Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

soobinkiki

  • 2 years ago

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

  • This Question is Closed
  1. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hi

  2. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hello

  6. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh product rule

  8. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you there? please help me

  10. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hi

  11. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what i have done so far is

  12. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my a = 1/2

  16. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    should i just plug in everything on the formula?

  17. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  18. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  19. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you RolyPoly but last quick question

  20. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Actually, I don't know too...

  22. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOLLLLLL ok! RolyPoly anyways thank you so much!

  23. genius12
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  24. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]

  25. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  26. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  27. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's why i was confused..

  28. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  31. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got it now ! thanks a lot!

  32. soobinkiki
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  33. genius12
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  34. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy