高等组合学笔记(十二):Bell多项式,二项式型多项式序列,Faà di Bruno公式

Bell 多项式

定义1: 部分Bell多项式Bn,k=Bn,k(x1,x2,⋯,xn−k+1)B_{n,k}=B_{n,k}(x_1,x_2,\cdots,x_{n-k+1})Bn,k=Bn,k(x1,x2,⋯,xn−k+1) (指数型)
Φ(t,u)=∑n,k≥0Bn,ktnn!uk=exp⁡(u∑m≥1xmtmm!),\Phi(t,u)=\sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k=\exp\left(u\sum_{m\geq1}x_m\frac{t^m}{m!}\right), Φ(t,u)=n,k≥0∑Bn,kn!tnuk=exp(um≥1∑xmm!tm),
或者
Φk(t)=∑n≥kBn,ktnn!=1k!(∑m≥1xmtmm!)k.\Phi_k(t)=\sum_{n\geq k}B_{n,k}\frac{t^n}{n!}=\frac1{k!}\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right)^k. Φk(t)=n≥k∑Bn,kn!tn=k!1(m≥1∑xmm!tm)k.

定义2: 完全Bell多项式Yn=Yn(x1,x2,⋯,xn)Y_n=Y_n(x_1,x_2,\cdots,x_{n})Yn=Yn(x1,x2,⋯,xn)
Yn=∑k=1nBn,k,Y0=1.Y_n=\sum_{k=1}^nB_{n,k},\ Y_0=1. Yn=k=1∑nBn,k, Y0=1.

1+∑n≥1Yn(x1,x2,⋯,xn)tnn!=exp⁡(∑m≥1xmtmm!).1+\sum_{n\geq1}Y_n(x_1,x_2,\cdots,x_{n})\frac{t^n}{n!}=\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right). 1+n≥1∑Yn(x1,x2,⋯,xn)n!tn=exp(m≥1∑xmm!tm).

定理1: 部分Bell多项式为kkk次齐次整系数多项式, 且
Bn,k(x1,x2,⋯,xn−k+1)=∑{c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0n!c1!c2!⋯(1!)c1(2!)c2⋯x1c1x2c2⋯B_{n,k}(x_1,x_2,\cdots,x_{n-k+1})=\sum_{\stackrel{ c_1,c_2,\cdots\geqslant0}{\begin{cases}c_1+c_2+\cdots+c_n=k\\ c_1+2c_2+\cdots+nc_n=n\end{cases}}}\frac{n!}{c_1!c_2!\cdots(1!)^{c_1}{(2!)}^{c_2}\cdots}x_1^{c_1}x_2^{c_2}\cdots Bn,k(x1,x2,⋯,xn−k+1)={c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0∑c1!c2!⋯(1!)c1(2!)c2⋯n!x1c1x2c2⋯
共有p(n,k)p(n,k)p(n,k)项(方程组有p(n,k)p(n,k)p(n,k)个解).

证明:

∑n,k≥0Bn,k(x1,x2,⋯,xn−k+1)tnn!uk=exp⁡(u∑m≥1xmtmm!)=∑k≥0ukk!(∑m≥1xmtmm!)k=∑k≥0ukk!(∑ci≥0c1+c2+⋯=k(kc1,c2,⋯)(x1t11!)c1(x2t22!)c2⋯)k=∑c1,c2,⋯≥0uc1+c2+⋯tc1+2c2+⋯c1!c2!⋯(1!)c1(2!)c2⋯x1c1x2c2⋯\begin{aligned} &\sum_{n,k\geq0}B_{n,k}(x_1,x_2,\cdots,x_{n-k+1})\frac{t^n}{n!}u^k\\ =&\exp\left(u\sum_{m\geq1}x_m\frac{t^m}{m!}\right)\\ =&\sum_{k\geq0}\frac{u^k}{k!}\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right)^k\\ =&\sum_{k\geq0}\frac{u^k}{k!}\left(\sum_{\stackrel{c_1+c_2+\cdots=k}{\small c_i\geq0}}\binom k{c_1,c_2,\cdots}\left(x_1\frac{t^1}{1!}\right)^{c_1}\left(x_2\frac{t^2}{2!}\right)^{c_2}\cdots\right)^k\\ =&\sum_{c_1,c_2,\cdots\geq0}\frac{u^{c_1+c_2+\cdots}t^{c_1+2c_2+\cdots}}{c_1!c_2!\cdots(1!)^{c_1}(2!)^{c_2}\cdots}x_1^{c_1}x_2^{c_2}\cdots \end{aligned} ====n,k≥0∑Bn,k(x1,x2,⋯,xn−k+1)n!tnukexp(um≥1∑xmm!tm)k≥0∑k!uk(m≥1∑xmm!tm)kk≥0∑k!uk⎝⎜⎜⎛ci≥0c1+c2+⋯=k∑(c1,c2,⋯k)(x11!t1)c1(x22!t2)c2⋯⎠⎟⎟⎞kc1,c2,⋯≥0∑c1!c2!⋯(1!)c1(2!)c2⋯uc1+c2+⋯tc1+2c2+⋯x1c1x2c2⋯
比较两端[tnn!uk]\left[\dfrac{t^n}{n!}u^k\right][n!tnuk], 得到
Bn,k(x1,x2,⋯,xn−k+1)=∑{c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0n!c1!c2!⋯(1!)c1(2!)c2⋯x1c1x2c2⋯B_{n,k}(x_1,x_2,\cdots,x_{n-k+1})=\sum_{\stackrel{ c_1,c_2,\cdots\geqslant0}{\begin{cases}c_1+c_2+\cdots+c_n=k\\ c_1+2c_2+\cdots+nc_n=n\end{cases}}}\frac{n!}{c_1!c_2!\cdots(1!)^{c_1}{(2!)}^{c_2}\cdots}x_1^{c_1}x_2^{c_2}\cdots Bn,k(x1,x2,⋯,xn−k+1)={c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0∑c1!c2!⋯(1!)c1(2!)c2⋯n!x1c1x2c2⋯

