Question

I NEED HELP.. Taylor series for x^2e^x and expanded around the point a=1/2

Answer 1

hi

Answer 2

Let f(x) = x^2 e^x
\[f(x) = f(a) + f'(a) (x-a) + \frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+... +\frac{f^{(n)}(a)}{n!}(x-a)^{n} +...\]
So, you need to keep differentiating f(x)...
Reference: http://mathworld.wolfram.com/TaylorSeries.html

Answer 3

ok i will work on it and i will put what i did

Answer 4

is f ' (x) = 2xe^x?

Answer 5

hello

Answer 6

\[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]

Answer 7

oh product rule

Answer 8

how many time do i have to do differentiation? (how do i know?)

Answer 9

are you there? please help me

Answer 10

hi

Answer 11

what i have done so far is

Answer 12

i took differentiation of f(x)=x^2e^x

Answer 13

so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x

Answer 14

and i know that taylor series formula f(x)=f(a)+f'(a)(x-a)+ - - - what don't know what to do next..

Answer 15

my a = 1/2

Answer 16

should i just plug in everything on the formula?

Answer 17

so my f(a) should be 1/4e^1/2 ??

Answer 18

f(a) is right.
Check this: http://www.wolframalpha.com/input/?i=+Taylor+series+x%5E2e%5Ex+at+x%3D1%2F2
It's not something nice

Answer 19

thank you RolyPoly but last quick question

Answer 20

how do i know how many times should i take differentiation of f(x)?

Answer 21

Actually, I don't know too...

Answer 22

LOLLLLLL ok! RolyPoly anyways thank you so much!

Answer 23

Is that (x^2)*(e^x)?

@soobinkiki

Answer 24

\[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]

Answer 25

hey rolypoly can you click the website that you sent me?

Answer 26

it says f''(1/2) is 17/8(e^1/2)

Answer 27

that's why i was confused..

Answer 28

The second term is \[\frac{f''(a)}{2!}(x-a)^2\]So, it is \[\frac{\frac{17}{4}\sqrt{e}}{2!}(x-a)^2\]that is \[\frac{17}{8}(x-a)^2\]

Answer 29

shoot...... how i missed that....

Answer 30

\[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x-\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x-\frac{1}{2})^2\]

Answer 31

i got it now ! thanks a lot!

Answer 32

and the series of this problem is pretty long.......................

Answer 33

Obviously it's long, it's a series with infinite terms.. -.-