Editing Laguerre polynomials (section)

=== Recurrence relations ===
The addition formula for Laguerre polynomials:<ref>{{Cite web |title=DLMF: §18.18 Sums ‣ Classical Orthogonal Polynomials ‣ Chapter 18 Orthogonal Polynomials |url=https://dlmf.nist.gov/18.18 |access-date=2025-03-18 |website=dlmf.nist.gov}}</ref>
<math display="block">L^{(\alpha_{1}+\dots+\alpha_{r}+r-1)}_{n}\left(x_{1}+\dots+x_{r}\right)=\sum_{m_{1}+\dots+m_{r}=n}L^{(\alpha_{1})}_{m_{1}}\left(x_{1}\right)\cdots L^{(\alpha_{r})}_{m_{r}}\left(x_{r}\right).</math>Laguerre's polynomials satisfy the recurrence relations
<math display="block">L_n^{(\alpha)}(x)= \sum_{i=0}^n L_{n-i}^{(\alpha+i)}(y)\frac{(y-x)^i}{i!},</math>
in particular
<math display="block">L_n^{(\alpha+1)}(x)= \sum_{i=0}^n L_i^{(\alpha)}(x)</math>
and
<math display="block">L_n^{(\alpha)}(x)= \sum_{i=0}^n {\alpha-\beta+n-i-1 \choose n-i} L_i^{(\beta)}(x),</math>
or
<math display="block">L_n^{(\alpha)}(x)=\sum_{i=0}^n {\alpha-\beta+n \choose n-i} L_i^{(\beta- i)}(x);</math>
moreover
<math display="block">\begin{align}
L_n^{(\alpha)}(x)- \sum_{j=0}^{\Delta-1} {n+\alpha \choose n-j} (-1)^j \frac{x^j}{j!}&= (-1)^\Delta\frac{x^\Delta}{(\Delta-1)!} \sum_{i=0}^{n-\Delta} \frac{{n+\alpha \choose n-\Delta-i}}{(n-i){n \choose i}}L_i^{(\alpha+\Delta)}(x)\\[6pt]
&=(-1)^\Delta\frac{x^\Delta}{(\Delta-1)!} \sum_{i=0}^{n-\Delta} \frac{{n+\alpha-i-1 \choose n-\Delta-i}}{(n-i){n \choose i}}L_i^{(n+\alpha+\Delta-i)}(x)
\end{align}</math>

They can be used to derive the four 3-point-rules
<math display="block">\begin{align}
L_n^{(\alpha)}(x) &= L_n^{(\alpha+1)}(x) - L_{n-1}^{(\alpha+1)}(x) = \sum_{j=0}^k {k \choose j}(-1)^j L_{n-j}^{(\alpha+k)}(x), \\[10pt]
n L_n^{(\alpha)}(x) &= (n + \alpha )L_{n-1}^{(\alpha)}(x) - x L_{n-1}^{(\alpha+1)}(x), \\[10pt]
& \text{or } \\
\frac{x^k}{k!}L_n^{(\alpha)}(x) &= \sum_{i=0}^k (-1)^i {n+i \choose i} {n+\alpha \choose k-i} L_{n+i}^{(\alpha-k)}(x), \\[10pt]
n L_n^{(\alpha+1)}(x) &= (n-x) L_{n-1}^{(\alpha+1)}(x) + (n+\alpha)L_{n-1}^{(\alpha)}(x) \\[10pt]
x L_n^{(\alpha+1)}(x) &= (n+\alpha)L_{n-1}^{(\alpha)}(x)-(n-x)L_n^{(\alpha)}(x);
\end{align}</math>

combined they give this additional, useful recurrence relations<math display="block">\begin{align}
L_n^{(\alpha)}(x)&= \left(2+\frac{\alpha-1-x}n \right)L_{n-1}^{(\alpha)}(x)- \left(1+\frac{\alpha-1}n \right)L_{n-2}^{(\alpha)}(x)\\[10pt]
&= \frac{\alpha+1-x}n  L_{n-1}^{(\alpha+1)}(x)- \frac x n L_{n-2}^{(\alpha+2)}(x)
\end{align}</math>

Since <math>L_n^{(\alpha)}(x)</math> is a monic polynomial of degree <math>n</math> in <math>\alpha</math>,
there is the [[partial fraction decomposition]]
<math display="block">\begin{align}
\frac{n!\,L_n^{(\alpha)}(x)}{(\alpha+1)_n} 
&= 1- \sum_{j=1}^n (-1)^j \frac{j}{\alpha + j} {n \choose j}L_n^{(-j)}(x) \\
&= 1- \sum_{j=1}^n \frac{x^j}{\alpha + j}\,\,\frac{L_{n-j}^{(j)}(x)}{(j-1)!} \\
&= 1-x \sum_{i=1}^n \frac{L_{n-i}^{(-\alpha)}(x) L_{i-1}^{(\alpha+1)}(-x)}{\alpha +i}.
\end{align}</math>
The second equality follows by the following identity, valid for integer ''i'' and {{mvar|n}} and immediate from the expression of <math>L_n^{(\alpha)}(x)</math> in terms of [[Charlier polynomials]]:
<math display="block"> \frac{(-x)^i}{i!} L_n^{(i-n)}(x) = \frac{(-x)^n}{n!} L_i^{(n-i)}(x).</math>
For the third equality apply the fourth and fifth identities of this section.