# How do you find the Taylor Series Expansion for f(x)=e^(-3x) at c = 2?

Jul 16, 2015

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:

$\textcolor{g r e e n}{{f}^{\left(0\right)} \left(x\right)} = f \left(x\right) = \textcolor{g r e e n}{{e}^{- 3 x}}$
$\textcolor{g r e e n}{f ' \left(x\right) = - 3 {e}^{- 3 x}}$
$\textcolor{g r e e n}{f ' ' \left(x\right) = 9 {e}^{- 3 x}}$
$\textcolor{g r e e n}{f ' ' ' \left(x\right) = - 27 {e}^{- 3 x}}$
$\textcolor{g r e e n}{f ' ' ' ' \left(x\right) = 81 {e}^{- 3 x}}$

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 + ...

$= {e}^{- 6} + \left(- 3 {e}^{- 6}\right) \left(x - 2\right) + \frac{9 {e}^{- 6}}{2} {\left(x - 2\right)}^{2} + \frac{- 27 {e}^{- 6}}{6} {\left(x - 2\right)}^{3} + \frac{81 {e}^{- 6}}{24} {\left(x - 2\right)}^{4} + \ldots$

$= {e}^{- 6} - 3 {e}^{- 6} \left(x - 2\right) + \frac{9 {e}^{- 6}}{2} {\left(x - 2\right)}^{2} - \frac{27 {e}^{- 6}}{6} {\left(x - 2\right)}^{3} + \frac{81 {e}^{- 6}}{24} {\left(x - 2\right)}^{4} - \ldots$

$= \textcolor{b l u e}{\frac{1}{e} ^ 6 - \frac{3}{e} ^ 6 \left(x - 2\right) + \frac{9}{2 {e}^{6}} {\left(x - 2\right)}^{2} - \frac{9}{2 {e}^{6}} {\left(x - 2\right)}^{3} + \frac{27}{8 {e}^{6}} {\left(x - 2\right)}^{4} - \ldots}$