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The 2011th derivative

  1. Aug 8, 2011 #1
    Hi everyone,

    I came across this problem that I can't solve, but I'm sure there's a simple elegant solution.

    Let [itex]f(x) = \frac{x}{(1-(x^2))^2}[/itex].

    Find [itex]f^{2011}(0)[/itex].
  2. jcsd
  3. Aug 8, 2011 #2
    Begin by taking the first few derivatives and hope to see a pattern emerging. So, what are the first, second, third, etc. derivatives?
  4. Aug 8, 2011 #3
    I think you'll have better luck trying to construct the Taylor series expansion than trying to compute the derivatives manually. By Taylor's theorem, the coefficient of x^2011 will be f^(2011)(0)/2011!, so multiply it by 2011! to get your answer.
  5. Aug 9, 2011 #4
    Thanks for responding everyone.

    Micromass, I got up to the 3rd derivative but I unfortunately did not see any nice pattern. :(

    Citan Uzuki, I see what you mean. So I basically got that

    [itex]\frac{x}{(1-(x^2))} = \sum_{n=1} n x^{2n-1}[/itex]

    by playing around with the geometric series. Now, you're saying that this is also the Taylor Series expansion? In which case, the answer I get is 1005 * 2011!

    See, I thought of this earlier, but I didn't realize that every nice function has a unique power series expansion. That is, the Taylor Series is also the geometric series.
  6. Aug 9, 2011 #5
    Yep, that's the approach exactly! Just one thing - for the exponent to be 2011, n must be 1006, not 1005. But otherwise, you got it exactly right.
  7. Aug 9, 2011 #6
    Also x/(1-x^2)^2 = (1/2)(d/dx)(1/(1-x^2))
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