多項式

（重定向自常数多项式

定義

${\displaystyle a_{0}+a_{1}X+\cdots +a_{n-1}X^{n-1}+a_{n}X^{n}}$

${\displaystyle p_{0}(X_{1})+p_{1}(X_{1})X_{2}+\cdots +p_{n_{2}-1}(X_{1})X_{2}^{n_{2}-1}+p_{n_{2}}(X_{1})X_{2}^{n_{2}}}$

${\displaystyle -5X^{3}Y}$

多项式的升幂及降幂排列

${\displaystyle \ 2X^{5}Y^{2}+7X^{3}Y^{4}+8X^{1}Y^{6}}$

${\displaystyle 2Y^{2}X^{5}+7Y^{4}X^{3}+8Y^{6}X^{1}}$

多项式的运算

多项式的加法

{\displaystyle {\begin{aligned}{\color {BrickRed}P}&={\color {BrickRed}3X^{2}-2X+5XY-2}\\{\color {RoyalBlue}Q}&={\color {RoyalBlue}-3X^{2}+3X+4Y^{2}+8}\end{aligned}}}

${\displaystyle {\color {BrickRed}P}+{\color {RoyalBlue}Q}=({\color {BrickRed}3X^{2}-2X+5XY-2})\;+\;({\color {RoyalBlue}-3X^{2}+3X+4Y^{2}+8})}$

${\displaystyle P+Q=X+5XY+4Y^{2}+6}$

多项式的减法

${\displaystyle P-Q=(36-5)x^{5}+(7+73)x^{4}+(66+11)x^{3}+(36+11)x^{2}+(66-5)x+(6-3)=31x^{5}+80x^{4}+77x^{3}+47x^{2}+61x+3}$

多项式乘法

{\displaystyle {\begin{aligned}\color {BrickRed}P&=\color {BrickRed}{2X+3Y+5}\\\color {RoyalBlue}Q&=\color {RoyalBlue}{2X+5Y+XY+1}\end{aligned}}}

${\displaystyle {\begin{array}{rccrcrcrcr}{\color {BrickRed}P}{\color {RoyalBlue}Q}&{=}&&({\color {BrickRed}2X}\cdot {\color {RoyalBlue}2X})&+&({\color {BrickRed}2X}\cdot {\color {RoyalBlue}5Y})&+&({\color {BrickRed}2X}\cdot {\color {RoyalBlue}XY})&+&({\color {BrickRed}2X}\cdot {\color {RoyalBlue}1})\\&&+&({\color {BrickRed}3Y}\cdot {\color {RoyalBlue}2X})&+&({\color {BrickRed}3Y}\cdot {\color {RoyalBlue}5Y})&+&({\color {BrickRed}3Y}\cdot {\color {RoyalBlue}XY})&+&({\color {BrickRed}3Y}\cdot {\color {RoyalBlue}1})\\&&+&({\color {BrickRed}5}\cdot {\color {RoyalBlue}2X})&+&({\color {BrickRed}5}\cdot {\color {RoyalBlue}5Y})&+&({\color {BrickRed}5}\cdot {\color {RoyalBlue}XY})&+&({\color {BrickRed}5}\cdot {\color {RoyalBlue}1})\end{array}}}$

${\displaystyle PQ=4X^{2}+21XY+2X^{2}Y+12X+15Y^{2}+3XY^{2}+28Y+5}$

多项式除法

${\displaystyle A=BQ+R}$

${\displaystyle {\begin{matrix}\qquad \quad \;\,X^{2}\;-9X\quad -27\\\qquad \quad X-3{\overline {\vert X^{3}-12X^{2}+0X-42}}\\\;\;{\underline {\;\;X^{3}-\;\;3X^{2}}}\\\qquad \qquad \quad \;-9X^{2}+0X\\\qquad \qquad \quad \;{\underline {-9X^{2}+27X}}\\\qquad \qquad \qquad \qquad \qquad -27X-42\\\qquad \qquad \qquad \qquad \qquad {\underline {-27X+81}}\\\qquad \qquad \qquad \qquad \qquad \qquad \;\;-123\end{matrix}}}$

多项式的矩阵算法

乘法

${\displaystyle f(x)=\sum _{k=0}^{n}a_{k}x^{k},g(x)=\sum _{k=0}^{m}b_{k}x^{k},f(x)g(x)=\sum _{k=0}^{n+m}c_{k}x^{k}}$

${\displaystyle {\begin{pmatrix}c_{0}&c_{1}&\cdots &c_{n+m}\end{pmatrix}}={\begin{pmatrix}a_{0}&a_{1}&\cdots &a_{n}\end{pmatrix}}{\begin{pmatrix}b_{0}&b_{1}&\cdots &b_{m}&0&\cdots &0\\0&b_{0}&\cdots &b_{m-1}&b_{m}&\cdots &0\\\cdots &\cdots &\cdots &\cdots &\cdots &\cdots &\cdots \end{pmatrix}}}$

