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Posted

Well, this isn't a homework problem or anything, but we're incorporating factorials into our Calc II now..and I hate them to be honest :)

 

I was just wondering what would be 4!!!! or if it's even possible to calculate. I did 3!!! which made my head hurt :P

Posted

One of my lecturers recons we should be able to remember up to 6!

 

4 is just 24 surely?

 

5 is 120 I think

 

and 6 is then 720

 

7 is 5040

 

etc... 7*6*5*4*3*2*1

Posted

4! is 24, but 4!!!! isn't! 4!! is 24!, for example. And 4!!!! is 24!!!. In other words, we're talking about an obscenely, obnoxiously big number. I'm not going to try to calculate it.

Posted
4! is 24, but 4!!!! isn't! 4!! is 24!, for example. And 4!!!! is 24!!!. In other words, we're talking about an obscenely, obnoxiously big number. I'm not going to try to calculate it.

 

A good point, serves me right for not reading it correctly. My excuse is it's 12:30am I've been awake since 6:30am

Posted

4!!!! = 24!!! > (24!/9!)!! > 10^10 !! > (10^10 ! / 10^9 !) ! > 10^90 ! > 10^90 ! / 10^89 ! > 10^(90*89) = 10^8010.

 

In other words: 4!!!! has more than eight thousand digits (probably a lot more). You´d better be prepared to need a large pice of paper if you want to calculate it.

 

EDIT: To emphasize that above was only a very rough lower limit: 24! = 6.2 *10^23 already, which already is 13 orders of magnitude greater than my approximation. And as a side-effect, we now know how to write the Avogadro constant in a way that no one will understand :D.

EDIT2: You (abskebabs) are right. I somehow though it was 6.22*10^23 when I wrote it; only remembered the ...22 .. ^23 part and forgot about the zero, somehow.

Posted

Sorry if I'm being hopelessly ignorant here, but; couldn't the gamma function be used and adapted to solve this problem? The answer would still be ridiculous tho....

EDIT: Atheist, I'm sure you know this but 24! is not avagadro's constant, sry if thats pedantic...

Posted

It looks like a pain for multiple !'s, but you can use Sterling's formula as a pretty good approximation for n! for n>10.

 

Sterling's forumla is:

[math]n! \sim \sqrt{2 \pi n}n^{n}e^{-n}[/math]

Posted

I love the factorial calculators, they just say infinity.

Haha, I wouldn't expect anyone to calculate such a massive number..

I just wondered if it was possible..which technically, it is.

thanks :P

Posted

Just a really quite, pedantic note. You should be careful when using multiple factorial symbols, as 4!! is generally regarded as the double factorial:

 

[math]n!! = \begin{cases} n(n-2) \cdots 5 \cdot 3 \cdot 1, & n \text{ odd} \\ n(n-2) \cdots 4 \cdot 2, & n \text{ even} \end{cases}[/math]

 

Indeed, multiple factorial symbols are generally regarded as multifactorials. Just so you know :)

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