Polyfit minus one term

Hi, and happy new year.

I need a 3rd order polynomiial approximation for the file Data001 in attached.

I can use polyfit but I need a polinomium without the term x^2.

My polinomium must be:

x^3+x^1+x^0.

How can I do that?

Answer 1

[Star Strider]

Use the mldivide,\ function with a specific design matrix (since this is a linear problem):

D = load('Data001.mat');

x = D.Data001(:,1);
y = D.Data001(:,2);

DesignMatrix = @(x) [x.^3 x ones(size(x))];
B = DesignMatrix(x) \ y;

xv = linspace(min(x), max(x), 1000);
yv = DesignMatrix(xv(:)) * B;

figure
plot(x, y, '.')
hold on
plot(xv(:), yv(:), '-r')
hold off
grid
xlabel('x')
ylabel('y')
legend('Data001', 'Regression Fit')

producing:

The ‘B’ vector are the respective regression coefficients, so that y(x) = B(1)*x^3 + B(2)*x + B(3).

[Paul Rogers]

oh my god, thank you. One thing I asked wrong,

I need x^3+x^2+x^0. The one I don't need is the therm x^1

[Star Strider]

My pleasure!

I was confused by your original Question, since you only left out ‘x^2’.

To leave out ‘x^1’, change ‘DesignMatrix’ to:

DesignMatrix = @(x) [x.^3 x.^2 ones(size(x))];

Also, your vectors are column vectors here, and this will only work with column vectors. To enforce that if you have row vectors, the ‘B’ calculation becomes:

B = DesignMatrix(x(:)) \ y(:);

since the ‘(:)’ will force ‘x’ and ‘y’ to become column vectors.

With that change, the ‘B’ vector are the respective regression coefficients, so that y(x) = B(1)*x^3 + B(2)*x^2 + B(3).

[Star Strider]

If my Answer helped you solve your problem, please Accept it!

Answer 2

[dpb]

Without toolbox, use backslash operator, \

X=[x.^[3 2 1 0]]; X(:,2)=0;    % design matrix straight calculation
b=X\y;                         % solve for coefficients

If have Stat or Curve Fitting TB, use one of the linear model fitting routines with custom model.