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Multiple Derivatives

  1. Mar 25, 2009 #1
    1. Find the first ten values of f (k)(0) and determine whether or not there is an equation for the fkth(0) term.



    f(x) = 1/(x2+1)
    x = 1, 2, 3,....

    Find f(k)(0) when k = 1, 2, 3,... for the first 10 values of k.


    I got that
    f'(x) = 0
    f''(x) = -2
    f'''(x) = 0
    f4(x) = 24
    f5(0) = 0

    And...that is as far as I got.
     
  2. jcsd
  3. Mar 25, 2009 #2
    Are you allowed to dip into complex numbers?
     
  4. Mar 25, 2009 #3

    Dick

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    Why? Write out the geometric series expansion of 1/(1-(-x^2)). Match that up with the taylor series expansion of 1/(1+x^2) with the x^k*f^(k)(x)/k! things in it. That will give you a formula for f^(k).
     
    Last edited: Mar 25, 2009
  5. Mar 26, 2009 #4
    Could you please elaborate on this a little bit more? I'm not familiar with the Taylor series.
     
  6. Mar 26, 2009 #5

    HallsofIvy

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    Then look at what you have so far:
    f'(0)= 0
    f"(0)= -2= -2!
    f"'(0)= 0
    f""(0)= 24= 4!
    f""'(0)= 0
    It should be easy to see what f(n)(0) is for n odd!

    I might guess that f"""(0)= -720= (-1)37!. Can you check that?
     
  7. Mar 26, 2009 #6

    Dick

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    The geometric series expansion is 1-x^2+x^4-x^6+... The Taylor series expansion is f(0)+f'(0)x+f''(0)*x^2/2!+f'''(0)x^3/3!+f''''(0)x^4/4!+... Match up equal powers of x and read off the derivatives of f.
     
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