Taylor Series Expansion: f(x)=e^-2x at a=1

[Brainiachotd]

I have been trying to derive an equation for this problem: f(x)=e^-2x at a=1.
I have been solving this problem for 3 hours and i am confused on the general equation used for the Taylor Series but i got a pattern in which the values were:
f '(x) = -2e^-2x
f"(x) = 4e^-2x
f'"(x) = -8e^-2x and so on...........
So, i think that the pattern is (-2)^n * e^-2x. I just do not how to incorporate this knowledge into the derivation of the expansion equation itself......

 

[Deleted member 4993]

I have been trying to derive an equation for this problem: f(x)=e^-2x at a=1.
I have been solving this problem for 3 hours and i am confused on the general equation used for the Taylor Series but i got a pattern in which the values were:
f '(x) = -2e^-2x
f"(x) = 4e^-2x
f'"(x) = -8e^-2x and so on...........
So, i think that the pattern is (-2)^n * e^-2x. I just do not how to incorporate this knowledge into the derivation of the expansion equation itself......

Please write the general expression for expansion of f(x+h), using TAylor's series.

 

[pka]

I have been trying to derive an equation for this problem: f(x)=e^-2x at a=1.
I have been solving this problem for 3 hours and i am confused on the general equation used for the Taylor Series but i got a pattern in which the values were:
f '(x) = -2e^-2x
f"(x) = 4e^-2x
f'"(x) = -8e^-2x and so on...........
So, i think that the pattern is (-2)^n * e^-2x. I just do not how to incorporate this knowledge into the derivation of the expansion equation itself......

Do you know what constitutes a Taylor Series expansion of a function? Please follow that link.
\(\displaystyle\large\sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(a)}}{{n!}}{{\left( {x - a} \right)}^n}} \)