How to prove the half number factorial formula?

[laker88116]

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

 

[shmoe]

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

 

[HallsofIvy]

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

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

 

[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.

 

[CRGreathouse]

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.

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

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

 

[laker88116]

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

 

[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.

 

[laker88116]

Alright, that makes sense, thanks.

 

[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.

 

[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.

 

[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).

$$\int {n}({1-x^2})^{n-1} {(-2x) x } dx$$ = $${x(1 – x^2 )^n } - \int {(({1-x^2})^{n})} dx$$
[tex]{-2n}\int {x^2 (1 – x^2 )^(n -1) } dx = {x(1 – x2 )n }- \int { (1 – x^2 )^n} dx [\tex]<br /> <br /> <br /> <br /> Add $$2\int {(1 – x^2 )^(n -1 ) dx$$ to both sides,<br /> <br /> $$2n \int({1-x^2})^{n}dx = {x}{({1-x^2})^{n} - \int({1-x^2})^{n}dx + 2n\int{({1-x^2})^{n-1} dx$$<br /> <br /> (2n+ 1)∫ (1 – x2 )n dx = x( 1 - x2 )n + 2n∫ (1 – x2 )n -1 dx <br /> <br /> <br /> Therefore, ∫ (1 – x2 )n dx = (x( 1 - x2 )n + (2n)∫ (1 – x2 )n-1 dx) /(2n +1)<br /> <br /> Set limits from 0 to +1,<br /> <br /> ∫01 (1 – x2 )n dx = (2n/(2n +1))∫01 (1 – x2 )n-1 dx<br /> <br /> Repeating the recurrence n – 1 more times, <br /> <br /> ∫01 (1 – x2 )n dx = [( 2n n! x / (2n+1)(2n -1)(2n-3)(2n-5)…)]01<br /> <br /> = (( 2n n! 2n n!/ (2n+1)! ) <br /> <br /> = (( 22n (n!)2 / (2n+1)! ) <br /> <br /> Make (n!) the subject,<br /> <br /> (n!) = √(( ∫01 (1 – x2 )n dx )/ ( 22n / (2n+1)! ))<br /> <br /> Put n = -½<br /> <br /> (-½) ! = √(( ∫01 (1 – x2 )-½ dx )/ ( 22(-½) / (2(-½)+1)! ))<br /> <br /> = √ (2( ∫01 (1 – x2 )-½ dx) ) = √(2( ∫01 (1 – x2 ) -½ dx ) ) =√(2( [sin-1 x ] 01) )= √(π).[/tex]

 

[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.

 

[leepakkee]

Please advise:

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

Thanks

 

[HallsofIvy]

Please advise:

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

Thanks

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 $u= t^{1/2}$.

 

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

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

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