Scilab/C4/Solving-Non-linear-Equations/English - Revision history

Nancyvarkey at 13:15, 12 February 2014

2014-02-12T13:15:55Z

Lavitha Pereira at 11:50, 12 February 2014

2014-02-12T11:50:38Z

Lavitha Pereira: Created page with ''''Title of script''': '''Solving Nonlinear Equations using Numerical Methods''' '''Author: Shamika''' '''Keywords: Nonlinear equation, root, zero''' {| style="border-spacing…'

2014-02-12T07:51:10Z

Created page with ''''Title of script''': '''Solving Nonlinear Equations using Numerical Methods''' '''Author: Shamika''' '''Keywords: Nonlinear equation, root, zero''' {| style="border-spacing…'

New page

'''Title of script''': '''Solving Nonlinear Equations using Numerical Methods'''

'''Author: Shamika'''

'''Keywords: Nonlinear equation, root, zero'''

{| style="border-spacing:0;"
! <center>Visual Cue</center>

! <center>Narration</center>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 1
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Dear Friends,

Welcome to the spoken tutorial on “'''Solving Nonlinear Equations using Numerical Methods'''”

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 2,3 -Objectives
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| At the end of this tutorial, you will learn how to:

* Solve '''nonlinear equations''' using numerical methods
* The methods we will be studying are
* Bisection method
* and Secant method

* We will also develop '''Scilab''' code to solve '''nonlinear equations '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 4-System Requirements
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| To record this tutorial, I am using

* '''Ubuntu 12.04''' as the operating system
* and '''Scilab 5.3.3''' version

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 5- Prerequisites
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Before practising this tutorial, a learner should have

* basic knowledge of '''Scilab '''and
* '''nonlinear equations'''

For Scilab, please refer to the '''Scilab''' tutorials available on the '''Spoken Tutorial '''website.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 6
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| For a given '''function f''' , we have to find the value of''' x''' for which '''f of x is equal to zero.'''

This solution x is called '''root of equation or zero of function f'''

This process is called '''root finding or zero finding'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 7
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We begin by studying '''Bisection Method'''.

* In '''bisection method,''' we calculate the initial bracket of the '''root'''.
* Then we iterate through the bracket and halve its length.
* We repeat this proces until we find the solution of the equation.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 8
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us solve this function using '''Bisection method'''

Given

'''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. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open '''Bisection.sci''' on Scilab Editor
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Open Bisection dot sci on Scilab editor
* Let us look at the code for Bisection method
* We define the function Bisection with input arguments a b f and tol
* Here a is the lower limit of the interval
* b is the upper limit of the interval
* f is the function to be solved
* and tol is th etolerance level
* We specify the maximum number of iterations to be equal to hundred.
* We find the midpoint of the interval and iterate till the value calculated is within the specified '''tolerance range'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us solve the problem using this code.

Save and execute the file

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab console

On Scilab Console type,

a=-5

Press enter

b=-3

Press enter

<nowiki>deff('[y]=f(x)','y=(2*(sin(x))-((%e^x)/4)-1')</nowiki>

Press enter

Tol=10^-5

Press enter

Bisection(a,b,f,Tol)

Press enter
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to Scilab console

Let us define the interval.

Let a be equal to minus five

Press enter

Let b be equal to minus three.

Press enter

Define the function using '''deff function'''.

We type

'''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'''

To know more about '''deff function''', type help '''deff'''

'''Press enter'''

Let '''tol '''be equal to '''10 to the power of minus five'''

'''Press enter'''

To solve the problem, type

'''Bisection open paranthesis a comma b comma f comma tol close paranthesis'''

'''Press enter'''

The root of the function is '''shown on the console'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 12
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us study Secant's method.

In '''Secan't method, '''the derivative is approximated by finite

difference using two successive Pls use a simpler word. This word

will pose a problem for our translators.''Reply to nancy (12/23/2013, 11:33): "..."''

Changed. iteration values.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 13
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us solve this example using '''Secant method'''

The function is '''f equal to x square minus six'''.

The two starting guesses are , '''p zero '''equal to two and '''p one '''equal to three.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open Secant.sci on Scilab Editor
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Before we solve the problem, let us look at the code for '''Secant meethod. '''

Open Secant dot sci on Scilab editor

We define the function secant with input arguments a, b and f

a is first starting guess for the root

b is the second starting guess and f is the function to be solved.

We find the difference between the value at the current point and the previous point.

We apply '''Secant's method '''and find the value of the root.

Finally we end the function

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let me save and execute the code.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab console

Type on Scilab console

clc

Press enter

a=2

Press enter

b=3

Press enter

<nowiki>deff('[y]=g(x)','y=(x^2)-6')</nowiki>

Press enter

Secant(a,b,g)

Press enter
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to Scilab console

Type clc

Press enter

Let me define the initial guesses for this example. Type

'''a '''equal to 2

Press enter

Then type

'''b '''equal to 3

Press enter

We define the function using '''deff function. '''

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'''

'''Press enter'''

We call the function by typing

'''Secant open paranthesis a comma b comma g close paranthesis. '''

'''Press enter'''

The value of the root is shown on the console

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 14
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us summarize this tutorial

In this tutorial we have learnt to:

* Find the roots of '''nonlinear equation'''
* Develop''' Scilab '''code for different solving methods

