伯努利數(shù)（2）——一些應(yīng)用

2021-12-18 13:09 作者:子瞻Louis 0人讀過 | 我要投稿

伯努利數(shù)第一期：自然數(shù)的等冪和——伯努利數(shù)

如上一期結(jié)尾所說，第二類伯努利數(shù)比原始的伯努利數(shù)應(yīng)用更多，而上一期是由原始伯努利數(shù)得到的一系列結(jié)論，這里有必要用第二類伯努利數(shù)改良一下：

為了區(qū)別，就暫時先用 $%5Cbeta_n$ 表示原始的伯努利數(shù)吧

$%5Csum_%7Bk%3D0%7D%5En%20%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20n%2B1%20%5C%5C%20k%20%5Cend%7Barray%7D%20%5Cright)B_k%3D0$
$B_n%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20n%20%5C%5C%20k%20%5Cend%7Barray%7D%20%5Cright)B_k$
$%5Cfrac%20x%7Be%5Ex-1%7D%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20B_n%5Cfrac%7Bx%5En%7D%7Bn!%7D$
$S_k(n)%3D%5Cfrac1%7Bk%2B1%7D%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%2B1%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)(-1)%5EjB_jn%5E%7Bk-j%2B1%7D$
$%5Cforall%20k%5Cgeq0%2CB_%7B2k%7D%3D%5Cbeta_%7B2k%7D%EF%BC%8CB_%7B2k%2B1%7D%3D-%5Cbeta_%7B2k%2B1%7D$
$%5Cforall%20k%5Cgeq1%2CB_%7B2k%2B1%7D%3D0$
$B_k%3D(-1)%5E%7Bk%7D%5Cbeta_k$

正切函數(shù)的Maclaurin級數(shù)

我們在微積分中已經(jīng)知道了一些函數(shù)的Maclaurin級數(shù)展開，

$e%5Ex%3D1%2Bx%2B%5Cfrac1%7B2%7Dx%5E2%2B%5Cfrac16x%5E3%2B%E2%80%A6%2Cx%5Cin%5Cmathbb%20C$

$%5Csin%20x%3D1-%5Cfrac1%7B3!%7Dx%5E3%2B%5Cfrac1%7B5!%7Dx%5E5-%E2%80%A6%2Cx%5Cin%5Cmathbb%20C$

這些都是在零點處的n階導(dǎo)數(shù)的規(guī)律比較好找到的函數(shù)，但是正切函數(shù)和余切函數(shù)的規(guī)律就沒有這么容易找到了，但是我們還是有辦法將它展開成冪級數(shù)

根據(jù)伯努利數(shù)的生成函數(shù)定義：

$%5Cfrac%7B2x%7D%7Be%5E%7B2x%7D-1%7D%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%202%5EnB_n%5Cfrac%7Bx%5En%7D%7Bn!%7D$

把第一項拎出來，則后面所有的奇數(shù)項都等于0了，即

$%5Cfrac%7B2x%7D%7Be%5E%7B2x%7D-1%7D%3D-x%2B%5Csum_%7Bn%3D0%7D%5E%5Cinfty%204%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n%7D%7D%7B(2n)!%7D$

$%5CRightarrow%20%5Cfrac%7B2x%7D%7Be%5E%7B2x%7D-1%7D%2Bx%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%204%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n%7D%7D%7B(2n)!%7D$

等式右邊又可以化為：

$%5Cbegin%7Baligned%7D%20%5Cfrac%7B2x%7D%7Be%5E%7B2x%7D-1%7D%2Bx%20%26%3Dx%5Cfrac%7B2%7D%7Be%5E%7B2x%7D-1%7D%2Bx%5Cfrac%7Be%5E%7B2x%7D-1%7D%7Be%5E%7B2x%7D-1%7D%20%5C%5C%20%26%3Dx%5Cfrac%7Be%5E%7B2x%7D%2B1%7D%7Be%5E%7B2x%7D-1%7D%20%5C%5C%20%26%3Dx%5Cfrac%7Be%5Ex%2Be%5E%7B-x%7D%7D%7Be%5Ex-e%5E%7B-x%7D%7D%3Dx%5Ccoth%20x%20%5Cend%7Baligned%7D$

