calculate coefficients with linear regression

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Antrea Plastira
Antrea Plastira 2022-10-12
评论： Image Analyst 2022-10-12
i have a set of data (xj, yj) which describe a function f(x) = a0 + a1x +a2 x^2 +a3 e^x
how can i find the coefficents a0, a1,a2,a3 modeling with linear regression and using quatratic programming.
so far i have found the slope and intercept.
my code is:
clc
clear
load('DataEx3(1).mat');
% calculating the mean for the x and y variable
x_mean = mean(xj);
y_mean = mean(yj);
% number of data points
n = length(xj);
%cross deviation of x and y
dev_xy = sum(xj*yj.') - (n*x_mean*y_mean);
%square deviation of x
dev_x = sum(xj*xj.') - (n*x_mean*x_mean);
% calculation of optimum coefficient a
a = dev_xy/dev_x;
% calculation of optimum intercept (c)
c = y_mean - (a*x_mean);
disp('value of coffiecent m is: '); disp(a);
disp('value of optimum intercept is: '); disp(c);
% best line has the form y = mx+c

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采纳的回答

Torsten
Torsten 2022-10-12
编辑：Torsten 2022-10-12
A = [ones(numel(xj),1),xj,xj.^2,exp(xj)];
b = yj;
sol = A\b;
a0 = sol(1)
a1 = sol(2)
a2 = sol(3)
a3 = sol(4)
where xj and yj are used as column vectors.
  2 个评论
Antrea Plastira
Antrea Plastira 2022-10-12
when i tried this way, it appears the following message:
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
Error in assignment1 (line 31)
A = [ones(numel(xj),1),xj,xj.^2,exp(xj)];
i forgot to mention that the xj and yj data provided are [1x401] double
Torsten
Torsten 2022-10-12
编辑：Torsten 2022-10-12
As I wrote, you have to turn xj and yj into column vectors.
Your xj and yj are row vectors.

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更多回答（1 个）

Image Analyst
Image Analyst 2022-10-12
Why do you want to use linear regression for that very non-linear function? Why not use fitnlm to do a non-linear fit to your equation? Demos attached.
  2 个评论
Torsten
Torsten 2022-10-12
The function is linear in the parameters to be fitted. Thus a linear regression suffices.
Image Analyst
Image Analyst 2022-10-12
@Torsten Oh (sound of hand slapping forehead) you're right.
However if the x in the exponential had a coefficient a4,
f(x) = a0 + a1 * x + a2 * x^2 + a3 * exp(a4 *x)
then we could use fitnlm. Not sure why that last x doesn't have a coefficient, which would allow for a "slope" or different steepnesses of the exponential term. I think that would give a more flexible model. You could always put in a4, and if it's 1, then it's 1, and a linear estimation is fine. But if a4 is not 1, it might be a better model by including a4.

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