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

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(Created page with '{| Border=1 || Time || Narration |- | 00.01 |Dear Friends, |- | 00.02 |Welcome to the spoken tutorial on ''' “Solving Nonlinear Equations using Numerical Methods” ''' …')
 
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{| Border=1
 
{| Border=1
  
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|| Time
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| '''Time'''
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|'''Narration'''
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|| Narration
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|-
 
|-
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| 00.01
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| 00:01
 
|Dear Friends,  
 
|Dear Friends,  
  
 
|-
 
|-
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| 00.02
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| 00:02
 
|Welcome to the spoken tutorial on  ''' “Solving Nonlinear Equations using Numerical Methods” '''
 
|Welcome to the spoken tutorial on  ''' “Solving Nonlinear Equations using Numerical Methods” '''
  
  
 
|-
 
|-
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| 00.10.
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| 00:10.
 
| At the end of this tutorial, you will learn how to:   
 
| At the end of this tutorial, you will learn how to:   
  
 
|-
 
|-
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|00.13
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|00:13
 
|Solve '''nonlinear equations''' using numerical methods
 
|Solve '''nonlinear equations''' using numerical methods
  
 
|-
 
|-
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|00.18
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|00:18
 
|The methods we will be studying are  
 
|The methods we will be studying are  
  
  
 
|-
 
|-
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| 00.20
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| 00:20
 
|'''Bisection method and '''
 
|'''Bisection method and '''
  
 
|-
 
|-
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|00.22
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|00:22
 
|'''Secant method'''  
 
|'''Secant method'''  
  
  
 
|-
 
|-
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| 00.23
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| 00:23
 
|We will also develop '''Scilab''' code to solve '''nonlinear equations.'''  
 
|We will also develop '''Scilab''' code to solve '''nonlinear equations.'''  
  
 
|-
 
|-
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| 00.30
+
| 00:30
 
| To record this tutorial, I am using  
 
| To record this tutorial, I am using  
  
 
|-
 
|-
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|00.32
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|00:32
 
| '''Ubuntu 12.04 '''as the operating system and  
 
| '''Ubuntu 12.04 '''as the operating system and  
  
 
|-
 
|-
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|00.36
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|00:36
 
|'''Scilab 5.3.3''' version  
 
|'''Scilab 5.3.3''' version  
  
 
|-
 
|-
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|00.40
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|00:40
 
| Before practising this tutorial, a learner should have  
 
| Before practising this tutorial, a learner should have  
  
 
|-
 
|-
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| 00.43
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| 00:43
 
| basic knowledge of '''Scilab''' and  
 
| basic knowledge of '''Scilab''' and  
  
 
|-
 
|-
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| 00.46
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| 00:46
 
| '''nonlinear equations'''
 
| '''nonlinear equations'''
  
 
|-
 
|-
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| 00.48
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| 00:48
 
| For '''Scilab''', please refer to the '''Scilab''' tutorials available on the '''Spoken Tutorial''' website.  
 
| For '''Scilab''', please refer to the '''Scilab''' tutorials available on the '''Spoken Tutorial''' website.  
  
 
|-
 
|-
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|00.55
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|00:55
 
|For a given '''function f''', we have to find the value of '''x''' for which '''f of x''' is equal to zero.  
 
|For a given '''function f''', we have to find the value of '''x''' for which '''f of x''' is equal to zero.  
  
 
|-
 
|-
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|01.04
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|01:04
 
|This solution '''x'''  is called '''root of equation ''' or ''' zero of function f.'''  
 
|This solution '''x'''  is called '''root of equation ''' or ''' zero of function f.'''  
  
 
|-
 
|-
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|01.11
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|01:11
 
| This process is called ''' root finding''' or '''zero finding.'''  
 
| This process is called ''' root finding''' or '''zero finding.'''  
  
  
 
|-
 
|-
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|01.16
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|01:16
 
|We begin by studying '''Bisection Method. '''
 
|We begin by studying '''Bisection Method. '''
  
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|-
 
|-
  
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|01.20
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|01:20
  
 
|In '''bisection method''' we calculate the '''initial bracket''' of the '''root.'''  
 
|In '''bisection method''' we calculate the '''initial bracket''' of the '''root.'''  
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|-
 
|-
  
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|01.25
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|01:25
  
 
|Then we iterate through the '''bracket''' and halve its length.  
 
|Then we iterate through the '''bracket''' and halve its length.  
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|-
 
|-
  
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| 01.31
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| 01:31
  
 
|We repeat this process until we find the solution of the equation.  
 
