Question

How do you find the power series for `f(x)=e^(-4x)` and determine its radius of convergence?

Answer 1

`e^(-4x) = 1 - 4x + 8x^2-32/3x^3 + 32/3x^4 + ...`
`" " = sum_(n=0)^oo (-4x)^n/(n!) " " AA x in RR`

Explanation:

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

`e^x = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ...`
`\ \ \ = sum_(n=0)^oo x^n/(n!)`

So then replacing `x` with `-4x` we have:

`e^(-4x) = 1 + (-4x) + (-4x)^2/(2!)+(-4x)^3/(3!) + (-4x)^4/(4!) + ...`
`" " = 1 - 4x + (4^2x^2)/(2!)-(4^3x^3)/(3!) + (4^4x^4)/(4!) + ...`
`" " = 1 - 4x + (16x^2)/(2)-(64x^3)/(6) + (256x^4)/(24) + ...`
`" " = 1 - 4x + 8x^2-32/3x^3 + 32/3x^4 + ...`
`" " = sum_(n=0)^oo (-1)^n(4x)^n/(n!)`

The initial series converges `AA x in RR` and so therefore dos the modified series.