Kepler's Equation

Author

Eric W. Weisstein
September 21, 2007

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Algebra/KeplersEquation.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/KeplersEquation.html.

©2007 Wolfram Research, Inc. except for portions noted otherwise

Plot

MakeBoxes[xlabel,TraditionalForm]:="E"
MakeBoxes[ylabel,TraditionalForm]:=RowBox[{"M","(","E",")"}]

Plot[Evaluate[Table[x-i Sin[x],{i,0,1,.2}]],{x,0,4Pi},
AxesLabel->TraditionalForm/@{xlabel,ylabel},
PlotStyle->Hue/@Range[0,1,.17]
]

Laplace Limit

Solve[((x Exp[Sqrt[1+x^2]])/(1+Sqrt[1+x^2]))==1,x]

x/.FindRoot[((x Exp[Sqrt[1+x^2]])/(1+Sqrt[1+x^2]))==1,{x,.6},WorkingPrecision->50]

Stieltjes Constant

FindRoot

eqn=E^x(x-1)==Exp[-x](x+1);

Solve[eqn,x]

c=x/.FindRoot[eqn,{x,1},WorkingPrecision->110]

Solve[x==Coth[x],x]

FullSimplify[E^x(x-1)-Exp[-x](x+1),x>0]

FindRoot[x==Coth[x],{x,1},WorkingPrecision->110]

Transcendental Root

r1=Reduce[E^x(x-1)==Exp[-x](x+1)&&0<x<2,x][[2]]

r2=Reduce[Coth[x]==x&&0<x<2,x][[2]]

FullSimplify[r1-r2]

N[%]

N[r1-r2,100]

PossibleZeroQ[r1-r2]

Sum

(Subscript[a, n])==(1/((2^(n-1)) n!))(∑_(k=0)^(Floor[n/2]))(((-1))^k) (((n-2 k))^(n-1)) (Binomial[n, k]) sin(M (n-2 k))

(1/((2^(n-1))n!))Sum[(((-1))^k)Binomial[n,k](((n-2k))^(n-1))Sin[(n-2k)M],{k,0,Floor[n/2]}]

E==(∑_n^∞)((2 (BesselJ[n, n e]) sin(M n))/n)+M

M+Sum[(2/n)BesselJ[n,n e]Sin[n M],{n,∞}]