樂府古題要解
作者：吳兢‧唐

舊本題唐吳兢撰。兢有《貞觀政要》，已著錄。考《崇文總目》載《古樂府古題要解》共十二卷。晁公武《讀書志》稱兢纂采漢、魏以來古樂府詞凡十卷。又於傳記及諸家文集中采樂府所起本義，以釋解古題。觀《崇文總目》稱二書共十二卷，而《讀書志》稱古樂府十卷，則所餘二卷為《樂府古題要解》矣。卷數與今本相合。《崇文總目》又載《樂府解題》，稱不著撰人名氏。與吳兢所撰《樂府古題》頗同，以《江南曲》為首。其後所解差異。此本為毛晉津逮秘書所刊。後有晉跋，稱今人以兢所撰與《樂府解題》混為一書。又稱太原郭氏諸敘中，輒引《樂府解題》不及《古題要解》。今考郭茂倩《樂府詩集》所引《樂府解題》，自漢鐃歌《上之回》篇始，乃明題吳兢之名。則混為一書，已不始於近代。然茂倩所引，其文則與此書全同，不過偶刪一二句，或增入樂府本詞一二句，不應互相剿襲至此。疑兢書久佚，好事者因《崇文總目》有「《樂府解題》與吳兢所撰樂府頗同」語，因捃拾郭茂倩所引《樂府解題》，偽為兢書。而不知王堯臣等所謂與樂府頗同者，乃指其解說古題體例相近，非謂其文全同。觀下文即雲以《江南曲》為首，其後所解差異，是二書不同之明證。安有兩家之書如出一口者乎？且樂府自樂府，雜詩自雜詩，卷末乃載及建除諸體，並及於字謎之類，其為捃拾以足兩卷之數，灼然可知矣。《晉跋》稱是書凡三本，一得之廣山楊氏，一得之錫山顏氏，最後乃得一元板。然則是書為元人所贗造也。

序

樂府之興，肇於漢魏。歷代文士，篇詠實繁。或不睹於本章，便斷題取義。贈夫利涉，則述《公無度河》；慶彼載誕，乃引《烏生八九子》；賦雉斑者，但美繡錦臆；歌天馬者，唯敘驕馳亂蹋。類皆若茲，不可勝載。遞相祖習，積用為常，欲令後生，何以取正？余頃因涉閱傳記，用諸家文集，每有所得，輒疏記之。歲月積深，以成卷軸，向編次之，目為《古題要解》雲爾。

水仙操

右：舊說伯牙學鼓琴於成連先生，三年而成。至於精神寂寞，情誌專一，尚未能也。成連雲『吾理由子春在海中，能移人情。』乃與伯牙延望，無人。至蓬萊山，留伯牙曰：『吾將迎吾師。』刺船而去，旬時不返，但聞海上水汩汲漰澌之聲。山林窅冥，群鳥悲號，愴然嘆曰：『先生將移我情。』乃援琴而歌之。曲終，成連刺船而還。伯牙遂為天下妙手。

歇後語說︰水仙不開花，裝蒜。貌似未必真，贗造常是假。思想起伯牙和鍾子期，可能世上知音希。因著歐拉求和公式註釋多！於是暫且放下克萊因先生文字註釋？與其左右為難，又不能照顧前後？不如先回到白努利多項式基本性質的了？！

假設 $B_n(x)$ 為白努利多項式，那麼依定義

$B_n(x) = \sum \limits_{k=0}^{n} \binom {n}k B_k x^{n-k}$ 。

已求過 $B_n(x)$ 之生成函數為

$\frac{t e^{xt}}{e^t -1} = \sum \limits_{k=0}^{\infty} B_k (x) \frac{t^k}{k!} \ \ \ \ \ (1)$ 。

將 $(1)$ 式兩邊對 $x$ 微分，右邊得

$\sum \limits_{k=0}^{\infty} B_k^{'} (x) \frac{t^k}{k!}$ ，由於

$\lim \limits_{t \to 0} \frac{t e^{xt}}{e^t -1} = 1 = B_0 (x)$ ，所以

$B_0^{'} (x) = 0$ ，因此右邊可寫成

$\sum \limits_{k=1}^{\infty} B_k^{'} (x) \frac{t^k}{k!}$ 。左邊得

${\left( \frac{t e^{xt}}{e^t -1} \right)}^{'} = \frac{t^2 e^{xt}}{e^t -1} = t \times \frac{t e^{xt}}{e^t -1}$

