阿多米安多项式一种参数化算法

来源:Applicable Analysis and Discrete Mathematics 10(01):168-185

方法

\begin{equation} A_n(u_0,u_1,\cdots,u_n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}N\left(\sum_{k=0}^{n}u_ke^{ik\lambda} \right)e^{-in\lambda}\mathrm d\lambda \label{sm} \end{equation}

积分限制了此方法的应用,可采用技巧变化出$N\left(\sum_{k=0}^{n}u_ke^{ik\lambda} \right)=N_0(u_0)N_1\left(\sum_{k=1}^{n}u_ke^{ik\lambda} \right)$,将$N_1(x)$展开成麦克劳林级数,保留至第$n$项,令$x=\sum_{k=1}^{n}u_ke^{ik\lambda}$,取出$e^{in\lambda}$前的系数$C_n$,则$A_n=C_nN_0(u_0)$,即

\begin{equation} \begin{split} A_n(u_0,u_1,\cdots,u_n)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}N\left(\sum_{k=0}^{n}u_ke^{ik\lambda} \right)e^{-in\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}N_0(u_0)N_1\left(\sum_{k=1}^{n}u_ke^{ik\lambda} \right)e^{-in\lambda}\mathrm d\lambda \\ =& \frac{N_0(u_0)}{2\pi}\int_{-\pi}^{\pi}\sum_{j=0}^nN_1^{(j)}\frac{\left(\sum_{k=1}^{n}u_ke^{ik\lambda} \right)^j}{j!}e^{-in\lambda}\mathrm d\lambda \\ =& N_0(u_0) \mathrm{Coefficient[e^{in\lambda}]} \end{split} \label{smexe} \end{equation}

该方法由如下性质:
设$A_{1_n}$、$A_{2_n}$、$\cdots$、$A_{m_n}$分别是算符$N_{1}$、$N_{2}$、$\cdots$、$N_{m}$的阿多米安多项式,我们有:

1 $N(u)=\sum_{k=1}^m\alpha_kN_k(u)$的阿多米安多项式:

\begin{equation} A_n=\sum_{k=1}^m \alpha_k A_{k_n} \label{prop1} \end{equation}

2 $N(u)=\Pi_{k=1}^m N_k(u)$的阿多米安多项式:

\begin{equation} A_n=\sum_{\sum_{j=1}^mk_j=n}\Pi_{k=1}^m A_{j_{k_j}} \label{prop2} \end{equation}

特别地,$N(u)= N_1(u)N_2(u)$,由\eqref{prop2}式,得阿多米安多项式:

\begin{equation} A_n=\sum_{k=0}^{n}A_{1_{k}}A_{2_{n-k}} \label{prop2p} \end{equation}

3 $N(u)= N_1(u)/N_2(u)$的阿多米安多项式:

\begin{equation} \begin{split} A_0=&A_{1_0}/A_{2_0} \\ A_n=&\frac{1}{A_{2_0}}\left (A_{1_{n}}-\sum_{k=1}^{n}A_{1_{k}}A_{2_{n-k}} \right ) \end{split} \label{prop3} \end{equation}

4 $N(u)= N_1^p(u),p\in \mathbb N$的阿多米安多项式:

\begin{equation} \begin{split} A_0=&A_{1_0}^p \\ A_n=&\frac{1}{nA_{1_0}}\sum_{k=1}^{n}(kp-n+k)A_{1_{k}}A_{n-k} \end{split} \label{prop4} \end{equation}

5 $N(u)=N_1(N_2(u))$的阿多米安多项式:

\begin{equation} \begin{split} A_0=&N_1(A_{2_0}) \\ A_n=&\sum_{\sum_{j=1}^njk_j=n,k_j\ge 0}N_1^{(\sum_{j=1}^nk_j)}(A_{2_0})\Pi_{j=1}^n\frac{A_{2_j}^{k_j}}{k_j!} \end{split} \label{prop5} \end{equation}

下面解释一下公式中的求和

比如$n=3$,满足$\sum_{j=1}^njk_j=n,k_j\ge 0$的情形只有三种:$k_1=0,k_2=0,k_3=1$,$k_1=1,k_2=1,k_3=0$,$k_1=3,k_2=0,k_3=0$,于是,

\begin{equation} \begin{split} A_3=&\sum_{\sum_{j=1}^njk_j=3,k_j\ge 0}N_1^{(\sum_{j=1}^nk_j)}(A_{2_0})\Pi_{j=1}^n\frac{A_{2_j}^{k_j}}{k_j!} \\ =& N_1^{(3)}(A_{2_0})A_{2_3}+N_1^{(2)}(A_{2_0})A_{2_1}A_{2_2}+N_1^{(3)}(A_{2_0})\frac{A_{2_1}^3}{3!} \end{split} \label{A3} \end{equation}

