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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. $t$.

hasExams
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  • First: There is a small typo in your equation, it should be $xL_n'(x) = n L_n(x) - n L_{n-1}(x)$. Second and more important: your link and the code expresse $L_n$ in terms of $L_n'$ and $L_{n-1}'$ and not $L_n'$ in terms of $L_n$ and $L_{n-1}$! – gammatester Sep 10 '13 at 08:32
  • @gammatester thanks, then I need to show $ L_n'(x) - L_{n-1}'(x) = L_{n-1}(x)$ – hasExams Sep 10 '13 at 10:41
  • There is a sign error, it should be $\quad L_n'(x)−L_{n−1}'(x) = - L_{n−1}'(x), \quad$ see e.g. Lebedev, Special Functions (4.18.2) – gammatester Sep 10 '13 at 10:58
  • @gammatester yes yes sorry again :(( – hasExams Sep 10 '13 at 10:59
  • @gammatester you could have posted as an answer. – hasExams Sep 10 '13 at 11:06

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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:

$$ L_n'(x) - L_{n-1}'(x)= \frac{ -x L_{n-1}(x)}{x} = -L_{n-1}(x)$$

gammatester
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