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

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|Welcome to the spoken tutorial on ''' “Solving Nonlinear Equations using Numerical Methods” '''
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|Welcome to the spoken tutorial on '''Solving Nonlinear Equations using Numerical Methods'''.
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+
  
 
|-
 
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| 00:10.
 
| 00:10.
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| At the end of this tutorial, you will learn how to:  
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| At the end of this tutorial, you will learn how to   
  
 
|-
 
|-
 
|00:13
 
|00:13
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|Solve '''nonlinear equations''' using numerical methods
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|solve '''nonlinear equations''' using numerical methods.
  
 
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|-
 
|00:18
 
|00:18
 
|The methods we will be studying are  
 
|The methods we will be studying are  
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| 00:20
 
| 00:20
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|'''Bisection method and '''
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|'''Bisection method''' and
  
 
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|00:22
 
|00:22
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|'''Secant method'''  
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|'''Secant method'''.
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|00:36
 
|00:36
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|'''Scilab 5.3.3''' version  
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|'''Scilab 5.3.3''' version.
  
 
|-
 
|-
 
|00:40
 
|00:40
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| Before practising this tutorial, a learner should have  
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| Before practicing this tutorial, a learner should have  
  
 
|-
 
|-
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| 00:46
 
| 00:46
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| '''nonlinear equations'''
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| '''nonlinear equations'''.
  
 
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|01:04
 
|01:04
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|This solution '''x''' is called '''root of equation ''' or ''' zero of function f.'''  
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|This solution '''x''' is called '''root of equation''' or '''zero of function f.'''  
  
 
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|01:11
 
|01:11
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| This process is called ''' root finding''' or '''zero finding.'''  
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| This process is called '''root finding''' or '''zero finding.'''  
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+
  
 
|-
 
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|01:16
 
|01:16
 
|We begin by studying '''Bisection Method. '''
 
|We begin by studying '''Bisection Method. '''
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|01:20
 
|01:20
  
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|In '''bisection method''' we calculate the '''initial bracket''' of the '''root.'''  
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|In '''bisection method''', we calculate the '''initial bracket''' of the '''root.'''  
  
 
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| 01:41
 
| 01:41
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|| Given
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|| Given:
  
 
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| '''Open Bisection dot sci on Scilab editor. '''
 
| '''Open Bisection dot sci on Scilab editor. '''
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|-
 
|-
 
|02:00
 
|02:00
 
| Let us look at the code for '''Bisection method.'''  
 
| Let us look at the code for '''Bisection method.'''  
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|02:14
 
|02:14
 
|'''b''' is the upper limit of the '''interval'''  
 
|'''b''' is the upper limit of the '''interval'''  
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| 02:16
 
| 02:16
 
||'''f''' is the function to be solved  
 
||'''f''' is the function to be solved  
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| 02:54
 
| 02:54
 
|Let '''b''' be equal to minus three.  
 
|Let '''b''' be equal to minus three.  
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| 02:56
 
| 02:56
 
| Press '''Enter. '''
 
| Press '''Enter. '''
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|To know more about '''deff function''' type '''help deff'''
 
|To know more about '''deff function''' type '''help deff'''
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||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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||Press '''Enter.'''  
 
||Press '''Enter.'''  
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| To solve the problem, type  
 
| To solve the problem, type  
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| '''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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|04:07
 
|04:07
 
| Press '''Enter.'''  
 
| Press '''Enter.'''  
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| The root of the function is shown on the console.  
 
| The root of the function is shown on the console.  
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||Let us study '''Secant's method.'''  
 
||Let us study '''Secant's method.'''  
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| 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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| Let us solve this example using '''Secant method. '''
 
| Let us solve this example using '''Secant method. '''
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|The function is '''f equal to x square minus six. '''
 
|The function is '''f equal to x square minus six. '''
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| 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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| 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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||Open '''Secant dot sci''' on '''Scilab editor.'''   
 
||Open '''Secant dot sci''' on '''Scilab editor.'''   
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|'''b''' is the second starting guess and  
 
|'''b''' is the second starting guess and  
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|'''f''' is the function to be solved.  
 
|'''f''' is the function to be solved.  
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|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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| 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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| Finally we end the function.  
 
| Finally we end the function.  
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|| Let me save and execute the code.  
 
|| Let me save and execute the code.  
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| Press '''Enter'''
 
| Press '''Enter'''
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| Type  '''a''' equal to 2  
 
| Type  '''a''' equal to 2  
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| Press '''Enter. '''
 
| Press '''Enter. '''
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| Then type  '''b''' equal to 3  
 
| Then type  '''b''' equal to 3  
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| 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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| Press '''Enter'''
 
| Press '''Enter'''
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|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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| It summarises the Spoken Tutorial project  
 
| It summarises the Spoken Tutorial project  
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||Conducts workshops using spoken tutorials  
 
||Conducts workshops using spoken tutorials  
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|-
 
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||Gives certificates to those who pass an online test  
 
||Gives certificates to those who pass an online test  
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||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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Revision as of 14:26, 27 February 2015

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 practicing 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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