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RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Let f(x) = x^2 e^x \[f(x) = f(a) + f'(a) (xa) + \frac{f''(a)}{2!}(xa)^2+\frac{f^{(3)}(a)}{3!}(xa)^3+... +\frac{f^{(n)}(a)}{n!}(xa)^{n} +...\] So, you need to keep differentiating f(x)... Reference: http://mathworld.wolfram.com/TaylorSeries.html

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0ok i will work on it and i will put what i did

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0is f ' (x) = 2xe^x?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1\[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0how many time do i have to do differentiation? (how do i know?)

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0are you there? please help me

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0what i have done so far is

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0i took differentiation of f(x)=x^2e^x

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0and i know that taylor series formula f(x)=f(a)+f'(a)(xa)+    what don't know what to do next..

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0should i just plug in everything on the formula?

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0so my f(a) should be 1/4e^1/2 ??

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1f(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

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0thank you RolyPoly but last quick question

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0how do i know how many times should i take differentiation of f(x)?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Actually, I don't know too...

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0LOLLLLLL ok! RolyPoly anyways thank you so much!

genius12
 one year ago
Best ResponseYou've already chosen the best response.0Is that (x^2)*(e^x)? @soobinkiki

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1\[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0hey rolypoly can you click the website that you sent me?

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0it says f''(1/2) is 17/8(e^1/2)

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0that's why i was confused..

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1The second term is \[\frac{f''(a)}{2!}(xa)^2\]So, it is \[\frac{\frac{17}{4}\sqrt{e}}{2!}(xa)^2\]that is \[\frac{17}{8}(xa)^2\]

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0shoot...... how i missed that....

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1\[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x\frac{1}{2})^2\]

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0i got it now ! thanks a lot!

soobinkiki
 one year ago
Best ResponseYou've already chosen the best response.0and the series of this problem is pretty long.......................

genius12
 one year ago
Best ResponseYou've already chosen the best response.0Obviously it's long, it's a series with infinite terms.. .
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