切比雪夫多項式

根據三角恒等式

$$cos(n+1)\theta +cos(n-1)\theta =2cos\theta \cdot cos(n\theta)$$

令$\theta=arccost$,則對于函數

$${P}_{n}(t)=cos((n-1)arccost),-1\leqslant t\leqslant 1\text{ }\left( n=1,2,\cdots \right)$$

有遞推關系

$${P}_{n+1}(t)=2t \cdot{P}_{n}(t) -{P}_{n-1}(t)$$

由上式推出的多項式為第一類切比雪夫多項式,其前幾個形式為

$$\left\{\begin{matrix}
\\ {P}_{1}(t)=1
\\ {P}_{2}(t)=t
\\ {P}_{3}(t)=2{{t}^{2}}-1
\\ {P}_{4}(t)=4{{t}^{2}}-3t
\\ {P}_{5}(t)=8{{t}^{4}}-8{{t}^{2}}+1
\\ {P}_{6}(t)=16{{t}^{5}}-20{{t}^{3}}+5t
\\ {P}_{7}(t)=32{{t}^{6}}-48{{t}^{4}}+18{{t}^{2}}-1
\\\vdots
\end{matrix}\right.$$

—個n次多項式可以表示為多個切比雪夫多項式的加權和,在NumPy中,使用Chebyshev類表示由切比雪夫多項式組成的多項式:

$$f\left( t \right)=\sum\limits_{i=0}^{n}{{{a}_{i}}}{{P}_{i}}\left( t \right)$$

${{P}_{i}}\left( t \right)$多項式可以通過Chebyshev.basis(i)獲得,通過多項式類的conveit()方法可以在不同類型的多項式之間相互轉換,轉換的目標類型由kind參數指定,多項式各項的系數按照從小到大的順序排列。

from numpy.polynomial import Polynomial, ChebyshevP1 = Chebyshev.basis(1).convert(kind=Polynomial)P6 = Chebyshev.basis(6).convert(kind=Polynomial)

P1: poly([0. 1.])P6: poly([ -1.   0.  18.   0. -48.   0.  32.])

上所示即為多項式的系數,如果想得到表達式,如下

from numpy.polynomial import Polynomial, Chebyshevfrom sympy import *P6 = Chebyshev.basis(6).convert(kind=Polynomial)t = Symbol('t')P6(t)

1.0*t**2*(1.0*t**2*(32.0*t**2 - 48.0) + 18.0) - 1.0

如上所示即為第六個切比雪夫不等式。

${a}_{i}$的獲得則需要知道$f\left( t \right)$的坐標點,假設已知其五個坐標點(2,6)、(3,9)、(4,10)、(5,15)、(6,19),由于有五個點,故x最高次數為4次方,所以

$$f\left( t \right)=\sum\limits_{i=0}^{4}{{{a}_{i}}}{{P}_{i}}\left( t \right)$$

將其五個點坐標代入,

$$f\left( 2 \right)={{a}_{0}}{{P}_{0}}\left( 2 \right)+{{a}_{1}}{{P}_{1}}\left( 2 \right)+{{a}_{2}}{{P}_{2}}\left( 2 \right)+{{a}_{3}}{{P}_{3}}\left( 2 \right)+{{a}_{4}}{{P}_{4}}\left( 2 \right)=6$$

$$f\left( 3 \right)={{a}_{0}}{{P}_{0}}\left( 3 \right)+{{a}_{1}}{{P}_{1}}\left( 3 \right)+{{a}_{2}}{{P}_{2}}\left( 3 \right)+{{a}_{3}}{{P}_{3}}\left( 3 \right)+{{a}_{4}}{{P}_{4}}\left( 3 \right)=9$$

$$f\left( 4 \right)={{a}_{0}}{{P}_{0}}\left( 4 \right)+{{a}_{1}}{{P}_{1}}\left( 4 \right)+{{a}_{2}}{{P}_{2}}\left( 4 \right)+{{a}_{3}}{{P}_{3}}\left( 4 \right)+{{a}_{4}}{{P}_{4}}\left( 4 \right)=10$$

$$f\left( 5 \right)={{a}_{0}}{{P}_{0}}\left( 5 \right)+{{a}_{1}}{{P}_{1}}\left( 5 \right)+{{a}_{2}}{{P}_{2}}\left( 5 \right)+{{a}_{3}}{{P}_{3}}\left( 5 \right)+{{a}_{4}}{{P}_{4}}\left( 5 \right)=15$$

$$f\left( 6 \right)={{a}_{0}}{{P}_{0}}\left( 6 \right)+{{a}_{1}}{{P}_{1}}\left( 6 \right)+{{a}_{2}}{{P}_{2}}\left( 6 \right)+{{a}_{3}}{{P}_{3}}\left( 6 \right)+{{a}_{4}}{{P}_{4}}\left( 6 \right)=19$$

矩陣表示為:

由上述方程組即可解得${a}_{i}$。Numpy庫里numpy.polynomial.chebyshev模塊提供了chebvander函數可以自動完成上述計算過程,稱虛范德蒙矩陣。

import numpy as npimport numpy.polynomial.chebyshev as chebyshevimport numpy.linalg as linalgx = np.array([2, 3, 4, 5, 6])y = np.array([5, 9, 10, 15, 19])deg = len(x) - 1R = chebyshev.chebvander(x, deg)A = linalg.solve(R, y) 

R:  [[1.0000e+00 2.0000e+00 7.0000e+00 2.6000e+01 9.7000e+01] 　　[1.0000e+00 3.0000e+00 1.7000e+01 9.9000e+01 5.7700e+02] 　　[1.0000e+00 4.0000e+00 3.1000e+01 2.4400e+02 1.9210e+03] 　　[1.0000e+00 5.0000e+00 4.9000e+01 4.8500e+02 4.8010e+03] 　　[1.0000e+00 6.0000e+00 7.1000e+01 8.4600e+02 1.0081e+04]]

A:  [-1.23937500e+02  1.24958333e+02 -2.40000000e+01  2.04166667e+00 -6.25000000e-02]

A即為所需求系數${a}_{i}$。