How to get decimal answers of a factorial

Question

I have had a familiar understanding of how factorials work within mathematics for some time now. However, I have recently discovered that you are able to get decimal factorials.

So far under my research and calculations, I have been able to deduce that:

$\left(t+\frac{1}{2}\right)!=\frac{\sqrt{\pi }}{2}\cdot \prod _{n=1}^(t\frac{2n+1}{2}$

t is a variable

Additionally:

$\left(t-\frac{1}{2}\right)!=\frac{\sqrt{\pi }}{2}\cdot \prod _{n=1}^{t-1}\frac{2n+1}{2}$

t is a variable

This so far will only work for half values and therefore I have not found a way for it to work with other fractions e.g.

$\frac{1}{3},\ \frac{1}{4},\ \frac{1}{5},\ ...$

$\left(t+\frac{1}{v}\right)!= ?$

t and v are both variables

If anyone has any intel on how this would work please free to leave a comment below or any solutions. It would be as always, much appreciated. I am aware that there is a calculus solution but I am looking for a non-calculus solution. Anyone interested in the calculus solution it is:

$Γ\left(t\right)=\int _0^∞x^{\left(t-1\right)}\cdot e^{-x}dx$

t is a variable

the gamma function is a generaisation of the factorial function to all real and even all complex numbers () except the negative integers, there it has (and should have) poles — Jens Renders, Aug 01 '17 at 14:38
Wikipedia's Gamma function article has some remarks about how to approximate values of the gamma function under the heading "Euler's definition as an infinite product". — hmakholm left over Monica, Aug 01 '17 at 14:58
@JensRenders "there it has (and should have) poles" – why? — BarbaraKwarc, Jan 03 '18 at 08:24

hmakholm left over Monica · Answer 1 · 2017-08-01T15:56:56.253

First, let's seek $\Gamma(x)$ where $x=n+\varepsilon$ where $n$ is a large integer and $|\varepsilon|<1$.

Let $f(x) = \log(\Gamma(x))$. Since $f(x+1) = f(x) + \log(x)$ we have $$ f'(x) \sim \log x $$ $\log x$ doesn't change very quickly compared to its magnitude when $x$ is large, so let's approximate $$ f(n+\varepsilon) \approx f(n) + \varepsilon f'(n) \approx f(n) + \varepsilon \log n $$ which is the same as $$ \Gamma(n+\varepsilon) \approx \Gamma(n) \cdot n^\varepsilon = (n-1)!\cdot n^\varepsilon$$ This approximation is of course exact for $\varepsilon=0$ and $\varepsilon=1$. To get an idea of the error we risk, let's set $\varepsilon=-1$ and compare: $$ \frac{\Gamma(n-1)_{\rm approx}}{\Gamma(n-1)} = \frac{\Gamma(n) / n }{\Gamma(n) / (n-1) } = 1 + \frac{1}{n-1} $$

So now we can compute $\Gamma(n+\varepsilon)$ to a relative precision of about $\frac1n$.

If we want better precision than that, choose a larger $n$ (but the same $\varepsilon$) and work our way down to the original $n$ by repeatedly applying $$ \Gamma(x-1) = \frac{\Gamma(x)}{x-1} $$ which preserves the relative precision all the way down (assuming the divisions are exact).

A slightly more careful derivation yields that $$ \Gamma(n+\varepsilon) \approx (n-1)! (n-\tfrac12+\tfrac\varepsilon2)^\varepsilon \quad \text{with }\varepsilon\in[-\tfrac12,\tfrac12] $$ gives a relative error of at most $\frac{1}{8n^2}$.

score 1 · Answer 2 · answered Aug 01 '17 at 17:17

For the exact values of $x!$ for $x=a/b$ and $120\equiv0\pmod b$, see this paper. However, the solutions are extraordinarily complicated.

There are a few approaches, however, to dealing with the factorials of fractional values. The first is the multiplication formula:

$$(2\pi)^{\frac{k-1}2}k^{\frac12-kz}\Gamma(kz)=\prod_{i=0}^{k-1}\Gamma\left(z+\frac ik\right)$$

For example, with $k=2$, we get

$$\sqrt{2\pi}k^{\frac12-2z}\frac{\Gamma(2z)}{\Gamma(z)}=\Gamma\left(z+\frac12\right)$$

However, this formula does not really provide closed forms for any other fractional values. At best, it will give you $\Gamma(z+\frac13)$ in terms of $\Gamma(z+\frac23)$, for example..

