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Homework Help: Compute a derivative using MacLaurin series

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the 9th derivative of
    [tex]f(x) = \frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}[/tex]
    at x=(0)
    2. Relevant equations
    [tex]f(x)=\sum^{\infty}_{n=0} \frac{f^{(n)}(c)}{n!}x^n[/tex]
    [tex]cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}[/tex]

    3. The attempt at a solution
    The correct answer is 1224720.
    I've already read this thread(https://www.physicsforums.com/showthread.php?t=362369&highlight=Maclaurin), and I think I know how to do these kind of problems. However, I have a problem with that one since it has [tex]{-1}}[/tex]

    What I did is write the power series for cos(x),
    [tex]cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}[/tex]

    Substitute [tex] 3 x^{4}[/tex] , substract 1, and divide by [tex] x^{7}[/tex].
    And getting:
    [tex]\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=0} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}-1}{x^7}[/tex]
    But after that I don't know.
  2. jcsd
  3. Dec 12, 2009 #2


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    The only term that will produce a nonzero derivative when evaluated at x=0 will have a power of x^9. What is that term and what is it's derivative?
  4. Dec 13, 2009 #3


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    Nope. You can't move the 1 inside the sum like that. Sum(something) -1 is not equal to sum(something - 1).

    What you should do is recognize that the n = 0 term of the cosine series is 1, so this cancels with your minus 1. Your sum is then from n = 1 to infinity, and you can safely divide the x^7 through. Then I think you probably know what to do from there!
  5. Dec 13, 2009 #4


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    But the n= 0 term for the MacLaurin series for cosine is just "1". The "-1" term just cancels that.

    [tex]\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=1} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}}{x^7}[/tex]
  6. Dec 13, 2009 #5
    Thank you, it worked. I knew there was something wrong with the -1. I don't know why I thought I could do that operation on a sum. I had also simplified [tex] \frac{x^{8n}}{x^{7}}[/tex] to [tex] {x^{n}}[/tex] , which is totally wrong. I don't why I did this.
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