Identity for the Odd Double Factorial July 7, 2023
It holds
( d d z + z ) n ∣ z = 0 = { ( n − 1 ) ! ! ≡ Product of all odd numbers up to n-1 n even 0 n odd ,
\left(\frac{\text{d}}{\text{d}z} + z \right)^n \vert_{z=0} = \begin{cases} (n-1)!! \equiv \text{Product of all odd numbers up to n-1} & n \quad \text{even} \\
0 & n \quad \text{odd},
\end{cases} ( d z d + z ) n ∣ z = 0 = { ( n − 1 )!! ≡ Product of all odd numbers up to n-1 0 n even n odd , which I personally do not find obvious.
Proof ¶ To see this, consider Wick’s theorem ( D z = d d z ) \left(D_z = \frac{\text{d}}{\text{d}z}\right) ( D z = d z d )
D z n e z 2 / 2 ∣ z = 0 = { ( n − 1 ) ! ! n even 0 n odd D_z^n e^{z^2/2} \vert_{z=0} = \begin{cases} (n-1)!! \qquad & n \quad \text{even} \\
0 \qquad & n \quad \text{odd}
\end{cases} D z n e z 2 /2 ∣ z = 0 = { ( n − 1 )!! 0 n even n odd This can be checked by expanding the exponential and differentiating term by term. We have for sufficiently well-behaved functions f = f ( z ) , p = p ( z ) f=f(z),p = p(z) f = f ( z ) , p = p ( z )
D z f e p = ( f D z p + D z f ) ⋅ e p = ( ( D z p ) + D z ) f ⋅ e p . D_z f e^{p} = (f D_zp + D_zf) \cdot e^{p} = ( \left(D_z p\right) + D_z)f \cdot e^{p}. D z f e p = ( f D z p + D z f ) ⋅ e p = ( ( D z p ) + D z ) f ⋅ e p . Note that the differential operator does not act on e p e^{p} e p
D z n f e p = ( ( D z p ) + D z ) n f ⋅ e p . D_z^n f e^{p} = ( \left(D_z p \right) + D_z)^nf \cdot e^{p}. D z n f e p = ( ( D z p ) + D z ) n f ⋅ e p . Choosing f = 1 , p = z 2 / 2 f=1, p = z^2/2 f = 1 , p = z 2 /2 n + 1 2 n + \frac{1}{2} n + 2 1
Γ ( n + 1 2 ) = π 2 n ( D z + z ) 2 n ∣ z = 0 . \Gamma(n + \frac{1}{2}) = \frac{\sqrt{\pi}}{2^n} \left( D_z + z \right)^{2n}\vert_{z=0}. Γ ( n + 2 1 ) = 2 n π ( D z + z ) 2 n ∣ z = 0 . Check ¶ Check this for some even n n n f ( 0 ) ≡ f ∣ z = 0 f(0) \equiv f\vert_{z=0} f ( 0 ) ≡ f ∣ z = 0
( D z + z ) 2 ( 0 ) = ( D z + z ) ( D z + z ) 1 ( 0 ) = ( D z + z ) z ( 0 ) = ( 1 + z 2 ) ( 0 ) = 1 = ( 2 − 1 ) ! ! ( D z + z ) 4 ( 0 ) = D z 2 z 2 + ( D z z ) 2 = 3 = ( 4 − 1 ) ! ! ( D z + z ) 6 ( 0 ) = D z 3 z 3 + D z 2 z D z z 2 + 2 D z 2 z 2 D z z + ( D z z ) 3 = 6 + 4 + 2 ⋅ 2 + 1 = 15 = ( 6 − 1 ) ! ! \begin{align}
&(D_z + z)^2 (0) = (D_z + z)(D_z + z)1 (0) = (D_z + z)z (0) = (1 + z^2)(0) = 1 = (2-1)!! \\
&(D_z + z)^4 (0) = D_z^2 z^2 + (D_z z)^2 = 3 = (4-1)!!\\
&(D_z + z)^6 (0) = D_z^3 z^3 + D_z^2 z D_z z^2 + 2 D_z^2 z^2 D_z z + (D_z z)^3 = 6 + 4 + 2 \cdot 2 + 1 = 15 = (6-1)!!
\end{align} ( D z + z ) 2 ( 0 ) = ( D z + z ) ( D z + z ) 1 ( 0 ) = ( D z + z ) z ( 0 ) = ( 1 + z 2 ) ( 0 ) = 1 = ( 2 − 1 )!! ( D z + z ) 4 ( 0 ) = D z 2 z 2 + ( D z z ) 2 = 3 = ( 4 − 1 )!! ( D z + z ) 6 ( 0 ) = D z 3 z 3 + D z 2 z D z z 2 + 2 D z 2 z 2 D z z + ( D z z ) 3 = 6 + 4 + 2 ⋅ 2 + 1 = 15 = ( 6 − 1 )!! For odd n n n D z D_z D z z z z