How can I prove the following recurrence relation for Laguerre polynomials eqn $(11)$.
$$xL_n'(x) = n L_n(x) - n L_{n-1}(x)$$
I managed to show that the following which seems to be true. I put all things on one side and this yields zero.
$$(x-n) L_n'(x) = n L_n(x) - nL_{n-1}(x) $$
Mathematica code
Simplify[Table[
n LaguerreL[n, x] + (n - x) D[LaguerreL[n, x], {x, 1}] -
n D[LaguerreL[n - 1, x], {x, 1}], {n, 1, 5}]]
This has been bugging me for a while. Did I do anything wrong or that recurrence relation on math-world site is wrong?
Added:: my equation was $\displaystyle (x-n) L_n'(x) = n L_n(x) - nL_{n-1}'(x) $ Now I need to show $\displaystyle L_n'(x) - L_{n-1}'(x) = -L_{n-1}(x)$ holds, and that it does, can be shown via differentiating generating function w.r.t. $t$.
I will prove $L_n'(x) - L_{n-1}'(x) = -L_{n-1}(x).\quad $ We have the recurrence formula $$ n L_n(x) = (2n-1-x) L_{n-1}(x) - (n-1)L_{n-2}(x) $$ and therefore $$ -xL_{n-1}(x) = n L_n(x) - (2n-1) L_{n-1}(x) + (n-1)L_{n-2}(x) $$ Then we use $$x L_n'(x) = n L_n(x) - n L_{n-1}(x)$$ or rewritten: $$ L_n'(x) = \frac{ n L_n(x) - n L_{n-1}(x)}{x}$$ And the same or for $(n-1)$ $$ L_{n-1}'(x) = \frac{ (n-1) L_{n-1}(x) - (n-1) L_{n-2}(x)}{x}$$ Subtracting the last two gives $$ L_n'(x) - L_{n-1}'(x)= \frac{ n L_n(x) - n L_{n-1}(x) - (n-1)L_{n-1}(x) + (n-1)L_{n-2}(x)}{x}$$
$$ L_n'(x) - L_{n-1}'(x)= \frac{ n L_n(x) - (2n-1) L_{n-1}(x) + (n-1)L_{n-2}(x)}{x}$$ Now substitute the rewritten recurrence formula:
$$ L_n'(x) - L_{n-1}'(x)= \frac{ -x L_{n-1}(x)}{x} = -L_{n-1}(x)$$