从上述定理,可以得到:
Bn,k(abx1,ab2x2,⋯)=akbnBn,k(x1,x2,⋯),B_{n,k}(abx_1,ab^2x_2,\cdots)=a^kb^nB_{n,k}(x_1,x_2,\cdots), Bn,k(abx1,ab2x2,⋯)=akbnBn,k(x1,x2,⋯),

Bn,kB_{n,k}Bn,k的一些简单的值:
B0,0=1,B1,1=x1,B2,1=x2,⋯,Bn,1=xn,B2,2=x12,B3,2=3x1x2,B3,3=x13,⋯,Bn,n=x1nBn,0=0,(n>1),Bn,k=0,(k>n)\begin{aligned} B_{0,0}=1,\quad B_{1,1}&=x_1,\quad B_{2,1}=x_2,\quad\cdots,\quad B_{n,1}=x_n,\\ B_{2,2}&=x_1^2,\quad B_{3,2}=3x_1x_2,\quad B_{3,3}=x_1^3,\quad\cdots,\quad B_{n,n}=x_1^n\\ B_{n,0}=0,\ (n>1&),\ B_{n,k}=0,\ (k>n) \end{aligned} B0,0=1,B1,1B2,2Bn,0=0, (n>1=x1,B2,1=x2,⋯,Bn,1=xn,=x12,B3,2=3x1x2,B3,3=x13,⋯,Bn,n=x1n), Bn,k=0, (k>n)

定理2 Bn,kB_{n,k}Bn,k满足的递推关系式:(n⩾1n\geqslant1n⩾1)

Bn,k=∑l=k−1n−1(n−1l)xn−lBl,k−1,B_{n,k}=\sum_{l=k-1}^{n-1}\binom{n-1}{l}x_{n-l}B_{l,k-1}, Bn,k=l=k−1∑n−1(ln−1)xn−lBl,k−1,
kBn,k=∑l=k−1n−1(nl)xn−lBl,k−1,kB_{n,k}=\sum_{l=k-1}^{n-1}\binom nlx_{n-l}B_{l,k-1}, kBn,k=l=k−1∑n−1(ln)xn−lBl,k−1,
Bn,k(x1,x2,⋯)=∑l=0k(nk)x1lBn−l,k−1(0,x2,x3,⋯)=∑l=0kn!(n−k)!l!x1lBn−k,k−l(x22,x33,⋯)\begin{aligned} B_{n,k}(x_1,x_2,\cdots)&=\sum_{l=0}^k\binom nkx_1^lB_{n-l,k-1}(0,x_2,x_3,\cdots)\\ &=\sum_{l=0}^k\frac{n!}{(n-k)!l!}x_1^lB_{n-k,k-l}\left(\frac{x_2}2,\frac{x_3}3,\cdots\right) \end{aligned} Bn,k(x1,x2,⋯)=l=0∑k(kn)x1lBn−l,k−1(0,x2,x3,⋯)=l=0∑k(n−k)!l!n!x1lBn−k,k−l(2x2,3x3,⋯)

