How do you find a power series representation for #e^(-x^2)# and what is the radius of convergence?

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Answer 1 · 2015-10-24T19:40:15

Use the power series for #e^t# and substitution to find:

#e^(-x^2) = sum_(n=0)^oo (-1)^n/(n!) x^(2n)#

with infinite radius of convergence.

#e^t = sum_(n=0)^oo t^n/(n!)#

with infinite radius of convergence.

Substitute #t = -x^2# to find:

#e^(-x^2) = sum_(n=0)^oo (-x^2)^n/(n!)=sum_(n=0)^oo (-1)^n/(n!) x^(2n)#

Which will converge for any #x in RR#, so has an infinite radius of convergence.