g(x) = f(x)-R
look for zero of g(x)
Xn+1 = Xn -g(Xn)/g'(Xn)
here we have
Xn+1 = Xn -(f(Xn)-R)/f'(Xn)
Xn+1 =Xn -f(Xn)/f'(Xn) + R/f'(Xn)
so here
f(x)/f'(x) = cos x sin x
R/f'(x) = - R x cos^2 x
so
f' = 1/(x cos^2x)
f = cos x sin x * 1/(x cos^2 x) = sin x/x cos x
I am not sure I have all your parentheses right and it does not quite check out but that is the idea.
the iteration formula x_(n+1)=x_(n) - (cos(x_(n)))(sin(x_(n))) + R(cos(x_(n))^2)x_n, where R is a positive constant, was obtained by applying Newton's Method to some function f(x). What was f(x)? What can this formula be used for?
1 answer