Newton Interpolation - matlab

Question

Hi I have this function to calculate the coefficient list for the Newton polynomial:
function p = polynom(x,y,c)
m = length(x);
p = c(m)*ones(size(y));
for k = m-1:-1:1
p = p.*(y-x(k)) + c(k);
end
I already have another program that finds the divided differences c correctly. For x=[3 1 5 6], y=[1 -3 2 4] I get c=[1.0000 2.0000 -0.3750 0.1750] which is correct.
But when I use the above function it gives as a result:
p =
-3.0000 -53.6000 -0.1000 1.3500
But the correct answer should be:
p =
0.1750 -1.9500 7.5250 -8.7500
What is wrong with my function?

Answer 1

This is the code I've used. I've used it long time ago, so I'll post whole code. I hope this is what you need.
table.m
%test table
X = [1 1.05 1.10 1.15 1.2 1.25];
Y = [0.68269 0.70628 0.72867 0.74986 0.76986 0.78870];
position.m
%M-file position.m with function position(x)
%that for argument x returns
%value 1 if x<X(2),
%value 2 if x>X(n-1)
%else 0
function position=position(x)
table;
n=length(X);
if x<X(2)
position=1;
else
if x>X(n-1)
position=2;
else position=0;
end
end
Newton.m
function Newton=Newton(x)
table;
%position(x);
n=length(X);
A=zeros(n,n+1);
% table of final differences - upper triangle
for i=1:n
A(i,1)=X(i);
A(i,2)=Y(i);
end
for i=3:n+1
for j=1:(n-i+2)
A(j,i)=A(j+1,i-1)-A(j,i-1);
end
end
%showing matrix of final differences, A
A
h=X(2)-X(1);
q1=(x-X(1))/h;
q2=(x-X(n))/h;
s1=q1;
s2=q2;
%making I Newton polynomial
if (position(x)==1)
p1=A(1,2);
f=1;
for i=1:n-1
f=f*i;
p1=p1+(A(1,i+2)*s1)/f;
s1=s1*(q1-i);
end
Newton=p1;
else
%making II Newton polynomial
if (position(x)==2)
p2=A(n,2);
f1=1;
for i=1:n-1
f1=f1*i;
p2=p2+(A(n-i,i+2)*s2)/f1;
s2=s2*(q2+i);
end
Newton=p2;
%else, printing error
else Newton='There are more suitable methods than Newton(I,II)';
end
end