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NEWTON-RAPSON METHOD FOR KEPLER’S ECCENTRICITY

##Constants and initializations
ecc=0.1; ## Eccentricity
M=24.851090; ## Mean anomaly(degrees)
Mr=M*pi/180; ## Mean anomaly (rads)
f=[]; ## The fundamental eqn we want to solve
## Kepler's Equation's zero function f(E)=E-esinE-M=0
fp=[]; ## 1st derivative of f(E). fp(E)=1+ecosE
Ed=[]; ## Eccentric anomaly(degrees)
Er=[]; ## Eccentric anomaly(rads)
Ed(1)=M+0.85*sign(sin(Mr))*ecc; ## The eccentric anomaly initial
## guess for Danby's method(degrees)
Er(1)=Ed(1)*pi/180; ## rads
for i=1:100
f(i)=Ed(i)-ecc*sin(Er(i))-M;
fp(i)=1+ecc*cos(Er(i));
Ed(i+1)=Ed(i)-f(i)/fp(i); ## The iteration for an updated eccentric anomaly E(i+1)
## based on a current value E(i) by Newton-Rapson Method
Er(i+1)=Ed(i+1)*pi/180;
endfor
save -text NEWTONRAPSON.dat
printf('We find E to be %f. This is the answer of Q1(a)\n',Ed(i+1));

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## of the 8th degree Legendre Polynomial
## P8(x)=(1/128)(6435x^8-12012x^6+6930x^4-1260x^2+35)
## where -1<=x<=1.
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## all the terms except the last two by truncated
## the function to be zero and find
## x=0.167 as the initial smallest non negative
## root.
##Constants and initializations
x=[]; ## Empty array for the iterated x roots
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## smallest non-negative root.
L8=[]; ## Empty array for the Legendre polynomial
L8p=[]; ## Empty array for the derivative of the Legendre polynomial
for i=1:100
##The value of the function at x
L8(i)=(1/128)*(6435*x(i)^8-12012*x(i)^6+6930*x(i)^4-1260*x(i)^2+35);
##The value of the derivative of the function at x
L8p(i)=(1/128)*(6435*8*x(i)^7-12012*6*x(i)^5+6930*4*x(i)^3-1260*2*x(i));
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endfor
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