|We repeat this process until we find the solution of the equation.  
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|-
 
|-
  
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| 01.36
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| 01:36
 
||Let us solve this function using '''Bisection method.'''  
 
||Let us solve this function using '''Bisection method.'''  
  
 
|-
 
|-
  
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| 01.41
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| 01:41
 
|| Given
 
|| Given
  
 
|-
 
|-
  
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|01.42
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|01:42
  
 
|| '''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'''
 
|| '''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'''
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|-
 
|-
  
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|01.54
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|01:54
  
 
| '''Open Bisection dot sci on Scilab editor. '''
 
| '''Open Bisection dot sci on Scilab editor. '''
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|-
 
|-
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|02.00
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|02:00
 
| Let us look at the code for '''Bisection method.'''  
 
| Let us look at the code for '''Bisection method.'''  
  
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|-
 
|-
  
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|02.03
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|02:03
  
 
|We define the function '''Bisection''' with input arguments '''a b f''' and '''tol.'''  
 
|We define the function '''Bisection''' with input arguments '''a b f''' and '''tol.'''  
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|-
 
|-
  
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|02.10
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|02:10
  
 
|| Here '''a''' is the lower limit of the '''interval'''  
 
|| Here '''a''' is the lower limit of the '''interval'''  
  
 
|-
 
|-
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|02.14
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|02:14
 
|'''b''' is the upper limit of the '''interval'''  
 
|'''b''' is the upper limit of the '''interval'''  
  
  
 
|-
 
|-
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| 02.16
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| 02:16
 
||'''f''' is the function to be solved  
 
||'''f''' is the function to be solved  
  
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|-
 
|-
  
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| 02.19
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| 02:19
 
||and '''tol''' is the '''tolerance level'''  
 
||and '''tol''' is the '''tolerance level'''  
  
 
|-
 
|-
  
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|02.22
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|02:22
  
 
|| We specify the maximum number of iterations to be equal to hundred.  
 
|| We specify the maximum number of iterations to be equal to hundred.  
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|-
 
|-
  
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|02.28
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|02:28
  
 
| We find the '''midpoint of the interval''' and iterate till the value calculated is within the specified '''tolerance range.'''  
 
| We find the '''midpoint of the interval''' and iterate till the value calculated is within the specified '''tolerance range.'''  
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|-
 
|-
  
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|02.37
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|02:37
  
 
| Let us solve the problem using this code.  
 
| Let us solve the problem using this code.  
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|-
 
|-
  
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| 02.40
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| 02:40
  
 
|| Save and execute the file.  
 
|| Save and execute the file.  
  
 
|-
 
|-
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| 02.43
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| 02:43
 
| Switch to '''Scilab console'''  
 
| Switch to '''Scilab console'''  
  
 
|-
 
|-
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|02.47
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|02:47
 
| Let us define the '''interval.'''  
 
| Let us define the '''interval.'''  
  
 
|-
 
|-
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|02.50
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|02:50
 
| Let '''a''' be equal to minus five.  
 
| Let '''a''' be equal to minus five.  
  
 
|-
 
|-
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| 02.52
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| 02:52
 
| Press '''Enter.'''  
 
| Press '''Enter.'''  
  
 
|-
 
|-
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| 02:54
 
|Let '''b''' be equal to minus three.  
 
|Let '''b''' be equal to minus three.  
  
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|-
 
|-
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| 02.56
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| 02:56
 
| Press '''Enter. '''
 
| Press '''Enter. '''
  
  
 
|-
 
|-
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| 02.58
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| 02:58
 
|Define the function using '''deff function.'''  
 
|Define the function using '''deff function.'''  
  
 
|-
 
|-
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|03.01
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|03:01
 
| We type  
 
| We type  
  
 
|-
 
|-
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| 03.02
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| 03:02
 
| '''deff open paranthesis 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 paranthesis open paranthesis percentage e to the power of x close paranthesis divided by four close paranthesis minus one close single quote close paranthesis'''
 
| '''deff open paranthesis 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 paranthesis open paranthesis percentage e to the power of x close paranthesis divided by four close paranthesis minus one close single quote close paranthesis'''
  
 
|-
 
|-
  
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| 03.41
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| 03:41
  
 
|To know more about '''deff function''' type '''help deff'''
 
|To know more about '''deff function''' type '''help deff'''
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|-
 
|-
  
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| 03.46
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| 03:46
 
|| Press '''Enter.'''  
 
|| Press '''Enter.'''  
  
