Recurrence relations of the Laguerre Polynomials

The recurrence relations of the Laguerre polynomials describe how each polynomial in the sequence can be obtained from its predecessors.

Some useful recurrence relations of the Laguerre polynomials include

$(n+1)L_{n+1}=(2n+1-x)L_n-nL_{n-1}\;\;\;\;\;\;\;\;434$

$L_{n-1}=L_{n-1}'-L_{n}'\;\;\;\;\;\;\;\;435$

$xL_{n}'=nL_{n}-nL_{n-1}\;\;\;\;\;\;\;\;436$

To derive eq434, differentiate eq430 with respect to $t$ to give

$\frac{\partial G}{\partial t}=\frac{1-x-t}{(1-t)^3}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{\infty}nL_nt^{n-1}$

Substituting eq430 gives

$(1-x-t)\sum_{n=0}^{\infty}L_nt^{n}=(1-t)^2\sum_{n=0}^{\infty}nL_nt^{n-1}$

Equating the coefficients of $t^n$ yields

$(1-x)L_{n}-L_{n-1}=(n+1)L_{n+1}-2nL_{n}+(n-1)L_{n-1}$

which rearranges to eq434.

To derive eq435, differentiate eq430 with respect to $x$ to give

$\frac{\partial G}{\partial t}=-\frac{t}{(1-t)^2}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{\infty}L_n't^{n}$

Substituting eq430 gives

$-t\sum_{n=0}^{\infty}L_nt^{n}=(1-t)\sum_{n=0}^{\infty}L_n't^n$

Equating the coefficients of $t^n$ yields eq435. To derive eq436, differentiate eq434 with respect to $x$ to yield

$(n+1)L_{n+1}'=(2n+1-x)L_n'-L_n-nL_{n-1}'$

Substituting eq435 gives

$(n+1)L_{n+1}'=(2n+1-x)L_n'-L_n-n(L_{n-1}+L_n')$

Letting $n=n+1$ in eq435, substituting the result in the above equation and rearranging yields eq436.