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

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RolyPolyBest ResponseYou've already chosen the best response.1
Let 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
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
ok i will work on it and i will put what i did
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
is f ' (x) = 2xe^x?
 11 months ago

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

soobinkikiBest ResponseYou've already chosen the best response.0
how many time do i have to do differentiation? (how do i know?)
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
are you there? please help me
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
what i have done so far is
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
i took differentiation of f(x)=x^2e^x
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x
 11 months ago

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

soobinkikiBest ResponseYou've already chosen the best response.0
should i just plug in everything on the formula?
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
so my f(a) should be 1/4e^1/2 ??
 11 months ago

RolyPolyBest ResponseYou've already chosen the best response.1
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
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
thank you RolyPoly but last quick question
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
how do i know how many times should i take differentiation of f(x)?
 11 months ago

RolyPolyBest ResponseYou've already chosen the best response.1
Actually, I don't know too...
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
LOLLLLLL ok! RolyPoly anyways thank you so much!
 11 months ago

genius12Best ResponseYou've already chosen the best response.0
Is that (x^2)*(e^x)? @soobinkiki
 11 months ago

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

soobinkikiBest ResponseYou've already chosen the best response.0
hey rolypoly can you click the website that you sent me?
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
it says f''(1/2) is 17/8(e^1/2)
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
that's why i was confused..
 11 months ago

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

soobinkikiBest ResponseYou've already chosen the best response.0
shoot...... how i missed that....
 11 months ago

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

soobinkikiBest ResponseYou've already chosen the best response.0
i got it now ! thanks a lot!
 11 months ago

soobinkikiBest ResponseYou've already chosen the best response.0
and the series of this problem is pretty long.......................
 11 months ago

genius12Best ResponseYou've already chosen the best response.0
Obviously it's long, it's a series with infinite terms.. .
 11 months ago
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