🔎 Legendre Polynomial - Definition and Interpretations

an introduction

Legendre . polynomial

The Legendre . polynomial are solutions of the Legendre differential equation, and form the simplest example of a sequence orthogonal polynomials (In mathematics, a sequence of orthogonal polynomials is an infinite sequence of…).

Legendre equation

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

Thus we define the Polynomial Legendre (Legendre polynomials are solutions of the differential equation for…) $s Not$ (for everything (All is understood as all that exists most often as the world or …) all natural (In mathematics, a natural number is a positive number (or zero) that basically allows…) 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 $P_n = P_n^{(0,0)}$ where $P_n^{(\alpha,\beta)}$ You know? polynomial (A polynomial, in mathematics, is the linear summation of the products of…) Jacobi from index Not associated with the parameters α and .

Some polynomials

The first polynomials are:

first 20 Legendre . polynomial (Legendre polynomials are y solutions of the Legendre differential equation 🙂.

Other definitions

bonnet repetition formula

$P_0 (x) = 1, \P_1 (x) = x,$ and for everyone Not> 0

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

Formula Rodriguez (Rodrigues is the smallest of the three islands in the Mascarene Archipelago.)

Define the polynomial $s Not$ (for all natural integers Not) by :

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

identification (Definition is a letter that says what the thing is or what the name means. Hence…) Analytical

We can also define this sequence of polynomials by its generating function:

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

The theory (A theory is a proposition that can be demonstrated mathematically, i.e….) Then give the residue:

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

where is the Outline (COMET Nucleus TOUR (CONTOUR) is a NASA space probe that is part of the program…) Surrounds the original and gets caught in the meaning (SENS (Neglected Engineering Aging Strategies) is a scientific project aimed at…) Furniture Calculation.

Tariffs as a total

We define this polynomial in two ways in the form of a sum:

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

(infer $P_{2n}(0)=\frac{1}{2^{2n}}(-1)^n\binom{2n}{n}\,$ )

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

Sequential decomposition of Legendre polynomials

Holomorphic function decomposition

Any function f, holomorphic inside an ellipse of foci -1 and +1, can be written as a series that converges uniformly inside the ellipse:

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

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

Decomposition of the Lipschitz function

note $\ tilde {P_n}$ Polynomial quotient $s Not$ by it Basic (Standard, from the Latin norma (“square, ruler”) designating…).

This is it $F$ Continuous application on [-1,1]. Everything is natural Not is put

$c_n(f) = \int \ limits _{- 1}^1 f (x) \ tilde P_n (x) \, dx,$

Then $c_n(f)\,$ who is he Field (A square is a regular four-sided polygon. This means that…) In short, and makes it possible to explain the orthogonal projection of $F$ on me $\ R_n[X]$ :

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

We also have:

$\ forall x \ in[-1,1], \; S_nf (x) = \ int \ limits _ { - 1} ^ 1 K_n (x, \; y) f (y) \, dy$ with $K_n(x,\;y) = \frac{n+1}{2}\frac{\tilde P_{n+1}(x)\tilde P_n(y) -\tilde P_{n+1}(y) )\tilde P_n(x)}{xy};$
$S_nf (x) -f (x) = \int \ limits _{- 1}^1 K_n(x, \;y) (f (y) -f (x)) \, dy.$

Further suppose that f is a Lipschitz function. Then we have the additional property:^{[réf. souhaitée]}

$\ forall x \ in]-1,1[,\;\lim_{n\to\infty}S_nf(x)=f(x).$

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 job weight (Weight is the gravitational force, the gravitational force and the origin of inertia, which it exerts…) is simply a constant function of the value 1: these polynomials are orthogonal with respect to scalar product (In vector geometry, the point product is an algebraic operation…) set to $\ s[X]$ in relation:

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

identification $s Not$ indicates that clean vector (In mathematics, the eigenvector concept is an algebraic concept that applies to…) to special value (In mathematics, the eigenvector concept is an algebraic concept that applies to…) -n(n+1) to resemble:

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

But this similarity is the same with the product scalar (The real scalar is a number independent of the choice of base chosen to express…) Above, because integration by parts shows this

: $<u (P)، Q> = \int_{- 1}^{+1}u(P)(x)Q(x)\,\mathrm{d} x = – \int_{-1}^{+1}(1-x ^ 2) P’ (x) Q’ (x) \, \mathrm{d} x = <P، u (Q)>src=”http://upload.wikimedia.org/math/9/e/d/9ed0d590d11227a26d6896d3771044e5.png”/>.</li> </ul> <p>Since they are eigenvectors associated with distinct eigenvalues, the Legendre family of polynomials is orthogonal.</p> </p></div> </p></div> </p></div> <h3> <span class=$ Basic
base square, in The²([-1,1]), he is

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

In fact, for all n > 1, we can establish the relationship

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

From which we deduce (use that for everyone KAnd the $s “ K -1$ class K-2k orthogonal to that $s K$ and implementing integration by parts):

$<P_n، (2n + 1) P_n> = <P_n، P '_ {n + 1} -P' _ {n-1}> = <P_n، P '_ {n + 1}> =[P_nP_{n+1}]_{-1}^{\1}- <P'_n، P_ {n + 1}> =[P_nP_{n+1}]_{-1}^{\1}.”src=”http://upload.wikimedia.org/math/4/2/6/426624b97534f713fd273a54bec16b8b.png”/></dd> </dl> <p>as such <span class=$ s_Nots_{Not + 1} It’s weird for everyone KAnd the $s K (1) = 1$ thus we arrive at $(2 Not + 1) | | s Not | | 2 = 2.$

Carly Burrows

“Evil thinker. Music scholar. Hipster-friendly communicator. Bacon geek. Amateur internet enthusiast. Introvert.”

Your Decommissioning News

🔎 Legendre Polynomial – Definition and Interpretations