Question

use substitution and power series to find the taylor series about x=0 of e^-(x/2)

Answer 1

\[e ^{-x/2}\]

Answer 2

\(e^u = 1 + u +\frac{u^2}{2!} + \dots = \sum_{k=0}^\infty \frac{u^k}{k!}\)
replace \(u\) by \(-x/2\). it gives
\(\sum_{k=0}^\infty \frac{(-x/2)^k}{k!}\) which you can rewrite a little bit.

Answer 3

ahhh thanks!