Matrix is close to singular or badly scaled. HELP!

Hello guys, I am new here, I am doing some project on interpolation and approximation. I have been given values for x and y, now I have to get the polynomials for them. I've done something, but the values of coefficients for approximation and those for interpolation are totally different, so something's wrong. Can someone just help me? Here is what I did till now:

% APPROXIMATION

%________________________________________________________________________

format long g;

% Input values for x and y

x1 = 0.131568108472; y1 = 258.433497034; x2 = 0.199557064651; y2 = 94.8235556162; x3 = 0.596426092301; y3 = 46.4773492701; x4 = 10.1301082178; y4 = 4.40548687035; x5 = 350.467757284; y5 = 0.770454788749; x6 = 1271.93067998; y6 = 0.468660756742; x7 = 12162.8432144; y7 = 0.409611101849; x8 = 59113.7385534; y8 = 0.458959722808; x9 = 287130.854195; y9 = 0.414685133805;

%Matrix form of approximation % A * x = b;

b = [y1; y2; y3; y4; y5; y6; y7; y8; y9];

v1 = [x1^2, x1, 1]; v2 = [x2^2, x2, 1]; v3 = [x3^2, x3, 1]; v4 = [x4^2, x4, 1]; v5 = [x5^2, x5, 1]; v6 = [x6^2, x6, 1]; v7 = [x7^2, x7, 1]; v8 = [x8^2, x8, 1]; v9 = [x9^2, x9, 1];

A = [v1;v2;v3;v4;v5;v6;v7;v8;v9]; %Augmented matrix:

 AugA = A'*A; 
 Augy = A'*b;
% 
 x = inv(AugA)*Augy; % the coefficients of the quadratic equation

a = x(1); b = x(2); c = x(3); %approximated_function = a*x^2 + b*x + c;

%________________________________________________________________________

% INTERPOLATION

x = [x1, x2, x3, x4, x5, x6, x7, x8, x9]; y = [y1, y2, y3, y4, y5, y6, y7, y8, x9];

Dy(1) = (y(2)-y(1))/(x(2)-x(1)); Dy(2) = (y(3)-y(2))/(x(3)-x(2));

Dy2(1) = (Dy(2) - Dy(1))/(x(3)-x(1));

a1 = y1; a2 = Dy(1); a3 = Dy2(1);

a, b, and c are coefficients for approximation a1, a2, a3 are those for interpolation

Why are they so different, I think something is wrong with approximation as MATLAB gives me info on error:

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 9.495734e-022. > In approximation at 43

Answer 1

[Sean de Wolski]

range(AugA(:))

You'll see that the difference is on the order of 10^21. Double precision floating point values only have an accurate range of 16 decimal digits.

http://en.wikipedia.org/wiki/IEEE_754-2008

You'll need to either:

• use the symbolic toolbox
• use John's vpi (variable precision integers (constrained to integers only))
• reformat your problem

[Ensar Jusufovic]

I will try, thank you for your response.