代入回上式就能得到雙曲余切函數(shù)的Lauaent級數(shù)展開：

$x%5Ccoth%20x%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%204%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n%7D%7D%7B(2n)!%7D$

$%5CRightarrow%20%5Ccoth%20x%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%204%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D$

又有 $%5Ccot%20x%3Di%5Cfrac%7Be%5E%7Bix%7D%2Be%5E%7B-ix%7D%7D%7Be%5E%7Bix%7D-e%5E%7B-ix%7D%7D%3Di%5Ccoth%20ix$ ，則

$%5Ccot%20x%3Di%5Csum_%7Bn%3D0%7D%5E%5Cinfty%204%5EnB_%7B2n%7D%5Cfrac%7B(ix)%5E%7B2n-1%7D%7D%7B(2n)!%7D$

$%5CRightarrow%20%5Ccot%20x%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20(-1)%5En4%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D$

又得到了余切函數(shù)的Lauaent級數(shù)展開

再根據(jù)以下：

$%5Cbegin%7Baligned%7D%20%5Ccot%20x-2%5Ccot%202x%20%26%3D%7B%5Ccos%20x%5Cover%5Csin%20x%7D-%7B2%5Ccos%202x%5Cover%5Csin2x%7D%20%5C%5C%20%26%3D%7B%5Ccos%5E2x-%5Ccos%5E2x%2B%5Csin%5E2x%5Cover%5Csin%20x%5Ccos%20x%7D%20%5C%5C%20%26%3D%7B%5Csin%20x%5Cover%5Ccos%20x%7D%3D%5Ctan%20x%20%5Cend%7Baligned%7D$

于是，

$%5Cbegin%7Baligned%7D%5Ctan%20x%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20(-1)%5En4%5EnB_%7B2n%7D%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D-2%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20(-1)%5En4%5EnB_%7B2n%7D%5Cfrac%7B(2x)%5E%7B2n-1%7D%7D%7B(2n)!%7D%20%5C%5C%20%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty(-1)%5En4%5EnB_%7B2n%7D(1-2%5E%7B2n%7D)%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D%20%5Cend%7Baligned%7D$

再將它美化一下：

$%5Ctan%20x%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty(-1)%5E%7Bn%2B1%7D2%5E%7B2n%7D(2%5E%7B2n%7D-1)B_%7B2n%7D%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D$

Zeta函數(shù)的偶數(shù)處值

根據(jù)上一期Weierstrass分解定理得到的等式：

$%5Csin%20z%3Dz%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1-%5Cfrac%7Bz%5E2%7D%7Bn%5E2%5Cpi%5E2%7D%5Cright)$

以及：

$%5Cfrac%7Be%5Ez-e%5E%7B-z%7D%7D2%3D-i%5Cfrac%7Be%5E%7Bi(iz)%7D-e%5E%7B-i(iz)%7D%7D%7B2i%7D%3D-i%5Csin%20iz$
$e%5E%7B2z%7D-1%3D2e%5Ez%5Ccdot%20%5Cfrac%7Be%5Ez-e%5E%7B-z%7D%7D2$

可得：

$e%5E%7B2z%7D-1%3D2e%5Ezz%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%7Bz%5E2%7D%7Bn%5E2%5Cpi%5E2%7D%5Cright)$

而剛好右邊是個整函數(shù)，根據(jù)上一期的定理可得這個等式是成立的

用 $%5Cfrac%20z2$ 替換 $z$ ：

$e%5E%7Bz%7D-1%3De%5E%7Bz%2F2%7Dz%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%7Bz%5E2%7D%7B4n%5E2%5Cpi%5E2%7D%5Cright)$

取對數(shù)導(dǎo)數(shù)：

$%5Cfrac%7Be%5Ez%7D%7Be%5Ez-1%7D%3D%5Cfrac1z%2B%5Cfrac12%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B2z%7D%7Bz%5E2%2B4n%5E2%5Cpi%5E2%7D$

$%5CRightarrow%20%5Cfrac%7Bze%5Ez%7D%7Be%5Ez-1%7D%3D1%2B%5Cfrac12z%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Ccolor%7Bblue%7D%7B%5Cfrac%7B2z%5E2%7D%7Bz%5E2%2B4n%5E2%5Cpi%5E2%7D%7D$

