LAGRANGE INTERPOLATION POLYNOMIAL

$\backslash$table$[[$Name:$,$Grade $(5)],[$ID Number:,$],[$Lab Section:,$]]$

Objectives

Knowing how to perform an interpolation with a Lagrange polynomial.

To understand the MATLAB implementation of the Lagrange Interpolation method.

To study the effect of extrapolation.

Algorithm

Suppose we formulate a linear interpolating polynomial as the weighted average of the two values that we are connecting by a straight line:

$f(x)=L_{1}f(x_1)+L_{2}f(x_2)$

where the L$\text{'}$s are the weighting coefficients. It is logical that the first weighting coefficient is the straight line that is equal to $1$ at $x_{\text{I}}$ and $0$ at $x_2$ :

$L_1=\frac{x-x_2}{x_1-x_2}$

Similarly, the second coefficient is the straight line that is equal to $1$ at $x_2$ and $0$ at $x_1$ :

$L_2=\frac{x-x_1}{x_2-x_1}$

Substituting these coefficients into the first equation yields the straight line that connects the points:

$f_1(x)=\frac{x-x_2}{x_1-x_2}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)$

where the nomenclature $f1(x)$ designates that this is a first$-$order polynomial. This equation is referred to as the linear Lagrange interpolating polynomial.

The same strategy can be employed to fit a parabola through three points. For this case three parabolas would be used with each one passing through one of the points and equaling zero at the other two. Their sum would then represent the unique parabola that connects the three points. Such a second$-$order Lagrange interpolating polynomial can be written as:

$1$

$f_2(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}f(x_1)+\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}f(x_2)$

$+\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}f(x_3)$

Notice how the first term is equal to $f(x_l)$ at $x_l$ and is equal to zero at $x_2$ and $x_3.$ The other terms work in a similar fashion.

Both the first$-$ and second$-$order versions as well as higher$-$order Lagrange polynomials can be represented concisely as

$f_{n-1}(x)={\displaystyle \underset{i=1}{\overset{n}{}}}{L}_i(x)f(x_i)$

where

$]=[1$

where $n=$ the number of data points and I designates the "product of$."$

Practical Work

Preliminary Work $($to be done before lab session$)$

Use a Lagrange interpolating polynomial to find the solution $y$ at $x=15,$ based on the following values of xa and ya:

$xa=\left[\begin{array}{cccc}-40& 0& 20& 50\end{array}\right]$

$ya=\left[\begin{array}{cccc}1\mathrm{.}52& 1\mathrm{.}29& 1\mathrm{.}2& 1\mathrm{.}09\end{array}\right]$

Lab Work

Given the following Algorithm of LAGRANGE Interpolation formula:

Step $1.$ Start the Program

Step $2.$ Input number of terms $n$

Step $3.$ Input the array ax

Step $4.$ Input the array ay

Step $5.$ for $nr=1dr=1i=0$;$i$

Step $6.nr=1$

Step $7.dr=1$

$2$

$$

$$

Step$8.$ for $j$inr$=nr**(x-ax[j])dr=dr**(ax[i]-ax[j])Y=(n\frac{r}{d}r{)}^{**}ix,yj=0$;$j$

Step $9.Ifji$

$a.nr=nr**(x-ax[j])$

$b.dr=dr**(ax[i]-ax[j])$

Step $10.$ End Loop $j$

Step $11.Y=(n\frac{r}{d}r{)}^{**}ayi$

Step $12.$ End Loop $i$

Step $13.$ Print Output $x,y$

Step $14.$ End $of$ Program

Write a well commented MATLAB Program for the given Algorithm

$3$

Use your program to find the value of $y$ at $x=15,$ based on the following values of $xa$ and ya:

$xa=\left[\begin{array}{cccc}-40& 0& 20& 50\end{array}\right]$

$ya=\left[\begin{array}{cccc}1\mathrm{.}52& 1\mathrm{.}29& 1\mathrm{.}2& 1\mathrm{.}09\end{array}\right]$

Assume you have the following function $f(x)=x.{?}^{**}sin(2^{**}x).$ Plot the function $y=f(x)$ at $x=-1$:$0\mathrm{.}01$:$+1.$ Then use the Lagrange algorithm to find the values of $y$ at the given values of $x,$ if the given points to be used are:

$xp=[0,\frac{}{4},\frac{}{2},3**\frac{}{4},]$;

$yp=f(xp)$;

Plot the results on the same graph to detect the interpolation and extrapolation from the graph. $($Hint: Use the command hold on to plot more than one function at the same graph$)$

$4$

LAGRANGE INTERPOLATION POLYNOMIAI

$\backslash$table$[[$Name:$,$Grade $(5)],[$ID Number:,$],[$Lab Section:,$]]$

Objectives

Knowing how to perform an interpolation with a Lagrange polynomial,

To understand the MATLAB implementation of the Lagrange Interpolation method.

To study the effect of extrapolation.

Algorithm

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