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

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

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

Latest revision as of 11:53, 8 May 2019

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@@ Line 6: / Line 6: @@
 |-
 | 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
-|solve '''nonlinear equations''' using numerical methods.
+|Solve '''nonlinear equations''' using numerical methods
 |-
@@ Line 26: / Line 22: @@
 |-
 | 00:20
-|'''Bisection method''' and
+|'''Bisection method and '''
 |-
 |00:22
-|'''Secant method'''.
+|'''Secant method'''
 |-
@@ Line 46: / Line 42: @@
 |-
 |00:36
-|'''Scilab 5.3.3''' version.
+|'''Scilab 5.3.3''' version
 |-
 |00:40
-| Before practicing this tutorial, a learner should have
+| Before practising this tutorial, a learner should have
 |-
@@ Line 58: / Line 54: @@
 |-
 | 00:46
-| '''nonlinear equations'''.
+| '''nonlinear equations'''
 |-
@@ Line 70: / Line 66: @@
 |-
 |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
-| This process is called '''root finding''' or '''zero finding.'''
+| This process is called ''' root finding''' or '''zero finding.'''
 |-
@@ Line 81: / Line 77: @@
 |-
 |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
 |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:
+|| 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. '''
@@ Line 125: / Line 109: @@
 |-
 |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
 || Here '''a''' is the lower limit of the '''interval'''
@@ Line 145: / Line 125: @@
 |-
 | 02:19
-||and '''tol''' is the '''tolerance level'''
+||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.'''
-| 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.
@@ Line 210: / Line 181: @@
 |-
 | 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'''
@@ Line 249: / Line 209: @@
 |-
 | 04:09
 | The root of the function is shown on the console.
 |-
 |04:14
+||Let us study '''Secant method.'''
-||Let us study '''Secant's method.'''
 |-
 |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
 | 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.
-| 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.'''
-| 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
@@ Line 394: / Line 309: @@
 |-
 | 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.
-| 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.