定义3 普通型Bell多项式B^n,k\hat B_{n,k}B^n,k,
∑n,k≥0B^n,ktnukk!=exp⁡(u∑m≥1xmtm),\sum_{n,k\geq0}\hat B_{n,k}t^n\frac{u^k}{k!}=\exp\left(u\sum_{m\geq1}x_m{t^m}\right), n,k≥0∑B^n,ktnk!uk=exp(um≥1∑xmtm),
或者
∑n,k≥0B^n,ktn=(∑m≥1xmtm)k,\sum_{n,k\geq0}\hat B_{n,k}t^n=\left(\sum_{m\geq1}x_m{t^m}\right)^k, n,k≥0∑B^n,ktn=(m≥1∑xmtm)k,

定理2: B^n,k\hat B_{n,k}B^n,k也为kkk次齐次整系数多项式
B^n,k(x1,x2,⋯)=∑{c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0k!c1!c2!⋯x1c1x2c2⋯\hat B_{n,k}(x_1,x_2,\cdots)=\large\sum_{\stackrel{\huge {c_1,c_2,\cdots\geqslant0}}{\large\begin{cases}c_1+c_2+\cdots+c_n=k\\ c_1+2c_2+\cdots+nc_n=n\end{cases}}}\frac{k!}{c_1!c_2!\cdots}x_1^{c_1}x_2^{c_2}\cdots B^n,k(x1,x2,⋯)=⎩⎨⎧c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0∑c1!c2!⋯k!x1c1x2c2⋯
更一般地, 还有 ωn\omega_nωn型Bell多项式:

∑n,k≥0B~n,ktnωn=1ωk(∑m≥1xmtmωm)k,\sum_{n,k\geq0}\tilde B_{n,k}\frac{t^n}{\omega_n}=\frac1{\omega_k}\left(\sum_{m\geq1}x_m\frac{t^m}{\omega_m}\right)^k, n,k≥0∑B~n,kωntn=ωk1(m≥1∑xmωmtm)k,
特别地,

ωn=n!\omega_n=n!ωn=n!: 指数型Bell多项式;
ωn=1\omega_n=1ωn=1: 普通型Bell多项式.

定理1 : Bn,kB_{n,k}Bn,k的nnn个特殊值:

Bn,k(1,1,⋯)=S(n,k)B_{n,k}(1,1,\cdots)=S(n,k)Bn,k(1,1,⋯)=S(n,k)(第二类Stirling数);
Bn,k(1!,2!,3!,⋯)=(n−1k−1)n!k!B_{n,k}(1!,2!,3!,\cdots)=\binom {n-1}{k-1}\dfrac{n!}{k!}Bn,k(1!,2!,3!,⋯)=(k−1n−1)k!n!(Lah数);
Bn,k(0!,1!,2!,⋯)=∣s(n,k)∣=∣sˉ(n,k)∣B_{n,k}(0!,1!,2!,\cdots)=|s(n,k)|=|\bar s(n,k)|Bn,k(0!,1!,2!,⋯)=∣s(n,k)∣=∣sˉ(n,k)∣(第一类无符号Stirling数);
Bn,k(1,2,3,⋯)=(nk)kn−kB_{n,k}(1,2,3,\cdots)=\binom {n}{k}k^{n-k}Bn,k(1,2,3,⋯)=(kn)kn−k(幂等数);

证明: (通过Bn,kB_{n,k}Bn,k的生成函数证明)
∑n,k≥0Bn,ktnn!uk=exp⁡(u∑m≥1xmtmm!),\sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k=\exp\left(u\sum_{m\geq1}x_m\frac{t^m}{m!}\right), n,k≥0∑Bn,kn!tnuk=exp(um≥1∑xmm!tm),

取xi=1x_i=1xi=1, 得到
∑n,k≥0Bn,ktnn!uk=exp⁡(u(et−1))⇒Bn,k(1,1,⋯)=S(n,k).\sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k=\exp\left(u(e^t-1)\right)\Rightarrow B_{n,k}(1,1,\cdots)=S(n,k). n,k≥0∑Bn,kn!tnuk=exp(u(et−1))⇒Bn,k(1,1,⋯)=S(n,k).

