hi can u help me write out the formula for the equation below using the Newton-Raphson method?

f(x)= cos x - 1.6 x^3

We want to solve f(x)=0 for x.

f'(x)=-sin x -1.6 (3x^2)

f'(x) = -sin x -4.8 x^2

Start the process with an initial approximation x0=1 (you may choose any reasonable value)

First approximation:

x1 = x0- f(x0) / f'(x0)

x1 = 1 - f(1) / f'(1)

f(1)= cos(1)-1.6(1)^3 =-1.0596976941

f'(1)=-sin(1)-4.8(1)^2 =-3.958529015

x1= 1 - [-1.0596976941] /[-3.958529015]

x1 = 0.81215933

Repeat the process with x1

x2= x1 - f(x1) / f'(x1)

and continue until the difference between two successive approximations is less than, say .00001

x = 1.00000000 -1.0596976941

x = 0.81215933 -0.1691912247

x = 0.76868631 -0.0078959742

x = 0.76645039 -0.0000202264

x = 0.76644463 -1e-10.0000000000

The process converges to 0.76644463

x=0.76644463

x= 0.76644463

f(x) = cos x - 1.6 x^1.3

f '(x) = - sin x -1.6*1.3 x^.3 = - sin x - 2.08 x^.3

xn+1 = xn - f(xn) / f '(xn)

xn __________________ f(xn) ______ f '(xn)

1 __________________ -1.05970 ___ -2.92147

0.637272559049036 ___ -0.08700 ___ -2.41203

0.601202693915652 ___ -0.00109 ___ -2.35117

0.600738500498314 ___ -0/2 _______ -2.35038

0.600738421844613 ___ +0 ________ -2.35038

0.600738421844611 ___ 0 _________ -2.35038

0.600738421844611 ___ 0 _________ -2.35038

0.600738421844611 ___ 0 _________ -2.35038

0.600738421844611 ___ 0 _________ -2.35038