How to interpolate a set of data with the cubic method?

I tried looking it up, and it comes out that i should use the spline command, but when i use it nothing comes out. I need to plot the data and the interpolant. I also need to find x for when y= 600

x = [74 29 21 12 8 5.7 4.4 3.6 2.1 1.8 1.5 1.0 0.7];
y = [80 131 189 270 320 407 450 530 620 686 740 900 1095];
s = spline(x, y, 600);

[KSSV]

600 falls outside the range of x.....this is extrapolation and cnnot be trusted. Are you sure you want for x = 600?

[Pipe]

You are right, i meant y=600

[John D'Errico]

Note that this is rather noisy looking data, and it has large changes in slope. That makes a cubic spline a terribly poor choice to model that curve, even if it appears it survived the process and produced a monotonic curve. Using a higher order interpolant to model noise is often a bad idea.

Instead, use pchip, a tool designed not to introduce spurious extrema and non-monotonic behavior into a problem. The curve will still come out looking smooth.

Answer 1

[KSSV]

x = [74 29 21 12 8 5.7 4.4 3.6 2.1 1.8 1.5 1.0 0.7];
y = [80 131 189 270 320 407 450 530 620 686 740 900 1095];

yi = linspace(min(y),max(y)) ;
xi = interp1(y,x,yi,'spline') ;

plot(x,y,'*r',xi,yi,'b')

x_600 = interp1(y,x,600,'spline') ;

[John D'Errico]

Note that this is rather noisy looking data, and it has large changes in slope. That makes a cubic spline a terribly poor choice to model that curve, even if it appears it survived the process and produced a monotonic curve. Using a higher order interpolant to model noise is often a bad idea.

Instead, use pchip, a tool designed not to introduce spurious extrema and non-monotonic behavior into a problem. The curve will still come out looking smooth.