# Two dimensional interpolation polynomial

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Jaime De La Mota Sanchis
on 25 Sep 2019

Answered: Sharon Dolberg
on 24 Dec 2022

Hello everyone.

I have a file which size is 681*441, it contains values measured at 300321 points. I want to find the interpolation polynomial. I have looked for functions which return what I am seeking; but I have found none. griddedinterpolant does not return the polynomial in question, but only values in a more precise grid. On the other hand, the function polyfit does provide with the nodes of a polynomial of any desired order, but it seems to work only in one dimension.

Can someone please tell me if there is a way to do as in the function polyfit but returning a two dimensional polynomial?

Thanks.

Jaime.

##### 4 Comments

David K.
on 25 Sep 2019

Hm, actually I do not know if I understand your question. I thought you had data and was looking for the polynomial equation that fit it, but you already have said equation. Is this just a test example and you plan to apply to actual data to find the equations for?

You are also mentioning interpolation and I am not sure why gridded interpolant ot interp2 would not work with what you need it for.

### Accepted Answer

David K.
on 25 Sep 2019

First off, I looked a little bit on wikis to find https://en.wikipedia.org/wiki/Multivariate_interpolation which might have what you want but I am not sure. However, the way I wanted to do it is with fmincon. I attached the objective function I created. Basically it creates a function of the form

a1x^2+a2y^2+a3xy+a4x+a5x+a6

then it loads in the X,Y, and wind data that you have and determines how closely it fits by a basic difference and summing.

This is utilized by fmincon as such:

a = fmincon(@objfun, rand(6,1), eye(6),ones(6,1).*20)

I decided to make the initial conditions random just because. Then, because this is a constrained minimum search, I just made it so that no value could go over 20, it would also likely work with large constraints as well but I doubt they will come into affect much. I tried using other optimization algorithms that did not require constraints but they did not work as well. For the example you gave the output should be

a = [0 1 0 1 0 0]

It generally gets pretty close, sometimes the constant value messes up though.

##### 4 Comments

David K.
on 30 Sep 2019

### More Answers (2)

John D'Errico
on 26 Sep 2019

I don't think you realize that an interpolation polynomial that passes exctly through 300321 points will be impossible to evaluate in double precision arithmetic. In fact, it may require a precision that is on the order of many thousand of decimal digits to get any thing out if it. Possibly hundreds of thousands of digits will be required.

And then, expect complete crapola anyway. Why? Because the polynomial might actually predict the points, but give you meaningless garbage between the points!

##### 0 Comments

Sharon Dolberg
on 24 Dec 2022

I wrote a multidimensional polyval function, a bit different from the original polyval. May it assist the users:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [a,ErrV]=polyvalN(MatIn,TargetIn)

[n,m]=size(MatIn);

MatOut=sum(pagemtimes(reshape(MatIn',m,1,n),reshape(MatIn',1,m,n)),3);

TargetOut=sum(MatIn.*TargetIn)';

a=MatOut\TargetOut;

ErrV=MatIn*a-TargetIn;

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

This function takes in a matrix MatIn which already includes the approximated structure for a linear approximated function (polynom) which the coefficient (a) of the approximated function needs to be found according to the target values TargetIn

for example:

------------

We have 8 points on a box

X=[-1;1;-1;1;-1;1;-1;1];

Y=[-1;-1;1;1;-1;-1;1;1]*2;

Z=[-1;-1;-1;-1;1;1;1;1]*0.5;

The approximated function which we intend to find the coefficients for is:

MatIn=[ones(8,1), X , Y , Z , X.*Y , X.*Z , Y.*Z ];

The value in each point is given. We can use rand or any known values in each point. Just in order to check the function polyvalN, we'll use known a's

intentionally:

TargetIn= 7 + 6*X + 5*Y + 4*Z + 3*X.*Y + 2*X.*Z + 1*Y.*Z;

Now, is we pose MatIn,TargetIn to polyvalN , we expect to get a's

[7;6;5;...1]

-------------

Notes:1. MatIn should be nXm wherein n>=m

TargetIn should be nx1

2. The point represented by X,Y,Z are not necessarily be in a box

structure as in the example. Actually, they can also represent any

single-dimension/multidimensional location, as long as note 1 is kept, and that MatIn includes

the vectors in the approximation.

##### 0 Comments

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