Re: [微積] Legendre polynomail遞迴式證明

[Vulpix (Sebastian)]

※ 引述《Desperato (Farewell)》之銘言：
: https://goo.gl/xhxzHn
: 其實網路上有答案ow o
: 不過這是我硬幹 4 小時卻完全做不出來之後了，只好抄答案了qw q
: 令 D = d/dx = D, y = x^2-1
: 關鍵是要把遞迴式中的 x 塞進微分裡面
: (Lemma) Let f = f(x) be smooth.
:         Then D^n (x f) = x D^n f + n D^(n-1) f
:         Hence x D^n f = D^n (x f) - n D^(n-1) f
:    (pf) Using general Leibniz rule.
: p.s. 原文是使用 commutator, [D, x] = 1 and [D^n, x] = n D^(n-1)
: 所以就有 x D^n f = D^n (x f) - n D^(n-1) f
: 原式 (n+1) P_(n+1) = (2n+1) x P_n - n P_(n-1)
: 整式同乘 1/(2^(n+1) n!) 可得
: D^(n+1) y^(n+1) = 2(2n+1) x D^n y^n - 4 n^2 D^(n-1) y^(n-1)
: 注意到中間那項 x D^n y^n = D^n (x y^n) - n D^(n-1) y^n
: 因此可以計算
: LHS - RHS
: = D^(n-1) [ D^2 y^(n+1) - 2(2n+1)D(x y^n) + 2(2n+1)n y^n + 4n^2 y^(n-1)]
: = D^(n-1) [ 2(n+1) y^n + 4n(n+1)x^2 y^(n-1)
:             - 2(2n+1)y^n - 4n(2n+1)x^2 y^(n-1) + 2n(2n+1)y^n + 4n^2 y^(n-1)]
: = D^(n-1) { y^(n-1) [ 2(n+1)y - 4n^2 x^2 - 2(2n+1)y + 2n(2n+1)y + 4n^2 ] }
: = D^(n-1) { y^(n-1) [ 0 ] } = 0
: 從上面計算可以看出，想直接從 LHS 推到 RHS 會非常有技術性...
: 至少我是寫不出來啦(攤手)

It's not that difficult.

Derived directly from Rodrigue's formula,

        P_n(x) = Σ_{0}^{n} (-1)^k C(2n-2k,n)C(n,k) x^{n-2k} /2^n

               = Σ_{-∞}^{∞} (-1)^k C(2n-2k,n)C(n,k) x^{n-2k} /2^n

               (where 1/(-m)! is taken to be 1/Γ(1-m)=0 for natural number m

                and Γ(y+1)=yΓ(y) for all real number y

                this can be realized by a simple ε-trick)

               = Σ (-1)^k C(2n-2k,n)C(n,k) x^{n-2k} /2^n

               (the domain of summation is omitted for simplicity)

We wish to express (n+1)P_{n+1}(x)

by a linear combination of lower degree Legendre polynomials,

and the coefficient ring is supposed to be R[x].

WLOG, we assume that there is a P_n term carrying out the x^{n+1} term.

Since (n+1)C(2n+2,n+1)/2^{n+1} / [C(2n,n)/2^n] = 2n+1,

the P_n term is (2n+1)xP_n.

  (the "x" with no constant term is chosen to stick with parity)

The rest of the calculation is shown below,

  (n+1)P_{n+1}(x) - (2n+1)xP_n(x)

= Σ (-1)^k x^{n+1-2k}/2^{n+1} *

     [ (n+1)C(2n+2-2k,n+1)C(n+1,k) - 2(2n+1)C(2n-2k,n)C(n,k) ]

= Σ (-1)^k x^{n+1-2k}/2^{n+1} * C(2n-2k,n)C(n,k) *

     [ (n+1)(2n+2-2k)(2n+1-2k)/[(n+1-2k)(n+1-k)] - 2(2n+1) ]

= Σ (-1)^k x^{n+1-2k}/2^n * C(2n-2k,n)C(n,k) *

     [ (n+1)(2n+1-2k)/(n+1-2k) - (2n+1) ]

= Σ (-1)^k x^{n+1-2k}/2^n * C(2n-2k,n-2k)C(n,k) * [ 2nk/(n+1-2k) ]

= nΣ (-1)^k x^{n+1-2k}/2^{n-1} * C(2n-2k,n)nC(n-1,k-1) / (n+1-2k)

= nΣ (-1)^k x^{n+1-2k}/2^{n-1} * C(2n-2k,n-1)C(n-1,k-1)

= nΣ (-1)^{K+1} x^{n-1-2K}/2^{n-1} * C(2n-2K-2,n-1)C(n-1,K)    (where k=K+1)

= -nP_{n-1}(x)


Just play with factorials!

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※ 編輯: Vulpix (111.243.92.106), 11/25/2018 00:53:55
推 Lanjaja     : 謝謝V大 可以再解釋一下due to parity這段嗎?
In fact, we may have recurrence relation with some extra terms,
but we can still reduce the coefficients to a decent form.
Let (n+1)P_{n+1}(x) = Σ_{k<n+1} f_k(x)P_k(x).
Since P_{n+1}(-x) = (-1)^{n+1} P_{n+1}(x),
we have (n+1)P_{n+1}(x) = (-1)^{n+1}Σ_{k<n+1} (-1)^k f_k(-x)P_k(x),
and (n+1)P_{n+1}(x) = Σ_{k<n+1} [ f_k(x) + (-1)^{n+k+1} f_k(-x) ]/2 P_k(x).
The coefficient [f_k(x) + (-1)^{n+k+1} f_k(-x)]/2 has a certain parity.
推 Desperato   : 先推 原來gamma倒數為0可以解釋這個
Just for the sake of convenience,
for I don't want to waste time checking the constant term, etc.
※ 編輯: Vulpix (111.243.92.106), 11/25/2018 19:41:42