soobinkiki
I NEED HELP.. Taylor series for x^2e^x and expanded around the point a=1/2
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soobinkiki
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hi
RolyPoly
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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
soobinkiki
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ok i will work on it and i will put what i did
soobinkiki
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is f ' (x) = 2xe^x?
soobinkiki
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hello
RolyPoly
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\[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]
soobinkiki
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oh product rule
soobinkiki
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how many time do i have to do differentiation? (how do i know?)
soobinkiki
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are you there? please help me
soobinkiki
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hi
soobinkiki
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what i have done so far is
soobinkiki
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i took differentiation of f(x)=x^2e^x
soobinkiki
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so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x
soobinkiki
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and i know that taylor series formula f(x)=f(a)+f'(a)(x-a)+ - - - what don't know what to do next..
soobinkiki
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my a = 1/2
soobinkiki
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should i just plug in everything on the formula?
soobinkiki
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so my f(a) should be 1/4e^1/2 ??
soobinkiki
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thank you RolyPoly but last quick question
soobinkiki
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how do i know how many times should i take differentiation of f(x)?
RolyPoly
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Actually, I don't know too...
soobinkiki
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LOLLLLLL ok! RolyPoly anyways thank you so much!
genius12
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Is that (x^2)*(e^x)?
@soobinkiki
RolyPoly
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\[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]
soobinkiki
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hey rolypoly can you click the website that you sent me?
soobinkiki
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it says f''(1/2) is 17/8(e^1/2)
soobinkiki
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that's why i was confused..
RolyPoly
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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 \[\frac{17}{8}(x-a)^2\]
soobinkiki
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shoot...... how i missed that....
RolyPoly
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\[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x-\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x-\frac{1}{2})^2\]
soobinkiki
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i got it now ! thanks a lot!
soobinkiki
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and the series of this problem is pretty long.......................
genius12
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Obviously it's long, it's a series with infinite terms.. -.-