Problem:
Using Newton's method with the initial point x = -2, find the root of the equation

f(x) = x2.

Visualization:
Using Newton Slide Show:

After running the program newton.bat, press at the initial information screen and then press at the main menu. You can use the following keys to move through the program:

• To move to the next slide.
• To move to the first slide.
• To move to the last slide.
• To move to the previous slide.
• To immediately quit the program.
The following are some of the pictures in this slide show:

The other slide shows contained in this program are:

• B. Newton's method applied to the function

f(x) = sin(x)

using the initial point x0 = 11 /32.

The following slide shows demonstrate some of the problems encountered with the use of Newton's method.
• C. Newton's method applied to the function

f(x) = sin(x)

using the initial point x0 = /2 - 0.02. If you have several roots then you may not get the root which you expected.

• D. Newton's method applied to the function

f(x) = sin(x)

using the initial point x0 = /2. The tangent line is horizontal and the method fails to find a second estimate for x.

• E. Newton's method applied to

f(x) = x3 - x

using the initial point x0 = 1/ pi5. The method is trapped in a periodic cycle.

• F. Newton's method applied to

f(x) = |x|

using the initial point x0 = 1. Again, the method is trapped in a periodic cycle.

• G. Newton's method applied to

f(x) = x e- x

using the initial point x0 = 2 . This time the method fails since the values diverge.

• H. Newton's method applied to

f(x) = x4 - 20 x3 - 25 x2 + 500 x + 1

using various initial points.