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Help with Taylor Series/Maclaurin Series Question

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Problem is attached in this post.

    2. Relevant equations

    Problem is attached in this post.

    3. The attempt at a solution

    I've tried using Maclaurin Series for e^x, and get the term -x^10/5!, however f(0) = 0 which is not the correct answer. Also taking 10 derivatives seems too burdensome of a task etc. Is there any other method to solve this problem?

    The answer is D. -10!/5!
     

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  3. Apr 28, 2014 #2

    SammyS

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    attachment.php?attachmentid=69190&d=1398730870.png

    [itex]\displaystyle f^{(10)}(x) = \frac{d^{10}}{dx^{10}}f(x)\ .[/itex]



    It's not [itex]\displaystyle \left(f(x)\right)^{10}[/itex]
     
  4. Apr 28, 2014 #3
    I don't understand what you mean?
     
  5. Apr 28, 2014 #4

    SammyS

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    It's the tenth derivative ... evaluated at x = 0 .
     
  6. Apr 28, 2014 #5
    I know that, but solving for 10 derivatives seems like an impractical way to solve this problem. Is there any other method to solve this problem?
     
  7. Apr 28, 2014 #6

    SammyS

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    Solving for 10 derivatives is not really that difficult -- using the MacLaurin Series.

    What is the 10th derivative of x6, for example?
     
  8. Apr 28, 2014 #7
    10th derivative of x^6 =0, I've tried Maclaurin, and got my 10th powered term as -x^10/5!, but f(0)=0 and the answer is -10!/5!
     
  9. Apr 28, 2014 #8

    SammyS

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    What is the 10th derivative of x10 ?
     
  10. Apr 28, 2014 #9
    3,628,800
     
  11. Apr 28, 2014 #10

    SammyS

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    Yes, that's 10! . Right.

    Even if x = 0.
     
  12. Apr 28, 2014 #11
    So do I take the derivative of -x^10/5!?
     
  13. Apr 28, 2014 #12

    SammyS

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    What is the MacLaurin series for ##e^{-x^2} \ ? ##


    The 10th derivative of of the first 5 terms, those with power, 0, 2, 4, 6, 8, are all zero, right?

    ...
     
  14. Apr 28, 2014 #13
    Yes:

    1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^10/5! ...... etc.
     
  15. Apr 28, 2014 #14

    SammyS

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    ##1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^{10}/5!+x^{12}/6! \dots ##

    The 10th derivative of that ... evaluated at x = 0 ?
     
  16. Apr 28, 2014 #15
    I understand it now, thanks for the help. (Since the f^(10)(0)/10! = -1/5! etc.)
     
  17. Apr 28, 2014 #16

    SammyS

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    Good.

    That notation is a bit strange.


    If ##f(x) = 1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^{10}/5!+x^{12}/6!\ \dots##

    then the 10th derivative is:

    ##f^{(10)}(x) = - 10!/5!+(12!/2)x^{2}/6!\ \dots##

    Then ##f^{(10)}(0) = - 10!/5!+(12!/2)0^{2}/6!\ \dots =- 10!/5!##
     
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