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

From Script | Spoken-Tutorial
Jump to: navigation, search
 
(6 intermediate revisions by 2 users not shown)
Line 6: Line 6:
 
|-
 
|-
 
| 00:01
 
| 00:01
−
|Dear Friends,  
+
|Dear Friends, Welcome to the spoken tutorial on  ''' “Solving Nonlinear Equations using Numerical Methods” '''
  
 
|-
 
|-
−
| 00:02
+
| 00:10
−
|Welcome to the spoken tutorial on '''Solving Nonlinear Equations using Numerical Methods'''.
+
| At the end of this tutorial, you will learn how to:  
−
 
+
−
|-
+
−
| 00:10.
+
−
| At the end of this tutorial, you will learn how to   
+
  
 
|-
 
|-
 
|00:13
 
|00:13
−
|solve '''nonlinear equations''' using numerical methods.
+
|Solve '''nonlinear equations''' using numerical methods
  
 
|-
 
|-
 
|00:18
 
|00:18
−
|The methods we will be studying are:
+
|The methods we will be studying are  
  
 
|-
 
|-
 
| 00:20
 
| 00:20
−
|'''Bisection method''' and
+
|'''Bisection method and '''
  
 
|-
 
|-
 
|00:22
 
|00:22
−
|'''Secant method'''.
+
|'''Secant method'''  
  
 
|-
 
|-
Line 46: Line 42:
 
|-
 
|-
 
|00:36
 
|00:36
−
|'''Scilab 5.3.3''' version.
+
|'''Scilab 5.3.3''' version  
  
 
|-
 
|-
 
|00:40
 
|00:40
−
| Before practicing this tutorial, a learner should have  
+
| Before practising this tutorial, a learner should have  
  
 
|-
 
|-
Line 58: Line 54:
 
|-
 
|-
 
| 00:46
 
| 00:46
−
| '''nonlinear equations'''.
+
| '''nonlinear equations'''
  
 
|-
 
|-
Line 70: Line 66:
 
|-
 
|-
 
|01:04
 
|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.'''  
  
 
|-
 
|-
 
|01:11
 
|01:11
−
| This process is called '''root finding''' or '''zero finding.'''  
+
| This process is called ''' root finding''' or '''zero finding.'''  
  
 
|-
 
|-
Line 81: Line 77:
  
 
|-
 
|-
−
 
 
|01:20
 
|01:20
−
 
+
|In '''bisection method''' we calculate the '''initial bracket''' of the '''root.'''  
−
|In '''bisection method''', we calculate the '''initial bracket''' of the '''root.'''  
+
  
 
|-
 
|-
−
 
 
|01:25
 
|01:25
−
 
 
|Then we iterate through the '''bracket''' and halve its length.  
 
|Then we iterate through the '''bracket''' and halve its length.  
  
 
|-
 
|-
−
 
 
| 01:31
 
| 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.  
  
 
|-
 
|-
−
 
 
| 01:36
 
| 01:36
 
||Let us solve this function using '''Bisection method.'''  
 
||Let us solve this function using '''Bisection method.'''  
  
 
|-
 
|-
−
 
 
| 01:41
 
| 01:41
−
|| Given:
+
|| Given
  
 
|-
 
|-
−
 
 
|01:42
 
|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'''
  
 
|-
 
|-
−
 
 
|01:54
 
|01:54
−
 
+
| '''Open Bisection dot sci on Scilab editor. '''
−
| '''Open Bisection dot sci''' on '''Scilab editor. '''
+
  
 
|-
 
|-
Line 125: Line 109:
 
   
 
   
 
|-
 
|-
−
 
 
|02:03
 
|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.'''  
+
  
 
|-
 
|-
−
 
 
|02:10
 
|02:10
−
 
+
|| Here '''a''' is the lower limit of the '''interval'''  
−
|| Here '''a''' is the lower limit of the '''interval''',
+
  
 
|-
 
|-
 
|02:14
 
|02:14
−
|'''b''' is the upper limit of the '''interval''',
+
|'''b''' is the upper limit of the '''interval'''  
  
 
|-
 
|-
 
| 02:16
 
| 02:16
−
||'''f''' is the function to be solved,
+
||'''f''' is the function to be solved  
  
 
|-
 
|-
−
 
 
| 02:19
 
| 02:19
−
||and '''tol''' is the '''tolerance level'''.
+
||and '''Tol''' is the '''tolerance level'''  
  
 
|-
 
|-
−
 
 
|02:22
 
|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.  
  
 
|-
 
|-
−
 
 
|02:28
 
|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.'''  
+
 
   
 
   
 
|-
 
|-
−
 
 
|02:37
 
|02:37
−
 
 
| Let us solve the problem using this code.  
 
| Let us solve the problem using this code.  
  
 
|-
 
|-
−
 
 
| 02:40
 
| 02:40
−
 
 
|| Save and execute the file.  
 
|| Save and execute the file.  
  
Line 199: Line 170:
 
|-
 
|-
 
| 02:58
 
| 02:58
−
|Define the function using '''deff''' function.
+
|Define the function using '''deff function.'''  
  
 
|-
 
|-
 
|03:01
 
|03:01
−
| We type:
+
| We type  
  
 
|-
 
|-
 
| 03:02
 
| 03:02
−
| '''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'''
+
| '''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
 
| 03:41
−
 
+
|To know more about '''deff function''' type '''help deff'''
−
|To know more about '''deff''' function, type '''help deff'''
+
  
 
|-
 
|-
−
 
 
| 03:46
 
| 03:46
 
|| Press '''Enter.'''  
 
|| Press '''Enter.'''  
  
 
|-
 
|-
−
 
 
|03:48
 
|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.  
  
 
|-
 
|-
−
 
 
|03:53
 
|03:53
−
 
 
||Press '''Enter.'''  
 
||Press '''Enter.'''  
  
 
|-
 
|-
−
 
 
| 03:56
 
| 03:56
−
 
 
| To solve the problem, type  
 
| To solve the problem, type  
  
 
|-
 
|-
−
 
 
| 03:58
 
| 03:58
−
 
+
| '''Bisection open paranthesis a comma b comma f comma tol close paranthesis'''  
−
| '''Bisection open parenthesis a comma b comma f comma tol close parenthesis'''  
+
  
 
|-
 
|-
Line 249: Line 209:
  
 
|-
 
|-
−
 
 
| 04:09
 
| 04:09
−
 
 
| The root of the function is shown on the console.  
 
| The root of the function is shown on the console.  
  
 
|-
 
|-
−
 
 
|04:14
 
|04:14
−
 
+
||Let us study '''Secant method.'''  
−
||Let us study '''Secant's method.'''  
+
  
 
|-
 
|-
−
 
 
|04:17
 
|04:17
−
 
+
| In '''Secant 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.
+
  
 
|-
 
|-
−
 
 
| 04:27
 
| 04:27
−
 
+
| Let us solve this example using '''Secant method. '''
−
| Let us solve this example using '''Secant method.'''
+
  
 
|-
 
|-
−
 
 
| 04:30
 
| 04:30
−
 
 
|The function is '''f equal to x square minus six. '''
 
|The function is '''f equal to x square minus six. '''
  
 
|-
 
|-
−
 
 
| 04:36
 
| 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.  
  
 
|-
 
|-
−
 
 
| 04:44
 
| 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.'''
+
  
 
|-
 
|-
−
 
 
| 04:50
 
| 04:50
−
 
 
||Open '''Secant dot sci''' on '''Scilab editor.'''   
 
||Open '''Secant dot sci''' on '''Scilab editor.'''   
  
 
|-
 
|-
−
 
 
| 04:54
 
| 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.'''  
+
  
 
|-
 
|-
−
 
 
| 05:01
 
| 05:01
−
 
+
||'''a''' is first starting guess for the root  
−
||'''a''' is first starting guess for the root,
+
  
 
|-
 
|-
−
 
 
| 05:04
 
| 05:04
−
 
 
|'''b''' is the second starting guess and  
 
|'''b''' is the second starting guess and  
  
 
|-
 
|-
−
 
 
| 05:07
 
| 05:07
−
 
 
|'''f''' is the function to be solved.  
 
|'''f''' is the function to be solved.  
  
 
|-
 
|-
−
 
 
|05:10
 
|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.  
  
 
|-
 
|-
−
 
 
|05:15
 
|05:15
−
 
+
| We apply '''Secant method ''' and find the value of the root.  
−
| We apply '''Secant's method ''' and find the value of the root.  
+
  
 
|-
 
|-
−
 
 
| 05:21
 
| 05:21
−
 
 
| Finally we end the function.  
 
| Finally we end the function.  
  
 
|-
 
|-
−
 
 
|05:24
 
|05:24
−
 
 
|| Let me save and execute the code.  
 
|| Let me save and execute the code.  
  
 
|-
 
|-
−
 
 
| 05:27
 
| 05:27
 +
| Switch to '''Scilab console.'''
  
