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

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

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shmoe

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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!)

(Schmo already told you that- my point is you should think about it!)

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CRGreathouse

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[tex]\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}[/tex]laker88116 said:

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.

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I'm not understanding that. What does it have to do with my equation I listed.

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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.

[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.

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Alright, that makes sense, thanks.

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mathwonk

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

this is a little industry today.

- #10

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(n+1/4)!

And particularly one that does not involve the gamma function.

- #11

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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).

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).

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- #12

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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.

- #13

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Please advise:

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

Thanks

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

Thanks

- #14

HallsofIvy

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1. What is thePlease advise:

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

Thanks

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].

- #15

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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

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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