Legendre polynomial ($P_l(x)$)

The polynomial solution of polynomial degree $l$ normalized by $P_l(1)=1$ is the Legendre polynomial $P_l$.

For orthonormal Legendre polynomials on $[-1,1]$, the polynomial $q_n$ represents evaluation at zero on polynomials of degree at most $n$: $⟨r,q_n⟩=r(0)$. This follows by expanding $r$ in the orthonormal basis, or from the Riesz representation theorem. With $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$, orthonormality givesThe last identity telescopes using $c_m/c_{m-1}=(2m-1)/(2m)$. The norms are unbounded: $c_m\ge 1/(2m+1)$ makes the sum dominate a divergent harmonic series. Thus the evaluation identities hold eventually for every fixed polynomial but cannot extend to convergence for every $L^2$ vector, by the Uniform boundedness principle. Point evaluation is also not defined on $L^2$ equivalence classes.

For normalized Legendre polynomials $p_n=\sqrt{(2n+1)/2}{P}_n$, parity makes the sign function orthogonal to every even index. Integrating the Legendre differential equation gives ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. The differentiated generating function $z(1+z^2{)}^{-3/2}$ gives the derivatives at zero and hence the displayed odd coefficients. The Parseval identity and squared norm two then evaluate ${\sum }_{m\ge 0}((2m)!{)}^2(4m+3)/[2^{4m}(m!(m+1)!{)}^2]=4$.

This instance of Rodrigues' formula constructs the Legendre polynomial for every nonnegative integer $l$. Put $v=(x^2-1{)}^l$; then $(x^2-1)v^{\prime }=2lxv$. Differentiating $l+1$ times by the Leibniz rule gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$ for $y=v^{(l)}$, which is the Legendre differential equation after reversing signs. The leading term of the normalized derivative is $(2l)!x^l/[2^l(l!{)}^2]$, proving its polynomial degree is exactly $l$. At $x=1$, the product $(x+1{)}^l(x-1{)}^l$ contributes only the term with all $l$ derivatives on the second factor, giving $v^{(l)}(1)=2^{l}l!$. Thus the normalization is $P_l(1)=1$.

Differentiating the Legendre differential equation gives $((1-x^2{)}^2{R}_n^{\prime }{)}^{\prime }+(n-1)(n+2)(1-x^2)R_n=0$ for $R_n=P_n^{\prime }$. Multiply the equations for two distinct indices by the opposite polynomial and subtract. The endpoint term vanishes because of $(1-x^2{)}^2$, giving the displayed weighted orthogonality. The zero polynomial $R_0$ is not a nonzero eigenfunction.

With $P_n(1)=1$, the generating function satisfies ${\partial }_x(x^2{\partial }_{x}G)+{\partial }_t((1-t^2){\partial }_{t}G)=0$. Expanding in Legendre polynomials gives $x^2{a}_n^{\prime \prime }+2x{a}_n^{\prime }-n(n+1)a_n=0$. Boundedness at zero removes $x^{-n-1}$, and the value at $t=1$ makes $a_n=x^n$. Orthogonality then gives ${\int }_{-1}^1{G}^{2}dt={\sum }_n{k}_n{x}^{2n}$. The integral is $x^{-1}\log ((1+x)/(1-x))$, so $k_n=2/(2n+1)$. Uniform analyticity on compact subdiscs $|x|<1$ justifies the generating expansion and termwise calculations.

The Legendre polynomials are orthogonal on $[-1,1]$ with unit weight:

The three-term recurrenceturns multiplication by $x$ into coupling between adjacent Legendre multipoles.

For a contour enclosing $t$,

For fixed $0<\theta <\pi$,

For fixed $\xi >0$,

The endpoint Laplace method for a Legendre polynomial ceases to localize when $n\sqrt{x-1}=O(1)$. On the scale $x=1+\nu /n^2$, its real integral representation givesThe Modified Bessel function of the first kind equals one at $\nu =0$ and has large-$\nu$ behavior $e^{\sqrt{2\nu }}/\sqrt{2\pi \sqrt{2\nu }}$, matching the Hyperbolic Debye asymptotic for Legendre polynomials. This boundary layer resolves an increasingly steep function at an endpoint where its value remains fixed.