Newton Forward Difference Interpolating Polynomials

Hi guys, I have a quick question. I'm building a Newton Forward Difference method in Matlab, but I don't get how to set up an equations.

Here is what I have so far;

 function yi = Newton_FD(x, y, xi)

% this function computes the interpolating polynomials
% for the given data, x and y, using Newton's forward-
% difference formula. The polynomials of degree
% 1, 2, ..., n are computed, where n is one less than the
% number of data points. The polynomials are then evaluated
% at xi.
% n is the maximum degree possible with the data
n=length(y)-1;
% Initialize difference table
d=y;
for i=1:n
    for j=n+1:-1:i+1,
        d(j)=(d(j-1)-d(j))/(x(j-1)-x(j));
    end
    % use nested multiplication to evaluate the
    % polynomial at xi
    yi=d(i+1);
    for j=i:-1:1,
        yi=d(j)+(xi-x(j))*yi;
    end
    % display coefficients of Newton form of interpolating
    % polynomial, and the value at xi
    disp(sprintf('Degree %d:',i))
    fprintf('\n')
    fprintf('Coefficients=')
    fprintf('\n\n')
    Coefficients=d(1:i+1);
    disp(Coefficients)
    fprintf('Value=')
    fprintf('\n\n')
    Value=yi;
    disp(Value)
end

and given values are:

x=[1;2;3;4;5;6;7;8;9;10];
y=[1;0.4444;0.2632;0.1818;0.1373;0.1096;0.0929;0.0775;0.0675;0.0597];

But, somehow it is giving me a wrong values for coefficients. Any suggestions?

Answer 1

[Sourabh ahlawat]

function yint = Newtint(x,y,xx)

% Newtint: Newton interpolating polynomial

% yint = Newtint(x,y,xx): Uses an (n - 1)-order Newton

% interpolating polynomial based on n data points (x, y)

% to determine a value of the dependent variable (yint)

% at a given value of the independent variable, xx.

% input:

% x = independent variable

% y = dependent variable

% xx = value of independent variable at which

% interpolation is calculated

% output:

% yint = interpolated value of dependent variable

% compute the finite divided differences in the form of a

% difference table

n = length(x);

if length(y)~=n, error('x and y must be same length'); end

b = zeros(n,n);

% assign dependent variables to the first column of b.

b(:,1) = y(:); % the (:) ensures that y is a column vector.

for j = 2:n

for i = 1:n-j+1

b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i));

end

end

% use the finite divided differences to interpolate

xt = 1;

yint = b(1,1);

for j = 1:n-1

xt = xt*(xx-x(j));

yint = yint+b(1,j+1)*xt;

end

Answer 2

[gizem]

Do you have any suggestions on what kind of code I should write so that the coefficients are given in front of the x?

For example;ax^5+bx^4+cx^3...............

[Aissam]

if true
  % code
function p = newtint(x, y)
n=length(y);
d=zeros(n);
d(:,1)=y;
for j=2:n
  for i=j:n
      d(i,j)=(d(i,j-1)-d(i-1,j-1))/(x(i)-x(i-j+1));
  end
end
d
pstr="";
pstr=string(d(1,1));
for i=2:n
  pstr=pstr+"+";
  for j=1:i-1
      pstr=pstr+"(x-"+string(x(j))+")*";
  end
  pstr=pstr+string(d(i,i));
end
syms x p(x);
p(x)=simplify(str2sym(pstr));
end
Example:
x=[1:5]'
y=log(x)
p=newtint(x,y)
p(1.5)

Answer 3

[sakshi kumari]

function yi = Newton_FD(x, y, xi)

d=y;

for i=1:n

for j=n+1:-1:i+1,

d(j)=(d(j-1)-d(j))/(x(j-1)-x(j));

end

% use nested multiplication to evaluate the

% polynomial at xi

yi=d(i+1);

for j=i:-1:1,

yi=d(j)+(xi-x(j))*yi;

end

% display coefficients of Newton form of interpolating

% polynomial, and the value at xi

disp(sprintf('Degree %d:',i))

fprintf('\n')

fprintf('Coefficients=')

fprintf('\n\n')

Coefficients=d(1:i+1);

disp(Coefficients)

fprintf('Value=')

fprintf('\n\n')

Value=yi;

disp(Value)

end

x=[1.0;1.1;1.2;1.3;1.4;1.5;1.6];

y=[7.989;8.403;8.781;9.129;9.451;9.750;10.031];