Define the double factorial as being a continous, non-hybrid function

[Saracen Rue]

TL;DR

Define $x!!$ as being a continuous, non-hybrid function over the domain of $[-1, \infty)$

The double factorial, $n!$ (not to be confused with $(n!)!$), can be defined for positive integer values like so:
$$n!=n(n−2)(n−4)(n−6)...(n-a)$$
Where $(n−a)=1$ if $n$ is odd or $(n−a)=2$ if $n$ is even. Additionally, the definition of the double factorial extends such that $(-1)!=0!=1$

I'm curious as to if it would be possible to definite a function which not only accurately evaluates all integer values for $n!$, $n \geq -1$, but also allows you to calculate the value of any real value of $n$ over the domain $[-1, \infty)$

 

[trurle]

Will be a discontinuity in second and higher derivatives at n=1. Therefore, single (not piece-wise) interpolation is not possible.

 

[Infrared]

You can also continuously extend a function defined on a discrete set, but here's a concrete approach. I assume you're familiar with the gamma function $\Gamma(z)=\int_0^{\infty} x^{z-1}e^{-x}dx$ which satisfies $\Gamma(n)=(n-1)!$ when $n$ is a nonnegative integer.

There is the duplication formula $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi} \ \Gamma(2z)$. When $z=n$ is a non-negative integer, this gives

$\Gamma(n+1/2)=\sqrt{\pi} \ 2^{1-2n}\frac{(2n-1)!}{(n-1)!}=\sqrt{\pi} \ 2^{-n} \frac{(2n-1)!}{2^{n-1}(n-1)!}=\sqrt{\pi} \ 2^{-n} (2n-1)!$

Rearranging, we find that the function $\frac{1}{\sqrt{\pi}}2^{(1+z)/2}\Gamma(z/2+1)$ is equal to $z!$ when $z$ is an odd (positive) integer. As far as I can tell, this is the standard analytic continuation of $(2n-1)!$. Unfortunately, this formula is false when $z$ is even.

I found a stackexchange answer (https://math.stackexchange.com/ques...ion-of-factorial-be-done-for-double-factorial)
which gives the function $2^{(1+2z-\cos(\pi z))/4}\pi^{(\cos(\pi z)-1)/4}\Gamma(z/2+1).$ When $z$ is odd, this is just the function I gave above. When $z=2n$ is even, it gives $2^nn!=(2n)!$. I don't think this is standard though.

 

[fresh_42]

Unfortunately, this formula is false when $z$ is even.

For even $z$ one could use $(2n)!=2^nn!$. A similar easy function is $(2n-1)!=\dfrac{(2n)!}{2^nn!}$. That leaves us with the problem to find one function for both cases, in which case we could simply substitute the gamma function.

 

[Saracen Rue]

You can also continuously extend a function defined on a discrete set, but here's a concrete approach. I assume you're familiar with the gamma function $\Gamma(z)=\int_0^{\infty} x^{z-1}e^{-x}dx$ which satisfies $\Gamma(n)=(n-1)!$ when $n$ is a nonnegative integer.

There is the duplication formula $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi} \ \Gamma(2z)$. When $z=n$ is a non-negative integer, this gives

$\Gamma(n+1/2)=\sqrt{\pi} \ 2^{1-2n}\frac{(2n-1)!}{(n-1)!}=\sqrt{\pi} \ 2^{-n} \frac{(2n-1)!}{2^{n-1}(n-1)!}=\sqrt{\pi} \ 2^{-n} (2n-1)!$

Rearranging, we find that the function $\frac{1}{\sqrt{\pi}}2^{(1+z)/2}\Gamma(z/2+1)$ is equal to $z!$ when $z$ is an odd (positive) integer. As far as I can tell, this is the standard analytic continuation of $(2n-1)!$. Unfortunately, this formula is false when $z$ is even.

I found a stackexchange answer (https://math.stackexchange.com/ques...ion-of-factorial-be-done-for-double-factorial)
which gives the function $2^{(1+2z-\cos(\pi z))/4}\pi^{(\cos(\pi z)-1)/4}\Gamma(z/2+1).$ When $z$ is odd, this is just the function I gave above. When $z=2n$ is even, it gives $2^nn!=(2n)!$. I don't think this is standard though.

Thank you, a function like $f(z)=2^{(1+2z-\cos(\pi z))/4}\pi^{(\cos(\pi z)-1)/4}\Gamma(z/2+1)$ is exactly what I was looking for. I wonder if it would be able to define a similar such function for the tripple factorial.