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 = 0So 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) .
________________________________ | | | Y1=X^3-2X-5 | | Y2=3X^2-2 | | Y3= | | Y4= | | Y5= | | | | | | | | | |________________________________|
________________________________ | | | 2->X:X-Y1/Y2 | | | | | | | | | | | | | | | | | |________________________________|
________________________________ | | | 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.