Question

0. Approximate a root of f using (a) a fisx nplo 1 Consider the function f)r method. and (b) Newton's Method

numerical analysis Newton and fixed point iteration method

Answer 1

a) We keep applying the function
cos(z) o(z) COS T
for the initial point
= 0.5
to get the below sequence of iterations:

n	x_n
0	0.5
1	0.877583
2	0.639012
3	0.802685
4	0.694778
5	0.768196
6	0.719165
7	0.752356
8	0.730081
9	0.74512
10	0.735006

Thus, an approximate fixed point (i.e. solution to
cos(x) =x COS T
) is
0,735

b) Using $f(x)=\cos(x)-x\Rightarrow f'(x)=-\sin(x)-1$

We get the iteration scheme:

cos(x) - 2r ㅡ sin (2)-1 rn+1 = o(%) where o(x) = x-f(r) f'(x) Tn +1-o(m) where ψ(z

That is,

cos(x) - ir o(r) = r + sin(z) +1

Which means

r sin() r +cos(x) - r TTCOST sin(x) 1 sin(r) +

Which is

0(1)-!sinir) tools(r sin(r) + 1

Applying this iteration scheme, we get:

n	x_n
0	0.5
1	0.755222
2	0.739142
3	0.739085
4	0.739085
5	0.739085

So that the root is

0.739085