欧拉-麦克劳林求和公式

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。 设 f ( x ) {\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}} 为一至少 k + 1 {\displaystyle {\begin{smallmatrix}k+1\end{smallmatrix}}} 阶可微的函数， a , b ∈ Z {\displaystyle {\begin{smallmatrix}a,b\in \mathbb {Z} \end{smallmatrix}}} ，则 ∑ a < n ≤ b f ( n ) = ∫ a b f ( t ) d t + ∑ r = 0 k ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) k ( k + 1 ) ! ∫ a b B ¯ k + 1 ( t ) f ( k + 1 ) ( t ) d t {\displaystyle {\begin{aligned}\sum _{a<n\leq 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{aligned}}} 其中 n ! := 1 × 2 × . . . × n {\displaystyle {\begin{smallmatrix}n!:=1\times 2\times ...\times n\end{smallmatrix}}} 表示 n {\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}} 的阶乘 f ( n ) ( x ) {\displaystyle {\begin{smallmatrix}f^{(n)}(x)\end{smallmatrix}}} 表示 f ( x ) {\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}} 的 n {\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}} 阶导函数 B ¯ n ( x ) = B n ( ⟨ x ⟩ ) {\displaystyle {\begin{smallmatrix}{\bar {B}}_{n}(x)=B_{n}(\left\langle x\right\rangle )\end{smallmatrix}}} ，其中 B n ( x ) {\displaystyle {\begin{smallmatrix}B_{n}(x)\end{smallmatrix}}} 表示第 n {\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}} 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： { B 0 ( x ) ≡ 1 B r ′ ( x ) ≡ r B r − 1 ( x ) ( r ≥ 1 ) ∫ 0 1 B r ( x ) d x = 0 ( r ≥ 1 ) {\displaystyle {\begin{cases}B_{0}(x)\equiv 1\\B'_{r}(x)\equiv rB_{r-1}(x)\quad (r\geq 1)\\\int _{0}^{1}B_{r}(x)\,\mathrm {d} x=0\quad (r\geq 1)\end{cases}}} ⟨ x ⟩ {\displaystyle {\begin{smallmatrix}\left\langle x\right\rangle \end{smallmatrix}}} 表示 x {\displaystyle {\begin{smallmatrix}x\end{smallmatrix}}} 的小数部分 B n := B n ( 0 ) = B ¯ n ( 0 ) {\displaystyle {\begin{smallmatrix}B_{n}:=B_{n}(0)={\bar {B}}_{n}(0)\end{smallmatrix}}} 为第 n {\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}} 个伯努利数 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 f ( x ) {\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}} 的可微次数足够大， a , b ∈ Z {\displaystyle {\begin{smallmatrix}a,b\in \mathbb {Z} \end{smallmatrix}}} 。 为了方便，将原式的各项用不同颜色表示： ∑ a < n ≤ b f ( n ) = ∫ a b f ( t ) d t + ∑ r = 0 k ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) k ( k + 1 ) ! ∫ a b B ¯ k + 1 ( t ) f ( k + 1 ) ( t ) d t {\displaystyle \sum _{a<n\leq 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}} 容易算出 B ¯ 1 ( t ) = ⟨ t ⟩ − 1 2 {\displaystyle {\bar {B}}_{1}(t)={\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}}} ∑ a < n ≤ b f ( n ) = ∫ a b f ( t ) d ⌊ t ⌋ = ∫ a b f ( t ) d t − ∫ a b f ( t ) d ⟨ t ⟩ = ∫ a b f ( t ) d t − ∫ a b f ( t ) d ( ⟨ t ⟩ − 1 2 ) = ∫ a b f ( t ) d t − ∫ a b f ( t ) d B 1 ¯ ( t ) {\displaystyle {\begin{aligned}\sum _{a<n\leq 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{aligned}}} 其中橙色的项通过分部积分可化为 ∫ a b f ( t ) d B 1 ¯ ( t ) = ( f ( t ) B 1 ¯ ( t ) ) | t = a t = b − ∫ a b B 1 ¯ ( t ) d f ( t ) = f ( b ) B 1 ( ⟨ b ⟩ ) − f ( a ) B 1 ( ⟨ a ⟩ ) − ∫ a b B 1 ¯ ( t ) f ′ ( t ) d t = B 1 ⋅ ( f ( b ) − f ( a ) ) − ∫ a b B 1 ¯ ( t ) f ′ ( t ) d t {\displaystyle {\begin{aligned}{\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{aligned}}} ∑ a < n ≤ b f ( n ) = ∫ a b f ( t ) d t + ∑ r = 0 n − 1 ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) n − 1 n ! ∫ a b B ¯ n ( t ) f ( n ) ( t ) d t {\displaystyle \sum _{a<n\leq 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}} ( − 1 ) n − 1 n ! ∫ a b B ¯ n ( t ) f ( n ) ( t ) d t = ( − 1 ) n − 1 n ! ∫ a b B ′ ¯ n + 1 ( t ) n + 1 f ( n ) ( t ) d t = ( − 1 ) n − 1 ( n + 1 ) ! ∫ a b B ′ ¯ n + 1 ( t ) f ( n ) ( t ) d t = ( − 1 ) n − 1 ( n + 1 ) ! ∫ a b f ( n ) ( t ) d B ¯ n + 1 ( t ) = ( − 1 ) n − 1 ( n + 1 ) ! ( ( f ( n ) ( t ) B n + 1 ¯ ( t ) ) | t = a t = b − ∫ a b B ¯ n + 1 ( t ) d f ( n ) ( t ) ) = ( − 1 ) n − 1 ( n + 1 ) ! ( f ( n ) ( b ) B n + 1 ( ⟨ b ⟩ ) − f ( n ) ( a ) B n + 1 ( ⟨ a ⟩ ) − ∫ a b B ¯ n + 1 ( t ) f ( n + 1 ) ( t ) d t ) = ( − 1 ) n − 1 B n + 1 ( n + 1 ) ! ⋅ ( f ( n ) ( b ) − f ( n ) ( a ) ) − ( − 1 ) n − 1 ( n + 1 ) ! ∫ a b B ¯ n + 1 ( t ) f ( n + 1 ) ( t ) d t ) = ( − 1 ) n + 1 B n + 1 ( n + 1 ) ! ⋅ ( f ( n ) ( b ) − f ( n ) ( a ) ) + ( − 1 ) n ( n + 1 ) ! ∫ a b B ¯ n + 1 ( t ) f ( n + 1 ) ( t ) d t ) {\displaystyle {\begin{aligned}{\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{aligned}}} ∑ a < n ≤ b f ( n ) = ∫ a b f ( t ) d t + ∑ r = 0 n − 1 ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) n − 1 n ! ∫ a b B ¯ n ( t ) f ( n ) ( t ) d t = ∫ a b f ( t ) d t + ∑ r = 0 n − 1 ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) n + 1 B n + 1 ( n + 1 ) ! ⋅ ( f ( n ) ( b ) − f ( n ) ( a ) ) + ( − 1 ) n ( n + 1 ) ! ∫ a b B ¯ n + 1 ( t ) f ( n + 1 ) ( t ) d t ) = ∫ a b f ( t ) d t + ∑ r = 0 n ( − 1 ) r + 1 B r + 1 ( r + 1 ) ! ⋅ ( f ( r ) ( b ) − f ( r ) ( a ) ) + ( − 1 ) ( n ) ( n + 1 ) ! ∫ a b B ¯ n + 1 ( t ) f ( n + 1 ) ( t ) d t {\displaystyle {\begin{aligned}\sum _{a<n\leq 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{aligned}}} 得到想要的结果。 欧拉-麦克劳林求和公式的精确度通常不一定随着 k {\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}} 的增加而增加，相反地，如果 k {\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}} 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 k {\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}} 对应的积分项的绝对值： 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： ζ ( s ) = ∑ n = 1 N − 1 n − s + N 1 − s s − 1 + 1 2 N − s + B 2 2 s N − s − 1 + . . . + B 2 ν ( 2 ν ) ! s ( s + 1 ) . . . ( s + 2 ν − 2 ) N ( − s − 2 ν + 1 ) + R 2 ν {\displaystyle {\begin{aligned}\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}}sN^{-s-1}+...+{\frac {B_{2\nu }}{(2\nu )!}}s(s+1)...(s+2\nu -2)N^{(-s-2\nu +1)}+R_{2\nu }\end{aligned}}} 其中 R 2 ν = − s ( s + 1 ) . . . ( s + 2 ν − 1 ) ( 2 ν ) ! ∫ N ∞ B ¯ 2 ν ( x ) x − s − 2 ν d x {\displaystyle 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} 欧拉-麦克劳林求和公式有时也被写成如下形式： ∑ y < n ≤ x f ( n ) = ∫ y x f ( t ) d t + ∫ y x ( t − ⌊ t ⌋ ) f ′ ( t ) d t + f ( x ) ( ⌊ x ⌋ − x ) − f ( y ) ( ⌊ y ⌋ − y ) {\displaystyle \sum _{y<n\leq 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)} 这是欧拉给出的原始形式。 Gérald Tenenbaum. 解析与概率数论导引. 高等教育出版社. 2011年1月: 5 [2015-05-03]. ISBN 978-7-04-029467-5 （中文）.  H.M.Edwards. Riemann's Zeta Function. Dover Publications. 2001: 114. ISBN 978-0-486-41740-0 （英语）.  使用|accessdate=需要含有|url= (帮助) Tom M.Apostol. Introduction to Analytic Number Theory. 世界图书出版社. 2012: 54. ISBN 978-7-5100-4062-7 （英语）.  使用|accessdate=需要含有|url= (帮助)