Question

1 Interpolation methods 1.1 The Lagrange polynomial function (y dot,p] = my Lagrange (x, y,x_dot) Type Description Input 1xM double 1xM double A row vector with the x data that you should fit your Lagrange polynomial to A row vector with the y data that you should fit your Lagrange polynomial to A row vector of target points that you should evaluate your Lagrange polynomial at x_dot 1xN double Output y-dot 1xN double The value of your Lagrange polynomial at the given points The polynomial coefficients of the Lagrange polynomial (from highest to lowest order) 1xM double Details The Lagrange polynomial for a given data set is defined as the polynomial of least degree which passes through all the points of the data set. For a set of M data points, the Lagrange polynomial is guaranteed to be of degree (M-1) or lower, and can therefore be written in the following form: L(x) = am-12M-1 +am-22M-2 + ... +ar+ao where x is the independent variable and ai, i = 0,1,..., M-1 are the coefficients of the poly- nomial.

I'm not allowed to use polyfit or polyval functions.

The hint the homework gives is setting up a linear system for the polynomial coefficients and solving it.

The test case x = [1,2,3,4]

y = [5,-2,3,0]

x_dot = [1.5,2.5]

should result in

y_dot = -1.2500 0.2500

p = -3.333 26.0000 -61.6667 44.0000

Thanks to anyone who can answer this within the next day or so!

Answer 1

`Hey,

Note: If you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

clc%clears screen
clear all%clears history
close all%closes all files
format long
x = [1,2,3,4];
y = [5,-2,3,0];
x_dot = [1.5,2.5];
[y_dot,p]=myLagrange(x,y,x_dot)

function [y_dot,p]=myLagrange(x,y,x_dot)
f=y;
X=x;
Y=f;
N = length(X);
pvals = zeros(N,N);
for i = 1:N
% the polynomial with roots may be values of X other than this one
pp = poly(X( (1:N) ~= i));
pvals(i,:) = pp ./ polyval(pp, X(i));
end
P = Y*pvals;
p=P;
y_dot=polyval(P,x_dot);
end

Kindly revert for any queries

Thanks.