How do you find the power series for #f(x)=e^(-4x)# and determine its radius of convergence?
1 Answer
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 = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ... #
# \ \ \ = sum_(n=0)^oo x^n/(n!) #
So then replacing
# 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
