Difference between revisions of "Associated Laguerre L"

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Let $\alpha \in \mathbb{R}$. The associated Laguerre polynomials, $L_n^{(\alpha)}(x)$ are solutions of the differential equation
+
Let $\lambda \in \mathbb{R}$. The associated Laguerre polynomials, $L_n^{(\lambda)}(x)$ are solutions of the differential equation
$$x\dfrac{d^2y}{dx^2} + (1-x)\dfrac{dy}{dx} + ny=0.$$
+
$$x\dfrac{d^2y}{dx^2} + (\lambda+1-x)\dfrac{dy}{dx} + ny=0.$$
  
 
The first few Laguerre polynomials are given by
 
The first few Laguerre polynomials are given by
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
L_0^{(\alpha)}(x) &= 1 \\
+
L_0^{(\lambda)}(x) &= 1 \\
L_1^{(\alpha)}(x) &= -x+\alpha+1 \\
+
L_1^{(\lambda)}(x) &= -x+\lambda+1 \\
L_2^{(\alpha)}(x) &= \dfrac{x^2}{2} -(\alpha+2)x+\dfrac{(\alpha+2)(\alpha+1)}{2} \\
+
L_2^{(\lambda)}(x) &= \dfrac{x^2}{2} -(\lambda+2)x+\dfrac{(\lambda+2)(\lambda+1)}{2} \\
L_3^{(\alpha)}(x) &= -\dfrac{x^3}{6} + \dfrac{(\alpha+3)x^2}{2} - \dfrac{(\alpha+2)(\alpha+3)x}{2} + \dfrac{(\alpha+1)(\alpha+2)(\alpha+3)}{6} \\
+
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
 
\vdots
 
\end{array}$$
 
\end{array}$$
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$L_n^{(\alpha)}(x) = \displaystyle\sum_{k=0}^n (-1)^k {n+\alpha \choose n-k} \dfrac{x^k}{k!}.$$
+
$$L_n^{(\lambda)}(x) = \displaystyle\sum_{k=0}^n (-1)^k {n+\lambda \choose n-k} \dfrac{x^k}{k!}.$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$L_n^{(\alpha+\beta+1)}(x+y) = \displaystyle\sum_{k=0}^n L_k^{(\alpha)}(x)L_{n-k}^{(\beta)}(x).$$
+
$$L_n^{(\lambda+\beta+1)}(x+y) = \displaystyle\sum_{k=0}^n L_k^{(\lambda)}(x)L_{n-k}^{(\beta)}(x).$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$L_n^{(\alpha)}(x) = L_n^{(\alpha+1)}(x)-L_{n-1}^{(\alpha+1)}(x).$$
+
$$L_n^{(\lambda)}(x) = L_n^{(\lambda+1)}(x)-L_{n-1}^{(\lambda+1)}(x).$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$nL_n^{(\alpha+1)}(x) = (n-x)L_{n-1}^{(\alpha+1)}(x)-(n+\alpha)L_{n-1}^{(\alpha)}(x).$$
+
$$nL_n^{(\lambda+1)}(x) = (n-x)L_{n-1}^{(\lambda+1)}(x)-(n+\lambda)L_{n-1}^{(\lambda)}(x).$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$xL_n^{(\alpha+1)}(x)= (n+\alpha)L_{n-1}^{(\alpha)}(x)-(n-x)L_n^{(\alpha)}(x).$$
+
$$xL_n^{(\lambda+1)}(x)= (n+\lambda)L_{n-1}^{(\lambda)}(x)-(n-x)L_n^{(\lambda)}(x).$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
$$\dfrac{d^k}{dx^k} L_n^{(\alpha)}(x) = (-1)^kL_{n-k}^{(\alpha+k)}(x).$$
+
$$\dfrac{d^k}{dx^k} L_n^{(\lambda)}(x) = (-1)^kL_{n-k}^{(\lambda+k)}(x).$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 03:40, 25 October 2014

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 $$\begin{array}{ll} L_0^{(\lambda)}(x) &= 1 \\ L_1^{(\lambda)}(x) &= -x+\lambda+1 \\ L_2^{(\lambda)}(x) &= \dfrac{x^2}{2} -(\lambda+2)x+\dfrac{(\lambda+2)(\lambda+1)}{2} \\ L_3^{(\lambda)}(x) &= -\dfrac{x^3}{6} + \dfrac{(\lambda+3)x^2}{2} - \dfrac{(\lambda+2)(\lambda+3)x}{2} + \dfrac{(\lambda+1)(\lambda+2)(\lambda+3)}{6} \\ \vdots \end{array}$$

Associatedlaguerrealpha=1.png

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: