数学物理方程 第8次作业

Chasse_neige

1.试求

(a) ${\mathrm{P}}_{2n}(0)$, ${\mathrm{P}}_{2n+1}^{\prime }(0)$

$$\begin{array}{c}{P}_n(x)=\frac{1}{2^{n}n!}\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}(x^2-1{)}^n\\ =\frac{1}{2^{n}n!}\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}\sum _{m=0}^n\frac{(-1{)}^{m}n!}{m!(n-m)!}{x}^{2(n-m)}\\ =\frac{1}{2^{n}n!}\sum _{m=0}^{[\frac{n}{2}]}\frac{(-1{)}^{m}n!}{m!(n-m)!}\frac{(2n-2m)!}{(n-2m)!}{x}^{n-2m}\end{array}$$$$P_n^{\prime }(x)=\frac{1}{2^{n}n!}\sum _{m=0}^{[\frac{n-1}{2}]}\frac{(-1{)}^{m}n!}{m!(n-m)!}\frac{(2n-2m)!}{(n-2m-1)!}{x}^{n-2m-1}$$$$P_n^{″}(x)=\frac{1}{2^{n}n!}\sum _{m=0}^{[\frac{n}{2}]-1}\frac{(-1{)}^{m}n!}{m!(n-m)!}\frac{(2n-2m)!}{(n-2m-2)!}{x}^{n-2m-2}$$

所以

$$P_{2n}(0)=\frac{1}{2^{2n}(2n)!}\frac{(-1{)}^n(2n)!}{(n!{)}^2}(2n)!=\frac{(-1{)}^n(2n)!}{2^{2n}(n!{)}^2}$$$$P_{2n+1}^{\prime }(0)=\frac{1}{2^{2n+1}(2n+1)!}\frac{(-1{)}^n(2n+1)!}{n!(n+1)!}(2n+2)!=\frac{(-1{)}^n(2n+1)!}{2^{2n}(n!{)}^2}$$

(b) ${\mathrm{P}}_n^{\prime }(\pm 1)$, ${\mathrm{P}}_n^{\prime \prime }(\pm 1)$

利用生成函数

$$\frac{1}{\sqrt{1-2xt+t^2}}=\sum _{l=0}^{\mathrm{\infty }}{P}_l(x)t^l$$

带入 $x=1$ ，得到

$$\sum _{l=0}^{\mathrm{\infty }}{P}_l(1)t^l=\frac{1}{\sqrt{1-2t+t^2}}=\frac{1}{1-t}=\sum _{l=0}^{\mathrm{\infty }}{t}^l$$

比较两侧系数，得到

$$P_l(1)=1$$

带入 $x=-1$ ，得到

$$\sum _{l=0}^{\mathrm{\infty }}{P}_l(-1)t^l=\frac{1}{1+t}=\sum _{l=0}^{\mathrm{\infty }}(-1{)}^l{t}^l$$

比较系数，得到

$$P_l(-1)=(-1{)}^l$$

两侧同时对 $x$ 偏导并且带入 $x=\pm 1$

$$\frac{t}{(1-2xt+t^2{)}^{\frac{3}{2}}}=\sum _{l=0}^{\mathrm{\infty }}{P}_l^{\prime }(x)t^l$$$$\frac{t}{(1\mp 2t+t^2{)}^{\frac{3}{2}}}=\sum _{l=0}^{\mathrm{\infty }}{P}_l^{\prime }(\pm 1)t^l$$$$\frac{t}{(1\mp t{)}^3}=\sum _{l=0}^{\mathrm{\infty }}{P}_l^{\prime }(\pm 1)t^l=\sum _{l=2}^{\mathrm{\infty }}t(1\pm 3t+(-1{)}^l\frac{(l+1)(l+2)}{2}(\mp t{)}^l)$$

对比系数，得到

$$\begin{array}{c}{P}_0^{\prime }(\pm 1)=0\\ P_1^{\prime }(\pm 1)=1\\ P_2^{\prime }(\pm 1)=\pm 3\\ P_l^{\prime }(\pm 1)=(\pm 1{)}^{l+1}\frac{l(l+1)}{2}\end{array}$$

