Question

What is the Maclaurin series for `e^(-x)`?

Answer 1

`e^(-x) = 1 -x +(x^2)/(2!) - (x^3)/(3!) + x^4/(4!) - ...`

Explanation:

We can start with the well known series for `e^(x)`

`e^x = 1 +x +(x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + ...`

If we replace `x` in the above series by `-x` we get:

`e^(-x) = 1 +(-x) +((-x)^2)/(2!) + ((-x)^3)/(3!) + ((-x)^4)/(4!) + ...`
`\ \ \ \ \ \ = 1 -x +(x^2)/(2!) - (x^3)/(3!) + x^4/(4!) - ...`