Help!!!Use three iterations of the secant method to find an approximate solution of the equation?

Question

1.8X^2=5cos(X)

inital guess: x0=2.3, x1=3.2

find the next three iterate?

can someone check if i have did this correctly, here are my answers

x0=12.85338011

x1=23.42347388

x2=1.205587562

x3=1.132276999

x4=1.111309839

thank you..

Answer 1

If the values x0=2.3 and x1=3.2 determine the secant line, then x2 will be given by

x_2 = x_1 - f(x_1)*f(x_0 - x_1)/ ( f(x_0 - x_1) )

x_2 = 1.205587561978509

Where I am using f(x) = 5cos(x) - 1.8x^2.

I am not sure where you are getting x0=12.85338011 and x1=23.42347388. You can find x_3 the same way:

x_3 = x_2 - f(x_2)*f(x_1 - x_2)/ ( f(x_2 - x_2) )

x_3 =  1.132276998704147

Using a quick MATLAB program, the results of several more iteration are:

x_4 = 1.111309839473516

x_5 = 1.110656932694117

x_6 = 1.110652116257178

x_7 = 1.110652115169982

It is kind of interesting that the procedure converged so quickly. According to my program x_7 is accurate to within at least  1X10^-10 of the actual root. That is pretty good.

Anyway, your last three numbers are correct, but I have no idea where the first two came from, as the results of the next three iterations after x0=2.3 and x1=3.2 are simply

x_2 = 1.205587561978509

x_3 =  1.132276998704147

x_4 = 1.111309839473516