TI-83 Example: Approximate a root to the equation x³-5x+3=0, starting with an initial guess of x=2.

Solution: 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

\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.

TI-85 Example: Approximate a root to the equation x³-5x+3=0, starting with an initial guess of x=2.

Solution: 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

To teach this formula to the calculator, we go to the GRAPH screen, and then into the y(x) editor. We set y1 to

We then return to the home screen and enter our initial guess:

where "->" is the STO key, which shows up as a right arrow on the screen.

Now to get the next iterate, we need only enter y1 (using 2nd-alpha-y to get the lower-case y) and ENTER. Then we store that value back into x and repeat the process. In fact, we can do all this at once by entering

The calculator shows us the next iterate and then stores it into x. And now, thanks to a feature of this particular calculator, if we simply press ENTER again, the calculator behaves as though we had entered y1 -> x again. So to carry out further iterations of Newton's method, we need only press ENTER repeatedly.

Doing this, we get the following sequence of approximations:

Since the last two iterates agree to all displayed decimal positions, we assume that they are correct, and to an accuracy of 10^-10, the root is 1.83424318431.

Excel Example: Approximate a root to the equation x³-5x+3=0, starting with an initial guess of x=2.

Solution: We begin by opening a new Worksheet document in Excel. (A blank worksheet usually comes up when you first start Excel.) We highlight cell A1 and type in our initial guess: 2.

The general iteration rule for Newton's method is

To apply the same iteration rule to cells A2 and beyond, we need only copy the rule into those cells. We do this by highlighting cell A2, and then carefully picking up the small black box at the cell's lower right corner and dragging it down the column for a dozen or so cells.

The result should be a list of the x_n from Newton's method. To see more than the default number of digits of precision, use the mouse to widen column A (drag the right edge of the "A" box in the row of column headers), then highlight column A and choose Format, Cells to increase the number of digits displayed.

Here's what the worksheet might look like:

To 14 digits of precision, the root appears to be 1.834243184313920.

Comments: The given function has two other roots. To find them, we need only change our initial guess. That is, we just type some other number into cell A1. An initial guess of -3 should give us the value of the negative root, and an initial guess of 1 should give us the root between 0 and 1.