anonymous
  • anonymous
I NEED HELP.. Taylor series for x^2e^x and expanded around the point a=1/2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hi
anonymous
  • anonymous
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
anonymous
  • anonymous
ok i will work on it and i will put what i did

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anonymous
  • anonymous
is f ' (x) = 2xe^x?
anonymous
  • anonymous
hello
anonymous
  • anonymous
\[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]
anonymous
  • anonymous
oh product rule
anonymous
  • anonymous
how many time do i have to do differentiation? (how do i know?)
anonymous
  • anonymous
are you there? please help me
anonymous
  • anonymous
hi
anonymous
  • anonymous
what i have done so far is
anonymous
  • anonymous
i took differentiation of f(x)=x^2e^x
anonymous
  • anonymous
so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x
anonymous
  • anonymous
and i know that taylor series formula f(x)=f(a)+f'(a)(x-a)+ - - - what don't know what to do next..
anonymous
  • anonymous
my a = 1/2
anonymous
  • anonymous
should i just plug in everything on the formula?
anonymous
  • anonymous
so my f(a) should be 1/4e^1/2 ??
anonymous
  • anonymous
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
anonymous
  • anonymous
thank you RolyPoly but last quick question
anonymous
  • anonymous
how do i know how many times should i take differentiation of f(x)?
anonymous
  • anonymous
Actually, I don't know too...
anonymous
  • anonymous
LOLLLLLL ok! RolyPoly anyways thank you so much!
anonymous
  • anonymous
Is that (x^2)*(e^x)? @soobinkiki
anonymous
  • anonymous
\[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]
anonymous
  • anonymous
hey rolypoly can you click the website that you sent me?
anonymous
  • anonymous
it says f''(1/2) is 17/8(e^1/2)
anonymous
  • anonymous
that's why i was confused..
anonymous
  • anonymous
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 \[\frac{17}{8}(x-a)^2\]
anonymous
  • anonymous
shoot...... how i missed that....
anonymous
  • anonymous
\[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x-\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x-\frac{1}{2})^2\]
anonymous
  • anonymous
i got it now ! thanks a lot!
anonymous
  • anonymous
and the series of this problem is pretty long.......................
anonymous
  • anonymous
Obviously it's long, it's a series with infinite terms.. -.-

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