Difference between revisions of "Scilab/C4/Solving-Non-linear-Equations/English-timed"
From Script | Spoken-Tutorial
Sandhya.np14 (Talk | contribs) |
Sandhya.np14 (Talk | contribs) |
||
| Line 75: | Line 75: | ||
|01:11 | |01:11 | ||
| This process is called ''' root finding''' or '''zero finding.''' | | This process is called ''' root finding''' or '''zero finding.''' | ||
| − | |||
|- | |- | ||
|01:16 | |01:16 | ||
|We begin by studying '''Bisection Method. ''' | |We begin by studying '''Bisection Method. ''' | ||
| − | |||
|- | |- | ||
| Line 121: | Line 119: | ||
| '''Open Bisection dot sci on Scilab editor. ''' | | '''Open Bisection dot sci on Scilab editor. ''' | ||
| − | |||
|- | |- | ||
|02:00 | |02:00 | ||
| Let us look at the code for '''Bisection method.''' | | Let us look at the code for '''Bisection method.''' | ||
| − | |||
|- | |- | ||
| Line 143: | Line 139: | ||
|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 | ||
| − | |||
|- | |- | ||
| Line 198: | Line 192: | ||
| 02:54 | | 02:54 | ||
|Let '''b''' be equal to minus three. | |Let '''b''' be equal to minus three. | ||
| − | |||
| − | |||
|- | |- | ||
| 02:56 | | 02:56 | ||
| Press '''Enter. ''' | | Press '''Enter. ''' | ||
| − | |||
|- | |- | ||
| Line 223: | Line 214: | ||
|To know more about '''deff function''' type '''help deff''' | |To know more about '''deff function''' type '''help deff''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 236: | Line 225: | ||
||Let '''tol''' be equal to 10 to the power of minus five. | ||Let '''tol''' be equal to 10 to the power of minus five. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 244: | Line 231: | ||
||Press '''Enter.''' | ||Press '''Enter.''' | ||
| − | |||
|- | |- | ||
| Line 251: | Line 237: | ||
| To solve the problem, type | | To solve the problem, type | ||
| − | |||
|- | |- | ||
| Line 258: | Line 243: | ||
| '''Bisection open paranthesis a comma b comma f comma tol close paranthesis''' | | '''Bisection open paranthesis a comma b comma f comma tol close paranthesis''' | ||
| − | |||
|- | |- | ||
|04:07 | |04:07 | ||
| Press '''Enter.''' | | Press '''Enter.''' | ||
| − | |||
|- | |- | ||
| Line 270: | Line 253: | ||
| The root of the function is shown on the console. | | The root of the function is shown on the console. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 278: | Line 259: | ||
||Let us study '''Secant's method.''' | ||Let us study '''Secant's method.''' | ||
| − | |||
|- | |- | ||
| Line 285: | Line 265: | ||
| 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. | ||
| − | |||
|- | |- | ||
| Line 292: | Line 271: | ||
| Let us solve this example using '''Secant method. ''' | | Let us solve this example using '''Secant method. ''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 300: | Line 277: | ||
|The function is '''f equal to x square minus six. ''' | |The function is '''f equal to x square minus six. ''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 308: | Line 283: | ||
| 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. | ||
| − | |||
|- | |- | ||
| Line 315: | Line 289: | ||
| 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. ''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 323: | Line 295: | ||
||Open '''Secant dot sci''' on '''Scilab editor.''' | ||Open '''Secant dot sci''' on '''Scilab editor.''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 343: | Line 313: | ||
|'''b''' is the second starting guess and | |'''b''' is the second starting guess and | ||
| − | |||
|- | |- | ||
| Line 350: | Line 319: | ||
|'''f''' is the function to be solved. | |'''f''' is the function to be solved. | ||
| − | |||
|- | |- | ||
| Line 357: | Line 325: | ||
|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. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 365: | Line 331: | ||
| We apply '''Secant's method ''' and find the value of the root. | | We apply '''Secant's method ''' and find the value of the root. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 373: | Line 337: | ||
| Finally we end the function. | | Finally we end the function. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 381: | Line 343: | ||
|| Let me save and execute the code. | || Let me save and execute the code. | ||
| − | |||
|- | |- | ||
| Line 399: | Line 360: | ||
| Press '''Enter''' | | Press '''Enter''' | ||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| Line 414: | Line 372: | ||
| Type '''a''' equal to 2 | | Type '''a''' equal to 2 | ||
| − | |||
|- | |- | ||
| Line 421: | Line 378: | ||
| Press '''Enter. ''' | | Press '''Enter. ''' | ||
| − | |||
| − | |||
|- | |- | ||
| Line 429: | Line 384: | ||
| Then type '''b''' equal to 3 | | Then type '''b''' equal to 3 | ||
| − | |||
| − | |||
|- | |- | ||
| Line 445: | Line 398: | ||
| 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 ''' | ||
| − | |||
|- | |- | ||
| Line 470: | Line 422: | ||
| Press '''Enter''' | | Press '''Enter''' | ||
| − | |||
|- | |- | ||
| Line 507: | Line 458: | ||
|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. | ||
| − | |||
| − | |||
|- | |- | ||
| Line 520: | Line 469: | ||
| It summarises the Spoken Tutorial project | | It summarises the Spoken Tutorial project | ||
| − | |||
| − | |||
|- | |- | ||
| Line 540: | Line 487: | ||
||Conducts workshops using spoken tutorials | ||Conducts workshops using spoken tutorials | ||
| − | |||
|- | |- | ||
| Line 547: | Line 493: | ||
||Gives certificates to those who pass an online test | ||Gives certificates to those who pass an online test | ||
| − | |||
|- | |- | ||
| Line 554: | Line 499: | ||
||For more details, please write to conatct@spoken-tutorial.org | ||For more details, please write to conatct@spoken-tutorial.org | ||
| − | |||
|- | |- | ||
| Line 561: | Line 505: | ||
|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 | ||
| − | |||
| − | |||
|- | |- | ||
Revision as of 10:03, 26 December 2017
| 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 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'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. |
