Question

Use the recurrence relation (35)and P₀(x₁)= 1,P₁(X)=xto generate the next six Legendre polynomials.

Short answer

The next six Legendre polynomials are,

$$\begin{gathered} P_2\left(x\right)=\frac{3}{2}{x}^2-\frac{1}{2} \\ {P}_3\left(x\right)=\frac{5}{2}{x}^3-\frac{3}{2} \\ {P}_4\left(x\right)=\frac{35}{8}{x}^4-\frac{15}{4}{x}^2+\frac{3}{8} \\ {P}_5\left(x\right)=\frac{63}{8}{x}^5-\frac{35}{4}{x}^3+\frac{15}{8} \\ {P}_6\left(x\right)=\frac{231}{16}{x}^6-\frac{315}{16}{x}^4+\frac{105}{16}{x}^2-\frac{5}{16} \\ {P}_7\left(x\right)=\frac{429}{16}{x}^7-\frac{693}{16}{x}^5+\frac{315}{16}{x}^3-\frac{35}{16} \end{gathered}$$

Step-by-step solution

Step 1:DefineRecurrence relation.

Recurrence relations that relate Legendre polynomialso f different degrees are also important in some aspects of their applications.

_{$(\mathrm{k}+1){\mathrm{P}}_{\mathrm{k}+1}\left(\mathrm{x}\right)-(2\mathrm{k}+1){\mathrm{xP}}_{\mathrm{k}}\left(\mathrm{x}\right)+{\mathrm{kP}}_{\mathrm{k}-1}\left(\mathrm{x}\right)=0..............\left(1\right)$}

Derive the next six Legendre polynomials.

Let the values are$P_0\left(x\right)=1$and$P_1\left(x\right)=x$.

For$k=1$, then (1) becomes,

$\begin{array}{c}2{\mathrm{P}}_2\left(\mathrm{x}\right)=3{\mathrm{xP}}_1\left(\mathrm{x}\right)-{\mathrm{P}}_0\left(\mathrm{x}\right)\\ =3\mathrm{x}\times \mathrm{x}-1\times 1\\ =3{\mathrm{x}}^2-1\\ {\mathrm{P}}_2\left(\mathrm{x}\right)=\frac{3}{2}{\mathrm{x}}^2-\frac{1}{2}\end{array}$

For $k=2$, then (1) becomes,

$\begin{array}{c}3{\mathrm{P}}_3\left(\mathrm{x}\right)=5{\mathrm{xP}}_2\left(\mathrm{x}\right)-2{\mathrm{P}}_1\left(\mathrm{x}\right)\\ =5\mathrm{x}\left[\frac{3}{2}{\mathrm{x}}^2-\frac{1}{2}\right]-2\mathrm{x}\\ =\frac{15}{2}{\mathrm{x}}^3-\frac{5}{2}\mathrm{x}-2\mathrm{x}\\ =\frac{15}{2}{\mathrm{x}}^3-\frac{9}{2}\mathrm{x}\\ {\mathrm{P}}_3\left(\mathrm{x}\right)=\frac{5}{2}{\mathrm{x}}^3-\frac{3}{2}\mathrm{x}\end{array}$

For $k=3$, then (1) becomes,

$\begin{array}{c}4P_4\left(x\right)=7x{P}_3\left(x\right)-3P_2\left(x\right)\\ P_4\left(x\right)=\frac{7}{4}x\left[\frac{5}{2}{x}^3-\frac{3}{2}x\right]-\frac{3}{4}\left[\frac{3}{2}{x}^2-\frac{1}{2}\right]\\ =\frac{35}{8}{x}^4-\frac{21}{8}{x}^2-\frac{9}{8}{x}^2+\frac{3}{8}\\ =\frac{35}{8}{x}^4-\frac{15}{4}{x}^2+\frac{3}{8}\end{array}$

For$k=4$, then (1) becomes,

$\begin{array}{c}5{\mathrm{P}}_5\left(\mathrm{x}\right)=9{\mathrm{xP}}_4\left(\mathrm{x}\right)-4{\mathrm{P}}_3\left(\mathrm{x}\right)\\ =\frac{63}{8}{\mathrm{x}}^5-\frac{27}{4}{\mathrm{x}}^3+\frac{27}{40}\mathrm{x}-2{\mathrm{x}}^3+\frac{6}{5}\mathrm{x}\\ {\mathrm{P}}_5\left(\mathrm{x}\right)=\frac{63}{5}{\mathrm{x}}^5-\frac{35}{4}{\mathrm{x}}^3+\frac{15}{8}\mathrm{x}\end{array}$

For $k=5$, then (1) becomes,

$\begin{array}{c}6P_6\left(x\right)=11x{P}_5\left(x\right)-5P_4\left(x\right)\\ P_6\left(x\right)=\frac{11}{6}x\left[\frac{63}{8}{x}^5-\frac{35}{4}{x}^3+\frac{15}{8}x\right]-\frac{5}{6}\left[\frac{35}{8}{x}^4-\frac{15}{4}{x}^2+\frac{3}{8}\right]\\ =\frac{231}{16}{x}^6-\frac{385}{24}{x}^4+\frac{55}{16}{x}^2-\frac{175}{48}{x}^4+\frac{25}{8}{x}^2-\frac{15}{48}\\ =\frac{231}{16}{x}^6-\frac{315}{16}{x}^4+\frac{105}{16}{x}^2-\frac{5}{16}\end{array}$

For $k=6$, then (1) becomes,

$\begin{array}{c}7P_7\left(x\right)=13x{P}_6\left(x\right)-6P_5\left(x\right)\\ P_7\left(x\right)=\frac{13}{7}x\left[\frac{231}{16}{x}^6-\frac{315}{16}{x}^4+\frac{105}{16}{x}^2-\frac{5}{16}\right]-\frac{6}{7}\left[\frac{63}{8}{x}^5-\frac{35}{4}{x}^3+\frac{15}{8}x\right]\\ =\frac{429}{16}{x}^7-\frac{585}{16}{x}^5+\frac{195}{16}{x}^3-\frac{65}{112}x-\frac{27}{4}{x}^5+\frac{15}{2}{x}^3-\frac{45}{28}x\\ =\frac{429}{16}{x}^7-\frac{693}{16}{x}^5+\frac{315}{16}{x}^3-\frac{35}{16}x\end{array}$