对于 $2$ 阶导，在两边继续对 $x$ 偏导得到

$$\frac{3t^2}{(1-2xt+t^2{)}^{\frac{5}{2}}}=\sum _{l=0}^{\mathrm{\infty }}{P}_l^{″}(x)t^l$$$$\begin{array}{c}\frac{3t^2}{(1\mp t{)}^5}=\sum _{l=0}^{\mathrm{\infty }}{P}_l^{″}(\pm 1)t^l\\ =\sum _{l=4}^{\mathrm{\infty }}3t^2(1\pm 5t+15(\mp t{)}^2+35(\pm t{)}^3+(-1{)}^l\frac{(l+1)(l+2)(l+3)(l+4)}{24}(\mp t{)}^l)\end{array}$$

对比两边系数，得到

$$\begin{array}{c}{P}_0^{″}(\pm 1)=P_1^{″}(\pm 1)=0\\ P_2^{″}(\pm 1)=3\\ P_3^{″}(\pm 1)=\pm 15\\ P_4^{″}(\pm 1)=45\\ P_l^{″}(\pm 1)=(\pm 1{)}^l\frac{(l-1)l(l+1)(l+2)}{8}\end{array}$$

2.证明

$${\int }_x^1{\mathrm{P}}_k(x){\mathrm{P}}_l(x)\mathrm{d}x=\frac{(1-x^2)[{\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)-{\mathrm{P}}_l^{\prime }(x){\mathrm{P}}_k(x)]}{k(k+1)-l(l+1)},k\ne l$$

勒让德多项式 ${\mathrm{P}}_k(x)$ 和 ${\mathrm{P}}_l(x)$ 满足勒让德微分方程

$$(1-x^2){\mathrm{P}}_k^{″}(x)-2x{\mathrm{P}}_k^{\prime }(x)+k(k+1){\mathrm{P}}_k(x)=0$$$$(1-x^2){\mathrm{P}}_l^{″}(x)-2x{\mathrm{P}}_l^{\prime }(x)+l(l+1){\mathrm{P}}_l(x)=0$$

令 $u={\mathrm{P}}_k(x)$, $v={\mathrm{P}}_l(x)$。考虑表达式

$$\frac{d}{dx}[(1-x^2)(u^{\prime }v-u{v}^{\prime })]$$

计算导数

$$\frac{d}{dx}[(1-x^2)(u^{\prime }v-u{v}^{\prime })]=(1-x^2)\frac{d}{dx}(u^{\prime }v-u{v}^{\prime })+(-2x)(u^{\prime }v-u{v}^{\prime })$$

其中

$$\frac{d}{dx}(u^{\prime }v-u{v}^{\prime })=u^{″}v+u^{\prime }{v}^{\prime }-u^{\prime }{v}^{\prime }-u{v}^{″}=u^{″}v-u{v}^{″}$$

所以

$$\frac{d}{dx}[(1-x^2)(u^{\prime }v-u{v}^{\prime })]=(1-x^2)(u^{″}v-u{v}^{″})-2x(u^{\prime }v-u{v}^{\prime })$$

从微分方程代入 $u^{″}$ 和 $v^{″}$

$$(1-x^2)u^{″}=2x{u}^{\prime }-k(k+1)u,(1-x^2)v^{″}=2x{v}^{\prime }-l(l+1)v$$

所以

$$(1-x^2)(u^{″}v-u{v}^{″})=[2x{u}^{\prime }-k(k+1)u]v-u[2x{v}^{\prime }-l(l+1)v]=2x(u^{\prime }v-u{v}^{\prime })-[k(k+1)-l(l+1)]uv$$

因此

$$\begin{array}{c}\frac{d}{dx}[(1-x^2)(u^{\prime }v-u{v}^{\prime })]=2x(u^{\prime }v-u{v}^{\prime })-[k(k+1)-l(l+1)]uv-2x(u^{\prime }v-u{v}^{\prime })\\ =-[k(k+1)-l(l+1)]uv\end{array}$$

即：

$$\frac{d}{dx}[(1-x^2)({\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)-{\mathrm{P}}_l^{\prime }(x){\mathrm{P}}_k(x))]=-[k(k+1)-l(l+1)]{\mathrm{P}}_k(x){\mathrm{P}}_l(x)$$

