Editing Rodrigues' formula (section)

==Statement==
Let <math>(P_n(x))_{n=0}^\infty</math> be a [[sequence]] of orthogonal polynomials on the interval <math>[a, b]</math> with respect to [[weight function]] <math>w(x)</math>. That is, they have degrees <math>deg(P_n) = n</math>, satisfy the orthogonality condition
<math display="block">\int_a^b P_m(x) P_n(x) w(x) \, dx = K_n \delta_{m,n}</math>
where <math>K_n</math> are nonzero constants depending on <math>n</math>, and <math>\delta_{m,n}</math> is the [[Kronecker delta]]. The interval <math>[a, b]</math> may be infinite in one or both ends.

{{Math theorem
| name = Rodrigues' type formula
| note = 
| math_statement = If
<math display="block">w(x)=W(x)/B(x), \quad \frac{W'(x)}{W(x)} = \frac{A(x)}{B(x)},</math>
where <math>A(x)</math> is a [[polynomial]] with [[degree of a polynomial|degree]] at most 1 and <math>B(x)</math> is a polynomial with degree at most 2, and
<math display="block">\lim_{x \to a} x^k W(x) = 0, \qquad \lim_{x \to b}  x^k W(x) = 0.</math>
for any <math>k = 0, 1, 2, \dots</math>.

Then, if <math>\frac{d^n}{dx^n} \!\left[ B(x)^n w(x)\right] \neq 0</math> for all <math>n = 0, 1, 2, \dots</math>, then
<math display="block">P_n(x) = \frac{c_n}{w(x)} \frac{d^n}{dx^n} \!\left[ B(x)^n w(x)\right],</math>
for some constants <math>c_n</math>.
}}

{{Math proof|title=Proof|proof= 

Let <math display="inline">F_k := \frac 1w D_x^k(B^n w)</math>, then <math display="inline">F_k = B^{n-k} p_k</math> for all <math display="inline">k \in 0:n</math> for some polynomials <math display="inline">p_k</math>, such that <math display="inline">deg(p_k) \leq k</math>. Proven by induction on <math display="inline">k</math>: <math display="block">
 F_{k+1} = B^{n-k-1}(B p_k' + (n-k)B' p_k + (A-B')p_k)
 </math>

Let <math display="inline">Q_n := \frac 1w D_x^n(B^n w)</math>. We have shown that <math display="inline">Q_n</math> is a polynomial of degree <math>\leq n</math>. With integration by parts, we have for all <math display="inline">n > m</math>, <math display="block">
 \int_a^b Q_m Q_n w dx = \int_a^b B^n w (D_x^n Q_m) dx = 0
 </math> since <math display="inline">D_x^n Q_m=0</math>. Thus, <math display="inline">Q_0, Q_1, \dots</math> make up an orthogonal polynomial series with respect to <math display="inline">w</math>. Thus, <math display="inline">P_n = c_n Q_n</math> for some constants <math display="inline">c_n</math>.
}}

{{Math theorem
| name = Differential equation
| note = 
| math_statement = <math display="block">B(x) \frac{d^2}{dx^2} P_n(x) + A(x) \frac{d}{dx}  P_n(x) + \lambda_n P_n(x) = 0</math>

<math display="block">\lambda_n = -\frac{1}{2}n(n-1)B''-nA'</math>
}}

{{Math proof|title=Proof|proof= 
When <math>n = 0</math>, it is trivial. When <math>n = 1</math>, it simplifies to <math>AP_1' = A'P_1</math>, which is true since <math>P_1 = \frac{c_1}{w}(Bw)' = c_1A</math>. So assume <math>n \geq 2</math>. Define <math>I_n(x) = \frac{d^n}{dx^n}(B^n(x) w(x))</math>, then by direct computation and simplification, the equation to be proven is equivalent to

<math display="block">\frac{d^2}{dx^2} (B(x) I_n(x)) - \frac{d}{dx} (A(x) I_n(x)) + \lambda_n I_n(x) = 0</math>

By Leibniz differentiation rule, we have

<math display="block">B(x) \frac{d^n}{dx^n} y = \frac{d^n}{dx^n} (B(x) y) - n \frac{d^{n-1}}{dx^{n-1}} (B'(x) y) + \frac{n(n-1)}{2} \frac{d^{n-2}}{dx^{n-2}} (B'' y)</math>

<math display="block">A(x) \frac{d^n}{dx^n} y = \frac{d^n}{dx^n} (A(x) y) - n \frac{d^{n-1}}{dx^{n-1}} (A' y)</math>

for arbitrary <math>y</math>. This allows us to move <math>A(x), B(x)</math> to the other side of the <math>n</math>-th derivative. Set <math>y = B^n(x) w(x) </math>, and define

<math display="block">J(x) = \frac{d^2}{dx^2} (B(x) y(x)) - n \frac{d}{dx} (B'(x) y(x)) + \frac{n(n-1)}{2} B'' y(x)</math>

<math display="block">K(x) = -\frac{d}{dx} (A(x) y(x)) + n A' y(x)</math>

<math display="block">L(x) = \lambda_n y(x)</math>

Then the equation simplifies to <math>\frac{d^n}{dx^n} (J+K + L) = 0</math>

<math>J(x)</math> has three terms, call them in order <math>J_1(x), J_2(x), J_3(x)</math>. <math>K(x)</math> has two terms, call them in order <math>K_1(x), K_2(x)</math>. 

<math>J_3(x) + K_2(x) + L(x) = (\lambda_n + \frac{n(n-1)}{2} B'' + n A')y=0</math>.

That <math>J_1(x) + J_2(x) + K_1(x) = 0</math>. follows from first writing <math>J_1(x)</math> as

<math>J_1(x) = \frac{d^2}{dx^2} \left(B^n(x) \int \exp\left(\frac{A(x)}{B(x)}\right)dx \right)</math>

and then taking the innermost first derivative to obtain

<math>J_1(x) = \frac{d}{dx}\left[\bigg(nB'(x)B^{n-1}(x) + A(x)B^{n-1}(x)\bigg)\int \exp\left(\frac{A(x)}{B(x)}\right)dx\right]</math>

and then rewriting this as

<math>J_1(x) = \frac{d}{dx}\Big(nB'(x)B^{n}(x)w(x)+ A(x)B^{n}(x)w(x)\Big)</math>

The first term is the negative of <math>J_2(x)</math> and the second term is the negative of <math>K_1(x)</math>.
}}

More abstractly, this can be viewed through [[Sturm–Liouville theory]]. Define an operator <math>Lf := - \frac{1}{w} (Wf')'</math>, then the differential equation is equivalent to <math>LP_n = \lambda_n P_n</math>. Define the functional space <math>X = L^2([a,b], w(x)dx)</math> as the Hilbert space of functions over <math>[a, b]</math>, such that <math>\langle f, g\rangle := \int_a^b fgw</math>. Then the operator <math>L</math> is self-adjoint on functions satisfying certain boundary conditions, allowing us to apply the [[spectral theorem]].