# 切比雪夫多项式

${\displaystyle (1-x^{2})\,y''-x\,y'+n^{2}\,y=0}$

${\displaystyle (1-x^{2})\,y''-3x\,y'+n(n+2)\,y=0}$

## 定义

${\displaystyle T_{0}(x)=1\,}$
${\displaystyle T_{1}(x)=x\,}$
${\displaystyle T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x).\,}$

${\displaystyle \sum _{n=0}^{\infty }T_{n}(x)t^{n}={\frac {1-tx}{1-2tx+t^{2}}}.}$

${\displaystyle U_{0}(x)=1\,}$
${\displaystyle U_{1}(x)=2x\,}$
${\displaystyle U_{n+1}(x)=2xU_{n}(x)-U_{n-1}(x).\,}$

${\displaystyle \sum _{n=0}^{\infty }U_{n}(x)t^{n}={\frac {1}{1-2tx+t^{2}}}.}$

## 从三角函数定义

${\displaystyle T_{n}(\cos(\theta ))=\cos(n\theta )\,}$

${\displaystyle T_{n}(x)={\begin{cases}\cos(n\arccos(x)),&\ x\in [-1,1]\\\cosh(n\,\mathrm {arccosh} (x)),&\ x\geq 1\\(-1)^{n}\cosh(n\,\mathrm {arccosh} (-x)),&\ x\leq -1\\\end{cases}}}$

${\displaystyle {\begin{matrix}T_{n}(x)&=&\cos(n\arccos(x))\\&=&\mathrm {cosh} (n\,\mathrm {arccosh} (x))\end{matrix}}\ ,\quad \forall x\in \mathbb {R} .}$

${\displaystyle U_{n}(\cos(\theta ))={\frac {\sin((n+1)\theta )}{\sin \theta }}.}$

## 以佩尔方程定义

${\displaystyle T_{i}^{2}-(x^{2}-1)U_{i-1}^{2}=1\,\!}$

${\displaystyle T_{i}+U_{i-1}{\sqrt {x^{2}-1}}=(x+{\sqrt {x^{2}-1}})^{i}.\,\!}$

## 递归公式

${\displaystyle T_{0}(x)=1}$
${\displaystyle U_{-1}(x)=1}$
${\displaystyle T_{n+1}(x)=xT_{n}(x)-(1-x^{2})U_{n-1}(x)}$
${\displaystyle U_{n}(x)=xU_{n-1}(x)+T_{n}(x)}$

${\displaystyle T_{n+1}(x)=T_{n+1}(\cos \vartheta )={}}$ ${\displaystyle \cos((n+1)\vartheta )={}}$ ${\displaystyle \cos(n\vartheta )\cos \vartheta -\sin(n\vartheta )\sin \vartheta ={}}$ ${\displaystyle T_{n}(\cos \vartheta )\cos \vartheta -U_{n-1}(\cos \vartheta )\sin ^{2}\vartheta ={}}$ ${\displaystyle xT_{n}(x)-(1-x^{2})U_{n-1}(x)}$

## 正交性

TnUn 都是区间[−1,1] 上的正交多项式系.

${\displaystyle {\frac {1}{\sqrt {1-x^{2}}}},}$

${\displaystyle \int _{-1}^{1}T_{n}(x)T_{m}(x)\,{\frac {dx}{\sqrt {1-x^{2}}}}=\left\{{\begin{matrix}0&:n\neq m~~~~~\\\pi &:n=m=0\\\pi /2&:n=m\neq 0\end{matrix}}\right.}$

${\displaystyle {\sqrt {1-x^{2}}}}$

${\displaystyle \int _{-1}^{1}U_{n}(x)U_{m}(x){\sqrt {1-x^{2}}}\,dx={\begin{cases}0&:n\neq m\\\pi /2&:n=m\end{cases}}}$

## 基本性质

${\displaystyle n\geq 1}$ 时，${\displaystyle T_{n}}$  的最高次项系数为 ${\displaystyle 2^{n-1}}$ ${\displaystyle n=0}$ 时系数为${\displaystyle 1}$

