Difference between revisions of "Rodrigues formula for Meixner polynomial"
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<strong>[[Rodrigues formula for Meixner polynomial|Theorem]]:</strong> ([[Rodrigues Formula]]) The following formula holds: | <strong>[[Rodrigues formula for Meixner polynomial|Theorem]]:</strong> ([[Rodrigues Formula]]) The following formula holds: | ||
$$\dfrac{\beta^{\overline{x}}c^x}{x!} M_n(x;\beta,c)= \nabla^n \left[ \dfrac{(\beta+n)^{\overline{x}}}{x!}c^x \right],$$ | $$\dfrac{\beta^{\overline{x}}c^x}{x!} M_n(x;\beta,c)= \nabla^n \left[ \dfrac{(\beta+n)^{\overline{x}}}{x!}c^x \right],$$ | ||
| − | where $\beta^{\overline{x}}$ denotes a [[rising factorial]] and $M_n$ is a [[Meixner polynomial]]. | + | where $\nabla$ denotes the backwards difference operator $\nabla f = f(x)-f(x-1)$, $\beta^{\overline{x}}$ denotes a [[rising factorial]] and $M_n$ is a [[Meixner polynomial]]. |
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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
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Revision as of 14:39, 20 May 2015
Theorem: (Rodrigues Formula) The following formula holds: $$\dfrac{\beta^{\overline{x}}c^x}{x!} M_n(x;\beta,c)= \nabla^n \left[ \dfrac{(\beta+n)^{\overline{x}}}{x!}c^x \right],$$ where $\nabla$ denotes the backwards difference operator $\nabla f = f(x)-f(x-1)$, $\beta^{\overline{x}}$ denotes a rising factorial and $M_n$ is a Meixner polynomial.
Proof: █
