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”
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| 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
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| 00.20 | Bisection method and |
| 00.22 | Secant method
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| 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.
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| 01.16 | We begin by studying Bisection Method.
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| 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.
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| 02.00 | Let us look at the code for Bisection method.
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| 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
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| 02.16 | f is the function to be solved
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| 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.
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| 02.56 | Press Enter.
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| 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
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| 03.46 | Press Enter. |
| 03.48 | Let tol be equal to 10 to the power of minus five.
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| 03.53 | Press Enter.
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| 03.56 | To solve the problem, type
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| 03.58 | Bisection open paranthesis a comma b comma f comma tol close paranthesis
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| 04.07 | Press Enter.
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| 04.09 | The root of the function is shown on the console.
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| 04.14 | Let us study Secant's method.
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| 04.17 | In Secant's method, the derivative is approximated by finite difference using two successive iteration values.
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| 04.27 | Let us solve this example using Secant method.
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| 04.30 | The function is f equal to x square minus six.
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| 04.36 | The two starting guesses are , p zero equal to two and p one equal to three.
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| 04.44 | Before we solve the problem, let us look at the code for Secant method.
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| 04.50 | Open Secant dot sci on Scilab editor.
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| 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
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| 05.07 | f is the function to be solved.
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| 05.10 | We find the difference between the value at the current point and the previous point.
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| 05.15 | We apply Secant's method and find the value of the root.
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| 05.21 | Finally we end the function.
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| 05.24 | Let me save and execute the code.
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| 05.27 | Switch to Scilab console. |
| 05.30 | Type clc. |
| 05.32 | Press Enter
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| 05.34 | Let me define the initial guesses for this example. |
| 05.38 | Type a equal to 2
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| 05.40 | Press Enter.
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| 05.42 | Then type b equal to 3
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| 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
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| 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
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| 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
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| 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
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| 07.10 | Gives certificates to those who pass an online test
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| 07.14 | For more details, please write to conatct@spoken-tutorial.org
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| 07.21 | Spoken Tutorial Project is a part of the Talk to a Teacher project
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| 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. |
