Quadratic interpolation of an N dim array

Hello I would to know how I can perform a quadratic interpolation of an array using matlab ? it is worth mentioning that I am using interp1 for now. thank you!

Answer 1

[Ameer Hamza]

You can use polyfit(), and fit a polynomial of order two.

coefficients = polyfit(x, y, 2);

Then to interpolate the value use

 interpolatedValues = polyval(coefficients, xq); % xq vector of points where you want to interpolate

[Ano]

thank you for your reply! although it seems what I need but I have some troubles in getting the results as y is for me a matrix. any suggestions ?!

    Error using polyfit (line 47)
    X and Y vectors must be the same size.

[Ameer Hamza]

Yes, for interpolation. You first need a training set of x and y. They must have the same length. Can you show what are values in you X and Y variables?

[Ano]

Please find bellow how x and y should look like:

    x = [1 3 5];
    y(:,:,1) = [0.5 0.6 0.9 ; -1 -3 5;0.2 0.01 -3+1i];

it is worth mentioning that x is the third dim of y and they will not have the same length always. thank you!

[Ameer Hamza]

In training phase of interpolation you need to vector x and y of equal length. Because by definition of interpolation, you are trying to estimate a function f: x -> y, which will best (e.g. in the least square sense) model the relation between x and y. This requires training dataset to have the same length of x and y.

After training is complete, you can use f to estimate y value at points xq for which you don't know the actual value of y, using f(xq).

Answer 2

[Walter Roberson]

A = [x(:).^2, x(:), ones(numel(x),1)];
b = y;
coefficients = (A\b).';

This code is fine with x being either a row vector or column vector. The code expects that you are doing the quadratic fitting over each column of y independently (not over the rows.)

The array of coefficients that results will be N x 3 where N is the number of columns of y. The order of the coefficients will be the x^2 coefficient, then the x coefficient, then the constant.