Difference between revisions of "Scilab/C4/Solving-Non-linear-Equations/English-timed"

|Dear Friends, Welcome to the spoken tutorial on '''Solving Nonlinear Equations using Numerical Methods'''.

| At the end of this tutorial, you will learn how to   

|solve '''nonlinear equations''' using numerical methods.

|The methods we will be studying are:

|'''Bisection method''' and

|'''Secant method'''.

|'''Scilab 5.3.3''' version.

| Before practicing this tutorial, a learner should have  

| '''nonlinear equations'''.

|This solution '''x''' is called '''root of equation''' or '''zero of function f.'''  

| This process is called '''root finding''' or '''zero finding.'''  

 

|In '''bisection method''', we calculate the '''initial bracket''' of the '''root.'''  

|| Given: '''function f equal to two sin x minus e to the power of x divided by four minus one in the interval minus five and minus three'''

 

| '''Open Bisection dot sci''' on '''Scilab editor. '''

 

|We define the function '''Bisection''' with input arguments '''a b f''' and '''tol.'''  

 

|| Here '''a''' is the lower limit of the '''interval''',

|'''b''' is the upper limit of the '''interval''',

||'''f''' is the function to be solved,

||and '''tol''' is the '''tolerance level'''.

 

| We find the '''midpoint of the interval''' and iterate till the value calculated is within the specified '''tolerance range.'''  

|Define the function using '''deff''' function.

| We type: '''deff open parenthesis open single quote open square bracket y close square bracket equal to f of x close single quote comma open single quote y equal to two asterisk sin of x minus open parenthesis open parenthesis percentage e to the power of x close parenthesis divided by four close parenthesis minus one close single quote close parenthesis'''

 

|To know more about '''deff''' function, type '''help deff'''

 

| '''Bisection open parenthesis a comma b comma f comma tol close parenthesis'''  

 

||Let us study '''Secant's method.'''  

 

| In '''Secant's method,''' the '''derivative''' is approximated by finite difference using two successive iteration values.

 

| Let us solve this example using '''Secant method.'''

 

| Before we solve the problem, let us look at the code for '''Secant method.'''

 

||We define the function '''Secant''' with input arguments '''a, b''' and '''f.'''  

 

||'''a''' is first starting guess for the root,

 

| We apply '''Secant's method ''' and find the value of the root.  

 

|Type '''clc'''.

 

| Press '''Enter'''.

 

| Type  '''a''' equal to 2.

 

| Then type  '''b''' equal to 3.

|We define the function using '''deff''' function.

| Type '''deff open parenthesis open single quote open square bracket y close square bracket equal to g of x close single quote comma open single quote y equal to open parenthesis x to the power of two close parenthesis minus six close single quote close parenthesis '''

| Press '''Enter'''.

|06:20

| '''Secant open parenthesis a comma b comma g close parenthesis.'''

| Press '''Enter'''.

| The value of the root is shown on the '''console'''.

|Find the roots of '''nonlinear equation '''.

| Watch the video available at the link shown below.

| It summarizes the Spoken Tutorial project.

||If you do not have good bandwidth, you can download and watch it.

||The spoken tutorial project Team:

||Conducts workshops using spoken tutorials.

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||For more details, please write to conatct@spoken-tutorial.org.

|Spoken Tutorial Project is a part of the Talk to a Teacher project.

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|This is Ashwini Patil, signing off.