查看“欧拉-麦克劳林求和公式”的源代码

[[File:Colin Maclaurin color.jpg|thumb|right|150px|[[科林·麦克劳林]]是欧拉-麦克劳林求和公式的提出者之一]]
[[File:Leonhard Euler.jpg|thumb|right|150px|[[莱昂哈德·欧拉]]是欧拉-麦克劳林求和公式的提出者之一]]
'''欧拉-麦克劳林求和公式'''在1735年由[[莱昂哈德·欧拉]]与[[科林·麦克劳林]]分别独立发现，该公式提供了一个联系积分与求和的方法，由此可以导出一些渐进展开式。

==公式==
<ref> {{Cite book | author = Gérald Tenenbaum | title = 解析与概率数论导引 | publisher = 高等教育出版社 | date = 2011年1月 | pages = 5 | ISBN = 978-7-04-029467-5 | accessdate = 2015-05-03 | language = 中文 }} </ref>
设{{Smallmath|f=  f(x)}}为一至少{{Smallmath|f=  k+1}}阶可微的函数，{{Smallmath|f=  a,b \in \mathbb{Z} }}，则<br>
<math> 
\begin{align}
\sum_{a<n \le b} f(n) & = \int_{a}^{b} f(t) \, \mathrm{d}t \\
& \quad + \sum_{r=0}^{k} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a)) \\
& \quad + \frac{(-1)^k}{(k+1)!} \int_{a}^{b} \bar{B}_{k+1}(t)f^{(k+1)}(t)dt \\
\end{align}
</math>
<br>
其中
*{{Smallmath|f=  n!:=1 \times 2\times ... \times n}}表示{{Smallmath|f=  n}}的阶乘
*{{Smallmath|f=  f^{(n)}(x)}}表示{{Smallmath|f=  f(x)}}的{{Smallmath|f=  n}}阶导函数
*{{Smallmath|f=  \bar{B}_{n}(x) = B_{n}(\left \langle x \right \rangle)}}，其中
**{{Smallmath|f=  B_{n}(x)}}表示第{{Smallmath|f=  n}}个[[伯努利多项式]]
***[[伯努利多项式]]是满足以下条件的多项式序列：
***<math>
\begin{cases}
B_0(x) \equiv 1 \\
B'_r(x) \equiv r B_{r-1}(x) \quad (r \ge 1) \\
\int_{0}^{1} B_r(x) \, \mathrm{d} x = 0 \quad (r \ge 1)
\end{cases}
</math>
**{{Smallmath|f=  \left \langle x \right \rangle}}表示{{Smallmath|f=  x}}的小数部分
*{{Smallmath|f=  B_{n}:=B_{n}(0)=\bar{B}_{n}(0)}}为第{{Smallmath|f=  n}}个[[伯努利数]]

