Associated Legendre Function of the First Kind/Examples/m gt n

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Example of Associated Legendre Function of the First Kind

Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.

Let $m > n$.

Then:

$\map { {P_n}^m} x = 0$


Proof

From Derivative Form of Associated Legendre Function of the First Kind:

$\map { {P_n}^m} x = \dfrac {\paren {1 - x^2}^{m / 2} } {2^n n!} \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n$

We have that:

$\paren {x^2 - 1}^n = x^{2 n} + \map Q x$

where $\map Q x$ is a polynomial of degree $2 n - 2$.


Next we note that $m > n \implies m + n > 2 n > 2 n - 2$.


Hence:

\(\ds \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n\) \(=\) \(\ds \map {\dfrac {\d^{m + n} } {\d x^{m + n} } } {x^{2 n} + \map Q x}\)
\(\ds \) \(=\) \(\ds 0\)

Hence:

$\dfrac {\paren {1 - x^2}^{m / 2} } {2^n n!} \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n = 0$

and the result follows.

$\blacksquare$


Sources

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