### -0.02x3 +0.2x2-0.4x +1.28.

Polinómios de Lagrange,

 Polinómios de Lagrange Dados n+1 nós de interpolação x0 , ... , xn, definimos para cada i = 0, ..., n o polinómio de Lagrange li(x) de grau n tal que : Podemos deduzir uma expressão explícita dos polinómios de Lagrange. Fixando i e variando j = 0, ..., n , obtemos: E a constante Ci pode determinar-se, pois: Consequentemente: para i = 0, ..., n .

Agora, basta considerar a Fórmula Interpoladora de Lagrange:

 pn( x ) = f0 l0(x) + ... + fn ln(x)

que nos dá a expressão do polinómio interpolador, pois é fácil verficar que pn ( xi ) = fi .

### o código

14.  arquivo lagran.m no editor de MATLAB

function [C,L]=lagran(X,Y)

%Input  - X é o vetor com as abscisas dos pontos
%       - Y é o vetor com as ordenadas dos pontos
%Output - C é a matriz com os coeficientes com is a matrix that contains the coefficents of
%         the Lagrange interpolatory polynomial
%       - L is a matrix that contains the Lagrange
%         coefficient polynomials

15.  Thus to call this function we set up the vectors X and Y with the x and y coefficients of the interpolating points.  Then call the function to return the interpolating polynomial in C and the Lagrange coefficients for that polynomial in L.  For our above example, it would be:

» X = [1 2 2.5]
» Y = [3 3 3.3]
» [C L] = lagran(X,Y)

We see that the same answer is returned as the one we computed step by step above.

Look at how the coefficients are computed in the body of the function

for k=1:n+1    %  Calculate each of n+1 Lagrange coefficient
V=1;        %  Accumulate computations in V temporarily
for j=1:n+1
% Multiply by (x - X(j))/(X(k) - X(j))
if k~=j  %  Be sure to skip the k'th one
V=conv(V,poly(X(j)))/(X(k)-X(j));
end
end
L(k,:)=V;    %  Store Lagrange coefficient in kth row of L
end

The final step is to compute the interpolating polynomial:

C = Y*L

This uses matrix multiplication to compute C =  Y * L. It is not difficult to verify using the rules for matrix multiplication that this gives the correct polynomial.  For example, the first entry in C will be
Y(1)*L(1,1) + Y(2)*L(2,1) + Y(3)*L(3,1)  which is the correct coefficient of x^2 term.

16.  Exercise:  Use lagran to calculate the third degree Lagrange polynomial for cos(x) at the evenly spaced interpolating points for the x-values (abscissas) of 0.0, 0.4, 0.8, and 1.2  (see page 210-211, Example 4.7, part b), text for the computations and page 210 figure 4.12b)  comparing the graph of this polynomial to the graph for  function cos(x).

» X = [0.0 0.4 0.8 1.2]

» Y = cos(X)

» [C , L ] = lagran(X,Y)

Compare the graph of  y = cos(x) to your computed polynomial with coefficients stored in C with the following commands

» SP = poly2sym(C)
» clf
» ezplot('cos(x)',0,2)
» hold on
» ezplot(SP,0,2)

We see they look quite close at least through the interpolating range 0 to 1.2.

Use the error formula of Theorem 4.4 (see page 213 text) to compute an upper bound of the error for these nodes (interpolating points).
h = distance between evenly spaced nodes = .4 in this case.
M4 is the maximum of the absolute value of the  4th derivative of f(x) = cos(x)  on the interpolation range -- we can use 1 for this.   Thus our error should be <= (.4)^4/25:

» (.4)^4/24

Does this appear to be true? --Calculate some of the errors  with the following commands  (and compare to table 4.7, page 216)

» cos(.1)
» polyval(C,.1)
» cos(.1)-polyval(C,.1)

### Laboratory Exercise to Turn In:   Page 220 Algorithms and Programs 2 a) b) c)

Print your graph for c). Write on the print out what the approximating polynomial is returned by lagran for part a)
For part b)  also write on your printout, what the integral polynomial was, and your answer for the average.
Turn in by next lab period, Friday Feb 24th.

For part a)   Use the lagran function.  Set up your X, Y vectors   ( X is time values,  Y is temp values ).

For part c)  Use the following commands to plot your data points and the corresponding Lagrange polynomial stored in C
» SP = poly2sym(C)
» clf                         Clear any previous figures
» plot(X,Y,'r*')           Plot the original data points in blue starts
» hold on;                Retain the graph for next plot also.
» ezplot(SP,0,7)

For part b -- (See Theorem 1.10 -- Mean Value Theorem for Integrals, page 6).   Average temperature is the integral of the temperature function (your approximating polynomial for temperature)  over the limits of the time interval  divided by the length of the time interval.

Thus you need to: