Question

Find the Taylor Series for f(x) = x^4e^(-3x^2) about x = 0

Answer 1

\[f(0)+f'(0)x+\frac{f''(0)}{2}x^2+...\]

Answer 2

It's really long though, can't I use something else to solve o_O for it.
Since I know the series for e^(x). Or am I thinking too into it

Answer 3

\[f(0)=0\]
\[f'(x)=e^{-3 x^2} (4 x^3-6 x^5)\]
\[f'(0)=-2\]

Answer 4

go get 'em joe. don't see a quick gimmick do you?

Answer 5

i was wondering that too. Maybe you would be able to plug in (-3x^2) into the taylor series for e^x, then multiply everything by x^4?

i dont know if that would give you the same answer though =/

Answer 6

oh next one is 0

Answer 7

shouldnt f'(0) be 0?

Answer 8

is this odd?

Answer 9

oh right, sorry.

Answer 10

Yeah that was what I was thinking. I may have written it poorly, it's x^4 * e^(-3x^2). Can't pull out the x^4 then find the series for just the e^(-3x-2) function

Answer 11

i think the first non-zero term you should end up with is the 4th derivative.

Answer 12

yeah maybe and then multiply

Answer 13

i dunno though, i hate calculus =/

Answer 14

oh of course. write the series for
\[e^{-3x^2}\] and then multiply by
\[x^4\] doh

Answer 15

gotta run but first write the series for
\[e^x\] then replace x by -3x^2, then multiply by x^4
that is it

Answer 16

kk the I gots it, I got my answer to be the sum from 0 to infinity of (-3)^n * x^(2n+4) all over n!