Hermitian (Hermite) interpolation and its MATLAB program

%hermite.m% of Hermitian polynomial and error estimate matlab Main program% input amount: X is n+1 node (x_i,y_i) (i =, ..., n+1) horizontal vector, y is the ordinate vector,% with F ' (x_i) =y ' _i (i =,...
, n+1) is the vector Y1 of the element; %x is an M interpolation point entered as a vector, M satisfies |f~ (2n+2) (x) |≤m% in [A, b]: f~ (2n+2) (x) represents the 2n+2 derivative of f (x), the amount of the output: vector y is the interpolation at vector x, error limit r,2n+1 order Hermitian interpolation polynomial h_
K and its coefficients vector%h_c, error formula Wcgs and its coefficient vector cw.
FUNCTION[Y,R,HC,HK,WCGS,CW] = Hermite (x,y,y1,x,m) n = length (X);


m = length (x);
For t = 1:m z = x (t);
H = 0;
Q = 1;

C1 = 1;
    For k = 1:n s = 0;
    V = 1;
            For i = 1:n if k ~= i s = s + (1/(X (k)-X (i)));
        V = CONV (V,poly (x (i)))/(x (k)-X (i));
    End end h = Poly (X (k));
    g = ([0 1]-2 * H * s);% note Do not write 1-2*h*s, because it is a polynomial subtraction, the lower order polynomial front must be filled with 0, the book error, Wasting my night time g = g * Y (k) + H * Y1 (k);
    H = h + conv (G,conv (v,v));%hermite interpolation polynomial B = poly (X (k));
    b2 = conv (b,b);
Q = conv (Q,B2);
End Hc = H;
Hk = Poly2sym (H);
Q = poly2sym (q);
For i = 1:2*n C1 = C1 * I;
End WCGS = M * Q/C1;
Cw = M * Q/C1;
Y (t) = Polyval (hc,x (t));
R (t) = Polyval (cw,x (t)); End

%hermite.m% of Hermitian polynomial and error estimate matlab Main program% input amount: X is n+1 node (x_i,y_i) (i =, ..., n+1) horizontal vector, y is the ordinate vector,% with F ' (x_i) =y ' _i (i =,...
, n+1) is the vector Y1 of the element; %x is an M interpolation point entered as a vector, M satisfies |f~ (2n+2) (x) |≤m% in [A, b]: f~ (2n+2) (x) represents the 2n+2 derivative of f (x), the amount of the output: vector y is the interpolation at vector x, error limit r,2n+1 order Hermitian interpolation polynomial h_
K and its coefficients vector%h_c, error formula Wcgs and its coefficient vector cw.
FUNCTION[Y,R,HC,HK,WCGS,CW] = Hermite (x,y,y1,x,m) n = length (X);


m = length (x);
For t = 1:m z = x (t);
H = 0;
Q = 1;

C1 = 1;
    For k = 1:n s = 0;
    V = 1;
            For i = 1:n if k ~= i s = s + (1/(X (k)-X (i)));
        V = CONV (V,poly (x (i)))/(x (k)-X (i));
    End end h = Poly (X (k));
    g = ([0 1]-2 * H * s);% note Do not write 1-2*h*s, because it is a polynomial subtraction, the lower order polynomial front must be filled with 0, the book error, Wasting my night time g = g * Y (k) + H * Y1 (k);
    H = h + conv (G,conv (v,v));%hermite interpolation polynomial B = poly (X (k));
    b2 = conv (b,b);
Q = conv (Q,B2);
End Hc = H;
Hk = Poly2sym (H);
Q = poly2sym (q);
For i = 1:2*n C1 = C1 * I;
End WCGS = M * Q/C1;
Cw = M * Q/C1;
Y (t) = Polyval (hc,x (t));
R (t) = Polyval (cw,x (t));
 End

From this picture is better than Lagrange and Newton interpolation, but when the interpolation is more, that is, the number of Hermitian polynomial is high after the error will become very large. Don't believe us to try.

We still use Sinx, but this time we use 5 points.

X 0 PI/6 Pi/4 PI/3 Pi/2

Y 0 0.5 0.7071 0.8660                       1 Y1 1 0.8660 0.7071 0.5 0

>> X = [0 pi/6 pi/4 PI/3 PI/2]; Y = [0 0.5 0.7071 0.8660 1]; Y1 = [1 0.8660 0.7071 0.5 0];
>> M = 1;
>> x = linspace (0, Pi, +);
>> [Y,R,HC,HK,WCGS,CW] = Hermite (x,y,y1,x,m)
>> y1 = sin (X);  
>> ErrorBar (X,y,r, '. G ')  
>> hold on  
>> plot (x, Y, ' or ', x, Y, '. k ', X, Y1, '-B ');  
>> legend (' Error ', ' Sample point ', ' Hermitian interpolation estimate ', ' sin (x) ');  

It can be seen that the fitted polynomial deviates from the Sinx from the x=2.5, and the error formula also fails. This is the high-level interpolation oscillation problem we need to talk about later.