Scilab/C4/Solving-Non-linear-Equations/English-timed

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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.

Contributors and Content Editors

PoojaMoolya, Pratik kamble, Sandhya.np14