How to get polynomial equation of polyfit?

Hello. I have the following code and the data. I need to find the equation of the polynomial and write it on the graph. Also, I will be changing the x and y datas so, it needs to be specific for a set of datas. Thank you for your help! It will be appreciated!!

clc;
clear;
y = [4.0432,4.1073,4.0899,4.1319,4.2885,4.4305,4.5249,4.6172,4.6962,4.7235,4.5987,4.7927,4.895,4.9079];
x = [1999,2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012];
pf = polyfit(x,y,2);%Calculating the corresponding polynomial function
disp('The coefficients of the desired polynomial:');
fprintf('%3.3f\n',pf) %Showing the coefficients of the fitted polynomial function
pv = polyval(pf,x);%Calculating the values of polynomial function at x's
disp('The values of polynomial at given set of x values:');
fprintf('\n %3.3f',pv)
scatter(x,y)
hold on
plot(x,pv)
hold off

[Rik]

You already have the vector with the coefficients. You are also aware of how to use fprintf, so the step to sprintf is minimal. What exactly is the output you would like?

[Berk Cem Yamanlar]

This is the simple linear regression line for my data and there is the equation of the line. I would like to find the equation of second order polynomial and show the equation on the figure as below.

Answer 1

[Mathieu NOE]

hello

my suggestion below :

the equation is given by the string eqn

code :

clc;
clear;
y = [4.0432,4.1073,4.0899,4.1319,4.2885,4.4305,4.5249,4.6172,4.6962,4.7235,4.5987,4.7927,4.895,4.9079];
x = [1999,2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012];

% Fit a polynomial p of degree "degree" to the (x,y) data:
degree = 2;
p = polyfit(x,y,degree);

% Evaluate the fitted polynomial p and plot:
f = polyval(p,x);
eqn = poly_equation(flip(p)); % polynomial equation (string)
Rsquared = my_Rsquared_coeff(y,f); % correlation coefficient

figure(3);plot(x,y,'o',x,f,'-')
legend('data',eqn)
title(['Data fit - R squared = ' num2str(Rsquared)]);

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

function Rsquared = my_Rsquared_coeff(data,data_fit)
    % R2 correlation coefficient computation
    
    % The total sum of squares
    sum_of_squares = sum((data-mean(data)).^2);
    
    % The sum of squares of residuals, also called the residual sum of squares:
    sum_of_squares_of_residuals = sum((data-data_fit).^2);
    
    % definition of the coefficient of correlation is
    Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
    
    
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function eqn = poly_equation(a_hat)
eqn = " y = "+a_hat(1);
for i = 2:(length(a_hat))
    if sign(a_hat(i))>0
        str = " + ";
    else
        str = " ";
    end
    if i == 2
        eqn = eqn+str+a_hat(i)+"*x";
    else
        eqn = eqn+str+a_hat(i)+"*x^"+(i-1)+" ";
    end
end
eqn = eqn+" ";
end
        

[Berk Cem Yamanlar]

Thank you sir! It was very helpful

[Mathieu NOE]

My pleasure !

[Mathieu NOE]

version 2

clc;
clear;
y = [4.0432,4.1073,4.0899,4.1319,4.2885,4.4305,4.5249,4.6172,4.6962,4.7235,4.5987,4.7927,4.895,4.9079];
x = [1999,2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012];

% Fit a polynomial p of degree "degree" to the (x,y) data:
degree = 2;
p = polyfit(x,y,degree);

% Evaluate the fitted polynomial p and plot:
f = polyval(p,x);
eqn = sprintf('%f*Year^2+%f*Year+%f\n',p);

Rsquared = my_Rsquared_coeff(y,f); % correlation coefficient

figure(3);plot(x,y,'o',x,f,'-')
legend('data',eqn)
title(['Data fit - R squared = ' num2str(Rsquared)]);

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

function Rsquared = my_Rsquared_coeff(data,data_fit)
    % R2 correlation coefficient computation
    
    % The total sum of squares
    sum_of_squares = sum((data-mean(data)).^2);
    
    % The sum of squares of residuals, also called the residual sum of squares:
    sum_of_squares_of_residuals = sum((data-data_fit).^2);
    
    % definition of the coefficient of correlation is
    Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
    
    
end

        

[Steven Lord]

fprintf lets you automatically include the sign of the variable if you include the flag + in your format specifier. The leading + in the result is a little awkward looking, but the command itself is quite simple and you could post-process the string to trim a leading + and to replace x^1 and x^0 with x and nothing if desired.

coeffs = [3 -4 5];
powers = 2:-1:0;
s = sprintf('%+dx^%d', [coeffs; powers])

s = '+3x^2-4x^1+5x^0'

if s(1) == '+'
    s = s(2:end);
end
s = replace(s, 'x^1', 'x');
s = replace(s, 'x^0', '')

s = '3x^2-4x+5'

Answer 2

[KSSV]

fprintf('%f*Year^2+%f*YEar+%f\n',pf)

[Berk Cem Yamanlar]

Thank you sir! It was very helpful