\eqref{prop5}式也是计算阿多米安多项式的通用公式。令$N_1=N$,$N_2(u)=u$,代入\eqref{prop5}式,得计算阿多米安多项式的通用公式

\begin{equation} \begin{split} A_0=&N(u_{0}) \\ A_n=&\sum_{\sum_{j=1}^njk_j=n,k_j\ge 0}N^{(\sum_{j=1}^nk_j)}(u_{0})\Pi_{j=1}^n\frac{u_{j}^{k_j}}{k_j!} \end{split} \label{AdoGen} \end{equation}

举例

例1 $N(u)=\sin u$

\begin{equation*} A_0(u_0)=\sin u_0 \end{equation*}

\begin{equation*} \begin{split} A_1(u_0,u_1)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin \left(u_0+u_1e^{i\lambda} \right)e^{-i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}[\sin u_0 \cos(u_1e^{i\lambda})+\cos u_0 \sin(u_1e^{i\lambda})]e^{-i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\left [\sin u_0 \left(1-\frac{u_1^2e^{2i\lambda}}{2!}+\cdots\right)+\cos u_0 \left(u_1e^{i\lambda}-\frac{u_1^3e^{3i\lambda}}{3!}+\cdots\right)\right]e^{-i\lambda}\mathrm d\lambda \\ = &u_1\cos u_0 \end{split} \end{equation*}

\begin{equation*} \begin{split} A_2(u_0,u_1,u_2)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin \left(u_0+u_1e^{i\lambda}+u_2e^{2i\lambda} \right)e^{-2i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}[\sin u_0 \cos(u_1e^{i\lambda}+u_2e^{2i\lambda})+\cos u_0 \sin(u_1e^{i\lambda}+u_2e^{2i\lambda})]e^{-2i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\left [\sin u_0 \left(1-\frac{\left(u_1e^{i\lambda}+u_2e^{2i\lambda} \right)^2}{2!}+\cdots\right)+\cos u_0 \left(u_1e^{i\lambda}+u_2e^{2i\lambda}-\frac{\left(u_1e^{i\lambda}+u_2e^{2i\lambda} \right)^3}{3!}+\cdots\right)\right]e^{-3i\lambda}\mathrm d\lambda \\ = &u_2\cos u_0-\frac{1}{2}u_1^2 \sin u_0 \end{split} \end{equation*}

\begin{equation*} \begin{split} A_3(u_0,u_1,u_2,u_3)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin \left(u_0+u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda} \right)e^{-3i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}[\sin u_0 \cos(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda} )+\cos u_0 \sin(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda}) ]e^{-3i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\left [\sin u_0 \left(1-\frac{\left(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda} \right)^2}{2!}+\cdots\right)+\cos u_0 \left(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda}-\frac{\left(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda} \right)^3}{3!}+\cdots\right)\right]e^{-3i\lambda}\mathrm d\lambda \\ = &u_3\cos u_0-\frac{1}{6}u_1^3 \cos u_0-u_1u_2\sin u_0 \end{split} \end{equation*}

\begin{equation*} \begin{split} A_4(u_0,u_1,u_2,u_3,u_4)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin \left(u_0+u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda}+u_4e^{4i\lambda} \right)e^{-4i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\sin u_0 \cos(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda}+u_4e^{4i\lambda} )e^{-4i\lambda}\mathrm d\lambda +\\ & \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos u_0 \sin(u_1e^{i\lambda}+u_2e^{2i\lambda}+u_3e^{3i\lambda}+u_4e^{4i\lambda} )e^{-4i\lambda}\mathrm d\lambda \\ =& \sin u_0 \mathrm{Coefficient[e^{4i\lambda}]} + \cos u_0 \mathrm{Coefficient[e^{4i\lambda}]} \end{split} \end{equation*}

其中,$\mathrm{Coefficient[e^{4i\lambda}]}$为泰勒级数展开式中$e^{4i\lambda}$项的系数,由Mathematica 计算,结果如下:

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In[58]:= Coefficient[
Normal[Series[Cos[x], {x, 0, 4}]] /.
x -> (Subscript[u, 1]*Exp[I*x] + Subscript[u, 2]*Exp[2*I*x] +
Subscript[u, 3]*Exp[3*I*x] + Subscript[u, 4]*Exp[4*I*x]),
Exp[I*x], 4]
<p>Out[58]=<br />
!(*SubsuperscriptBox[(u), (1), (4)])/24 -<br />
!(*SubsuperscriptBox[(u), (2), (2)])/2 -<br />
Subscript[u, 1] Subscript[u, 3]<br />
In[59]:= Coefficient[<br />
Normal[Series[Sin[x], {x, 0, 4}]] /.<br />
x -&gt; (Subscript[u, 1]<em>Exp[I</em>x] + Subscript[u, 2]<em>Exp[2</em>I<em>x] +<br />
Subscript[u, 3]</em>Exp[3<em>I</em>x] + Subscript[u, 4]<em>Exp[4</em>I<em>x]),<br />
Exp[I</em>x], 4]</p>
<p>Out[59]= -(1/2)<br />
!(*SubsuperscriptBox[(u), (1), (2)]) Subscript[u,<br />
2] + Subscript[u, 4]</p>

