Difference between revisions of "Scilab/C4/Solving-Non-linear-Equations/English-timed"
From Script | Spoken-Tutorial
Sandhya.np14 (Talk | contribs) |
Sandhya.np14 (Talk | contribs) |
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| − | |The methods we will be studying are | + | |The methods we will be studying are: |
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|01:54 | |01:54 | ||
| − | | '''Open Bisection dot sci on Scilab editor. ''' | + | | '''Open Bisection dot sci''' on '''Scilab editor. ''' |
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|02:10 | |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 | |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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| 02:19 | | 02:19 | ||
| − | ||and '''tol''' is the '''tolerance level''' | + | ||and '''tol''' is the '''tolerance level'''. |
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| 02:58 | | 02:58 | ||
| − | |Define the function using '''deff | + | |Define the function using '''deff''' function. |
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|03:01 | |03:01 | ||
| − | | We type | + | | We type: |
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| 03:02 | | 03:02 | ||
| − | | '''deff open | + | | '''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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| 03:41 | | 03:41 | ||
| − | |To know more about '''deff | + | |To know more about '''deff''' function, type '''help deff''' |
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| 03:58 | | 03:58 | ||
| − | | '''Bisection open | + | | '''Bisection open parenthesis a comma b comma f comma tol close parenthesis''' |
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|04:17 | |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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| 04:27 | | 04:27 | ||
| − | | Let us solve this example using '''Secant method. ''' | + | | Let us solve this example using '''Secant method.''' |
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| 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.''' |
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| 04:54 | | 04:54 | ||
| − | ||We define the function ''' | + | ||We define the function '''Secant''' with input arguments '''a, b''' and '''f.''' |
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| 05:01 | | 05:01 | ||
| − | ||'''a''' is first starting guess for the root | + | ||'''a''' is first starting guess for the root, |
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| 05:30 | | 05:30 | ||
| − | |Type '''clc | + | |Type '''clc'''. |
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| 05:32 | | 05:32 | ||
| − | | Press '''Enter''' | + | | Press '''Enter'''. |
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| 05:38 | | 05:38 | ||
| − | | Type '''a''' equal to 2 | + | | Type '''a''' equal to 2. |
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| 05:42 | | 05:42 | ||
| − | | Then type '''b''' equal to 3 | + | | Then type '''b''' equal to 3. |
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| 05:46 | | 05:46 | ||
| − | |We define the function using '''deff | + | |We define the function using '''deff''' function. |
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| 05:49 | | 05:49 | ||
| − | | Type '''deff open | + | | 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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| 06:15 | | 06:15 | ||
| − | | Press '''Enter''' | + | | Press '''Enter'''. |
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| 06:20 | | 06:20 | ||
| − | | '''Secant open | + | | '''Secant open parenthesis a comma b comma g close parenthesis.''' |
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| 06:27 | | 06:27 | ||
| − | | Press '''Enter''' | + | | Press '''Enter'''. |
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| 06:30 | | 06:30 | ||
| − | | The value of the root is shown on the '''console''' | + | | The value of the root is shown on the '''console'''. |
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| 06:45 | | 06:45 | ||
| − | |Find the roots of '''nonlinear equation ''' | + | |Find the roots of '''nonlinear equation '''. |
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|06:55 | |06:55 | ||
| − | | 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 | ||
| − | | It | + | | It summarizes the Spoken Tutorial project. |
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|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. |
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|07:05 | |07:05 | ||
| − | ||The spoken tutorial project Team | + | ||The spoken tutorial project Team: |
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|07:07 | |07:07 | ||
| − | ||Conducts workshops using spoken tutorials | + | ||Conducts workshops using spoken tutorials. |
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|07:10 | |07:10 | ||
| − | ||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 | ||
| − | ||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 | ||
| − | |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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| 07:32 | | 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. |
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| − | |This is Ashwini Patil signing off. | + | |This is Ashwini Patil, signing off. |
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Revision as of 16:33, 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 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. |
