查看“︁伯努利多項式”︁的源代码

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[[Image:Bernoulli polynomials no title.svg|thumb|right|伯努利多項式]]

在[[數學]]中，'''伯努利多項式'''在對多種[[特殊函數]]特別是[[黎曼ζ函數]]和[[赫尔维茨ζ函数]]的研究中出現。作為[[阿佩爾序列]]的一種，與[[正交多項式]]不同的是，伯努利多項式的函數圖像與''x''軸在單位長度區間內的交點數目並不會隨著多項式次數的增加而增長。當多項式的次數趨近無窮大的時候，伯努利多項式的函數形狀類似于[[三角函數]]。

==表示法==

伯努利多項式 ''B''<sub>''n''</sub> 有多種表示法，可視情況選用。

===代數法===
當 ''n'' &ge; 0 時，
:<math>B_n(x) = \sum_{k=0}^n {n \choose k} b_k x^{n-k},</math>
其中''b''<sub>''k''</sub> 則為 [[伯努利數]]

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==其他表示法==

另一個伯努利多項式的代數表示法

:<math>B_m(x)=
\sum_{n=0}^m \frac{1}{n+1}
\sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m.</math>

Note the remarkable similarity to the globally convergent series expression for the [[Hurwitz zeta function]]. Indeed, one has

:<math>B_n(x) = -n \zeta(1-n,x)</math>

where ''ζ''(''s'',&nbsp;''q'') is the Hurwitz zeta; thus, in a certain sense, the Hurwitz zeta generalizes the Bernoulli polynomials to non-integer values of&nbsp;''n''.

The inner sum may be understood to be the ''n''th [[forward difference]] of ''x''<sup>''m''</sup>; that is,

:<math>\Delta^n x^m = \sum_{k=0}^n (-1)^{n-k} {n \choose k} (x+k)^m</math>

where Δ is the [[forward difference operator]]. Thus, one may write

:<math>B_m(x)= \sum_{n=0}^m \frac{(-1)^n}{n+1} \Delta^n x^m. </math>

This formula may be derived from an identity appearing above as follows. Since the forward difference operator Δ equals
:<math>\Delta = e^D - 1</math>
where ''D'' is differentiation with respect to ''x'', we have, from the [[Mercator series]]

:<math>{D \over e^D - 1} = {\log(\Delta + 1) \over \Delta} = \sum_{n=0}^\infty {(-\Delta)^n \over n+1}.</math>

As long as this operates on an ''m''th-degree polynomial such as ''x''<sup>''m''</sup>, one may let ''n'' go from 0 only up to&nbsp;''m''.

An integral representation for the Bernoulli polynomials is given by the [[Nörlund&ndash;Rice integral]], which follows from the expression as a finite difference.

An explicit formula for the Euler polynomials is given by

:<math>E_m(x)=
\sum_{n=0}^m \frac{1}{2^n}
\sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m\,.</math>

This may also be written in terms of the [[Euler number]]s ''E''<sub>''k''</sub> as

:<math>E_m(x)=
\sum_{k=0}^m {m \choose k} \frac{E_k}{2^k}
\left(x-\frac{1}{2}\right)^{m-k} \,.</math>

-->

===函數法===
伯努利多項式的[[母函數]]是

:<math>\frac{t e^{xt}}{e^t-1}= \sum_{n=0}^\infty B_n(x) \frac{t^n}{n!}.</math>

歐拉多項式母函數是
:<math>\frac{2 e^{xt}}{e^t+1}= \sum_{n=0}^\infty E_n(x) \frac{t^n}{n!}.</math>

===微分法===

伯努利多項式亦可表示為微分的形式

:<math>B_n(x)={D \over e^D -1} x^n</math>

其中 ''D'' = ''d''/''dx'' 是一個關於x的[[微分]]式，上述分式可以展開得到[[形式幂级数]]
:<math>\int _a^x  B_n (u) ~du = \frac{B_{n+1}(x) - B_{n+1}(a)}{n+1}   ~.</math>
===積分法===


伯努利多項式的多項式''f''可以通此[[積分方程]]求得

:<math>\int_x^{x+1} B_n(u)\,du = x^n.</math>

通過[[积分变换]]得到

:<math>(Tf)(x) = \int_x^{x+1} f(u)\,du</math>

多項式''f'' 等價于 
:<math>
\begin{align}
(Tf)(x) = {e^D - 1 \over D}f(x) & {} = \sum_{n=0}^\infty {D^n \over (n+1)!}f(x) \\
& {} = f(x) + {f'(x) \over 2} + {f''(x) \over 6} + {f'''(x) \over 24} + \cdots  ~.
\end{align}
</math>
這可以用來求解伯努利多項式的[[反函數]]。

[[Category:特殊函数|B]]
[[Category:多项式|B]]