Numerical analysis 4 (Fixed-Point Iteration). Consider the problem of finding the root of thefunctionf(x) = 3.294a – cos(x?)in [0, 1]. Observe thatcos(2')3.294x – cos(æ²) = 0 + x =3.294on 0, 1), and thus the problem is reduced to a fixed-point problem forthe functioncos (2)9(x) =3.294on [0, 1]. Find (for your own use) the first g' (x) and the secondderivative g" (x) of the function g(x).(i) We claim that the iteration function g(x) takes the interval (0, 1|into itself. Indeed, the derivative g' (x) of the function g(x) isnegativepositiveat every point x € (0, 1). In effect, the function g(x) is strictlyincreasingdecreasingon [0, 1], and hence its minimum value on [0, 1] is equal tog(0)9(1)and its maximum value on 0, 1 is equal tog(0)g(1)We complete the argument by observing that (here and below, roundyour result to a 7-digit floating point number)g(0) =andg(1) =which demonstrates thatg(10, 1) 드 [0, 1],as claimed(iii) Next, by analyzing the second derivative g" (r) on (0, 1), we seethat for every x E (0,1),|g' (2)| < |g'(7)),where%3Dandlg'(?D =(iv) Accordingly, by (ii) and (iii), the corresponding fixed-point iteration. (enter a correct word) towards the fixed point of the function g(x) in [0, 1].(ii) Use the fixed-point iteration to find an approximation PN of thefixed-point p of the function g(x) in [0, 1], the root of the functionf(x) in [0, 1], satisfying RE(pN PN-1) < 10-6 by takingPo = 1 as the initial approximation. All calculations are to be carriedout in the FPA7.(a) Enter the termsP1, P2, . .., PNyou will generate in the process of obtaining the requiredapproximation in the corresponding input fields below; type theasterisks * in the input fields for the terms Pk with k > N.Pi =P2 =P3 =P4 =P5 =P6 =P7 =Ps =Pg =(b) The stopping criterion has been triggered at the N-th step, sincethe corresponding relative error (round the value to a 7-digit floating-point number)RE(PN - PN-1) =became less than the tolerance for the first time.

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(Fixed-Point Iteration). Consider the problem of finding the root of the function f(x) = 3.294a – cos(x?) in [0, 1]. Observe that cos(a?) 3.294x – cos(æ) = 0 + x = 3.294 on (0, 1), and thus the problem is reduced to a fixed-point problem for the function cos(22) g(x) 3.294 on [0, 1]. Find (for your own use) the first g' (æ) and the second derivative g" (x) of the function g(æ). (i) We claim that the iteration function g(x) takes the interval 0, 1| into itself. Indeed, the derivative g' (x) of the function g(x) is negative positive at every point x € (0, 1). In effect, the function g(x) is strictly increasing decreasing on [0, 1], and hence its minimum value on [0, 1] is equal to g(0) g(1) and its maximum value on [0, 1] is equal to g(0) g(1) We complete the argument by observing that (here and below, round your result to a 7-digit floating point number) g(0) = and g(1) = which demonstrates that 9([0, 1) C [0, 1], as imed (iii) Next, by analyzing the second derivative g" (a) on (0, 1), we see that for every x E (0,1), |g' (x)| < |g'(7), where %3D and |9'(2) =