Question

find power series (maclaurin series):
integral [ e^(-x)^(2) ] dx

Answer 1

well you know the series of e^x

Answer 2

is that correct for e^x

Answer 3

i meant e^-x

Answer 4

are you looking for the series to

\[\int e^{-x^2}dx\]

Answer 5

affirmative.

Answer 6

find the series to \[\Large e^{-x^2}\]

then integrate term by term

Answer 7

can you elaborate

Answer 8

can you find the series of \(e^{-x^2}\)?

Answer 9

no

Answer 10

replace x in the expansion of \(e^x\) with \(-x^2\)

Answer 11

okay...

Answer 12

then integrate each term

Answer 13

in this case i would prefer to only eat for a day

Answer 14

\[\int e^{-x^2}dx=\int\sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}dx=\int\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\int\frac{(-1)^nx^{2n}}{n!}dx\]

Answer 15

very illuminating, thank you very much!

Answer 16

you still need to integrate...I'll leave that to you