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Expressing double factorial for odd integers

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data


    Express [itex]\frac{1}{(2n+1)!!}[/itex] as the following

    [itex]\frac{(-2)^{n}n!x^{2n+1}}{(2n+1)!}[/itex]

    where 0 <= n <= infinity

    2. Relevant equations

    The double factorial for odd integers is

    (2n+1)!! = (2n+1)(2n-1)(2n-3)....1 where 0 <= n <= infinity


    3. The attempt at a solution

    Visited this website http://mathworld.wolfram.com/DoubleFactorial.html and scrolled
    down to the expression for

    [itex](-2n-1)!! = \frac{(-1)^{n}2^{n}n!}{2n!}[/itex]

    performing the sign inverse, i got

    [itex](2n+1)!! = \frac{(1)^{n}-2^{n}n!}{-2n!}[/itex]
     
  2. jcsd
  3. Sep 17, 2011 #2

    SammyS

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    It's not at all clear what the x2n+1 is doing in there.

    As for (2n+1)!!, start by working with (2n+1)! and see what needs to be canceled out.

    [itex]\displaystyle (2n+1)!=(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\dots(5)(4)(3)(2)(1)[/itex]
    [itex]\displaystyle =\{(2n+1)(2n-1)(2n-3)\dots(5)(3)(1)\}\{(2n)(2n-2)\dots(4)(2)\}[/itex]

    [itex]\displaystyle =\{(2n+1)!!\}\{(2^n)(n!)\}[/itex]​

    Solve for (2n+1)!! .

    As for [itex]\displaystyle \frac{(-2)^{n}n!x^{2n+1}}{(2n+1)!}\,,[/itex] that's the same as [itex]\displaystyle \frac{2^nn!}{(2n+1)!}(-1)^{n}\,x^{2n+1}\,.[/itex]
     
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