Bonnet's Recursion Formula

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Theorem

Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.

Bonnet's Recursion Formula states:

$\paren {n + 1} \map {P_{n + 1} } x = \paren {2 n + 1} x \map {P_n} x - n \map {P_{n - 1} } x$


Proof

From Generating Function for Legendre Polynomials, the generating function for $P_n$ is:

$(1): \quad \ds \frac 1 {\sqrt {1 - 2 x t + t^2} } = \sum_{n \mathop = 0}^\infty \map {P_n} x t^n$

Differentiating both sides of $(1)$ with respect to $t$:

\(\ds \map {\dfrac \d {\d t} } {\paren {1 - 2 x t + t^2 }^{-1/2} }\) \(=\) \(\ds \map {\dfrac \d {\d t} } {\sum_{n \mathop = 0}^\infty \map {P_n} x t^n}\)
\(\ds \leadsto \ \ \) \(\ds -\dfrac 1 2 \paren {-2 x + 2 t} \paren {1 - 2 x t + t^2 }^{-3/2}\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \map {\dfrac \d {\d t} } {\map {P_n} x t^n}\) Derivative of Power, Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \dfrac {x - t} {\sqrt {1 - 2 x t + t^2} }\) \(=\) \(\ds \paren {1 - 2 x t + t^2} \sum_{n \mathop = 1}^\infty n \map {P_n} x t^{n - 1}\) Derivative of Power and multiplying both sides by $1 - 2 x t + t^2$
\(\ds \leadsto \ \ \) \(\ds \paren {x - t} \sum_{n \mathop = 0}^\infty \map {P_n} x t^n\) \(=\) \(\ds \paren {1 - 2 x t + t^2} \sum_{n \mathop = 1}^\infty n \map {P_n} x t^{n - 1}\) substituting $\ds \sum_{n \mathop = 0}^\infty \map {P_n} x t^n$ for $\dfrac 1 {\sqrt {1 - 2 x t + t^2} }$ from $(1)$
\(\ds \leadsto \ \ \) \(\ds \paren {x - t} \sum_{n \mathop = 0}^\infty \map {P_n} x t^n\) \(=\) \(\ds \paren {1 - 2 x t + t^2} \sum_{n \mathop = 0}^\infty \paren {n + 1} \map {P_{n + 1} } x t^n\) Translation of Index Variable of Summation in right hand side


Equating coefficients of $t^n$:

\(\ds x \map {P_n} x - \map {P_{n - 1} } x\) \(=\) \(\ds \paren {n + 1} \map {P_{n + 1} } x - 2 x n \map {P_n} x + \paren {n - 1} \map {P_{n - 1} } x\)
\(\ds \leadsto \ \ \) \(\ds \paren {n + 1} \map {P_{n + 1} } x\) \(=\) \(\ds x \map {P_n} x - \map {P_{n - 1} } x + 2 x n \map {P_n} x - \paren {n - 1} \map {P_{n - 1} } x\) rearranging
\(\ds \) \(=\) \(\ds \paren {2 n + 1} x \map {P_n} x - n \map {P_{n - 1} } x\) simplifying

Hence the result.

$\blacksquare$


Also presented as

Bonnet's recursion formula can also be presented in the form:

$\paren {n + 1} \map {P_{n + 1} } x - \paren {2 n + 1} x \map {P_n} x + n \map {P_{n - 1} } x = 0$


Source of Name

This entry was named for Pierre Ossian Bonnet.


Sources

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