xi=i!x_i=i!xi=i!, 得到
∑n,k≥0Bn,ktnn!uk=exp⁡(ut1−t)=∑k≥0(ut)kk!(1−t)−k=∑k≥0(ut)kk!∑l≥0(−kl)(−t)l=∑k≥0(ut)kk!∑l≥0⟨k⟩ll!tl=∑k,l≥0⟨k⟩lk!l!uktl+k=∑k,n≥0⟨k⟩nk!n!uktn+k\begin{aligned} \sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k&=\exp\left(\frac{ut}{1-t}\right)\\ &=\sum_{k\geq0}\frac{(ut)^k}{k!}(1-t)^{-k}\\ &=\sum_{k\geq0}\frac{(ut)^k}{k!}\sum_{l\geq0}\binom{-k}l(-t)^l\\ &=\sum_{k\geq0}\frac{(ut)^k}{k!}\sum_{l\geq0}\frac{\langle k\rangle_l}{l!}t^l\\ &=\sum_{k,l\geq0}\frac{\langle k\rangle_l}{k!l!}u^kt^{l+k}=\sum_{k,n\geq0}\frac{\langle k\rangle_n}{k!n!}u^kt^{n+k} \end{aligned} n,k≥0∑Bn,kn!tnuk=exp(1−tut)=k≥0∑k!(ut)k(1−t)−k=k≥0∑k!(ut)kl≥0∑(l−k)(−t)l=k≥0∑k!(ut)kl≥0∑l!⟨k⟩ltl=k,l≥0∑k!l!⟨k⟩luktl+k=k,n≥0∑k!n!⟨k⟩nuktn+k
比较[tnn!uk]\left[\dfrac{t^n}{n!}u^k\right][n!tnuk],
Bn,k(1!,2!,3!,⋯)=n!k!(n−k)!⟨k⟩n−k=(n−1k−1)n!k!.B_{n,k}(1!,2!,3!,\cdots)=\frac{n!}{k!(n-k)!}\langle k\rangle_{n-k}=\binom {n-1}{k-1}\dfrac{n!}{k!}. Bn,k(1!,2!,3!,⋯)=k!(n−k)!n!⟨k⟩n−k=(k−1n−1)k!n!.

取xi=(i−1)!x_i=(i-1)!xi=(i−1)!,
∑n,k≥0Bn,ktnn!uk=exp⁡(−uln⁡(1−t))=(1−t)−u\begin{aligned} \sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k&=\exp\left(-u\ln{(1-t)}\right)\\ &=(1-t)^{-u} \end{aligned} n,k≥0∑Bn,kn!tnuk=exp(−uln(1−t))=(1−t)−u
于是
Bn,k(1!,2!,3!,⋯)=∣s(n,k)∣=∣sˉ(n,k)∣.B_{n,k}(1!,2!,3!,\cdots)=|s(n,k)|=|\bar s(n,k)|. Bn,k(1!,2!,3!,⋯)=∣s(n,k)∣=∣sˉ(n,k)∣.

xi=ix_i=ixi=i,
∑n,k≥0Bn,ktnn!uk=exp⁡(utet)=∑k≥0uktkk!ekt=∑k≥0uktkk!∑l≥0kltll!=∑k,l≥0uktk+lklk!l!\begin{aligned} \sum_{n,k\geq0}B_{n,k}\frac{t^n}{n!}u^k&=\exp\left(ute^t\right)=\sum_{k\geq0}\frac{u^kt^k}{k!}e^{kt}\\ &=\sum_{k\geq0}\frac{u^kt^k}{k!}\sum_{l\geq0}\frac{k^lt^l}{l!}\\ &=\sum_{k,l\geq0}\frac{u^kt^{k+l}k^l}{k!l!} \end{aligned} n,k≥0∑Bn,kn!tnuk=exp(utet)=k≥0∑k!uktkekt=k≥0∑k!uktkl≥0∑l!kltl=k,l≥0∑k!l!uktk+lkl
比较[tnn!uk]\left[\dfrac{t^n}{n!}u^k\right][n!tnuk], 取l=n−kl=n-kl=n−k, 得到
Bn,k(1,2,3,⋯)=(nk)kn−k.B_{n,k}(1,2,3,\cdots)=\binom {n}{k}k^{n-k}. Bn,k(1,2,3,⋯)=(kn)kn−k.

二项式型多项式序列

记为φn(x)\varphi_n(x)φn(x), 通过下面的式子定义:
φn(x+y)=∑k=0n(nk)φk(x)φn−k(y).\varphi_n(x+y)=\sum_{k=0}^n\binom nk\varphi_{k}(x)\varphi_{n-k}(y). φn(x+y)=k=0∑n(kn)φk(x)φn−k(y).
例如:
φn(x)=xn,φn(x)=(x)n,φn(x)=⟨x⟩n,\varphi_n(x)=x^n,\quad\varphi_n(x)=(x)_n,\quad\varphi_n(x)=\langle x\rangle_n, φn(x)=xn,φn(x)=(x)n,φn(x)=⟨x⟩n,

