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How to prove the half number factorial formula?

  1. Dec 1, 2005 #1
    Any ideas on how to prove this?
    [tex] (n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2} [/tex]
     
  2. jcsd
  3. Dec 1, 2005 #2

    shmoe

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    You've presumably met the Gamma function. Have a look through identites involving Gamma that you know, you should find something useful.
     
  4. Dec 1, 2005 #3

    HallsofIvy

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    How do you define (n+ 1/2)!???

    (Schmo already told you that- my point is you should think about it!)
     
  5. Dec 1, 2005 #4
    Problem is, I don't know what Gamma is other than a greek letter. I can use the formula, that's not the problem. I just was curious if there was a way to prove it. I was messing with my calculator and I noticed that half numbers have factorials and other decimals don't. So, I looked this up. I am not sure what level math it is. I am through Calc 2. If you could let me know what these identies are, I would appreciate it.
     
  6. Dec 1, 2005 #5

    CRGreathouse

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    [tex]\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}[/tex]

    For nonnegative integers, [tex]\Gamma(x+1)=x![/tex]. It is sensible to expand the notation x! to all real numbers other than the nonpositive integers.
     
  7. Dec 1, 2005 #6
    I'm not understanding that. What does it have to do with my equation I listed.
     
  8. Dec 1, 2005 #7

    shmoe

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    Okie, you know that (and how) factorial n! is defined on the integers. We define Gamma as:

    [tex]\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt[/tex]

    for all x>0 real numbers. Then you, having completed calc 2, can prove the following:

    -the integral in the definition of Gamma does indeed converge when x>0
    -for all x>0, [tex]\Gamma(x+1)=x\Gamma(x)[/tex], the "Functional equation" of gamma
    -if n is a non-negative integer, then [tex]n!=\Gamma(n+1)[/tex], which justifies calling Gamma an extension of factorial.

    Some calculators will use Gamma to make sense of arguments of x! that are not non-negative integers, and are essentially defining [tex]x!=\Gamma(x+1)[/tex] when x is a real number greater than -1. I find it odd that yours does half integral values but not other decimals. I believe you, it just seems like an odd thing to do.

    To prove your formula, you can use the identity involving sine CRGreathouse supplied to find [tex]\Gamma(1/2)[/tex], then use the functional equation above to find Gamma at 3/2, 5/2, ... n/2 (using induction) then translate back to factorial to get the equation in your first post. Alternatively, look directly at the integral definition of [tex]\Gamma(1/2)[/tex] and try to relate it back to the Gaussian probability integral.
     
    Last edited: Dec 1, 2005
  9. Dec 1, 2005 #8
    Alright, that makes sense, thanks.
     
  10. Dec 1, 2005 #9

    mathwonk

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    Riemann observed in one of his early papers that this expression for factorials of non integers allows differentiation to non integral orders, since by the cauchy integral formula differentiation to a non integral order t, simply requires one to integrate a non integral power, no problem, and determine the appropriate non integral factorial to multiply the integral by.

    so you can take the 1/2 derivative of something, e.g. or even the ith, I suppose.

    this is a little industry today.
     
  11. Aug 13, 2008 #10
    I have a somewhat related question. Is there/ does anyone know of a similar equation to the first one shown on this page for quarter integers. Basically:

    (n+1/4)!

    And particularly one that does not involve the gamma function.
     
  12. Dec 25, 2009 #11
    HALF INTEGER FACTORIAL

    The result [tex]({-½ })! = \sqrt{\pi}[/tex] is a well known result of half integer factorials.and can be proved easily using the gamma functiion.I have been searching over the internet for a proof using high school calculus but not successful. Therefore I came up with a short proof to share. Please advise.

    To prove: [tex](-½ )! = √(\pi)[/tex]

    First prove the relation:
    [tex]\int_{0}^{1}({1-x^2})^{n}dx =((2n/(2n +1)) \int_{0}^{1}({1-x^2})^{n-1}dx[/tex]

    Using integration by parts,
    [tex]\int{ v’}{ u } dx = uv - \int {v}{ u’ } dx [/tex]

    Let v = (1 – x2 )n , u = x. Then v’ = n(1 – x2 )n -1 (-2x).

    [tex]\int {n}({1-x^2})^{n-1} {(-2x) x } dx[/tex] = [tex]{x(1 – x^2 )^n } - \int {(({1-x^2})^{n})} dx [/tex]
    [tex]{-2n}\int {x^2 (1 – x^2 )^(n -1) } dx = {x(1 – x2 )n }- \int { (1 – x^2 )^n} dx [\tex]



    Add [tex] 2\int {(1 – x^2 )^(n -1 ) dx [/tex] to both sides,

    [tex]2n \int({1-x^2})^{n}dx = {x}{({1-x^2})^{n} - \int({1-x^2})^{n}dx + 2n\int{({1-x^2})^{n-1} dx [/tex]

    (2n+ 1)∫ (1 – x2 )n dx = x( 1 - x2 )n + 2n∫ (1 – x2 )n -1 dx


    Therefore, ∫ (1 – x2 )n dx = (x( 1 - x2 )n + (2n)∫ (1 – x2 )n-1 dx) /(2n +1)

    Set limits from 0 to +1,

    ∫01 (1 – x2 )n dx = (2n/(2n +1))∫01 (1 – x2 )n-1 dx

    Repeating the recurrence n – 1 more times,

    ∫01 (1 – x2 )n dx = [( 2n n! x / (2n+1)(2n -1)(2n-3)(2n-5)…)]01

    = (( 2n n! 2n n!/ (2n+1)! )

    = (( 22n (n!)2 / (2n+1)! )

    Make (n!) the subject,

    (n!) = √(( ∫01 (1 – x2 )n dx )/ ( 22n / (2n+1)! ))

    Put n = -½

    (-½) ! = √(( ∫01 (1 – x2 )-½ dx )/ ( 22(-½) / (2(-½)+1)! ))

    = √ (2( ∫01 (1 – x2 )-½ dx) ) = √(2( ∫01 (1 – x2 ) -½ dx ) ) =√(2( [sin-1 x ] 01) )= √(π).
     