這里假定 $%5Cvert%20z%5Cvert%20%EF%BC%9C2%5Cpi$ ，將藍色部分展開為冪級數(shù)

$%5Cbegin%7Baligned%7D%5Cfrac%7Bze%5Ez%7D%7Be%5Ez-1%7D%26%3D1%2B%5Cfrac12z%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20(-1)%5E%7Bm-1%7D%5Cfrac%7B2z%5E%7B2m%7D%7D%7B(2n%5Cpi)%5E%7B2m%7D%7D%20%5C%5C%20%26%3D1%2B%5Cfrac12z%2B%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20(-1)%5E%7Bm-1%7D%5Cfrac%7B2%7D%7B(2%5Cpi)%5E%7B2m%7D%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E%7B2m%7D%7Dz%5E%7B2m%7D%20%5C%5C%20%26%3D1%2B%5Cfrac12z%2B%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20%5Ccolor%7Bpurple%7D%7B(-1)%5E%7Bm-1%7D%5Cfrac%7B2(2m)!%5Czeta(2m)%7D%7B(2%5Cpi)%5E%7B2m%7D%7D%7D%5Cfrac%7Bz%5E%7B2m%7D%7D%7B(2m)!%7D%5Cend%7Baligned%7D$

而根據(jù)前面的一期專欄又有

$%5Cbegin%7Baligned%7D%5Cfrac%7Bze%5Ez%7D%7Be%5Ez-1%7D%26%3D%5Csum_%7Bm%3D0%7D%5E%5Cinfty%20(-1)%5EmB_m%5Cfrac%7Bz%5Em%7D%7Bm!%7D%20%5C%5C%20%26%3D1%2B%5Cfrac12z%2B%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20%5Ccolor%7Bpurple%7D%7BB_%7B2m%7D%7D%5Cfrac%7Bz%5E%7B2m%7D%7D%7B(2m)!%7D%5Cend%7Baligned%7D$

于是：

$(-1)%5E%7Bm-1%7D%5Cfrac%7B2(2m)!%5Czeta(2m)%7D%7B(2%5Cpi)%5E%7B2m%7D%7D%3DB_%7B2m%7D$

$%5CRightarrow%20%5Czeta(2m)%3D(-1)%5E%7Bm%2B1%7D%5Cfrac%7BB_%7B2m%7D(2%5Cpi)%5E%7B2m%7D%7D%7B2(2m)!%7D%2Cm%5Cin%5Cmathbb%20N$

至此，就可以利用伯努利數(shù)來解決更廣義的Basel問題了：

$%5Cbegin%7Baligned%7D%5Czeta(2)%26%3D1%2B%5Cfrac1%7B2%5E2%7D%2B%5Cfrac1%7B3%5E2%7D%2B%E2%80%A6%3D%5Cfrac%7B%5Cpi%5E2%7D6%20%5C%5C%20%5Czeta(4)%26%3D1%2B%5Cfrac1%7B2%5E4%7D%2B%5Cfrac1%7B3%5E4%7D%2B%E2%80%A6%20%3D%5Cfrac%7B%5Cpi%5E4%7D%7B90%7D%5C%5C%20%5Czeta(6)%26%3D1%2B%5Cfrac1%7B2%5E6%7D%2B%5Cfrac1%7B3%5E6%7D%2B%E2%80%A6%3D%5Cfrac%7B%5Cpi%5E6%7D%7B945%7D%20%5C%5C%26%E2%80%A6%E2%80%A6%5Cend%7Baligned%7D$

伯努利多項式

前面我們已經(jīng)得到了

$S_k(n)%3D%5Cfrac1%7Bk%2B1%7D%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%2B1%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)%5Cbeta_jn%5E%7Bk-j%2B1%7D$

其中 $S_k(n)%3D1%5Ek%2B2%5Ek%2B%E2%80%A6%2Bn%5Ek$ ，右邊括號則為二項式系數(shù)，

不妨再將其化簡，

? $S_k(n)%3D%5Cfrac1%7Bk%2B1%7D%5Csum_%7Bj%3D0%7D%5Ek%5Cfrac%7B(k%2B1)!%7D%7Bj!(k-j%2B1)%7D%5Cbeta_jn%5E%7Bn-j%2B1%7D$

