Let us be given the function f with f(x) = e^{A x + B}.
Determine the Taylor polynomial
at x_0 = X0 of degree three.
\displaystyle \color{teal}T_3 (x)
=A0 + " + " + A1 + " (x- " + X0 + ") + " + fractionReduce(A2,2) + " (x- " + X0 + ")^2 + " + fractionReduce(A3,6) + " (x- " + X0 + ")^3"We are looking for \displaystyle \color{teal}T_3 (x) =
f \left(X0 \right) + f'\left(X0\right )\left(x- X0\right )
+ \frac 12 f''\left(X0\right )\left(x- X0\right )^2 + \frac 16 f'''\left(X0\right )\left(x- X0\right )^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 \left(x- X0\right ) +
fractionReduce(A2,2) \left(x- X0\right ) ^2 +
fractionReduce(A3,6) \left(x- X0\right ) ^3.