⊛Bn(x)=∑k=0nS(n,k)xk⇒Bn(x+y)=∑k=0n(nk)Bk(x)Bn−k(y)\circledast \qquad B_n(x)=\sum_{k=0}^nS(n,k)x^k \Rightarrow B_n(x+y)=\sum_{k=0}^n\binom nkB_k(x)B_{n-k}(y) ⊛Bn(x)=k=0∑nS(n,k)xk⇒Bn(x+y)=k=0∑n(kn)Bk(x)Bn−k(y)

计算生成函数:
∑n≥0Bn(x)tnn!=exp⁡(x(et−1)).\sum_{n\geq0}B_{n}(x)\frac{t^n}{n!}=\exp\left(x(e^t-1)\right). n≥0∑Bn(x)n!tn=exp(x(et−1)).
其中Bn(1)=bnB_n(1)=b_nBn(1)=bn(Bell数).
⊛⊛\circledast\circledast⊛⊛Abel多项式 An(x,z)=x(x−nz)n−1A_n(x,z)=x(x-nz)^{n-1}An(x,z)=x(x−nz)n−1.
An(x+y,z)=∑k=0n(nk)Ak(x,z)An−k(y,z),A_n(x+y,z)=\sum_{k=0}^n\binom nkA_k(x,z)A_{n-k}(y,z), An(x+y,z)=k=0∑n(kn)Ak(x,z)An−k(y,z),

∑n≥0An(x,z)tnn!=exp⁡(xfˉ(t)),fˉ(t)=∑k≥1(−kz)k−1tkk!.\sum_{n\geq0}A_n(x,z)\frac{t^n}{n!}=\exp{(x\bar f(t))},\qquad\bar f(t)=\sum_{k\geq1}(-kz)^{k-1}\frac{t^k}{k!}. n≥0∑An(x,z)n!tn=exp(xfˉ(t)),fˉ(t)=k≥1∑(−kz)k−1k!tk.

定理2 若φn(x)\varphi_n(x)φn(x)是一个二项式型多项式, 则

Bn,k(φ0(x),2φ1(x),3φ2(x),⋯)=(nk)φn−k(kx),B_{n,k}(\varphi_0(x),2\varphi_1(x),3\varphi_2(x),\cdots)=\binom nk\varphi_{n-k}(kx), Bn,k(φ0(x),2φ1(x),3φ2(x),⋯)=(kn)φn−k(kx),
Bn,k(φ1(x),φ2(x),φ3(x),⋯)=1k!∑j=0k(−1)k−j(kj)φn(jx),B_{n,k}(\varphi_1(x),\varphi_2(x),\varphi_3(x),\cdots)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\varphi_{n}(jx), Bn,k(φ1(x),φ2(x),φ3(x),⋯)=k!1j=0∑k(−1)k−j(jk)φn(jx),

证明:
因为φn(x)\varphi_n(x)φn(x)是一个二项式型多项式,
Ψ(x,t)=∑m≥0φm(x)tmm!=exf(t)\varPsi(x,t)=\sum_{m\geq0}\varphi_m(x)\frac{t^m}{m!}=e^{xf(t)} Ψ(x,t)=m≥0∑φm(x)m!tm=exf(t)
于是
Ψk(x,t)=ekxf(t)=Ψ(kx,t),\varPsi^k(x,t)=e^{kxf(t)}=\varPsi(kx,t), Ψk(x,t)=ekxf(t)=Ψ(kx,t),

1k!(tΨ(x,t))k=1k!tk∑m≥0φm(kx)tmm!=∑m≥0φm(kx)tm+kk!m!=∑n≥k(nk)φn−k(kx)tnn!\begin{aligned} \frac1{k!}(t\varPsi(x,t))^k&=\frac1{k!}t^k\sum_{m\geq0}\varphi_m(kx)\frac{t^m}{m!}\\ &=\sum_{m\geq0}\varphi_m(kx)\frac{t^{m+k}}{k!m!}\\ &=\sum_{n\geq k}\binom nk\varphi_{n-k}(kx)\frac{t^n}{n!} \end{aligned} k!1(tΨ(x,t))k=k!1tkm≥0∑φm(kx)m!tm=m≥0∑φm(kx)k!m!tm+k=n≥k∑(kn)φn−k(kx)n!tn