$= t \times \sum \limits_{k=0}^{\infty} B_k (x) \frac{t^k}{k!}$

$= \sum \limits_{k=0}^{\infty} B_k (x) \frac{t^{k+1}}{k!}$

$= \sum \limits_{k=1}^{\infty} B_{k-1} (x) \frac{t^k}{(k-1)!}$

$\therefore \frac{B_k^{'} (x)}{k!} = \frac{B_{k-1} (x)}{(k-1)!}$ 。故得

$B_k^{'} (x) = k \cdot B_{k-1} (x), \ k \ge 1$ ☆

此 $k$ 描述無窮白努利多項式間前後兩者之關係，實不同於某有既定關係函數序列 $y_n = A \cdot {(x + \alpha)}^n$ ， $y_n = (x+\alpha) \cdot y_{n-1}$ 。

若因剛好 $\frac{d y_n}{dx} = n \left( A {(x + \alpha)}^{n-1} \right) = n \cdot y_{n-1}$ ，如是就以為如果 $y_0 = 1$ ，那麼 $A=1$ ，要是 $y_1 = x - \frac{1}{2}$ ，那麼 $\alpha = - \frac{1}{2}$ ，得到了 $y_n = {(x-\frac{1}{2})}^n$ ，當真是白努利多項式耶？？豈不會陷於淺盤深水乎！！

此所以有人講『陰影微積分』夫？？！！

Umbral calculus

In mathematics before the 1970s, the term umbral calculus referred to the surprising similarity between seemingly unrelated polynomial equations and certain shadowy techniques used to ‘prove’ them. These techniques were introduced by John Blissard (1861) and are sometimes called Blissard’s symbolic method. They are often attributed to Édouard Lucas (or James Joseph Sylvester), who used the technique extensively.^[1]

In the 1930s and 1940s, Eric Temple Bell attempted to set the umbral calculus on a rigorous footing.

In the 1970s, Steven Roman, Gian-Carlo Rota, and others developed the umbral calculus by means of linear functionals on spaces of polynomials. Currently, umbral calculus refers to the study of Sheffer sequences, including polynomial sequences of binomial type and Appell sequences, but may encompass in its penumbra systematic correspondence techniques of the calculus of finite differences.

The 19th-century umbral calculus

The method is a notational procedure used for deriving identities involving indexed sequences of numbers by pretending that the indices are exponents. Construed literally, it is absurd, and yet it is successful: identities derived via the umbral calculus can also be properly derived by more complicated methods that can be taken literally without logical difficulty.

An example involves the Bernoulli polynomials. Consider, for example, the ordinary binomial expansion (which contains a binomial coefficient):

(y+x)^{n}=\sum _{{k=0}}^{n}{n \choose k}y^{{n-k}}x^{k}

and the remarkably similar-looking relation on the Bernoulli polynomials:

B_{n}(y+x)=\sum _{{k=0}}^{n}{n \choose k}B_{{n-k}}(y)x^{k}.

Compare also the ordinary derivative

{\frac {d}{dx}}x^{n}=nx^{{n-1}}

to a very similar-looking relation on the Bernoulli polynomials:

{\frac {d}{dx}}B_{n}(x)=nB_{{n-1}}(x).

These similarities allow one to construct umbral proofs, which, on the surface, cannot be correct, but seem to work anyway. Thus, for example, by pretending that the subscript n − k is an exponent:

B_{n}(x)=\sum _{{k=0}}^{n}{n \choose k}b^{{n-k}}x^{k}=(b+x)^{n},

and then differentiating, one gets the desired result:

B_{n}'(x)=n(b+x)^{{n-1}}=nB_{{n-1}}(x).\,

In the above, the variable b is an “umbra” (Latin for shadow).

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每日彙整: 2017-04-02

時間序列︰生成函數‧漸近展開︰白努利 □○《十前》

Umbral calculus

The 19th-century umbral calculus

輕。鬆。學。部落客