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Calc 2 - Taylor Expansion Series of x^(1/2)

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = [tex]\sqrt{x}[/tex], a = 4

    2. Relevant equations

    f(x) = [tex]\sum[/tex]f[tex]^{n}[/tex](a)/n! (x-a)[tex]^{n}[/tex]

    3. The attempt at a solution

    f(x) = x[tex]^{1/2}[/tex]
    f[tex]^{'}[/tex](x) = [tex]\frac{1}{2}[/tex]x[tex]^{1/2}[/tex]
    f[tex]^{2}[/tex](x) = -[tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]x[tex]^{-3/2}[/tex]
    f[tex]^{3}[/tex](x) = [tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]*[tex]\frac{3}{2}[/tex]x[tex]^{-5/2}[/tex]
    f[tex]^{4}[/tex](x) = -[tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]*[tex]\frac{3}{2}[/tex]*[tex]\frac{5}{2}[/tex]x[tex]^{-7/2}[/tex]

    f[tex]^{n}[/tex](x) = (-1)[tex]^{n+1}[/tex]*[tex]\frac{1}{2}[/tex][tex]^{n}[/tex]*x[tex]^{-[(2n-1)/2]}[/tex]*?????


    The problem I am having here is with identifying the pattern. I am able to describe everything except the numbers in the numerator(1, 1*1, 1*1*3, 1*1*3*5.....). Any help is greatly appreciated!
     
  2. jcsd
  3. Dec 13, 2009 #2
    I have no idea why it looks like that...
     
  4. Dec 13, 2009 #3
    shouldn't the number's in the numerator be (1*-1*-3*-5*...)?
     
  5. Dec 14, 2009 #4
    I left out the signs to simplify and becaus I already identified the pattern w/ (-1)^(n+1)... so if i kept the signs in, the #'s in the numerator would be (1, -1*1, -3*-1*1, -5*-3*-1*1)
     
  6. Dec 14, 2009 #5
    how are you trying to explain it then?
     
  7. Dec 14, 2009 #6
    n!! denotes the double factorial of n and is defined recursively for odd numbers,,
    eg: 9!! = 1 × 3 × 5 × 7 × 9 = 945

    does that help?
     
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