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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i will work on it and i will put what i did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you there? please help me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what i have done so far is

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

anonymous
 3 years 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

anonymous
 3 years 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..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should i just plug in everything on the formula?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you RolyPoly but last quick question

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, I don't know too...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOLLLLLL ok! RolyPoly anyways thank you so much!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey rolypoly can you click the website that you sent me?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's why i was confused..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The second term is \[\frac{f''(a)}{2!}(xa)^2\]So, it is \[\frac{\frac{17}{4}\sqrt{e}}{2!}(xa)^2\]that is \[\frac{17}{8}(xa)^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0shoot...... how i missed that....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x\frac{1}{2})^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got it now ! thanks a lot!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the series of this problem is pretty long.......................

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Obviously it's long, it's a series with infinite terms.. .
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