 
|-
 
|-
  
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|03.48
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|03:48
  
 
||Let '''tol''' be equal to 10 to the power of minus five.  
 
||Let '''tol''' be equal to 10 to the power of minus five.  
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|-
 
|-
  
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|03.53
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|03:53
  
 
||Press '''Enter.'''  
 
||Press '''Enter.'''  
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|-
 
|-
  
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| 03.56
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| 03:56
  
 
| To solve the problem, type  
 
| To solve the problem, type  
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|-
 
|-
  
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| 03.58
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| 03:58
  
 
| '''Bisection open paranthesis a comma b comma f comma tol close paranthesis'''  
 
| '''Bisection open paranthesis a comma b comma f comma tol close paranthesis'''  
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|-
 
|-
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|04.07
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|04:07
 
| Press '''Enter.'''  
 
| Press '''Enter.'''  
  
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|-
 
|-
  
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| 04.09
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| 04:09
  
 
| The root of the function is shown on the console.  
 
| The root of the function is shown on the console.  
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|-
 
|-
  
−
|04.14
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|04:14
  
 
||Let us study '''Secant's method.'''  
 
||Let us study '''Secant's method.'''  
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|-
 
|-
  
−
|04.17
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|04:17
  
 
| In '''Secant's method,''' the derivative is approximated by finite difference using two successive iteration values.
 
| In '''Secant's method,''' the derivative is approximated by finite difference using two successive iteration values.
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|-
 
|-
  
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| 04.27
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| 04:27
  
 
| Let us solve this example using '''Secant method. '''
 
| Let us solve this example using '''Secant method. '''
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|-
 
|-
  
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| 04.30
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| 04:30
  
 
|The function is '''f equal to x square minus six. '''
 
|The function is '''f equal to x square minus six. '''
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|-
 
|-
  
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| 04.36
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| 04:36
  
 
| The two '''starting guesses''' are , '''p zero''' equal to two and '''p one''' equal to three.  
 
| The two '''starting guesses''' are , '''p zero''' equal to two and '''p one''' equal to three.  
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|-
 
|-
  
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| 04.44
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| 04:44
  
 
| Before we solve the problem, let us look at the code for '''Secant method. '''
 
| Before we solve the problem, let us look at the code for '''Secant method. '''
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|-
 
|-
  
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| 04.50
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| 04:50
  
 
||Open '''Secant dot sci''' on '''Scilab editor.'''   
 
||Open '''Secant dot sci''' on '''Scilab editor.'''   
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|-
 
|-
  
−
| 04.54
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| 04:54
  
 
||We define the function '''secant''' with input arguments '''a, b and f.'''  
 
||We define the function '''secant''' with input arguments '''a, b and f.'''  
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|-
 
|-
  
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| 05.01
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| 05:01
  
 
||'''a''' is first starting guess for the root  
 
||'''a''' is first starting guess for the root  
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|-
 
|-
  
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| 05.04
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| 05:04
  
 
|'''b''' is the second starting guess and  
 
|'''b''' is the second starting guess and  
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|-
 
|-
  
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| 05.07
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| 05:07
  
 
|'''f''' is the function to be solved.  
 
|'''f''' is the function to be solved.  
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|-
 
|-
  
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|05.10
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|05:10
  
 
|We find the difference between the value at the current point and the previous point.  
 
|We find the difference between the value at the current point and the previous point.  
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|-
 
|-
  
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|05.15
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|05:15
  
 
| We apply '''Secant's method ''' and find the value of the root.  
 
| We apply '''Secant's method ''' and find the value of the root.  
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|-
 
|-
  
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| 05:21
  
 
| Finally we end the function.  
 
| Finally we end the function.  
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|-
 
|-
  
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|05.24
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|05:24
  
 
|| Let me save and execute the code.  
 
|| Let me save and execute the code.  
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|-
 
|-
  
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| 05.27
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| 05:27
  
 
| Switch to '''Scilab console.'''  
 
| Switch to '''Scilab console.'''  
 
|-
 
|-
  
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| 05.30
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| 05:30
  
 
|Type '''clc. '''
 
|Type '''clc. '''
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|-
 
|-
  
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| 05.32
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| 05:32
  
 
| Press '''Enter'''
 
| Press '''Enter'''
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|-
 
|-
  
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| 05.34
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| 05:34
  
 
|Let me define the initial guesses for this example.  
 
