Numerical Analysis problem Newton's method

[carrab]

View attachment 6617

Can anyone help in the solution of this problem? how can i determine the zero x*??

 

[MarkFL]

Hello, and welcome to MHB, carrab! (Wave)

To find the zero, I would equate the function $f$ to zero, and solve for $x$:

$$f(x)=0$$

$$x^3-8=0$$

$$x^3-2^3=0$$

When you factor as the difference of cubes, what do you find?

 

[HallsofIvy]

"Newton's method" is a numerical method for solving an equation that basically replaces the function at a given value of x by the tangent function at that point. Here the function is f(x)= x^3- 8 which has derivative f'(x)= 3x^2, the derivative at x_0 f(x_0)= 3x_0^2 while the value of the function is x_0^3- 8. So the tangent function at x= x_0 is y= 3x_0^2(x- x_0)+ x_0^3- 8. Setting that equal to 0, 3x_0^2(x- x_0)+ x_0^3- 8= 0, 3x_0^2(x- x_0)= 8- x_0^3, x- x_0= \frac{8- x_0^3}{3x_0^2}, and x= x_0+ \frac{8- x_0^3}{3x_0^2}.

Start with some reasonable value for x_0 and calculate the next value for x: with, say, x_0= 1, x= 1+ \frac{8- 1}{3}= 1+ \frac{7}{3}= \frac{10}{3}. Now take x_0= \frac{10}{3} and calculate the next value for x. Repeat until you get two consecutive values for x that are closer together than your allowable error.

x_0= 0 is not a "good" starting value (in fact, it is impossible) because the denominator, 3x_0^2, would be 0.