$%3D%5Csum_%7Bj%3D0%7D%5Ek%5Cfrac%7Bk!%7D%7Bj!(k-j)!%7D%5Cbeta_j%5Ccolor%7Bblue%7D%7B%5Cfrac%7Bn%5E%7Bn-j%2B1%7D%7D%7Bk-j%2B1%7D%7D$

根據(jù)Newton-Leibniz?formula（牛頓-萊布尼茲公式），得到

$S_k(n)%3D%5Csum_%7Bj%3D0%7D%5Ek%5Cfrac%7Bk!%7D%7Bj!(k-j)!%7D%5Cbeta_j%5Ccolor%7Bblue%7D%7B%5Cint_%7B0%7D%5Ent%5E%7Bk-j%7D%5Cmathrm%20dt%7D$

$%3D%5Cint_0%5En%5Csum_%7Bj%3D0%7D%5Ek%5Cfrac%7Bk!%7D%7Bj!(k-j)!%7D%5Cbeta_jt%5E%7Bk-j%7D%5Cmathrm%20dt$

$%3D%5Cint_0%5En%5Ccolor%7Bred%7D%7B%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)%5Cbeta_jt%5E%7Bk-j%7D%7D%5Cmathrm%20dt$

用 $%5Cbeta_n(t)$ 來表示紅色部分，就能得到一個比較方便的式子

$S_k(n)%3D%5Cint_0%5En%5Cbeta_k(t)%5Cmathrm%20dt$

利用導(dǎo)數(shù)的性質(zhì)，有

$%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20t%7DS_k(t)%3D%5Cbeta_k(t)$

上一期中我們得到了 $S_k(t)$ 的母函數(shù)為

$e%5Ex%5Cfrac%7Be%5E%7Btx%7D-1%7D%7Be%5Ex-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20S_k(t)%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

取導(dǎo)數(shù)，

$%5CRightarrow%20%5Cfrac%7Bxe%5E%7B(t%2B1)x%7D%7D%7Be%5Ex-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20%5Cbeta_k(t)%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

用-t替換t，有

$%5Cfrac%7Bxe%5E%7B(1-t)x%7D%7D%7Be%5Ex-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20%5Cbeta_k(-t)%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

$%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%5Cleft(%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)%5Cbeta_j(-t)%5E%7Bk-j%7D%5Cright)%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

$%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%5Cleft(%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)%5Ccolor%7Bred%7D%7B(-1)%5E%7Bj%7D%5Cbeta_j%7Dt%5E%7Bk-j%7D%5Cright)%5Cfrac%7B(-x)%5Ek%7D%7Bk!%7D$

紅色部分就是第二類伯努利數(shù)，再用-x替換x，得到

$%5Cfrac%7Bxe%5E%7Btx%7D%7D%7Be%5Ex-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%5Ccolor%7Bgreen%7D%7B%5Cleft(%5Csum_%7Bj%3D0%7D%5Ek%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20k%20%5C%5C%20j%20%5Cend%7Barray%7D%20%5Cright)B_jt%5E%7Bk-j%7D%5Cright)%7D%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

用 $B_k(t)$ 表示綠色部分，這才是耳熟能詳?shù)?strong>伯努利多項式

可以得到以下性質(zhì)：

$B_n(0)%3DB_n$
$B_n(x%2B1)%3D%5Cbeta_n(x)$
$B_n(1-x)%3D(-1)%5EnB_n(x)$
$B_n(1)%3D%5Cbeta_n$
$n%E2%89%A01%E6%97%B6%2CB_n(1)%3DB_n(0)%3DB_n$

1,2兩個可以由生成函數(shù)直接得到，3可以在2中用-x替換x得到，4在3中代入x=0就可以得到，5則可以由4直接推出

微分性質(zhì)

對伯努利多項式求導(dǎo)，得到

$%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7DB_n(x)%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20n%20%5C%5C%20k%20%5Cend%7Barray%7D%20%5Cright)B_k(n-k)x%5E%7Bn-k-1%7D$

$%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cfrac%7Bn(n-1)!%7D%7Bk!(n-1-k)!%7DB_kx%5E%7Bn-1-k%7D$

