MATLAB: Evaluating polynomial at a certain x point to find y

anonanonymous functionfittingfor looppolyfitpolynomialpolyval

Hi all, so I have a script that evaluates the 1st through 4th degree polynomials of a given dataset of volume of liquid vs vol. of ice. I already did everything and plotted it just fine, the only thing I want to do is evaluate it for when x = 1.5 (vol. of liquid) for the 4th degree polynomial.

The script: https://pastebin.com/tcuM5bBk

What I wanted/tried to do is just evaluate the anonymous function, but it's within the for-loop and I don't know to pull the 4th iteration so I can evaluate it at 1.5. Any tips?

Best Answer

As you have (x,y) data in hand....you can use interpolation. Read about interp1.

iwant = interp1(x,y,1.5) ;

Related Solutions

MATLAB: Polyfit not showing a curve

polyfit only get you the coefficients, you then have to actually evaluate some values to get the curve. An easy alternative is to use the function polyval (https://de.mathworks.com/help/matlab/ref/polyval.html), which will evaluate the given polynom for the given points. The following code should give you the result you would expect:

% store coordinates
x = [5.2 5.5 5.6 5.8];
y = [6.408 16.125 19.816 27.912];
 
% use polyfit to get coefficients of cubic
p = polyfit(x, y, 3);
yfit = polyval(p,x); % Evaluate polynomial coefficients p in values x
scatter(x, y, 'black');
hold on
plot(x,yfit, 'green');

MATLAB: Using roots with vectorization of a polynomial vs. the function it’s self

Look at what the roots function is documented to do.

r = roots(p) returns the roots of the polynomial represented by p as a column vector. Input p is a vector containing n+1 polynomial coefficients, starting with the coefficient of xn.

It does not accept a vector of values of your function and find the zeros of the function that takes those values. In other words, since x has 201 elements so does y and so r is the roots of a 200th degree polynomial not the 5th degree polynomial you evaluated to compute y. Zero crossings in a vector is better done by comparing the sign of the elements in that vector, or perhaps interpolation and fzero.

On the other hand, y1 does not represent the polynomial you evaluated to compute y. You're missing the constant term, which has to be present in the coefficient vector even if it's 0. But it's closer.

Best Answer

Related Solutions

MATLAB: Polyfit not showing a curve

MATLAB: Using roots with vectorization of a polynomial vs. the function it’s self

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