Question

Use newton raphson method to Find an approximate root of x3

-5x+3 = 0 that lies in

[0,1]. Find the approximation to three decimal places.

Answer 1

$x^3-5x+3=0\\ \left[0,1\right]\\ f(x)=x^3-5x+3\\ f'(x)=3x^2-5\\ f(0)=3, f'(0)=-5\\ f(1)=1-5+3=-1, f'(1)=3-5=-2\\ a_0=0\\ a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}\\ a_1=a_0-\frac{f(a_0)}{f'(a_0)}=0-\frac{3}{-5}=0.6$

$f(a_1)=f(0.6)=0.6^3-5\cdot0.6+3=0.216\\ f'(a_1)=f'(0.6)=3\cdot0.6^2-5=-3.92\\ a_2=a_1-\frac{f(a_1)}{f'(a_1)}=0.6-\frac{0.216}{-3.92}=0.6551\\ |a_2-a_1|=0.0551$

$f(a_2)=f(0.6551)=0.6551^3-5\cdot0.6551+3=\\=0.0056\\ f'(a_2)=f'(0.6551)=3\cdot0.6551^2-5=-3.7125\\ a_3=a_2-\frac{f(a_2)}{f'(a_2)}=0.6551-\frac{0.0056}{-3.7125}=0.6566\\ |a_3-a_2|=0.0015$

$f(a_3)=f(0.6566)=0.6566^3-5\cdot0.6566+3=\\=0.00008\\ f'(a_3)=f'(0.6566)=3\cdot0.6566^2-5=-3.7067\\ a_4=a_3-\frac{f(a_3)}{f'(a_3)}=0.6566-\frac{0.00008}{-3.7067}=0.6566\\ |a_4-a_3|=0\\ x=a_4=0.6566\approx0.657$