what is the mean of factorial(-1/2)??


by coki2000
Tags: factorial1 or 2
coki2000
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#1
Oct16-09, 02:29 PM
P: 91
Hi, in gamma function gamma(1/2)=factorial(-1/2)=(pi)^(1/2) but how did it? i don't understand that because the factorial of 5 equals to 4*(4-1)*(4-2)*(4-3) but the factorial of (-1/2) doesn't equals to (-1/2)*(-1/2-1)*(-1/2-2)*(...) please explain that to me
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lurflurf
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#2
Oct16-09, 04:01 PM
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we have gamma(x+1)=x gamma(x)
so your products use gamma(1)=1
to extend gamma we need the other values in say (0,1)
we end up requireing gamma have a nice property like log gamma is convex
and finish with something like
[tex]\Gamma(x)=\int_0^\infty t^x e^{-t} \frac{dt}{t}[/tex]
HallsofIvy
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#3
Oct16-09, 08:52 PM
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Quote Quote by coki2000 View Post
Hi, in gamma function gamma(1/2)=factorial(-1/2)=(pi)^(1/2) but how did it? i don't understand that because the factorial of 5 equals to 4*(4-1)*(4-2)*(4-3) but the factorial of (-1/2) doesn't equals to (-1/2)*(-1/2-1)*(-1/2-2)*(...) please explain that to me
The "factorial of -1/2" doesn't equal anything at all because n! is only defined for n an non-negative integer. gamma(n)= (n-1)! is only true for n a positive integer.

Mathjunkie
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#4
Oct17-09, 06:10 PM
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what is the mean of factorial(-1/2)??


The factorial notation only defined for positive interger arguments. It is a special case of the Gamma function and is only defined for positive integers where the two coincide. So, as someone else has replied (-1/2)! is meaningless on its own, so don't try to understand it as an infinite product. However, because n!=GAMMA(n+1), frequently the factorial notation is extended to non-(positive) integral values of the argument when people get lazy. So, with this (shorthand) understanding, (-1/2)!=GAMMA(1/2) in the same way as n!=GAMMA(n+1). It is well-known that GAMMA(1/2)=sqrt(Pi). You can find derivations in many places.
lurflurf
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#5
Oct18-09, 04:33 AM
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^^
Why is it lazy, it is a perfectly reasonable definition. I think we should only define sin(x) when x/pi is rational and then we can have a new funtion pik'jhkrdwgfjh;swakjlh(x) that is equal to sin(x) when x/pi is rational. That way if some one says "What is sin(1)" we can say "sin(1) is not defined but pik'jhkrdwgfjh;swakjlh(1)~0.8414709848078965066525023216303". Many textbooks and refrences simply state x!:=gamm(x+1). It seems some newer books I have not seen have this ".5! is not defined" stuff, but I don't see the point.
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Oct18-09, 09:10 AM
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thank you for your helps
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Oct18-09, 09:48 AM
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Quote Quote by lurflurf View Post
^^
Why is it lazy, it is a perfectly reasonable definition. I think we should only define sin(x) when x/pi is rational and then we can have a new funtion pik'jhkrdwgfjh;swakjlh(x) that is equal to sin(x) when x/pi is rational. That way if some one says "What is sin(1)" we can say "sin(1) is not defined but pik'jhkrdwgfjh;swakjlh(1)~0.8414709848078965066525023216303". Many textbooks and refrences simply state x!:=gamm(x+1). It seems some newer books I have not seen have this ".5! is not defined" stuff, but I don't see the point.
You certainly can do that but I don't believe it is (yet) standard practice. But maybe I am just showing my age!

(I wasn't the one who called it "lazy"!)
Mathjunkie
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#8
Oct18-09, 04:44 PM
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Reply to lurflurf:

In the following n is an integer, x is a continuous variable (complex if you want).

That is exactly the point, and it is a very important one that comes up many places - quantum field theory being a prime example. The business of generalizing a function over a new region of the independant variable is frought with danger, and much care is required. Your example (define sin(x) only when it is rational and say it is something else when it isn't) is an example of the ambiguity that exists. There is a theorem that says that a function can be UNIQUELY "analytically continued" from a region where the independent variable is continuous, no matter how small that region is. The rationals are NOT continuous; neither are the positive integers. Therefore, any function that is defined only over the integers (or rationals) can be extended (i.e. generalized, NOT "analytically continued" which is something very different) in an infinite number of ways. Simply setting n=x doesn't always work, although it is usually the best choice.

Example: A(x) is any function you care to name that is finite when x=n:

generalize as follows: x! = gamma(x+1) + A(x)*sin(Pi*x).

Then n! will still equal gamma(n+1), but anyone who uses the usual definition will get a different result when x.ne.n.

So, everyone has to come to some agreement, and the agreement is that x!=GAMMA(x+1).
But the fact that someone had to ask the question suggests that not everyone knows the convention. So, if you find a textbook that refers to x!, somewhere in that book it should be written that x!=GAMMA(x+1). Otherwise the author is lazy, and worse, mathematically imprecise.
g_edgar
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#9
Oct18-09, 08:21 PM
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Writing x! for non-integer (or even complex) x is something I see primarily in 19th-century papers.
lurflurf
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Oct19-09, 08:08 PM
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Quote Quote by Mathjunkie View Post
Reply to lurflurf:

In the following n is an integer, x is a continuous variable (complex if you want).

That is exactly the point, and it is a very important one that comes up many places - quantum field theory being a prime example. The business of generalizing a function over a new region of the independant variable is frought with danger, and much care is required. Your example (define sin(x) only when it is rational and say it is something else when it isn't) is an example of the ambiguity that exists. There is a theorem that says that a function can be UNIQUELY "analytically continued" from a region where the independent variable is continuous, no matter how small that region is. The rationals are NOT continuous; neither are the positive integers. Therefore, any function that is defined only over the integers (or rationals) can be extended (i.e. generalized, NOT "analytically continued" which is something very different) in an infinite number of ways. Simply setting n=x doesn't always work, although it is usually the best choice.

