Other examples, figures, tables and equations mentioned in the main example, are included for referencein this and following pages

Modified Statement for Example 6.2 (as it may appear in exams)

Consider the following function:

where is the independent variable

As you learned in class, the Newton-Raphson method for computing the root of a function, , allows to estimate the next ‘guess’ value for the root, , based on the previous ‘guess’ value, , with the following equation:

′

For the function given above, and the derivative of the function that you are required to compute, perform 3 iterations of the Newton-Raphson method to estimate the root. Assume an initial guess value = 0. For each of the 3 iterations, compute the true relative error (assume that the exact value of the root is 0.5671433) and the approximate relative error .

Note: -You are expected to know how to compute the derivative ′ of the given function to implement the Newton-Raphson method, so the derivative will not be given in exams. -No equations to compute the errors and will be provided in exams. You are expected to remember the equations to compute these errors.

Solution of Example 6.2 (with modified statement as it may appear in exams)

Given function

f x( ) ex

x

First derivative of given function

der_f x( ) ex

1

x_new x_old( ) x_oldf x_old( )

der_f x_old( ) <-- this is equation (6.6) in the textbook

x_sol 0.5671433 <-- root solution (given in problem statement)

x_0 0 <-- initial guess value (given in problem statement)

εt_0x_sol x_0

x_sol

εt_0 100 %

Iteration #1

x_1 x_new x_0( ) 0.5 Note that in the software MATCHAD used to type this document,

x_new x_0( ) 0.5

gives the same result as

x_0e

x_0x_0

ex_0

1 0.5

εt_1x_sol x_1

x_sol

εa_1x_1 x_0

x_1

εt_1 11.8389 % εa_1 100 %

Iteration #2

x_2 x_new x_1( ) 0.5663

εt_2x_sol x_2

x_sol εa_2

x_2 x_1

x_2

εa_2 11.7093 %εt_2 0.1468 %

Iteration #3

x_3 x_new x_2( ) 0.5671

εt_3x_sol x_3

x_sol εa_3

x_3 x_2

x_3

εa_3 0.1467 %εt_3 2.3797 10

5 %