How can we find factorials in decimal form?

Question

I've heard of factorials such as $5!$ and $3!$, which work like this: $5!=5\times4\times3\times2\times1=120$ and $3!=3\times2\times1=6$. At least this is what we get. Also, surprisingly, $0!=1$, but how can we figure out values for factorials in decimal form such as $(2.5)!$ and $(3.34)!$? Maybe we should use gamma for part of it? What else? I also know that we can draw graphs for factorials. I just want to know how to find values for factorials in decimal form?

Answer 1

There are many functions $F\colon[0,\infty)$ such that $f(n)=n!$ for $n=0,1,2\dots$. How to choose the right one? It seems reasonable to require that $F$ be continuous, but this is not enough. An important property of the factorial is $$ (n+1)!=(n+1)\,n!. $$ We then add the requirement $$ F(x+1)=(x+1)\,F(x),\quad x>0. $$ There are still many functions satisfying all these requirements. The Gamma function, defined as $$ \Gamma(x)=\int_0^\infty x^n\,e^{-x}\,dx, $$ Note: this is not the $\Gamma$ function, which is defined as $\Gamma(x)=\int_0^\infty x^{n-1}\,e^{-x}\,dx$, so that $n!=\Gamma(n+1)$.

is the unique that satisfies the additional requirement that $\ln F(x)$ is a convex function. To summarize, $\Gamma$ is the only function such that

1. $F(0)=1$ and $F(x+1)=(x+1)\,F(x)$ if $x>0$.
2. $\ln F(x)$ is convex.

It turns out that $\Gamma(z)$ can be extended to complex numbers $z$ which are not a negative integer.

Answer 2

This is not a question that could be easily answered, as there are many ways (infinite ways of drawing a curve through the integer factorials) of defining the factorial function over the real numbers.

One of the most used functions that extends the factorial to the reals, and the one that you yourself suggested is the gamma function, for which the equality $$\Gamma(n+1)=n!$$ for integer $n$ exists. The problem is that the gamma function cannot be expresed without using the language of calculus, which I assume you have yet to learn.

Currently it would be impossible for you to calculate the values of non-integer factorials.

Answer 3

$0!=1$ because that makes the functional equation $n!=(n-1)!*n$ hold (and because it is generally smart to define the empty product to be $1$.

To make any sense of things like $(\frac12)!$, we need to find some way to define factorial of non-integers, and the common way to do that is to observe that $\Gamma(n+1)=n!$ for integers and turn that into the definition of factorials for non-integers.

There are other functions that satisfy the functional equation, and agree with $n!$ on the integers, but $\Gamma$ is the most useful.

Answer 4

I just want to know how to find values for factorials in decimal form?

Hi;

According to Meck in Numerical Analysis you can use this truncated series

$$f(x)=\log(x!)$$

$$\log(x!)\approx\left(x+\frac{1}{2}\right) \log (x)-x+\frac{1}{2} \log (2 \pi \ )+\frac{1}{12x}\left ( 1-\frac{1}{x^2}\left ( \frac{1}{30}-\frac{1}{x^2}\left ( \frac{1}{105}-\frac{1}{x^2}\left ( \frac{1}{140}-\frac{1}{x^2}\left ( \frac{1}{99} \right ) \right ) \right ) \right ) \right )$$

because the above series yields about 8 digits if $x\geq5.5$ we use the relation $$\text{B) }3.34!=\frac{5.34!}{5.34\cdot4.34}$$

Now we just plugin to f:

$$ f (5.34) = 5.37784377585049$$

$$5.34!\approx e^{5.37784377585049}\approx 216.554830892397$$

Now into B:

$$3.34!\approx\frac{216.554830892397}{5.34\cdot 4.34}\approx9.34408735447613$$

which is off by one in the 10th decimal place.