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Legendre Rodrigues equation

Statement

For $l \in N = {0, 1, 2, \dots}$ define the function $P_{l} : R \to R$ by the Rodrigues formula (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set)

$P_{l} (t) : = \frac{1}{2^{l} l!} \frac{d^{l}}{d t^{l}} [(t^{2} - 1)^{l}] .$

Then:

(a) each $P_{l}$ is the polynomial function of the polynomial $p_{l} (t) = \frac{1}{2^{l} l!} \sum_{k : 2 k \geq l}^{l} (- 1)^{l - k} (\binom{l}{k}) \frac{(2 k)!}{(2 k - l)!} t^{2 k - l} \in R [t]$ (so the two expressions agree at every real $t$ , Evaluation and roots of a polynomial in a commutative target ring), $p_{l}$ has degree $l$ with leading coefficient $\frac{(2 l)!}{2^{l} (l!)^{2}} \neq 0$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree);

(b) $P_{l} (1) = 1$ and $P_{l} (- 1) = (- 1)^{l}$ ;

(c) $P_{l}$ satisfies Legendre's differential equation on $R$ : $(1 - t^{2}) P_{l}^{''} (t) - 2 t P_{l}^{'} (t) + l (l + 1) P_{l} (t) = 0 (t \in R);$

(d) $P_{0} = 1$ , $P_{1} (t) = t$ and $P_{2} (t) = \frac{1}{2} (3 t^{2} - 1)$ .

No probability, measure or physical interpretation is attached to $P_{l}$ ; this item constructs the polynomials that later items use as angular functions.

Facts & Assumptions

Given: $l \in N$ ; the function $F_{l} (t) : = (t^{2} - 1)^{l}$ ; the function $P_{l}$ defined by the Rodrigues formula; the polynomial $p_{l}$ displayed in (a).

[F1]

The derivative of $t^{0} = 1$ is $0$ ; for $n \geq 1$ , $\frac{d}{d t} t^{n} = n t^{n - 1}$ . Iterating this formula and the factorial recursion gives $\frac{d^{m}}{d t^{m}} t^{n} = \frac{n!}{(n - m)!} t^{n - m}$ for $0 \leq m \leq n$ while $\frac{d^{m}}{d t^{m}} t^{n} = 0$ for $m > n$ ; derivatives are linear and satisfy the product and power rules (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F2]

The general Leibniz rule: for $m$ -times differentiable $f, g$ , $(f g)^{(m)} = \sum_{j = 0}^{m} (\binom{m}{j}) f^{(j)} g^{(m - j)}$ (The general Leibniz rule for the $n$ -th derivative of a product).

[F3]

The binomial coefficients satisfy $(\binom{l}{k}) = \frac{l!}{k! (l - k)!}$ , $(\binom{l}{0}) = (\binom{l}{l}) = 1$ , and $n! = (n - 1)! n$ for $n \geq 1$ , with $0! = 1$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ , The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The natural numbers $N$ (von Neumann)).

[F4]

Polynomial evaluation is a ring homomorphism $R [t] \to R$ , $p \mapsto p (t)$ , compatible with addition and multiplication, and a nonzero polynomial of degree $d$ has leading coefficient the coefficient of $t^{d}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, Evaluation and roots of a polynomial in a commutative target ring, Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[F5]

The binomial theorem over a commutative ring: $(x + y)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) x^{k} y^{n - k}$ in every commutative ring and for all elements $x, y$ (The binomial theorem over an arbitrary commutative ring).