另一方面,
1k!(tΨ(x,t))k=1k!(t∑m≥0φm(kx)tmm!)k=1k!(∑m≥1mφm−1(kx)tmm!)k=∑n≥kBn,k(φ0(x),2φ1(x),⋯)tnn!\begin{aligned} \frac1{k!}(t\varPsi(x,t))^k&=\frac1{k!}\left(t\sum_{m\geq0}\varphi_m(kx)\frac{t^m}{m!}\right)^k\\ &=\frac1{k!}\left(\sum_{m\geq1}m\varphi_{m-1}(kx)\frac{t^m}{m!}\right)^k\\ &=\sum_{n\geq k}B_{n,k}\left(\varphi_0(x),2\varphi_1(x),\cdots\right)\frac{t^n}{n!} \end{aligned} k!1(tΨ(x,t))k=k!1(tm≥0∑φm(kx)m!tm)k=k!1(m≥1∑mφm−1(kx)m!tm)k=n≥k∑Bn,k(φ0(x),2φ1(x),⋯)n!tn

由
∑n≥kBn,k(φ1(x),φ2(x),φ3(x),⋯)tnn!=1k!(∑m≥1φm(x)tmm!)k=1k!(Ψ(x,t)−1)k=1k!∑j=0k(−1)k−j(kj)Ψ(jx,t)=1k!∑j=0k(−1)k−j(kj)∑n≥0φn(jx)tnn!\begin{aligned} \sum_{n\geq k}B_{n,k}(\varphi_1(x),\varphi_2(x),\varphi_3(x),\cdots)\frac{t^n}{n!}&=\frac1{k!}\left(\sum_{m\geq1}\varphi_m(x)\frac{t^m}{m!}\right)^k\\ &=\frac1{k!}(\varPsi(x,t)-1)^k\\ &=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\varPsi(jx,t)\\ &=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\sum_{n\geq0}\varphi_{n}(jx)\frac{t^n}{n!}\\ \end{aligned} n≥k∑Bn,k(φ1(x),φ2(x),φ3(x),⋯)n!tn=k!1(m≥1∑φm(x)m!tm)k=k!1(Ψ(x,t)−1)k=k!1j=0∑k(−1)k−j(jk)Ψ(jx,t)=k!1j=0∑k(−1)k−j(jk)n≥0∑φn(jx)n!tn
比较两端[tnn!]\left[\dfrac{t^n}{n!}\right][n!tn], 得到
Bn,k(φ1(x),φ2(x),φ3(x),⋯)=1k!∑j=0k(−1)k−j(kj)φn(jx).B_{n,k}(\varphi_1(x),\varphi_2(x),\varphi_3(x),\cdots)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\varphi_{n}(jx). Bn,k(φ1(x),φ2(x),φ3(x),⋯)=k!1j=0∑k(−1)k−j(jk)φn(jx).

特殊的二项式型多项式序列, 代入上述定理验证.

φn(x)=xn\varphi_n(x)=x^nφn(x)=xn,
Bn,k(1,2x,3x2,⋯)=(nk)(kx)n−kBn,k(x,x2,x3,⋯)=1k!∑j=0k(−1)k−j(kj)(jx)n\begin{aligned} B_{n,k}(1,2x,3x^2,\cdots)&=\binom nk(kx)^{n-k}\\ B_{n,k}(x,x^2,x^3,\cdots)&=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj(jx)^n \end{aligned} Bn,k(1,2x,3x2,⋯)Bn,k(x,x2,x3,⋯)=(kn)(kx)n−k=k!1j=0∑k(−1)k−j(jk)(jx)n
令x=1x=1x=1, 代入即得到定理1的①.
φn(x)=⟨x⟩n\varphi_n(x)=\langle x\rangle_nφn(x)=⟨x⟩n,
Bn,k(1,2x,3x(x+1),⋯)=(nk)⟨kx⟩n−kBn,k(x,x(x+1),x(x+1)(x+2),⋯)=1k!∑j=0k(−1)k−j(kj)⟨jx⟩n\begin{aligned} B_{n,k}(1,2x,3x(x+1),\cdots)&=\binom nk\langle kx\rangle_{n-k}\\ B_{n,k}(x,x(x+1),x(x+1)(x+2),\cdots)&=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\langle jx\rangle_n \end{aligned} Bn,k(1,2x,3x(x+1),⋯)Bn,k(x,x(x+1),x(x+1)(x+2),⋯)=(kn)⟨kx⟩n−k=k!1j=0∑k(−1)k−j(jk)⟨jx⟩n
令x=1x=1x=1得到定理1的②.
φn(x)=∑k=0nS(n,k)xk\varphi_n(x)=\sum\limits_{k=0}^nS(n,k)x^kφn(x)=k=0∑nS(n,k)xk, 令x=1x=1x=1得到φn(1)=bn\varphi_n(1)=b_nφn(1)=bn,
Bn,k(b0,2b1,3b2,⋯)=(nk)∑j=0n−kS(n−k,j)kjBn,k(b1,b2,b3,⋯)=1k!∑j=0k(−1)k−j(kj)∑i=0nS(n,i)ji=∑i=0nS(n,i)⋅1k!∑j=0k(−1)k−j(kj)ji=∑i=0nS(n,i)⋅S(i,k)\begin{aligned} B_{n,k}(b_0,2b_1,3b_2,\cdots)&=\binom nk\sum_{j=0}^{n-k}S(n-k,j)k^j\\ B_{n,k}(b_1,b_2,b_3,\cdots)&=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj\sum_{i=0}^nS(n,i)j^i\\ &=\sum_{i=0}^nS(n,i)\cdot \frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom kj j^i\\ &=\sum_{i=0}^nS(n,i)\cdot S(i,k) \end{aligned} Bn,k(b0,2b1,3b2,⋯)Bn,k(b1,b2,b3,⋯)=(kn)j=0∑n−kS(n−k,j)kj=k!1j=0∑k(−1)k−j(jk)i=0∑nS(n,i)ji=i=0∑nS(n,i)⋅k!1j=0∑k(−1)k−j(jk)ji=i=0∑nS(n,i)⋅S(i,k)

