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[ "article:topic", "generating function", "authorname:nwalet", "license:ccbyncsa", "showtoc:no", "Legendre polynomials", "Rodrigues\u2019 Formula" ]
Mathematics LibreTexts

11.2: Properties of Legendre Polynomials

  • Page ID
    8323
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    Let \(F(x,t)\) be a function of the two variables \(x\) and \(t\) that can be expressed as a Taylor’s series in \(t\), \(\sum_{n} c_n(x) t^{n}\). The function \(F\) is then called a generating function of the functions \(c_{n}\).

    Exercise \(\PageIndex{1}\)

    Show that \(F(x,t) = \frac{1}{1-xt}\) is a generating function of the polynomials \(x^{n}\).

    Answer

    Look at \[\frac{1}{1-xt} = \sum_{n=0}^{\infty} x^{n}t^{n}\;\;(|xt|<1). \nonumber\]

    Exercise \(\PageIndex{2}\)

    Show that \(F(x,t) = \exp\left(\frac{tx-t}{2t}\right)\) is the generating function for the Bessel functions,

    \[F(x,t) = \exp\left(\frac{tx-t}{2t}\right) = \sum_{n=0}^{\infty} J_{n}(x)t^{n}.\nonumber\]

    Answer

    TBA

    Exercise \(\PageIndex{4}\)

    (The case of most interest here) \[F(x,t) =\frac{1}{\sqrt{1-2xt+t^{2}}} = \sum_{n=0}^{\infty} P_{n}(x) t^{n}.\nonumber\]

    Answer

    TBA

    Rodrigues’ Generating Formula

    \[P_{n}(x) = \frac{1}{2^{n} n!} \frac{d^{n}}{dx^{n}} (x^2-1)^n. \label{Rodrigues}\]

    properties of Legendre Polynomials

    1. \(P_{n}(x)\) is even or odd if \(n\) is even or odd.
    2. \(P_{n}(1)=1\).
    3. \(P_{n}(-1)=(-1)^{n}\).
    4. \((2n+1) P_{n}(x) = P'_{n+1}(x)-P'_{n-1}(x)\).
    5. \((2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x)\).
    6. \(\int_{-1}^{x} P_n(x') dx'= \frac{1}{2n+1} \left[P_{n+1}(x)-P_{n-1}(x)\right]\).

    Let us prove some of these relations, first Rodrigues’ formula (Equation \ref{Rodrigues}). We start from the simple formula

    \[(x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}=0,\]

    which is easily proven by explicit differentiation. This is then differentiated \(n+1\) times,

    \[\begin{aligned} { \frac{d^{n+1}}{dx^{n+1}}\left[ (x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}\right]} \nonumber\\ &= n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2(n+1) x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &-2n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n - 2n x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n \nonumber\\ &=-n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2 x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &= -\left[\frac{d}{dx}(1-x^2)\frac{d}{dx}\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n \right\} +n(n+1)\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n\right\}\right] =0.\end{aligned}\]

    We have thus proven that \(\frac{d^n}{dx^n}(x^2-1)^n\) satisfies Legendre’s equation. The normalization follows from the evaluation of the highest coefficient,

    \[\frac{d^n}{dx^n} x^{2n} = \frac{2n!}{n!} x^n,\]

    and thus we need to multiply the derivative with \(\frac{1}{2^n n!}\) to get the properly normalized \(P_n\).

    Let’s use the generating function to prove some of the other properties: 2.: \[F(1,t) = \frac{1}{1-t} = \sum_n t^n\] has all coefficients one, so \(P_n(1)=1\). Similarly for 3.: \[F(-1,t) = \frac{1}{1+t} = \sum_n (-1)^nt^n .\] Property 5. can be found by differentiating the generating function with respect to \(t\):

    \[\begin{aligned} \frac{d}{dt} \frac{1}{\sqrt{1-2tx +t^2}} &= \frac{d}{dt} \sum_{n=0}^{\infty} t^n P_n(x)\nonumber\\ \frac{x-t}{(1-2tx+t^{2})^1.5} &= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \frac{x-t}{1-2xt +t^{2}} \sum_{n=0}^{\infty} t^n P_n(x)&= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \sum_{n=0}^{\infty} t^n x P_n(x)- \sum_{n=0}^{\infty} t^{n+1} P_n(x) &= \sum_{n=0}^{\infty} nt^{n-1} P_n(x) - 2\sum_{n=0}^{\infty} nt^n xP_n(x) +\sum_{n=0}^{\infty} nt^{n+1} P_n(x)\nonumber\\ \sum_{n=0}^{\infty} t^n(2n+1)x P_n(x) &= \sum_{n=0}^{\infty} (n+1)t^n P_{n+1}(x) + \sum_{n=0}^{\infty} n t^{n} P_{n-1}(x)\end{aligned}\]

    Equating terms with identical powers of \(t\) we find \[(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x).\]

    Proofs for the other properties can be found using similar methods.