$%3Dn%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%20n-1%20%5C%5C%20k%20%5Cend%7Barray%7D%20%5Cright)B_kx%5E%7Bn-1-k%7D%3DnB_%7Bn-1%7D(x)$

利用該性質(zhì)，可得

$%5Cint_0%5E1B_n(t)%5Cmathrm%20dt%3D%5Cfrac%7BB_%7Bn%2B1%7D(1)-B_%7Bn%2B1%7D(0)%7D%7Bn%2B1%7D%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A0%20%26%20n%E2%89%A00%2Cn%5Cin%20%5Cmathbb%20N%20%5C%5C%201%20%20%26%20n%3D0%20%0A%5Cend%7Barray%7D%5Cright.$

于是，就可以再次美化一下等冪和公式，對k≠0

$S_k(n)%3D%5Cint_0%5En%5Cbeta_k(t)%5Cmathrm%20dt%3D%5Cint_0%5EnB_k(t%2B1)%5Cmathrm%20dt$

$%5CRightarrow%20S_k(n)%3D%5Cint_0%5E%7Bn%2B1%7DB_k(t)%5Cmathrm%20dt$

差分性質(zhì)

為了方便計算伯努利多項式的差分，不妨利用生成函數(shù)

$%5Cfrac%7Bxe%5E%7B(t%2B1)x%7D-%7Bxe%5E%7Btx%7D%7D%7D%7Be%5Ex-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty(B_k(t%2B1)-B_k(t))%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

$%5CRightarrow%20xe%5E%7Btx%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty(B_k(t%2B1)-B_k(t))%5Cfrac%7Bx%5Ek%7D%7Bk!%7D$

將右式展為Maclaurin級數(shù)，

$xe%5E%7Btx%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20t%5Ek%5Cfrac%7Bx%5E%7Bk%2B1%7D%7D%7Bk!%7D%3D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%20t%5E%7Bk-1%7D%5Cfrac%7Bx%5E%7Bk%7D%7D%7B(k-1)!%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20kt%5E%7Bk-1%7D%5Cfrac%7Bx%5E%7Bk%7D%7D%7Bk!%7D$

對比系數(shù)，就可以得到

$B_k(t%2B1)-B_k(t)%3Dkt%5E%7Bk-1%7D$

這樣我們就又可以通過伯努利多項式的差分性質(zhì)來得到等冪和公式，

Euler-Maclaurin求和公式

伯努利多項式的微分性質(zhì)可以用來導(dǎo)出Euler-Maclaurin求和公式：?

可以計算得到

$B_1(x)%3Dx-%5Cfrac12$

對兩邊微分，則

$%5Cmathrm%20dB_1(x)%3D%5Cmathrm%20dx%20$

設(shè) $F(x)$ 是在[0,1]上N階可導(dǎo)的函數(shù)

對它在[0,1]上積分，有

$%5Cint_0%5E1F(x)%5Cmathrm%20dx%3D%5Cint_0%5E1F(x)%5Cmathrm%20dB_1(x)$

用分部積分法可得：

$%5Cbegin%7Baligned%7D%5Cint_0%5E1F(x)%5Cmathrm%20dx%26%3DF(1)B_1(1)-F(0)B_1(0)-%5Cint_0%5E1F'(x)B_1(x)%5Cmathrm%20dx%20%5C%5C%20%26%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cint_0%5E1F'(x)%5Ccolor%7Bblue%7D%7BB_1(x)%5Cmathrm%20dx%7D%5Cend%7Baligned%7D$

根據(jù)微分性質(zhì)，對整數(shù)k≥0，有

$B_k(x)%5Cmathrm%20dx%3D%5Cfrac1%7Bk%2B1%7D%5Cmathrm%20dB_%7Bk%2B1%7D(x)$

代入到藍色部分，得到

$%5Cint_0%5E1F(x)%5Cmathrm%20dx%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cfrac12%5Cint_0%5E1F'(x)%5Cmathrm%20dB_2(x)$

于是我們又可以對后面的積分用分部積分法

$%5Cint_0%5E1F(x)%5Cmathrm%20dx%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cleft%5B%5Cfrac12B_2(x)F(x)%5Cright%5D_0%5E1%2B%5Cfrac12%5Cint_0%5E1F''(x)%5Ccolor%7Bgreen%7D%7BB_2(x)%5Cmathrm%20dx%7D$

