- #26

- 442

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n!=n*(n-1)*(n-2)*...*(1), 0!=1 (my fault for stopping at 2 earlier)

we defined it that way and it allows us to interpret the gamma function as a "continuous factorial" using real numbers. It just so happens that Gamma(x)=(x-1)Gamma(x-1), but this doesn't define the factorial for real numbers (because the factorial is already defined in the positive integers), although it does generalize what it would be like.

"for a real argument x, Gamma(x)=(x-1)Gamma(x-1)

If x is an integer n = 1, 2, 3, ..., then

Gamma(n)=(n-1)Gamma(n-1)=(n-2)(n-1)Gamma(n-2)=(n-1)(n-2)...1=(n-1)!

so the gamma function reduces to the factorial for a positive integer argument." - http://mathworld.wolfram.com/GammaFunction.html

Here, they are using the definition of the factorial as stated above, they didn't derive the definition of the factorial. It was defined the way it was so that it can be used in helping define functions like the gamma function.

if you use the definition n! with real numbers x, you get x!=Gamma(x+1), Gamma(0)=1. If x! wasn't defined the way it is, then x! could equal Gamma(x+1). The lim (as x approaches zero from the right) of Gamma(x+1) = 0 not 1. so the way the factorial is defined means it is discontinuous at zero so the extra constraint that Gamma(0) = 1 must be accounted for. All of this is based on if you were to define the factorial with Gamma using real numbers and allowing one integer downward into the negative reals approaching -1 as Gamma(x+1) approaches zero. If we don't allow an integer to be subtracted from x such that -1>x-1<0 (as when defining x!=Gamma(x+1), Gamma(0)=1 for reals), then x!'s domain changes from [0,infinity) to [1, infinity). In this conception the iterated stepping downwards an integer at a time stops when it reaches bottom, and bottom is zero, there is no more "stepping down" if you're at the bottom. Then the function is even more discontinuous, it's defined at 0 as 1, and defined from 1 to infinity, with a gap along 0>x>1. I plotted the x! function and the Gamma(x+1) in mathematica from x=0 to x=1.5 to be sure and saw no differance, they both approach zero comming from the right, between in the interval (0,1), x! is negative and concave up.

I guess the point is that we define functions and axioms according to our interpretation of them when we consider how they can fit together easier with other ones in an effort to render more powerful functions and axioms.

I've always interpreted zero to be nothing, but through this exploration I think that maybe it's the wrong interpretation (of course the word nothing may mean something different to you than to me). Rather, zero represents nothing and it is something. If this is how I should interpret zero, then I can treat it as an entity and I can add up how many of them I have, but it doesn't matter to me, because no matter how many I have the total value it represents is nothing. I guess I gotta get over my distaste for treating zero as something rather than nothing.

I always figured that if I have nothing, then I can't put anything in any particular arrangement, so there are no combinations of arrangements of anything. But the factorial is defined as if there is a single possible arrangement of nothing. But what about two nothings? three? hehe...