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 15- Assignment
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Solve this problem on your own using the two methods we learnt today.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 16'''

'''Title: About the Spoken Tutorial Project'''

* Watch the video available at [http://spoken-tutorial.org/What_is_a_Spoken_Tutorial http://spoken-tutorial.org/What_is_a_Spoken_Tutorial]

* It summarises the Spoken Tutorial project

* If you do not have good bandwidth, you can download and watch it

| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Watch the video available at the following link
* It summarises the Spoken Tutorial project
* If you do not have good bandwidth, you can download and watch it <br/>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 17'''

'''Title: Spoken Tutorial Workshops'''

The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact@spoken-tutorial.org

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact at spoken hyphen tutorial dot org

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 18'''

'''Title: Acknowledgement'''

* Spoken Tutorial Project is a part of the Talk to a Teacher project

* It is supported by the National Mission on Education through ICT, MHRD, Government of India

* More information on this Mission is available at

* [http://spoken-tutorial.org/NMEICT-Intro http://spoken-tutorial.org/NMEICT-Intro]

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Spoken Tutorial Project is a part of the Talk to a Teacher project
* It is supported by the National Mission on Education through ICT, MHRD, Government of India
* More information on this Mission is available at
* spoken hyphen tutorial dot org slash NMEICT hyphen Intro

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''On previous slide'''
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| This is Ashwini Patil signing off. Thanks for joining.

|}

@@ Line 25: / Line 25: @@
 * Solve '''nonlinear equations''' using numerical methods
 * The methods we will be studying are
-* Bisection method
+* '''Bisection method'''
-* and Secant method
+* and '''Secant method'''
 * We will also develop '''Scilab''' code to solve '''nonlinear equations '''
@@ Line 63: / Line 63: @@
-* In '''bisection method,''' we calculate the initial bracket of the '''root'''.
+* In '''bisection method,''' we calculate the '''initial bracket''' of the '''root'''.
-* Then we iterate through the bracket and halve its length.
+* Then we iterate through the '''bracket''' and halve its length.
 * We repeat this proces until we find the solution of the equation.
@@ Line 80: / Line 80: @@
 |-
 | style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open '''Bisection.sci''' on Scilab Editor
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Open Bisection dot sci on Scilab editor
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Open '''Bisection dot sci '''on '''Scilab editor'''
-* Let us look at the code for Bisection method
+* Let us look at the code for '''Bisection method'''
-* We define the function Bisection with input arguments a b f and tol
+* We define the function '''Bisection''' with input arguments '''a b f '''and '''tol'''
-* Here a is the lower limit of the interval
+* Here '''a''' is the lower limit of the interval
-* b is the upper limit of the interval
+* '''b '''is the upper limit of the interval
-* f is the function to be solved
+* '''f''' is the function to be solved
-* and tol is th etolerance level
+* and '''tol''' is the''' tolerance level'''
 * We specify the maximum number of iterations to be equal to hundred.
-* We find the midpoint of the interval and iterate till the value calculated is within the specified '''tolerance range'''.
+* We find the '''midpoint of the interval '''and iterate till the value calculated is within the specified '''tolerance range'''.
@@ Line 97: / Line 97: @@
-Save and execute the file
+'''Save and execute '''the file
 |-
@@ Line 126: / Line 126: @@
 Press enter
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to Scilab console
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to '''Scilab console'''
@@ Line 132: / Line 132: @@
-Let a be equal to minus five
+Let '''a '''be equal to minus five
 Press enter
-Let b be equal to minus three.
+Let '''b''' be equal to minus three.
 Press enter
@@ Line 154: / Line 154: @@
-Let '''tol '''be equal to '''10 to the power of minus five'''
+Let '''tol '''be equal to 10 to the power of minus five
 '''Press enter'''
@@ Line 171: / Line 171: @@
 |-
 | style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 12
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us study Secant's method.
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us study '''Secant's method'''.
 In '''Secan't method, '''the derivative is approximated by finite
-difference using two successive Pls use a simpler word. This word
+difference using two successive '''iteration values.'''
-−
-will pose a problem for our translators.''Reply to nancy (12/23/2013, 11:33): "..."''
-−
-Changed. iteration values.
 |-
@@ Line 189: / Line 185: @@
-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 196: / Line 192: @@
-Open Secant dot sci on Scilab editor
+Open '''Secant dot sci '''on '''Scilab editor'''
-We define the function secant with input arguments a, b and f
+We define the function '''secant '''with input arguments '''a, b '''and '''f'''
-a is first starting guess for the root
+'''a''' is first starting guess for the root
-b is the second starting guess and f is the function to be solved.
+'''b''' is the second starting guess and f is the function to be solved.
@@ Line 218: / Line 214: @@
 |-
 | style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let me save and execute the code.
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let me '''save and execute '''the code.
 |-
@@ Line 249: / Line 245: @@
 Press enter
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to Scilab console
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to '''Scilab console'''
-Type clc
+Type '''clc'''
-Press enter
+Press '''enter'''
@@ Line 261: / Line 257: @@
 '''a '''equal to 2
-Press enter
+Press '''enter'''
@@ Line 268: / Line 264: @@
 '''b '''equal to 3
-Press enter
+Press '''enter'''
@@ Line 287: / Line 283: @@
-The value of the root is shown on the console
+The value of the root is shown on the '''console'''