Difference between revisions of "Rodrigues formula for Meixner polynomial"
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− | + | ==Theorem== | |
− | + | 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]]. |
− | + | ||
− | + | ==Proof== | |
− | + | ||
− | + | ==References== | |
+ | |||
+ | [[Category:Theorem]] | ||
+ | [[Category:Unproven]] |
Latest revision as of 02:39, 21 December 2016
Theorem
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.