Legendre polynomials. Use the Gram–Schmidt procedure (Problem A.4) to orthonormalize the functions 1 , x , x^2 , and x^3 on the interval -1 ≤ x ≤ 1 . You may recognize the results—they are (apart from normalization)39 Legendre polynomials (Problem 2.64 and Table 4.1).

Question

Legendre polynomials. Use the Gram–Schmidt procedure (Problem A.4) to orthonormalize the functions 1 , x , x² , and x³ on the interval -1 ≤ x ≤ 1 . You may recognize the results—they are (apart from normalization)39 Legendre polynomials (Problem 2.64 and Table 4.1).

Table 4.1: The first few Legendre polynomials,[latex] P_{\ell}(x) [/latex] : (a) functional form, (b) graph
	[latex] P_{0}=1 [/latex]
	[latex] P_{1}=x [/latex]
	[latex] P_{2}=\frac{1}{2}\left(3 x^{2}-1\right) [/latex]
	[latex] P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right) [/latex]
	[latex] P_{4}=\frac{1}{8}\left(35 x^{4}-30 x^{2}+3\right) [/latex]
	[latex] P_{5}=\frac{1}{8}\left(63 x^{5}-70 x^{3}+15 x\right) [/latex]
(b)	(a)

Accepted Answer

[latex] \left|e_{1}\right\rangle=1 ;\left\langle e_{1} \mid e_{1}\right\rangle=\int_{-1}^{1} 1 d x=2 [/latex]. So [latex] \left|e_{1}^{\prime}\right\rangle=\frac{1}{\sqrt{2}} [/latex].

[latex] \left|e_{2}\right\rangle=x ;\left\langle e_{1}^{\prime} \mid e_{2}\right\rangle=\frac{1}{\sqrt{2}} \int_{-1}^{1} x d x=0 ;\left\langle e_{2} \mid e_{2}\right\rangle=\int_{-1}^{1} x^{2} d x=\left.\frac{x^{3}}{3}\right|_{-1} ^{1}=\frac{2}{3} [/latex]. So [latex] \left|e_{2}^{\prime}\right\rangle=\sqrt{\frac{3}{2}} x [/latex].

[latex] \left|e_{3}\right\rangle=x^{2} ;\left\langle e_{1}^{\prime} \mid e_{3}\right\rangle=\frac{1}{\sqrt{2}} \int_{-1}^{1} x^{2} d x=\frac{1}{\sqrt{2}} \frac{2}{3} ;\left\langle e_{2}^{\prime} \mid e_{3}\right\rangle=\sqrt{\frac{2}{3}} \int_{-1}^{1} x^{3} d x=0 [/latex].

So (Problem A.4): [latex] \left|e_{3}^{\prime \prime}\right\rangle=\left|e_{3}\right\rangle-\frac{1}{\sqrt{2}} \frac{2}{3}\left|e_{1}^{\prime}\right\rangle=x^{2}-\frac{1}{3} [/latex].

[latex] \left\langle e_{3}^{\prime \prime} \mid e_{3}^{\prime \prime}\right\rangle=\int_{-1}^{1}\left(x^{2}-\frac{1}{3}\right)^{2} d x=\left.\left(\frac{x^{5}}{5}-\frac{2}{3} \cdot \frac{x^{3}}{3}+\frac{x}{9}\right)\right|_{-1} ^{1}=\frac{2}{5}-\frac{4}{9}+\frac{2}{9}=\frac{8}{45} [/latex]. So

[latex] \left|e_{3}^{\prime}\right\rangle=\sqrt{\frac{45}{8}}\left(x^{2}-\frac{1}{3}\right)=\mid \sqrt{\frac{5}{2}}\left(\frac{3}{2} x^{2}-\frac{1}{2}\right) [/latex].

[latex] \left|e_{4}\right\rangle=x^{3} . \quad\left\langle e_{1}^{\prime} \mid e_{4}\right\rangle=\frac{1}{\sqrt{2}} \int_{-1}^{1} x^{3} d x=0 ; \quad\left\langle e_{2}^{\prime} \mid e_{4}\right\rangle=\sqrt{\frac{3}{2}} \int_{-1}^{1} x^{4} d x=\sqrt{\frac{3}{2}} \cdot \frac{2}{5} [/latex];

[latex] \left\langle e_{3}^{\prime} \mid e_{4}\right\rangle=\sqrt{\frac{5}{2}} \int_{-1}^{1}\left(\frac{3}{2} x^{5}-\frac{1}{2} x^{3}\right) d x=0 . \quad\left|e_{4}^{\prime \prime}\right\rangle=\left|e_{4}\right\rangle-\left\langle e_{2}^{\prime} \mid e_{4}\right\rangle\left|e_{2}^{\prime}\right\rangle=x^{3}-\sqrt{\frac{3}{2}} \frac{2}{5} \sqrt{\frac{3}{2}} x=x^{3}-\frac{3}{5} x [/latex];

[latex] \left\langle e_{4}^{\prime \prime} \mid e_{4}^{\prime \prime}\right\rangle=\int_{-1}^{1}\left(x^{3}-\frac{3}{5} x\right)^{2} d x=\left.\left[\frac{x^{7}}{7}-\frac{2 \cdot 3}{5} \frac{x^{5}}{5}+\frac{9}{25} \frac{x^{3}}{3}\right]\right|_{-1} ^{1}=\frac{2}{7}-\frac{12}{25}+\frac{18}{75}=\frac{8}{7 \cdot 25} [/latex];

[latex] \left|e_{4}^{\prime}\right\rangle=\frac{5}{2} \sqrt{\frac{7}{2}}\left(x^{3}-\frac{3}{5} x\right)=\sqrt{\frac{7}{2}}\left(\frac{5}{2} x^{3}-\frac{3}{2} x\right) [/latex].

Table 4.1: The first few Legendre polynomials, $P_{\ell}(x)$ : (a) functional form, (b) graph
	$P_{0}=1$
	$P_{1}=x$
	$P_{2}=\frac{1}{2}\left(3 x^{2}-1\right)$
	$P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right)$
	$P_{4}=\frac{1}{8}\left(35 x^{4}-30 x^{2}+3\right)$
	$P_{5}=\frac{1}{8}\left(63 x^{5}-70 x^{3}+15 x\right)$
(b)	(a)

Question 3.31: Legendre polynomials. Use the Gram–Schmidt procedure (Proble...

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