- #1

- 43

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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

- Thread starter Belgium 12
- Start date

- #1

- 43

- 0

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

- #2

- 13,001

- 550

Factorial (and double factorial) is defined for natural numbers only.

- #3

lurflurf

Homework Helper

- 2,432

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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

- 48

- 2

-Junaid

- #5

- 798

- 34

gamma(2(n+3/2))=gamma(2n+1) is false because 2(n+3/2)=2n+3 which is not 2n+1

- #6

- 48

- 2

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

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

-J

- #7

- 798

- 34

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.

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

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

-J

- #8

- 48

- 2

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

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

OK.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

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

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Thanks.

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