Consider the function f with f(x) = e^{Fx^2 - (A+B)*F x + A*B*F}.
Determine the second-degree Taylor polynomial at x_0 = X0.
\displaystyle \color{teal}T_2 (x) =
A0 + " + " + A1 + " (x- " + X0 + ") + " + fractionReduce(A2,2) + " (x- " + X0 + ")^2"
We are looking for \displaystyle \color{teal}T_2 (x) = f(X0) + f'(X0)(x- X0) + \frac{1}{2} f''(X0)(x- X0)^2.
A direct computation shows that \displaystyle f(X0) = 1, and we calculate the two missing values \displaystyle f'(X0) and \displaystyle f''(X0) as follows:
We first get \displaystyle f'(x) = (2*Fx - (A+B)*F)\cdot f(x) and therefore \displaystyle f'(X0) = A1.
By differentiating this again, we obtain \displaystyle f''(x) = (4*F*Fx^2 - 4*F * (A+B)*F x + 2* F + (A+B)*(A+B)*F*F)\cdot f(x), so \displaystyle f''(X0) = A2.
Substituting now yields \displaystyle T_2(x) = A0 + A1 (x- X0) + fractionReduce(A2,2) (x- X0)^2.