※ 引述《Frobenius (▽．(▽×▽φ)=0)》之銘言：
: ※ [本文轉錄自 Math 看板]
: 作者: Frobenius (▽．(▽×▽φ)=0) 看板: Math
: 標題: [物數] Factorial Function
: 時間: Wed Dec 12 08:29:23 2007
: Mathematical Methods For Physicists 5th ( Arfken and Weber )
: Chapter 10   The Gamma Function ( Factorial Function )
: Exercises 10.1.3
: Show that
:                                  n-s
:                  (s - n)!     (-1)  (2n - 2s)!
:                ────── = ────────
:                 (2s - 2n)!        (n - s)!
: Here s and n are integers with s < n. This result can be used to avoid
: negative factorials such as in the series representations of the spherical
: Neumann funtions and the Legendre functions of the second kind.
: 我認為前式在 s > n 適用，後式在 s < n 適用，視情況可互相轉換，
: 不過我一直推導不出來，希望版上高手能幫我解決這個問題，謝謝^^

Γ(z)Γ(1-z) = π/sin(zπ)

Γ(1-z) = π/(Γ(z)sin(zπ))

let k = n - s => s - n = - k ； 2s - 2n = - 2k

(s - n)! = (- k)! = Γ(1-k) = π/(Γ(k)sin(kπ))

(2s - 2n)! = (- 2k)! = Γ(1-2k) = π/(Γ(2k)sin(2kπ))

                                = π/(Γ(2k)2sin(kπ)cos(kπ))

     (s - n)!          π/(Γ(k)sin(kπ))        cos(kπ) 2Γ(2k)
   ────── = ────────────── = ────────
    (2s - 2n)!    π/(Γ(2k)2sin(kπ)cos(kπ))        Γ(k)

       k                  k                  k             n-s
   (-1) (2k) Γ(2k)   (-1) Γ(2k + 1)    (-1) (2k)!    (-1)  (2n - 2s)!
 = ──────── = ──────── = ───── =  ────────
      (k) Γ(k)          Γ(k + 1)           k!            (n - s)!

感謝 timlintt 大 ^^

不過這下新的問題又出來了，Γ(z)Γ(1-z) = π/sin(zπ) 又是怎麼得出來的 XD

好像會牽扯到無窮乘積 orz

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推 Linderman:Γ(z)Γ(1-z)是很出名的結果,我記得王竹溪那本應該有吧   02/24 17:47
→ Linderman:以前我們物數用這本和Arfk,這個式子記得好像還蠻好用的   02/24 17:49