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

1 Answer
Steve M
May 10, 2017

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.

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