积分从 $x$ 到 $1$

$${\int }_x^1{\mathrm{P}}_k(t){\mathrm{P}}_l(t)dt=-\frac{1}{k(k+1)-l(l+1)}{\int }_x^1\frac{d}{dt}[(1-t^2)({\mathrm{P}}_k^{\prime }(t){\mathrm{P}}_l(t)-{\mathrm{P}}_l^{\prime }(t){\mathrm{P}}_k(t))]dt$$$${\int }_x^1\frac{d}{dt}[\cdots ]dt={[(1-t^2)({\mathrm{P}}_k^{\prime }(t){\mathrm{P}}_l(t)-{\mathrm{P}}_l^{\prime }(t){\mathrm{P}}_k(t))]}_{t=x}^{t=1}$$

在 $t=1$ 时，$1-t^2=0$，所以该项为零。在 $t=x$ 时，值为 $-(1-x^2)({\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)-{\mathrm{P}}_l^{\prime }(x){\mathrm{P}}_k(x))$。因此

$${\int }_x^1{\mathrm{P}}_k(t){\mathrm{P}}_l(t)dt=-\frac{1}{k(k+1)-l(l+1)}[0-(1-x^2)({\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)-{\mathrm{P}}_l^{\prime }(x){\mathrm{P}}_k(x))]$$$$=\frac{(1-x^2)[{\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)-{\mathrm{P}}_l^{\prime }(x){\mathrm{P}}_k(x)]}{k(k+1)-l(l+1)}$$

4.计算下列积分

(a)

$${\int }_{-1}^1(1-x^2)\frac{\mathrm{d}{P}_k(x)}{\mathrm{d}x}\frac{\mathrm{d}{P}_l(x)}{\mathrm{d}x}\mathrm{d}x$$

勒让德微分方程为

$$\frac{d}{dx}((1-x^2)\frac{d{P}_n}{dx})+n(n+1)P_n=0$$

令 $L=\frac{d}{dx}((1-x^2)\frac{d}{dx})$，则 $L{P}_n=-n(n+1)P_n$

对于 $u=P_k(x)$, $v=P_l(x)$，有

$${\int }_{-1}^{1}uLvdx={\int }_{-1}^{1}u\frac{d}{dx}((1-x^2)\frac{dv}{dx})dx$$

分部积分：

$$={[u(1-x^2)\frac{dv}{dx}]}_{-1}^1-{\int }_{-1}^1\frac{du}{dx}(1-x^2)\frac{dv}{dx}dx$$

在 $x=\pm 1$ 时，$1-x^2=0$，边界项为零，所以

$${\int }_{-1}^{1}uLvdx=-{\int }_{-1}^1(1-x^2)\frac{du}{dx}\frac{dv}{dx}dx$$

代入 $Lv=-l(l+1)v$

$$-{\int }_{-1}^1(1-x^2)\frac{du}{dx}\frac{dv}{dx}dx=-l(l+1){\int }_{-1}^{1}uvdx$$

利用正交性 ${\int }_{-1}^1{P}_k{P}_{l}dx=\frac{2}{2l+1}{\delta }_{kl}$

$${\int }_{-1}^1(1-x^2)P_k^{\prime }{P}_l^{\prime }dx=l(l+1)\frac{2}{2l+1}{\delta }_{kl}=\frac{2l(l+1)}{2l+1}{\delta }_{kl}$$

由于 ${\delta }_{kl}$，可写为

$${\int }_{-1}^1(1-x^2)\frac{\mathrm{d}{P}_k(x)}{\mathrm{d}x}\frac{\mathrm{d}{P}_l(x)}{\mathrm{d}x}\mathrm{d}x=\frac{2k(k+1)}{2k+1}{\delta }_{kl}$$

(b)

$${\int }_{-1}^1\frac{\mathrm{d}{P}_k(x)}{\mathrm{d}x}\frac{\mathrm{d}{P}_l(x)}{\mathrm{d}x}\mathrm{d}x$$

不失一般性，假设 $k<l$

$${\int }_{-1}^1\frac{d{P}_k(x)}{dx}\frac{d{P}_l(x)}{dx}\mathrm{d}x={P_l(x)\frac{d{P}_k(x)}{dx}|}_{-1}^1-{\int }_{-1}^1{P}_l(x)P_k^{″}(x)\mathrm{d}x$$

由于

$$P_k^{″}(x)=\frac{1}{2^{k}k!}\sum _{m=0}^{[\frac{k}{2}]-1}\frac{(-1{)}^{m}k!}{m!(k-m)!}\frac{(2k-2m)!}{(k-2m-2)!}{x}^{k-2m-2}$$

