state and prove Rodrigue's formula
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hey mate
In mathematics, Rodrigues' formula (formerly called the Ivory–Jacobi formula) is a formula for the Legendre polynomials independently introduced by Olinde Rodrigues (1816), Sir James Ivory (1824) and Carl Gustav Jacobi (1827). The name "Rodrigues formula" was introduced by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the first to discover it. The term is also used to describe similar formulas for other orthogonal polynomials. Askey (2005) describes the history of the Rodrigues formula in detail.
If n is integer
dndxn(x2−1)n==dndxn[∑k=0n(−1)kn!k!(n−k)!x2n−2k]∑k=0n(−1)kn!k!(n−k)!(2n−2k)!(n−2k)!xn−2k.
The sum above does not go up to k=n, since after k=[n/2], the derivatives are 0 then we write
dndxn(x2−1)n=∑k=0[n/2](−1)kn!k!(n−k)!(2n−2k)!(n−2k)!xn−2k.
It follows from the infinite series truncated to the Legendre polynmial that
Pn(x)=12nn!dndxn(x2−1)n.
The approach followed here is in reverse order. We started with Rodriguez's formula and showed that it corresponds to a Legendre polynomial. A more intuitive approach is to start at the polynomials
y(x)=(1−x2)n.
and take derivates, and verifty that the derivatives taken n times will get you to the Legendre differential equation. That is, we have that
y′=−2nx(1−x2)n−1
which we can write as
(1−x2)y′+2nxy=0.
and starts looking a bit like a Legendre differential equation.
We want to differentiate this equation k times and use the Leibniz rule. That is, if we call u=1−x2,
dkdxk[uy′]=∑j=0k(kj)u(j)y(k−j+1)
Given that u is a second order polynomial only three terms of this sum will survive. That is
dkdxk[uy′]==uy(k+1)+ku′y(k)+k(k−1)u(2)y(k−1)(1−x2)y(k+1)−2kxy(k)−2k(k−1)2y(k−1)=0
Likewise we use the Leibniz rule for the product 2nxy where only two terms will survive. That is
dkdxk[2nxy]=2nxy(k)+2nky(k−1),
we combine the two results above to find
(1−x2)y(k+1)−2kxy(k)−k(k−1)y(k−1)+2nxy(k)+2nky(k−1)=0
At this point we observe that if k=n+1, we find
(1−x2)y(n+2)−2(n+1)xy(n+1)−n(n+1)y(n)+2nxy(n+1)+2n(n+1)y(n)=0
which simplifies to
(1−x2)y(n+2)−2xy(n+1)+n(n+1)y(n)=0.
and this is the Lagrange differential equation with yn=Pn. We then showed that
dndxn(1−x2)n
satisfies the Lagrange differential equation. The factor 1/(2nn!) is included to make P(1)=1.
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