We are looking for \displaystyle \color{teal}T_3 (x) = f \left(X0 \right) + f'\left(X0\right )(x- X0) + \frac{1}{2} f''\left(X0\right )(x- X0)^2 + \frac{1}{6} f'''\left(X0\right )(x- X0)^3.

We compute \displaystyle f\left(X0\right ) = 1, and we calculate the three missing values \displaystyle f'\left(X0\right ), \displaystyle f''\left(X0\right ), and \displaystyle f'''\left(X0\right ) as follows:

We first get \displaystyle f'(x) = A f(x), \displaystyle f''(x) = A*A f(x), and \displaystyle f'''(x) = A*A*A f(x).

Substituting, we obtain \displaystyle f'\left(X0\right ) = A1, f''\left(X0\right ) = A2, and \displaystyle f'''\left(X0\right ) = A3.

Substituting once again gives \displaystyle T_3(x) = A0 + A1 (x- X0) + fractionReduce(A2,2) (x- X0)^2 + fractionReduce(A3,6) (x- X0)^3.