|Let me define the initial guesses for this example.  
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|-
 
|-
  
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| 05:38
  
 
| Type  '''a''' equal to 2  
 
| Type  '''a''' equal to 2  
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|-
 
|-
  
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| 05.40
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| 05:40
  
 
| Press '''Enter. '''
 
| Press '''Enter. '''
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|-
 
|-
  
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| 05:42
  
 
| Then type  '''b''' equal to 3  
 
| Then type  '''b''' equal to 3  
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|-
 
|-
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| 05.44
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| 05:44
 
| Press ''' Enter.'''  
 
| Press ''' Enter.'''  
  
 
|-
 
|-
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| 05.46
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| 05:46
 
|We define the function using '''deff function. '''
 
|We define the function using '''deff function. '''
  
 
|-
 
|-
  
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| 05.49
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| 05:49
  
 
| Type '''deff open paranthesis 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 paranthesis x to the power of two close paranthesis minus six close single quote close paranthesis '''
 
| Type '''deff open paranthesis 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 paranthesis x to the power of two close paranthesis minus six close single quote close paranthesis '''
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|-
 
|-
  
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| 06:15
  
 
| Press '''Enter'''
 
| Press '''Enter'''
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|-
 
|-
  
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| 06:18
  
 
| We call the function by typing  
 
| We call the function by typing  
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|-
 
|-
  
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| 06:20
  
 
| '''Secant open paranthesis a comma b comma g close paranthesis.'''
 
| '''Secant open paranthesis a comma b comma g close paranthesis.'''
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|-
 
|-
  
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| 06:27
  
 
| Press '''Enter'''
 
| Press '''Enter'''
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|-
 
|-
  
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| The value of the root is shown on the '''console'''
 
| The value of the root is shown on the '''console'''
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|-
 
|-
  
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| 06.35
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| 06:35
  
 
| Let us summarize this tutorial.  
 
| Let us summarize this tutorial.  
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|-
 
|-
  
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| 06:38
  
 
| In this tutorial we have learnt to:  
 
| In this tutorial we have learnt to:  
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|-
 
|-
  
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|Develop '''Scilab''' code for different solving methods  
 
|Develop '''Scilab''' code for different solving methods  
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|-
 
|-
  
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| 06:45
  
 
|Find the roots of '''nonlinear equation '''
 
|Find the roots of '''nonlinear equation '''
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|-
 
|-
  
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| 06:48
  
 
|Solve this problem on your own using the two methods we learnt today.  
 
|Solve this problem on your own using the two methods we learnt today.  
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|-
 
|-
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|06.55
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|06:55
  
 
| Watch the video available at the link shown below  
 
| Watch the video available at the link shown below  
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|-
 
|-
  
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| 06:58
  
 
| It summarises the Spoken Tutorial project  
 
| It summarises the Spoken Tutorial project  
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|-
 
|-
  
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||If you do not have good bandwidth, you can download and watch it  
 
||If you do not have good bandwidth, you can download and watch it  
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|-
 
|-
  
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||The spoken tutorial project Team
 
||The spoken tutorial project Team
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|-
 
|-
  
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|07:07
  
 
||Conducts workshops using spoken tutorials  
 
||Conducts workshops using spoken tutorials  
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|-
 
|-
  
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|07:10
  
 
||Gives certificates to those who pass an online test  
 
||Gives certificates to those who pass an online test  
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|-
 
|-
  
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|07.14
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|07:14
  
 
||For more details, please write to conatct@spoken-tutorial.org  
 
||For more details, please write to conatct@spoken-tutorial.org  
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|-
 
|-
  
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|Spoken Tutorial Project is a part of the Talk to a Teacher project  
 
|Spoken Tutorial Project is a part of the Talk to a Teacher project  
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|-
 
|-
  
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| It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.  
 
| It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.  
 
|-
 
|-
  
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| 07:32
  
 
|More information on this mission is available at  http://spoken-tutorial.org/NMEICT-Intro
 
|More information on this mission is available at  http://spoken-tutorial.org/NMEICT-Intro
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|-
 
|-
  
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|This is Ashwini Patil signing off.
 
|This is Ashwini Patil signing off.
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|-
 
|-
  
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|07:41
  
 
| Thank you for joining.
 
| Thank you for joining.

