Can the value of $(-9!)$ be found

Question

I saw this question on an fb page and I couldn't solve it.

Question:

What is the value of $(-9!)$?

a)$362800$

b)$-362800$

c) Can not be calculated

The first options seems to be incorrect,which leaves $c$ but I can't justify it.Does it have something to do with gamma function which asks for$\int _{ 0 }^{ \infty }{ { x }^{ -10 } } { e }^{ -x }dx$. Why can't it be calculated?

Update:

I have been given answers that "using the Gamma function, it can't be evaluated". Isn't there some other way to do so?

Answer 1

The factorial function from $\mathbb{N}$ to $\mathbb{N}$ is a special case of the gamma function from $\mathbb{C}$ to $\mathbb{C}$:

$$n! = \Gamma(n+1) = \int\limits_{0}^{\infty}{x}^{n}{e}^{-x}\,{\rm{d}}x$$

Unfortunately, this function is defined for all complex numbers except negative integers and zero.

Answer 2

The above two are correct about operator precedence and so (b) would be my answer. But might I also add that the gamma function form also yields undefined values for negative integer factorials.

Answer 3

If you mean $-(9!)$, which will often be written simply as $-9!$, then the answer would be $b$, because it is simply $(-1)*(9!)$, or -362800. However, if you mean $(-9)!$, then the answer would be $c$, because the gamma function has poles at the negative integers and zero, because $\Gamma(x)$ can be defined as $\frac{\Gamma(x+k)}{\prod_{j=0}^{k-1}x+j}$ for any $k$ where $x+k>0$. Therefore if $x$ is a negative integer, than one of the factors of the denominator will be zero, and so there will be a pole at all negative integers.

Answer 4

99.9 % of peopel wil get dis wrong!!!

But in all seriousness:

You don't have to define the factorial function in terms of the gamma function nor the gamma function in terms of the factorial function!!!! In this light, the other answers (in my humble opinion) are very wrong.

However, the gamma function is nice because it $interpolates$ the factorial function nicely for positive integers. Physicists often like to do other things such as extend the factorial function to be able to compute things like $\frac{1}{2}! = \sqrt{\frac{\pi}{4}}$

You should realize that you CAN extend the factorial function so long as your audience is clear that you are defining something a certain way. For example, the factorial function is usually defined as $0!=1$ and $n!=(n-1)!$ then it might be natural to do the following for the factorial function on negative integers, let $(-n!)=(-(n-1)!)$ and $0!=1$, so that for even $n$ we have $n!=(-n!)$ and that for odd $n$ we have that $-(n!)=(-n!)$

For example, $$-1!=-1$$ $$-2!=2$$ $$-3!=-6$$ $$-4!=24$$ and so on; of course, this is just one way to define it.... and as the post you found (probably) said itself, $99$% of people get this wrong.

Answer 5

We have the following property: $n!=\dfrac{(n+1)!}{n+1}$ . Hence, $0!=\dfrac{1!}1=1;~(-1)!=\dfrac{0!}{0_\pm}=\pm\infty$, etc.