5.1.3 勒让德多项式的正交性及相应的广义傅里叶级数

标签：5.1 cos int 多项式 dx quad theta 性及 equiv

勒让德多项式的正交性

对于不同项的勒让德多项式：

\[\begin{cases} (1-x^2)P_n''(x)-2xP_n'(x)+n(n+1)P_n(x) \equiv 0, \quad (1)\\ (1-x^2)P_m''(x)-2xP_m'(x)+m(m+1)P_m(x) \equiv 0, \quad (2) \end{cases}\]

证明其正交性：
\(\int_{-1}^1 \{(1) \times P_m(x) - (2) \times P_n(x)\}dx\)

\(\int_{-1}^1 \left\{P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)] - P_n(x)\frac{d}{dx}[(1-x^2)P_m'(x)]\right\}dx+[n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x)dx \equiv 0\)

\((1-x^2)[P_m(x)P_n'(x)-P_n(x)P_m'(x)]_{-1}^1 + \int_{-1}^1(1-x^2)[P_m'(x)P_n'(x)-P_n'(x)P_m'(x)]dx+[n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x)dx \equiv 0\)

\[\int_{-1}^1 P_n(x)P_m(x)dx = 0, \quad m \neq n \]

不同阶勒让德多项式正交！

\(\{P_l(x), \quad l=0,1,2,\cdots\}\) — \([-1,1]\)上连续函数空间的正交完备函数基底

\[f(x)=\sum_{l=0}^{\infty} A_lP_l(x), \quad A_l = \frac{1}{\int_{-1}^1[P_l]^2dx}\int_{-1}^1 f(x)P_l(x)dx \]

广义傅里叶级数展开

\[\int_0^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta d\theta \equiv 0, \quad m \neq n \]

\[f(\theta) = \sum_{l=0}^{\infty} A_lP_l(\cos\theta), \quad A_l = \frac{1}{\int_0^\pi [P_l(\cos\theta)]^2\sin\theta d\theta}\int_0^\pi f(\theta)P_l(\cos\theta)\sin\theta d\theta \]

\(\theta \in [0,\pi]\)

标签：5.1,cos,int,多项式,dx,quad,theta,性及,equiv
From： https://www.cnblogs.com/RES-HON/p/18546879