2 ! = 2 and 3 ! = 6 .
We know thatFind the value of ( 2 . 5 ) ! to 3 decimal places.
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Exactly how I did it :)
Yeah inspired by @Chew-Seong Cheong
I think a factorial operation can only be done on a positive INTEGER
Technically, yes; however, for non-natural arguments (ie not 0, 1, 2, 3, etc), s ! is conventionally taken as Γ ( s + 1 ) .
n ! = 0 ∫ ∞ x n e − x d x
2 . 5 ! = 0 ∫ ∞ x 2 . 5 e − x d x
Applied calculator. I apologize for wanting convenience. Gamma (3.5) = 15 Sqrt (Pi)/ 8 = 3.323
2.5!=Γ(3.5) 2.5!=2.5⋅Γ(2.5) 2.5!=2.5⋅1.5⋅Γ(1.5) 2.5!=2.5⋅1.5⋅.5⋅Γ(.5) 2.5!=2.5⋅1.5⋅.5⋅ √π 2.5!=3.323
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I tried my best to upload an easy and elegant solution
For Help about Gamma function click here
Its easy first of all
⇒ n ! = n × ( n − 1 ) !
⇒ Γ n = ( n − 1 ) !
⇒ n ! = n Γ n ⇒ Γ ( 2 5 ) = Γ ( 1 + 2 3 )
⇒ ∴ as we know Γ ( 1 + n ) = n Γ ( n )
⇒ Γ ( 1 + 2 3 ) = 2 3 Γ ( 2 3 )
Similarly ⇒ Γ ( 2 3 ) = Γ ( 1 + 2 1 )
⇒ ∴ Γ ( 2 3 ) = 2 1 Γ ( 2 1 )
and we know ⇒ Γ ( 2 1 ) = π
so ⇒ Γ ( 2 3 ) = 2 1 π = 0 . 8 8 6 4
⇒ ∴ Γ ( 2 5 ) = 2 3 Γ ( 2 3 ) = 1 . 3 2 9
⇒ ∴ 2 5 ! = 2 5 ⋅ 1 . 3 2 9
the answer is ⇒ 3 . 3 2 3