利用Legendre多项式和低次项的正交性

$${\int }_{-1}^1{x}^k{P}_l(x)\mathrm{d}x=0(k<l)$$

所以后一项为 $0$，得到

$$\begin{array}{c}{\int }_{-1}^1\frac{d{P}_k(x)}{dx}\frac{d{P}_l(x)}{dx}\mathrm{d}x={P_l(x)\frac{d{P}_k(x)}{dx}|}_{-1}^1=P_l(1)P_k^{\prime }(1)-P_l(-1)P_k^{\prime }(-1)\\ =\frac{k(k+1)}{2}(1+(-1{)}^{l+k})\end{array}$$

5.计算下列积分

(a)

$${\int }_{-1}^1(1+x{)}^k{\mathrm{P}}_l(x)\mathrm{d}x$$

• 如果 $k<l$，则由于 $P_l(x)$ 与所有次数小于 $l$ 的多项式正交，而 $(1+x{)}^k$ 是次数为 $k$ 的多项式，积分值为 $0$。
• 如果 $k\ge l$，则经过非常复杂的分部积分运算，可以得到（大体过程就是先把 Legendre 多项式中的 $l$ 阶导全部转移到另一边，然后再对得到的 $(x+1{)}^k(x-1{)}^l$ 项利用 Beta 积分得出结果）

$${\int }_{-1}^1(1+x{)}^k{\mathrm{P}}_l(x)\mathrm{d}x=\frac{2^{k+1}(k!{)}^2}{(k+l+1)!(k-l)!}$$

(b)

$${\int }_0^1{\mathrm{P}}_k(x){\mathrm{P}}_l(x)\mathrm{d}x$$

• 如果 $k$ 和 $l$ 同奇偶性，则 ${\int }_0^1{P}_k{P}_{l}dx=\frac{1}{2}{\int }_{-1}^1{P}_k{P}_{l}dx=\frac{1}{2}\cdot \frac{2}{2l+1}{\delta }_{kl}=\frac{1}{2l+1}{\delta }_{kl}$，即当 $k\ne l$ 时积分值为 $0$
• 如果 $k$ 和 $l$ 奇偶性不同，则利用 Legendre 方程的结论计算 (不失一般性，假设 $k=2k^{\prime },l=2l^{\prime }+1$)

$$\begin{array}{c}{\int }_0^1{\mathrm{P}}_k(x){\mathrm{P}}_l(x)\mathrm{d}x=\frac{1}{l(l+1)-k(k+1)}{(1-x^2)(P_k^{\prime }(x)P_l(x)-P_l^{\prime }(x)P_k(x))|}_0^1\\ =\frac{P_l^{\prime }(0)P_k(0)-P_k^{\prime }(0)P_l(0)}{l(l+1)-k(k+1)}=\frac{1}{(2l^{\prime }+1)(2l^{\prime }+2)-2k^{\prime }(2k^{\prime }+1)}\left(\frac{(-1{)}^{l^{\prime }}(2l^{\prime }+1)!}{2^{2l^{\prime }}(l^{\prime }!{)}^2}\frac{(-1{)}^{k^{\prime }}(2k^{\prime })!}{2^{2k^{\prime }}(k^{\prime }!{)}^2}\right)\\ =\frac{(-1{)}^{l^{\prime }+k^{\prime }}}{(2l^{\prime }+1)(2l^{\prime }+2)-2k^{\prime }(2k^{\prime }+1)}\frac{(2l^{\prime }+1)!(2k^{\prime })!}{2^{2(l^{\prime }+k^{\prime })}(l^{\prime }!{)}^2(k^{\prime }!{)}^2}\end{array}$$

(c)

$${\int }_{-1}^1{\mathrm{P}}_l(x)\ln (1-x)\mathrm{d}x$$

当 $l=0$ 时，

$${\int }_{-1}^1{\mathrm{P}}_0(x)\ln (1-x)\mathrm{d}x=2\ln 2-2$$

当$l\ne 0$ 时，直接利用和 (a) 类似的思路，分部积分得到

$$\begin{array}{c}{\int }_{-1}^1{\mathrm{P}}_l(x)\ln (1-x)\mathrm{d}x\\ =-\frac{(l-1)!}{2^{l}l!}{\int }_{-1}^1(1+x{)}^l\mathrm{d}x=-\frac{2}{l(l+1)}\end{array}$$

