Recently, I had assigned a project to my class where they needed to regress *n* number of *x-y* data points to a nonlinear regression model *y*=exp(*b***x*). However, they were NOT allowed to transform the data, that is, transform data such that linear regression formulas can be used to find the constant of regression *b*. They had to do it the new-fashioned way: Find the sum of the square of the residuals and then minimize the sum with respect to the constant of regression *b*.

To do this, they conducted the following steps

- setup the equation by declaring
*b*as a syms variable, - calculate the sum of the square of the residuals using a loop,
- use the
*diff*command to set up the equation, - use the
*solve*command.

However, the *solve* command gave some odd answer like log(z1)/5 + (2*pi*k*i)/5. The students knew that the equation has only one real solution – this was deduced from the physics of the problem.

We did not want to set up a separate function mfile to use the numerical solvers such as *fsolve*. To circumvent the setting up of a separate function mfile, we approached it as follows. If dbsr=0 is the equation you want to solve, use

F = vectorize(inline(char(dbsr)))

fsolve(F, -2.0)

What *char* command does is to convert the function dbsr to a string, inline constructs it to an inline function, vectorize command vectorizes the formula (I do not fully understand this last part myself or whether it is needed).

Pingback: Does the solve command in MATLAB not give you an answer? | The Numerical Methods Guy