Cyanide Posted March 14, 2007 Share Posted March 14, 2007 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 Link to comment Share on other sites More sharing options...
Klaynos Posted March 15, 2007 Share Posted March 15, 2007 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 Link to comment Share on other sites More sharing options...
Sisyphus Posted March 15, 2007 Share Posted March 15, 2007 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. Link to comment Share on other sites More sharing options...
Klaynos Posted March 15, 2007 Share Posted March 15, 2007 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 Link to comment Share on other sites More sharing options...
timo Posted March 15, 2007 Share Posted March 15, 2007 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 . 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. Link to comment Share on other sites More sharing options...
abskebabs Posted March 15, 2007 Share Posted March 15, 2007 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... Link to comment Share on other sites More sharing options...
Bignose Posted March 15, 2007 Share Posted March 15, 2007 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] Link to comment Share on other sites More sharing options...
Cyanide Posted March 15, 2007 Author Share Posted March 15, 2007 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 Link to comment Share on other sites More sharing options...
Dave Posted March 15, 2007 Share Posted March 15, 2007 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 Link to comment Share on other sites More sharing options...
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