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Using Maclaurin series to find 2005-order derivative

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]f(x) = \arctan(\frac{1+x}{1-x})[/itex]
    Find [itex]f^{2005}[/itex](0)

    2. Relevant equations

    I'm guessing this has to do with maclaurin's?

    3. The attempt at a solution
    ...
    [itex]
    f(x) = \pi /4 + \sum^∞_{n = 0} \frac{(-1)^n}{2n+1}x^{2n+1}
    [/itex]

    [itex]\sum^∞_{n = 0}\frac{f^n(0)x^n}{n!} = \pi /4 + \sum^∞_{n = 0} \frac{(-1)^n}{2n+1}x^{2n+1}[/itex]

    So anyone knows how I go about from here? The answer is 2004!(factorial)
    How do you compare two infinite series can you cancel them out?
     
  2. jcsd
  3. Apr 14, 2012 #2
    (d/dx)xn = nxn-1

    (dp/dxp) xn = n!/(n-p)! xn-p
    (can you see where this formula comes from?)

    Also, (d/dx)C=0

    They want the value of the 2005th derivative at 0. If you think about it, the term involving x2005 will become a constant after 2005 differentiations. All terms before x2005 will disappear due to (d/dx)C=0, and all terms after are irrelevant since we are calculating at 0.

    The term at x2005 is supplied by the formula you found:

    Ʃ (-1)n/(2n+1) x2n+1

    Of course, the π/4 term also disappears.
     
    Last edited: Apr 14, 2012
  4. Apr 14, 2012 #3
    You can compare the two series since they are polynomials in x, and what you are interested in is only one of the terms of the polynomial. f^(2000) occurs exactly in the term of x^(2005), so you can ignore all other terms and compare the coefficients of x^2005 on both sides.
     
  5. Apr 14, 2012 #4
    Ah...i get it now so,
    [itex]f^{2005}(0)=\frac{2005!}{2005} * (-1)^{1002}[/itex]

    Thanks for your help!
     
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