## 最小零偏差

${\displaystyle n\geq 1}$ ，在所有最高次项系数为1的${\displaystyle n}$ 次多项式中 ， ${\displaystyle f(x)={\frac {1}{2^{n-1}}}T_{n}(x)}$  对零的偏差最小，即它是使得${\displaystyle f(x)}$ ${\displaystyle [-1,1]}$  上绝对值的最大值最小的多项式。 其绝对值的最大值为${\displaystyle {\frac {1}{2^{n-1}}}}$  ， 分别在${\displaystyle -1}$ ${\displaystyle 1}$ ${\displaystyle f}$  的其他 ${\displaystyle n-1}$  个极值点上达到 。

## 两类切比雪夫多项式间的关系

${\displaystyle {\frac {d}{dx}}\,T_{n}(x)=nU_{n-1}(x){\mbox{ , }}n=1,\ldots }$
${\displaystyle T_{n}(x)={\frac {1}{2}}(U_{n}(x)-\,U_{n-2}(x)).}$
${\displaystyle T_{n+1}(x)=xT_{n}(x)-(1-x^{2})U_{n-1}(x)\,}$
${\displaystyle T_{n}(x)=U_{n}(x)-x\,U_{n-1}(x).}$

${\displaystyle 2T_{n}(x)={\frac {1}{n+1}}\;{\frac {d}{dx}}T_{n+1}(x)-{\frac {1}{n-1}}\;{\frac {d}{dx}}T_{n-1}(x){\mbox{ , }}\quad n=1,2,\ldots }$

## 例子

${\displaystyle T_{0}(x)=1\,}$
${\displaystyle T_{1}(x)=x\,}$
${\displaystyle T_{2}(x)=2x^{2}-1\,}$
${\displaystyle T_{3}(x)=4x^{3}-3x\,}$
${\displaystyle T_{4}(x)=8x^{4}-8x^{2}+1\,}$
${\displaystyle T_{5}(x)=16x^{5}-20x^{3}+5x\,}$
${\displaystyle T_{6}(x)=32x^{6}-48x^{4}+18x^{2}-1\,}$
${\displaystyle T_{7}(x)=64x^{7}-112x^{5}+56x^{3}-7x\,}$
${\displaystyle T_{8}(x)=128x^{8}-256x^{6}+160x^{4}-32x^{2}+1\,}$
${\displaystyle T_{9}(x)=256x^{9}-576x^{7}+432x^{5}-120x^{3}+9x.\,}$

${\displaystyle U_{0}(x)=1\,}$
${\displaystyle U_{1}(x)=2x\,}$
${\displaystyle U_{2}(x)=4x^{2}-1\,}$
${\displaystyle U_{3}(x)=8x^{3}-4x\,}$
${\displaystyle U_{4}(x)=16x^{4}-12x^{2}+1\,}$
${\displaystyle U_{5}(x)=32x^{5}-32x^{3}+6x\,}$
${\displaystyle U_{6}(x)=64x^{6}-80x^{4}+24x^{2}-1.\,}$

${\displaystyle T_{n}'(1)=n^{2}\,}$
${\displaystyle T_{n}'(-1)=-(-1)^{n}*n^{2}\,}$
${\displaystyle T_{n}''(1)=(n^{4}-n^{2})/3\,}$
${\displaystyle T_{n}''(-1)=(-1)^{n}*(n^{4}-n^{2})/3\,}$

## 按切比雪夫多项式的展开式

${\displaystyle p(x)=\sum _{n=0}^{N}a_{n}T_{n}(x)}$

## 切比雪夫根

${\displaystyle x_{i}=\cos \left({\frac {2i-1}{2n}}\pi \right){\mbox{ , }}i=1,\ldots ,n.}$

${\displaystyle x_{i}=\cos \left({\frac {i}{n+1}}\pi \right){\mbox{ , }}i=1,\ldots ,n.}$