Obtaining the constant that makes the integral equal to zero in Matlab

Question

I'm trying to code a MATLAB program and I have arrived at a point where I need to do the following. I have this equation:

I must find the value of the constant "Xcp" (greater than zero), that is the value that makes the integral equal to zero.

In order to do so, I have coded a loop in which the the value of Xcp advances with small increments on each iteration and the integral is performed and checked if it's zero, if it reaches zero the loop finishes and the Xcp is stored with this value.

However, I think this is not an efficient way to do this task. The running time increases a lot, because this loop is long and has the to perform the integral and the integration limits substitution every time.

Is there a smarter way to do this in Matlab to obtain a better code efficiency?

P.S.: I have used conv() to multiply both polynomials. Since cl(x) and (x-Xcp) are both polynomials.

EDIT: Piece of code.

p = [1 -Xcp];    % polynomial (x-Xcp)
Xcp=0.001;
i=1;
found=false;
while(i<=x_te && found~=true)      % Xcp is upper bounded by x_te
   int_cl_p = polyint(conv(cl,p));
   Cm_cp=(-1/c^2)*diff(polyval(int_cl_p,[x_le,x_te]));
   if(Cm_cp==0)
     found=true;
   else
     Xcp=Xcp+0.001;
   end
end

This is the code I used to run this section. Another problem is that I have to do it for different cases (different cl functions), for this reason the code is even more slow.

Answer 1

As far as I understood, you need to solve the equation for X_CP. I suggest using symbolic solver for this. This is not the most efficient way for large polynomials, but for polynomials of degree 20 it takes less than 1 second. I do not claim that this solution is fastest, but this provides generic solution to the problem. If your polynomial does not change every iteration, then you can use this generic solution many times and not spend time for calculating integral.

So, generic symbolic solution in terms of xLE and xTE is obtained using this:

syms xLE xTE c x xCP
a = 1:20;
%//arbitrary polynomial of degree 20
cl = sum(x.^a.*randi([-100,100],1,20));
tic
eqn = -1/c^2 * int(cl * (x-xCP), x, xLE, xTE) == 0;
xCP = solve(eqn,xCP);
pretty(xCP)
toc

Elapsed time is 0.550371 seconds.

You can further use matlabFunction for finding the numerical solutions:

xCP_numerical = matlabFunction(xCP);
%// we then just plug xLE = 10 and xTE = 20 values into function
answer = xCP_numerical(10,20)
answer =  
        19.8038

The slight modification of the code can allow you to use this for generic coefficients.

Hope that helps