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

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Homework Statement



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

Homework Equations



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

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!
 
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shouldn't the number's in the numerator be (1*-1*-3*-5*...)?
 
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)
 
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?