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.