Revision as of 17:47, 10 July 2014

Time Narration
00:01 Dear Friends,
00:02 Welcome to the spoken tutorial on “Solving Nonlinear Equations using Numerical Methods”


00:10. At the end of this tutorial, you will learn how to:
00:13 Solve nonlinear equations using numerical methods
00:18 The methods we will be studying are


00:20 Bisection method and
00:22 Secant method


00:23 We will also develop Scilab code to solve nonlinear equations.
00:30 To record this tutorial, I am using
00:32 Ubuntu 12.04 as the operating system and
00:36 Scilab 5.3.3 version
00:40 Before practising this tutorial, a learner should have
00:43 basic knowledge of Scilab and
00:46 nonlinear equations
00:48 For Scilab, please refer to the Scilab tutorials available on the Spoken Tutorial website.
00:55 For a given function f, we have to find the value of x for which f of x is equal to zero.
01:04 This solution x is called root of equation or zero of function f.
01:11 This process is called root finding or zero finding.


01:16 We begin by studying Bisection Method.


01:20 In bisection method we calculate the initial bracket of the root.
01:25 Then we iterate through the bracket and halve its length.
01:31 We repeat this process until we find the solution of the equation.
01:36 Let us solve this function using Bisection method.
01:41 Given
01:42 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
01:54 Open Bisection dot sci on Scilab editor.


02:00 Let us look at the code for Bisection method.


02:03 We define the function Bisection with input arguments a b f and tol.
02:10 Here a is the lower limit of the interval
02:14 b is the upper limit of the interval


02:16 f is the function to be solved


02:19 and tol is the tolerance level
02:22 We specify the maximum number of iterations to be equal to hundred.
02:28 We find the midpoint of the interval and iterate till the value calculated is within the specified tolerance range.
02:37 Let us solve the problem using this code.
02:40 Save and execute the file.
02:43 Switch to Scilab console
02:47 Let us define the interval.
02:50 Let a be equal to minus five.
02:52 Press Enter.
02:54 Let b be equal to minus three.


02:56 Press Enter.


02:58 Define the function using deff function.
03:01 We type
03:02 deff open paranthesis 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 paranthesis open paranthesis percentage e to the power of x close paranthesis divided by four close paranthesis minus one close single quote close paranthesis
03:41 To know more about deff function type help deff


03:46 Press Enter.
03:48 Let tol be equal to 10 to the power of minus five.


03:53 Press Enter.


03:56 To solve the problem, type


03:58 Bisection open paranthesis a comma b comma f comma tol close paranthesis


04:07 Press Enter.


04:09 The root of the function is shown on the console.


04:14 Let us study Secant's method.


04:17 In Secant's method, the derivative is approximated by finite difference using two successive iteration values.


04:27 Let us solve this example using Secant method.


04:30 The function is f equal to x square minus six.


04:36 The two starting guesses are , p zero equal to two and p one equal to three.


04:44 Before we solve the problem, let us look at the code for Secant method.


04:50 Open Secant dot sci on Scilab editor.


04:54 We define the function secant with input arguments a, b and f.
05:01 a is first starting guess for the root
05:04 b is the second starting guess and


05:07 f is the function to be solved.


05:10 We find the difference between the value at the current point and the previous point.


05:15 We apply Secant's method and find the value of the root.


05:21 Finally we end the function.


05:24 Let me save and execute the code.


05:27 Switch to Scilab console.
05:30 Type clc.
05:32 Press Enter



05:34 Let me define the initial guesses for this example.
05:38 Type a equal to 2


05:40 Press Enter.


05:42 Then type b equal to 3


05:44 Press Enter.
05:46 We define the function using deff function.
05:49 Type deff open paranthesis 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 paranthesis x to the power of two close paranthesis minus six close single quote close paranthesis


06:15 Press Enter
06:18 We call the function by typing
06:20 Secant open paranthesis a comma b comma g close paranthesis.
06:27 Press Enter


06:30 The value of the root is shown on the console
06:35 Let us summarize this tutorial.
06:38 In this tutorial we have learnt to:
06:41 Develop Scilab code for different solving methods
06:45 Find the roots of nonlinear equation
06:48 Solve this problem on your own using the two methods we learnt today.


06:55 Watch the video available at the link shown below
06:58 It summarises the Spoken Tutorial project


07:01 If you do not have good bandwidth, you can download and watch it
07:05 The spoken tutorial project Team
07:07 Conducts workshops using spoken tutorials


07:10 Gives certificates to those who pass an online test


07:14 For more details, please write to conatct@spoken-tutorial.org


07:21 Spoken Tutorial Project is a part of the Talk to a Teacher project


07:24 It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.
07:32 More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro
07:39 This is Ashwini Patil signing off.
07:41 Thank you for joining.

Contributors and Content Editors

PoojaMoolya, Pratik kamble, Sandhya.np14

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