Let us be given 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 12 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 yeilds \displaystyle T_2(x) = A0+ A1 (x- X0) + fractionReduce(A2,2) (x- X0) ^2
.