Tristan
20031125 01:41:23 
.5!
I was using my TI83 plus to graph y=x!. To my deep surprise, the calculator said that there is a .5! Wow! Besides that, it also seemed that there was a solution for (.5)! and every other multiple of 1/2 above these. The calculator gave irrational numbers for all of them. I've been puzzling over this over the last few days. Can someone explain why you can take the factorial of halves, or, in the likeliness that it is much too complex, verify that it isn't just a bug in the calculator or my mind playing tricks on me? 
fwaff
20031125 06:28:33 
Re: .5!
Given that the definition of n! is 1x2x3x...xn, then 0.5! is undefined. What answer did your calculator give? If the calculator gave 0.5 as the answer, then it could be working on the basis of 0.5! = (1/2)! = 1!/2! = 1/2 (=0.5) What solutions does it give for other multiples of 1/2? eg does 2.5! = 60 (=5!/2!)
You could always have a look at the instructions to see if it gives any clue. 
Popstar Dave
20031125 07:01:23 
Re: .5!
I've got a TI83 as well, and for me I'm getting:
(1/2)! = 0.8862269255 (x)
(3/2)! = 1.329340388 (y)
(5/2)! = 3.32335097 (z)
As a note:
x/y = 2/3
y/z = 2/5
x/z = 4/15
So, yeah...
As for factorials of negatives, I get an error for all negative integer values and negative fractional values EXCEPT:
(1/2)! = 1.772453851
Using the values calculated above, this value of (1/2)! is double the value for (1/2)!.
I'm stumped. 
Charlie
20031125 08:32:16 
Re: .5!
The version of factorials extended to the real domain (and even the complex domain) exists in the Gamma function. For integers, n! = Γ(n+1). A graph of the Gamma function can be found at Mathworld. So the calculator in this instance is giving you Γ(n+1). As (n1)! = n!/n, (1/2)!, defined in this manner, is (1/2)!/(1/2), or 2 (1/2)!. 
Charlie
20031125 08:50:09 
Re: .5!
And of course the factorials for negative integers do not exist, as 1! would be 0!/0, or 1/0. 
Brian Smith
20031125 09:49:57 
Re: .5!
The exact value of (1/2)! = Gamma(3/2) = (sqrt(pi))/2
The exact value of (1/2)! = Gamma(1/2) = (Gamma(3/2))/(1/2) = sqrt(pi)

Tristan
20031125 18:15:59 
Re: .5!
I thought it might be related somehow to advanced calculus! As long as it's not a bug, I'm happy. Thanks! 
Charlie
20031125 18:38:28 
Re: .5!
I can't figure out why they'd limit the fractional possibilities to multiples of 1/2 though. 
Tristan
20031125 19:12:09 
Re: .5!
You mean in the calculator? I was wondering that too. Maybe the real bug in calculator is that it can't calculate the factorial of smaller fractions. lol 
Professor K at MCC
20040304 19:19:15 
Re: .5!
The reason that the TI83 and probably others allows for increments of .5 is that it is required to calculate the Probability Density function for a tdistribution. In the formula one encounters both (n/2)! and [(n1)/2]! Where n can be any positive integer. So of the two expressions in the factorials must be of the form #.5 and thus the TI wants to know that value and to allow people to calculate it if they have to. As was mentioned previously the Gamma function is what we use to extend the factorial and Gamma(n+1)=n!. Hope this helps. 
Grade 11 student
20040312 12:07:36 
Re: .5!
I always thought it went like this:
4,5! = 4,5 * 3,5 * 2,5 * 1,5 * (0,5)!
The (0,5)! could be replaced with either (sqrt(ð))/2 or just 0,886226925 if you wanted.
ð = pi so you don't confuse it with n. 
Steve Royer
20040617 21:20:35 
Re: .5!
my 20 year old casio fx570 gives an error for 0.5!
(i think its on its 4th set of batteries) my son had
to get a ti graphing job for school math, batteries last
about 3 months go figure 
brianjn
20040626 01:13:28 
Re: .5!
In Excel the factorial function is FACT(#), where # is the upper multiplier.
It appears to treat any decimal like the INT function before doing the factorial calculation; I note the same occurs if I use vulgar fractions eg (3 + 1/2). 
Jer
20041008 12:44:07 
Re: .5!
So what about .6!?
Does that have a simply expressable exact value?
Does any rational number?

Brian Smith
20041008 22:09:13 
Re: .5!
First, .6! = Gamma(1.6)
One definition of the gamma function is: Gamma(z) = Integral {0,1} (ln(1/t))^(z1) dt
So .6! = Integral {0,1} (ln(1/t))^.6 dt = .89351
I gave only five places since with my calculator, this formula calculated .5! to only 5 correct decimal places due to the behavior of ln(t) at the t=0 end of the integral.

Aspiring Novice
20051031 15:37:11 
Re: .5!
I am a sophmore in high school but my math teacher just mentioned the exact same problem, noting how it takes many math ideas to create a new one. The upper calculus 2 class is asked to solve (1/2)! he wouldn't tell us why but the answer is precisly sqrt(2)/2. So they do have a precise value. You are p[robably much more aware than i am of why this is but i think it should help. 
Aspiring Novice
20051031 15:38:40 
Re: .5!
oops didn't read the above post, i guess some of ou were a bit ahead of me
