question about i

Originally posted by sonhouse
So if I input 5.5 it should give me 4.5! right? But it goes 287.88 and change. I wish I knew just what it was computing. If I do 5! it gives 120.

The Gamma function has the property:

Gamma(x + 1) = x * Gamma (x)

So, if we define non-integer factorials in terms of it then we can assign an answer to your example of 5.5!

5.5! = Gamma(6.5) = 5.5 * Gamma(5.5) = ... = 5.5 * 4.5 * 3.5 * 2.5 * 1.5 * 0.5 * Gamma(0.5)

Gamma(1/2) = √π so,

(11/2)! = (11*9*7*5*3*1)/(2*2*2*2*2*2) Gamma(1/2) = (10395/64) Gamma(1/2) = 10395 * √π / 64

Giving 5.5! = 287.885

I was working from memory earlier and gave the wrong result for Gamma(1/2). Your calculator probably uses a series for the log Gamma function and then exponentials.

Originally posted by DeepThought
The Gamma function has the property:

Gamma(x + 1) = x * Gamma (x)

So, if we define non-integer factorials in terms of it then we can assign an answer to your example of 5.5!

5.5! = Gamma(6.5) = 5.5 * Gamma(5.5) = ... = 5.5 * 4.5 * 3.5 * 2.5 * 1.5 * 0.5 * Gamma(0.5)

Gamma(1/2) = √π so,

(11/2)! = (11*9*7*5*3*1)/(2*2*2*2*2*2) Gamma(1/2) = (1 ...[text shortened]... (1/2). Your calculator probably uses a series for the log Gamma function and then exponentials.

One part I would like clarified: 11/2! as 11*10etc/2*2*2*2*2*2, why did you have to do the 2*2*2 part to come up with 64? How does that fit into the gamma or factorial?

Oh, just notice it was odd numbers from 11, 9, 7 etc. Why that AND why 2*2 series?

Originally posted by sonhouse
One part I would like clarified: 11/2! as 11*10etc/2*2*2*2*2*2, why did you have to do the 2*2*2 part to come up with 64? How does that fit into the gamma or factorial?

Oh, just notice it was odd numbers from 11, 9, 7 etc. Why that AND why 2*2 series?

I was just rewriting 5.5*4.5*3.5*2.5*1.5 as (11/2)*(9/2)*(7/2)*(5/2)*(3/2)*(1/2).

Originally posted by DeepThought
I was just rewriting 5.5*4.5*3.5*2.5*1.5 as (11/2)*(9/2)*(7/2)*(5/2)*(3/2)*(1/2).

Ah, thanks.