    Last edited: Dec 26, 2009
  13. Dec 25, 2009 #12
    Very easy induction if you know that [tex] (\frac{1}{2})! = \frac{\sqrt{\pi}}{2} [/tex].

    We'll use induction. So, true for the case n = 0.

    Assume true up to some integer n. Prove n + 1.

    So, given,

    [tex]
    (n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}
    [/tex]

    Multiply both sides by [tex] (n + 1 + \frac{1}{2}) [/tex]

    [tex]
    (n+ 1 + \tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2} (n + 1 + \frac{1}{2}) = \sqrt{\pi} \prod_{k=0}^{n+1}\frac{2k+1}{2}[/tex]

    As requested.
     
  14. Dec 25, 2009 #13
    Please advise:

    How do you prove (1/2)! = sqt (pi)/2

    Thanks
     
  15. Dec 26, 2009 #14

    HallsofIvy

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    1. What is the definition of "(1/2)!"?

    2. All you need has already been given. In particular, use [itex]x!= \gamma(x+1)[/itex]
    and [tex]\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt[/tex].

    What is [itex]\int_0^\infty t^{1/2}e^{-t}dt[/itex]? I suggest the change of variable [itex]u= t^{1/2}[/itex].
     
  16. Dec 28, 2009 #15
    I am sorry that I messed up the LATEX in my previous post. I hope that this one would work.

    The result [tex] (-\tfrac{1}{2})! = \sqrt{\pi}[/tex] is a well known result of half integer factorials, and can be proved easily using the gamma functiion.I have been searching over the internet for a proof using high school calculus WITHOUT USING THE GAMMA FUNCTION but have not been successful. Therefore I came up with a short proof to share. Please advise.

    To prove:

    [tex] (-\tfrac{1}{2})! = \sqrt{\pi}[/tex]

    First prove the relation:
    [tex] \int_{0}^{1}(1-x^2 )^n dx = (( 2^{2n} (n!)^2 / (2n+1)! ) [/tex]

    Using integration by parts,
    [tex]\int ( \frac {dv}{dx}) u dx = uv - \int (\frac {du}{dx})v dx [/tex]

    Let [tex] v = (1-x^2 )^n , u = x, then \frac {dv}{dx} = n((1-x^2)^{(n-1)}) (-2x) [/tex].

    [tex]\int n(1-x^2 )^{(n-1)} (-2x) x dx = x(1-x^2 )^n - \int (1-x^2 )^n dx [/tex]
    [tex] -2n \int x^2 (1-x^2 )^{(n-1)} dx = x(1-x^2)^n - \int (1-x^2 )^n dx [/tex]

    Add [tex] 2n \int (1-x^2 )^{(n-1) } dx [/tex] to both sides,

    [tex] 2n \int (1-x^2 )^n dx = x(1-x^2 )^n - \int (1-x^2 )^n dx + 2n\int (1-x^2 )^{(n-1)} dx [/tex]

    [tex] (2n+ 1) \int (1-x^2 )^n dx = x( 1-x^2 )^n + 2n \int (1-x^2 )^{(n-1) } dx [/tex]


    Therefore, [tex] \int (1-x^2 )^n dx = (x( 1-x^2 )^n + (2n) \int (1-x^2 )^{n-1} dx) /(2n +1) [/tex]

    Set limits from 0 to +1,

    [tex] \int_{0}^{1}(1-x^2 )^n dx = (2n/(2n +1))\int_{0}^{1} (1-x^2 )^{n-1} dx [/tex]

    Repeating the recurrence n – 1 more times,

    [tex] \int_{0}^{1} (1-x^2 )^n dx = [( 2^n n! x / (2n+1)(2n -1)(2n-3)(2n-5)\cdots]_{0}^{1} [/tex]

    [tex] = (( 2^n n! 2^n n!/ (2n+1)! ) [/tex]

    [tex] = (( 2^{2n} (n!)^2 / (2n+1)! ) [/tex]

    Make [tex](n!) [/tex]the subject,

    [tex] (n!) = \sqrt (( \int_{0}^{1} (1-x^2 )^n dx )/ ( 2^{2n} / (2n+1)! )) [/tex]

    Put [tex] n =(-\tfrac{1}{2}) [/tex]

    [tex] (n!) =(-\tfrac{1}{2}) ! = \sqrt ( \int_{0}^{1} (1-x^2 )^{-\tfrac{1}{2}}dx / ( 2^{2(-\tfrac{1}{2})} / (2((-\tfrac{1}{2}))+1)! )) [/tex]

    [tex] = \sqrt (2( \int_{0}^{1} (1-x^2 )^{-\tfrac{1}{2}} dx) ) = \sqrt (2( \int_{0}^{1} (1-x^2 ) ^{-\tfrac{1}{2}} dx ) ) =\sqrt (2( [arcsin x ] _{0}^{1}) )= \sqrt {\pi} [/tex].
     
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