User:Egm6322.s12.team2.Xia/RP9.8

=R* 9.8 -Verify the second homogeneous solutions of Legendre equation =

Given
Part 1 Recurrence relation RR2: $$\displaystyle (n+1)P_{n+1}-(2n+1) \mu P_n+nP_{n-1}=0 $$ (9.8.1) with $$\displaystyle P_0(\mu)=1 $$ and $$ \displaystyle P_1(\mu)= \mu$$ Part 2 Define: $$ \displaystyle A(\mu ,\rho):=1-2\mu\rho+\rho^2=:1-x$$ $$\displaystyle -x:=-2\mu\rho+\rho^2$$ and the binomial theorem: $$\displaystyle (1-x)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\alpha_{i}x^{i}$$ (9.8.3) $$\displaystyle \alpha_{i}=\frac{1\cdot3\cdots(2i-1)}{2\cdot4\cdots(2i)}$$ (9.8.4)

Find
Finding $$ \displaystyle \{P_{2},\cdots,P_{6}\}$$ in terms of condition 1, and 2 respectively. and compare the result

Solution
Part 1 From 9.8.1, we have $$\displaystyle P_{n+1}=\frac{2n+1}{n+1}\mu P_{n}-\frac{n}{n+1}P_{n-1}$$ (9.8.5) with $$\displaystyle P_0(\mu)=1 $$ and $$ \displaystyle P_1(\mu)= \mu$$ then $$\displaystyle \begin{array}{l} \\ P_{2}=\frac{3}{2}\mu P_{1}-\frac{1}{2}P_{0}=\frac{1}{2}(3\mu^{2}-1)\\ \\ P_{3}=\frac{5}{3}\mu P_{2}-\frac{2}{3}P_{1}=\frac{1}{2}(5\mu^{3}-3\mu)\\ \\ P_{4}=\frac{7}{4}\mu P_{3}-\frac{3}{4}P_{2}=\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\\ \\ P_{5}=\frac{9}{5}\mu P_{4}-\frac{4}{5}P_{3}=\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\\ \\ P_{6}=\frac{11}{6}\mu P_{5}-\frac{5}{6}P_{4}=\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5) \end{array}$$

Part 2 Since $$\displaystyle x:=2\mu\rho-\rho^{2}=\rho(2\mu-\rho)$$ From the 9.8.3 and 9.8.4 we have $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{\infty}\alpha_{i}x^{i}=\sum_{i=0}^{\infty}\alpha_{i}\rho^{i}(2\mu-\rho)^{i}$$ when computing for $$\displaystyle P_{6}$$, we can see that the highest order is $$\displaystyle i=6$$，thus $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{6}\alpha_{6}x^{6}=\sum_{i=0}^{6}\alpha_{6}\rho^{6}(2\mu-\rho)^{6}$$ $$\displaystyle =\frac{231{\left(2\mu\rho-{\rho}^{2}\right)}^{6}}{1024}+\frac{63{\left(2\mu\rho-{\rho}^{2}\right)}^{5}}{256}+\frac{35{\left(2\mu\rho-{\rho}^{2}\right)}^{4}}{128}+\frac{5{\left(2\mu\rho-{\rho}^{2}\right)}^{3}}{16}+\frac{3{\left(2\mu\rho-{\rho}^{2}\right)}^{2}}{8}+\frac{2\mu\rho-{\rho}^{2}}{2}+1$$ $$\displaystyle =\frac{231\,{\rho}^{12}}{1024}-\frac{693\,\mu\,{\rho}^{11}}{256}+\frac{3465\,{\mu}^{2}\,{\rho}^{10}}{256}-\frac{63\,{\rho}^{10}}{256}-\frac{1155\,{\mu}^{3}\,{\rho}^{9}}{32}+\frac{315\,\mu\,{\rho}^{9}}{128}$$ $$\displaystyle +\frac{3465\,{\mu}^{4}\,{\rho}^{8}}{64}-\frac{315\,{\mu}^{2}\,{\rho}^{8}}{32}+\frac{35\,{\rho}^{8}}{128}-\frac{693\,{\mu}^{5}\,{\rho}^{7}}{16}+\frac{315\,{\mu}^{3}\,{\rho}^{7}}{16}-\frac{35\,\mu\,{\rho}^{7}}{16}$$ $$\displaystyle +\frac{231\,{\mu}^{6}\,{\rho}^{6}}{16}-\frac{315\,{\mu}^{4}\,{\rho}^{6}}{16}+\frac{105\,{\mu}^{2}\,{\rho}^{6}}{16}-\frac{5\,{\rho}^{6}}{16}+\frac{63\,{\mu}^{5}\,{\rho}^{5}}{8}-\frac{35\,{\mu}^{3}\,{\rho}^{5}}{4}+\frac{15\,\mu\,{\rho}^{5}}{8}$$ $$\displaystyle +\frac{35\,{\mu}^{4}\,{\rho}^{4}}{8}-\frac{15\,{\mu}^{2}\,{\rho}^{4}}{4}+\frac{3\,{\rho}^{4}}{8}+\frac{5\,{\mu}^{3}\,{\rho}^{3}}{2}-\frac{3\,\mu\,{\rho}^{3}}{2}+\frac{3\,{\mu}^{2}\,{\rho}^{2}}{2}-\frac{{\rho}^{2}}{2}+\mu\,\rho+1$$ Since the highest order we consider is 6，then $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=1+\mu\rho+\frac{1}{2}(3\mu^{2}-1)\rho^{2}+\frac{1}{2}(5\mu^{3}-3\mu)\rho^{3}+\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\rho^{4}$$ $$\displaystyle +\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\rho^{5}+\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5)\rho^{6}$$ Therefore $$ \displaystyle \begin{align} P_0 &= 1\\ P_1 &= \mu \\ P_2 &= \frac{1}{2}(3\mu^2-1) \\ P_3 &= \frac{1}{2}(5\mu^3-3\mu) \\ P_4 &= \frac{1}{8}(35\mu^4-30\mu^2+3) \\ P_5 &= \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \\ P_6 &= \frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5) \end{align} $$ So, using two different methods, we obtained the same result.