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Math Help - formula for the n:th derivative

  1. #1
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    formula for the n:th derivative

    Find the formula for the n:th derivative y^(n) for the function y=1/x^(1/2)'

    I started by finding out the formula for the functions up to the fourth derivative.
    and i came as far as
    y = y=1/x^(1/2)
    y' = y= -1/2x^(3/2)
    y'' =3/4x^(5/2)
    y'''=-15/8x^(1/2)
    y''''=105/16x^(9/2)

    And the pattern which i found is that y^n where n starts from 0. so 0 is the original function.
    The nth derivative y^(n) = (-1)^n / ((2^n)*x^(0.5 +n)). The problem here is that i dont find anything that can fix the numerator. I realise the pattern is 1*3*5*7...n. But i dont know how to write a value for this since i cant use ! because its 1*2*3*4 etc. Is it possible to write a factorial equation that skipps whole numbers and only takes odd numbers? And if so, how do i write it mathematically?
    Any help is appreciated.
    Last edited by Zamzen; October 9th 2010 at 02:20 PM.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Zamzen View Post
    Find the formula for the n:th derivative y^(n) for the function y=1/x^(1/2)'

    I started by finding out the formula for the functions up to the fourth derivative.
    and i came as far as
    y = y=1/x^(1/2)
    y' = y= -1/2x^(3/2)
    y'' =3/4x^(5/2)
    y'''=-15/8x^(1/2)
    y''''=105/16x^(9/2)

    And the pattern which i found is that y^n where n starts from 0. so 0 is the original function.
    The nth derivative y^(n) = (-1)^n / ((2^n)*x^(0.5 +n)). The problem here is that i dont find anything that can fix the top. I realise the pattern is 1*3*5*7...n. But i dont know how to write a value for this since i cant use ! because its 1*2*3*4 etc. Is it possible to write a factorial equation that skipps whole numbers and only takes odd numbers? And if so, how do i write it mathematically?
    Any help is appreciated.
    (2n)!! = 2n * (2n-2) *(2n-4) * .... == even numbers
    (2n+1)!! = (2n+1)*(2n-1) *(2n-3) * ..... == odd numbers
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  3. #3
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    Quote Originally Posted by yeKciM View Post
    (2n)!! = 2n * (2n-2) *(2n-4) * .... == even numbers
    (2n+1)!! = (2n+1)*(2n-1) *(2n-3) * ..... == odd numbers
    More info

    Double Factorial -- from Wolfram MathWorld
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  4. #4
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    If you want a more "standard" formula, for the product of even numbers, you can factor out "2"s:
    2(4)(6)(8)\cdot\cdot\cdot (2(n-1))(2n)= 2^n(1(2)(3)(4)\cdot\cdot\cdot (n-1)(n))= 2^n n!.

    For the product of odd numbers, put in the even numbers:
    1(3)(5)(7)\cdot\cdot\cdot (2n-1)(2n+1)= \frac{1(2)(3)(4)(5)(6)(7)\cdot\cdot\cdot\(2n-2)(2n)(2n+1)}{2(4)(6)\cdot\cdot\cdot (2n-2)(2n)}= \frac{(2n+1)!}{2^n n!}
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