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

1. soobinkiki

hi

2. RolyPoly

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

3. soobinkiki

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

4. soobinkiki

is f ' (x) = 2xe^x?

5. soobinkiki

hello

6. RolyPoly

$f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)$

7. soobinkiki

oh product rule

8. soobinkiki

how many time do i have to do differentiation? (how do i know?)

9. soobinkiki

10. soobinkiki

hi

11. soobinkiki

what i have done so far is

12. soobinkiki

i took differentiation of f(x)=x^2e^x

13. soobinkiki

so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x

14. soobinkiki

and i know that taylor series formula f(x)=f(a)+f'(a)(x-a)+ - - - what don't know what to do next..

15. soobinkiki

my a = 1/2

16. soobinkiki

should i just plug in everything on the formula?

17. soobinkiki

so my f(a) should be 1/4e^1/2 ??

18. RolyPoly

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

19. soobinkiki

thank you RolyPoly but last quick question

20. soobinkiki

how do i know how many times should i take differentiation of f(x)?

21. RolyPoly

Actually, I don't know too...

22. soobinkiki

LOLLLLLL ok! RolyPoly anyways thank you so much!

23. genius12

Is that (x^2)*(e^x)? @soobinkiki

24. RolyPoly

$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}$

25. soobinkiki

hey rolypoly can you click the website that you sent me?

26. soobinkiki

it says f''(1/2) is 17/8(e^1/2)

27. soobinkiki

that's why i was confused..

28. RolyPoly

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 $\frac{17}{8}(x-a)^2$

29. soobinkiki

shoot...... how i missed that....

30. RolyPoly

$=\frac{\frac{17}{4}\sqrt{e}}{2!}(x-\frac{1}{2})^2$$=\frac{17}{8}\sqrt{e}(x-\frac{1}{2})^2$

31. soobinkiki

i got it now ! thanks a lot!

32. soobinkiki

and the series of this problem is pretty long.......................

33. genius12

Obviously it's long, it's a series with infinite terms.. -.-