I created the next code:
f := M+2*e*sin(M)+(5/4)*e^2*sin(2*M)+(13/12)*e^3*sin(3*M)-(1/4)*e^3*sin(M)+e^4*((103/96)*sin(4*M)-(11/24)*sin(2*M))
fb := Mb+2*eb*sin(Mb)+(5/4)*eb^2*sin(2*Mb)+(13/12)*eb^3*sin(3*Mb)-(1/4)*eb^3*sin(Mb)+eb^4*((103/96)*sin(4*Mb)-(11/24)*sin(2*Mb))
x := cos(wb-w+fb-f)
rho1 := ab*(1-eb^2)/(1+eb*cos(fb))
rho2 := a*(1-e^2)/(1+e*cos(f))
P3 = (5/2)*x^3-(3/2)*x
R3 = GM3*(rho1/rho2)^3*P3/rho2
When I do
R3exp := mtaylor(R3, [e, eb], 5);
it returns
R3exp := R3
When I do
> R3temp := series(R3, e = 0, 5);
> R3exp := series(R3temp, eb = 0, 5);
> convert(R3exp, polynom);
it returns
R3
Could you tell me how I can expand in Taylor R3 around e=0 and eb=0 ?