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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

hi
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
ok i will work on it and i will put what i did

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question