不難注意到又能將綠色部分用微分性質(zhì)，

$%5Cbegin%7Baligned%7D%5Cint_0%5E1F(x)%5Cmathrm%20dx%26%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cleft%5B%5Cfrac12B_2(x)F(x)%5Cright%5D_0%5E1%2B%5Cfrac1%7B2%5Ccdot3%7D%5Cint_0%5E1F''(x)%5Cmathrm%20dB_3(x)%5C%5C%26%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cleft%5B%5Cfrac12B_2(x)F(x)%5Cright%5D_0%5E1%2B%5Cleft%5B%5Cfrac1%7B2%5Ccdot3%7DF''(x)B_3(x)%5Cright%5D_0%5E1%20%5C%5C%20%26%20-%5Cfrac1%7B2%5Ccdot3%7D%5Cint_0%5E1F'''(x)B_3(x)%5Cmathrm%20dx%5Cend%7Baligned%7D$

重復(fù)上述操作N次，可得

$%5Cint_0%5E1F(x)%5Cmathrm%20dx%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Cleft%5B%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7B(-1)%5E%7Bn%2B1%7D%7D%7B(n%2B1)!%7DB_%7Bn%2B1%7D(x)F%5E%7B(n)%7D(x)%5Cright%5D_0%5E1%2B%5Cfrac%7B(-1)%5EN%7D%7BN!%7D%5Cint_0%5E1B_N(x)F%5E%7B(N)%7D(x)%5Cmathrm%20dx$ ? ?當n≥1時， $(-1)%5E%7Bn%2B1%7DB_%7Bn%2B1%7D(1)%3D(-1)%5E%7Bn%2B1%7DB_%7Bn%2B1%7D(0)%3DB_%7Bn%2B1%7D$

$%5Cleft%5B%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7B(-1)%5E%7Bn%2B1%7D%7D%7B(n%2B1)!%7DB_%7Bn%2B1%7D(x)F%5E%7B(n)%7D(x)%5Cright%5D_0%5E1%3D%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7BB_%7Bn%2B1%7D%7D%7B(n%2B1)!%7D%5Cleft(F%5E%7B(n)%7D(1)-F%5E%7B(n)%7D(0)%5Cright)$

代回到上式中，

$%5Cint_0%5E1F(x)%5Cmathrm%20dx%3D%5Cfrac%7BF(1)%2BF(0)%7D2-%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7BB_%7Bn%2B1%7D%7D%7B(n%2B1)!%7D%5Cleft(F%5E%7B(n)%7D(1)-F%5E%7B(n)%7D(0)%5Cright)%2B%5Cfrac%7B(-1)%5EN%7D%7BN!%7D%5Cint_0%5E1B_N(x)F%5E%7B(N)%7D(x)%5Cmathrm%20dx$

不知道看到這里的人眼睛還好嗎（#￣▽￣#）

令 $f(x%2Bt)%3DF(x)%2Ct%5Cin%5Cmathbb%20N$ ，則有

$%5Cint_t%5E%7Bt%2B1%7Df(x)%5Cmathrm%20dx%3D%5Cfrac%7Bf(t%2B1)%2Bf(t)%7D2-%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7BB_%7Bn%2B1%7D%7D%7B(n%2B1)!%7D%5Cleft(f%5E%7B(n)%7D(t%2B1)-f%5E%7B(n)%7D(t)%5Cright)%2B%5Cfrac%7B(-1)%5EN%7D%7BN!%7D%5Cint_t%5E%7Bt%2B1%7DB_N(x)f%5E%7B(N)%7D(x)%5Cmathrm%20dx$