==证明==
证明使用[[数学归纳法]]以及[[黎曼－斯蒂尔杰斯积分]]，下文中假设{{Smallmath|f=  f(x)}}的可微次数足够大，{{Smallmath|f=  a,b \in \mathbb{Z} }}。<br>
为了方便，将原式的各项用不同颜色表示：<br>
<math> 
\sum_{a<n \le b} f(n) = {\color{red} \int_{a}^{b} f(t) \, \mathrm{d}t} + {\color{OliveGreen} \sum_{r=0}^{k} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a))} + {\color{blue} \frac{(-1)^k}{(k+1)!} \int_{a}^{b} \bar{B}_{k+1}(t)f^{(k+1)}(t)dt}
</math>
<br>
==={{Smallmath|f=  k=0}}的情形===
容易算出<br>
<math>\bar{B}_1(t)={\color{Purple}\left \langle t \right \rangle - \frac{1}{2}}</math><br>
<math> 
\begin{align}
\sum_{a<n \le b} f(n) & = \int_{a}^{b} f(t) \, \mathrm{d}\left \lfloor t \right \rfloor \\
& = {\color{red}\int_{a}^{b} f(t) \, \mathrm{d} t} - \int_{a}^{b} f(t) \, \mathrm{d}\left \langle t \right \rangle \\
& = {\color{red}\int_{a}^{b} f(t) \, \mathrm{d} t} - \int_{a}^{b} f(t) \, \mathrm{d}({\color{Purple}\left \langle t \right \rangle - \frac{1}{2}}) \\
& = {\color{red}\int_{a}^{b} f(t) \, \mathrm{d} t} - {\color{BurntOrange}\int_{a}^{b} f(t) \, \mathrm{d}\bar{B_1}(t)} \\
\end{align}
</math>
<br>
其中橙色的项通过分部积分可化为<br>
<math> 
\begin{align}
{\color{BurntOrange}\int_{a}^{b} f(t) \, \mathrm{d}\bar{B_1}(t)} & = (f(t)\bar{B_1}(t))|_{t=a}^{t=b} - \int_{a}^{b} \bar{B_1}(t) \, \mathrm{d}f(t) \\
& = f(b)B_1(\left \langle b \right \rangle)-f(a)B_1(\left \langle a \right \rangle) - {\color{blue}\int_{a}^{b} \bar{B_1}(t)f'(t) \, \mathrm{d}t} \\
& = {\color{OliveGreen}B_1 \cdot (f(b)-f(a))} - {\color{blue}\int_{a}^{b} \bar{B_1}(t)f'(t) \, \mathrm{d}t} \\
\end{align}
</math>

===假设{{Smallmath|f=  k=n-1}}时原式成立===
<math> 
\sum_{a<n \le b} f(n) = {\color{red} \int_{a}^{b} f(t) \, \mathrm{d}t} + {\color{OliveGreen} \sum_{r=0}^{n-1} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a))} + {\color{blue} \frac{(-1)^{n-1}}{n!} \int_{a}^{b} \bar{B}_{n}(t)f^{(n)}(t) \, \mathrm{d}t }
</math>

===处理积分（蓝色项）===
<math>
\begin{align}
{\color{blue} \frac{(-1)^{n-1}}{n!} \int_{a}^{b} \bar{B}_{n}(t)f^{(n)}(t) \, \mathrm{d}t} & = \frac{(-1)^{n-1}}{n!} \int_{a}^{b} \frac{\bar{B'}_{n+1}(t)}{n+1}f^{(n)}(t) \, \mathrm{d}t \\
& = \frac{(-1)^{n-1}}{(n+1)!} \int_{a}^{b} \bar{B'}_{n+1}(t)f^{(n)}(t) \, \mathrm{d}t \\
& = \frac{(-1)^{n-1}}{(n+1)!} \int_{a}^{b} f^{(n)}(t) \, \mathrm{d}\bar{B}_{n+1}(t) \\
& = \frac{(-1)^{n-1}}{(n+1)!} ((f^{(n)}(t)\bar{B_{n+1}}(t))|_{t=a}^{t=b} - \int_{a}^{b} \bar{B}_{n+1}(t) \, \mathrm{d}f^{(n)}(t)) \\
& = \frac{(-1)^{n-1}}{(n+1)!} (f^{(n)}(b)B_{n+1}(\left \langle b \right \rangle)-f^{(n)}(a)B_{n+1}(\left \langle a \right \rangle) - \int_{a}^{b} \bar{B}_{n+1}(t)f^{(n+1)}(t) \, \mathrm{d}t) \\
& = \frac{(-1)^{n-1}B_{n+1}}{(n+1)!}\cdot (f^{(n)}(b)-f^{(n)}(a)) - \frac{(-1)^{n-1}}{(n+1)!}\int_{a}^{b} \bar{B}_{n+1}(t)f^{(n+1)}(t) \, \mathrm{d}t) \\
& = {\color{OliveGreen}\frac{(-1)^{n+1}B_{n+1}}{(n+1)!}\cdot (f^{(n)}(b)-f^{(n)}(a))} + {\color{blue}\frac{(-1)^{n}}{(n+1)!}\int_{a}^{b} \bar{B}_{n+1}(t)f^{(n+1)}(t) \, \mathrm{d}t)} \\
\end{align}
</math>

