A negative, and a non integer factorial!

Calculus Level 3

( 1 2 ) ! = ? \large \left (-\frac{1}{2} \right ) != \ ?


This problem is a part of the set Easy Factorials .
π 2 \pi^2 π \pi π \sqrt{\pi} undefined

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4 solutions

Caleb Townsend
Mar 26, 2015

n ! = Γ ( n + 1 ) ( 1 / 2 ) ! = Γ ( 1 / 2 ) Γ ( 1 / 2 ) = 0 x 1 / 2 e x d x n! = \Gamma(n+1) \\ (-1/2)! = \Gamma(1/2) \\ \Gamma(1/2) = \int_0^\infty x^{-1/2}e^{-x} dx Let x = u 2 . x = u^2. Then Γ ( 1 / 2 ) = 0 2 e u 2 d u = e u 2 d u \Gamma(1/2) = \int_0^\infty 2e^{-u^2} du \\ = \int_{-\infty}^{\infty} e^{-u^2} du This is a famous integral that eluded early attempts to solve, since there is no elementary antiderivative. The definite integral above, though, can be evaluated as π . \boxed{\sqrt{\pi}.}

By the way, even today, factorial is typically defined only for nonnegative integers, so I think the "undefined" answer should be replaced with a different answer so that people do not think you are referring to the classical definition k = 1 n k \prod_{k=1}^n k which is undefined for n = 1 / 2. n = -1/2.

I completely agree that the undefined answer should be removed. Factorial is not the same as gamma; that's like saying that the floor function is the same as f ( x ) = x f(x) = x . Just because they spit out the same values when x is natural doesn't mean they're the same. It felt a little bit like a trick to me.

Brock Brown - 5 years, 9 months ago
Otto Bretscher
Mar 29, 2015

By definition, we have n ! = Γ ( n + 1 ) n!=\Gamma(n+1) , the Gamma Function. Now ( 1 2 ) ! = Γ ( 1 2 ) = π (-\frac{1}{2})!=\Gamma(\frac{1}{2})=\sqrt{\pi} by Euler's Reflection Formula, Γ ( z ) Γ ( 1 z ) = π sin ( π z ) \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)} .

Trishit Chandra
Mar 27, 2015

Check the note in my profile😍✌️

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