The Taylor Series Expansion is written as:
#sum_(n=0)^(N) f^((n))(a)/(n!)(x-a)^n#
Thus, we need to find the #n#th derivative of the function. Let's say we go to #n = N = 4#. Then:
#color(green)(f^((0))(x)) = f(x) = color(green)(e^(-3x))#
#color(green)(f'(x) = -3e^(-3x))#
#color(green)(f''(x) = 9e^(-3x))#
#color(green)(f'''(x) = -27e^(-3x))#
#color(green)(f''''(x) = 81e^(-3x))#
Now we have:
#sum_(n=0)^(N) f^((n))(a)/(n!)(x-a)^n#
#= (e^(-3a))/(0!)(x-a)^0 + (-3e^(-3a))/(1!)(x-a)^1 + (9e^(-3a))/(2!)(x-a)^2 + (-27e^(-3a))/(3!)(x-a)^3 + (81e^(-3a))/(4!)(x-a)^4 + ...#
#= e^(-6) + (-3e^(-6))(x-2) + (9e^(-6))/(2)(x-2)^2 + (-27e^(-6))/(6)(x-2)^3 + (81e^(-6))/(24)(x-2)^4 + ...#
#= e^(-6) -3e^(-6)(x-2) + (9e^(-6))/(2)(x-2)^2 - (27e^(-6))/6(x-2)^3 + (81e^(-6))/(24)(x-2)^4 - ...#
#= color(blue)(1/e^6 -3/e^6(x-2) + 9/(2e^6)(x-2)^2 - 9/(2e^6)(x-2)^3 + 27/(8e^6)(x-2)^4 - ...)#
