# How to prove the half number factorial formula?

1. Dec 1, 2005

### laker88116

Any ideas on how to prove this?
$$(n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}$$

2. Dec 1, 2005

### shmoe

You've presumably met the Gamma function. Have a look through identites involving Gamma that you know, you should find something useful.

3. Dec 1, 2005

### HallsofIvy

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

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

4. Dec 1, 2005

### laker88116

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.

5. Dec 1, 2005

### CRGreathouse

$$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$$

For nonnegative integers, $$\Gamma(x+1)=x!$$. It is sensible to expand the notation x! to all real numbers other than the nonpositive integers.

6. Dec 1, 2005

### laker88116

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

7. Dec 1, 2005

### shmoe

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

$$\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt$$

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, $$\Gamma(x+1)=x\Gamma(x)$$, the "Functional equation" of gamma
-if n is a non-negative integer, then $$n!=\Gamma(n+1)$$, 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 $$x!=\Gamma(x+1)$$ 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 $$\Gamma(1/2)$$, 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 $$\Gamma(1/2)$$ and try to relate it back to the Gaussian probability integral.

Last edited: Dec 1, 2005
8. Dec 1, 2005

### laker88116

Alright, that makes sense, thanks.

9. Dec 1, 2005

### mathwonk

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.

10. Aug 13, 2008

### rman144

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.

11. Dec 25, 2009

### leepakkee

HALF INTEGER FACTORIAL

The result $$({-½ })! = \sqrt{\pi}$$ 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: $$(-½ )! = √(\pi)$$

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

Using integration by parts,
$$\int{ v’}{ u } dx = uv - \int {v}{ u’ } dx$$

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

$$\int {n}({1-x^2})^{n-1} {(-2x) x } dx$$ = $${x(1 – x^2 )^n } - \int {(({1-x^2})^{n})} dx$$
$${-2n}\int {x^2 (1 – x^2 )^(n -1) } dx = {x(1 – x2 )n }- \int { (1 – x^2 )^n} dx [\tex] Add [tex] 2\int {(1 – x^2 )^(n -1 ) dx$$ to both sides,

$$2n \int({1-x^2})^{n}dx = {x}{({1-x^2})^{n} - \int({1-x^2})^{n}dx + 2n\int{({1-x^2})^{n-1} dx$$

(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
12. Dec 25, 2009

### l'Hôpital

Very easy induction if you know that $$(\frac{1}{2})! = \frac{\sqrt{\pi}}{2}$$.

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

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

So, given,

$$(n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}$$

Multiply both sides by $$(n + 1 + \frac{1}{2})$$

$$(n+ 1 + \tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2} (n + 1 + \frac{1}{2}) = \sqrt{\pi} \prod_{k=0}^{n+1}\frac{2k+1}{2}$$

As requested.

13. Dec 25, 2009

### leepakkee

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

Thanks

14. Dec 26, 2009

### HallsofIvy

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

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

What is $\int_0^\infty t^{1/2}e^{-t}dt$? I suggest the change of variable $u= t^{1/2}$.

15. Dec 28, 2009

### leepakkee

I am sorry that I messed up the LATEX in my previous post. I hope that this one would work.

The result $$(-\tfrac{1}{2})! = \sqrt{\pi}$$ 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:

$$(-\tfrac{1}{2})! = \sqrt{\pi}$$

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

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

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

$$\int n(1-x^2 )^{(n-1)} (-2x) x dx = x(1-x^2 )^n - \int (1-x^2 )^n dx$$
$$-2n \int x^2 (1-x^2 )^{(n-1)} dx = x(1-x^2)^n - \int (1-x^2 )^n dx$$

Add $$2n \int (1-x^2 )^{(n-1) } dx$$ to both sides,

$$2n \int (1-x^2 )^n dx = x(1-x^2 )^n - \int (1-x^2 )^n dx + 2n\int (1-x^2 )^{(n-1)} dx$$

$$(2n+ 1) \int (1-x^2 )^n dx = x( 1-x^2 )^n + 2n \int (1-x^2 )^{(n-1) } dx$$

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

Set limits from 0 to +1,

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

Repeating the recurrence n – 1 more times,

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

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

$$= (( 2^{2n} (n!)^2 / (2n+1)! )$$

Make $$(n!)$$the subject,

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

Put $$n =(-\tfrac{1}{2})$$

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

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