于是,

\begin{equation*} A_4(u_0,u_1,u_2,u_3,u_4)=\left(\frac{u_1^4}{24}-u_3 u_1-\frac{u_2^2}{2}\right)\sin u_0 +\left(u_4-\frac{1}{2} u_1^2 u_2\right)\cos u_0 \end{equation*}

类似,可求$A_5$、$A_6$、$\cdots$

例2 $N(u)=e^u$

\begin{equation*} A_0(u_0)=e^{ u_0} \end{equation*}

\begin{equation*} \begin{split} A_1(u_0,u_1)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\exp \left(u_0+u_1e^{i\lambda} \right)e^{-i\lambda}\mathrm d\lambda \\ =& \frac{e^{u_0}}{2\pi}\int_{-\pi}^{\pi}\exp \left(u_1e^{i\lambda} \right)e^{-i\lambda}\mathrm d\lambda \\ =& e^{u_0} \mathrm{Coefficient[e^{i\lambda}]}\\ = &u_1e^{u_0} \end{split} \end{equation*}

最后一个等号由Mathematica计算得到:

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In[62]:= Coefficient[
Normal[Series[Exp[x], {x, 0, 1}]] /. x -> (Subscript[u, 1]*Exp[I*x]),
Exp[I*x], 1]
<p>Out[62]= Subscript[u, 1]</p>

\begin{equation*} \begin{split} A_2(u_0,u_1,u_2)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\exp \left(u_0+u_1e^{i\lambda}+u_2e^{2i\lambda} \right)e^{-i\lambda}\mathrm d\lambda \\ =& \frac{e^{u_0}}{2\pi}\int_{-\pi}^{\pi}\exp \left(u_1e^{i\lambda}+u_2e^{2i\lambda} \right)e^{-2i\lambda}\mathrm d\lambda \\ =& e^{u_0} \mathrm{Coefficient[e^{2i\lambda}]}\\ = &\left(\frac{u_1^2}{2}+u_2 \right)e^{u_0} \end{split} \end{equation*}

最后一个等号由Mathematica计算得到:
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In[65]:= Coefficient[
Normal[Series[Exp[x], {x, 0, 2}]] /.
x -> (Subscript[u, 1]*Exp[I*x] + Subscript[u, 2]*Exp[2*I*x]),
Exp[I*x], 2]
<p>Out[65]=<br />
!(*SubsuperscriptBox[(u), (1), (2)])/2 + Subscript[u, 2]</p>

例3 $N(u)=\ln(u)$

\begin{equation*} A_0(u_0)=\ln u_0 \end{equation*}

\begin{equation*} \begin{split} A_1(u_0,u_1)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\ln \left(u_0+u_1e^{i\lambda} \right)e^{-i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\ln \left [ u_0 \left(1+\frac{u_1e^{i\lambda}}{u_0} \right)\right ]e^{-i\lambda}\mathrm d\lambda \\ =& \frac{1}{2\pi}\int_{-\pi}^{\pi}\left [\ln u_0+\ln \left(1+\frac{u_1e^{i\lambda}}{u_0} \right)\right ]e^{-i\lambda}\mathrm d\lambda \\ =& \mathrm{Coefficient[e^{i\lambda}]}\\ = &\frac{u_1}{u_0} \end{split} \end{equation*}

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In[69]:= Coefficient[
Normal[Series[Log[1 + x], {x, 0, 1}]] /.
x -> (Subscript[u, 1]*Exp[I*x]/Subscript[u, 0]), Exp[I*x], 1]
<p>Out[69]= Subscript[u, 1]/Subscript[u, 0]</p>

同理:

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In[70]:= Coefficient[
Normal[Series[Log[1 + x], {x, 0, 2}]] /.
x -> (Subscript[u, 1]*Exp[I*x]/Subscript[u, 0] +
Subscript[u, 2]*Exp[2*I*x]/Subscript[u, 0]), Exp[I*x], 2]
<p>Out[70]= -(<br />
!(*SubsuperscriptBox[(u), (1), (2)])/(2<br />
!(*SubsuperscriptBox[(u), (</p>

\begin{equation*} A_2(u_0,u_1,u_2)=\frac{u_2}{u_0}-\frac{u_1^2}{2 u_0^2} \end{equation*}

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In[71]:= Coefficient[
Normal[Series[Log[1 + x], {x, 0, 3}]] /.
x -> (Subscript[u, 1]*Exp[I*x]/Subscript[u, 0] +
Subscript[u, 2]*Exp[2*I*x]/Subscript[u, 0] +
Subscript[u, 3]*Exp[3*I*x]/Subscript[u, 0]), Exp[I*x], 3]
<p>Out[71]=<br />
!(*SubsuperscriptBox[(u), (1), (3)])/(3<br />
!(*SubsuperscriptBox[(u), (</p>

\begin{equation*} A_3(u_0,u_1,u_2,u_3)=\frac{u_1^3}{3 u_0^3}-\frac{u_2 u_1}{u_0^2}+\frac{u_3}{u_0} \end{equation*}

以上各例,也可以由\eqref{AdoGen}式来计算。

标签: 阿多米安分解法

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