Polynomial curve fitting - MATLAB polyfit - MathWorks United Kingdom (2024)

Generate 10 points equally spaced along a sine curve in the interval [0,4*pi].

x = linspace(0,4*pi,10);y = sin(x);

Use polyfit to fit a 7th-degree polynomial to the points.

p = polyfit(x,y,7);

Evaluate the polynomial on a finer grid and plot the results.

x1 = linspace(0,4*pi);y1 = polyval(p,x1);figureplot(x,y,'o')hold onplot(x1,y1)hold off

Create a vector of 5 equally spaced points in the interval [0,1], and evaluate $$y(x)=(1+x{)}^{-1}$$ at those points.

x = linspace(0,1,5);y = 1./(1+x);

Fit a polynomial of degree 4 to the 5 points. In general, for n points, you can fit a polynomial of degree n-1 to exactly pass through the points.

p = polyfit(x,y,4);

Evaluate the original function and the polynomial fit on a finer grid of points between 0 and 2.

x1 = linspace(0,2);y1 = 1./(1+x1);f1 = polyval(p,x1);

Plot the function values and the polynomial fit in the wider interval [0,2], with the points used to obtain the polynomial fit highlighted as circles. The polynomial fit is good in the original [0,1] interval, but quickly diverges from the fitted function outside of that interval.

figureplot(x,y,'o')hold onplot(x1,y1)plot(x1,f1,'r--')legend('y','y1','f1')

First generate a vector of x points, equally spaced in the interval [0,2.5], and then evaluate erf(x) at those points.

x = (0:0.1:2.5)';y = erf(x);

Determine the coefficients of the approximating polynomial of degree 6.

p = polyfit(x,y,6)

p = 1×7 0.0084 -0.0983 0.4217 -0.7435 0.1471 1.1064 0.0004

To see how good the fit is, evaluate the polynomial at the data points and generate a table showing the data, fit, and error.

f = polyval(p,x);T = table(x,y,f,y-f,'VariableNames',{'X','Y','Fit','FitError'})

T=26×4 table X Y Fit FitError ___ _______ __________ ___________ 0 0 0.00044117 -0.00044117 0.1 0.11246 0.11185 0.00060836 0.2 0.2227 0.22231 0.00039189 0.3 0.32863 0.32872 -9.7429e-05 0.4 0.42839 0.4288 -0.00040661 0.5 0.5205 0.52093 -0.00042568 0.6 0.60386 0.60408 -0.00022824 0.7 0.6778 0.67775 4.6383e-05 0.8 0.7421 0.74183 0.00026992 0.9 0.79691 0.79654 0.00036515 1 0.8427 0.84238 0.0003164 1.1 0.88021 0.88005 0.00015948 1.2 0.91031 0.91035 -3.9919e-05 1.3 0.93401 0.93422 -0.000211 1.4 0.95229 0.95258 -0.00029933 1.5 0.96611 0.96639 -0.00028097 ⋮

In this interval, the interpolated values and the actual values agree fairly closely. Create a plot to show how outside this interval, the extrapolated values quickly diverge from the actual data.

x1 = (0:0.1:5)';y1 = erf(x1);f1 = polyval(p,x1);figureplot(x,y,'o')hold onplot(x1,y1,'-')plot(x1,f1,'r--')axis([0 5 0 2])hold off

Create a table of population data for the years 1750 - 2000 and plot the data points.

year = (1750:25:2000)';pop = 1e6*[791 856 978 1050 1262 1544 1650 2532 6122 8170 11560]';T = table(year, pop)

T=11×2 table year pop ____ _________ 1750 7.91e+08 1775 8.56e+08 1800 9.78e+08 1825 1.05e+09 1850 1.262e+09 1875 1.544e+09 1900 1.65e+09 1925 2.532e+09 1950 6.122e+09 1975 8.17e+09 2000 1.156e+10

plot(year,pop,'o')

Use polyfit with three outputs to fit a 5th-degree polynomial using centering and scaling, which improves the numerical properties of the problem. polyfit centers the data in year at 0 and scales it to have a standard deviation of 1, which avoids an ill-conditioned Vandermonde matrix in the fit calculation.

[p,~,mu] = polyfit(T.year, T.pop, 5);

Use polyval with four inputs to evaluate p with the scaled years, (year-mu(1))/mu(2). Plot the results against the original years.

f = polyval(p,year,[],mu);hold onplot(year,f)hold off

Fit a simple linear regression model to a set of discrete 2-D data points.

Create a few vectors of sample data points (x,y). Fit a first degree polynomial to the data.

x = 1:50; y = -0.3*x + 2*randn(1,50); p = polyfit(x,y,1); 

Evaluate the fitted polynomial p at the points in x. Plot the resulting linear regression model with the data.

f = polyval(p,x); plot(x,y,'o',x,f,'-') legend('data','linear fit') 

Fit a linear model to a set of data points and plot the results, including an estimate of a 95% prediction interval.

Create a few vectors of sample data points (x,y). Use polyfit to fit a first degree polynomial to the data. Specify two outputs to return the coefficients for the linear fit as well as the error estimation structure.

x = 1:100; y = -0.3*x + 2*randn(1,100); [p,S] = polyfit(x,y,1)

p = 1×2 -0.3142 0.9614

S = struct with fields: R: [2x2 double] df: 98 normr: 22.7673 rsquared: 0.9407

Evaluate the first-degree polynomial fit in p at the points in x. Specify the error estimation structure as the third input so that polyval calculates an estimate of the standard error. The standard error estimate is returned in delta.

[y_fit,delta] = polyval(p,x,S);

Plot the original data, linear fit, and 95% prediction interval $\mathit{y}\pm 2\Delta$.

plot(x,y,'bo')hold onplot(x,y_fit,'r-')plot(x,y_fit+2*delta,'m--',x,y_fit-2*delta,'m--')title('Linear Fit of Data with 95% Prediction Interval')legend('Data','Linear Fit','95% Prediction Interval')

p — Least-squares fit polynomial coefficients
vector

S — Error estimation structure
structure

mu — Centering and scaling values
two-element vector