other examples, figures, tables and equations mentioned in
TRANSCRIPT
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 %