🔎 Legendre Polynomial – Definition and Explanations

Introduction

Legendre polynomials

The Legendre polynomials are options of Legendre’s differential equation, and represent the best instance of a sequence of orthogonal polynomials (In arithmetic, a sequence of orthogonal polynomials is an infinite sequence of…).

Legendre’s equation

We name Legendre’s equation the equation: $frac{textrm{d}}{textrm{d}x}[(1-x^{2})frac{textrm{d}y}{textrm{d}x}]+n(n+1)y=0$

We thus outline the Legendre polynomial (The Legendre polynomials are options of the differential equation of…) $P not$ (to pay every little thing (The entire understood as the entire of what exists is usually interpreted because the world or…) entire pure (In arithmetic, a pure quantity is a optimistic (or zero) quantity that principally permits…) not):

$frac{textrm{d}}{textrm{d}x}[(1-x^{2})frac{textrm{d}P_n(x)}{textrm{d}x}]+n(n+1)P_n(x)=0,qquad P_n(1)=1.$

So we have now $P_n=P_n^{(0,0)}$ the place $P_n^{(alpha,beta)}$ means the polynomial (A polynomial, in arithmetic, is the linear mixture of the merchandise of…) by Jacobi of indexes not related to the parameters α and β.

Some polynomials

The primary polynomials are:

The primary 20 Legendre polynomials (Legendre polynomials are options y of Legendre’s differential equation:).

Different definitions

Bonnet recurrence method

$P_0(x)=1, P_1(x)=x,$ and for all not>0

$(n+1)P_{n+1}(x)=(2n+1)xP_n(x) - nP_{n-1}(x).,$

Components of Rodrigues (Rodrigues is the smallest of the three islands within the Mascarene archipelago.)

We outline the polynomial $P not$ (for all pure integers not) by :

$P_n(x)=frac{1}{n! 2^n}frac{textrm{d}^n}{textrm{d}x^n}left((x^2-1)^nright)$

Definition (A definition is a discourse that claims what a factor is or what a reputation means. Therefore the…) analytic

We will additionally outline this sequence of polynomials by its producing operate:

$frac{1}{sqrt{1-2xz+z^2}} = sum_{n=0}^infty P_n(x) z^n.$

The theorem (A theorem is a proposition that may be mathematically demonstrated, that’s, a…) residues then provides:

$P_{n}(x)=frac{1}{2pi i}point(1-2xz+z^2)^{-1/2}z^{-n-1}textrm{d} z$

the place the define (COmet Nucleus TOUR (CONTOUR) is a NASA house probe that’s a part of the Program…) surrounds the origin and is caught within the that means (SENS (Methods for Engineered Negligible Senescence) is a scientific undertaking which goals to…) trigonometric.

Definitions as a sum

We outline this polynomial in two methods within the type of a sum:

$P_{n}(x)=frac{1}{2^n}sum_{k=0}^{E(n/2)} (-1)^k some{n}{k}some {2n-2k}{n}x^{n-2k}$

(we will deduce $P_{2n}(0)=frac{1}{2^{2n}}(-1)^nbiname{2n}{n} ,$ )

$P_{n}(x)=frac{1}{2^n}sum_{k=0}^{n}binom{n}{k}^2(x-1)^{nk}(x +1)^{k}$

Sequence decomposition of Legendre polynomials

Decomposition of a holomorphic operate

Any operate f, holomorphic inside an ellipse of foci -1 and +1, will be written as a sequence which converges uniformly contained in the ellipse:

$f(x)=sum_{n=0}^infty lambda_n P_n(x)$

with $forall n in mathbb{N}, lambda_n in mathbb{C}.$

Decomposition of a Lipschitz operate

Comment $tilde{P_n}$ the quotient of the polynomial $P not$ the couple mentioned requirements (A norm, from the Latin norma (“sq., ruler”) designates a…).

