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soobinkiki

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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  1. soobinkiki
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    hi

    • 11 months ago
  2. 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

    • 11 months ago
  3. soobinkiki
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    ok i will work on it and i will put what i did

    • 11 months ago
  4. soobinkiki
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    is f ' (x) = 2xe^x?

    • 11 months ago
  5. soobinkiki
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    hello

    • 11 months ago
  6. RolyPoly
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    \[f'(x) = e^x \frac{d}{dx}(x^2) + x^2\frac{d}{dx}(e^x)\]

    • 11 months ago
  7. soobinkiki
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    oh product rule

    • 11 months ago
  8. soobinkiki
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    how many time do i have to do differentiation? (how do i know?)

    • 11 months ago
  9. soobinkiki
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    are you there? please help me

    • 11 months ago
  10. soobinkiki
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    hi

    • 11 months ago
  11. soobinkiki
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    what i have done so far is

    • 11 months ago
  12. soobinkiki
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    i took differentiation of f(x)=x^2e^x

    • 11 months ago
  13. soobinkiki
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    so i got f '(x)=x^2e^x+2xe^x f ''(x)=4xe^x+x^2e^x+2e^x

    • 11 months ago
  14. 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..

    • 11 months ago
  15. soobinkiki
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    my a = 1/2

    • 11 months ago
  16. soobinkiki
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    should i just plug in everything on the formula?

    • 11 months ago
  17. soobinkiki
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    so my f(a) should be 1/4e^1/2 ??

    • 11 months ago
  18. RolyPoly
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    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
  19. soobinkiki
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    thank you RolyPoly but last quick question

    • 11 months ago
  20. soobinkiki
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    how do i know how many times should i take differentiation of f(x)?

    • 11 months ago
  21. RolyPoly
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    Actually, I don't know too...

    • 11 months ago
  22. soobinkiki
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    LOLLLLLL ok! RolyPoly anyways thank you so much!

    • 11 months ago
  23. genius12
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    Is that (x^2)*(e^x)? @soobinkiki

    • 11 months ago
  24. RolyPoly
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    \[f'(x) = 2xe^x+x^2e^x\]\[f''(x)=2(e^x+xe^x) + 2xe^x+x^2e^x=e^x(2+4x+x^2)\]\[f''(\frac{1}{2})=e^{\frac{1}{2}}(2+4(\frac{1}{2})+(\frac{1}{2})^2)=\sqrt{e}(2+2+\frac{1}{4}) = \frac{17}{4}\]

    • 11 months ago
  25. soobinkiki
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    hey rolypoly can you click the website that you sent me?

    • 11 months ago
  26. soobinkiki
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    it says f''(1/2) is 17/8(e^1/2)

    • 11 months ago
  27. soobinkiki
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    that's why i was confused..

    • 11 months ago
  28. 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 \[\frac{17}{8}(x-a)^2\]

    • 11 months ago
  29. soobinkiki
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    shoot...... how i missed that....

    • 11 months ago
  30. RolyPoly
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    \[=\frac{\frac{17}{4}\sqrt{e}}{2!}(x-\frac{1}{2})^2\]\[=\frac{17}{8}\sqrt{e}(x-\frac{1}{2})^2\]

    • 11 months ago
  31. soobinkiki
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    i got it now ! thanks a lot!

    • 11 months ago
  32. soobinkiki
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    and the series of this problem is pretty long.......................

    • 11 months ago
  33. genius12
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    Obviously it's long, it's a series with infinite terms.. -.-

    • 11 months ago
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