# TI-83 Examples

## Newton's Method

The goal of this tutorial is to use the TI-83 to do the computations in two of the examples in the text.

Example 1 (page 296 in Stewart)

The problem is to use Newton's method to approximate the solution to the equation given below.

x3 - 2x - 5 = 0

So that you can picture what we are working on, here is a graph of the function.

We will enter the function f(x) = x3-2x-5, together with its derivative f '(x) = 3x2-2. Then we will compute successive approximations, using Newton's formula

xn+1 = xn - f(xn) / f '(xn) .

### Step 1

Use the key to get to the function editor.
Enter the function as y1 = x3 - 2x - 5 and its derivative as y2 = 3x2 - 2.
On your calculator, the entries for Y1 and Y2 in the function editor should look like this:
```     ________________________________
|                                |
| Y1=X^3-2X-5                    |
| Y2=3X^2-2                      |
| Y3=                            |
| Y4=                            |
| Y5=                            |
|                                |
|                                |
|                                |
|                                |
|________________________________|
```

### Step 2

Store the value 2 in the variable X, by using the keystrokes .
Use the keystrokes to enter a " : " .
Use the and keys to enter Newton's formula as X - Y1 / Y2.
(See the "Function Evaluation" tutorial if you are unsure how to call the Y-variables.)
Your screen should look like this:
```     ________________________________
|                                |
| 2->X:X-Y1/Y2                    |
|                                |
|                                |
|                                |
|                                |
|                                |
|                                |
|                                |
|                                |
|________________________________|
```

### Step 3

Now press the key to do the calculation.
To calculate the next approximation, use the keystrokes to recall the previous entry.
Now you can use the cursor keys to go back to the 2 and replace it with .
Your screen will show ANS, and the effect is to store your previous answer in the variable X.
The keystrokes , followed by , will repeat the process.
Your screen should look like this.
```     ________________________________
|                                |
| 2->X:X-Y1/Y2                    |
|                           2.1  |
| ANS->X:X-Y1/Y2                  |
|                   2.094568121  |
| ANS->X:X-Y1/Y2                  |
|                   2.094551482  |
| ANS->X:X-Y1/Y2                  |
|                   2.094551482  |
|                                |
|________________________________|
```
Notice that the last two answers are the same, so you have done the calculation to the limits of the calculator.

Example 2 (page 296 in Stewart)

This example finds the positive root of x6 - 2 = 0.

Use the function editor to enter the function as y1 = x6 - 2 and its derivative as y2 = 6x5.
This time we will use a slightly different method.
On the main screen, you need to first store the starting value x1 = 1 in X.
Then enter the expression x - y1 / y2, stored in X.
When you use the key, the calculator returns the next value.
Your screen should look like this.

```     ________________________________
|                                |
| 1->X                        1  |
| X-Y1/Y2->X                     |
|                   1.166666667  |
|                   1.126443678  |
|                   1.122497067  |
|                   1.122462051  |
|                   1.122462048  |
|                                |
|                                |
|________________________________|
```
This carries out the calculation to one more significant digit than the calculation in the text.