Faà di Bruno公式

复合函数求导

定理1: (Faà di Bruno公式, 复合函数高阶求导公式)

设fff和ggg为两个形式TaylorTaylorTaylor级数,
f=∑k≥0fkukk!,g=∑m≥0gmtmm!,(g0=0).f=\sum_{k\geq0}f_k\frac{u^k}{k!}, \quad g=\sum_{m\geq0}g_m\frac{t^m}{m!},(g_0=0). f=k≥0∑fkk!uk,g=m≥0∑gmm!tm,(g0=0).
令hhh为fff和ggg复合的形式级数,
h=∑n≥0hntnn!=f∘g=f(g),h=\sum_{n\geq0}h_n\frac{t^n}{n!}=f\circ g=f(g), h=n≥0∑hnn!tn=f∘g=f(g),
则hhh的系数可由下式给出(Bn,kB_{n,k}Bn,k为部分Bell多项式)
h0=f0,hn=∑k=1nfkBn,k(g1,g2,⋯,gn−k+1).h_0=f_0,\qquad h_n=\sum_{k=1}^nf_kB_{n,k}(g_1,g_2,\cdots,g_{n-k+1}). h0=f0,hn=k=1∑nfkBn,k(g1,g2,⋯,gn−k+1).

证明:

由
h=∑n≥0hntnn!=f(g)=∑k≥0fkgkk!,h=\sum_{n\geq0}h_n\frac{t^n}{n!}=f(g)=\sum_{k\geq0}f_k\frac{g^k}{k!}, h=n≥0∑hnn!tn=f(g)=k≥0∑fkk!gk,
得到hnh_nhn是fkf_kfk的线性组合, 即
hn=∑k=0nAn,kfkh_n=\sum_{k=0}^nA_{n,k}f_k hn=k=0∑nAn,kfk
(An,kA_{n,k}An,k仅依赖于g1,g2,...g_1,g_2,...g1,g2,..., 与fkf_kfk无关)

选取特殊的fkf_kfk, 令
f(u)=eau=∑k≥0akukk!,fk=ak,f(u)=e^{au}=\sum_{k\geq0}a^k\frac{u^k}{k!},\qquad f_k=a^k, f(u)=eau=k≥0∑akk!uk,fk=ak,
得到
h=f∘g=f(g)=eag=exp⁡(a∑m≥1gmtmm!)=∑n,k≥0Bn,k(g1,g2,...)tnn!ak\begin{aligned} h&=f{\small\circ}\, g=f(g)=e^{ag}=\exp(a\sum_{m\geq1}g_m\frac{t^m}{m!})\\ &=\sum_{n,k\geq0}B_{n,k}(g_1,g_2,...)\frac{t^n}{n!}a^k \end{aligned} h=f∘g=f(g)=eag=exp(am≥1∑gmm!tm)=n,k≥0∑Bn,k(g1,g2,...)n!tnak
又由
h=∑n≥0hntnn!=∑n≥0(∑k=0nAn,kfk)tnn!=∑n,k≥0An,ktnn!akh=\sum_{n\geq0}h_n\frac{t^n}{n!}=\sum_{n\geq0}\left(\sum_{k=0}^nA_{n,k}f_k\right)\frac{t^n}{n!}=\sum_{n,k\geq0}A_{n,k}\frac{t^n}{n!}a^k h=n≥0∑hnn!tn=n≥0∑(k=0∑nAn,kfk)n!tn=n,k≥0∑An,kn!tnak
比较[tnn!ak]\left[\dfrac{t^n}{n!}a^k\right][n!tnak], 得到:
An,k=Bn,k(g1,g2,...),A_{n,k}=B_{n,k}(g_1,g_2,...), An,k=Bn,k(g1,g2,...),
又因为B0,0=1B_{0,0}=1B0,0=1, 得到h0=f0h_0=f_0h0=f0, Bn,0=0B_{n,0}=0Bn,0=0,(n>1n>1n>1) 得到∑k=1nBn,k(g1,g2,...)fk\sum_{k=1}^nB_{n,k}(g_1,g_2,...)f_k∑k=1nBn,k(g1,g2,...)fk.