Example: A(x) is any function you care to name that is finite when x=n:

generalize as follows: x! = gamma(x+1) + A(x)*sin(Pi*x).

Then n! will still equal gamma(n+1), but anyone who uses the usual definition will get a different result when x.ne.n.

So, everyone has to come to some agreement, and the agreement is that x!=GAMMA(x+1).
But the fact that someone had to ask the question suggests that not everyone knows the convention. So, if you find a textbook that refers to x!, somewhere in that book it should be written that x!=GAMMA(x+1). Otherwise the author is lazy, and worse, mathematically imprecise.
How did quantum field theory or analytically contination get into this? It seemed as if you meant it was (bad) lazy to make the definition x! = gamma(x+1). If you only meant the definition should be clearly stated, that is fine. Though it is tedious to clearly define everything at all times. There is really not a competing definition of x!, x! = gamma(x+1) + A(x)*sin(Pi*x) having not caught on. No confusion results. As I already stated it is desirable that x! be log convex. Speaking of being (bad) lazy one does not normally speak of sets like N, Q, and R as being continuous or not. When one speaks of a function being continuous, one specifies a topology or equivelently type of limit that defines the continuity.
If we define f(x)=sin(x*pi) when x is rational (that is Q is the domain of f)
Theorem 1: f(x) is (rational) continuous for all x
because sin(x+h)-sin(x) is defined and small for all x and small h (where x+h and x are in the domain of f)
Theorem 2: f(x) is an algebraic number for all x

So Theorem 1 will remain true, but theorem 2 will be ruined by the generalization to real numbers.
lurflurf
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#11
Oct19-09, 08:29 PM
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Quote Quote by g_edgar View Post
Writing x! for non-integer (or even complex) x is something I see primarily in 19th-century papers.
There good old days. I would like to know when and why that stopped.
Mathjunkie
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#12
Oct20-09, 04:55 PM
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Quote Quote by lurflurf View Post
How did quantum field theory or analytically contination get into this? It seemed as if you meant it was (bad) lazy to make the definition x! = gamma(x+1). If you only meant the definition should be clearly stated, that is fine. Though it is tedious to clearly define everything at all times. There is really not a competing definition of x!, x! = gamma(x+1) + A(x)*sin(Pi*x) having not caught on. No confusion results. As I already stated it is desirable that x! be log convex. Speaking of being (bad) lazy one does not normally speak of sets like N, Q, and R as being continuous or not. When one speaks of a function being continuous, one specifies a topology or equivelently type of limit that defines the continuity.
If we define f(x)=sin(x*pi) when x is rational (that is Q is the domain of f)
Theorem 1: f(x) is (rational) continuous for all x
because sin(x+h)-sin(x) is defined and small for all x and small h (where x+h and x are in the domain of f)
Theorem 2: f(x) is an algebraic number for all x

So Theorem 1 will remain true, but theorem 2 will be ruined by the generalization to real numbers.
As you have written: "If you only meant the definition should be clearly stated, that is fine." which is exactly what I meant. And the reason that I meant it, is because the usual definition is not always clearly understood - otherwise we would not be having this discussion because the question would not have been asked.

As for QFT - this same issue crops up almost everywhere in QFT. Typically, after a long calculation, an integral appears that is infinite, and the question becomes what to do. The usual answer is to change one of the (integral) parameters of the problem from an integer to something continuous; the integral becomes tractable, and the final answer is obtained after the parameter is reset to its initial integral value. Books and careers have been devoted to methods of doing this - it is called "regularization" when the integer is generalized, and "analytic continuation" when the general value is returned to an integer. There are an infinite number os ways of doing this.

Example: a four dimensional integral is infinite. So change the dimensionality from "4", an integer, to something continuous (never mind how - as long as the two coincide when the dimensionality is 4, its OK), the "regularized" integral is no longer infinite and the calculation proceeds. Finally, reset the dimensionality back to 4. This is exactly the same issue as generalizing n!=GAMMA(n+1) to x!=GAMMA(x+1), so it is important to understand this simple case.

I don't understand your example: You write "f(x) is (rational) continuous for all x".

"Rational" means that it is the ratio of integers, and it is well-known that the rationals are not continuous - there are in infinity of irrational numbers between every rational one- and that is prior to considering transcendentals.
lurflurf
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#13
Oct20-09, 06:54 PM
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Quote Quote by Mathjunkie View Post
I don't understand your example: You write "f(x) is (rational) continuous for all x".

"Rational" means that it is the ratio of integers, and it is well-known that the rationals are not continuous - there are in infinity of irrational numbers between every rational one- and that is prior to considering transcendentals.
I award thee one point for entertaining use of ".
It is not usual (to me at least) to say "the rationals are not continuous", please define. I would guess you are either making a statement about functions of the rationals or referring to what I would call complete. To define a continuous function (using limits) one must define a way of approach called a direction, though this is often omitted due to laziness. Each direction defines a type of continuity. It therefore makes sense to say a function with rational domain is (rational) continuous because our direction only calls the function with rational arguments.
ie a function f is rational continuous at (at rational number) x if given any rational (or real) 0<ε there exist 0<δ such
|f(x+h)-f(x)|<ε
for all rational h for which -δ<h<δ

in particular it does not matter if f is defined for any irrational numbers, or if defined what values it takes
example the function
f=cos(x) x rational,sin(x) x irrational
is rational continuous (all rational x), irrational continuous (all irrational x), and nowhere real continuous (all x for which sin(x)=cos(x).


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