(d)

$${\int }_{-1}^1{\mathrm{P}}_l(x)(1-x{)}^{\alpha }\mathrm{d}x,-1<\alpha <0$$

先进行分部积分，得到

$${\int }_{-1}^1{\mathrm{P}}_l(x)(1-x{)}^{\alpha }\mathrm{d}x=\frac{(-1{)}^l}{2^{l}l!}\frac{\mathrm{\Gamma }(\alpha +1)}{\mathrm{\Gamma }(\alpha -l+1)}{\int }_{-1}^1(1-x{)}^{\alpha }(1+x{)}^l\mathrm{d}x$$

这正好是 Beta 积分的形式，得到

$${\int }_{-1}^1{P}_l(x)(1-x{)}^{\alpha }dx=(-1{)}^l\frac{2^{\alpha +1}\mathrm{\Gamma }(\alpha +1)\mathrm{\Gamma }(\alpha +1)}{\mathrm{\Gamma }(\alpha +l+2)\mathrm{\Gamma }(\alpha -l+1)}$$

(e)

$${\int }_{-1}^1{x}^2{\mathrm{P}}_l(x){\mathrm{P}}_{l+2}(x)\mathrm{d}x$$

利用递推关系 $x{P}_l(x)=\frac{l+1}{2l+1}{P}_{l+1}(x)+\frac{l}{2l+1}{P}_{l-1}(x)$ 及正交性。计算得

$${\int }_{-1}^1{x}^2{P}_l{P}_{l+2}dx=\frac{2(l+1)(l+2)}{(2l+1)(2l+3)(2l+5)}$$

(f)

$${\int }_{-1}^1[x{\mathrm{P}}_l(x){]}^2\mathrm{d}x$$

利用 $x{P}_l(x)=\frac{l+1}{2l+1}{P}_{l+1}(x)+\frac{l}{2l+1}{P}_{l-1}(x)$，及正交性

$${\int }_{-1}^1[x{P}_l(x){]}^{2}dx={\left(\frac{l+1}{2l+1}\right)}^2\frac{2}{2l+3}+{\left(\frac{l}{2l+1}\right)}^2\frac{2}{2l-1}$$

简化得

$${\int }_{-1}^1[x{P}_l(x){]}^{2}dx=\frac{2(2l^2+2l-1)}{(2l-1)(2l+3)(2l+1)}$$

6.把下列函数按照勒让德多项式展开

(a) $f(x)=x^2$

利用 $P_2(x)=\frac{3x^2-1}{2}$，解得 $x^2=\frac{2}{3}{P}_2(x)+\frac{1}{3}{P}_0(x)$，所以：

$$x^2=\frac{1}{3}{P}_0(x)+\frac{2}{3}{P}_2(x)$$

(b) $f(x)=|x|$

$|x|$ 为偶函数，展开仅含偶次项

$$|x|=\sum _{n=0}^{\mathrm{\infty }}{a}_{2n}{P}_{2n}(x),a_{2n}=\frac{4n+1}{2}{\int }_{-1}^1|x|P_{2n}(x)dx$$$$\begin{array}{c}\frac{2}{4n+1}{a}_{2n}=-{\int }_{-1}^0{P}_1(x)P_{2n}(x)\mathrm{d}x+{\int }_0^1{P}_1(x)P_{2n}(x)\mathrm{d}x\\ =2{\int }_0^1{P}_1(x)P_{2n}(x)\mathrm{d}x\end{array}$$

直接带入上一题中的结果

$${\int }_0^1{P}_1(x)P_{2n}(x)\mathrm{d}x=\frac{(-1{)}^{n+1}(2n)!}{2^{2n+1}n!(n+1)!(2n-1)}$$

得到

$$a_{2n}=\frac{(-1{)}^{n+1}(2n)!}{2^{2n+1}n!(n+1)!}\frac{4n+1}{2n-1}$$

(d) $f(x)=\sqrt{1-2xt+t^2}$

假设 $f(x)=\sum _n{a}_n{P}_n(x)$

利用生成函数的表达式

$$\begin{array}{c}\sqrt{1-2xt+t^2}=(1-2xt+t^2)\sum _{l=0}^{\mathrm{\infty }}{P}_l(x)t^l\\ =\sum _l(1+t^2)t^l{P}_l(x)-\sum _{l}2t^{l+1}(\frac{l+1}{2l+1}{P}_{l+1}(x)+\frac{l}{2l+1}{P}_{l-1}(x))\end{array}$$

