How do I use the ppval, pp and spline in this situation?

I have the following in a code:

pp=spline(diameter,nbdensity);

all(i,j) = ((diameter(i,1)^3)-(diameter(j,1)^3))^(1/3);       
finp(i,j) = ppval(pp,all(i,j));

I need to find the interpolated value of "nbdensity" at a given value of "all". How would I express it using the above? Would the following be correct?

nbdensity(i)*finp(i,j)

[Ameer Hamza]

ppval(pp,all(i,j)) IS the interpolated value you are looking for

[Sakina]

How sure about this are you? My future depends on your answer.

[Ameer Hamza]

Yes, this is correct. You can read the explanation in my answer below.

[Sakina]

Okay, thank you very much!

[Ameer Hamza]

You are welcome.

Answer 1

[Ameer Hamza]

Given data vectors x and y. f = spline(x, y) function returns a piecewise-polynomial which models the best relation (in the least square sense) f:x->y. After estimating the polynomial, if you want to find the value of y at some arbitrary value of x, you can just evaluate piecewise polynomial f at. In MATLAB you can use ppval function to evaluate the piecewise polynomial f.