That’s $F$ steady software on [-1,1]. For all pure not posing

$c_n(f)=intlimits_{-1}^1 f(x)tilde P_n(x),dx,$

So subsequent $c_n(f),$ is of sq. (A sq. is a daily polygon with 4 sides. Which means that its…) summable, and makes it potential to elucidate the orthogonal projection of $F$ on $R_n[X]$ :

$S_nf=sum_{k=0}^n c_k(f)tilde P_k.$

We even have:

$for all xin[-1,1],;S_nf(x)=intlimits_{-1}^1 K_n(x,;y)f(y),dy$ with $K_n(x,;y)=frac{n+1}{2}frac{tilde P_{n+1}(x)tilde P_n(y)-tilde P_{n+1}(y )tilde P_n(x)}{xy};$
$S_nf(x)-f(x)=intlimits_{-1}^1 K_n(x,;y)(f(y)-f(x)),dy.$

Suppose additional that f is a Lipschitz operate. We then have the extra property:^{[réf. souhaitée]}

$for all xin]-1,1[,;lim_{ntoinfty}S_nf(x)=f(x).$

autrement dit, l’égalité

$f=sum_{n=0}^infty c_n(f)tilde P_n$

est vraie non seulement au sens L² mais au sens de la convergence simple (La convergence simple ou ponctuelle est un critère de convergence dans un espace fonctionnel,…) sur ]-1.1[.

Propriétés

Degré (Le mot degré a plusieurs significations, il est notamment employé dans les domaines…)

Le polynôme $P n$ est de degré n.

Parité

Les polynômes de Legendre suivent la parité de n. On peut exprimer cette propriété par :

$P_n(-x)=(-1)^nP_n(x).,$

(en particulier, $P n ( - 1) = ( - 1) n$ et $P 2 n + 1 (0) = 0$ ).

Orthogonalité (En mathématiques, l’orthogonalité est un concept d’algèbre linéaire…)

Les polynômes orthogonaux les plus simples sont les polynômes de Legendre pour lesquels l’intervalle d’orthogonalité est [−1, 1] and the operate weight (Weight is the drive of gravity, of gravitational and inertial origin, exerted by the…) is solely the fixed operate of worth 1: these polynomials are orthogonal with respect to the scalar product (In vector geometry, the dot product is an algebraic operation…) set to $R[X]$ by the connection:

$<P,Q>= int_{-1}^{+1} P(x) Q(x), mathrm{d}x” src=”http://add.wikimedia.org/math/a/b/0/ ab0346c74def670bd9db9597c27a4abe.png”/>.</dd> <dd><img class=$

The very definition of $P not$ reveals that it’s a clear vector (In arithmetic, the idea of eigenvector is an algebraic notion making use of to a…) for the personal worth (In arithmetic, the idea of eigenvector is an algebraic notion making use of to a…) -n(n+1) of the endomorphism:

$P in R[X] to u(P)= frac{textrm{d}}{textrm{d}x}[(1-x^{2})frac{textrm{d}P}{textrm{d}x}]$

However this endomorphism is symmetric for the product scalar (A real scalar is a quantity that’s unbiased of the selection of base chosen to precise the…) above, since an integration by elements reveals that

: $<u(P),Q>= int_{-1}^{+1} u(P)(x) Q(x), mathrm{d}x = – int_{-1}^{+ 1} (1-x^2 ) P'(x) Q'(x), mathrm{d}x = <P,u(Q)>” src=”http://add.wikimedia.org/math/9/e/d/9ed0d590d11227a26d6896d3771044e5.png”/>.</li> </ul> <p>As they’re eigenvectors related to distinct eigenvalues, the household of Legendre polynomials is orthogonal.</p> </p></div> </p></div> </p></div> <h3> <span class=$ Requirements
The sq. of the norm, in L²([-1,1]), is

$||P_n||^2=frac{2}{2n+1}.$

Certainly, for all n>1, we will set up the relation

$P'_{n+1}-P'_{n-1}=(2n+1)P_n, ,$

from which we deduce (utilizing that for all ok, $P ‘ ok -1$ is of diploma ok-2k so is orthogonal to $P ok$ and performing an integration by elements):

$<P_n,(2n+1)P_n>=<P_n,P'_{n+1}-P'_{n-1}>=<P_n,P'_{n+1}>=[P_nP_{n+1}]_{-1}^{1}-<P'_n,P_{n+1}>=[P_nP_{n+1}]_{-1}^{1}.”src=”http://add.wikimedia.org/math/4/2/6/426624b97534f713fd273a54bec16b8b.png”/></dd> </dl> <p>As <span class=$ P_notP_{not + 1} is odd and for all ok, $P ok (1) = 1$ we thus arrive at $(2 not + 1) | | P not | | 2 = 2.$

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