Scilab/C4/Solving-Non-linear-Equations/English-timed
From Script | Spoken-Tutorial
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. |