Newton's method on a TI-83

Here's an example showing how to use a TI-83 to carry out Newton's method.

We'll approximate a root to the equation x3-5x+3=0, starting with an initial guess of x=2.

We let f(x)=x3-5x+3, so that f'(x)=3x2-5. The iteration rule for Newton's approximation is

so in this case we get

xn+1 = xn - (x3-5x+3)/(3x2-5). 

\Y1=X-(X^3-5X+3)
/(3X^2-5)
\Y2=
\Y3=
\Y4=
\Y5=
\Y6=
To teach this formula to the calculator, we go to the GRAPH screen, and then into the Y= editor. We set Y1 to our iteration rule. The screen looks like this:

We push [2nd][QUIT] to get back to the algebra screen and enter our initial guess as 2 [STO>] [X]. 

VARS Y-VARS
1:Function...
2:Parametric...   
3:Polar...
4:On/Off
Now to get the next iterate, we ask the calculator to evaluate the function Y1. To do this, we push [VARS], then push the right-arrow key to highlight Y-VARS. We get the screen as pictured at right, with Y-VARS and 1: highlighted. 

2->X
                2
Y1
We hit ENTER twice to select Y1 and return to the algebra screen, which now looks like this: 

                2
Y1->X 
      1.857142857
       1.83478735
      1.834243504
      1.834243184
      1.834243184
Since we want the value of Y1 stored back into X, we enter [STO>][X] and then press the [ENTER] key. From now on, each time we press [ENTER], the calculator takes the value of Y1, displays it, and stores it back into X. After a few iterations, the screen looks like this:
Since none of the displayed digits changed at the last iteration, we assume all displayed digits are correct. The root, to nine decimal places, is 1.834243184.