Question

e^(-x^2) interval of convergence of expansion in power of x is (-?,?)

Answer 1

what is the taylor series of e^(-x^2) ?

Answer 2

it's \[\sum_{n=0}^{\infty}((-1^{n})(x ^{2n}))\div(n!)\]

Answer 3

good\[\sum_{n=0}^\infty{(-1)^nx^{2n}\over n!}\]
now apply the ratio test for convergence

Answer 4

do you know how to do that?

Answer 5

i'm looking my textbook now\

Answer 6

we want to check when\[\lim_{n\to\infty}\left|{a_{n+1}\over a_n}\right|<1\]

Answer 7

n! is infinite and i think x^(2n) is infinite, too

Answer 8

so the answer should be (-infinite, infinite) ?

Answer 9

i got -(x^2)/(n+1) what is the answer?

Answer 10

yes yo have it
because the limit coverges to zero the radius of covergence is infinite, and the interval of valididty is \((-\infty,\infty)\)

Answer 11

thank you !