Legendre Polynomial/Examples/P1

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Example of Legendre Polynomial

The $1$st Legendre polynomial is:

$\map {P_1} x = x$


Proof

From Generating Function for Legendre Polynomials, $\map {P_1} x$ is the coefficient of the term in $t^1$ of the generating function:

$\map G t = \dfrac 1 {\sqrt {1 - 2 x t + t^2} }$


Thus:

\(\ds \map G t\) \(=\) \(\ds \dfrac 1 {\sqrt {1 - 2 x t + t^2} }\)
\(\ds \) \(=\) \(\ds \paren {1 - 2 x t + t^2}^{-1/2}\)
\(\ds \) \(=\) \(\ds 1 + \paren {-\dfrac 1 2 } \paren {-2 x t + t^2} + \paren {\dfrac 1 {2!} } \paren {-\dfrac 1 2} \paren {-\dfrac 3 2} \paren {-2 x t + t^2}^2 + \cdots\) General Binomial Theorem
\(\ds \) \(=\) \(\ds 1 + \paren {x t^1 - 2 t^2} + \dfrac 3 8 \paren {4 x^2 t^2 - 4 x t^3 + t^4} + \cdots\) Square of Sum and simplifying

Further terms generate terms in $t$ of higher powers.

Hence the term in $t^1$ is seen to be $x t$.

Hence, by definition of generating function:

$\map {P_1} x = x$

$\blacksquare$


Sources

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