===将处理后的积分代入===
<math> 
\begin{align}
\sum_{a<n \le b} f(n) & = {\color{red} \int_{a}^{b} f(t) \, \mathrm{d}t} + {\color{OliveGreen} \sum_{r=0}^{n-1} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a))} + {\color{blue} \frac{(-1)^{n-1}}{n!} \int_{a}^{b} \bar{B}_{n}(t)f^{(n)}(t) \, \mathrm{d}t } \\
& = {\color{red} \int_{a}^{b} f(t) \, \mathrm{d}t} + {\color{OliveGreen} \sum_{r=0}^{n-1} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a))} + {\color{OliveGreen}\frac{(-1)^{n+1}B_{n+1}}{(n+1)!}\cdot (f^{(n)}(b)-f^{(n)}(a))} + {\color{blue}\frac{(-1)^{n}}{(n+1)!}\int_{a}^{b} \bar{B}_{n+1}(t)f^{(n+1)}(t) \, \mathrm{d}t)} \\
& = {\color{red} \int_{a}^{b} f(t) \, \mathrm{d}t} + {\color{OliveGreen} \sum_{r=0}^{n} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} \cdot (f^{(r)}(b)-f^{(r)}(a))} + {\color{blue} \frac{(-1)^{(n)}}{(n+1)!} \int_{a}^{b} \bar{B}_{n+1}(t)f^{(n+1)}(t) \, \mathrm{d}t } \\
\end{align}
</math>
<br>
得到想要的结果。

==余项（积分项）估计==
欧拉-麦克劳林求和公式的精确度通常不一定随着{{Smallmath|f=  k}}的增加而增加，相反地，如果{{Smallmath|f=  k}}相当大，则积分项也会很大。右图是在计算[[调和级数]]的前100项时用[[Mathematica]]算出不同的{{Smallmath|f=  k}}对应的积分项的[[绝对值]]：
[[File:ErrorTeamOfEuler–MaclaurinFormula.png|thumb|600px|计算[[调和级数]]时的误差项]]
<br>

==应用==
通过欧拉-麦克劳林求和公式可以给出[[黎曼ζ函数]]的渐进式：<ref> {{Cite book | author = H.M.Edwards | title = Riemann's Zeta Function | publisher = Dover Publications | date = 2001 | pages = 114 | ISBN = 978-0-486-41740-0 | accessdate = 2015-05-03 | language = en }} </ref><br>
<math>
\begin{align}
\zeta(s) & = \sum_{n=1}^{N-1} n^{-s} + \frac{N^{1-s}}{s-1} + \frac{1}{2}N^{-s} \\
& \quad + \frac{B_2}{2}s N^{-s-1} + ... + \frac{B_{2 \nu}}{(2 \nu)!} s(s+1)...(s+2 \nu -2) N^{(-s-2 \nu +1)}+R_{2 \nu}
\end{align}
</math><br>
其中<br>
<math>R_{2 \nu}= -\frac{s(s+1)...(s+2 \nu -1)}{(2 \nu)!} \int_{N}^{\infty} \bar{B}_{2 \nu}(x)x^{-s-2 \nu} \, \mathrm{d}x</math>

==其他形式==
欧拉-麦克劳林求和公式有时也被写成如下形式：<ref> {{Cite book | author = Tom M.Apostol | title = Introduction to Analytic Number Theory | publisher = 世界图书出版社 | date = 2012 | pages = 54 | ISBN = 978-7-5100-4062-7 | accessdate = 2015-05-03 | language = en }} </ref><br>
<math>\sum_{y<n \le x} f(n)= \int_{y}^{x} f(t) \, \mathrm{d}t + \int_{y}^{x} (t-\left \lfloor t \right \rfloor)f'(t) \, \mathrm{d}t + f(x)(\left \lfloor x \right \rfloor - x)-f(y)(\left \lfloor y \right \rfloor - y)</math><br>
这是欧拉给出的原始形式。

==参考文献==

[[Category:求和法]]