Let fx cos x and n exists in Naturals a Find the Taylor pol

Let f(x) = cos x and n exists in Naturals.

(a) Find the Taylor polynomial P2n : = P 2n to f ,0

(b) Prove that if x is in [-1,1], then |cos x - P2n (x)| < = 1/(2n+1)!

(c) Find an n so large that P2n approximates cos x on [-1, 1] to seven decimal places.

Solution

Let g be a continuous function on the interval [a, b]. If g(x) [a, b] for each x [a, b], then g has a fixed point in [a, b]. Given a continuous function g that is known to have a fixed point in an interval [a, b], we can try to find this fixed point by repeatedly evaluating g at points in [a, b] until we find a point x for which g(x) = x. This is the essence of the method of fixed-point iteration, the implementation of which we now describe. Algorithm (Fixed-Point Iteration) Let g be a continuous function de- fined on the interval [a, b]. The following algorithm computes a number x (a, b) that is a solution to the equation g(x) = x. Choose an initial guess x0 in [a, b]. for k = 0, 1, 2, . . . do xk+1 = g(xk) if |xk+1 xk| is sufficiently small then x = xk+1 return x end end Under what circumstances will fixed-point iteration converge to a fixed point x ? We say that a function g that is continuous on [a, b] satisfies a Lipschitz condition on [a, b] if there exists a positive constant L such that |g(x) g(y)| L|x y|, x, y [a, b]. The constant L is called a Lipschitz constant. If, in addition, L < 1, we say that g is a contraction on [a, b]. If we denote the error in xk by ek = xk x , we can see from the fact that g(x ) = x that if xk [a, b], then |ek+1| = |xk+1 x | = |g(xk) g(x )| L|xk x | L|ek| < |ek|. Therefore, if g is a contraction on [a, b], then fixed-point iteration is convergent; that is, xk converges to x if g satisfies the conditions of the Brouwer Fixed-Point Theorem, and x0 [a, b]. Furthermore, the fixed point x must be unique, for if there exist two distinct fixed points x and y in [a, b], then, by the Lipschitz condition, we have 0 < |x y | = |g(x ) g(y )| L|x y | < |x y |,