Arman G. answered • 10/21/19
Aerospace Engineer
Newton-Raphson Method: Xn+1= Xn - (F(Xn)/F'(Xn)).
F(x) = X^2 -7, F'(x) = 2x,
1st iteration: X1= X0 - (F(X0)/F'(X0)). = X1= 3 - F(3)/F'(3) = 3 - (1/3) = 8/3.
2nd iteration: X2 = X1 - (F(X1)/F'(X1)) = X2 = (8/3) - F(8/3)/F'(8/3) = (8/3) - (1/9)/(16/3) = 381/144
3rd iteration; X3 = X2 - (F(X2)/F'(X2)) = X3 = (381/144) - F(381/144)/F'(381/144) = 2024/765. = sqrt(7)
Notice that 2024/765 is approximately equal to 2.6475 which is the square root of 7. Also notice that if you solve for the root by adding 7 to both sides and taking the square root, you get x is equal to the square root of 7 as well, so this verifies that the Newton-Raphson method works!
Doug C.
I think typo in calc for x_2 (the 8/3 became a 3 in last step)10/21/19