這樣就得到了[t,t+1]區(qū)間上 $f(x)$ 的伯努利數(shù)加積分的表達式，用同樣的方法也可以在其他這樣的區(qū)間上得到上面這種表達式

設(shè)a，b都是整數(shù)，對t從a加到b-1，

$%5Cbegin%7Baligned%7D%5Csum_%7Bt%3Da%7D%5E%7Bb-1%7D%5Cfrac%7Bf(t%2B1)%2Bf(t)%7D2%26%3D%5Csum_%7Bt%3Da%2B1%7D%5E%7Bb%7D%5Cfrac%7Bf(t)%7D2%2B%5Csum_%7Bt%3Da%7D%5E%7Bb-1%7D%5Cfrac%7Bf(t)%7D2%20%5C%5C%20%26%3D%5Csum_%7Bt%3Da%7D%5E%7Bb%7D%5Cfrac%7Bf(t)%7D2-%5Cfrac%7Bf(a)%7D2%2B%5Csum_%7Bt%3Da%7D%5E%7Bb%7D%5Cfrac%7Bf(t)%7D2-%5Cfrac%7Bf(b)%7D2%20%5C%5C%20%26%3D%5Csum_%7Bt%3Da%7D%5E%7Bb%7Df(t)-%5Cfrac%7Bf(b)%2Bf(a)%7D2%5Cend%7Baligned%7D$

而其余的部分可以根據(jù)差分求和規(guī)則和積分區(qū)間的可加性化簡，得到

$%5Cint_a%5E%7Bb%7Df(x)%5Cmathrm%20dx%3D%5Csum_%7Bt%3Da%7D%5E%7Bb%7Df(t)-%5Cfrac%7Bf(b)%2Bf(a)%7D2-%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7BB_%7Bn%2B1%7D%7D%7B(n%2B1)!%7D%5Cleft(f%5E%7B(n)%7D(b)-f%5E%7B(n)%7D(a)%5Cright)%2B%5Cfrac%7B(-1)%5EN%7D%7BN!%7D%5Cint_a%5E%7Bb%7DB_N(%5Cleft%5C%7Bx%5Cright%5C%7D)f%5E%7B(N)%7D(x)%5Cmathrm%20dx$

最后移項，就能得到大名鼎鼎的Euler-Maclaurin求和公式：

$%5Csum_%7Bn%3Da%7D%5Ebf(n)%3D%5Cint_%7Ba%7D%5E%7Bb%7Df(x)%5Cmathrm%20dx%2B%5Cfrac%7Bf(b)%2Bf(a)%7D2%2B%5Csum_%7Bn%3D1%7D%5E%7BN-1%7D%5Cfrac%7BB_%7Bn%2B1%7D%7D%7B(n%2B1)!%7D%5Cleft(f%5E%7B(n)%7D(b)-f%5E%7B(n)%7D(a)%5Cright)-%5Cfrac%7B(-1)%5EN%7D%7BN!%7D%5Cint_a%5E%7Bb%7DB_N(%5Cleft%5C%7Bx%5Cright%5C%7D)f%5E%7B(N)%7D(x)%5Cmathrm%20dx$

若 $f(x)$ 無窮階可導(dǎo)，令 $N%5Crightarrow%20%5Cinfty$ ，后面那個積分就趨于零，于是可以把它抬走，而在大于1的奇數(shù)上的伯努利數(shù)恒為零，就有

$%5Csum_%7Bn%3Da%7D%5Ebf(n)%3D%5Cint_%7Ba%7D%5E%7Bb%7Df(x)%5Cmathrm%20dx%2B%5Cfrac%7Bf(b)%2Bf(a)%7D2%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7BB_%7B2n%7D%7D%7B(2n)!%7D%5Cleft(f%5E%7B(2n-1)%7D(b)-f%5E%7B(2n-1)%7D(a)%5Cright)$

總結(jié)

這一期我們得到了以下結(jié)論

正切函數(shù)的Maclaurin級數(shù)：

$%5Ctan%20x%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty(-1)%5E%7Bn%2B1%7D2%5E%7B2n%7D(2%5E%7B2n%7D-1)B_%7B2n%7D%5Cfrac%7Bx%5E%7B2n-1%7D%7D%7B(2n)!%7D$

zeta函數(shù)在偶數(shù)上的值：

$%5Czeta(2m)%3D(-1)%5E%7Bm%2B1%7D%5Cfrac%7BB_%7B2m%7D(2%5Cpi)%5E%7B2m%7D%7D%7B2(2m)!%7D$

Euler-Maclaurin求和公式：

以上就是本期專欄的全部內(nèi)容了，掰掰

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