun Feb 24
04:07
Herbert Eberle: imaginary unit i with ()
#13 by Jon E. Schoenfield at Sat Feb 23 18:47:42 EST 2019
STATUS

editing

proposed

#12 by Jon E. Schoenfield at Sat Feb 23 18:46:16 EST 2019
NAME

Numerator coefficients of the bivariate Maclaurin series ("inverse Kepler equation") developped developed as Lagrange inversion E=KeplerInv(e,M) of Kepler's equation M = Kepler(e,E) = E - e*sin(E).

COMMENTS

Coefficients of the numerator polynomials of the bivariate Maclaurin series ("inverse Kepler equation") developped developed as Lagrange inversion E = KeplerInv(e,M) of Kepler's equation M = Kepler(e,E) = E - e*sin(E), where e=numeric eccentricity, M=mean anomaly, E=eccentric anomaly. The series is: KeplerInv(e,M) = M/(1-e) +sum_ Sum_{n>=1..infinity} (-1)^n*(sum_Sum_{j=1..n} a(n,j)*e^j)/(1-e)^(3n+1)*M^(2n+1)/(2n+1)!(factorial) = M/(1-e) - (e/(1-e)^4)*M^3/3! + ((e+9*e^2)/(1-e)^7)*M^5/5! - + ... .

The element a(n,n) with highest index in each row (the diagonal element) has the form prod_Product_{j=1..n} (2*j+1)^2.

The derivative dKepler/dE = 1 - e*cos(E) goes to zero at E = i*ArcCosharccosh(1/e) in the complex plane. Thus dKeplerInv/dM goes to infinity at M = i*ArcCosharccosh(1/e) - sqrt(1-e^2), so that the radius of convergence of KeplerInv(e,M) is ArcCosharccosh(1/e) - sqrt(1-e^2). KeplerInv(e,M) converges linearly within the circle of convergence |M| <ArcCosh arccosh(1/e) - sqrt(1-e^2).

FORMULA

While M = E - e*sin(E) = E*(1-e) - e*Sum_{n>=1} (-1)^n*E^(2n+1)/(2n+1)! the formal power series of the compositional inverse KeplerInv(e,M) is as above according to A111785 and A304462.

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proposed

editing

#11 by Michel Marcus at Sat Feb 23 16:09:33 EST 2019
STATUS

editing

proposed

Discussion
Sat Feb 23
16:10
Michel Marcus: ok ?
#10 by Michel Marcus at Sat Feb 23 16:09:24 EST 2019
FORMULA

While M = E-e*sin(E) = E*(1-e)-e*sum_Sum_{n>=1..infinity} (-1)^n*E^(2n+1)/(2n+1)! the formal power series of the compositional inverse KeplerInv(e,M) is as above according to A111785 and A304462.

EXAMPLE

1, 9;

1, 54, 225;

1, 243, 4131, 11025;

1, 1008, 50166, 457200, 893025;

1, 4077, 520218, 11708154, 70301925, 108056025;

1, 16362, 5020623, 243313164, 3274844175, 14427513450, 18261468225;

1,65511,46789461,4535570691,119537963811,1107456067125,3821273720775,4108830350625;

...

CROSSREFS

generated Generated by A111785 or A304462, diagonal elements are in A001818.

STATUS

proposed

editing

#9 by Herbert Eberle at Sat Feb 23 13:39:27 EST 2019
STATUS

editing

proposed