Scilab/C4/Interpolation/English

Title of script: Numerical Interpolation

Author: Shamika

Keywords: Interpolation, Lagrange method, Newton divided difference method

Visual Cue	Narration
Slide 1	Dear Friends, Welcome to the Spoken Tutorial on “Numerical Interpolation”
Slide 2 -Learning Objective Slide	At the end of this tutorial, you will learn how to: Develop Scilab code for different Numerical Interpolation algorithms Calculate new value of function from given data points
Slide 3-System Requirement slide	To record this tutorial, I am using Ubuntu 12.04 as the operating system and Scilab 5.3.3 version
Slide 4- Prerequisites slide	To practise this tutorial, a learner should have basic knowledge of Scilab and should know Numerical Interpolation To learn Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.
Slide 5- Numerical Interpolation	Numerical interpolation is a method of constructing new data points within the range of a discrete set of known data points. We can solve interpolation problems using numerical methods.
Slide 6- Lagrange Interpolation	In Lagrange interpolation, we pass a polynomial of degree N – 1 through N points. Then we find the unique N order polynomial y of x which interpolates the data samples.
Slide 7- Example	We are given the natural logarithm values for nine, nine point five and eleven. We have to find the value ofIs this correct? Pls check.Reply to nancy (10/06/2013, 16:20): "..." Change made. natural logarithm of nine point two. Let us solve this problem using Lagrange interpolation method.
Show Lagrange.sci code on Scilab editor	Let us look at the code for Lagrange interpolation.
Highlight Lagrange(x0, x,f, n)	We define the function Lagrange with arguments x zero, x, f and n. X zero is the unknown interpolation point. x is the vector containing the data points. f is the vector containing the values of the function at correspoding data points. And n is the order of the interpolating polynomial.
Highlight m = n + 1; N = ones(1,m);	We use n to initialize m and vector N. The order of the interpolating polynomail determines the number of nodes created.
Highlight for j = 1:m for k = 1:m if (k<>j) then N(j) = N(j)*(x0 - x(k)) D(j) = D(j)*(x(j) - x(k)) end end	Then we apply Lagrange interpolation formula to find the value of the numerator and denominator.
Highlight L(j) = N(j)/D(j); y = y + L(j)*f(j); end disp(L','L') disp(f,'f(x)')	Then we divide the numerator and denominator to get the value of L. We use L to find the value of the function y at the given data point. Finally we display the value of L and f of x.
Click on Execute and select Save and Execute	Let us save and execute the file.
Switch to Scilab console	Switch to Scilab console to solve the example problem.
Type on console x=[9.0,9.5,11.0]	Let us define the data points vector. On the console type, x equal to open square bracket nine point zero comma nine point five comma eleven point zero close square bracket. Press Enter
Type on console f=[2.1972,2.2513,2.3979]	Then type f equal to open square bracket two point one nine seven two comma two point two five one three comma two point three nine seven nine close square bracket Press Enter
Type on console x0=9.2	Then type x zero equal to nine point two Press Enter
Type on console n=2	Let us use a quadratic polynomial interpolating polynomial. Type n equal to two Press Enter
Type on console y = Lagrange(x0, x,f, n)	To call the function, type y equal to Lagrange open paranthesis x zero comma x comma f comma n close paranthesis Press Enter
Show console	The value of the function y at x equal to nine point two is displayed.
	Let us look at Newton's Divided Difference Method.
Slide 8- Newton's Divided Difference Method	In this method, divided differences recursive method is used. It uses lesser number of computation than Lagrange method. In spite of this, the same interpolating polynomial as in Lagrange method is generated.
Slide 9- Example	Let us solve this example using divided difference method. We are given the data points and the corresponding values of the function at those data points. We have to find the value of the function at x equal to three.
Switch to Scilab editor	Let us look at the code for Newton Divided difference method. Open the file Newton underscore divided dot sci on Scilab Editor.
Highlight Newton_Divided(x,f,x0)	We define the function Newton underscore Divided with arguments x, f and x zero. X is a vector containing the data points, f is the corresponding function value and x zero is the unknown interpolation point.
Highlight n = length(x);	We find the length of vector and then equate it to n.
Highlight a(1) = f(1);	The first value of vector is equated to a of one.
Highlight for k = 1 : n - 1 D(k, 1) = (f(k+1) - f(k))/(x(k+1) - x(k)); end for j = 2:n-1 for k = 1:n-j D(k, j) = (D(k+1, j-1) - D(k, j-1))/(x(k+j) - x(k)) end end disp(D, 'The Divided Difference Table')	Then we apply divided difference algorithm and compute the divided difference table
Highlight Df(1) = 1; c(1) = a(1); for j = 2 : n Df(j)=(x0 - x(j-1)).*Df(j-1); c(j) = a(j).*Df(j); end	Then we find the coefficient list of the Newton polynomial
Highlight IP = sum(c);	We sum the coefficient list to find the value of the function at given data point.
Click on Execute and select Save and Execute	Save and execute the file Newton underscore divided dot sci.
Switch to Scilab console	Switch to Scilab console
Type clc	Clear the screen by typing c l c. Press Enter
Type on console x=[2,2.5,3.25,4]	Let us enter the data points vector Type x equal to open square bracket two comma two point five comma three point two five comma four close square bracket Press Enter
Type on console f=[0.5,0.4,0.3077,0.25]	Then type values of the function Type f equal to open square bracket zero point five comma zero point four comma zero point three zero seven seven comma zero point two five close square bracket Press Enter
Type on console x0=3	Type x zero equal to three Press Enter
Type on console IP = Newton_Divided(x,f,x0)	Then call the function by typing i p equal to Newton underscore divided open paranthesis x comma f comma x zero close paranthesis Press Enter
Show console	The value of y at x equal to three is shown.
Slide 9- Summary	Let us summarize this tutorial. In this tutorial we have learnt to develop Scilab code for interpolation methods. We have also learnt to find the value of a function at new data point.
Slide 10- Assignment	Solve this problem on your own using Lagrange method and Newton's divided difference method.
Show Slide 11 Title: About the Spoken Tutorial Project Watch the video available at 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	* About the Spoken Tutorial Project Watch the video available at 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
Show Slide 12 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	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
Show Slide 13 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	* 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
	This is Ashwini Patil signing off. Thanks for joining.

Contributors and Content Editors

Lavitha Pereira, Nancyvarkey