Section 6.11

Legendre Polynomials and Related Topics

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Problem 6.84

Pn(x)P_n(x) is, apart from a constant factor, equal to the nnth derivative of (x2−1)n(x^2-1)^n:

Pn(x)=12nn!dndxn(x2−1)n(Rodrigues’ formula). P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^n \qquad (\textit{Rodrigues' formula}).
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Problem 6.85

We have

(1−t)nPn(1+t1−t)=1+(n1)2t+(n2)2t2+(n3)2t3+⋯+(nn−1)2tn−1+tn. (1-t)^n P_n\left(\frac{1+t}{1-t}\right) = 1 + \binom {n}{1}^2 t + \binom {n}{2}^2 t^2 + \binom {n}{3}^2 t^3 + \cdots + \binom {n}{n-1}^2 t^{n-1} + t^n.
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Problem 6.86

Pn(x)P_n(x) may be represented in the following integral form:

Pn(x)=1π∫0π(x+x2−1cos⁡φ)n dφ(Laplace’s formula). P_n(x) = \frac{1}{\pi }\int _0^{\pi } (x + \sqrt{x^2-1}\cos \varphi )^n \, d\varphi \qquad (\textit{Laplace's formula}).
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Problem 6.87

Between three successive Legendre polynomials there holds the following recurrence relation:

Pn(x)=2n−1nxPn−1(x)−n−1nPn−2(x),n=2,3,4,…. P_n(x) = \frac{2n-1}{n}xP_{n-1}(x) - \frac{n-1}{n}P_{n-2}(x), \qquad n = 2, 3, 4, \ldots .
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Problem 6.88

There exists a unique polynomial Sn(x)S_n(x) of nnth degree with real coefficients such that the equation

∫−11Sn(x)K(x) dx=K(1) \int _{-1}^{1} S_n(x)K(x) \, dx = K(1)

is satisfied for all polynomials K(x)K(x) of nnth degree. Express Sn(x)S_n(x) and (1−x)Sn(x)(1-x)S_n(x) as a linear combination of Legendre polynomials.

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Remark.
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Express K(x)K(x) similarly.

Problem 6.89

(Continuation of problem 88.) The polynomials

S0(x),S1(x),S2(x),…,Sn(x),… S_0(x), \quad S_1(x), \quad S_2(x), \quad \ldots , \quad S_n(x), \ldots

satisfy the orthogonality condition

∫−11(1−x)Sm(x)Sn(x) dx={0,if m≠n,n+12,if m=n;m,n=0,1,2,…. \int _{-1}^{1} (1-x)S_m(x)S_n(x)\, dx = \begin{cases} 0, & \text{if } m \neq n, \\ \dfrac {n+1}{2}, & \text{if } m=n; \quad m,n=0,1,2,\ldots . \end{cases}
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Problem 6.90

Pn(x)P_n(x) satisfies a homogeneous linear differential equation of second order

(1−x2)Pn′′(x)−2xPn′(x)+n(n+1)Pn(x)=0. (1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1)P_n(x) = 0.
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Problem 6.91

For sufficiently small ww, we have identically in ww and xx

11−2xw+w2=P0(x)+P1(x)w+P2(x)w2+⋯+Pn(x)wn+⋯ . \frac{1}{\sqrt{1-2xw+w^2}} = P_0(x) + P_1(x)w + P_2(x)w^2 + \cdots + P_n(x)w^n + \cdots .

(Generating function for the Legendre polynomials.)

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Problem 6.92

Deduce problems 84, 86, 87, 90 directly from problem 91. (Problem 91 may also be regarded as offering a definition of the Legendre polynomials, and in fact a ``centrally located'' definition: from here there are particularly convenient paths leading to most of the properties of the Legendre polynomials.)

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Problem 6.93

Pn(cos⁡ϑ)P_n(\cos \vartheta ) is a cosine polynomial with non-negative coefficients. Determine these coefficients.

Deduce from this the following inequality:

∣Pn(x)∣≦1,−1≦x≦1. \left|P_n(x)\right| \leqq 1, \qquad -1 \leqq x \leqq 1.

For n>0n>0 the equality sign holds only if x=1x=1 or x=−1x=-1.

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Problem 6.94

Let x>1x>1. The sequence

P0(x),P1(x),P2(x),…,Pn(x),… P_0(x), \quad P_1(x), \quad P_2(x), \quad \ldots , \quad P_n(x), \ldots

increases monotonically.

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Problem 6.95

The sum of the first nn Legendre polynomials is non-negative in the interval −1≦x≦1-1 \leqq x \leqq 1:

P0(x)+P1(x)+P2(x)+⋯+Pn(x)≧0,−1≦x≦1. P_0(x)+P_1(x)+P_2(x)+\cdots +P_n(x) \geqq 0, \qquad -1 \leqq x \leqq 1.

The equality sign holds only if nn is odd and x=−1x=-1.

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Remark.
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See Problem 157 of Part III.

Problem 6.96

The sum of the first nn Legendre polynomials

P0(x)+P1(x)+P2(x)+⋯+Pn(x) P_0(x)+P_1(x)+P_2(x)+\cdots +P_n(x)

is positive for every value of xx if nn is even; it has its only change of sign at the point x=−1x=-1 if nn is odd.

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Remark.
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See problem 94.

Problem 6.97

The nnth Legendre polynomial Pn(x)P_n(x) has only real, simple zeros all of which lie in the interior of the interval −1,1-1, 1.

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Remark.
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See Problem 140 of Part II.

Problem 6.98

Generalize the theorems in problems 84--91, 97 for the Jacobi (hypergeometric) polynomials

P0(α,β)(x),P1(α,β)(x),P2(α,β)(x),…,Pn(α,β)(x),…,α,β>−1, P_0^{(\alpha ,\beta )}(x), \quad P_1^{(\alpha ,\beta )}(x), \quad P_2^{(\alpha ,\beta )}(x), \quad \ldots , \quad P_n^{(\alpha ,\beta )}(x), \ldots , \qquad \alpha , \beta > -1,

defined by the following conditions: (1), (2) Pn(α,β)(x)P_n^{(\alpha ,\beta )}(x) is of degree nn, has real coefficients and orthogonality property

∫−11(1−x)α(1+x)βPm(α,β)(x)Pn(α,β)(x) dx \int _{-1}^{1} (1-x)^\alpha (1+x)^\beta P_m^{(\alpha ,\beta )}(x)P_n^{(\alpha ,\beta )}(x)\, dx = \begin{cases} 0, & \text{if } m \neq n, \\ ]4pt] \dfrac {2^{\alpha +\beta +1}}{2n+\alpha +\beta +1}\dfrac {\Gamma (n+\alpha +1)\Gamma (n+\beta +1)}{\Gamma (n+1)\Gamma (n+\alpha +\beta +1)}, & \text{if } m=n; \quad m,n=0,1,2,\ldots \end{cases} \[ (for $n=0$ the above expression is to be taken as $2^{\alpha +\beta +1}\dfrac {\Gamma (\alpha +1)\Gamma (\beta +1)}{\Gamma (\alpha +\beta +2)}$); (3) the coefficient of $x^n$ in $P_n^{(\alpha ,\beta )}(x)$ is positive. Some cases already encountered, corresponding to special values of $\alpha $ and $\beta $, are collected in the following table: \] \begin{array}{lccccc} \alpha = & 0, & 1, & -\tfrac {1}{2}, & \tfrac {1}{2}, \\ \beta = & 0, & 0, & -\tfrac {1}{2}, & \tfrac {1}{2}, \\ \end{array} \[ \] P_n^{(\alpha ,\beta )}(x) = P_n(x), \quad \frac{2}{n+1}S_n(x), \quad \frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots 2n}T_n(x), \quad 2\frac{1\cdot 3\cdots (2n+1)}{2\cdot 4\cdots (2n+2)}U_n(x). \[ Cf.\ p.\ 85, problem 89, problem 4, problem 4. (The coefficient of $T_n(x)$ must be replaced by $1$ for $n=0$.) \]
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Problem 6.99