At $1/4$, we happen to have Gauss' constant,

$$\Gamma\left(\frac14\right)=\sqrt{G\sqrt{8\pi^3}}$$

$G$ can be easily calculated using the AM-GM form:

$$G=\frac1{\operatorname{agm}(1,\sqrt2)}$$

As a quick example,

$a_0=1,g_0=\sqrt2$

$a_1=\frac{1+\sqrt2}2\approx1.207,g_1=\sqrt[4]{2}\approx1.189$

$a_{n+1}=\frac{a_n+g_n}2,g_{n+1}=\sqrt{a_ng_n}$

Thus. we know that $G\approx1/1.195=0.83$

From there, it is easy to find $\Gamma\left(n+\frac14\right)$ from $\Gamma(x+1)=x\Gamma(x)$.

Values for $\Gamma\left(n+\frac34\right)$ may then be calculated from the $\Gamma\left(z+\frac12\right)$ formula.

Straight away approximations may also be calculated from the full Stirling approximation:

$$\ln(\Gamma(x))=x\ln(x)-x-\frac12\ln\left(\frac x{2\pi}\right)+\sum_{k=2}^\infty\frac{B_{k+1}}{k(k+1)z^k}$$

where $B_k$ are the Bernoulli numbers.

robjohn · Accepted Answer · 2017-08-01T17:58:12.540

1

In this answer, it is shown that $$ \frac1{\Gamma(1+z)}=e^{z\gamma}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} $$ Therefore, we have $$ z!=\Gamma(1+z)=e^{-z\gamma}\prod_{k=1}^\infty e^{\frac{z}{k}}\left(1+\frac{z}{k}\right)^{-1} $$ where $\gamma$ is the Euler-Mascheroni Constant.

edited Aug 01 '17 at 17:58

answered Aug 01 '17 at 17:29

robjohn

345,667

Sorry, what is z in this solution – Morgan Aug 01 '17 at 17:57
$z$ is a complex number. You can plug in any $z\in\mathbb{R}$. However, using $z\in[0,1]$ and using the duplication formula $\Gamma(1+z)=z\Gamma(z)$ to get $\Gamma(1+z)$ in other ranges works better. – robjohn Aug 01 '17 at 17:58
Oh, thats interesting, I always assumed that i or j was used to represent imaginary numbers, thanks for the insight. – Morgan Aug 01 '17 at 18:00
$i$ so that $i^2=-1$ is the imaginary unit. $z$ is simply a variable that takes a complex value. – robjohn Aug 01 '17 at 18:02
1

@Morgan $z$ is like the complex variable for complex analysis, just like $x,y$ are the real variables you often see in algebra. Of course, complex numbers also cover real numbers. :-) – Simply Beautiful Art Aug 01 '17 at 18:06
So can $z^t$ be possible is t is a variable as well? – Morgan Aug 01 '17 at 18:08
in terms of imaginary numbers, i feel that people give $e^{\pi \cdot i}$ too much credit when $e^{\tau \cdot i}$ gives you 1 – Morgan Aug 01 '17 at 18:09
@Morgan What $t$? – Simply Beautiful Art Aug 01 '17 at 18:09
Personally, I prefer the Bohr-Mollerup formula. Requires nothing really advanced:$$\Gamma(x)=\lim _{n\to \infty }{\frac {n^xn!}{(x+n)(x+n-1)\cdots (x+1)x}}$$ – Simply Beautiful Art Aug 01 '17 at 18:10
@Simply Beautiful Art $t$ is just another variable that I use because a lot of the alphabet is already used. – Morgan Aug 01 '17 at 18:10
@Morgan $k,n,i$ are often natural numbers. $a,b,c,t,u,v,x,y$ are often real numbers. $s,z$ are often complex. Often you will see things like $z=x+iy$ or $z=u+vi$. – Simply Beautiful Art Aug 01 '17 at 18:12
Let us continue this discussion in chat. – Morgan Aug 01 '17 at 18:12

How to get decimal answers of a factorial

3 Answers3