Proof

1.1givenF1F3F4F5algebra

[(a), expansion] By [F5] applied in the polynomial ring to $x = t^{2}$ and $y = - 1$ , $F_{l} (t) = \sum_{k = 0}^{l} (\binom{l}{k}) (- 1)^{l - k} t^{2 k}$ as polynomials; differentiating $l$ times termwise with [F1] and [F4] (the sum is finite, so linearity applies), every term with $2 k < l$ differentiates to $0$ and every term with $2 k \geq l$ contributes $(\binom{l}{k}) (- 1)^{l - k} \frac{(2 k)!}{(2 k - l)!} t^{2 k - l}$ ; hence $P_{l}$ agrees with the polynomial $p_{l}$ of the Statement. All exponents $2 k - l$ with $2 k \geq l$ are nonnegative, and the largest is $2 l - l = l$ , attained only at $k = l$ with coefficient $(2 l)! / l! \neq 0$ by [F3]; so $p_{l}$ has degree $l$ and leading coefficient $(2 l)! / (l! \cdot 2^{l} l!) = (2 l)! / (2^{l} (l!)^{2})$ , which is nonzero.

1.2givenF1F2F3algebra

[(b), endpoint values] By the factorisation $F_{l} = (t - 1)^{l} (t + 1)^{l}$ and [F2], the $l$ -th derivative at $t = 1$ is $\sum_{j = 0}^{l} (\binom{l}{j}) \frac{d^{j}}{d t^{j}} (t - 1)^{l} ∣_{t = 1} \cdot \frac{d^{l - j}}{d t^{l - j}} (t + 1)^{l} ∣_{t = 1}$ , and every term with $j < l$ vanishes because $\frac{d^{j}}{d t^{j}} (t - 1)^{l}$ still contains a factor $(t - 1)$ ; the surviving term is $l! \cdot (1 + 1)^{l} = l! 2^{l}$ by [F1], [F3], so $P_{l} (1) = \frac{l! 2^{l}}{2^{l} l!} = 1$ . The same computation at $t = - 1$ gives $l! \cdot (- 1 - 1)^{l} = (- 1)^{l} l! 2^{l}$ , so $P_{l} (- 1) = \frac{(- 1)^{l} l! 2^{l}}{2^{l} l!} = (- 1)^{l}$ .

1.3givenF1F2F3F4algebra

[(c), Legendre's equation] If $l = 0$ , then $P_{0} = 1$ and the equation is immediate. Suppose $l \geq 1$ . Differentiating $F_{l} = (t^{2} - 1)^{l}$ with [F1] gives $F_{l}^{'} = 2 l t (t^{2} - 1)^{l - 1}$ , hence $(t^{2} - 1) F_{l}^{'} = 2 l t F_{l}$ . Apply $\frac{d^{l + 1}}{d t^{l + 1}}$ to both sides using [F2]; on the left the third and higher derivatives of $t^{2} - 1$ vanish, so $(t^{2} - 1) F_{l}^{(l + 2)} + (l + 1) (2 t) F_{l}^{(l + 1)} + (\binom{l + 1}{2}) (2) F_{l}^{(l)}$ , and on the right $2 l (t F_{l}^{(l + 1)} + (l + 1) F_{l}^{(l)})$ , where $(\binom{l + 1}{2}) = \frac{l (l + 1)}{2}$ by [F3]; subtracting gives $(t^{2} - 1) F_{l}^{(l + 2)} + 2 t F_{l}^{(l + 1)} - l (l + 1) F_{l}^{(l)} = 0$ . Since $P_{l} = \frac{1}{2^{l} l!} F_{l}^{(l)}$ is exactly the $l$ -fold derivative, dividing by the nonzero constant and multiplying by $- 1$ yields $(1 - t^{2}) P_{l}^{''} - 2 t P_{l}^{'} + l (l + 1) P_{l} = 0$ at every real $t$ : the displayed polynomial identity holds for the polynomial $p_{l}$ by [F4].

2.1givenF1algebra∎

[(d)] For $l = 0$ , $P_{0} = 1$ ; for $l = 1$ , $P_{1} (t) = \frac{1}{2} \frac{d}{d t} (t^{2} - 1) = t$ ; for $l = 2$ , $P_{2} (t) = \frac{1}{8} \frac{d^{2}}{d t^{2}} (t^{4} - 2 t^{2} + 1) = \frac{1}{8} (12 t^{2} - 4) = \frac{3 t^{2} - 1}{2}$ by [F1].

The identities proved here are polynomial identities in one real variable; the item makes no angular, measure or spectral claim.

Legendre Rodrigues equation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources