Solution to Problem 9.3

Step 1: Get the equation into the form of f(y) = 0. Our situation is that we know what x is and we need to find what the corresponding y is to solve Kepler's equation. Newton-Raphson is a method for finding where a function crosses zero. So we need to get it into a form where finding where some function of y crosses zero gives us the solution for where y - e sin(y) is equal to x.

You're starting out with

   x  =  y  -  E sin(y)

where e is a constant. So just subtract the right-hand side from the left, and you have

   f(y)  =  x  -  y  +  E sin(y)  =  0

Now if you find the y that makes f(y) be zero, you have a solution.

Step 2: Find the derivative of f(y). You must assume x to be a constant. Why? Because the problem is that x is given -- hence it is a constant. So taking the derivative of f(y) with respect to y, you get

   f'(y)  =  -1  +  E cos(y)  =  E cos(y)  -  1

Step 3: Set up the Newton-Raphson formula. That is

y_n+1 = y_n -

f(y_n) f'(y_n)

So you put in the f(y) and the f'(y) you got in the previous steps.

y_n+1 = y_n -

x - y_n + E sin(y_n) E cos(y_n) - 1

That completes the setup part. All that's left is to...

Step 4: Iterate the formula with the numbers given. That was with E = 0.2 and x = 0.5 , using y₀ = 0. I get

   y₀    =    0.000000000000000
   y₁    =    0.625000000000000
   y₂    =    0.615474486571046
   y₃    =    0.615468169492719
   y₄    =    0.615468169489965
   y₅    =    0.615468169489965

After just five iterations, it has converged to within the resolution of my calculator. How well does y₅ work in the original equation getting back x = 0.5 when we plug in the numbers? Let's see.

   sin(0.615468169489965)                 =  0.577340847449826

   0.3 sin(0.615468169489965)             =  0.115468169489965

   0.615468169489965 - 0.115468169489965  =  0.500000000000000

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