Prove the analogues of problems 84, 85, 87--91, 97 for the generalized Laguerre polynomials

L0(α)(x),L1(α)(x),L2(α)(x),…,Ln(α)(x),…,α>−1, L_0^{(\alpha )}(x), \quad L_1^{(\alpha )}(x), \quad L_2^{(\alpha )}(x), \quad \ldots , \quad L_n^{(\alpha )}(x), \ldots , \qquad \alpha > -1,

defined by the following conditions: (1), (2) Ln(α)(x)L_n^{(\alpha )}(x) is of degree nn, has real coefficients and orthogonality property

\int _0^{\infty } e^{-x}x^\alpha L_m^{(\alpha )}(x)L_n^{(\alpha )}(x)\, dx = \begin{cases} 0, & \text{for } m \neq n, \\ ]4pt] \Gamma (\alpha +1)\dbinom {n+\alpha }{n}, & \text{for } m=n; \quad m,n=0,1,2,\ldots ; \end{cases} \[ (3) the coefficient of $x^n$ in $L_n^{(\alpha )}(x)$ has the sign $(-1)^n$. \]
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Problem 6.100

Prove the analogues of problems 84, 87, 88, 90, 91, 97 for the Hermite polynomials

H0(x),H1(x),H2(x),…,Hn(x),…, H_0(x), \quad H_1(x), \quad H_2(x), \quad \ldots , \quad H_n(x), \ldots ,

defined by the following conditions: (1), (2) Hn(x)H_n(x) is of degree nn, has real coefficients and orthogonality property

\int _{-\infty }^{\infty } e^{-\frac{x^2}{2}}H_m(x)H_n(x)\, dx = \begin{cases} 0, & \text{for } m \neq n, \\ ]4pt] \sqrt{2\pi }\, n!, & \text{for } m=n; \quad m,n=0,1,2,\ldots ; \end{cases} \[ (3) the coefficient of $x^n$ in $H_n(x)$ has the sign $(-1)^n$. \]
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Problem 6.101

If Pn(α,β)(x)P_n^{(\alpha ,\beta )}(x) and Ln(α)(x)L_n^{(\alpha )}(x) are the functions defined in problems 98 and 99, respectively, then we have

lim⁡β→+∞Pn(α,β)(1−ε)=Ln(α)(x), \lim _{\beta \to +\infty } P_n^{(\alpha ,\beta )}(1-\varepsilon ) = L_n^{(\alpha )}(x),

if ε\varepsilon converges to 00 as β\beta increases in such a manner that

lim⁡β→+∞εβ=2x. \lim _{\beta \to +\infty } \varepsilon \beta = 2x.
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Problem 6.102

If Ln(α)(x)L_n^{(\alpha )}(x) and Hn(x)H_n(x) are the functions defined in problems 99 and 100, respectively, then we have

H2q(x)=(−1)q1⋅3⋅5⋯(2q−1)Lq(−1/2)(x22), H_{2q}(x) = \frac{(-1)^q}{1\cdot 3\cdot 5\cdots (2q-1)}L_q^{(-1/2)}\left(\frac{x^2}{2}\right), H2q+1(x)=(−1)q+11⋅3⋅5⋯(2q+1)xLq(1/2)(x22),q=0,1,2,…. H_{2q+1}(x) = \frac{(-1)^{q+1}}{1\cdot 3\cdot 5\cdots (2q+1)}xL_q^{(1/2)}\left(\frac{x^2}{2}\right), \qquad q=0,1,2,\ldots .
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