Length of Legendre Polynomial
Jump to navigation
Jump to search
Theorem
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Let $\norm {\map {P_n} x}$ denote the length of $\map {P_n} x$.
Then:
- $\norm {\map {P_n} x} := \sqrt {\dfrac 2 {2 n + 1} }$
Proof
Applying Bonnet's Recursion Formula for $n - 1$:
- $n \map {P_n} x = \paren {2 n - 1} x \map {P_{n - 1} } x - \paren {n - 1} \map {P_{n - 2} } x$
so:
- $\map {P_n} x = \dfrac {2 n - 1} n x \map {P_{n - 1} } x - \dfrac {n - 1} n \map {P_{n - 2} } x$
Substituting for $\map {P_n} x$:
\(\ds \norm {\map {P_n} x}^2\) | \(=\) | \(\ds \int_{-1}^1 \map {P_n} x \map {P_n} x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{-1}^1 \map {P_n} x \paren {\frac {2 n - 1} n x \map {P_{n - 1} } x - \frac {n - 1} n \map {P_{n - 2} } x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} n \int_{-1}^1 x \map {P_n} x \map {P_{n - 1} } x \rd x - \frac {n - 1} n \int_{-1}^1 \map {P_n} x \map {P_{n - 2} } x \rd x\) | Linear Combination of Definite Integrals |
From Orthogonality of Legendre Polynomials:
- $\ds \int_{-1}^1 \map {P_n} x \map {P_m} x \rd x = 0 \iff n \ne m$
so:
- $\ds (1): \quad \norm {\map {P_n} x}^2 = \frac {2 n - 1} n \int_{-1}^1 x \map {P_n} x \map {P_{n - 1} } x \rd x$
From Bonnet's Recursion Formula:
- $x \map {P_n} x = \dfrac {n + 1} {2 n + 1} \map {P_{n + 1} } x + \dfrac n {2 n + 1} \map {P_{n - 1} } x$
Substituting for $x \map {P_n} x$ in $(1)$:
\(\ds \norm {\map {P_n} x}^2\) | \(=\) | \(\ds \frac {2 n - 1} n \frac {n + 1} {2 n + 1} \int_{-1}^1 \map {P_{n + 1} } x \map {P_{n - 1} } x \rd x + \frac {2 n - 1} {2 n + 1} \int_{-1}^1 \map {P_{n - 1} } x \map {P_{n - 1} } x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac{2 n - 1} {2 n + 1} \int_{-1}^1 \map {P_{n - 1} } x \map {P_{n - 1} } x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n + 1} \norm {\map {P_{n - 1} } x}^2\) |
Thus:
\(\ds \norm {\map {P_n} x}^2\) | \(=\) | \(\ds \frac {2 n - 1} {2 n + 1} \norm {\map {P_{n - 1} } x}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n + 1} \frac {2 n - 3} {2 n - 1} \norm {\map {P_{n - 2} } x}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n + 1} \frac {2 n - 3} {2 n - 1} \frac {2 n - 5} {2 n - 3} \dotsm \frac 3 5 \frac 1 3 \norm {\map {P_0} x}^2\) |
Most of this cancels out, leaving:
- $\norm {\map {P_n} x}^2 = \dfrac {\norm {\map {P_0} x}^2} {2 n + 1}$
It remains to compute the length of the first Legendre polynomial:
\(\ds \norm {\map {P_0} x}^2\) | \(=\) | \(\ds \int_{-1}^1 1 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \bigintlimits x {-1} 1\) | Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds 2\) |
Thus:
- $\norm {\map {P_n} x}^2 = \dfrac 2 {2 n + 1}$
and so taking the square root:
- $\norm {\map {P_n} x} = \sqrt {\dfrac 2 {2 n + 1} }$
$\blacksquare$