Piecewise Cubic Hermite Interpolating Polynomial (PCHIP)

Hello friends

          I am doing cubic interpolation for the data x = [5.8808    6.5137    7.1828    7.8953];   y = [31.2472 33.9977 36.7661 39.3567];
pp = pchip(x,y);

Now pp gives coefficients of the polynomials but why they are not satisfying the data points, I am not able to understand, why is it happening like that?

If anyone knows about it, please let me know

Thanks

Bhomik

Answer 1

[Richard Brown]

It works perfectly

x = [5.8808    6.5137    7.1828    7.8953];   
y = [31.2472 33.9977 36.7661 39.3567];
pp = pchip(x,y);

You could draw a picture

xs = linspace(5, 8, 200);
ys = ppval(pp, xs);
plot(xs, ys, 'b-', x, y, 'ro')

Or test the function values

ppval(pp, x) - y

Where's the problem?

[Bhomik Luthra]

Hi Richard

            I meant to say that if you try this command
x = [5.8808    6.5137    7.1828    7.8953];   
y = [31.2472 33.9977 36.7661 39.3567];
pp = pchip(x,y)
and now see pp. It gives pp as

    form: 'pp'
    breaks: [5.8808 6.5137 7.1828 7.8953]
     coefs: [3x4 double]
    pieces: 3
     order: 4
       dim: 1
and pp.coefs are -0.0112 -0.1529 4.4472 31.2472

-0.3613 0.0884 4.2401 33.9977

-0.0422 -0.3028 3.8731 36.7661

I am not able to understand what these coefficients are? I think these are the polynomials representing the three intervals [5.8808:6.5137],[6.5137:7.1828],[7.1828:7.8953]

I tried to check the value of this fitting polynomial at the data points but it gives negative y values for second polynomial . Even third polynomial do not seem to satisfy the points.

I used these commands for obtaining the values

(for second polynomial)

xs = linspace(6.5137, 7.1828, 200);

y = polyval(pp.coefs(2,:),xs);

plot(xs,y)

I am trying this way because I wanted to find the area under the curve of the polynomials for my project, I am trying to obtain the functions and then integrating it over the x range.

But due to this issue, I think, these coefficients are not correct. If you know any other way to find the area of curve using your method, i.e. just by having the y values and x values, please let me know.

Thanks

Bhomik Luthra