Difference between revisions of "Scilab/C4/Solving-Non-linear-Equations/English-timed"
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| − | |Dear Friends, | + | |Dear Friends, Welcome to the spoken tutorial on '''Solving Nonlinear Equations using Numerical Methods'''. |
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| 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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| − | || Given: | + | || 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''' |
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| − | | We type: | + | | 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''' |
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| 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 ''' | | 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 ''' | ||
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| Press '''Enter'''. | | Press '''Enter'''. | ||
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| We call the function by typing | | We call the function by typing | ||
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| '''Secant open parenthesis a comma b comma g close parenthesis.''' | | '''Secant open parenthesis a comma b comma g close parenthesis.''' | ||
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| Press '''Enter'''. | | Press '''Enter'''. | ||
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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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| 06:35 | | 06:35 | ||
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| Let us summarize this tutorial. | | Let us summarize this tutorial. | ||
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| 06:38 | | 06:38 | ||
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| In this tutorial we have learnt to: | | In this tutorial we have learnt to: | ||
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|Develop '''Scilab''' code for different solving methods | |Develop '''Scilab''' code for different solving methods | ||
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|Find the roots of '''nonlinear equation '''. | |Find the roots of '''nonlinear equation '''. | ||
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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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|06:55 | |06:55 | ||
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| Watch the video available at the link shown below. | | Watch the video available at the link shown below. | ||
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| 06:58 | | 06:58 | ||
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| It summarizes the Spoken Tutorial project. | | It summarizes the Spoken Tutorial project. | ||
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|07:01 | |07:01 | ||
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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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|07:05 | |07:05 | ||
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||The spoken tutorial project Team: | ||The spoken tutorial project Team: | ||
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||Conducts workshops using spoken tutorials. | ||Conducts workshops using spoken tutorials. | ||
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|07:10 | |07:10 | ||
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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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|07:14 | |07:14 | ||
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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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|07:21 | |07:21 | ||
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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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| 07:24 | | 07:24 | ||
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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 | | 07:32 | ||
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|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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| 07:39 | | 07:39 | ||
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|This is Ashwini Patil, signing off. | |This is Ashwini Patil, signing off. | ||
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|07:41 | |07:41 | ||
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| Thank you for joining. | | Thank you for joining. | ||
Revision as of 11:33, 10 March 2017
| 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 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: 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: 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 |
| 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 parenthesis a comma b comma f comma tol close parenthesis |
| 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 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 | Press Enter. |
| 06:18 | We call the function by typing |
| 06:20 | Secant open parenthesis a comma b comma g close parenthesis. |
| 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 summarizes 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. |