整理得到

$$a_0=(1+t^2)-\frac{2}{3}{t}^2=1+\frac{1}{3}{t}^2$$$$\begin{array}{c}{a}_n=(1+t^2)t^n-2t^n\frac{n}{2n-1}-2t^{n+2}\frac{n+1}{2n+3}\\ =-\frac{1}{2n-1}{t}^n+\frac{1}{2n+3}{t}^{n+2}\end{array}$$

8.利用勒让德多项式的生成函数证明

(a)

$${\mathrm{P}}_l(-\frac{1}{2})=\sum _{k=0}^{2l}{\mathrm{P}}_k(-\frac{1}{2}){\mathrm{P}}_{2l-k}\left(\frac{1}{2}\right)$$

证明： 生成函数在 $x=-\frac{1}{2}$ 和 $x=\frac{1}{2}$ 处

$$\frac{1}{\sqrt{1+t+t^2}}=\sum _m{P}_m(-\frac{1}{2})t^m,\frac{1}{\sqrt{1-t+t^2}}=\sum _n{P}_n\left(\frac{1}{2}\right)t^n$$

乘积

$$\frac{1}{\sqrt{1+t+t^2}}\cdot \frac{1}{\sqrt{1-t+t^2}}=\frac{1}{\sqrt{1+t^2+t^4}}=\sum _l{P}_l(-\frac{1}{2})t^{2l}$$

右边乘积的展开

$$\left(\sum _m{P}_m(-\frac{1}{2})t^m\right)\left(\sum _n{P}_n\left(\frac{1}{2}\right)t^n\right)=\sum _s\left(\sum _{k=0}^s{P}_k(-\frac{1}{2})P_{s-k}\left(\frac{1}{2}\right)\right)t^s$$

比较 $t^{2l}$ 的系数，得到

$${\mathrm{P}}_l(-\frac{1}{2})=\sum _{k=0}^{2l}{\mathrm{P}}_k(-\frac{1}{2}){\mathrm{P}}_{2l-k}\left(\frac{1}{2}\right)$$

(b)

$${\mathrm{P}}_l(\cos 2\theta )=\sum _{k=0}^{2l}(-1{)}^k{\mathrm{P}}_k(\cos \theta ){\mathrm{P}}_{2l-k}(\cos \theta )$$

证明： 生成函数在 $x=\cos 2\theta$ 处

$$\frac{1}{\sqrt{1-2\cos 2\theta t+t^2}}=\sum _n{P}_n(\cos 2\theta )t^n$$

但 $\cos 2\theta =2{\cos }^2\theta -1$，所以

$$1-2\cos 2\theta t+t^2=1-2(2{\cos }^2\theta -1)t+t^2=1-4{\cos }^2\theta t+2t+t^2$$

考虑生成函数在 $x=\cos \theta$ 与 $t$ 和 $-t$ 的乘积

$$\begin{array}{c}\frac{1}{\sqrt{1-2\cos \theta t+t^2}}\cdot \frac{1}{\sqrt{1-2\cos \theta (-t)+(-t{)}^2}}\\ =\frac{1}{\sqrt{1-2\cos \theta t+t^2}}\cdot \frac{1}{\sqrt{1+2\cos \theta t+t^2}}=\frac{1}{\sqrt{1-2\cos 2\theta t^2+t^4}}\end{array}$$

右边

$$\frac{1}{\sqrt{1-2\cos 2\theta t^2+t^4}}=\sum _l{P}_l(\cos 2\theta )t^{2l}$$

左边乘积的展开

$$(\sum _m{P}_m(\cos \theta )t^m)(\sum _n{P}_n(\cos \theta )(-t{)}^n)=\sum _s(\sum _{k=0}^s{P}_k(\cos \theta )P_{s-k}(\cos \theta )(-1{)}^{s-k})t^s$$

对于 $s=2l$，系数为 $\sum _{k=0}^{2l}{P}_k(\cos \theta )P_{2l-k}(\cos \theta )(-1{)}^{2l-k}=\sum _{k=0}^{2l}(-1{)}^k{P}_k(\cos \theta )P_{2l-k}(\cos \theta )$，与左边 $t^{2l}$ 的系数 $P_l(\cos 2\theta )$ 相等，即

$${\mathrm{P}}_l(\cos 2\theta )=\sum _{k=0}^{2l}(-1{)}^k{\mathrm{P}}_k(\cos \theta ){\mathrm{P}}_{2l-k}(\cos \theta )$$