1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook