calculate coefficients with linear regression

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

Answer 1

[Torsten]

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.

[Antrea Plastira]

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]

As I wrote, you have to turn xj and yj into column vectors.

Your xj and yj are row vectors.

Answer 2

[Image Analyst]

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.

[Torsten]

The function is linear in the parameters to be fitted. Thus a linear regression suffices.

[Image Analyst]

@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.