### Investigating Newton's method

One can probably do everything by "pure thought" and never look at a graph or any algebra, but I am not that smart. If the function f(x) is at all complicated and I want to see what happens when using Newton's method with different initial values, I usually look at a graph and almost always do some sample computations. You can do most of what I do below on a graphing calculator. These days I tend to use Maple because the computer screen is larger than the screen on a graphing calculator. I will investigate the "random" function f(x)=1/x2-5. I think that we can actually get the roots of the equation f(x)=0 explicitly: 1/x2-5=0 when x=+/-sqrt(5) (approximately +/-.44721).

Step 1
I would probably graph y=f(x). I actually tried various windows before I looked at the graph for x between -8 and 8 and y between -5 and 1. The picture to the right was produced by Maple in response to the instruction plot(1/x^2-5.,x=-8..8,y=-5..1,color=black,thickness=3); but you don't need to use Maple. A graphing calculator will give good results. Then I would look at the graph and try to make some guesses about the behavior of Newton's method. I would pick a starting point and slide down (or up!) the tangent line until it hit the x-axis, and then repeat again and again in my mind until I got some feeling about whether the iterates were converging to a root (which root?) or doing something else.

Step 2
(I'm a picture sort of person. If I were an algebra sort of person, this might have been step 1.) I would write f(x), and compute f'(x), and put it all together in the Newton's method iteration: x-[f(x)/f'(x)] and then simplify the iteration as much as possible. In the case of this f(x), f'(x) is -2/x3, and then (after doing all the algebra) the Newton's method iteration turns out to be "replace x by 1.5x-2.5x3". I would write a short program on my graphing calculator which would evaluate this, because I will want to repeatedly evaluate this function on its own output.

Here are some Maple instructions:

h:=x->1.5*x-2.5*x^3; This tells Maple to remember that h(x)=1.5x-2.5x3. You can write and store a short program for your Newton's method iteration on your calculator.
G:=x->seq((h@@n)(x),n=1..10); This instruction is rather weird and specialized. It tells Maple to make G(x) a function whose output is going to be a sequence of repeated function compositions. The rather strange @@ asks Maple to repeatedly compose the function h with itself. So (h@@3)(.7) will return h(h(h(.7))). G(x) will then return a sequence of 10 numbers: the first number will be h(x), the second number will be h(h(x)), etc.
You should be able to get the same effect as (h@@3)(.7) on your graphing calculator fairly easily if you have inserted the formula for h as a program.

One the left is part of my Maple session after I defined the functions h and G as above. My input typing is after the Maple prompt (which is >) and ends with a semicolon (;). Maple's response follows. I wrote some comments on the right.

 ```>h(.5); 0.4375 >G(.5); 0.4375, 0.4468994140, 0.4472132645, 0.4472135956, 0.4472135955, 0.4472135954, 0.4472135955, 0.4472135954, 0.4472135955, 0.4472135954 >G(10.); 11 33 97 -2485.0, 0.3836358157 10 , -0.1411553832 10 , 0.7031246842 10 , 291 874 2621 -0.8690345500 10 , 0.1640782962 10 , -0.1104316145 10 , 7861 23582 0.3366822908 10 , -0.9541152318 10 , 70747 0.2171413312 10 ``` So h(.5) is .4375 (this agrees with what it should be: 1.5(.5)-2.5(.5)3=.75-.3125=.4375 These G-values are the first 10 iterates of Newton's method, starting at .5: it seems that this starting point gives converges to the right-hand root.   The next series of G-values sees what happens with a start at 10. There is rather rapid divergence. So I can experiment easily. I believe that a graphing calculator can be set up to do the same thing.

Step 3
I would think and go back through steps 1 and 2 where necessary. That is, I would look at the graph and the algebra and begin to learn what happens. For example:

Some (but only some!) conclusions for this graph:

 #1 This part of the graph seems immediately attracted to the root on the right. #2 This part of the graph seems to be taken into part #1 by a Newton iteration, so it too gets attracted to the right-hand root. #3 This point gets taken to the origin, so it is out of the game.
And I would progress along. This function is fairly simple and has symmetry, so every time I do something on the right I also get information for the left. For example, the "next" chunk on the right (between the lines colored brown (#3) and pink (#4)) would get sent to a piece on the left which gets attracted towards the left-hand root. That seems hard to see using the algebraic formulation, but, on the other hand, locating the number colored brown can be done quite easily using algebra (write the equation with slope=the derivative and slope=the slope between the origin and the point on the curve, and have the computer or calculator solve the equation which results by setting the two formulas equal).

Comment
The function I've asked you to analyze is a great deal simpler than this one, and you should be able to get an idea of its behavior quite easily, and then write, draw, compute, and explain your conclusions in a convincing and logical manner.