Difference between revisions of "Associated Laguerre L"
m (Tom moved page Associated Laguerre to Associated Laguerre L) |
|||
| Line 61: | Line 61: | ||
</div> | </div> | ||
</div> | </div> | ||
| + | |||
| + | {{:Orthogonal polynomials footer}} | ||
Revision as of 21:54, 22 March 2015
Let $\lambda \in \mathbb{R}$. The associated Laguerre polynomials, $L_n^{(\lambda)}(x)$ are solutions of the differential equation $$x\dfrac{d^2y}{dx^2} + (\lambda+1-x)\dfrac{dy}{dx} + ny=0.$$
The first few Laguerre polynomials are given by $$\begin{array}{ll} L_0^{(\lambda)}(x) &= 1 \\ L_1^{(\lambda)}(x) &= -x+\lambda+1 \\ L_2^{(\lambda)}(x) &= \dfrac{x^2}{2} -(\lambda+2)x+\dfrac{(\lambda+2)(\lambda+1)}{2} \\ L_3^{(\lambda)}(x) &= -\dfrac{x^3}{6} + \dfrac{(\lambda+3)x^2}{2} - \dfrac{(\lambda+2)(\lambda+3)x}{2} + \dfrac{(\lambda+1)(\lambda+2)(\lambda+3)}{6} \\ \vdots \end{array}$$
Properties
Theorem: The following formula holds: $$L_n^{(\lambda)}(x) = \displaystyle\sum_{k=0}^n (-1)^k {n+\lambda \choose n-k} \dfrac{x^k}{k!}.$$
Proof: █
Theorem: The following formula holds: $$L_n^{(\lambda+\beta+1)}(x+y) = \displaystyle\sum_{k=0}^n L_k^{(\lambda)}(x)L_{n-k}^{(\beta)}(x).$$
Proof: █
Theorem: The following formula holds: $$L_n^{(\lambda)}(x) = L_n^{(\lambda+1)}(x)-L_{n-1}^{(\lambda+1)}(x).$$
Proof: █
Theorem: The following formula holds: $$nL_n^{(\lambda+1)}(x) = (n-x)L_{n-1}^{(\lambda+1)}(x)-(n+\lambda)L_{n-1}^{(\lambda)}(x).$$
Proof: █
Theorem: The following formula holds: $$xL_n^{(\lambda+1)}(x)= (n+\lambda)L_{n-1}^{(\lambda)}(x)-(n-x)L_n^{(\lambda)}(x).$$
Proof: █
Theorem: The following formula holds: $$\dfrac{d^k}{dx^k} L_n^{(\lambda)}(x) = (-1)^kL_{n-k}^{(\lambda+k)}(x).$$
Proof: █
