Question

1. Prove for any Xo E R that the iteration In+1 = g(xn) converges to a unique fix point a where g(x) = cos X. Find the value a to at least 14 decimal places.

Answer 1

1. Prove for any Xo E R that the iteration In+1 = g(xn) converges to a unique fix point a where g(x) = cos X. Find the value

prove for any XoER that the iteration xn--g(n) converges to a unique fix point a where gre) = cose. Find the value a at least 14 decimal places. Ans: Let cos denotes the n times composition of the cosine function. ie, cos²x = cos (cos (cosx)). Let La Lim cosha (assuming t exist) - (1) [Note that we are actually compating cosh, while compating cost, while fixed fermat iteration So co gives concl) = cos (ligne cosme) Lim (caste) . reos is a continous = function The assuming t exist the process gives a root to the ear na cost This limit to actually (if exists) lies between osi because cos'C) -0 =170 cos (1)-44-45 <0 (BY intermediate value the brem)

It exists because of cose being a Contraction in [-!1] .e, I cosx-cosyl? elx.pl x y E (+b) & Lis unique by Banach fixed point thesiem. pow getting back to fixed point iteration for the function x= g(x), if I is the derived root them. by taylor's theorem about t o go) = g(1) +g'(L) (x-x)+..... e govo-z = g'(a)(x-2) In case of an iteration anti = gren) (X* t) «g'ces (In-t> dve, iteration converges if igre)le, diverges if ig (2)|>1- The case g (n=1 corresponds to a very slow convergence da very slow divergence In our case since cose is a Contraction on : [-so] Since Icosat's! Ve So in the fixed point iteration after the first Stage we get back to region where cose in Contracts . &. So it will converge to that unique limit which we get by our iteration.