Bn,k(x1,x2,⋯,xn−k+1)=∑{c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0n!c1!c2!⋯(1!)c1(2!)c2⋯x1c1x2c2⋯B_{n,k}(x_1,x_2,\cdots,x_{n-k+1})=\large\sum_{\huge\stackrel{\huge c_1,c_2,\cdots\geqslant0}{\large\begin{cases}c_1+c_2+\cdots+c_n=k\\ c_1+2c_2+\cdots+nc_n=n\end{cases}}}\frac{n!}{c_1!c_2!\cdots(1!)^{c_1}{(2!)}^{c_2}\cdots}x_1^{c_1}x_2^{c_2}\cdots Bn,k(x1,x2,⋯,xn−k+1)=⎩⎨⎧c1+c2+⋯+cn=kc1+2c2+⋯+ncn=nc1,c2,⋯⩾0∑c1!c2!⋯(1!)c1(2!)c2⋯n!x1c1x2c2⋯
得到:
B0,0=1,Bn,1=xn,Bn,n=x1n.B_{0,0}=1, B_{n,1}=x_n,B_{n,n}=x_1^n. B0,0=1,Bn,1=xn,Bn,n=x1n.

定理2:

设F(y),G(x)F(y),G(x)F(y),G(x)为给定的实函数, 且G(x)G(x)G(x)在x=ax=ax=a处具有任意阶导数, F(y)F(y)F(y)在y=b=G(a)y=b=G(a)y=b=G(a)处也有任意阶导数, 令H(x)=F(G(x))H(x)=F(G(x))H(x)=F(G(x)),若设
gm=dmGdxm∣x=a,fk=dkFdyk∣y=b,hn=dnHdxn∣x=a,g_m=\left.\frac{\text{d}^mG}{\text{d}x^m}\right|_{x=a}, \qquad f_k=\left.\frac{\text{d}^kF}{\text{d}y^k}\right|_{y=b},\qquad h_n=\left.\frac{\text{d}^nH}{\text{d}x^n}\right|_{x=a}, gm=dxmdmG∣∣∣∣x=a,fk=dykdkF∣∣∣∣∣y=b,hn=dxndnH∣∣∣∣x=a,
其中:
g0=G(a),f0=F(b)=F(G(a)),h0=H(a)=F(G(a)),g_0=G(a),\qquad f_0=F(b)=F(G(a)),\qquad h_0=H(a)=F(G(a)), g0=G(a),f0=F(b)=F(G(a)),h0=H(a)=F(G(a)),
并定义相应的形式级数:
g(t)=∑m≥1gmtmm!,f(u)=∑k≥0fkukk!,h(t)=∑n≥0hntnn!,g(t)=\sum_{m\geq1}g_m\frac{t^m}{m!},\quad f(u)=\sum_{k\geq0}f_k\frac{u^k}{k!}, \quad h(t)=\sum_{n\geq0}h_n\frac{t^n}{n!}, g(t)=m≥1∑gmm!tm,f(u)=k≥0∑fkk!uk,h(t)=n≥0∑hnn!tn,
则在形式上有h=f∘gh=f{\small\circ}gh=f∘g.

定理3: 对函数F,G,HF,G,HF,G,H, 其中H=F(G)=F⋅GH=F(G)=F\cdot GH=F(G)=F⋅G, 则HHH在x=ax=ax=a处的nnn阶导数为(n⩾1n\geqslant1n⩾1)
hn=dnHdxn∣x=a=∑k=1nfkBn,k(g1,g2,...,gn−k+1).h_n=\left.\frac{\text{d}^nH}{\text{d}x^n}\right|_{x=a}=\sum_{k=1}^nf_kB_{n,k}(g_1,g_2,...,g_{n-k+1}). hn=dxndnH∣∣∣∣x=a=k=1∑nfkBn,k(g1,g2,...,gn−k+1).