Gamma function and factorial

  • Thread starter Belgium 12
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  • #1
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Hi members,

gamma (n+1)=n!

Now I have the following:

gamma(2(n+3/2))=gamma(2n+1)

What is factorial notation for this??

Thank you
 

Answers and Replies

  • #2
dextercioby
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Factorial (and double factorial) is defined for natural numbers only.
 
  • #3
lurflurf
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^I don't know why people say that.
It is perfectly reasonable (and common) to define factorial (and double factorial) for complex numbers.
(-.5)!=sqrt(pi) for example is perfectly sensible

any way gamma (n+1)=n! implies
gamma (x)=(x-1)!
so
gamma(2(n+3/2))=gamma(2n+1)
(2(n+3/2)-1)!=(2n+1-1)!
(2n+2)!=(2n)!
which is in no way helpful
Are you to solve it?
If so hint
(a+b*sqrt(c))/d
 
  • #4
Although (2n+2)! = (2n)! looks strange, it is important to remember that the gamma function is anything but one-to-one on the real line. I really like lurflurf's train of thought. Looks like the quadratic formula would be golden in this problem.

-Junaid
 
  • #5
798
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gamma(2(n+3/2))=gamma(2n+1) is false because 2(n+3/2)=2n+3 which is not 2n+1
 
  • #6
The gamma function is not invertible on the real line so the statement is not necessarily false.

Consider instead the function f(x) = x^2.

Then f(-2)=f(2) is true even though -2 ≠ 2.

-J
 
  • #7
798
34
The gamma function is not invertible on the real line so the statement is not necessarily false.

Consider instead the function f(x) = x^2.

Then f(-2)=f(2) is true even though -2 ≠ 2.

-J
Belgium 12 stated : gamma(2(n+3/2))=gamma(2n+1) in which 2(n+3/2) and (2n+1) are integers. Obviously, no complex are considered. If Belgium 12 was thinking on complex roots, he would have mention it. Moreover, he would have not use the notation "n", but "x" or "z". So the statement is false.
 
  • #8
I agree. If the solutions for n are restricted to the integers, then there is no solution. If the solutions are open to the real line, then there are in fact two real solutions. Just as a clarification, note that integers are complex numbers. There are two real, irrational, complex solutions to this equation. I do agree with you though that there are no integer solutions! :)

Also, note that:
gamma(n+1)=gamma(n) has a solution in the integers even though n+1 = n is a false statement. This really brings home the idea that you cannot remove the gamma on both sides of the equation as the gamma function is not invertible.

Junaid Mansuri
 
Last edited:
  • #9
798
34
I agree. If the solutions for n are restricted to the integers, then there is no solution. If the solutions are open to the real line, then there are in fact two real solutions. Just as a clarification, note that integers are complex numbers. There are two real, irrational, complex solutions to this equation. I do agree with you though that there are no integer solutions! :)

Also, note that:
gamma(n+1)=gamma(n) has a solution in the integers even though n+1 = n is a false statement. This really brings home the idea that you cannot remove the gamma on both sides of the equation as the gamma function is not invertible.

Junaid Mansuri
OK.
In fact, the sender of the question recognizes that there was a mistake in the wording of his question :
http://mathhelpforum.com/calculus/220899-gamma-function-factorial.html#post793851
 
  • #10
Thanks.
 

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