除法

${\displaystyle f(x)=1-x-2x^{2}+x^{3}+3x^{4}-x^{5},g(x)=3-x+x^{2}-x^{3}}$ ，f(x)除以g(x)

${\displaystyle f(x)=q(x)g(x)+r(x)}$ ，应用多项式乘法的矩阵算法

${\displaystyle {\begin{pmatrix}1&-1&-2&1&3&-1\end{pmatrix}}={\begin{pmatrix}q_{0}&q_{1}&q_{2}\end{pmatrix}}{\begin{pmatrix}3&-1&1&-1&0&0\\0&3&-1&1&-1&0\\0&0&3&-1&1&-1\end{pmatrix}}+{\begin{pmatrix}r_{0}&r_{1}&r_{2}&0&0&0\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}q_{0}&q_{1}&q_{2}\end{pmatrix}}={\begin{pmatrix}1&3&-1\end{pmatrix}}{\begin{pmatrix}-1&0&0\\1&-1&0\\-1&1&-1\end{pmatrix}}^{-1}={\begin{pmatrix}-4&-2&1\end{pmatrix}}}$

${\displaystyle q(x)=-4-2x+x^{2}}$

${\displaystyle {\begin{pmatrix}r_{0}&r_{1}&r_{2}\end{pmatrix}}={\begin{pmatrix}1&-1&-2\end{pmatrix}}-{\begin{pmatrix}-4&-2&1\end{pmatrix}}{\begin{pmatrix}3&-1&1\\0&3&-1\\0&0&3\end{pmatrix}}={\begin{pmatrix}13&1&-3\end{pmatrix}}}$

${\displaystyle r(x)=13+x-3x^{2}}$ [3]

因式分解

${\displaystyle P=(X+1)(X-1)(X^{4}-X^{2}+1)}$

${\displaystyle P=(X+1)(X-1)(X^{2}-{\sqrt {3}}X+1)(X^{2}+{\sqrt {3}}X+1),}$

${\displaystyle P=(X+1)(X-1)(X-{\frac {{\sqrt {3}}+i}{2}})(X-{\frac {{\sqrt {3}}-i}{2}})(X+{\frac {{\sqrt {3}}+i}{2}})(X+{\frac {{\sqrt {3}}-i}{2}})}$

多項式函數

${\displaystyle f_{P}:\;\;\mathbb {A} \longrightarrow \mathbb {A} }$
${\displaystyle x\;\;\mapsto a_{0}+a_{1}x+\cdots +a_{n}x^{n}=P(x)}$

多项式方程

${\displaystyle f_{P}(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}=0}$

${\displaystyle x^{3}+3x-4=0}$

字典排列法

${\displaystyle ax_{1}^{k_{1}}x_{2}^{k_{2}}\dots x_{n}^{k_{n}},bx_{1}^{l_{1}}x_{2}^{l_{2}}\dots x_{n}^{l_{n}}}$ 是两个不同的项

多項式的分析特性

${\displaystyle f_{P}(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}=\sum _{k=0}^{n}a_{k}x^{k}}$

${\displaystyle f_{P}'(x)=a_{1}+2a_{2}x+\cdots +na_{n}x^{n-1}=\sum _{k=1}^{n}ka_{k}x^{k-1}}$

${\displaystyle \int f_{P}(x)\;\mathrm {d} x=C+a_{0}x+{\frac {1}{2}}a_{1}x^{2}+\cdots +{\frac {1}{n+1}}a_{n}x^{n+1}=C+\sum _{k=0}^{n}{\frac {1}{k+1}}a_{k}x^{k+1}}$

${\displaystyle \mathrm {D} (P)=a_{1}+2a_{2}X+\cdots +na_{n}X^{n-1}=\sum _{k=1}^{n}ka_{k}X^{k-1}}$

${\displaystyle \mathrm {I} (P)=a_{0}X+{\frac {1}{2}}a_{1}X^{2}+\cdots +{\frac {1}{n+1}}a_{n}X^{n+1}=\sum _{k=0}^{n}{\frac {1}{k+1}}a_{k}X^{k+1}}$

参考文献

1. Edwards, Harold M. Linear Algebra. Springer. 1995: 47. ISBN 9780817637316.
2. ^ Salomon, David. Coding for Data and Computer Communications. Springer. 2006: 459. ISBN 9780387238043.
3. ^
4. ^ 郭龙先, 张毅敏, 何建琼. 高等代数. 科學出版社. 2011. ISBN 9787030315991.