−
| Switch to '''Scilab console.'''
 
 
|-
 
|-
−
 
 
| 05:30
 
| 05:30
−
 
+
|Type '''clc. '''
−
|Type '''clc'''.
+
  
 
|-
 
|-
−
 
 
| 05:32
 
| 05:32
−
 
+
| Press '''Enter'''
−
| Press '''Enter'''.
+
  
 
|-
 
|-
−
 
 
| 05:34
 
| 05:34
−
 
 
|Let me define the initial guesses for this example.  
 
|Let me define the initial guesses for this example.  
  
 
|-
 
|-
−
 
 
| 05:38
 
| 05:38
−
 
+
| Type  '''a''' equal to 2  
−
| Type  '''a''' equal to 2.
+
  
 
|-
 
|-
−
 
 
| 05:40
 
| 05:40
−
 
 
| Press '''Enter. '''
 
| Press '''Enter. '''
  
 
|-
 
|-
−
 
 
| 05:42
 
| 05:42
−
 
+
| Then type  '''b''' equal to 3  
−
| Then type  '''b''' equal to 3.
+
  
 
|-
 
|-
Line 391: Line 306:
 
|-
 
|-
 
| 05:46
 
| 05:46
−
|We define the function using '''deff''' function.
+
|We define the function using '''deff function. '''
  
 
|-
 
|-
−
 
 
| 05:49
 
| 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 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 '''
+
  
 
|-
 
|-
−
 
 
| 06:15
 
| 06:15
−
 
+
| Press '''Enter'''
−
| Press '''Enter'''.
+
  
 
|-
 
|-
−
 
 
| 06:18
 
| 06:18
−
 
 
| We call the function by typing  
 
| We call the function by typing  
  
 
|-
 
|-
−
 
 
| 06:20
 
| 06:20
−
 
+
| '''Secant open paranthesis a comma b comma g close paranthesis.'''
−
| '''Secant open parenthesis a comma b comma g close parenthesis.'''
+
  
 
|-
 
|-
−
 
 
| 06:27
 
| 06:27
−
 
+
| Press '''Enter'''
−
| Press '''Enter'''.
+
  
 
|-
 
|-
−
 
 
| 06:30
 
| 06:30
−
 
+
| The value of the root is shown on the '''console'''
−
| The value of the root is shown on the '''console'''.
+
  
 
|-
 
|-
−
 
 
| 06:35
 
| 06:35
−
 
 
| Let us summarize this tutorial.  
 
| Let us summarize this tutorial.  
  
 
|-
 
|-
−
 
 
| 06:38
 
| 06:38
−
 
 
| In this tutorial we have learnt to:  
 
| In this tutorial we have learnt to:  
  
 
|-
 
|-
−
 
 
| 06:41
 
| 06:41
−
 
 
|Develop '''Scilab''' code for different solving methods  
 
|Develop '''Scilab''' code for different solving methods  
  
 
|-
 
|-
−
 
 
| 06:45
 
| 06:45
−
 
+
|Find the roots of '''nonlinear equation '''
−
|Find the roots of '''nonlinear equation '''.
+
  
 
|-
 
|-
−
 
 
| 06:48
 
| 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.  
  
 
|-
 
|-
 
|06:55
 
|06:55
−
 
+
| Watch the video available at the link shown below  
−
| Watch the video available at the link shown below.
+
  
 
|-
 
|-
−
 
 
| 06:58
 
| 06:58
−
 
+
| It summarises the Spoken Tutorial project  
−
| It summarizes the Spoken Tutorial project.
+
  
 
|-
 
|-
−
 
 
|07:01
 
|07:01
−
 
+
||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.
+
  
 
|-
 
|-
−
 
 
|07:05
 
|07:05
−
 
+
||The spoken tutorial project Team
−
||The spoken tutorial project Team:
+
  
 
|-
 
|-
−
 
 
|07:07
 
|07:07
−
 
+
||Conducts workshops using spoken tutorials  
−
||Conducts workshops using spoken tutorials.
+
  
 
|-
 
|-
−
 
 
|07:10
 
|07:10
−
 
+
||Gives certificates to those who pass an online test  
−
||Gives certificates to those who pass an online test.
+
  
 
|-
 
|-
−
 
 
|07:14
 
|07:14
−
 
+
||For more details, please write to conatct@spoken-tutorial.org  
−
||For more details, please write to conatct@spoken-tutorial.org.
+
  
 
|-
 
|-
−
 
 
|07:21
 
|07:21
−
 
+
|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.
+
  
 
|-
 
|-
−
 
 
| 07:24
 
| 07:24
 +
| 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.
 
 
|-
 
|-
−
 
 
| 07:32
 
| 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.
+
  
 
|-
 
|-
−
 
 
| 07:39
 
| 07:39
−
 
+
|This is Ashwini Patil signing off.
−
|This is Ashwini Patil, signing off.
+
  
 
|-
 
|-
−
 
 
|07:41
 
|07:41
−
 
 
| Thank you for joining.
 
| Thank you for joining.

Latest revision as of 11:53, 8 May 2019

Time Narration
00:01 Dear Friends, 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 method.
04:17 In Secant 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 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

Retrieved from "https://script.spoken-tutorial.org/index.php?title=Scilab/C4/Solving-Non-linear-Equations/English-timed&oldid=47421"