• is an odd function when is odd. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \nonumber\]\[F(x,t) = \exp\left(\frac{tx-t}{2t}\right) = \sum_{n=0}^{\infty} J_{n}(x)t^{n}.\nonumber\]\[F(x,t) =\frac{1}{\sqrt{1-2xt+t^{2}}} = \sum_{n=0}^{\infty} P_{n}(x) t^{n}.\nonumber\]\[(x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}=0,\]\[\begin{aligned} \frac{d}{dt} \frac{1}{\sqrt{1-2tx +t^2}} &= \frac{d}{dt} \sum_{n=0}^{\infty} t^n P_n(x)\nonumber\\ \frac{x-t}{(1-2tx+t^{2})^1.5} &= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \frac{x-t}{1-2xt +t^{2}} \sum_{n=0}^{\infty} t^n P_n(x)&= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \sum_{n=0}^{\infty} t^n x P_n(x)- \sum_{n=0}^{\infty} t^{n+1} P_n(x) &= \sum_{n=0}^{\infty} nt^{n-1} P_n(x) - 2\sum_{n=0}^{\infty} nt^n xP_n(x) +\sum_{n=0}^{\infty} nt^{n+1} P_n(x)\nonumber\\ \sum_{n=0}^{\infty} t^n(2n+1)x P_n(x) &= \sum_{n=0}^{\infty} (n+1)t^n P_{n+1}(x) + \sum_{n=0}^{\infty} n t^{n} P_{n-1}(x)\end{aligned}\]\[(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x).\] The parity and normalization implicate the values at the boundaries An explicit expression for the shifted Legendre polynomials is given by (C.2) We write the solution for a particular value of nas Pn(x). The normalization follows from the evaluation of the highest coefficient,and thus we need to multiply the derivative with \(\frac{1}{2^n n! This relation, along with the first two polynomials The generating function approach is directly connected to the The differential equation admits another, non-polynomial solution, the In physical settings, Legendre's differential equation arises naturally whenever one solves That the polynomials are complete means the following. An especially compact expression for the Legendre polynomials is given by This formula enables derivation of a large number of properties of the where the last, which is also immediate from the recursion formula, expresses the Legendre polynomials by simple monomials and involves the The Legendre polynomials were first introduced in 1782 by Legendre polynomials are also useful in expanding functions of the form (this is the same as before, written a little differently): The Legendre polynomials can also be defined as the coefficients in a formal expansion in powers of Expansion to higher orders gets increasingly cumbersome, but is possible to do systematically, and again leads to one of the explicit forms given below. }\) to get the properly normalized \(P_n\).Let’s use the generating function to prove some of the other properties: 2.: Equating terms with identical powers of \(t\) we find Proofs for the other properties can be found using similar methods.\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)[ "article:topic", "generating function", "authorname:nwalet", "license:ccbyncsa", "showtoc:no", "Legendre polynomials", "Rodrigues\u2019 Formula" ][ "article:topic", "generating function", "authorname:nwalet", "license:ccbyncsa", "showtoc:no", "Legendre polynomials", "Rodrigues\u2019 Formula" ]\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\[\frac{1}{1-xt} = \sum_{n=0}^{\infty} x^{n}t^{n}\;\;(|xt|<1).
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, ∑ n c n (x) t n. The function F is then called a generating function of the functions c n. Example 11.1: or, with the alternative expression, which also holds at the endpoints 11.2.1 Generating function. They can be defined in many ways, and the various definitions highlight different aspects as well as suggest generalizations and connections to different mathematical structures and physical and numerical applic… \frac{d^{n}}{dx^{n}} (x^2-1)^n. This orthogonal series expansion is also known as a Fourier-Legendre Series expansion or a Generalized Fourier Series expansion.. Even/Odd Functions: Whether a Legendre Polynomial is an even or odd function depends on its degree .. Based on , • is an even function when is even. We want to hear from you.Let \(F(x,t)\) be a function of the two variables \(x\) and \(t\) that can be expressed as a Taylor’s series in \(t\), \(\sum_{n} c_n(x) t^{n}\). 11.2 Properties of Legendre polynomials. Al-though it’s not all that convenient for calculating the polynomials them-selves, it can be used to prove various properties about them.
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.