Chebyshev 多项式

Chebyshev 多项式的引入需要用关于 \(\cos \theta\) 的多项式表示 \(\cos n\theta\), 这可以通过 Euler 公式得到: \[ \cos n\theta + i\sin n \theta = e^{i n \theta} = (\cos \theta + i\sin \theta)^{n} = \sum_{k=0}^{n} C_{n}^{k}(i\sin \theta)^{k}(\cos\theta)^{n-k} \] 因此, \(\cos n\theta\) 等于上式右端的实数部分: \[ \begin{aligned} \cos n\theta &= \mathrm{Re}~ \sum_{k=0}^{n} C_{n}^{k}(i)^{k}(\sin \theta)^{k}(\cos\theta)^{n-k} = \sum_{2k\leq n} C_{n}^{2k}(-1)^{k}(\sin \theta)^{2k}(\cos\theta)^{n-2k} \\ &= \sum_{2k\leq n} C_{n}^{2k}(-1)^{k}(1 - \cos^2\theta)^{k}(\cos\theta)^{n-2k} \triangleq T_{n}(\cos \theta) \end{aligned} \tag{1} \] 而 Chebyshev 多项式就等于 \(T_n\), \(n=1,2,\ldots\)

计算 Chebyshev 多项式

\(T_0 = 1\), \(T_1 = x\), \(T_2 = 2x^2 - 1\)

\(n=3\) 时, \[ \cos 3\theta = (\cos\theta)^3 + 3(-1)(1-\cos^2\theta)\cos\theta = 4(\cos\theta)^3 - 3\cos\theta \\ \Rightarrow T_3 = 4x^3 -3x \] \(n = 4\) 时, \[ \cos 4\theta = (\cos\theta)^4 + 6(-1)(1-\cos^2\theta)\cos^2\theta + (1-\cos^2\theta)^2 = 8(\cos\theta)^4 - 8(\cos\theta)^2 + 1 \\ \Rightarrow T_4 = 8x^4 -8x^2 + 1 \]

接下来我们希望得到 Chebyshev 多项式的正交性, 由三角函数之间的正交性可以得到: \[ \frac{1}{\pi}\int_{0}^{2\pi} \cos n \theta \cos k \theta ~\mathrm{d}\theta = \begin{cases} 1, \quad n=k\\ 0, \quad n \neq k \end{cases} \] 根据式 (1), 将上式积分项写成关于 \(\cos\theta\) 函数的形式: \[ \frac{1}{\pi}\int_{0}^{2\pi} \cos n \theta \cos k \theta ~\mathrm{d}\theta = \frac{1}{\pi}\int_{0}^{2\pi} T_n(\cos\theta) T_k(\cos\theta) ~\mathrm{d}\theta = \frac{2}{\pi}\int_{0}^{\pi} T_n(\cos\theta) T_k(\cos\theta) ~\mathrm{d}\theta \] 作换元 \(x \doteq \cos\theta\), 那么 \[ \mathrm{d} x = -\sin\theta ~\mathrm{d}\theta = -\sqrt{1-x^2} ~\mathrm{d}\theta, \quad \theta|_{0}^{\pi} \rightarrow x|_{1}^{-1} \] 所以得到 \[ \frac{2}{\pi}\int_{0}^{\pi} T_n(\cos\theta) T_k(\cos\theta) ~\mathrm{d}\theta = \frac{2}{\pi}\int_{-1}^{1} T_n(x) T_k(x) \frac{\mathrm{d}x}{\sqrt{1-x^2}} \tag{$\star$} \] 式 (\(\star\)) 表示, Chebyshev 多项式是在带权重意义下积分正交, 权函数 \(w \triangleq 1/\sqrt{1-x^2}\).

另一种推导 Chebyshev 多项式的方法是, 将正交多项式看作是某个对称微分方程的特征函数. 类似于对称矩阵的特征向量是正交的, 对称微分方程的特征函数也是正交的. Chebyshev 多项式对应的微分方程为 \[ \frac{\mathrm{d}~}{\mathrm{d}x} \left( \sqrt{1-x^2} \frac{\mathrm{d}u}{\mathrm{d}x} \right) = \frac{\lambda}{\sqrt{1-x^2}} u, \quad\text{or}\quad \frac{\mathrm{d}~}{\mathrm{d}x} \left( w^{-1} \frac{\mathrm{d}u}{\mathrm{d}x} \right) = \lambda w u \tag{2} \]

验证 Chebyshev 多项式满足微分方程 （2）

\(T_0\) 对应特征值为 \(0\), \(T_1\) 对应特征值为 \(-1\).

将 \(u=T_2(x)\)​ 代入式 (2) 左端得到 \[ \begin{aligned} LHS &= \frac{\mathrm{d}~}{\mathrm{d}x} \left( \sqrt{1-x^2} \frac{\mathrm{d}~}{\mathrm{d}x}\underbrace{(2x^2-1)}_{T_2(x)} \right) = \frac{\mathrm{d}~}{\mathrm{d}x} \left( \sqrt{1-x^2} 4x \right)\\ &= \frac{-x}{\sqrt{1-x^2}}4x + \sqrt{1-x^2}4 = \frac{-4x^2 + 4 - 4x^2}{\sqrt{1-x^2}} \\ &= \frac{-4}{\sqrt{1-x^2}} \underbrace{(2x^2-1)}_{T_3(x)} \equiv RHS(\lambda=-4) \end{aligned} \] 将 \(u=T_3(x)\) 代入式 (2) 左端, 得到 \[ \begin{aligned} LHS &= \frac{\mathrm{d}~}{\mathrm{d}x} \left( \sqrt{1-x^2} \frac{\mathrm{d}~}{\mathrm{d}x}\underbrace{(4x^3-3x)}_{T_3(x)} \right) = \frac{\mathrm{d}~}{\mathrm{d}x} \left( \sqrt{1-x^2} (12x^2-3) \right)\\ &= \frac{-x}{\sqrt{1-x^2}}(12x^2 - 3) + \sqrt{1-x^2}24x = \frac{-12x^3 + 3x + 24x - 24x^3}{\sqrt{1-x^2}}\\ &= \frac{-9}{\sqrt{1-x^2}} \underbrace{(4x^3-3x)}_{T_3(x)} \equiv RHS(\lambda=-9) \end{aligned} \] 将 \(u=T_4(x)\) 代入式 (2) 左端得到 \[ LHS \equiv RHS(\lambda = -16) \]