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This question already has an answer here:

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?

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marked as duplicate by Namaste, C. Falcon, Juniven, Shailesh, JonMark Perry Mar 31 '17 at 6:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Maybe it depends on the meaning of the symbols and the definition you use? Btw. does it mean $-(9!)$ or $(-9)!$? $\endgroup$ – user144248 Jun 19 '14 at 15:16
  • $\begingroup$ Operator precedence would usually have $-9!$ be parsed as $-(9!)$. $\endgroup$ – Daniel Fischer Jun 19 '14 at 15:16
  • $\begingroup$ What Daniel said. Also, the integral you wrote is divergent at $0$. $\endgroup$ – PA6OTA Jun 19 '14 at 15:21
  • $\begingroup$ @DanielFischer Can't $(-9)!$ be calculated in any way? $\endgroup$ – Syn Jun 19 '14 at 15:25
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    $\begingroup$ @Bananarama After all this debate about the meaning of $(-9!)$, to replace it by $(-9)!$ seems misleading. $\endgroup$ – Did Jun 19 '14 at 17:16
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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.

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    $\begingroup$ Because it's not defined for negative integers and $-9$ is a negative integer.... let's put that together.... $\endgroup$ – Squirtle Jun 19 '14 at 15:28
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    $\begingroup$ @Squirtle: Well... I figured OP would notice that $-9$ was a negative integer by now... $\endgroup$ – barak manos Jun 19 '14 at 15:28
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    $\begingroup$ Although, I think that $\Gamma(1)=1$ so that the $n=0$ is allowed. Also, just because the $\Gamma$ function doesn't work doesn't mean we can just define $(-n)!$ to be $-(n!)$. For example, many number theorists define $-3$ as a "negative prime number". $\endgroup$ – Squirtle Jun 19 '14 at 15:31
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    $\begingroup$ ...... I am not questioning the definition of $-p$ being prime; merely, I am stating we can OFTEN very naturally define things for $-n$ the same way we do for $n \in \mathbb{N}$ if we are careful. There's a problem with the definition of $(-n)!:=-(n!)$. Namely, why is it better than the other definition: $(-n)!=(-n)(-(n-1))!$ Notice these differ if $-n$ is even. The former computes $(-4)!=-(4!)=-24$; while the latter computes $(-4)!=(-4)(-3)(-2)(-1)(0!)=24$ Also, I should point out your definition of a prime number is not canonical, as normally it's only defined for natural numbers. $\endgroup$ – Squirtle Jun 19 '14 at 16:42
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    $\begingroup$ Additionally, if the definition of prime were given over $\mathbb{Z}$ then we would have to say, "There are exactly 4 divisors for $n$ ($n,-n,-1,$ and $1$)" As this is not the definition, we can just agree (perhaps informally) that $-p$ is prime if $p$ is. $\endgroup$ – Squirtle Jun 19 '14 at 16:45
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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.

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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.

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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.

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    $\begingroup$ I shouldn't have to say this... but the opening line is a reference to the fact that this question was pulled from a fb post. $\endgroup$ – Squirtle Jun 19 '14 at 17:14
  • $\begingroup$ Somebody upvoted me and then someone else downvoted me.... @Downvoter, care to explain yourself? You know that comment you get when you downvote: "Please consider leaving a comment if you think the post can be improved." Well!? $\endgroup$ – Squirtle Jun 19 '14 at 17:32
  • $\begingroup$ I have never seen the factorial function defined this way. Hardly seems useful to have the factorial of any positive integer be the same value $\endgroup$ – ClassicStyle Jul 13 '14 at 21:51
  • $\begingroup$ I made a really bad typo on my phone and wasn't able to edit it there...... I wanted to say, "You can define something however you wish; then people can do whatever they want with it." $\endgroup$ – Squirtle Jul 15 '14 at 18:06
  • $\begingroup$ I'm not sure if you are trolling or not...the factorial function is never defined as $n!=(n-1)!$. I'm also not sure why you would say that it usually is $\endgroup$ – ClassicStyle Jul 16 '14 at 7:49
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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.

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  • $\begingroup$ That only if you (rather arbitrarily) want that to continue holding... $\endgroup$ – vonbrand Jun 19 '14 at 20:26
  • $\begingroup$ @vonbrand: It's just like extending exponents to non-natural values. Generalizations are done by keeping the basic functional properties intact. $\endgroup$ – Lucian Jun 19 '14 at 20:48
  • $\begingroup$ But some properties go out the window, i.e., no more powers of negative numbers when raising to a real. $\endgroup$ – vonbrand Jun 19 '14 at 20:50
  • $\begingroup$ @vonbrand: Exponentiation can be generalized even further to encompass complex numbers, and so can factorials. Either way, if you have an alternative generalization, based on something else, no one is stopping you from posting it as an answer. $\endgroup$ – Lucian Jun 19 '14 at 20:55

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